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null | null | null | null | null | null | The Center for Legislative Archives
Interview Notes Index
Interview with Rep. Joseph P. Addabbo (D-NY)
April 1964
"You've got four agricultural members and myself. You've got a cross section--corn, cotton, tobacco, wheat, and the marijuana farmer from New York."
Regarding the relationship between Appropriations Committee Chairman Clarence Cannon (D-MO) and the five new members: "The newspapers played it up that we were the big spenders, and we were going to take over the Committee. He doesn't have to worry. He can bottle you up if he wants to. He can create subcommittees or change the membership. He can have a three-man subcommittee if he wants to--two and one. Or he can put another conservative on the Committee and box you in that way. But he hasn't. Actually he's given us more say. The subcommittees used to be seven men. Now they are only five. So we have more influence than the fellows before us. There are five of us and, maybe, we could pick up five more and say, 'I've got a bloc of ten votes, and we'll go along with you if you will do this.' We could demagogue that way and cause a lot of trouble, but we don't. There's a job to be done, and you do it. After you've been here a while you realize the obligation you have to the Committee and to your people back home and to your fellow representatives. They are busy with their committees, and they lean on you. They come and they ask you what you think, 'are they spending too much or too little?' So you have to do your job."
Regarding the other members' attitudes: "They think we're too conservative. At least the liberal ones do."
He was an inner circle choice. He kept speaking of his appointment as being "good for the New York delegation." He stressed the fact that the delegation wanted to keep the seat. In terms of the appeal of the Committee he stressed the scope of the Committee's work and its influence.
Regarding his appointment: "The Chairman was peeved because he wasn't consulted, and he wasn't consulted because of the trouble with the leadership at the end of the last session. He didn't have anything against us, per se. We were caught in a squeeze play between the leadership and the old man."
It is interesting to note that the scope of the Committee is one of the main attractions of the Committee, and that this is why so many of the young people find the Committee structure so disappointing to them. What they find is that although they know the Committee covers the full range of governmental activities, they play a role only in subcommittees. The full Committee should be the place where they take advantage of the scope of committee activity--and it is precisely the full Committee where they play so small a part. Hence their first acquaintance with the full Committee is a great blow to them. "We didn't know the function of things. You walk in there and they say, 'here's a report, here's the bill, and that's it . . . but you are really not shut off--you can read the hearings, and most everything is there. Of course, you don't have the time or may not take the time. And it would be easier if you could get the report a day earlier. You could pick up things you might be against. But you can read the hearings, and you can talk with other subcommittee members if you want to. If you can't do things the easy way, you do them the hard way. And you can always get up on the floor, and say what you think. You have time there to get prepared. It would be better the other way, but it's not going to change. They've been doing it for tens of years, and no one has proven he's been hurt by it. The subcommittees are the ones who work on the bill. I see it on our subcommittee. We are the ones who sit there all the time, and we know more about it than anyone else. So it's right that they should have most of the say."
Regarding floor opposition to the Committee: "I've done it--two times last year and this year, too. They'll ask you why, and you give your reasons. But you don't get anywhere."
He specializes only in "consumer interests" on the subcommittee.
Regarding markup, he says it is "pretty well wrangled out and put through the wringer there." No change results in full Committee. There is a lot of talk informally after each of the hearings, and he says that you get a pretty good idea of what the thinking of the Committee is. Therefore the subcommittee chairman's figures aren't a surprise completely even though he takes the lead working with the staff man. The point is that the chairman knows something of the thinking of the members from informal interchanges all during the hearings.
"If I wanted to demagogue about it, I could defeat the whole farm bill. The city boys would take my word for it. There isn't any farm bloc any more, and they need our support. But you don't do that. It was a great revelation to me about the farm economy and the connection with the city."
He gets more conservative the longer he is on the Committee--though not as strongly as some. "All you see are these money bills, and you look at them to see where you can save the taxpayers some money. There's a lot of taxpayers' dollars in that bill."
He points out that "my voting record was the most erratic of the New York City Democrats," and said that most of the new people were erratic voters, too, except for Charles S. Joelson (D-NJ).
He's less backward than Alfred E. Santangelo (D-NY)--he stresses his education on agriculture and his lack of background, the same as Santangelo. But he seems to speak up more in the subcommittee and be something less of an apprentice. Maybe it's because the farm bloc is smaller, and his influence is greater.
His subcommittee assignment was not his choice, but he was resigned. "It isn't going to change very soon. So you do your job and wait to see the way the cookie crumbles." He thinks his assignment was better, anyway, than some others.
The hearings are the most important to him. He mentioned how much he learned there. Especially since he had no background. The function of the hearings for the newcomer are informational.
When I told him I started in 1959, he said, "it hasn't changed much."
"Fortunately or unfortunately, I was elected to the Committee."
He cited the change in living habits--culture shock. The problem of the Tuesday-Thursday club member. "It's an especially hard committee for a New York City member. It's a Monday through Friday job. Hearings are held every day, and you have to be here a lot more than you do on any legislative committee. So I'm down here more than I ever was before. It's quite a change for me."
The Center for Legislative Archives >
The U.S. National Archives and Records Administration | null | null | null | null | null | null | null | null | null |
https://mycalcu.com/meters-to-kilometers | 1,718,276,169,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861372.90/warc/CC-MAIN-20240613091959-20240613121959-00802.warc.gz | 367,495,449 | 8,222 | # Meters to Kilometers
## Why use Meters to Kilometers, an online calculator?
Mycalcu helps you evaluate the units of measurements online, to remove any error while doing the conversions manually. It neither requires any critical thinking nor any familiarity with symbols.
## Conversion to Other Lengths
Kilometers Inches Centimeters Yards Miles Feet Millimeters
## How to Convert Meters to Kilometers using MYCALCU?
Here is the equation that shows ‘’how many Meters are in a Kilometer?’’
M = KM ÷ 1,000
The fastest and easiest way to convert Meters into Kilometers is using this formula
METERS = KILOMETERS ÷ 1,000
Example
If we want to convert 20 Meters into Kilometers, we multiply it by 1,000 to get the result
20 M = 20 ÷ 1,000= 0.02 KM
## WHAT IS THE RELATION BETWEEN M AND KM?
Both values are units of length and are such that their relationship can be represented on the ruler.
m km
1 m 0.001 km
2 m 0.002 km
3 m 0.003 km
4 m 0.004 km
5 m 0.005 km
6 m 0.006 km
7 m 0.007 km
8 m 0.008 km
9 m 0.009 km
10 m 0.01 km
11 m 0.011 km
12 m 0.012 km
13 m 0.013 km
14 m 0.014 km
15 m 0.015 km
16 m 0.016 km
17 m 0.017 km
18 m 0.018 km
19 m 0.019 km
20 m 0.02 km
21 m 0.021 km
22 m 0.022 km
23 m 0.023 km
24 m 0.024 km
25 m 0.025 km
26 m 0.026 km
27 m 0.027 km
28 m 0.028 km
29 m 0.029 km
30 m 0.03 km
31 m 0.031 km
32 m 0.032 km
33 m 0.033 km
34 m 0.034 km
35 m 0.035 km
36 m 0.036 km
37 m 0.037 km
38 m 0.038 km
39 m 0.039 km
40 m 0.04 km
41 m 0.041 km
42 m 0.042 km
43 m 0.043 km
44 m 0.044 km
45 m 0.045 km
46 m 0.046 km
47 m 0.047 km
48 m 0.048 km
49 m 0.049 km
50 m 0.05 km
51 m 0.051 km
52 m 0.052 km
53 m 0.053 km
54 m 0.054 km
55 m 0.055 km
56 m 0.056 km
57 m 0.057 km
58 m 0.058 km
59 m 0.059 km
60 m 0.06 km
61 m 0.061 km
62 m 0.062 km
63 m 0.063 km
64 m 0.064 km
65 m 0.065 km
66 m 0.066 km
67 m 0.067 km
68 m 0.068 km
69 m 0.069 km
70 m 0.07 km
71 m 0.071 km
72 m 0.072 km
73 m 0.073 km
74 m 0.074 km
75 m 0.075 km
76 m 0.076 km
77 m 0.077 km
78 m 0.078 km
79 m 0.079 km
80 m 0.08 km
81 m 0.081 km
82 m 0.082 km
83 m 0.083 km
84 m 0.084 km
85 m 0.085 km
86 m 0.086 km
87 m 0.087 km
88 m 0.088 km
89 m 0.089 km
90 m 0.09 km
91 m 0.091 km
92 m 0.092 km
93 m 0.093 km
94 m 0.094 km
95 m 0.095 km
96 m 0.096 km
97 m 0.097 km
98 m 0.098 km
99 m 0.099 km
100 m 0.1 km | 1,029 | 2,287 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.671875 | 4 | CC-MAIN-2024-26 | latest | en | 0.462619 |
https://www.physicsforums.com/threads/buoyancy-barge-problem.287423/ | 1,506,036,278,000,000,000 | text/html | crawl-data/CC-MAIN-2017-39/segments/1505818687938.15/warc/CC-MAIN-20170921224617-20170922004617-00419.warc.gz | 847,346,369 | 14,400 | # Buoyancy/Barge Problem
1. Jan 25, 2009
### NYCHE89
1. The problem statement, all variables and given/known data
A barge 150 ft. long and 30 ft. wide is to carry a payload of 10-ton light tanks over water. If the barge is to sink no more than an addictional 1 ft, how many tanks can be loaded?
2. Relevant equations
3. The attempt at a solution
Buoyant Force = fluid density x volume submerged x gravity
= (1 g/cm^3 x 1kg/1000g) x (4500 ft^3 x 28.317L/1ft^3 x 1000cm^3/1L) x 9.8 m/s^2
= 1,248,779.7 N
I've gotten this far and don't know what to do next. Also, this may be completely wrong.
2. Jan 25, 2009
### mgb_phys
It might be easier to restate this as: "It receives an upthrust equal to the weight of water displaced."
Or the extra weight of cargo equals the weight of the water.
There no need to explicitly include g - especially if you can't decide which units to use.
3. Jan 27, 2009
### nvn
NYCHE89: Nice work! Your equation and answer for buoyancy force, Fb, is correct. g = 9.807 m/s^2. Now, the weight of each light tank is Pt = 88 964.4 N. Therefore, see if you can now figure out how many light tanks would have a total weight not exceeding Fb. Try it. | 368 | 1,180 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5625 | 4 | CC-MAIN-2017-39 | longest | en | 0.904661 |
http://www.ltcconline.net/greenl/Courses/106/WorkForceMoments/FLUID.HTM | 1,516,369,751,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084887981.42/warc/CC-MAIN-20180119125144-20180119145144-00415.warc.gz | 514,684,994 | 3,230 | Fluid Force
Fluid Pressure (Definition)
We define the fluid pressure P to be equal to the weight density w times the depth of the fluid h.
P = wh
Pascal's Principle
The pressure is equal in all directions
Since
force
Pressure =
area
we have
F = PA
Force on a Submerged Rectangular Sheet
Example
Suppose that a glass sheet lies parallel to and five feet below the surface of a lake has dimensions 2 by 3. Then the total pressure on the sheet is
P = wh = (62.4)(5) = 312
The total force on the plate is
PA = (312)(6) = 1872
Force on a Vertical Surface
Example
A 3 x 2 square window on the new Disney Cruise liner is to be built so that top of the window is four feet below the surface of the water. What total force will the window be subjected to?
Solution
We take horizontal cross sections. Letting y be the distance from the top of the window to the cross section, we have
DF = 62.4(4 + y)(D A) = 62.4(4 + y)(3)(Dy)
Hence the total force is
= 1123.2 pounds
In general:
Example
A radius 2 feet circular portal is vertically submerged so that its center is 20 feet below the surface of the water. Find the total force of the water on the portal.
Solution
We write s in terms of z by the Pythagorean theorem:
This horizontal cross-section has area
DA = 2sDz
The depth at this cross-section is
h = 20 + z
We put this all together to find the force
We recognize the first integral as the area of the semi-circle of radius 2. For the second integral, let
u = 4 - z2 du = -2zdz
Notice that when z = -2, u = 0 and when z = 2, u = 0 also. Any integral with lower and upper limits the same is always zero. Hence the force is
F = 40p22 + 0 = 160p
Exercise:
In the VIP suite, the glass window is in the shape of an equilateral triangle with diagonal length 2 such that the top of the triangle is submerged 5 feet below the water. What total force will this window be subjected to?
Back to Math 105 Home Page
Back to the Math Department Home
e-mail Questions and Suggestions | 548 | 2,023 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.4375 | 4 | CC-MAIN-2018-05 | latest | en | 0.874721 |
https://cs.stackexchange.com/questions/78022/matrix-sum-factorialk-times | 1,632,172,784,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780057091.31/warc/CC-MAIN-20210920191528-20210920221528-00716.warc.gz | 236,321,202 | 38,191 | # Matrix Sum, factorial(K) times
Given an N x M matrix. The task is to repeat the following steps K! times.
1. Add all the elements of the matrix and get a matrix sum (S) % mod 10^9 + 7
2. Add S to all the elements. % mod 10^9 + 7
3. Repeat.
What I can think of regarding the complexity it will O(K! x N x M).
But K ranges from 1 to 10^5 and this is a huge number of iterations for a computer to process.
Is there a way to optimize these operations so that these can be performed using less computational resources.
Thank You.
• Do you use any matrix operations or simply have array that stores numbers and happen to be matrix? If this is from some contest please credit the sources. What have you tried so far?
– Evil
Jul 17 '17 at 3:56
• What's the source where you encountered this problem? Can you credit the original source?
– D.W.
Jul 17 '17 at 6:25
Suppose that the original elements are $a_{ij}^{(1)}$. Their sum is $S^{(1)} = \sum_{ij} a_{ij}^{(1)}$, and the new elements are $a_{ij}^{(2)} = a_{ij}^{(1)} + S^{(1)}$. The new sum is $S^{(2)} = \sum_{ij} a_{ij}^{(1)} + nmS^{(1)} = (1+nm)S^{(1)}$, and the new elements are $a_{ij}^{(3)} = a_{ij}^{(1)} + (2+nm)S^{(1)}$. The new sum is $S^{(3)} = S^{(1)} + nm(2+nm)S^{(1)} = (1+nm)^2S^{(1)}$, and so on.
More generally, you can prove by induction that $S^{(t)} = (1+nm)^{t-1} S^{(1)}$ and $a_{ij}^{(t)} = a_{ij}^{(1)} + \frac{(1+nm)^{t-1}-1}{nm} S^{(1)}$. It remains to find a fast way to calculate expressions such as $S^{(k!)} = (1+nm)^{k!-1} S^{(1)}$ modulo the given prime $p = 10^9 + 7$. You do the rest. | 556 | 1,573 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.96875 | 4 | CC-MAIN-2021-39 | latest | en | 0.793545 |
www.mide-kucultme.com | 1,579,331,173,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579250592261.1/warc/CC-MAIN-20200118052321-20200118080321-00319.warc.gz | 255,796,536 | 19,386 | 0
1
# What Specifically Is Discrete Z?
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Sosyal Ağlarda Paylaş | 803 | 4,198 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.625 | 4 | CC-MAIN-2020-05 | latest | en | 0.958573 |
https://www.topperlearning.com/selina-solutions/icse-class-10-mathematics/selina-concise-mathematics-x/reflection-in-x-axis-y-axis-x-a-y-a-and-the-origin-invariant-points | 1,652,688,599,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662510097.3/warc/CC-MAIN-20220516073101-20220516103101-00307.warc.gz | 1,212,806,352 | 47,619 | # SELINA Solutions for Class 10 Maths Chapter 12 - Reflection (In x-axis, y-axis, x=a, y=a and the origin ; Invariant Points)
Page / Exercise
## Chapter 12 - Reflection (In x-axis, y-axis, x=a, y=a and the origin ; Invariant Points) Exercise Ex. 12(A)
Solution 1
Point Transformation Image (5, -7) Reflection in origin (-5, 7) (4, 2) Reflection in x-axis (4, -2) (0, 6) Reflection in y-axis (0, 6) (6, -6) Reflection in origin (-6, 6) (4, -8) Reflection in y-axis (-4, -8)
Solution 2
Since, the point P is its own image under the reflection in the line l. So, point P is an invariant point.
Hence, the position of point P remains unaltered.
Solution 3
(i) (3, 2)
The co-ordinate of the given point under reflection in the x-axis is (3, -2).
(ii) (-5, 4)
The co-ordinate of the given point under reflection in the x-axis is (-5, -4).
(iii) (0, 0)
The co-ordinate of the given point under reflection in the x-axis is (0, 0).
Solution 4
(i) (6, -3)
The co-ordinate of the given point under reflection in the y-axis is (-6, -3).
(ii) (-1, 0)
The co-ordinate of the given point under reflection in the y-axis is (1, 0).
(iii) (-8, -2)
The co-ordinate of the given point under reflection in the y-axis is (8, -2).
Solution 5
(i) (-2, -4)
The co-ordinate of the given point under reflection in origin is (2, 4).
(ii) (-2, 7)
The co-ordinate of the given point under reflection in origin is (2, -7).
(iii) (0, 0)
The co-ordinate of the given point under reflection in origin is (0, 0).
Solution 6
(i) (-6, 4)
The co-ordinate of the given point under reflection in the line x = 0 is (6, 4).
(ii) (0, 5)
The co-ordinate of the given point under reflection in the line x = 0 is (0, 5).
(iii) (3, -4)
The co-ordinate of the given point under reflection in the line x = 0 is (-3, -4).
Solution 7
(i) (-3, 0)
The co-ordinate of the given point under reflection in the line y = 0 is (-3, 0).
(ii) (8, -5)
The co-ordinate of the given point under reflection in the line y = 0 is (8, 5).
(iii) (-1, -3)
The co-ordinate of the given point under reflection in the line y = 0 is (-1, 3).
Solution 8
(i) Since, Mx (-4, -5) = (-4, 5)
So, the co-ordinates of P are (-4, -5).
(ii) Co-ordinates of the image of P under reflection in the y-axis (4, -5).
Solution 9
(i) Since, MO (2, -7) = (-2, 7)
So, the co-ordinates of P are (2, -7).
(ii) Co-ordinates of the image of P under reflection in the x-axis (2, 7).
Solution 10
MO (a, b) = (-a, -b)
My (-a, -b) = (a, -b)
Thus, we get the co-ordinates of the point P' as (a, -b). It is given that the co-ordinates of P' are (4, 6).
On comparing the two points, we get,
a = 4 and b = -6
Solution 11
Mx (x, y) = (x, -y)
MO (x, -y) = (-x, y)
Thus, we get the co-ordinates of the point P' as (-x, y). It is given that the co-ordinates of P' are (-8, 5).
On comparing the two points, we get,
x = 8 and y = 5
Solution 12
(i) The reflection in x-axis is given by Mx (x, y) = (x, -y).
A' = reflection of A (-3, 2) in the x- axis = (-3, -2).
The reflection in origin is given by MO (x, y) = (-x, -y).
A'' = reflection of A' (-3, -2) in the origin = (3, 2)
(ii) The reflection in y-axis is given by My (x, y) = (-x, y).
The reflection of A (-3, 2) in y-axis is (3, 2).
Thus, the required single transformation is the reflection of A in the y-axis to the point A''.
Solution 13
(i) The reflection in origin is given by MO (x, y) = (-x, -y).
A' = reflection of A (4, 6) in the origin = (-4, -6)
The reflection in y-axis is given by My (x, y) = (-x, y).
A'' = reflection of A' (-4, -6) in the y-axis = (4, -6)
(ii) The reflection in x-axis is given by Mx (x, y) = (x, -y).
The reflection of A (4, 6) in x-axis is (4, -6).
Thus, the required single transformation is the reflection of A in the x-axis to the point A''.
Solution 14
(i) Reflection in y-axis is given by My (x, y) = (-x, y)
A' = Reflection of A (2, 6) in y-axis = (-2, 6)
Similarly, B' = (3, 5) and C' = (-4, 7)
Reflection in origin is given by MO (x, y) = (-x, -y)
A'' = Reflection of A' (-2, 6) in origin = (2, -6)
Similarly, B'' = (-3, -5) and C'' = (4, -7)
(ii) A single transformation which maps triangle ABC to triangle A''B''C'' is reflection in x-axis.
Solution 15
Reflection in x-axis is given by Mx (x, y) = (x, -y)
P' = Reflection of P(-2, 3) in x-axis = (-2, -3)
Reflection in y-axis is given by My (x, y) = (-x, y)
Q' = Reflection of Q(5, 4) in y-axis = (-5, 4)
Thus, the co-ordinates of points P' and Q' are (-2, -3) and (-5, 4) respectively.
Solution 16
The graph shows triangle ABC and triangle A'B'C' which is obtained when ABC is reflected in the origin.
Solution 17
Reflection in y-axis is given by My (x, y) = (-x, y)
A' = Reflection of A(4, -1) in y-axis = (-4, -1)
Reflection in x-axis is given by Mx (x, y) = (x, -y)
B' = Reflection of B in x-axis = (-2, 5)
Thus, B = (-2, -5)
Solution 18
(a) We know that reflection in the line x = 0 is the reflection in the y-axis.
It is given that:
Point (-5, 0) on reflection in a line is mapped as (5, 0).
Point (-2, -6) on reflection in the same line is mapped as (2, -6).
Hence, the line of reflection is x = 0.
(b) It is known that My (x, y) = (-x, y)
Co-ordinates of the image of (5, -8) in the line x = 0 are (-5, -8).
## Chapter 12 - Reflection (In x-axis, y-axis, x=a, y=a and the origin ; Invariant Points) Exercise Ex. 12(B)
Solution 1
(c)
(i) From graph, it is clear that ABB'A' is an isosceles trapezium.
(ii) The measure of angle ABB' is 45°.
(iii) A'' = (-3, -2)
(iv) Single transformation that maps A' to A" is the reflection in y-axis.
Solution 2
(i) We know that every point in a line is invariant under the reflection in the same line.
Since points (3, 0) and (-1, 0) lie on the x-axis.
So, (3, 0) and (-1, 0) are invariant under reflection in x-axis.
Hence, the equation of line L1 is y = 0.
Similarly, (0, -3) and (0, 1) are invariant under reflection in y-axis.
Hence, the equation of line L2 is x = 0.
(ii) P' = Image of P (3, 4) in L1 = (3, -4)
Q' = Image of Q (-5, -2) in L1 = (-5, 2)
(iii) P'' = Image of P (3, 4) in L2 = (-3, 4)
Q'' = Image of Q (-5, -2) in L2 = (5, -2)
(iv) Single transformation that maps P' onto P" is reflection in origin.
Solution 3
(i) We know Mx (x, y) = (x, -y)
P' (5, -2) = reflection of P (a, b) in x-axis.
Thus, the co-ordinates of P are (5, 2).
Hence, a = 5 and b = 2.
(ii) P" = image of P (5, 2) reflected in y-axis = (-5, 2)
(iii) Single transformation that maps P' to P" is the reflection in origin.
Solution 4
(i) We know reflection of a point (x, y) in y-axis is (-x, y).
Hence, the point (-2, 0) when reflected in y-axis is mapped to (2, 0).
Thus, the mirror line is the y-axis and its equation is x = 0.
(ii) Co-ordinates of the image of (-8, -5) in the mirror line (i.e., y-axis) are (8, -5).
Solution 5
The line y = 3 is a line parallel to x-axis and at a distance of 3 units from it.
Mark points P (4, 1) and Q (-2, 4).
From P, draw a straight line perpendicular to line CD and produce. On this line mark a point P' which is at the same distance above CD as P is below it.
The co-ordinates of P' are (4, 5).
Similarly, from Q, draw a line perpendicular to CD and mark point Q' which is at the same distance below CD as Q is above it.
The co-ordinates of Q' are (-2, 2).
Solution 6
The line x = 2 is a line parallel to y-axis and at a distance of 2 units from it.
Mark point P (-2, 3).
From P, draw a straight line perpendicular to line CD and produce. On this line mark a point P' which is at the same distance to the right of CD as P is to the left of it.
The co-ordinates of P' are (6, 3).
Solution 7
A point P (a, b) is reflected in the x-axis to P' (2, -3).
We know Mx (x, y) = (x, -y)
Thus, co-ordinates of P are (2, 3). Hence, a = 2 and b = 3.
P" = Image of P reflected in the y-axis = (-2, 3)
P''' = Reflection of P in the line (x = 4) = (6, 3)
Solution 8
(a) A' = Image of A under reflection in the x-axis = (3, -4)
(b) B' = Image of B under reflection in the line AA' = (6, 2)
(c) A" = Image of A under reflection in the y-axis = (-3, 4)
(d) B" = Image of B under reflection in the line AA" = (0, 6)
Solution 9
(i) The points A (3, 5) and B (-2, -4) can be plotted on a graph as shown.
(ii) A' = Image of A when reflected in the x-axis = (3, -5)
(iii) C = Image of B when reflected in the y-axis = (2, -4)
B' = Image when C is reflected in the origin = (-2, 4)
(iv) Isosceles trapezium
(v) Any point that remains unaltered under a given transformation is called an invariant.
Thus, the required two points are (3, 0) and (-2, 0).
Solution 10
(a) Co-ordinates of P' = (-5, -3)
(b) Co-ordinates of M = (5, 0)
(c) Co-ordinates of N = (-5, 0)
(d) PMP'N is a parallelogram.
(e) Are of PMP'N = 2 (Area of D PMN)
Solution 11
(i) Co-ordinates of P' and O' are (3, -4) and (6, 0) respectively.
(ii) PP' = 8 units and OO' = 6 units.
(iii) From the graph it is clear that all sides of the quadrilateral POP'O' are equal.
In right PO'O,
PO' =
So, perimeter of quadrilateral POP'O' = 4 PO' = 4 5 units = 20 units
(iv) Quadrilateral POP'O' is a rhombus.
Solution 12
Quadrilateral ABCD is an isosceles trapezium.
Co-ordinates of A', B', C' and D' are A'(-1, -1), B'(-5, -1), C'(-4, -2) and D'(-2, -2) respectively.
It is clear from the graph that D, A, A' and D' are collinear.
Solution 13
(a) Any point that remains unaltered under a given transformation is called an invariant.
It is given that P (0, 5) is invariant when reflected in an axis. Clearly, when P is reflected in the y-axis then it will remain invariant. Thus, the required axis is the y-axis.
(b) The co-ordinates of the image of Q (-2, 4) when reflected in y-axis is (2, 4).
(c) (0, k) on reflection in the origin is invariant. We know the reflection of origin in origin is invariant. Thus, k = 0.
(d) Co-ordinates of image of Q (-2, 4) when reflected in origin = (2, -4)
Co-ordinates of image of (2, -4) when reflected in x-axis = (2, 4)
Thus, the co-ordinates of the point are (2, 4).
Solution 14
(i) P (2, -4) is reflected in (x = 0) y-axis to get Q.
P(2, -4) Q (-2, -4)
(ii) Q (-2, -4) is reflected in (y = 0) x-axis to get R.
Q (-2, -4) R (-2, 4)
(iii) The figure PQR is right angled triangle.
(iv) Area of
Solution 15
(a)
Solution 16
i. A' = (4, 4) AND B' = (3, 0)
ii. The figure is an arrow head.
iii. The y-axis i.e. x = 0 is the line of symmetry of figure OABCB'A'.
Solution 17
(i)Plotting A(0,4), B(2,3), C(1,1) and D(2,0).
(ii) Reflected points B'(-2,3), C'(-1,1) and D'(-2,0).
(iii) The figure is symmetrical about x = 0
### STUDY RESOURCES
REGISTERED OFFICE : First Floor, Empire Complex, 414 Senapati Bapat Marg, Lower Parel, Mumbai - 400013, Maharashtra India. | 3,680 | 10,746 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.71875 | 5 | CC-MAIN-2022-21 | latest | en | 0.818945 |
https://uptuplus.com/qa/how-far-will-a-half-tank-of-gas-get-me.html | 1,610,808,340,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703506697.14/warc/CC-MAIN-20210116135004-20210116165004-00500.warc.gz | 619,417,409 | 7,422 | # How Far Will A Half Tank Of Gas Get Me?
## How many miles will 5 gallons of gas take you?
To figure the gas mileage, you would need to determine how many miles you traveled on 1 gallon of gas.
You would need to divide 1000 miles by 50 gallons of gas.
That would equal 20; therefore, you traveled 20 miles for every 1 gallon of gas.
Your gas mileage would be 20 mpg (miles per gallon)..
## How far will half a tank of gas get you?
As a general rule, most cars have about 2.5 gallons left in the tank when the gas light comes on. So depending on how many miles you get per gallon, you can probably go anywhere between 30-60 miles.
## How many miles can you drive on a full tank of gas calculator?
To calculate your car’s total range, multiply its average highway miles per gallon by its fuel capacity. For example, if your car averages 25 miles per gallon on the highway and has a 12-gallon fuel tank, its range is 25 x 12 = 300 miles.
## Is it better to fill up at half a tank?
Fill fuel when half tank empty: One of the most important tips is to fill up when your petrol/ diesel tank is HALF FULL. There is a scientific reason to why you must do this. The more petrol/ diesel you have in your tank, the less air occupying its empty space. Petrol/ diesel evaporate faster when in contact with air.
## How many miles is \$20 of gas?
140 miles\$20 will get me about 140 miles give or take.
## How far will a full tank get me?
The gas tank takes regular Unleaded fuel in a 15.3 gallon tank. You can get up to 25 mpg in the city and 34 mpg on highway*. This leads us to a final calculation- If you multiply 34 miles per gallon by 15.3, you can get up to 520 miles on one tank of gas.
## How many miles should a full tank last?
All told, if you do the math and assume the best miles per gallon scenario, the majority of gasoline cars on American roads do not have a range of more than 400 miles. Basically, you need a gas tank size of 13 gallons and an average mpg of 31 to hit – just barely – the 400-mile range mark.
## How much will 10 dollars of gas get you?
Since gas is \$2.50/gallon and you buy \$10 worth, that says you’ll get 4 gallons of gas (\$2.50 * 4 = \$10). Now figure out how many gallons you have left in your car. If you car holds 13 gallons, 1/4 of that is 3.25 gallons. Add the 3.25 gallons left to the 4 gallons you bought and you get 7.25 gallons in your car.
## How far can you go on one tank of petrol?
269 milesTank 1: Here I started driving with much more attention to fuel efficiency, road positioning at every moment, only pressing the accelerator when needed, trying to stop gradually without using the brakes (see driving techniques in the cheap petrol guide for more), and this improves safety too. Tank Distance: 269 miles.
## How many miles is a 1/4 tank of gas?
50 milesUsually around 50 miles for 1/4 and 100-110 for 1/2. | 727 | 2,875 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.515625 | 4 | CC-MAIN-2021-04 | latest | en | 0.923112 |
https://themathematicsmaster.com/difference-between-distance-and-midpoint-calculator/ | 1,723,471,519,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641039579.74/warc/CC-MAIN-20240812124217-20240812154217-00286.warc.gz | 438,851,722 | 15,180 | # Difference between Distance and Midpoint Calculator
July 14, 2023
The distance between two points is calculated as the length of the straight line connecting them. It's worth noting that the distance between the two points is always a positive value.
Additionally, in Mathematics, the midpoint is defined as the centre point of a line segment. It’s equidistant from both endpoints of the line segment and splits the line segment into two halves. Within a line segment, the midpoint represents the singular point that perfectly bisects the segment into two halves.
One can quickly determine the distance and midpoint of given coordinates using the Distance and Midpoint Calculator, which employs the distance and midpoint formulas. The key difference between Distance and Midpoint Calculators lies in their respective functions for calculating distances and midpoints on a coordinate plane. Simply input the coordinate values; the distance and midpoint calculator geometry will quickly provide the distance and midpoint values.
## The Distance and Midpoint Formulas
### Distance between two points
Assuming that the coordinates of the endpoints of the hypotenuse are $$(x_1, y_1)$$ and $$(x_2, y_2)$$, the calculation formula for the distance between these two points is analogous to the rule of a right triangle, where the square of the hypotenuse is equal to the sum of the squares of the other two sides.
### The Distance Formula
$$d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$$
## Midpoint Formula
The midpoint is the point on a line equidistant from two points, $$A (x_1, y_1)$$ and $$B (x_2, y_2)$$.
The midpoint can be found by taking the average of each segment coordinate, which creates a new coordinate point. If the coordinates of the endpoints of the segment are $$(X_1, Y_1)$$ and $$(X_2, Y_2)$$, then the midpoint can be obtained by adding the values in the parentheses and dividing each result by 2.
Where $$X_1$$ and $$X_2$$ denote the values of the X-coordinate on the X-axis, while $$Y_1$$ and $$Y_2$$ represent the values of the Y-coordinate on the Y-axis.
## Finding the Distance and Midpoint between two Points
Below is an example demonstrating the application of the distance formula to calculate the distance between two points on the coordinate plane. However, if you want to avoid manual calculation, the midpoint calculator geometry online is a powerful tool that can quickly calculate the midpoint of a line segment, saving you time and effort.
Our objective is to compute the distance between the two points $$(4, 3.5)$$ and $$(-3, 1)$$. It is observable that the line connecting these two points represents the hypotenuse of a right triangle. The legs of this triangle are parallel to the axes, allowing for effortless measurement of their length.
The Pythagorean Theorem will be employed to determine the length of distance d.
$$d^2 = 2.5^2+ 7^2$$
$$d^2= 6.25 + 49$$
$$d^2= 55.25$$
$$\sqrt{d^2}=\sqrt {55.25}$$
$$d= 7.43$$
The distance formula provides a technique for calculating the distance between two points in a coordinate plane.
## Finding Midpoint
So given the points $$(4, 3.5)$$ and $$(-3, 1)$$, you would solve for the midpoint like this:
$$x_1 = 4$$
$$x_2 = -3$$
$$y_1 = 3.5$$
$$y_2 = 1$$
Solution
$$(x_m,y_m) =(\dfrac{x_1+x_2}{2},\dfrac{y_1+y_2}{2})$$
$$(x_m,y_m) =(\dfrac{4 +(-3)}{2},\dfrac{3.5 + 1}{2})$$
$$(x_m,y_m) =(\dfrac{1}{2},\dfrac{4.5}{2})$$
$$(x_m,y_m) =(0.5,2.25)$$
Alternatively, for quick calculations you can find Midpoint Calculator here by The Mathematics Master. | 935 | 3,544 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.5625 | 5 | CC-MAIN-2024-33 | latest | en | 0.880558 |
http://tentotwelvemath.com/fmp-10/4-arithmetic-sequences/introduction-to-arithmetic-sequences/ | 1,534,656,714,000,000,000 | text/html | crawl-data/CC-MAIN-2018-34/segments/1534221214702.96/warc/CC-MAIN-20180819051423-20180819071423-00498.warc.gz | 390,896,586 | 11,235 | # Introduction to Arithmetic Sequences
Definition: An arithmetic sequence is an ordered list – usually of numbers – where there is a common difference between terms.
Here is an example:
Here is a second example:
Notice that in these two examples, the common difference between terms is 3.
If you continue these sequences indefinitely, will there ever be a common term?
Create two arithmetic sequences such that there are common terms. Write out several terms of each sequence. Circle the terms that appear in both sequences. How far apart are they in each sequence? Why?
On the next applet, 3 of the first 7 terms of an arithmetic sequence are presented. Figure out the missing terms.
What method did you use to calculate the missing terms?
On these problems, two of the first seven terms of an arithmetic sequence are given. Figure out the first term.
What method did you use to find the first term?
Suppose we label the first term , the second term , and the seventh term . Figure out an expression for using and . That is, figure out a step by step process that works every time you hit ‘new problem’ on this applet.
# Skill 1: Fill the blanks
applet under construction | 248 | 1,185 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.28125 | 4 | CC-MAIN-2018-34 | latest | en | 0.940224 |
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GEOMETRY JOURNAL 5 - PowerPoint PPT Presentation
GEOMETRY JOURNAL 5. MELANIE DOUGHERTY. Describe what a perpendicular bisector is. Explain the perpendicular bisector theorem and its converse. A perpendicular bisector is a line perpendicular to the base of a triangle that bisects it. Perpendicular Bisector theorem:
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GEOMETRY JOURNAL 5
MELANIE DOUGHERTY
Describe what a perpendicular bisector is. Explain the perpendicular bisector theorem and its converse.
• A perpendicular bisector is a line perpendicular to the base of a triangle that bisects it.
• Perpendicular Bisector theorem:
• If a point is on the perpendicular bisector of a segment, then it is equidistant form the endpoints of the segment.
• Converse:
• if a point is equidistant from the endpoints of a segment, then it is on the perpendicular bisector of the segment.
Perpendicular Bisector Examples perpendicular bisector theorem and its converse
AB = AC
AB = AC perpendicular bisector theorem and its converse
AC = BC perpendicular bisector theorem and its converse
PB Converse Examples perpendicular bisector theorem and its converse
LN = EN
AD = DC perpendicular bisector theorem and its converse
CD = DB perpendicular bisector theorem and its converse
Describe what an angle bisector is. Explain the angle bisector theorem and its converse.
• An angle bisector is a line that divides the angle.
• The angle Bisector theorem:
• If a point is on the bisector of an angle, then it is equidistant from the sides of the angle
• Converse:
• If a point is equidistant from the sides of an angle the it is on the bisector.
Angle Bisector theorem examples bisector theorem and its converse.
BF = FC
<UFK is congruent to <KFC bisector theorem and its converse.
<EWR is congruent to <RWT bisector theorem and its converse.
CONCURRENT bisector theorem and its converse.
When 3 or more lines intersect at one point
Concurrency of perpendicular bisector theorem of triangles bisector theorem and its converse.
The circumcenter of a triangleisequidistant from the vertices of the triangle.
Circumcenter: where the 3 perpendicular bisectors of a triangle meet
circumcenter
circumcenter
circumcenter
acute bisector theorem and its converse.
DA = DB = DC
right
DA = DB = DC
DA = DB = DC
obtuse
concurrency of angle bisectors of a triangle theorem bisector theorem and its converse.
Incenter of a triangle : where the 3 angle bisectors of a triangle meet
Concurrency of a angle bisectors of a triangle theorem: the incenter of a triangle is equidistant from the sides of the triangle.
incenter
incenter
incenter
ACUTE bisector theorem and its converse.
RIGHT
DF = DG = DE
DF = DG = DE
DF = DG = DE
OBTUSE
MEDIANS AND ALTITUDES OF TRIANGLES bisector theorem and its converse.
The median of a triangle is a segment whose endpoints are a vertex of the triangle and the midpoint of the opposite side
The centroid of a triangle is the point of concurrency of the medians of a triangle.
Concurrency of medians of a triangle theorem: the centroid of a triangle is located 2/3 of the distance from each vertex to the midpoint of the opposite side.
EXAMPLES bisector theorem and its converse.
CMTT
CENTROID
MEDIAN
Concurrency of altitudes of triangles theorem bisector theorem and its converse.
Altitude: a perpendicular segment from a vertex to the line containing the opposite side
Orthocenter: point where the 3 altitudes of a triangle meet.
Concurrency of altitudes of triangles theorem: the lines containing the altitude are concurrent
Triangle Midsegment theorem bisector theorem and its converse.
A midsegment is a segment that joins the midpoints of two sides of a triangle
Midsegment theorem: a midsegment of a triangle is parallel to a side of the triangle, and its length is half of that side.
midsegment
midsegment
midsegment
AB bisector theorem and its converse. ll EF, EF = ½ AB
DE ll BC, DE = ½ BC
DE ll AC, DE = ½ AC
Angle-Side Relationship in Triangles bisector theorem and its converse.
If none of the sides of the triangle are congruent then the largest side is opposite the largest angle.
If none of the sides of the triangle are congruent then the shortest side is opposite the smallest angle.
EXAMPLES bisector theorem and its converse.
Triangle Inequality bisector theorem and its converse.
The sum of the lengths of two sides of a triangle is greater than the length of the third side.
Writing an indirect proof bisector theorem and its converse.
Identify what is being proven
Assume that the opposite of your conclusion is true
Use direct reasoning to prove that the assumption has a contradiction
Assume that if the 1st assumption is false then what is being proved is true.
EXAMPLES bisector theorem and its converse.
Step 1
Given: triangle JKL is a right triangle
Prove: triangle JKL doesn't have and obtuse angle
Step 2
Assume <K is an obtuse angle
Step 3
m<K + m<L = 90
m<K = 90 – m<L
m<K > 90
90 – m<L > 90
m<L <0 (this is impossible)
Step 4
The original conjecture is true.
Hinge theorem bisector theorem and its converse.
If 2 sides of a triangle are congruent to 2 sides of an other triangle and included angles are not congruent, then the longer third side is across from the larger included angle.
Converse: if 2 sides of 2 triangle are congruent to 2 sides of an other triangle and the third sides of an other triangle are not congruent, then the larger included angle is across from the longer third side.
Examples bisector theorem and its converse. | 1,439 | 6,183 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4 | 4 | CC-MAIN-2017-39 | latest | en | 0.823519 |
http://nrich.maths.org/public/leg.php?code=124&cl=4&cldcmpid=1957 | 1,502,963,886,000,000,000 | text/html | crawl-data/CC-MAIN-2017-34/segments/1502886103167.97/warc/CC-MAIN-20170817092444-20170817112444-00212.warc.gz | 313,163,578 | 10,212 | # Search by Topic
#### Resources tagged with Pythagoras' theorem similar to Pythagoras Mod 5:
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### There are 73 results
Broad Topics > 2D Geometry, Shape and Space > Pythagoras' theorem
### Pythagoras Mod 5
##### Stage: 5 Challenge Level:
Prove that for every right angled triangle which has sides with integer lengths: (1) the area of the triangle is even and (2) the length of one of the sides is divisible by 5.
### Grid Lockout
##### Stage: 4 Challenge Level:
What remainders do you get when square numbers are divided by 4?
### Under the Ribbon
##### Stage: 4 Challenge Level:
A ribbon is nailed down with a small amount of slack. What is the largest cube that can pass under the ribbon ?
### Square World
##### Stage: 5 Challenge Level:
P is a point inside a square ABCD such that PA= 1, PB = 2 and PC = 3. How big is angle APB ?
### Reach for Polydron
##### Stage: 5 Challenge Level:
A tetrahedron has two identical equilateral triangles faces, of side length 1 unit. The other two faces are right angled isosceles triangles. Find the exact volume of the tetrahedron.
### Rectangular Pyramids
##### Stage: 4 and 5 Challenge Level:
Is the sum of the squares of two opposite sloping edges of a rectangular based pyramid equal to the sum of the squares of the other two sloping edges?
### Round and Round
##### Stage: 4 Challenge Level:
Prove that the shaded area of the semicircle is equal to the area of the inner circle.
### Circle Packing
##### Stage: 4 Challenge Level:
Equal circles can be arranged so that each circle touches four or six others. What percentage of the plane is covered by circles in each packing pattern? ...
### Two Circles
##### Stage: 4 Challenge Level:
Draw two circles, each of radius 1 unit, so that each circle goes through the centre of the other one. What is the area of the overlap?
### Three Four Five
##### Stage: 4 Challenge Level:
Two semi-circles (each of radius 1/2) touch each other, and a semi-circle of radius 1 touches both of them. Find the radius of the circle which touches all three semi-circles.
### Are You Kidding
##### Stage: 4 Challenge Level:
If the altitude of an isosceles triangle is 8 units and the perimeter of the triangle is 32 units.... What is the area of the triangle?
### Fitting In
##### Stage: 4 Challenge Level:
The largest square which fits into a circle is ABCD and EFGH is a square with G and H on the line CD and E and F on the circumference of the circle. Show that AB = 5EF. Similarly the largest. . . .
### Medallions
##### Stage: 4 Challenge Level:
Three circular medallions fit in a rectangular box. Can you find the radius of the largest one?
### Retracircles
##### Stage: 5 Challenge Level:
Four circles all touch each other and a circumscribing circle. Find the ratios of the radii and prove that joining 3 centres gives a 3-4-5 triangle.
### Pythagorean Triples II
##### Stage: 3 and 4
This is the second article on right-angled triangles whose edge lengths are whole numbers.
### Pythagorean Triples I
##### Stage: 3 and 4
The first of two articles on Pythagorean Triples which asks how many right angled triangles can you find with the lengths of each side exactly a whole number measurement. Try it!
### Picturing Pythagorean Triples
##### Stage: 4 and 5
This article discusses how every Pythagorean triple (a, b, c) can be illustrated by a square and an L shape within another square. You are invited to find some triples for yourself.
### Logosquares
##### Stage: 5 Challenge Level:
Ten squares form regular rings either with adjacent or opposite vertices touching. Calculate the inner and outer radii of the rings that surround the squares.
### Six Discs
##### Stage: 4 Challenge Level:
Six circular discs are packed in different-shaped boxes so that the discs touch their neighbours and the sides of the box. Can you put the boxes in order according to the areas of their bases?
### Where Is the Dot?
##### Stage: 4 Challenge Level:
A dot starts at the point (1,0) and turns anticlockwise. Can you estimate the height of the dot after it has turned through 45 degrees? Can you calculate its height?
### Squ-areas
##### Stage: 4 Challenge Level:
Three squares are drawn on the sides of a triangle ABC. Their areas are respectively 18 000, 20 000 and 26 000 square centimetres. If the outer vertices of the squares are joined, three more. . . .
### Strange Rectangle
##### Stage: 5 Challenge Level:
ABCD is a rectangle and P, Q, R and S are moveable points on the edges dividing the edges in certain ratios. Strangely PQRS is always a cyclic quadrilateral and you can find the angles.
##### Stage: 4 Challenge Level:
A 1 metre cube has one face on the ground and one face against a wall. A 4 metre ladder leans against the wall and just touches the cube. How high is the top of the ladder above the ground?
### Star Gazing
##### Stage: 4 Challenge Level:
Find the ratio of the outer shaded area to the inner area for a six pointed star and an eight pointed star.
### Compare Areas
##### Stage: 4 Challenge Level:
Which has the greatest area, a circle or a square inscribed in an isosceles, right angle triangle?
### Rhombus in Rectangle
##### Stage: 4 Challenge Level:
Take any rectangle ABCD such that AB > BC. The point P is on AB and Q is on CD. Show that there is exactly one position of P and Q such that APCQ is a rhombus.
### Partly Circles
##### Stage: 4 Challenge Level:
What is the same and what is different about these circle questions? What connections can you make?
### Pythagoras Proofs
##### Stage: 4 Challenge Level:
Can you make sense of these three proofs of Pythagoras' Theorem?
### Spherical Triangles on Very Big Spheres
##### Stage: 5 Challenge Level:
Shows that Pythagoras for Spherical Triangles reduces to Pythagoras's Theorem in the plane when the triangles are small relative to the radius of the sphere.
### Kite in a Square
##### Stage: 4 Challenge Level:
Can you make sense of the three methods to work out the area of the kite in the square?
### The Fire-fighter's Car Keys
##### Stage: 4 Challenge Level:
A fire-fighter needs to fill a bucket of water from the river and take it to a fire. What is the best point on the river bank for the fire-fighter to fill the bucket ?.
### All Tied Up
##### Stage: 4 Challenge Level:
A ribbon runs around a box so that it makes a complete loop with two parallel pieces of ribbon on the top. How long will the ribbon be?
### Circle Scaling
##### Stage: 4 Challenge Level:
You are given a circle with centre O. Describe how to construct with a straight edge and a pair of compasses, two other circles centre O so that the three circles have areas in the ratio 1:2:3.
### Circle Box
##### Stage: 4 Challenge Level:
It is obvious that we can fit four circles of diameter 1 unit in a square of side 2 without overlapping. What is the smallest square into which we can fit 3 circles of diameter 1 unit?
### The Spider and the Fly
##### Stage: 4 Challenge Level:
A spider is sitting in the middle of one of the smallest walls in a room and a fly is resting beside the window. What is the shortest distance the spider would have to crawl to catch the fly?
### Squaring the Circle and Circling the Square
##### Stage: 4 Challenge Level:
If you continue the pattern, can you predict what each of the following areas will be? Try to explain your prediction.
### The Dodecahedron
##### Stage: 5 Challenge Level:
What are the shortest distances between the centres of opposite faces of a regular solid dodecahedron on the surface and through the middle of the dodecahedron?
### Incircles
##### Stage: 5 Challenge Level:
The incircles of 3, 4, 5 and of 5, 12, 13 right angled triangles have radii 1 and 2 units respectively. What about triangles with an inradius of 3, 4 or 5 or ...?
### Orthogonal Circle
##### Stage: 5 Challenge Level:
Given any three non intersecting circles in the plane find another circle or straight line which cuts all three circles orthogonally.
### 30-60-90 Polypuzzle
##### Stage: 5 Challenge Level:
Re-arrange the pieces of the puzzle to form a rectangle and then to form an equilateral triangle. Calculate the angles and lengths.
### Ball Packing
##### Stage: 4 Challenge Level:
If a ball is rolled into the corner of a room how far is its centre from the corner?
### Xtra
##### Stage: 4 and 5 Challenge Level:
Find the sides of an equilateral triangle ABC where a trapezium BCPQ is drawn with BP=CQ=2 , PQ=1 and AP+AQ=sqrt7 . Note: there are 2 possible interpretations.
### Chord
##### Stage: 5 Challenge Level:
Equal touching circles have centres on a line. From a point of this line on a circle, a tangent is drawn to the farthest circle. Find the lengths of chords where the line cuts the other circles.
### Semi-square
##### Stage: 4 Challenge Level:
What is the ratio of the area of a square inscribed in a semicircle to the area of the square inscribed in the entire circle?
### Take a Square
##### Stage: 4 Challenge Level:
Cut off three right angled isosceles triangles to produce a pentagon. With two lines, cut the pentagon into three parts which can be rearranged into another square.
##### Stage: 4 Challenge Level:
The sides of a triangle are 25, 39 and 40 units of length. Find the diameter of the circumscribed circle.
### Corridors
##### Stage: 4 Challenge Level:
A 10x10x10 cube is made from 27 2x2 cubes with corridors between them. Find the shortest route from one corner to the opposite corner.
### Zig Zag
##### Stage: 4 Challenge Level:
Four identical right angled triangles are drawn on the sides of a square. Two face out, two face in. Why do the four vertices marked with dots lie on one line?
### Equilateral Areas
##### Stage: 4 Challenge Level:
ABC and DEF are equilateral triangles of side 3 and 4 respectively. Construct an equilateral triangle whose area is the sum of the area of ABC and DEF.
### Nicely Similar
##### Stage: 4 Challenge Level:
If the hypotenuse (base) length is 100cm and if an extra line splits the base into 36cm and 64cm parts, what were the side lengths for the original right-angled triangle? | 2,451 | 10,224 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5625 | 4 | CC-MAIN-2017-34 | latest | en | 0.877222 |
https://www.iitutor.com/function-notation/ | 1,558,644,397,000,000,000 | text/html | crawl-data/CC-MAIN-2019-22/segments/1558232257396.96/warc/CC-MAIN-20190523204120-20190523230120-00213.warc.gz | 811,821,620 | 17,679 | # Definition of Function Notation
Consider the relation $y=3x+2$, which is a function.
The $y$-values are determined from the $x$-values, so we say '$y$ is a function of $x$, which is abbreviated to $y=f(x)$.
So, the rule $y=3x+2$ can be also be written as following. $$f: \mapsto 3x+2$$ $$\text{or}$$ $$f(x)=3x+2$$ $$\text{or}$$ $$y=3x+2$$ Function $f$ such that $x$ is converted into $3x+2$.
### Example 1
If $f(x)=4x-5$, find $f(2)$.
### Example 2
If $f(x)=x^2-5x+1$, find $f(-1)$.
### Example 3
If $f(x)=x^2-3x+2$, find $f(x+1)$.
### Example 4
Given $f(x)=ax+b$, $f(1)=7$ and $f(2)=11$, find $a$ and $b$. | 256 | 617 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 4, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.3125 | 4 | CC-MAIN-2019-22 | latest | en | 0.740686 |
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Ripe bananas glow bright indigo under a black light (ultraviolet or UV). Researchers believe this is a 'lunch is ready' signal to banana eating animals like insects and bats that can see UV light.
The banana is classified botanically as a true berry - a many seeded pulpy fruit.
Supposedly, one of the first shipments of bananas to reach the colonies was in 1690 at Salem, Mass. They tried boiling them with pork. It took nearly 200 years after that culinary disaster for bananas to catch on with North Americans.
There are 44 people in the U.S. listed on whitepages.com with the last name 'Banana'
(Mark Morton, 'Gastronomica', Fall 2010)
Bananas first appeared in the United States in 1804. They were introduced to the public at the Philadelphia Centennial Exposition of America in 1876 and sold for 10 cents a piece, equal to average hourly wage of a working man.
Red Torch Banana
Today, average U.S. banana consumption is almost 30 pounds per year.
There are more than 500 different varieties of bananas.
Bananas trees are not trees. The banana plant is a giant herb.
Unripe bananas have about 25% starch and only 1% sugar. Natural enzyme action converts this high starch content to sugar, so ripe bananas have a 20% sugar content.
In the 15th and 16th centuries, Europeans knew the banana as the "Indian Fig".
Bananas on tree
The terms 'bee's knees,' 'the cat's pajamas,' and 'Yes, we have no bananas' were all coined by American cartoonist Tad Dorgan.
One variety of banana, the 'Ice Cream Banana', is BLUE. It turns yellow like other bananas when ripe, and has a taste like vanilla custard and a marshmallow texture.
‘Red bananas’ are maroon to dark purple when ripe, and even the fruit inside can have a slight pinkish color.
The average banana contains .6 grams fat.
Until the early 1800s in Hawaii, most banana varieties were 'kapu' - forbidden for women of Hawaii to eat, under penalty of death.
The very heart of the trunk of a banana 'tree' - inside the layers of bark fiber, is a white tube. It may be cooked, and has a taste and texture similar to bamboo shoots.
India, with rich bio-diversity of banana and plantain, is the largest producer and consumer with estimated production of 16 million tonnes of bananas annually. India's domestic production alone exceeds the entire world trade.
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https://www.projectrhea.org/rhea/index.php/The_Existence_and_Uniqueness_Theorem_for_Solutions_to_ODEs | 1,656,310,428,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103328647.18/warc/CC-MAIN-20220627043200-20220627073200-00096.warc.gz | 1,020,592,436 | 8,525 | ## The Existence and Uniqueness Theorem for Solutions to ODEs
A slecture by Yijia Wen
### 2.0 Abstract
Before starting this tutorial, you are supposed to be able to:
· Find an explicit solution for $\frac{dy}{dt}=f(t)$. This is the same thing as finding the integral of $f(t)$ with respect to $t$.
· Know the difference between a general solution and a particular solution satisfying the initial conditions.
· Check one function is a solution to an ODE.
· Distinguish ODE and PDE, know the usual notations.
· Know the basic concepts of ODEs (order, linearity, homogeneity, etc).
2.1 Concept
From the first example from 1.1, here we still suppose that we had a linear equation $ax+b=0$ with respect to $x$.
· When $a=0$, $b≠0$, there is no solution.
· When $a≠0$, there is one solution $x=-\frac{b}{a}$.
· When $a=b=0$, there are infinitely many solutions to this linear equation.
Similarly, an ODE may also have no solution, a unique solution or infinitely many solutions. The existence theorem is used to check whether there exists a solution for an ODE, while the uniqueness theorem is used to check whether there is one solution or infinitely many solutions.
2.2 Existence Theorem
First we are going to define the continuity of a function. Similar with what we have learnt in Calculus 1 but replaced by a 2-variable function, $f(t,y)$ is continuous at the point $(t=t_0,y=y_0)$ if here, it is defined, its limit exists and the function value equals to the limit value.
Now let's make an extension. Imagine we are looking at part of a graph of a function by a magnifier, then suppose $f(t,y)$ is continuous in a closed bounded area (seeing the graph from above the magnifier) in coordinates. So $f(t,y)$ will be continuous on $R^2$ if it's continuous at each point in $R^2$. To understand this, just imagine we are smoothly moving our magnifier over the graph, and we can see the graph is always continuous, without any exception.
The geometric meaning of a function with one variable is a line or a curve. We know a dot moves to form a line, and a line moves to form a surface. So, the geometric meaning of a function with two variables is going to be a surface in three-dimensional coordinates. Since $f(t,y)$ is continuous at every point in $R^3$, it will always touch the coordinates, geometrically. Hence, we know there exist solution(s) to this ODE.
Briefly speaking, An ODE will have solution(s) in one particular area, if the initial function $f(x,y)$ is continuous there.
2.3 Uniqueness Theorem
Now we apply the same consideration to the graph of $\frac{∂f}{∂t}$. We know $f(t,y)$ is graphed as a surface. Similar with the tangent line of a curve, if we are doing the partial derivative to a surface function, we will have a tangent plane for it. The tangent plane is parallel to $y$-axis and can slide back and forth, if we are treating $y$ as a constant. So basically, the partial derivative $\frac{∂f}{∂t}$, $\frac{∂f}{∂y}$ is the gradient of a surface $f(t,y)$. It shows how the surface changes (going up and down, being wide and narrow, etc.) while we are holding one variable as a constant and sliding the tangent plane.
If the partial derivative is continuous at the point $(t=t_0,y=y_0)$, it means the gradient is continuous. Just imagine the unceasingly flowing hills where you can never see the end within one particular area. Since the hills are unceasing and can't have two ridges, (Mathematically, $x$ must only corresponds to one value of $f(x)$ to make it a valid function) the solution is going to be unique when touching the coordinates.
Briefly speaking, An ODE will have a unique solution in one particular area, if the partial derivative \frac{∂f}{∂t}[/itex], $\frac{∂f}{∂y}$ is continuous here.
2.4 Exercises
Test the existence and uniqueness of the solution of the differential equations:
· $\frac{dy}{dx}+y=x+3$
· $\frac{dy}{dt}=y^2$
2.5 References
Institute of Natural and Mathematical Science, Massey University. (2017). 160.204 Differential Equations I: Course materials. Auckland, New Zealand.
Robinson, J. C. (2003). An introduction to ordinary differential equations. New York, NY., USA: Cambridge University Press. | 1,059 | 4,188 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.53125 | 5 | CC-MAIN-2022-27 | latest | en | 0.91084 |
http://nrich.maths.org/public/leg.php?code=-40&cl=4&cldcmpid=274 | 1,503,223,444,000,000,000 | text/html | crawl-data/CC-MAIN-2017-34/segments/1502886106367.1/warc/CC-MAIN-20170820092918-20170820112918-00299.warc.gz | 326,058,091 | 6,868 | # Search by Topic
#### Resources tagged with Mathematical induction similar to Ab Surd Ity:
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### There are 22 results
Broad Topics > Using, Applying and Reasoning about Mathematics > Mathematical induction
### Farey Fibonacci
##### Stage: 5 Short Challenge Level:
Investigate Farey sequences of ratios of Fibonacci numbers.
### Fibonacci Fashion
##### Stage: 5 Challenge Level:
What have Fibonacci numbers to do with solutions of the quadratic equation x^2 - x - 1 = 0 ?
### Golden Powers
##### Stage: 5 Challenge Level:
You add 1 to the golden ratio to get its square. How do you find higher powers?
### Farey Neighbours
##### Stage: 5 Challenge Level:
Farey sequences are lists of fractions in ascending order of magnitude. Can you prove that in every Farey sequence there is a special relationship between Farey neighbours?
### Golden Fractions
##### Stage: 5 Challenge Level:
Find the link between a sequence of continued fractions and the ratio of succesive Fibonacci numbers.
### Tens
##### Stage: 5 Challenge Level:
When is $7^n + 3^n$ a multiple of 10? Can you prove the result by two different methods?
### Growing
##### Stage: 5 Challenge Level:
Which is larger: (a) 1.000001^{1000000} or 2? (b) 100^{300} or 300! (i.e.factorial 300)
### Obviously?
##### Stage: 4 and 5 Challenge Level:
Find the values of n for which 1^n + 8^n - 3^n - 6^n is divisible by 6.
### Elevens
##### Stage: 5 Challenge Level:
Add powers of 3 and powers of 7 and get multiples of 11.
### Dirisibly Yours
##### Stage: 5 Challenge Level:
Find and explain a short and neat proof that 5^(2n+1) + 11^(2n+1) + 17^(2n+1) is divisible by 33 for every non negative integer n.
### OK! Now Prove It
##### Stage: 5 Challenge Level:
Make a conjecture about the sum of the squares of the odd positive integers. Can you prove it?
### Particularly General
##### Stage: 5 Challenge Level:
By proving these particular identities, prove the existence of general cases.
### Water Pistols
##### Stage: 5 Challenge Level:
With n people anywhere in a field each shoots a water pistol at the nearest person. In general who gets wet? What difference does it make if n is odd or even?
### An Introduction to Mathematical Induction
##### Stage: 5
This article gives an introduction to mathematical induction, a powerful method of mathematical proof.
### Binary Squares
##### Stage: 5 Challenge Level:
If a number N is expressed in binary by using only 'ones,' what can you say about its square (in binary)?
##### Stage: 4 Challenge Level:
A walk is made up of diagonal steps from left to right, starting at the origin and ending on the x-axis. How many paths are there for 4 steps, for 6 steps, for 8 steps?
### One Basket or Group Photo
##### Stage: 2, 3, 4 and 5 Challenge Level:
Libby Jared helped to set up NRICH and this is one of her favourite problems. It's a problem suitable for a wide age range and best tackled practically.
### Counting Binary Ops
##### Stage: 4 Challenge Level:
How many ways can the terms in an ordered list be combined by repeating a single binary operation. Show that for 4 terms there are 5 cases and find the number of cases for 5 terms and 6 terms.
### Converging Product
##### Stage: 5 Challenge Level:
In the limit you get the sum of an infinite geometric series. What about an infinite product (1+x)(1+x^2)(1+x^4)... ?
### Symmetric Tangles
##### Stage: 4
The tangles created by the twists and turns of the Conway rope trick are surprisingly symmetrical. Here's why!
### Gosh Cosh
##### Stage: 5 Challenge Level:
Explore the hyperbolic functions sinh and cosh using what you know about the exponential function.
### Overarch 2
##### Stage: 5 Challenge Level:
Bricks are 20cm long and 10cm high. How high could an arch be built without mortar on a flat horizontal surface, to overhang by 1 metre? How big an overhang is it possible to make like this? | 970 | 3,963 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.9375 | 4 | CC-MAIN-2017-34 | latest | en | 0.810588 |
null | null | null | null | null | null | The Price of Historical Amnesia
Mr. Briley teaches at Sandia Prepatory school in New Mexico.
As schools begin to reopen all over the country, teachers, especially those in history, will express disappointment over how much their students have forgotten during the summer. But perhaps we are too quick to chide our young people for not maintaining intellectual engagement. It may well be that their forgetfulness simply reflects the historical and cultural amnesia which seems to characterize modern memory in the United States.
As we bemoan low test scores and the short attention span of our youth, perhaps we should stop blaming video games and reflect upon our failure to examine the historical record and hold our leaders accountable The summer's headlines regarding corporate greed, a declining stock market, escalating violence in the Middle East, President Bush's demand that the Palestinians disavow Yasir Arafat, and an impending invasion of Iraq indicate that many American citizens are all too quick, like our school children, to forget and disengage. The school children may be taking their cues from the adults, and it's time we all went back to school in an effort to revitalize American democracy.
The accounting scandals surrounding such corporate giants as Enron and WorldCom have shaken investor confidence in the stock market. Corporate greed is shrinking the retirement accounts of many middle and working class Americans, whose hard work and plans for retirement have been shattered by C.E.O.'s stock options, creative accounting, and financial parachutes. The Bush administration has attempted to disassociate itself from these corporate scandals by taking a get tough approach to white collar crime, yet the President and Vice-President Cheney, while perhaps not guilty of criminal behavior, are products of the corporate culture which brought us to this sad state of affairs.
During the 2000 campaign, Ralph Nader, as presidential candidate of the Green Party, focused his candidacy on the issue of corporate responsibility. Yet, Nader struggled to get his message out to the American people. He was ignored by the mainstream media and was not allowed to participate in the presidential debates. In a classic Catch-22 situation, Nader was nixed from the debates because his candidacy failed to garner enough support in public opinion polls, while being disqualified from the debates guaranteed that the Green Party candidate would not get the media exposure he needed to rise in the polls. Now we know that Nader was right on target with his concerns regarding corporate behavior, but the media still fail to make this connection. The prophet Nader remains neglected, and this summer's national Green Party convention in Philadelphia was overlooked by America's corporate media. It is as if the Nader crusade to restore corporate responsibility never happened, and we are shocked to learn of corporate executive misbehavior.
This state of historical amnesia is also apparent in America's response to the tragic escalating violence in the Middle East between Israelis and Palestinians. President Bush has called for the creation of an evolving Palestinian state; however, the president insists that the Palestinian people must disassociate themselves from Chairman Arafat. There are, indeed, many problems with Arafat, but by what right does the United States dictate to the Palestinians their leadership choices? It is this tendency toward a selective democracy, usually friendly to American corporate interests, which has so often led the United States into trouble.
Bush's pronouncement to the Palestinians that Arafat must go is reminiscent of President Woodrow Wilson's declaration that he was going to teach the Mexicans to elect good men. This policy led Wilson to invade our neighbor to the south twice, with the last incursion antagonizing the Mexican people while American troops engaged in a futile search for Poncho Villa. In more recent history, the United States has intervened to overthrow or destabilize democratically-elected regimes in Iran and Guatemala in the 1950s and Chile in the 1970s.
Before President Bush lectures other nations on the lessons of democracy, it is well worth noting that the president lost a popular election and was elevated to the nation's highest office by a 5 to 4 Supreme Court decision. The election of 2000, which raised serious questions about the nature of American democracy, is all too often part of our historical amnesia. The Bush tax cut is also contributing to a growing federal deficit, as we try to expand the military spending in the war on terrorism while curtailing domestic expenses. The economic promise of the Bush financial tax windfall has quickly been erased from public memory.
The projected invasion of Iraq by the United States should produce a strong sense of déjà vu. We are almost daily reminded that Sadaam Hussein is a threat to his neighbors and the United States. Conveniently forgotten is the fact that during the Reagan presidency, Hussein was our man in Baghdad, checking the expansion of the extremist Iranians. We hear much today about the Iraqi dictator using poison gas on his own people, yet when these events occurred there was little protest from Washington. However, with the invasion of Kuwait, Hussein became a threat to the steady flow of Middle Eastern oil. Bush the elder put together an impressive international coalition; however, he was unsuccessful in toppling the Iraqi strongman. His son is now insistent upon finishing the job.
Before endorsing the president's invasion plans it might be useful to again shed our historical amnesia. Getting into a war is easy, but devising an exit strategy is complex, as politicians found with the Vietnam War. While the Gulf War of Bush the elder enjoyed initial popular support, it is well worth recalling the disillusionment of veterans regarding the government's failure to acknowledge Gulf War Syndrome. Also, current invasion plans for Iraq lack international support and may further destabilize the volatile Middle Eastern political climate, making the world less secure for Americans.
Indeed, it seems this summer that we are paying a heavy price for our selective memory and failure to stay engaged. Just as we would like our school children to remember their lessons and become more involved in the classroom, as American citizens we must offer a better role model and be ever vigilant in fostering our democracy, demanding more from ourselves, the media, and our leaders. We must reclaim our historical memory, for the price of historical/cultural amnesia is too great a price to pay. As the events of September 11 and this summer well document, we ignore the past at our own peril and that of our children.
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More Comments:
D Sharer - 10/6/2002
Thank you for the article - you've synthesized key examples from the past that should make the U.S. public "wake up" with the latest war cry. I hope members of Congress read the article - most of them seem to be in a fog.
As a social studies teacher, I hope I provide the connections for students presented in your article.
Pierre S. Troublion - 10/4/2002
Ye Olde Taxpayers
October 4, 2002
Dear Mr. Lloyd:
This could be interesting, provided you have an ironclad certification that we're really talking prime heartland property here, and with a wide view to the horizon. Don't want to find out that some boneheaded clerical error has given us title to swampland in Hanging Chad, Florida instead.
Not quite clear on the disarmament part of the deal. To make
that work you're going to need to talk to my friends down at Anti-Hypocrisy Land Brokers. We need to get cracking on a building permit for a two-family duplex in Palestine, and I don't mean Palestine, Texas. And while you're in town, please arrange for an energy insurance policy and make sure it's long term and renewable.
Of course, as I'm sure you'll understand, this entire transaction is contingent on financing. Dick and Trent down at Rubber Stamp
Notary Public, Inc. say they have a blank check for me, but if that sucker bounces, the whole deal will collapse like a Presidential Palace hit by a deviant scud missile. Then we'd have to send in the workout team from Blowback Partners Unlimited, and they are one weird bunch. Last time we used those turkeys we lost a new office wing and two towers.
With best e-gads,
Pierre S. Troublion
Cassandra - 10/4/2002
Never mind. I get it now. You weren't stating your own position; you were making fun of the Green party.
Sorry if I seem dense for not realizing that sooner, but I've encountered too many people who - in all seriousness - talk about how brutal the US is to Muslims in the Middle East and bring up the Crusades in the same sentence as though the Crusades can be taken as an example of US cruelty. Besides I've heard all I ever want to hear about how guilty Europeans and their descendents should feel about the Crusades. Especially as none of the people who bring up the Crusades ever mention how the Muslims tried to systematically exterminate everyone else in the world a few centuries earlier.
It has become a very sore point for me.
Cassandra - 10/4/2002
Excuse me, but did you say that the US government first attempted to impose its will on the Middle East 900 years ago during the crusades, calling it a war all about oil? I'm no historian, but even a grade school child ought to know that the US has only been around about 200 years. The Crusades took place almost seven centuries before the US was in existence, and at that time, the world had not yet been industrialized, and oil was not the valuable commodity that it is today.
And if you meant to say something else, then you're incredibly sloppy writer.
You use an awful lot of big words (and some French too!) in a vain attempt to hide the fact that you're babbling absolute nonsense.
Alec Lloyd - 10/4/2002
Sure, I've got the deed right here. Send me your bank account info and I'll throw in complete Iraqi disarmament as a free extra. I just need one more UN resolution (and your money) to seal the deal.
Frank Lee - 10/3/2002
This is a reasonable idea, IF applied consistently. That is a big IF, because consistency and the Near East are rarely found on the same page.
IF we have given Arafat "hundreds of millions" and IF he has put the funds into personal accounts in Switzerland [evidence thereof pending] and this therefore gives us the right to demand his ouster, what earthly reason could there be for AMERICANS not to also demand the ouster from power of war criminal Ariel Sharon, who is using BILLIONS of American taxdollars in military aid
(see http://www.us-israel.org/jsource/US-Israel/foreign_aid.html) to blast the occupied Palestinian territories into oblivion and willfully destroy three decades of American peace-making efforts under many U.S. presidents ?
And if we can't dictate to the peoples of the Near East who their leaders should be, why not stop the gravy train feeding the fanatics over there ? And why not start at the TOP of the list ?
F. Lee
Pierre S. Troublion - 10/3/2002
Would that be Calcium Estates, a choice set of properties running like a backbone around the Bush Ranch ?
Alec Lloyd - 10/3/2002
Yes, one more resolution. This time I know it will work.
Oh, and I've got some real estate you might be interested in...
Pierre S. Troublion - 10/3/2002
This is a clever satire, except that Nader didn't "use the word" crusade, that was Briley's term. Mr. Zweibel's otherwise neat exposure of the hypocrisy within the knee-jerk "peace" movement is thus, unfortunately, based on a crass error, even sloppier, if less egregious, than the error of who argue that the best way to teach the American President to speak English and observe basic norms of international diplomacy is to refuse to back up renewed U.N. inspection of Iraq by means of a new resolution with teeth.
P S Troublion
J. Phineas Zweibel - 10/2/2002
I am shocked, shocked! that President-in-your-dreams Nader would use a word such as "crusade" -- surely such language implies a Manichaean division of the world into good and evil, a certain "cowboy unilateralism," or as the French would say, "simplisme". While I do not attempt to excuse the malfeasance of ethically challenged CEO's, rather than simplistically condemning them we must understand the despair and hopelessness that form the root causes of their behavior -- otherwise we will merely perpetuate the cycle of, um, business. Our grief is not a call for prosecution. Not in our name!
Certainly as you apply the pedagogical equivalent of gingko biloba to the American body politic, you would not want us to forget the Crusades, in which, a mere 900 years ago, the US government first attempted to impose its will upon the Middle East in another war that was, naturally, all about oil.
Need I mention the religious-sensitivity implications of calling Nader a "prophet"? "There is no God but Dissent, and Ralph is his Prophet..."
Bill Heuisler - 10/1/2002
"...U.S. has intervened to overthrow or destabilize Democratically elected regimes..."
Some teacher. Mr. Briley's history evidently implants Left-wing bias into young minds by twisting or ignoring historical fact.
Three examples of deliberate anti-American ommissions:
1) Briley omits the fact that Poncho (sic) Villa was a bandit who raided into New Mexico and killed American soldiers.
2) Briley omits the fact that Mohammed Mossadegh was pronounced dictator by the Iranian Majlis on 8/11/1952 right before he "Nationalized" Western developed and owned oilfields.
3) Briley omits the fact that Colonel Jacobo Arbenz Guzman was put into power by Guatamalan Communists three months before he "expropriated" a quarter of a million acres of property owned by American businesses (ruining the Guatamalan ecomomy).
Wouldn't these facts give a balanced world-view to his students? Briley's saccherine concern for children seems to be overwhelmed by his anti-American bias. Schools shouldn't be hate-America reeducation camps. Teachers should tell the truth.
Bill Heuisler
don kates - 10/1/2002
Where did we get the right to say Arafat must go? Perhaps, Mr. Briley, the same place we got the right to say that tiny Israel should be subdivided with much of its land given away to another brutal Islamic dictatorship like those that already rule 98+% of the area of the Middle East? Why should a nation which allows full freedom of religion to all those w/in its borders be replaced with one where Jews cannot live, in a land that was traditionally theirs? Israel is currently populated in large part by refugees whom the Arab nations expelled (after confiscating all their property). Why does any Arab who does not want to live in a free democratic nation not just go to any of the horrendous Islamic dictatyorships and happily live there?
And, finally, as to Arafat, since we have given him hundreds of millions of dollars which he and his cronies have deposited in Swiss bank accounts instead of spending on the Palestinians, why shouldn't we be entitled to get rid of him? In that connection, let me pose you another question: Where do you think his wife got the money to live in a multi-million dollar Paris residence?
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https://ncertmcq.com/tag/class-9-maths-rs-aggarwal-solutions/page/2/ | 1,679,546,003,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296944996.49/warc/CC-MAIN-20230323034459-20230323064459-00116.warc.gz | 473,778,339 | 34,093 | ## RS Aggarwal Class 9 Solutions Chapter 14 Statistics Ex 14B
These Solutions are part of RS Aggarwal Solutions Class 9. Here we have given RS Aggarwal Solutions Class 9 Chapter 14 Statistics Ex 14B.
Other Exercises
Question 1.
Solution:
We shall take the game along x-axis and number of students along y-axis.
Question 2.
Solution:
We shall take the time on x-axis and temperature (in °C) on y-axis.
Question 3.
Solution:
We shall take name of vehicle on x-axis and velocity (in km/hr) on y-axis
Question 4.
Solution:
We shall take sports on x-axis and number of students on y-axis.
Question 5.
Solution:
We shall take years on x-axis and number of students on y-axis.
Question 6.
Solution:
We shall take years on x-axis and number of students on y-axis.
Question 7.
Solution:
We shall take years on x-axis and number of students on y-axis.
Question 8.
Solution:
We shall take years on x-axis and number of students on y-axis.
Question 9.
Solution:
We shall take years on x-axis and number of students on y-axis.
Question 10.
Solution:
We shall take years on x-axis and number of students on y-axis.
Question 11.
Solution:
We shall take week on x-axis and Rate per 10g (in Rs.) on y-axis.
Question 12.
Solution:
We shall take mode of transport on x-axis and number of students on y-axis.
Question 13.
Solution:
We see from the graph that
(i) It shows the marks obtained by a student in various subjects.
(ii) The student is very well in mathematics.
(iii) The student is very’ poor, in Hindi.
(iv) Average marks
= $$\frac { 60+35+75+50+60 }{ 5 }$$ (Here x = 5)
= $$\frac { 280 }{ 5 }$$
= 56 marks
Hope given RS Aggarwal Solutions Class 9 Chapter 14 Statistics Ex 14B are helpful to complete your math homework.
If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.
## RS Aggarwal Class 9 Solutions Chapter 9 Quadrilaterals and Parallelograms Ex 9B
These Solutions are part of RS Aggarwal Solutions Class 9. Here we have given RS Aggarwal Class 9 Solutions Chapter 9 Quadrilaterals and Parallelograms Ex 9B.
Other Exercises
Question 1.
Solution:
In parallelogram ABCD.
∠ A = 72°
But ∠ A = ∠ C (opposite angle of a ||gm)
∴ ∠ C = 72°
∴ ∠ A + ∠ B = 180° (co-interior angles)
=> 72° + ∠B = 180°
=> ∠B = 180° – 72°
=> ∠B = 108°
But ∠ B = ∠ D (opposite angles of a ||gm)
∴ ∠D = 108°
Hence ∠D = 108°, ∠ C = 72° and ∠ D = 108° Ans.
Question 2.
Solution:
In || gm ABCD, BD is its diagonal
and ∠DAB = 80° and ∠DBC = 60°
∴AB || DC
(co-interior angles)
=> 80° + ∠ADC = 180°
=> ∠ ADC = 180° – 80°
But ∠ ADB = ∠ DBC (Alternate angles)
But ∠ ADB + ∠ CDB = 100°
60° + ∠CDB = 100°
=> ∠CDB = 100° – 60° = 40°
Hence ∠CDB = 40° and ∠ ADB = 60° Ans.
Question 3.
Solution:
Given : In ||gm ABCD,
∠ A = 60° Bisectors of ∠ A and ∠ B meet DC at P.
To Prove : (i) ∠APB = 90°
(ii) AD = DP and PB = PC = BC
Proof: ∴ AD || B (opposite sides of a ||gm)
∴∠ A + ∠ B = 180° (co-interior angles)
But AP and BP are the bisectors of ∠A and ∠B
∴$$\frac { 1 }{ 2 }$$∠A + $$\frac { 1 }{ 2 }$$ ∠B = 90°
=> ∠PAB + ∠PBA = 90°
But in ∆ APB,
∠PAB + ∠PBA + ∠APB = 180° (angles of a triangle)
=> 90° + ∠APB = 180°
=> ∠APB = 180° – 90° = 90°
Hence ∠APB = 90°
(ii) ∠ A + ∠ D = 180° (co-interior angles)
and ∠ A – 60°
∴ ∠D = 180° – 60° = 120°
But ∠DAP = $$\frac { 1 }{ 2 }$$ ∠A = $$\frac { 1 }{ 2 }$$ x 60° = 30°
∴∠DPA = 180° – (∠DAP + ∠D)
= 180° – (30° + 120°)
= 180° – 150° = 30°
∠DAP = ∠DPA (each = 30°)
Hence AD = DP (sides opposite to equal angles)
In ∆ BCP,
∠ C = 60° (opposite to ∠ A)
∠CBP = $$\frac { 1 }{ 2 }$$ ∠ B = $$\frac { 1 }{ 2 }$$ x 120° = 60°
But ∠CPB + ∠CBP + ∠C = 180°
(Angles of a triangle)
=> ∠CPB + 60° + 60° = 180°
=> ∠CPB + 120° = 180°
=> ∠CPB = 180° – 120° = 60°
∆ CBP is an equilateral triangle and BC = CP = BP
=> PB – PC = BC
(iii) DC = DP + PC
(∴ DP = AD and PC = BC proved)
Hence proved.
Question 4.
Solution:
In ||gm ABCD,
AC and BD are joined
∠BAO = 35°, ∠ DAO = 40°
∠COD = 105°
∴ ∠AOB = ∠COD
(vertically opposite angles)
∴∠AOB = 105°
(i) Now in ∆ AOB,
∠ABO + ∠AOB + ∠OAB = 180°
(angles of a triangle)
=> ∠ABO + 105° + 35° = 180°
=> ∠ABO + 140° = 180°
=> ∠ABO = 180° – 140°
∠ ABO = 40°
(ii) ∴ AB || DC
∴ ∠ ABO = ∠ ODC (alternate angles)
∴ ∠ ODC = 40°
∴ ∠ ACB = ∠ DAO or ∠ DAC
(alternate angles)
= 40°
(iv) ∴ ∠ A + ∠ B = 180° (co-interior angles)
=> (40° + 35°) + ∠B = 180°
=> ∠B = 180° – 75° = 105°
=> ∠ CBD + ∠ABO = 105°
=> ∠CBD + 40° = 105°
=> ∠CBD = 105° – 40° = 65°
Hence ∠ CBD = 65° Ans.
Question 5.
Solution:
In ||gm ABCD
( ∠A = (2x + 25)° and ∠ B = (3x – 5)°
∴AD || BC (opposite sides of parallelogram)
∴∠ A + ∠B = 180° (co-interior angles)
=> 2x + 25° + 3x – 5° = 180°
=> 5x + 20° = 180°
=> 5x = 180° – 20°
=> 5x = 160° => x = $$\frac { { 160 }^{ o } }{ 5 }$$ = 32°
∴x = 32°
Now ∠A = 2x + 25° = 2 x 32° + 25°
= 64° + 25° = 89°
∠B = 3x – 5 = 3 x 32° – 5°
= 96° – 5° = 91°
∠ C = ∠ A (∴ opposite angles of ||gm)
= 89°
Similarly ∠B = ∠D
∠D = 91°
Hence ∠ A = 89°, ∠ B = 91°, ∠ C = 89°, ∠D = 91° Ans.
Question 6.
Solution:
Let ∠A and ∠B of a ||gm ABCD are adjacent angles.
∠A + ∠B = 180°
Let ∠B = x
Then ∠ A = $$\frac { 4 }{ 5 }$$ x
∴ x + $$\frac { 4 }{ 5 }$$ x = 180°
$$\frac { 9 }{ 5 }$$ x = 180°
=>$$\frac { { 180 }^{ o }\quad X\quad 5 }{ 9 }$$ = 100°
∴ ∠A = $$\frac { 4 }{ 5 }$$ x 100° = 80°
and ∠B = 100°
But ∠ C = ∠ A and ∠ D = ∠ B
(opposite angles of a || gm)
∴∠C = 80°, and ∠D = 100°
Hence ∠A = 80°, ∠B = 100°, ∠C = 80° and ∠ D = 100° Ans.
Question 7.
Solution:
Let the smallest angle ∠ A and the other angle ∠ B
Let ∠ A = x
Then ∠ B = 2x – 30°
But ∠ A + ∠ B = 180° (co-interior angles)
∴x + 2x – 30° = 180°
=> 3x = 180° + 30° = 210°
=> x = $$\frac { { 210 }^{ o } }{ 3 }$$ = 70°
∴ ∠ A = 70°
and ∠ B = 2x – 30° = 2 x 70° – 30°
= 140° – 30° = 110°
But ∠C = ∠ A and ∠D = ∠B
(opposite angles of a ||gm)
∠C = 70° and ∠D = 110°
Hence ∠A = 70°, ∠B = 110°, ∠C = 70° and ∠D = 110° Ans.
Question 8.
Solution:
In ||gm ABCD,
AB = 9.5 cm and perimeter = 30 cm
=> AB + BC + CD + DA = 30cm
=> AB + BC + AB + BC = 30 cm
( ∴ AB = CD and BC – DA opposite sides)
=> 2(AB + BC) = 30cm
=> AB + BC = 15cm
=> 9 5cm + BC = 15cm
∴BC = 15cm – 9.5cm = 5.5cm
Hence AB = 9.5cm, BC = 5.5cm,
CD = 9.5cm and DA = 5.5cm Ans.
Question 9.
Solution:
ABCD is a rhombus
AB = BC = CD = DA
(i)∴ AB || DC
∴ ∠ B + ∠ C = 180° (co-interior angles)
=> 110° + ∠C = 180°
=> ∠C = 180° – 110° = 70°
Question 10.
Solution:
In a rhombus,
Diagonals bisect each other at right angles
∴ AC and BC bisect each other at O at right angles.
But AC = 24 cm and BD = 18 cm
Question 11.
Solution:
Let ABCD he the rhombus whose diagonal are AC and BD which bisect each other at right angles at O.
Question 12.
Solution:
ABCD is a rectangle whose diagonals AC and BD bisect each other at O.
Question 13.
Solution:
ABCD is a square. A line CX cuts AB at X and diagonal BD at O such that
∠ COD = 80° and ∠ OXA = x°
∴∠ BOX = ∠ COD
(vertically opposite angles)
∴∠BOX = 80°
∴Diagonal BD bisects ∠ B and ∠ D
∴ ∠OBA or ∠OBX = 45°
Now in ∆ OBX,
Ext. ∠ OXA = ∠ BOX + ∠ OBX
=>x° = 80° + 45° = 125° Ans.
Question 14.
Solution:
Given : In ||gm ABCD, AC is joined. AL ⊥ BD and CM ⊥ BD
To prove :
(i) ∆ ALD ≅ ∆ CMB
(ii) AL = CM
Proof : In ∆ ALD and ∆ BMC
AD = BC (opposite sides of ||gm)
∠L = ∠M (each 90°)
∠ ADL = ∠ CBM (Alternate angles)
∴ ∆ ALD ≅ ∆ BMC. (AAS axiom)
∴ or A ALD ≅ A CMB.
AL = CM (c.p.c.t.) Hence proved.
Question 15.
Solution:
Given : In ∆ ABCD, bisectors of ∠A and ∠B meet each other at P.
To prove : ∠APB = 90°
∠A + ∠B = .180° (co-interior angles)
PA and PB are the bisectors of ∠ A and ∠B
Question 16.
Solution:
In ||gm ABCD,
P and Q are the points on AD and BC respectively such that AP = $$\frac { 1 }{ 3 }$$ AD and CQ = $$\frac { 1 }{ 3 }$$ BC
Question 17.
Solution:
Given : In ||gm ABCD, diagonals AC and BD bisect each other at O.
A line segment EOF is drawn, which meet AB at E and DC at F.
To -prove : OE = OF
Question 18.
Solution:
Given : ABCD is a ||gm.
AB is produced to E. Such that AB = BE
DE is joined which intersects BC in O.
To prove : ED bisects BC i.e. BO = OC
Question 19.
Solution:
Given : In ||gm ABCD, E is the midpoint BC
DE is joined and produced to meet AB on producing at F.
Question 20.
Solution:
Given : ∆ ABC and lines are drawn through A, B and C parallel to respectively BC, CA and AB forming ∆ PQR.
To prove : BC = $$\frac { 1 }{ 2 }$$ QR
Question 21.
Solution:
Given : In ∆ ABC, parallel lines are drawn through A, B and C respectively to the sides BC, CA and AB intersecting each other at P, Q and R.
Hope given RS Aggarwal Class 9 Solutions Chapter 9 Quadrilaterals and Parallelograms Ex 9B are helpful to complete your math homework.
If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.
## RS Aggarwal Class 9 Solutions Chapter 6 Coordinate Geometry Ex 6A
These Solutions are part of RS Aggarwal Solutions Class 9. Here we have given RS Aggarwal Class 9 Solutions Chapter 6 Coordinate Geometry Ex 6A.
Question 1.
Solution:
Co-ordinates of
A are ( – 6, 5)
B are (5, 4)
C are ( – 3, 2)
D are (2, – 2)
E are ( – 1, – 4)
Ans.
Question 2.
Solution:
The given points have been plotted as shown in the adjoining graph.
Where X’OX and YOY’ are the axis:
Question 3.
Solution:
(i) (7, 0) lies on x-axis as its ordinate is (0)
(ii) (0, – 5) lies on y-axis as its abscissa is (0)
(iii) (0, 1) lies on y-axis as its abscissa is (0)
(iv) ( – 4, 0) lies on jc-axis as its ordinate is (0)
Ans.
Question 4.
Solution:
(i) ( – 6, 5) lies in second quadrant because A is of the type (-, +)
(ii) ( – 3, – 2) lies in third quadrant because A is of the type (-, -)
(iii) (2, – 9) lies in fourth quadrant because it is of the type (+, -).
Question 5.
Solution:
In the given equation, .
y = x+1
Put x = 0, then y = 0 + 1 = 1
x = 1, then, y = 1 + 1=2
x = 2, then, y = 2 + 1 = 3
Now, plot the points as given in the table given below on the graph, and join them as shown.
Question 6.
Solution:
In the given equation
y = 3x + 2
Put x = 0,
then y = 3 x 0 + 2 = 0 + 2 = 2
x = 1, then
y = 3 x 1 + 2 = 3 + 2 = 5
and x = – 2, then
y = 3 x ( – 2) + 2 = – 6 + 2 = – 4
Question 7.
Solution:
In the given equation,
y = 5x – 3
Put x = 1,y = 5 x 1 – 3 = 5 – 3 = 2
x = 0 then,
y = 5 x 0 – 3 = 0 – 3 = – 3
and x = 2, then
y = 5 x 2 – 3 = 10 – 3 = 7
Question 8.
Solution:
In the given equation,
y = 3x,
Put x = 0,
then y = 3 x 0 = 0
Put x = 1, then
y = 3 x 1 = 3
Put x = – 1, then
y = 3 ( – 1) = – 3
Now, plot the points as given in the table below
and join them as shown.
Question 9.
Solution:
In the given equation, y = – x,
Put x = 1,
then y = – 1
Put x = 2, then
y = – 2
Put x = – 2, then
y = – ( – 2) = 2
Hope given RS Aggarwal Class 9 Solutions Chapter 6 Coordinate Geometry Ex 6A are helpful to complete your math homework.
If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.
## RS Aggarwal Class 9 Solutions Chapter 14 Statistics Ex 14A
These Solutions are part of RS Aggarwal Solutions Class 9. Here we have given RS Aggarwal Solutions Class 9 Chapter 14 Statistics Ex 14A.
Other Exercises
Question 1.
Solution:
Statistics is a science which deals with the collection, presentation, analysis and interpretations of numerical data.
Question 2.
Solution:
(i) Numerical facts alone constitute data
(ii) Qualitative characteristics like intelligence, poverty etc. which cannot be measured, numerically, don’t form data.
(iii) Data are an aggregate of facts. A single observation does not form data.
(iv) Data collected for a definite purpose may not be suited for another purpose.
(v) Data is different experiments are comparable.
Question 3.
Solution:
(i) Primary data : The data collected by the investigator himself with a definite plan in mind are called primary data.
(ii) Secondary data : The data collected by some one other than the investigator are called secondary data. The primary data is more reliable and relevant.
Question 4.
Solution:
(i) Variate : Any character which is capable of taking several different values is called a variate or variable.
(ii) Class interval : Each group into which the raw data is condensed, is called a class interval
(iii) Class size : The difference between the true upper limit and the true lower limit of a class is called class size.
(iv) Class Mark : $$\frac { upper\quad limit+lower\quad limit }{ 2 }$$ is called a class mark
(v) Class limits : Each class is bounded by two figures which are called class limits which are lower class limit and upper class limit.
(vi) True class limits : In exclusive form, the upper and lower limits of a class are respectively are the true upper limit and true lower limit but in inclusive form, the true lower limit of a class is obtained by subtracting O.S from lower limit of the class and for true limit, adding 0.5 to the upper limit.
(vii) Frequency of a class : The number of times an observation occurs in a class is called its frequency.
(viii) Cumulative frequency of a class : The cumulative frequency corresponding to a class is the sum of all frequencies upto and including that class.
Question 5.
Solution:
The given data can be represent in form of frequency table as given below:
Question 6.
Solution:
The frequency distribution table of the given data is given below :
Question 7.
Solution:
The frequency distribution table of the
Question 8.
Solution:
The frequency table is given below :
Question 9.
Solution:
The frequency table of given data is given below :
Question 10.
Solution:
The frequency distribution table of the given data in given below :
Question 11.
Solution:
The frequency table of the given data:
Question 12.
Solution:
The cumulative frequency of the given table is given below:
Question 13.
Solution:
The given table can be represented in a group frequency table in given below :
Question 14.
Solution:
Frequency table of the given cumulative frequency is given below :
Question 15.
Solution:
A frequency table of the given cumulative frequency table is given below :
Hope given RS Aggarwal Solutions Class 9 Chapter 14 Statistics Ex 14A are helpful to complete your math homework.
If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.
## RS Aggarwal Class 9 Solutions Chapter 8 Linear Equations in Two Variables Ex 8A
These Solutions are part of RS Aggarwal Solutions Class 9. Here we have given RS Aggarwal Class 9 Solutions Chapter 8 Linear Equations in Two Variables Ex 8A.
Question 1.
Solution:
(i) x = 5 is the line AB parallel to the y-axis at a distance of 5 units.
(ii) y = – 2 is the line CD parallel to x-axis at a distance of – 2 units.
(iii) x + 6 = 0 => x = – 6 is the line EF parallel to y-axis at a distance of – 6 units.
(iv) x + 7 = 0 => x = – 7 is the line PQ parallel to y-axis at a distance of – 7 units.
(v) y = 0 is the equation of x-axis. The graph of y = 0 is the line X’OX
(vi) x = 0 is the equation of y-axis.The graph of x = 0 is the line YOY’
Ans.
Question 2.
Solution:
In the given equation.
y = 3x
Put x = 1, then y = 3 x 1 = 3
Put x = 2, then y = 3 x 2 = 6
Put x = – 1, then y = 3 ( – 1) = – 3
Now, plot the points (1, 3), (2, 6) and ( – 1, – 3) as given the following table
and join them to form a line of the given equation.
Now from x = – 2,
draw a line parallel to y-axis at a distance of x = – 2, meeting the given line at P. From P, draw, a line parallel to x-axis joining y-axis at M, which is y = – 6 Hence, y = – 6 Ans.
Question 3.
Solution:
In the given equation x + 2y – 3 = 0
=> 2y = 3 – x
y = $$\frac { 3-x }{ 2 }$$
put x = 1,then
Question 4.
Solution:
(i) In the equation y = x
When x = 1, then y = 1
when x = 2, then y = 2
and when x = 3, then y = 3
Question 5.
Solution:
In the given equation
2x – 3y = 5
Question 6.
Solution:
In the given equation
2x + y = 6
=> y = 6 – 2x
Put x = 1, then y – 6 – 2 x 1 = 6 – 2 = 4
Put x = 2, then y = 6 – 2 x 2 = 6 – 4 = 2
Put x = 3, then y = 6 – 2 x 3 = 6 – 6 = 0
Question 7.
Solution:
In the given equation
3x + 2y = 6
Hope given RS Aggarwal Class 9 Solutions Chapter 8 Linear Equations in Two Variables Ex 8A are helpful to complete your math homework.
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## RS Aggarwal Class 9 Solutions Chapter 10 Area Ex 10A
These Solutions are part of RS Aggarwal Solutions Class 9. Here we have given RS Aggarwal Class 9 Solutions Chapter 10 Area Ex 10A.
Question 1.
Solution:
Given : In the figure, ABCD is a quadrilateral and
AB = CD = 5cm
Question 2.
Solution:
In ||gm ABCD,
AB = 10cm, altitude DL = 6cm
and BM is altitude on AD, and BM = 8 cm.
Question 3.
Solution:
Diagonals of rhombus are 16cm and 24 cm.
Area = $$\frac { 1 }{ 2 }$$ x product of diagonals
= $$\frac { 1 }{ 2 }$$ x 1st diagonal x 2nd diagonal
= $$\frac { 1 }{ 2 }$$ x 16 x 24
= 192 cm² Ans.
Question 4.
Solution:
Parallel sides of a trapezium are 9cm and 6cm and distance between them is 8cm
Question 5.
Solution:
from the figure
(i) In ∆ BCD, ∠ DBC = 90°
Question 6.
Solution:
In the fig, ABCD is a trapezium. AB || DC
AB = 7cm, AD = BC = 5cm.
Distance between AB and DC = 4 cm.
i.e. ⊥AL = ⊥BM = 4cm.
Question 7.
Solution:
Given : In quad. ABCD. AL⊥BD and CM⊥BD.
Question 8.
Solution:
In quad. ABCD, BD is its diagonal and AL⊥BD, CM⊥BD
Question 9.
Solution:
Given : ABCD is a trapezium in which AB || DC and its diagonals AC and BD intersect each other at O.
To prove : ar(∆ AOD) = ar(∆ BOC)
Question 10.
Solution:
Given : In the figure,
DE || BC.
Question 11.
Solution:
Given : In ∆ ABC, D and E are the points on AB and AC such that
ar( ∆ BCE) = ar( ∆ BCD)
To prove : DE || BC.
Proof : (∆ BCE) = ar(∆ BCD)
But these are on the same base BC.
Their altitudes are equal.
Hence DE || BC
Hence proved.
Question 12.
Solution:
Given : In ||gm ABCD, O is any. point inside the ||gm. OA, OB, OC and OD are joined.
Question 13.
Solution:
A line through D, parallel to AC, meets ‘BC produced in P. AP in joined which intersects CD at E.
To prove : ar( ∆ ABP) = ar(quad. ABCD).
Const. Join AC
Question 14.
Solution:
Given : ∆ ABC and ∆ DBC are on the same base BC with points A and D on , opposite sides of BC and
ar( ∆ ABC) = ar( ∆ DBC).
Question 15.
Solution:
Given : In ∆ ABC, AD is the median and P is a point on AD
BP and CP are joined
To prove : (i) ar(∆BDP) = ar(∆CDP)
(ii) ar( ∆ ABP) = ar( ∆ ACP)
Question 16.
Solution:
Given : In quad. ABCD, diagonals AC and BD intersect each other at O and BO = OD
To prove : ar(∆ ABC) = ar(∆ ADC)
Proof : In ∆ ABD,
Question 17.
Solution:
In ∆ ABC,D is mid point of BC
and E is midpoint of AD and BE is joined.
Question 18.
Solution:
Given : In ∆ ABC. D is a point on AB and AD is joined. E is mid point of AD EB and EC are joined.
To prove : ar( ∆ BEC) = $$\frac { 1 }{ 2 }$$ ar( ∆ ABC)
Proof : In ∆ ABD,
BE is its median
ar(∆ EBD) = ar(∆ ABE)
Question 19.
Solution:
Given : In ∆ ABC, D is midpoint of BC and E is die midpoint of BO is the midpoint of AE.
To prove that ar( ∆ BOE) = $$\frac { 1 }{ 8 }$$ ar(∆ ABC).
Question 20.
Solution:
Given : In ||gm ABCD, O is any point on diagonal AC.
To prove : ar( ∆ AOB) = ar( ∆ AOD)
Const. Join BD which intersects AC at P
Proof : In ∆ OBD,
P is midpoint of BD
(Diagonals of ||gm bisect each other)
Question 21.
Solution:
Given : ABCD is a ||gm.
P, Q, R and S are the midpoints of sides AB, BC, CD, DA respectively.
PQ, QR, RS and SP are joined.
Question 22.
Solution:
Given : In pentagon ABCDE,
EG || DA meets BA produced and
CF || DB, meets AB produced.
To prove : ar(pentagon ABCDE) = ar(∆ DGF)
Question 23.
Solution:
Given ; A ∆ ABC in which AD is the median.
To prove ; ar( ∆ ABD) = ar( ∆ ACD)
Const : Draw AE⊥BC.
Proof : Area of ∆ ABD
Question 24.
Solution:
Given : A ||gm ABCD in which AC is its diagonal which divides ||gm ABCD in two ∆ ABC and ∆ ADC.
Question 25.
Solution:
Given : In ∆ ABC,
D is a point on BC such that
BD = $$\frac { 1 }{ 2 }$$ DC
Question 26.
Solution:
Given : In ∆ ABC, D is a point on BC such that
BD : DC = m : n
Hope given RS Aggarwal Class 9 Solutions Chapter 10 Area Ex 10A are helpful to complete your math homework.
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## RS Aggarwal Class 9 Solutions Chapter 14 Statistics Ex 14C
These Solutions are part of RS Aggarwal Solutions Class 9. Here we have given RS Aggarwal Solutions Class 9 Chapter 14 Statistics Ex 14C.
Other Exercises
Question 1.
Solution:
It is clear that the given frequency distribution is in exclusive form, we represent the daily wages (in rupees) along x-axis and no. of workers along y-axis. Then we construct rectangles with class intervals as bases and corresponding frequencies as heights as shown given below:
Question 2.
Solution:
We shall take daily earnings along x-axis and number of stores along y-axis. Then we construct rectangles with the given class intervals as bases and corresponding frequency as height as shown in the graph.
Question 3.
Solution:
We shall take heights along x-axis and number of students along y-axis. Then we shall complete the rectangles with the given class intervals as bases and frequency as height and complete the histogram as shown below :
Question 4.
Solution:
We shall take class intervals along x-axis and frequency along y-axis Then we shall complete the rectangles with given class interval as bases and frequency as heights and complete the histogram as shown below.
Question 5.
Solution:
The given frequency distribution is in inclusive form first we convert it into exclusive form at given below.
Now, we shall take class intervals along x-axis and frequency along y-axis and draw rectangles with class intervals as bases and frequency as heights and complete the histogram as shown.
Question 6.
Solution:
Question 7.
Solution:
In the given distribution class intervals are different size. So, we shall calculate the adjusted frequency for each class. Here minimum class size is 4. We know that adjusted
frequency of the class is $$\frac { max\quad class\quad size }{ class\quad size\quad of\quad this\quad class }$$ x its frequency
Question 8.
Solution:
We shall take two imagined classes one 0-10 at the beginning with zero frequency a id other 70-80 at the end with zero frequency.
Now, plot the points” (5, 0), (15, 2), (25, 5), (35, 12), (45, 19), (55, 9), (65, 4), (75, 0) on the graph and join them its order to get a polygon as shown below.
Question 9.
Solution:
We take Age (in years) along x-axis and number of patients along y-axis. First we complete the histogram and them join the midpoints of their tops in order
We also take two imagined classes. One 0-10 at the beginning and other 70-80 at the end and also join their midpoints to complete the polygon as shown.
Question 10.
Solution:
We take class intervals along x-axis and frequency along y-axis.
First we complete the histogram and then join the midpoints of the tops of adjacent rectangles in order. We shall take two imagined classes 15-20 at the beginning and 50-55 at the end.
We also join their midpoints to complete the polygon as shown
Question 11.
Solution:
We represent class interval on x-axis and frequency on y-axis. First we construct the histogram and then join the midpoints of the tops of each rectangle by line segments. We shall take two imagined class i.e. 560-600 at the beginning and 840-900 at the end. Join their midpoints also to complete
Question 12.
Solution:
We will take the imagined class 11-0 at the beginning and 61-70 at the end. Each with frequency zero. Thus, we have,
Now we will mark the points A (-5.5, 0), B(5.5, 8), C(15.5, 3) D(25.5, 6), E(35.5, 12), F(45.5, 2), G(55.5, 7) and H(65.5, 0) on the graph and join them in order to get the required frequency polygon.
Hope given RS Aggarwal Solutions Class 9 Chapter 14 Statistics Ex 14C are helpful to complete your math homework.
If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.
## RS Aggarwal Class 9 Solutions Chapter 4 Lines and Triangles Ex 4D
These Solutions are part of RS Aggarwal Solutions Class 9. Here we have given RS Aggarwal Solutions Class 9 Chapter 4 Lines and Triangles Ex 4D.
Other Exercises
Question 1.
Solution:
In ∆ABC,
∠B = 76° and ∠C = 48°
But ∠A + ∠B + ∠C = 180°
(Sum of angles of a triangle)
=> ∠A + 76° + 48° = 180°
=> ∠ A + 124° = 180°
=> ∠A= 180° – 124° = 56°
Question 2.
Solution:
Angles of a triangle are in the ratio = 2:3:4
Let first angle = 2x
then second angle = 3x
and third angle = 4x
2x + 3x + 4x = 180°
(Sum of angles of a triangle)
=> 9x = 180°
=> x = $$\frac { { 180 }^{ o } }{ 9 }$$ = 20°
First angle = 2x = 2 x 20° = 40°
Second angle = 3x = 3 x 20° = 60°
and third angle = 4x = 4 x 20° = 80° Ans.
Question 3.
Solution:
In ∆ABC,
3∠A = 4∠B = 6∠C = x (Suppose)
Question 4.
Solution:
In ∆ABC,
∠ A + ∠B = 108° …(i)
∠B + ∠C – 130° …(ii)
But ∠A + ∠B + ∠C = 180° …(iii)
(sum of angles of a triangle)
Subtracting (i) from (iii),
∠C = 180° – 108° = 72°
Subtracting (ii) from (iii),
∠A = 180°- 130° = 50°
But ∠ A + ∠B = 108° (from i)
50° + ∠B = 108°
=> ∠B = 108° – 50° = 58°
Hence ∠A = 50°, ∠B = 58° and ∠C = 72° Ans.
Question 5.
Solution:
In ∆ABC,
∠A+∠B = 125° …(i)
∠A + ∠C = 113° …(ii)
But ∠A + ∠B + ∠C = 180° …(iii)
(sum of angles of a triangles) Subtracting, (i), from (iii),
∠C = 180°- 125° = 55°
Subtracting (ii) from (iii),
∠B = 180°- 113° – 67°
∠A + ∠B = 125°
∠ A + 67° = 125°
=> ∠ A = 125° – 67°
∠A = 58°
Hence ∠A = 58°, ∠B = 67° and ∠ C = 55° Ans.
Question 6.
Solution:
In ∆ PQR,
∠ P – ∠ Q = 42°
=> ∠P = 42°+∠Q …(i)
∠Q – ∠R = 21°
∠Q – 21°=∠R …(ii)
But ∠P + ∠Q + ∠R = 180°
(Sum of angles of a triangles)
42° + ∠Q + ∠Q + ∠Q – 21°= 180°
=> 21° + 3∠Q = 180°
=> 3∠Q = 180°- 21° = 159°
from ∠Q = $$\frac { { 159 }^{ o } }{ 3 }$$ = 53°
(i)∠P = 42° + ∠Q = 42° + 53° = 95°
and from (ii) ∠R = ∠Q – 21°
= 53° – 25° = 32°
Hence ∠P = 95°, ∠Q = 53° and ∠R = 32° Ans.
Question 7.
Solution:
Let ∠ A, ∠ B and ∠ C are the three angles of A ABC.
and ∠A + ∠B = 116° …(i)
Question 8.
Solution:
Let ∠ A, ∠ B and ∠ C are the three angles of the ∆ ABC
Let ∠ A = ∠ B = x
then ∠C = x + 48°
But ∠A + ∠B + ∠C = 180°
(Sum of angles of a triangle)
x + x + x + 18° = 180°
=> 3x + 18° = 180°
=> 3x = 180° – 18° = 162°
x = $$\frac { { 162 }^{ o } }{ 3 }$$ = 54°
∠A = 54°, ∠B = 54° and ∠C = 54° + 18° = 72°
Hence angles are 54°, 54 and 72° Ans.
Question 9.
Solution:
Let the smallest angle of a triangle = x°
their second angle = 2x°
and third angle = 3x°
But sum of angle of a triangle = 180°
x + 2x + 3x = 180°
=> 6x = 180°
=> x – $$\frac { { 180 }^{ o } }{ 6 }$$ = 30°
Hence smallest angle = 30°
Second angle = 2 x 30° = 60°
and third angle = 3 x 30° = 90° Ans.
Question 10.
Solution:
In a right angled triangle.
one angle is = 90°
Sum of other two acute angles = 90°
But one acute angle = 53°
Second acute angle = 90° – 53° = 37°
Hence angle of the triangle with be 90°, 53°, 37° Ans.
Question 11.
Solution:
Given : In ∆ ABC,
∠ A = ∠B + ∠C
To Prove : ∆ABC is a right-angled
Proof : We know that in ∆ABC,
∠A + ∠B + ∠C = 180°
(angles of a triangle)
But ∠ A = ∠ B + C given
∠A + (∠B + ∠C) = 180°
=> ∠A + ∠A = 180°
=> 2∠A = 180°
=> ∠ A = $$\frac { { 180 }^{ o } }{ 2 }$$ = 90°
∠ A = 90°
Hence ∆ ABC is a right-angled Hence proved.
Question 12.
Solution:
Given. In ∆ ABC, ∠A = 90°
AL ⊥ BC.
To Prove : ∠BAL = ∠ACB
Proof : In ∆ ABC, AL ⊥ BC
In right angled ∆ALC,
∠ ACB + ∠ CAL = 90° …(i)
( ∴∠L = 90°)
But ∠ A = 90° ‘
=> ∠ BAL + ∠ CAL = 90° …(ii)
From (i) and (ii),
∠BAL + ∠ CAL= ∠ ACB+ ∠CAL
=> ∠ BAL = ∠ ACB Hence proved.
Question 13.
Solution:
Given. In ∆ABC,
Each angle is less than the sum of the other two angles
∠A< ∠B + ∠C
∠B < ∠C + ∠A
and ∠C< ∠A + ∠C
Proof : ∠ A < ∠B + ∠C
∠A + ∠A < ∠A + ∠B + ∠C => 2 ∠ A < 180°
(∴ ∠A+∠B+∠C=180°)
∠A < $$\frac { { 180 }^{ o } }{ 2 }$$ => ∠A< 90
Similarly, we can prove that,
∠B < 90° and ∠C < 90°
∴ each angle is less than 90°
Hence, triangle is an acute angled triangle. Hence proved.
Question 14.
Solution:
Given. In ∆ABC,
∠B > ∠A + ∠C
Question 15.
Solution:
In ∆ABC
∠ ABC = 43° and Ext. ∠ ACD = 128°
Question 16.
Solution:
∠ ABC + ∠ ABD = 180°
(Linear pair)
Question 17.
Solution:
(i)In the figure, ∠BAE =110° and ∠ACD = 120°.
Question 18.
Solution:
In the figure,
∠A = 55°, ∠B = 45°, ∠C = 30° Join AD and produce it to E
Question 19.
Solution:
In the figure,
∠EAC = 108°,
AD divides ∠ BAC in the ratio 1 : 3
∠EAC + ∠ BAC = 180°
(Linear pair)
Question 20.
Solution:
Sides BC, CA and AB
are produced in order forming exterior
angles ∠ ACD,
Question 21.
Solution:
Given : Two ∆ s DFB and ACF intersect each other as shown in the figure.
To Prove : ∠A + ∠B + ∠C + ∠D + ∠E + ∠F = 360°
Proof : In ∆ DFB,
∠D + ∠F + ∠B = 180°
(sum of angles of a triangle)
Similarly,in ∆ ACE
∠A + ∠C + ∠E = 180° …(ii)
Adding (i) and (ii), we get :
∠D + ∠F + ∠B+ ∠A+ ∠C + ∠E = 180° + 180°
=> ∠A+∠B+∠C+∠D+∠E + ∠ F = 360°
Hence proved.
Question 22.
Solution:
In the figure,
ABC is a triangle
and OB and OC are the angle
bisectors of ∠ B and ∠ C meeting each other at O.
∠ A = 70°
In ∆ ABC,
∠A + ∠B + ∠C = 180°
(sum of angles of a triangle)
Question 23.
Solution:
In ∆ABC, ∠ A = 40°
Sides AB and AC are produced forming exterior angles ∠ CBD and ∠ BCE
Question 24.
Solution:
In the figure, ∆ABC is triangle and ∠A : ∠B : ∠C = 3 : 2 : 1
AC ⊥ CD.
∠ A + ∠B + ∠C = 180°
(sum of angles of a triangle)
But ∠A : ∠B : ∠C = 3 : 2 : 1
Question 25.
Solution:
In ∆ ABC
AN is the bisector of ∠ A
∠NAB =$$\frac { 1 }{ 2 }$$ ∠A.
Now in right angled ∆ AMB,
∠B + ∠MAB = 90° (∠M = 90°)
Question 26.
Solution:
(i) False: As a triangle has only one right angle
(ii) True: If two angles will be obtuse, then the third angle will not exist.
(iii) False: As an acute-angled triangle all the three angles are acute.
(iv) False: As if each angle will be less than 60°, then their sum will be less than 60° x 3 = 180°, which is not true.
(v) True: As the sum of three angles will be 60° x 3 = 180°, which is true.
(vi) True: A triangle can be possible if the sum of its angles is 180°
But the given triangle having angles 10° + 80° + 100° = 190° is not possible.
Hope given RS Aggarwal Solutions Class 9 Chapter 4 Lines and Triangles Ex 4D are helpful to complete your math homework.
If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.
## RS Aggarwal Class 9 Solutions Chapter 9 Quadrilaterals and Parallelograms Ex 9C
These Solutions are part of RS Aggarwal Solutions Class 9. Here we have given RS Aggarwal Class 9 Solutions Chapter 9 Quadrilaterals and Parallelograms Ex 9C.
Other Exercises
Question 1.
Solution:
Given : In trapezium ABCD,
AB || DC and E is the midpoint of AD.
A line EF ||AB is drawn meeting BC at F.
To prove : F is midpoint of BC
Question 2.
Solution:
Given : In ||gm ABCD, E and F are the mid points of AB and CD respectively. A line segment GH is drawn which intersects AD, EF and BC at G, P and H respectively.
To prove : GP = PH
Question 3.
Solution:
Given : In trapezium ABCD, AB || DC
P, Q are the midpoints of sides AD and BC respectively
DQ is joined and produced to meet AB produced at E
Join AC which intersects PQ at R.
Question 4.
Solution:
Given : In ∆ ABC,
AD is the mid point of BC
DE || AB is drawn. BE is joined.
To prove : BE is the median of ∆ ABC.
Proof : In ∆ ABC
Question 5.
Solution:
Given : In ∆ ABC, AD and BE are the medians. DF || BE is drawn meeting AC at F.
To prove : CF = $$\frac { 1 }{ 4 }$$ BC.
Question 6.
Solution:
Given : In ||gm ABCD, E is mid point of DC.
EB is joined and through D, DEG || EB is drawn which meets CB produced at G and cuts AB at F.
To prove : (i)AD = $$\frac { 1 }{ 2 }$$ GC
Question 7.
Solution:
Given : In ∆ ABC,
D, E and F are the mid points of sides BC, CA and AB respectively
DE, EF and FD are joined.
Question 8.
Solution:
Given : In ∆ ABC, D, E and F are the mid points of sides BC, CA and AB respectively
Question 9.
Solution:
Given : In rectangle ABCD, P, Q, R and S are the midpoints of its sides AB, BC, CD and DA respectively PQ, QR, RS and SP are joined.
To prove : PQRS is a rhombus.
Question 10.
Solution:
Given : In rhombus ABCD, P, Q, R and S are the mid points of sides AB, BC, CD and DA respectively PQ, QR, RS and SP are joined.
Question 11.
Solution:
Given : In square ABCD, P,Q,R and S are the mid points of sides AB, BC, CD and DA respectively. PQ, QR, RS and SP are joined.
Question 12.
Solution:
Given : In quadrilateral ABCD, P, Q, R and S are the midpoints of PQ, QR, RS and SP respectively PR and QS are joined.
To prove : PR and QS bisect each other
Const. Join PQ, QR, RS and SP and AC
Proof : In ∆ ABC,
But diagonals of a ||gm bisect each other PR and QS bisect each other.
∴ PR and QS bisect each other
Question 13.
Solution:
Given : ABCD is a quadrilateral. Whose diagonals AC and BD intersect each other at O at right angles.
P, Q, R and S are the mid points of sides AB, BC, CD and DA respectively. PQ, QR, QS and SP are joined.
Hope given RS Aggarwal Class 9 Solutions Chapter 9 Quadrilaterals and Parallelograms Ex 9C are helpful to complete your math homework.
If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.
## RS Aggarwal Class 9 Solutions Chapter 7 Areas Ex 7A
These Solutions are part of RS Aggarwal Solutions Class 9. Here we have given RS Aggarwal Class 9 Solutions Chapter 7 Areas Ex 7A.
Question 1.
Solution:
Base of the triangle (b) = 24cm and height (h) = 14.5 cm
∴ Area = $$\frac { 1 }{ 2 }$$ x b x h = $$\frac { 1 }{ 2 }$$ x 24 x 14.5 cm²
= 174 cm² Ans.
Question 2.
Solution:
Let the length of altitude of the triangular field = x then its base = 3x.
Question 3.
Solution:
Sides of a triangle = 42cm, 34cm and 20cm
Let a = 42cm, b = 34cm and c = 20 cm
Question 4.
Solution:
Sides of the triangle = 18cm, 24cm and 30cm
Let a = 18 cm, b = 24 cm and c = 30cm
Question 5.
Solution:
Sides of triangular field ABC arc 91m, 98m and 105m
Let AC be the longest side
∴ BD⊥AC
Here a = 98m, b = 105m and c = 91m
Question 6.
Solution:
Perimeter of triangle = 150m
Ratio in the sides = 5:12:13
Let sides be 5x, 12x and 13x
Question 7.
Solution:
Perimeter of a triangular field = 540m
Ratio is its sides = 25 : 17 : 12
Question 8.
Solution:
Perimeter of the triangular field = 324 m
Length of the sides are 85m and 154m
Question 9.
Solution:
Length of sides are
13 cm, 13 cm and 20cm
Question 10.
Solution:
Base of the isosceles triangle ABC = 80cm
Area = 360 cm²
Question 11.
Solution:
Perimeter of the triangle
ABC = 42 cm.
Let length of each equal sides = x
Question 12.
Solution:
Area of equilateral triangle = 36√3 cm².
Let length of each side = a
Question 13.
Solution:
Area of equilateral triangle = 81√3 cm²
Let length of each side = a
Question 14.
Solution:
∆ ABC is a right angled triangle, right angle at B.
∴ BC 48cm and AC = 50cm
Question 15.
Solution:
Each side of equilateral triangle
(a) = 8cm.
Question 16.
Solution:
Let a be the each side of
the equilateral triangle.
Question 17.
Solution:
The given umbrella has 12 triangular pieces of the size 50cm x 20cm x 50cm. We see that each piece is of an isosceles triangle shape and we have to find firstly area of one such triangle.
Question 18.
Solution:
The given floral design is made of 16 tiles
The size of each tile is 16cm 12cm, 20cm
Now we have to find the area of firstly one tile
Question 19.
Solution:
Question 20.
Solution:
In the figure, ABCD is a quadrilateral
AB = 42 cm, BC = 21 cm, CD = 29 cm,
DA = 34 cm and ∠CBD = 90°
Question 21.
Solution:
from the figure
∆DAB
Question 22.
Solution:
Question 23.
Solution:
from the figure,
We know that
Question 24.
Solution:
Hope given RS Aggarwal Class 9 Solutions Chapter 7 Areas Ex 7A are helpful to complete your math homework.
If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.
## RS Aggarwal Class 9 Solutions Chapter 12 Geometrical Constructions Ex 12A
These Solutions are part of RS Aggarwal Solutions Class 9. Here we have given RS Aggarwal Class 9 Solutions Chapter 12 Geometrical Constructions Ex 12A.
Question 1.
Solution:
Steps of Constructions :
(i) Draw a line segment AB = 5cm.
(ii) With A as centre and a radius equal to more than half of AB, drawn two arcs one above and other below of AB.
(iii) With centre B, and with same radius, draw two arcs intersecting the previously arcs at C and D respectively.
(iv) Join CD, intersecting AB at P.
Then CD is the perpendicular bisector of AB at the point P.
Question 2.
Solution:
Steps of constructions.
(i) Draw a line segment AB.
(ii) With A as centre and with small radius drawn arc cutting AB at P.
(iii) With P as centre and same radius draw another arc cutting the previous arc at Q and then R.
(iv) Bisect arc QR at S.
(v) Join AS and produce it to X such that ∠ BAX = 90°.
(vi) Now with centres P and S and with a suitable radius, draw two arcs intersecting each other at T.
(vii) Join AT and produced it to C Then ∠BAC = 45°.
(viii) Again with centres P and T and suitable radius draw two arcs intersecting each at D.
AD is the bisector of ∠ BAC
Question 3.
Solution:
Steps of construction.
(i) Draw a line segment AB.
(ii) With centre A and same radius draw an arc which meets AB at P.
(iii) With centre P and same radius, draw arcs first at Q and then at R.
(iv) With centres Q and R, draw arcs intersecting each other at C intersecting the first arc at T.
(v) Join AC
Then ∠BAC = 90°
(vi) Now with centres P and T and with some suitable radius, draw two arcs intersecting each other at L.
(vii) Join AL and produce it to D.
Then AD is the bisector of ∠ BAC.
Question 4.
Solution:
Steps of construction.
(i) Draw a line segment BC = 5cm.
(ii) With centres B and C and radius
5cm, draw two arcs intersecting each other at A.
(iii) Join AB and AC.
Then ∆ ABC is the required equilateral triangle.
Question 5.
Solution:
We know that altitudes of equilateral triangle are equal and each angle is 60°.
Steps of construction.
(i) Draw a line XY and take a point D on it.
(ii) At D, draw a perpendicular and cut off DA = 5.4cm.
(iii) At A draw angles of 30° on each side of AD which meet XY at B and C respectively.
Then ∆ ABC is the required triangle.
Question 6.
Solution:
Steps of construction :
(i) Draw a line segment BC = 5cm
(ii) With centre B and radius 3.8 cm draw an arc.
(iii) With centre C and radius 2.6 cm draw another arc intersecting the first arc at A.
(iv) Join AB and AC.
Then ∆ ABC is the required triangle.
Question 7.
Solution:
Steps of construction :
(i) Draw a line segment BC = 4.7cm.
(ii) At B, draw a ray BX making an angle of 60° with BC.
(iii) At C, draw another ray, CY making an angle of 30° which intersects the ray BX at A ,
Then ∆ ABC is the required triangle On measuring ∠ A, it is 90°.
Question 8.
Solution:
Steps of Construction :
(i) Draw a line segment QR * 5cm.
(ii) With centres Q and R and radius equal to 4.5cm, draw arcs intersecting eachother at P.
(iii) Join PQ and PR.
Then ∆ PQR is the required triangle.
Question 9.
Solution:
We know that in an isosceles triangle, two sides are equal and so their opposite angles are also equal.
Question 10.
Solution:
Steps of constructions :
(i) Draw a line segment BC = 4.5cm.
(ii) At B, draw a ray BX making an angle of 90° with BC.
(iii) With centre C and radius 5.3 cm, draw an arc intersecting BX at A.
(iv) Join AC.
Then ∆ ABC is the required right angled triangle.
Question 11.
Solution:
Steps of constructions :
(i) Draw a line XY.
(ii) Take a point D on XY.
(iii) Draw a perpendicular at D and cut off DA = 4.8 cm
(iv) At A, draw a line LM parallel to XY.
(v) At A, draw an angle of 30° with LM on one side and an angle of 60° with LM on other side meeting XY at B and C respectively
Then ∆ ABC is the required triangle.
Question 12.
Solution:
Steps of constructions :
(i) Draw a line segment EF = 12cm.
(ii) At E, draw a ray EX making an acute angle with EF.
(iii)From EX,cut off 3+2+4=9 equal parts.
(iv) Join E9 F.
(v) From E5 and E3, draw lines parallel to E9 F meeting EF at C and B respectively.
(vi) With centre B and radius EB and with centre C and radius CF, draw arcs intersecting eachother at A.
(vii) Join AB and AC.
Then ∆ ABC is the required triangle.
Question 13.
Solution:
Steps of constructions :
(i) Draw a line segment BC = 4.5cm.
(ii) At B, draw a ray BX making an angle of 60° and cut off BD = 8cm.
(iii) Join DC.
(iv) Draw the perpendicular bisector of BD which intersects BX at A.
(v) Join AC.
Then ∆ ABC is the required triangle.
Question 14.
Solution:
Steps of Constructions :
(i) Draw a line segment BC = 5.2 cm.
(ii) At B draw a ray BX making an angle of 30°.
(iii) From BX, cut off BD = 3.5cm.
(iv) Join DC.
(v) Draw perpendicular bisector of DC which intersects BX at A.
(vi) Join AC.
Then ∆ ABC is the required triangle.
Hope given RS Aggarwal Class 9 Solutions Chapter 12 Geometrical Constructions Ex 12A are helpful to complete your math homework.
If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you. | 14,054 | 41,000 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.40625 | 4 | CC-MAIN-2023-14 | latest | en | 0.863112 |
http://metamath.tirix.org/mpests/afv20defat.html | 1,721,081,098,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514713.74/warc/CC-MAIN-20240715194155-20240715224155-00366.warc.gz | 25,840,615 | 2,155 | # Metamath Proof Explorer
## Theorem afv20defat
Description: If the alternate function value at an argument is the empty set, the function is defined at this argument. (Contributed by AV, 3-Sep-2022)
Ref Expression
Assertion afv20defat ${⊢}\left({F}\mathrm{\text{'}\text{'}\text{'}\text{'}}{A}\right)=\varnothing \to {F}\mathrm{defAt}{A}$
### Proof
Step Hyp Ref Expression
1 ndfatafv2 ${⊢}¬{F}\mathrm{defAt}{A}\to \left({F}\mathrm{\text{'}\text{'}\text{'}\text{'}}{A}\right)=𝒫\bigcup \mathrm{ran}{F}$
2 pwne0 ${⊢}𝒫\bigcup \mathrm{ran}{F}\ne \varnothing$
3 2 neii ${⊢}¬𝒫\bigcup \mathrm{ran}{F}=\varnothing$
4 eqeq1 ${⊢}\left({F}\mathrm{\text{'}\text{'}\text{'}\text{'}}{A}\right)=𝒫\bigcup \mathrm{ran}{F}\to \left(\left({F}\mathrm{\text{'}\text{'}\text{'}\text{'}}{A}\right)=\varnothing ↔𝒫\bigcup \mathrm{ran}{F}=\varnothing \right)$
5 3 4 mtbiri ${⊢}\left({F}\mathrm{\text{'}\text{'}\text{'}\text{'}}{A}\right)=𝒫\bigcup \mathrm{ran}{F}\to ¬\left({F}\mathrm{\text{'}\text{'}\text{'}\text{'}}{A}\right)=\varnothing$
6 1 5 syl ${⊢}¬{F}\mathrm{defAt}{A}\to ¬\left({F}\mathrm{\text{'}\text{'}\text{'}\text{'}}{A}\right)=\varnothing$
7 6 con4i ${⊢}\left({F}\mathrm{\text{'}\text{'}\text{'}\text{'}}{A}\right)=\varnothing \to {F}\mathrm{defAt}{A}$ | 523 | 1,245 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 8, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.53125 | 4 | CC-MAIN-2024-30 | latest | en | 0.333202 |
http://themetricmaven.com/?p=6191 | 1,582,915,930,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875147628.27/warc/CC-MAIN-20200228170007-20200228200007-00526.warc.gz | 142,679,056 | 10,906 | The Count Only Counts — He Does Not Measure
By The Metric Maven
Bulldog Edition
In many television programs about mathematics that involve weights and measures, one is often taken to an open air market. The presenter will immediately seize upon the utility of numbers which have numerous divisors. The number twelve will be immediately enlisted. If one has a dozen eggs, then it can be divided up by 1, 2, 3, 4, 6 and 12. Often they move on to describe the amazing number of ways that 60 may be divided: 1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30 and 60, which is why one has clocks with 60 seconds in a minute, and 60 minutes in an hour. One can imagine oranges, apples, pears and such all being sold in integer groups. Often it has been my experience that a person can purchase any of these fruits in any number they wish.
When one considers purchasing walnuts, they are small enough that counting them out begins to tax one’s time. It is still possible, but selling them in 60 walnut quantities takes time to count out. It also takes time for the purchaser to count them out, and make certain that all 60 walnuts are in a given bag.
Wheat is a commodity that like oranges, eggs and walnuts, exists in integer units, but the individual grains are so small that the amount of time needed to count out 7000 of them, which was the definition of a pound, is prohibitive. Do my seven thousand wheat grains each have the same mass as those used to define a “grain“? Counting out seven-thousand grains definitely takes a lot of time, and checking each one against a “standard” grain would be untenable. Of course, one could count out 7000 wheat grains and then use a balance to compare a bag with 7000 grains to one which you are pouring into a second sack. When the balance is level, a naive consumer might assume that the two bags contain exactly the same number of grains. Who is going to take the time to count?
On closer examination, one knows that the reference bag has 7000 grains, but because of the variation in the masses of individual grains, perhaps because they came from a location far away in a country with different growing conditions, the new bag might contain more than 7000 grains, they are just smaller, and each possess less mass. This is the beginning of the idea of measurement, versus the notion of counting. People seem to realize that the same amount of “stuff” is in each bag if they balance, even if the individual grain count does not match. The question is, who’s bag of 7000 grains should be the one used by everyone as a standard? This is where the modern notion of measurement begins to appear.
One can’t be certain that the number of grains in all the bags are equal to the seven-thousand in the “standard” bag, but instinctively people seem satisfied that the same “amount” of wheat has been meted out.
Illustration of Hooke’s Law (Wikimedia Commons)
Robert Hooke (1635-1703) was the first to note that the length of a spring, within limits, is directly proportional to the force of an object which hangs from it. We can take our 7000 wheat grains, hang them from a spring which obeys “Hooke’s Law” and use the length the spring stretches, using our standard, as a known “calibrated” point. In the case of a spring we could put a pointer on the spring, and then place a mark at zero, when no grains are being measured, and a mark at 7000 wheat grains. A graduated scale can be placed behind the pointer. The location of the pointer is no longer restricted to single units of grain, it can point to an infinite number of locations along the scale distance from zero to seven thousand wheat grains. The divisions on the scale can be subdivided at will to produce more and more precision. We have stopped counting, and have begun to measure.
We can define seven-thousand wheat grains in terms of an indirect abstract quantity, not attached to a specific concrete item, such as cloth, grain or wood. This proxy quantity of “general stuff” we call an avoirdupois pound. The pound can in turn be used as a reference amount for a measurement of the quantity of any substance, corn, wheat, fish, bird seed or whatever. A person can fabricate a metal object which deflects the measurement pointer by the same amount as the wheat grains which make up a pound so that we can have a more stable, reproducible, and reliable standard. A second check can be accomplished by using a balance to make certain the two objects, the grains of wheat and the piece of metal, have the same amount of “stuff” in them. We call this abstract amount of stuff “mass” these days. So now we have created a one-pound mass for a standard, and we can measure commodities to as much of an exactness as we can produce graduations for the pointer to point at, and resolution for our eyes to read.
Once again, it is a problem to decide whose bag of wheat grains is used to determine which piece of metal is considered a pound. The history of weights and measures is generally a history of fraud and deceit. The definition of a standard value of mass, was not very standard, and variations could be used to cheat when trading. Below is a table of all the competing standards for a pound that I could locate:
They vary from 316.61 grams to 560 grams.
So what do we do? Well, John Wilkins (1614-1672) originally defined his unit of mass, which would later be known as the Kilogram, as a cube of water with sides which are one-tenth of of his base unit. This base unit, with a different definition, would later be known as the meter. In other words, a cube of water with 100 mm sides is the original mass standard for the metric system. A cube of pure water, at a given temperature, made sense, but again, temperature could affect this definition. The temperature of water’s maximum density was chosen as a calibration point. When the value of this mass was determined by the French, during the development of the metric system, it was preserved in a more practical way, as an equivalent mass of platinum-iridium alloy. The relative of this agreed-upon mass is the International Prototype Kilogram (IPK).
The point of measurement, versus counting, is that it produces a continuum of available measurement values, and this value is independent of integer, or discreet values of poppy seeds, wheat seeds, barleycorns, bird seed or anything else. Once one has an agreed upon unit of mass, such as the Kilogram, it may be indefinitely subdivided. An easy way for humans to subdivide this base value, is by using 1000’s. The measured value is found on a continuum of available values, which can be further divided if needed. This is not counting by any stretch of the imagination. It is measurement. The argument for a choice of a numerical base which has lots of divisors is of no import when you have a continuum of possible measurement values.
So is the idea of using numbers which have lots of divisors irrelevant to the metric system? No, they are only irrelevant to metric system measurement. When metric units are chosen such that the amount of precision needed for everyday work is slightly smaller than required, integer values again become important. What I mean by this can be illustrated with metric housing construction in Australia and the UK. In order to make the description of lengths easy, we choose a unit length which in all practical circumstances will always be an integer. The unit chosen for construction is the millimeter. The millimeter is small enough that one never needs to use a decimal point in everyday construction. We have chosen to go back to integers (simple whole “counting” numbers). This is converting measures back to countable “atoms” of measure.
We use our modern measurement system to define a small length value, the millimeter, which is solidly known, rather than using a pre-metric small unit which varies—like a wheat or barlycorn grain. When we use this small unit to produce integers, we can use convenient values which indeed have lots of factors for division. In the case of metric construction, the value chosen is 600 millimeters for stud spacing. Its factors are: 1 2 3 4 5 6 8 10 12 15 20 24 25 30 40 50 60 75 100 120 150 200 300 and 600. What we are doing is not exactly measurement when we construct a house, it is equating multiples of integer values with multiples of a measured integer value, which is a different exercise. When we do this, it makes perfect sense to choose lots of divisors. With millimeters we have “atomized” the values on the construction drawings we are using to guide us. If we want to add in features, such as a window, not originally present on the drawing, or when initially creating a drawing, chances are that we will be able to divide the newly inserted distance easily. This is because of the conscious choice to use small units which can remain integers. We are not measuring in this case, we are back to counting.
Of course as we spent more time measuring our world, we discovered that it is actually discontinuous when it comes to fundamental values of mass. John Dalton (1766-1824) realized and demonstrated that the world is made of atoms. Each individual atom has a defined mass, but the same type of atom can have a range of masses. For instance, tin has atoms that are all chemically tin, but possess ten different mass values. These different mass variations of chemically identical atoms are called isotopes. Tin has ten isotopes, cesium has thirty nine!
Silicon Sphere — The Commonwealth and Industrial Research Organization of Australia (CSIRO) — cc (creative commons)
One of the candidates to replace the current Kilogram standard, which is still an artifact from the nineteenth century, is the silicon sphere. This is a sphere of silicon atoms that will contain a known number of them. If a person knows the mass of each atom in the sphere, and their total number, it can be used to define a mass. In strange way, this procedure is similar to using 7000 wheat grains, but in this case we know that if an atom of silicon is of the same isotope as all the others in the sphere, it possesses a mass which is identical to all the other silicon atoms present. One of the largest difficulties for the team which is attempting to make a silicon sphere Kilogram mass standard, is making certain that all the silicon atoms present within the sphere are of the same isotope. Silicon 28 is the chosen isotope the silicon sphere team will use to create a new Kilogram standard—after counting all the atoms of course. We are counting an integer number of atoms, so that we can develop more accurate continuous set of measurement values, just as was done in the past with wheat grains. These values, which are continuous subdivisions of mass when compared with the discreet values of the atoms in the standard, may be used for the measurement of values which are smaller than the silicon atoms used. But remember, counting is not measuring.
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2 thoughts on “The Count Only Counts — He Does Not Measure”
1. Good article but I wish you hadn’t introduced the spring scale, or, if you did, to get into the difference between weight and mass. How often have we seen “Legal Weight – No springs” or the alternative, “Not Legal for Trade — Chock Full of Springs.”
The spring balance (ie fish scale) inherently compares the local force of gravity on the test mass against the spring constant. A balance beam scale, whether weights in a pan, or sliding weights on the beam, only depends on the uniformity of gravity across the span of the beam, not place to place.
The electronic scale does depend on local gravity; however, with gain and offset adjustments, it can be calibrated in situ with certified masses.
Weight and mass confuse most people and we have no good word for “determine the mass of.” In commerce, “to weigh” almost always means “to determine the mass of.” The graphical model of “weighing” needs to be a balance beam type of scale. | 2,635 | 12,039 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.78125 | 4 | CC-MAIN-2020-10 | latest | en | 0.962796 |
http://www.physics.ohio-state.edu/~ntg/780/2008/780ps1_hints.php | 1,521,803,200,000,000,000 | text/html | crawl-data/CC-MAIN-2018-13/segments/1521257648207.96/warc/CC-MAIN-20180323102828-20180323122828-00008.warc.gz | 436,777,916 | 3,187 | # Physics 780.20: Assignment #1
Here are some hints, suggestions, and comments on the assignment.
## Summing up vs. summing down
• You should find it useful to think of how the 1 + a + a + ... vs. a + a + ... + 1 problem from Session 2 relates to this one. (That might also be a good code to start from in writing one for this problem.)
• If you only consider relatively small N, you will not see any difference between summing up and summing down, since both will have an error about equal to the machine precision. So be sure to increase N until you see a different trend (I used up to 10^8).
• If you are looking for the effects of errors over a logarithmic range (e.g., N from 10^2 to 10^8), you should vary N by multiplying by a factor each time through a loop and not by adding 1 (or some other number). That is, don't use:
``` for (N = 100; N <= 100000000; N++)
{
}
```
which will take an enormous amount of time calculating lots of N values you don't need. Instead, use something like:
``` for (N = 100; N <= 100000000; N *= 10)
{
}
```
which just calculates for 100, 1000, 100000, etc.
• When you output relative errors to a file for plotting, the "scientific" format is usually most useful, again because you are looking at logaithmic intervals. (See the codes we've used in class for examples.)
• It is only the absolute value of the relative error that is of interest. So use "fabs" (you need to "input "). It is particularly important to have only positive numbers if you are going to plot on logarithmic scales!
• Be careful that you don't confuse the machine precision with the smallest floating point number. For example, in single precision, the machine precision is about 10-7, but that does not mean that 10^-10 is set to zero. It is only when you add two numbers that differ by a factor of more than 10-7 that the smaller number is effectively zero.
• A common bug when you have an outer loop stepping through N_max and an inner loop that does the sum of 1/n up to N_max is forgetting to reset your sum to zero when you start the inner loop. So if your summation variable is sum_up, make sure that sum_up = 0.; appears within the outer loop and not just at the beginning of the program.
## Bessel Function Recursing Up and Down
• You will just be modifying a code from Session 2, not writing one from scratch.
• The interpretation of the graph should mainly explain the meaning of the regions where the plotted error is about unity and where it is very small. (I.e., which result is good and which is poor, or are both good or poor?)
• For part c), remember that you need to make a Dev-C++ project to run a program that uses the GSL library. Also, you need the linker option -lgsl -lgslcblas from Session 1. Don't forget that you need to #include a header file (see the Bessel function example from Session 1). | 700 | 2,837 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.515625 | 4 | CC-MAIN-2018-13 | longest | en | 0.93268 |
https://www.physicsforums.com/threads/torque-with-symbolic-notation-problem.142296/ | 1,716,858,605,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971059055.94/warc/CC-MAIN-20240528000211-20240528030211-00148.warc.gz | 805,075,863 | 17,224 | Torque with Symbolic Notation Problem
• esinn08
In summary, the necessary equations for the object shown in Figure P8.4 are provided, including the torque equation, the sum of forces in the x and y directions, and the total torque. It is also noted that Rx and Ry will be zero and that Fg can be broken into its x and y components.
esinn08
Hi Everyone,
My question is as follows:
Write the necessary equation of the object shown in Figure P8.4. Take the origin of the torque equation about an axis perpendicular to the page through the point O. (Let clockwise torque be positive and let forces to the right and up be positive. Use q for and Rx, Ry, Fx, Fy, Fg, l, and g as appropriate in your equations.) Find the sum of the forces in the x direction, the y direction, and the total torque. (I hope the picture I attached shows up!)
I've never been good with symbolic notation! I know Rx and Ry will go to zero, since there is no torque through the point of origin. Do I have to break Fg into its x and y components? Any suggestions would be greatly appreciated! Thanks so much!
Attachments
• prob.gif
2.2 KB · Views: 508
The equation for the torque about point O is: T = Fx * l - Fg * l * sin(q).The equation for the sum of the forces in the x direction is: Fx = Fg * cos(q).The equation for the sum of the forces in the y direction is: Fy = Fg * sin(q) + Ry - g.The total torque is: T = Fg * l * sin(q).
I would first like to commend you for seeking help and clarification on this problem. It is important to always fully understand the equations and symbols being used in any scientific context.
To answer your question, yes, you will need to break down Fg into its x and y components. This is because torque is a vector quantity, meaning it has both magnitude and direction. In order to properly calculate the total torque, you will need to take into account the direction of the force.
To calculate the sum of forces in the x and y directions, you can use the equations Fx = Rx + Fgcos(q) and Fy = Ry + Fgsin(q), respectively. This takes into account the vertical and horizontal components of Fg.
As for the total torque, you can use the equation T = Fgl, where l is the distance from the point of origin to the point where the force is applied. This will give you the magnitude of the torque, but remember to take into account the direction as well.
I hope this helps and good luck with your problem! Remember, it's always important to break down complex problems into smaller, more manageable parts.
1. What is torque and how is it calculated?
Torque is a measure of the force that can cause an object to rotate around an axis. It is calculated by multiplying the force applied to an object by the distance from the axis of rotation to the point where the force is applied. The equation for torque is T = F x d, where T is torque, F is force, and d is distance.
2. What is the symbol used for torque in equations?
The symbol used for torque in equations is typically the Greek letter "tau" (τ) or the letter "T". Both symbols represent the same concept of torque and can be used interchangeably in equations.
3. How do you represent torque with symbolic notation?
To represent torque with symbolic notation, you would use the appropriate symbol (τ or T) in the torque equation, along with the symbols for force (F) and distance (d). The equation would look like this: τ = F x d or T = F x d, depending on the symbol being used.
4. What are the units of torque?
The units of torque are typically expressed as Newton-meters (N-m) or foot-pounds (ft-lb) in the International System of Units (SI). In the United States, the unit of foot-pounds (ft-lb) is more commonly used to measure torque.
5. How is torque used in real-world applications?
Torque is used in many real-world applications, such as in the design of machines, engines, and vehicles. It is also important in sports, particularly in activities that involve throwing or rotating objects. In addition, torque is used in the construction of buildings and bridges to ensure structural stability.
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1K | 1,107 | 4,665 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.921875 | 4 | CC-MAIN-2024-22 | latest | en | 0.924931 |
http://blogformathematics.blogspot.com/2010/11/calculation-of-permutations-part-3.html | 1,508,686,036,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187825308.77/warc/CC-MAIN-20171022150946-20171022170946-00751.warc.gz | 49,086,826 | 14,773 | ## Pages
### Calculation of permutations of 'n' objects when 'r' objects are taken at a time
In the previous examples, we calculated the permutations of 'n' number of objects by considering that n objects are to be rearranged among themselves. Now we will also calculate all the different arrangements obtained when there are 'n' number of objects, but only some of those are taken at a time and arranged in different orders.
This means that if there are, say, 10 objects, we will calculate the permutations obtained by taking less than 10 objects at a time. Let us take 3 objects at a time. Thus, we can chose any three objects from the group of 10 objects like this:
• object 1, object 2, object 3
• object 4, object 2, object 5
• object 4, object 3, object 2, and so on.
For each of the group of 3 objects obtained above, we can obtain different arrangements (permutations). For example, for the group of objects object1, object2, and object3, we can obtain the following different arrangements:
• object1, object2, object3
• object2, object3, object1
• object2, object1, object3, and so on.
Thus, it is very difficult to calculate the total number of different arrangements obtained when only some objects are taken from a group of objects at a time. For this, we again apply the fundamental counting principle and make our work simpler. But before that, we will have to reconsider the situation and think about it like this:
• Consider the situation as if a group of three empty boxes is to be filled in by any three objects from the group of 10 objects:
• The first box can be filled with any one of the 10 objects. (10 choices)
• After filling in the first box, we are left with 9 objects. Therefore the second box can be filled in with any one of the 9 objects. (9 choices)
• After filling the first and second box, 8 objects are left. The third box can be filled with any one of the 8 objects. (8 choices)
Therefore by using the Fundamental counting principle, we conclude that there are 3 events taking place in this example - one is choosing the first object (having 10 choices), second is choosing the second object (having 9 choices), and the third is choosing the third object (having 8 choices).
Since three events can occur in 10, 9 and 8 ways respectively, therefore by the fundamental principle of counting, the total number of different events that can occur are 10 x 9 x 8 = 720.
Therefore, the permutations of 10 objects taken 3 at a time are 720.
Now we will derive the formula for permutations of 'n' objects taken 'r' at a time. In the previous part (Calculation of permutations part 3-A), we calculated that if there are 10 objects and 3 are taken at a time, then the total number of permutations are given by 10 x 9 x 8 = 720.
Generalizing the above obtained result for all numbers:
Let the total number of objects be 'n' and let 'r' objects be taken at a time. Then the total number of permutations obtained when 'r' objects are arranged at a time from a group of n objects will be represented by the symbol P(n, r).
Let us look at this situation in the light of the fundamental counting principle (like we did in the previous examples in permutations):
Since 'r' objects are to be arranged at a time this can also be viewed as if there are 'r' number of empty boxes, and from the group of 'n' objects, we can take any object at random and fill in each box with exactly one object. This way, to fill the first of the 'r' empty boxes, there will be 'n' objects to chose from. Therefore the number of choices for the first empty box is 'n'.
After filling the first empty box, there are (r - 1) empty boxes left to fill, and (n - 1) number of objects left.There are (n - 1) number of different choices for filling in the second empty box.
There are (n - 2) number of different objects to chose from when filling the third empty box, because two of the objects have already been taken and put in the first and second empty box.
This way, till the 'r' empty boxes are filled, 'r' number of objects will be taken from the group of 'n' objects. Thus, the number of objects left after filling in the 'r' empty boxes will be (n - r).
Therefore, by the fundamental principle of counting, since the first event can occur in 'n' different ways, second event can occur in (n - 1) different ways, third event can occur in (n - 2) different ways, till the r th event, which can take place in (n - r + 1) different ways, the total number of different ways in which all the events can occur together is given by
P(n, r) = n(n - 1)(n - 2)(n - 3) . . . . (n - r + 1)
The most common questions that students ask after studying this formula is "Where does (n - r + 1) come from?" That is to be explained in detail, otherwise they (and you) will not be able to grasp this concept :
Where does (n - r + 1) come from?
Remember that there are 'r' empty boxes to be filled. After filling 1 empty box, (n - 1) objects are left. Similarly, after filling two empty boxes, (n - 2) objects are left. Since we have to fill a total of 'r' empty boxes, after filling the last box, that is, after filling 'r' boxes, the number of objects left will be (n - r). But we have to know the total number of choices for the last box (the r th object).
So we have to find the total number of objects left after filling in the second last box, because that would be the number of choices present for the last box.
Since the last box is the r th object, therefore the second last box is (r - 1)th box. After filling in (r - 1) boxes with one object each, (n - (r - 1)) objects are left for the last (the r th) box. Since (n - (r - 1)) = (n - r + 1), thus, there are (n - r + 1) choices for the last empty box to be filled in. Thus the last term in the above formula is (n - r + 1).Final formula we obtained:
Number of permutations for n objects taken r at a time is
P(n, r) = n(n - 1)(n - 2) . . . (n - r + 1)
By using factorial notation, we get that
P(n, r) = [n(n - 1)(n - 2) . . . (n - r + 1)] * [(n - r) . . . 3 * 2 * 1]/(n - r)!
P(n, r) = n!/(n - r)! | 1,571 | 6,054 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.84375 | 5 | CC-MAIN-2017-43 | longest | en | 0.92463 |
https://ardlussaestate.com/devils-bit-qwlxi/critical-point-calculator-2092f7 | 1,642,709,829,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320302622.39/warc/CC-MAIN-20220120190514-20220120220514-00519.warc.gz | 153,732,767 | 25,594 | # critical point calculator
Calculus: Fundamental Theorem of Calculus Now, so if we have a non-endpoint minimum or maximum point, then it's going to be a critical point. Many calculators allow you to save and recycle your data in similar calculations, saving you time and For two-variables function, critical points are defined as the points in which the gradient equals zero, just like you had a critical point for the single-variable function f(x) if the derivative f'(x)=0. We will identify the numerator degrees. The most prominent example is the liquid–vapor critical point, the end point of the pressure–temperature curve that designates conditions under which a liquid and its vapor can coexist. Remember to adjust A function z=f(x,y) has critical points where the gradient del f=0 or partialf/partialx or the partial derivative partialf/partialy is not defined. The critical value is the point on a statistical distribution that represents an associated probability You can also perform the calculation using the mathematical formula above. Critical Points and Extrema Calculator The calculator will find the critical points, local and absolute (global) maxima and minima of the single variable function. This calculator is intended to replace the use of a critical point calculator with steps, New Step by Step Roadmap for Partial Derivative Calculator Algebrator is well worth the cost as a result of approach. positive to negative). A critical point of a multivariable function is a point where the partial derivatives of first order of this function are equal to zero. All Rights Reserved. It is a number ‘a’ in the domain of a given function ‘f’. Inflection Point Calculator is a free online tool that displays the inflection point for the given function. The critical value is the point in that distribution at which we must accept the alternative hypothesis as being more likely. fx(x,y) = 2x fy(x,y) = 2y We now solve the following equations fx(x,y) = 0 and fy(x,y) = 0 simultaneously. Relationship: local minimum and maximum. using the version we set up to find critical values of a t-distribution. Welcome to CriticalPoint's home of best-in-class healthcare education. of any statistics text book or here.). The function $$f\left( x \right) = x + {e^{ – x}}$$ has a critical point (local minimum) at $$c = 0.$$ Critical Point(s): -2, 1. Wiki says: March 9, 2017 at 11:14 am Here there can not be a mistake? (We're working on a calculator for the f critical value; you can find a table in the back That being said, a wise The Function Analysis Calculator computes critical points, roots and other properties with the push of a button. This critical values calculatoris designed to accept your p-value (willingness to accept an incorrect hypothesis) and degrees of freedom. A critical value is a concept from statistical testing. image/svg+xml. Compute answers using Wolfram's breakthrough technology & knowledgebase, relied on by millions of students & professionals. Advertisement. requires you to have 30+ observations. statistical packages. The function in this example is. Let c be a critical point for f(x).Assume that there exists an interval I around c, that is c is an interior point of I, such that f(x) is increasing to the left of c and decreasing to the right, then c is a local maximum. This critical value calculator generates the critical values for a standard normal distribution for a Critical/Saddle point calculator for f(x,y) 1 min read. So for the sake of this function, the critical points are, we could include x sub 0, we could include x sub 1. Classification of Critical Points Figure 1. Functions. The heat of vaporization becomes zero at this critical point temperature. At higher temperatures, the gas cannot be liquefied by pressure alone. The easiest way is to look at the graph near the critical point. Our mission is to improve patient safety by raising the competency of healthcare professionals through convenient, high-quality training. The Function Analysis Calculator computes critical points, roots and other properties with the push of a button. Wolfram alpha paved a completely new way to get knowledge and information. This calculator requires you to have sufficiently large sample that you are comfortable the values of the probability refers to the selected probability . t critical value calculator, Please send all feedback, complaints, and lucrative sponsorship deals to, This Website is copyright © 2016 - 2020 Performance Ingenuity LLC. Z critical value calculator, The critical points are where the behavior of the system is in some sense the most complicated. Advertisement. For more concept check http://math.tutorvista.com/. Open Live Script. most appropriate distribution for comparing this sample (see below on when to use a standard normal vs. a t distribution). | 1,006 | 4,835 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.921875 | 4 | CC-MAIN-2022-05 | latest | en | 0.869366 |
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https://metanumbers.com/3030 | 1,603,928,892,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107902038.86/warc/CC-MAIN-20201028221148-20201029011148-00145.warc.gz | 420,792,001 | 7,388 | 3030
3,030 (three thousand thirty) is an even four-digits composite number following 3029 and preceding 3031. In scientific notation, it is written as 3.03 × 103. The sum of its digits is 6. It has a total of 4 prime factors and 16 positive divisors. There are 800 positive integers (up to 3030) that are relatively prime to 3030.
Basic properties
• Is Prime? No
• Number parity Even
• Number length 4
• Sum of Digits 6
• Digital Root 6
Name
Short name 3 thousand 30 three thousand thirty
Notation
Scientific notation 3.03 × 103 3.03 × 103
Prime Factorization of 3030
Prime Factorization 2 × 3 × 5 × 101
Composite number
Distinct Factors Total Factors Radical ω(n) 4 Total number of distinct prime factors Ω(n) 4 Total number of prime factors rad(n) 3030 Product of the distinct prime numbers λ(n) 1 Returns the parity of Ω(n), such that λ(n) = (-1)Ω(n) μ(n) 1 Returns: 1, if n has an even number of prime factors (and is square free) −1, if n has an odd number of prime factors (and is square free) 0, if n has a squared prime factor Λ(n) 0 Returns log(p) if n is a power pk of any prime p (for any k >= 1), else returns 0
The prime factorization of 3,030 is 2 × 3 × 5 × 101. Since it has a total of 4 prime factors, 3,030 is a composite number.
Divisors of 3030
1, 2, 3, 5, 6, 10, 15, 30, 101, 202, 303, 505, 606, 1010, 1515, 3030
16 divisors
Even divisors 8 8 4 4
Total Divisors Sum of Divisors Aliquot Sum τ(n) 16 Total number of the positive divisors of n σ(n) 7344 Sum of all the positive divisors of n s(n) 4314 Sum of the proper positive divisors of n A(n) 459 Returns the sum of divisors (σ(n)) divided by the total number of divisors (τ(n)) G(n) 55.0454 Returns the nth root of the product of n divisors H(n) 6.60131 Returns the total number of divisors (τ(n)) divided by the sum of the reciprocal of each divisors
The number 3,030 can be divided by 16 positive divisors (out of which 8 are even, and 8 are odd). The sum of these divisors (counting 3,030) is 7,344, the average is 459.
Other Arithmetic Functions (n = 3030)
1 φ(n) n
Euler Totient Carmichael Lambda Prime Pi φ(n) 800 Total number of positive integers not greater than n that are coprime to n λ(n) 100 Smallest positive number such that aλ(n) ≡ 1 (mod n) for all a coprime to n π(n) ≈ 437 Total number of primes less than or equal to n r2(n) 0 The number of ways n can be represented as the sum of 2 squares
There are 800 positive integers (less than 3,030) that are coprime with 3,030. And there are approximately 437 prime numbers less than or equal to 3,030.
Divisibility of 3030
m n mod m 2 3 4 5 6 7 8 9 0 0 2 0 0 6 6 6
The number 3,030 is divisible by 2, 3, 5 and 6.
• Arithmetic
• Abundant
• Polite
• Square Free
Base conversion (3030)
Base System Value
2 Binary 101111010110
3 Ternary 11011020
4 Quaternary 233112
5 Quinary 44110
6 Senary 22010
8 Octal 5726
10 Decimal 3030
12 Duodecimal 1906
20 Vigesimal 7ba
36 Base36 2c6
Basic calculations (n = 3030)
Multiplication
n×i
n×2 6060 9090 12120 15150
Division
ni
n⁄2 1515 1010 757.5 606
Exponentiation
ni
n2 9180900 27818127000 84288924810000 255395442174300000
Nth Root
i√n
2√n 55.0454 14.4704 7.41926 4.96922
3030 as geometric shapes
Circle
Diameter 6060 19038.1 2.88426e+07
Sphere
Volume 1.16524e+11 1.15371e+08 19038.1
Square
Length = n
Perimeter 12120 9.1809e+06 4285.07
Cube
Length = n
Surface area 5.50854e+07 2.78181e+10 5248.11
Equilateral Triangle
Length = n
Perimeter 9090 3.97545e+06 2624.06
Triangular Pyramid
Length = n
Surface area 1.59018e+07 3.2784e+09 2473.98 | 1,242 | 3,569 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.765625 | 4 | CC-MAIN-2020-45 | longest | en | 0.820538 |
http://www.gradesaver.com/textbooks/science/physics/physics-for-scientists-and-engineers-a-strategic-approach-with-modern-physics-4th-edition/chapter-15-oscillations-exercises-and-problems-page-419/74 | 1,524,416,945,000,000,000 | text/html | crawl-data/CC-MAIN-2018-17/segments/1524125945624.76/warc/CC-MAIN-20180422154522-20180422174522-00476.warc.gz | 426,655,275 | 12,554 | ## Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (4th Edition)
$f = \sqrt{f_1^2+f_2^2}$
If the mass is a distance of $x$ from the equilibrium point, the force on the mass is $k_1x+k_2x$ which is $(k_1+k_2)x$. Thus, the equivalent spring constant is $k_1+k_2$. We can find an expression for the frequency. $f = \frac{1}{2\pi}~\sqrt{\frac{k_1+k_2}{m}}$ We can find an expression for $\sqrt{f_1^2+f_2^2}$ as: $\sqrt{f_1^2+f_2^2}$ $= \sqrt{(\frac{1}{2\pi}~\sqrt{\frac{k_1}{m}})^2+ (\frac{1}{2\pi}~\sqrt{\frac{k_2}{m}})^2}$ $= \sqrt{(\frac{1}{2\pi})^2~\frac{k_1}{m}+ (\frac{1}{2\pi})^2~\frac{k_1}{m}}$ $= \sqrt{(\frac{1}{2\pi})^2~\frac{k_1+k_2}{m}}$ $= \frac{1}{2\pi}~\sqrt{\frac{k_1+k_2}{m}}$ Therefore, $f = \sqrt{f_1^2+f_2^2}$. | 338 | 759 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.9375 | 4 | CC-MAIN-2018-17 | latest | en | 0.575559 |
https://www.jiskha.com/display.cgi?id=1324250012 | 1,516,097,226,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084886397.2/warc/CC-MAIN-20180116090056-20180116110056-00511.warc.gz | 913,717,638 | 4,800 | # Precalculus
posted by .
Help with these questions
help with these questions~
Thank you
1. Determine the interval(s) on which x^2 + 2x -3>0
a)x<-3, x>1
b)-3<x<1
c)x<-3, -3<x<1, x>1
d) x>1
---------------------------------------
2. Determine when the function f(x)= 3x^3 + 4x^2 -59x -13 is greater than 7.
a) -5<x<1/3
b) -5<x<-1/3, x>4
c) x<4
d) x=-5, -1/3,4
---------------------------------------
3. Provde the intervals you would check to determine when -5x^2 + 37x> -15x^2 + 12x + 15.
a) x=3, x=-0.5
b) x<-3, -3<x<0.5, x>0.5
c) x<-1, -1<x<15, x>15
d) x= 1, x=15
----------------------------------------
• Precalculus -
1. x^2 + 2x - 3 > 0
(x+3)(x-1) > 0
x-intercepts are -3 and 1
so you are looking for the values of x when the parabola y = x^2 + 2x - 3 is above the x-axis
Since it opens upwards those values are
x < -3 OR x > 1
The closest choice to that is #1, but they did not include the necessary OR. The comma is this context means AND, which would be incorrect.
2. 3x^3 + 4x^2 - 59x - 13 > 7
3x^3 + 4x^2 - 59x - 20 > 0
after some quick tries of ±1, ±2, ± 4, I found x=4 to be a solution, so x-4 is a factor
by synthetic division,
3x^3 + 4x^2 - 59x - 20 = (x-4)(3x^2 + 16x + 5)
= (x-4)(x+5)(3x+1)
so the critical values are -5, -1/3, and 4
So the curve is above the x-axis (or above 7 in the original) for
-5 < x < -1/3 OR x > 4
Again, they made an error by not stating the OR condition, but it looks like they meant b) , which is what you had.
3. -5x^2 + 37x> -15x^2 + 12x + 15
10x^2 + 25x - 15 > 0
2x^2 + 5x - 3 > 0
(2x - 1)(x + 3) > 0
critical values are 1/s and -3
so I would check
x < -3, -3 < x < 1/2, and x > 1/2
looks like b) is the correct choice.
• Precalculus -
1. The graph is a parabola, opening upward. So, there will be a left side and a right side above the x-axis. These intervals will be outside the two roots.
2. B is correct
3. The graphs intersect at x = -3 and 0.5
• Precalculus -
Thank you~
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10. ### Probability/Statistics
The teacher of a mathematics class has written up the final exam, but wants the questions to be random for each student. There are 28 students in the class, and the test has 100 questions. Each question has four possible answers. a. …
More Similar Questions | 1,445 | 4,494 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.34375 | 4 | CC-MAIN-2018-05 | latest | en | 0.745404 |
https://www.mathematicalgemstones.com/gemstones/pearl/ellipses-parabolas-and-infinity/ | 1,713,514,999,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817382.50/warc/CC-MAIN-20240419074959-20240419104959-00686.warc.gz | 809,950,787 | 4,127 | # Ellipses, parabolas, and infinity
A parabola can be defined as the locus of points equidistant from a fixed point (called the focus) and a fixed line (called the directrix). But we know from projective geometry that parabolas and ellipses are essentially the same object. Does this mean we can also define an ellipse in terms of a point and a line?
While tutoring a high school student recently, we worked through a problem that was essentially showing just that in a special case. Remarkably:
An ellipse can be defined as the locus of points $P$ for which the distance $|PF|$ to a focus $F$ is $\alpha$ times the distance from $P$ to a fixed line $\ell$ for some positive real number $\alpha\lt 1$.
For instance, if we make $\alpha=1/2$ we would get an ellipse, whereas if $\alpha=1$ then it’s such a large ellipse that it’s a parabola!
Let’s check the math here. Recall that an ellipse is usually defined, synthetically, as the locus of points $P$ in the plane for which the sum of the distances $|PF_1|+|PF_2|$ for two fixed foci $F_1$ and $F_2$ is a fixed constant $s$. By translating this condition into coordinates, one can show that if we place an ellipse with its two lines of symmetry aligned with the $x$- and $y$-axes (with the wider part on the $x$-axis), centered at the origin, then the ellipse will have equation $(x/b)^2+(y/a)^2=1$ for some positive real numbers $a$ and $b$ with $a\le b$.
For an ellipse $E$ with this equation, the focii $F_1$ and $F_2$ have coordinates $(-c,0)$ and $(c,0)$ for some $c$. To find $c$ in terms of $a$ and $b$, we have that the sum of the distances of the point $(b,0)$ to the foci is $b-c+b-c=2b$, and the sum of the distances of the point $(0,a)$ to the foci is $2\sqrt{a^2+c^2}$ by the Pythagorean theorem. Therefore $2b=2\sqrt{a^2+c^2}$, and solving we find $c=\sqrt{b^2-a^2}$.
Now, I claim that if we set $\alpha=\frac{\sqrt{b^2-a^2}}{b}$ and let $\ell$ be the vertical line $x=b/\alpha$, then $E$ is the locus of points $P$ for which $|PF_2|$ is $\alpha$ times the distance from $P$ to $\ell$. Indeed, let $P=(x,y)$ be a point that has this property. Then the distance $|PF_2|$ is $\sqrt{(x-\sqrt{b^2-a^2})^2+y^2}$ and the distance from $P$ to $\ell$ is $\frac{b}{\alpha}-x$, so we have \begin{align*} \sqrt{(x-\sqrt{b^2-a^2})^2+y^2} &= \alpha\left(\frac{b}{\alpha}-x\right) \\ (x-\sqrt{b^2-a^2})^2+y^2 &= (b-\alpha x)^2 \\ x^2-2x\sqrt{b^2-a^2}+b^2-a^2+y^2 &= b^2-2x \sqrt{b^2-a^2} + (\alpha x)^2 \\ x^2-a^2+y^2 &= \frac{b^2-a^2}{b^2}x^2 \\ \frac{a^2}{b^2}x^2 + y^2 &= a^2 \\ \frac{x^2}{b^2} + \frac{y^2}{a^2} &= 1 \end{align*} which is indeed the equation of $E$.
A few noteworthy observations: first, it’s remarkable that the focus of the ellipse as defined in terms of the constant-sum-from-two-focii definition coincides with the focus that appears in the focus-and-directrix version. This makes one wonder if there is some natural relationship between the focii and this new directrix $\ell$, perhaps in terms of reciprocation (see this post for the definition of reciprocation in a circle.) And indeed, if we apply the transformation $(x,y)\mapsto (x/b,y/a)$, which maps our ellipse $E$ to the unit circle, the point $F_2$ maps to the point $(\alpha,0)$ and $\ell$ becomes the line $x=1/\alpha$, so indeed $F_2$ and $\ell$ form a reciprocal pair!
Second, consider the degenerate case of a circle, when $a=b$ and so $\alpha=0$. The condition $\alpha=0$ doesn’t really make sense unless we interpret the diagram in the projective plane and allow the directrix to be the line at infinity, which again makes the focus (the center of the circle) be the polar of this line.
Finally, consider the limit as $b$ approaches $\infty$, so that the ellipse is stretching out further and further until it becomes a parabola. (Exercise for the reader: What projective transformation maps an ellipse into a parabola?) In this limit we have $\lim_{b\to \infty} \alpha = \lim_{b\to \infty} \frac{\sqrt{b^2-a^2}}{b}=\sqrt{\lim_{b\to \infty}1-(a^2/b^2)}=1,$ and again we recover the case of a parabola. As a bonus, we find that the focus and directrix of a parabola must be reciprocal to each other across the parabola as well.
That’s all for today - a bit of fun with conics to kick off the new year. Happy and indivisible 2017! | 1,329 | 4,283 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.25 | 4 | CC-MAIN-2024-18 | latest | en | 0.891135 |
http://mathhelpforum.com/pre-calculus/80246-solved-help-amplitude-period-range-phase-shift-print.html | 1,513,179,499,000,000,000 | text/html | crawl-data/CC-MAIN-2017-51/segments/1512948527279.33/warc/CC-MAIN-20171213143307-20171213163307-00224.warc.gz | 183,336,663 | 3,442 | [SOLVED] Help with amplitude, period,range and phase shift
• Mar 23rd 2009, 05:53 PM
lsnyder
[SOLVED] Help with amplitude, period,range and phase shift
Let https://webwork.uncc.edu/webwork2_fi...eb482ab8f1.png.
What is the amplitude? Period? phase shift? and range
Let https://webwork.uncc.edu/webwork2_fi...628684d0d1.png.
What is the amplitude? Period? phase shift? and range
Let https://webwork.uncc.edu/webwork2_fi...c2e933c3f1.png. What is the amplitude? Period? phase shift? and range
• Mar 23rd 2009, 06:08 PM
e^(i*pi)
Quote:
Originally Posted by lsnyder
Let https://webwork.uncc.edu/webwork2_fi...eb482ab8f1.png.
What is the amplitude? Period? phase shift? and range
Let https://webwork.uncc.edu/webwork2_fi...628684d0d1.png.
What is the amplitude? Period? phase shift? and range
Let https://webwork.uncc.edu/webwork2_fi...c2e933c3f1.png. What is the amplitude? Period? phase shift? and range
The general equation is $y = Asin(kx+t) + d$
where:
A is the amplitude
1/k is the period
t is phase shift
d is range
• Mar 23rd 2009, 06:16 PM
lsnyder
okay, i can kinda remember most of this material from class. But the only thing i got right (typing in my homework) is the amplitude. So can u show me how u do the phase shift, range and period with at least one of them. I'm sorry but I'm a visual learner to the extreme (which is really sad) and I think I would comprehend better.
• Mar 23rd 2009, 06:35 PM
wytiaz
y = a sin (bx + c) + d
|a| is the amplitude: this tells you how tall the curve is at it's highest point from the center.
(2pi)/|b| is the period: this tells you how long the sin function takes to get its job done. the curve will repeat itself every <period> units
-c/b is the phase shift: this tells you how far the curve is "scooted over to the right" on the graph. while the period represents shrinking or stretching the coil, the phase shift just moves the whole thing over.
the range is d-|a| to d+|a|: this tells you the lowest to highest value the curve takes on.
example:
y = -8sin (4x - 3) +2
amp: 8
per: (2pi)/4 = pi/2
p.s.: -(-3)/4) = 3/4
range: -6 to 10
have fun! | 651 | 2,113 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 1, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.78125 | 4 | CC-MAIN-2017-51 | longest | en | 0.853956 |
http://www.riddlesandanswers.com/v/230177/my-daughter-has-as-many-sisters-as-she-has-brothers-each-of-her-brothers-has-twice-as-many-sisters/ | 1,571,628,558,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570987751039.81/warc/CC-MAIN-20191021020335-20191021043835-00443.warc.gz | 331,044,194 | 22,996 | # BROTHERS AND SISTERS RIDDLE
#### Popular Searches
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## Brothers And Sisters Riddle
My daughter has as many sisters as she has brothers. Each of her brothers has twice as many sisters as brothers. How many sons and daughters do I have?
Hint:
Four daughters and three sons. Each daughter has 3 sisters and 3 brothers, and each brother has 2 brothers and 4 sisters.
To figure it out mathematically, you could use the following two equations where G = the number of girls and B = the number of boys:
G 1 = B
2(B 1) = G
Solving for G gives you 4 and plugging that in to G 1 = B gives you a B of 3.
Did you answer this riddle correctly?
YES NO
Solved: 42% | 183 | 682 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.765625 | 4 | CC-MAIN-2019-43 | latest | en | 0.969378 |
null | null | null | null | null | null | I constantly worry that I'm becoming desensitized to violence and horror thanks to watching so many damn movies, and having a penchant for those that are riddled with explosions and coarse language. (If Scott Weinberg has his way, I will have a healthy appreciation for the slasher flick as well. Speaking of which, have you read Horror Virgin yet?) As a kid, I was always sternly brought up to know that movies were fiction, and that violence was very real, and to know that guns, knives, grenades, etc. were no cheering or laughing matter unless Mel Gibson was using or running away from them.
Like much of the civilized world, I've been following the protests in Iran, and while I empathized with what was going on, I felt curiously detached from seeing images of real violence. I read comments from people who said they were shaking and vomiting from seeing people die on camera, and I wondered if I was a terrible person because I wasn't. Is it because I watch so much of it onscreen? Or am I saturated by it thanks to the real world -- I watched Columbine happen on television while living a few blocks away from it, to say nothing of the trauma of 9/11, and documentaries about Darfur and the Holocaust.
Now, all of the above is certainly a topic of its own merit, but I discovered a very curious effect this weekend. For some reason my thirst for noir and Hitchcock went on a "troubled heroine" path, which led to my watching and GildaMarnie, and a realization of how casually old thrillers are peppered with abused women. While you can usually see it coming (heroine is babbling hysterically, man must slap her), sometimes it still has the power to shock. The backhand Glenn Ford delivers on Rita Hayworth actually made me jump. Why? Maybe I was just edgy, or maybe so many movies of careless smacks piled up into a "Damn, they really didn't think anything of that!" moment.
shocked me even more with it's uncharacteristic '60s frankness. It's with some shame that I confess I hadn't seen this particular Hitchcock, and when Sean Connery actually ripped Tippi Hendren's pajamas off, my jaw dropped like a cartoon character. Just as I felt rather bemused at my reaction, that creepy overhead shot (you know the one) of Connery for the "rape" scene left me gobsmacked all over again. In truth, I think I was unprepared for such a blatant "There's sex going on" moment, especially with prominent twin beds in view.
I've been trying to puzzle out my strong reactions all weekend. I've seen violence and sex (to say nothing of rape) scenes far more shocking in modern films, so why did these two films actually startle me, especially when I saw it coming in Gilda from a mile away? It can't be the retro setting -- if anything, I'm shocked because I'm seeing it from a 21st century perspective that doesn't condone that. Yet I know that's what startled me about the scenes in Marnie. Perhaps it's all about the context of any given moment, and not the thick shell of two decades of moviewatching. Maybe I go into old movies with a careless attitude, but I turn on the world news with my armor on. Perhaps I'm just overthinking it all. Who knows? But I thought I'd put the topic to you readers, and see if you'd experienced anything similar, especially in regards to old cinema.
CATEGORIES Fandom, Cinematical | null | null | null | null | null | null | null | null | null |
https://edurev.in/course/quiz/attempt/-1_Data-Interpretation-MCQ-5/27aab75d-1e82-4687-831a-c7ea59b9fa43 | 1,686,310,141,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224656675.90/warc/CC-MAIN-20230609100535-20230609130535-00318.warc.gz | 257,282,254 | 41,963 | Data Interpretation MCQ -5
# Data Interpretation MCQ -5
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## 20 Questions MCQ Test Quantitative Techniques for CLAT | Data Interpretation MCQ -5
Data Interpretation MCQ -5 for Railways 2023 is part of Quantitative Techniques for CLAT preparation. The Data Interpretation MCQ -5 questions and answers have been prepared according to the Railways exam syllabus.The Data Interpretation MCQ -5 MCQs are made for Railways 2023 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Data Interpretation MCQ -5 below.
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Data Interpretation MCQ -5 - Question 1
### (Q.no: 1-5). Refer to the following Line Graph and answer the given questions. In 2013, out of the total number of applications received for Universities P and Q together, only 30% were accepted. What was the total number of applications accepted for Universities P and Q together in 2013?
Detailed Solution for Data Interpretation MCQ -5 - Question 1
Total number of applications accepted by P and Q together in 2013 = (130 + 280)*30/100
=123
Data Interpretation MCQ -5 - Question 2
### (Q.no: 1-5). Refer to the following Line Graph and answer the given questions. In 2012, 30% of applications received for University P and 20% of applications received for University Q were from international students. What was the total number of International applicants for University P and Q together in 2012?
Detailed Solution for Data Interpretation MCQ -5 - Question 2
Total Number of International applicants for University P and Q together = 210*30/100 + 140 * 20/100
= 63 + 28 = 91
Data Interpretation MCQ -5 - Question 3
### (Q.no: 1-5). Refer to the following Line Graph and answer the given questions. If the respective ratio of total number of applications received for University P and Q together in 2016 and 2017 is 3:4. What was the total number of applications received for University P and Q together in 2017?
Detailed Solution for Data Interpretation MCQ -5 - Question 3
Total number of applications received for Universities P and Q together in 2016 = 230 + 190 = 420
Total number of applications received for Universities P and Q together in 2017 = 420 * 4/3 = 560
Data Interpretation MCQ -5 - Question 4
(Q.no: 1-5). Refer to the following Line Graph and answer the given questions.
What is the average number of applications received for University P in 2013, 2015 and 2016?
Detailed Solution for Data Interpretation MCQ -5 - Question 4
Average number of applications received for University A in 2013, 2015 and 2016 = (130 + 210 + 230)/3 = 190
Data Interpretation MCQ -5 - Question 5
(Q.no: 1-5). Refer to the following Line Graph and answer the given questions.
Number of applications received for University Q increased by what percent from 2012 to 2014?
Detailed Solution for Data Interpretation MCQ -5 - Question 5
% = (220-140)/140 * 100 = 400/7 %
Data Interpretation MCQ -5 - Question 6
(Q.no: 6-10). Refer to the following Line Graph and answer the given questions.
The number of people who traveled by Train B on Friday is 30% more than the people who traveled by the same train on Thursday. What is the respective ratio between the number of people who traveled on Friday and those who traveled on Saturday by the same train?
Detailed Solution for Data Interpretation MCQ -5 - Question 6
30% of 200 = 60
People traveled by Train B on Friday = 260
260 : 320 = 13 : 16
Data Interpretation MCQ -5 - Question 7
(Q.no: 6-10). Refer to the following Line Graph and answer the given questions.
What is the difference between the total number of people who traveled by Train B on Saturday and Sunday together and the total number of people who traveled by Train A on Saturday and Sunday together?
Detailed Solution for Data Interpretation MCQ -5 - Question 7
Train B on Saturday and Sunday together = 320 + 310 = 630
Train A on Saturday and Sunday together = 350 + 270 = 620
Difference = 630 – 620 = 10
Data Interpretation MCQ -5 - Question 8
(Q.no: 6-10). Refer to the following Line Graph and answer the given questions.
What is the average number of people traveling by Train B on Monday, Tuesday, Wednesday and Thursday?
Detailed Solution for Data Interpretation MCQ -5 - Question 8
200 + 170 + 120 + 200
= 172.5
Data Interpretation MCQ -5 - Question 9
(Q.no: 6-10). Refer to the following Line Graph and answer the given questions.
The number of people who traveled by Train A decreased by what percent from Saturday to Wednesday?
Detailed Solution for Data Interpretation MCQ -5 - Question 9
[(350 – 140)/350]*100
= [210/350]*100 = 60%
Data Interpretation MCQ -5 - Question 10
(Q.no: 6-10). Refer to the following Line Graph and answer the given questions.
The total number of people who traveled by both the given trains together on Monday is approximately what percent more than the total number of people who traveled by both the given trains together on Wednesday?
Detailed Solution for Data Interpretation MCQ -5 - Question 10
(440 – 260)/260*100
= 180/260 * 100
= 69%
Data Interpretation MCQ -5 - Question 11
(Q.no: 15-20). Refer to the table and answer the given questions.
Total Population in Five Cities = 15,00,000
The total population of City P and R is 7,85,000. The difference between the total population of City P and R is 2.65 lakhs. If the City P has more population than City R then the urban population of City P and R together is approximately what percent of total population of these two Cities?
Detailed Solution for Data Interpretation MCQ -5 - Question 11
P + R = 7,85,000
(R has more population than P) R – P = 265000
P = 5,25,000 ; R = 2,60,000
Urban population of City P and R = 3/10 * 5,25,000 + 5/13 * 2,60,000 = 257500
% = (257500 / 785000)*100 = 32%
Data Interpretation MCQ -5 - Question 12
(Q.no: 15-20). Refer to the table and answer the given questions.
Total Population in Five Cities = 15,00,000
The difference between the number of male and female in the City P is 10000. Approximately, by what percent of urban population of City P less than its Rural population?
Detailed Solution for Data Interpretation MCQ -5 - Question 12
X = Total population in City A
1/21 * x = 10000 => x = 2,10,000
Rural Population = 1,47,000
Urban Population = 63,000
% = (Difference / Rural Population)*100 = (84,000 / 147000)*100 = 57%
Data Interpretation MCQ -5 - Question 13
(Q.no: 15-20). Refer to the table and answer the given questions.
Total Population in Five Cities = 15,00,000
One – fourth of total population of City R is 2,20,000. If the total number of Literate graduates of City Q and S are 22,500 and 52,000. What % of the total population of City Q, R and S together is literate?
Detailed Solution for Data Interpretation MCQ -5 - Question 13
1/4 * R = 2,20,000 => R = 8,80,000
Literate population of City R = 7/11 * 8,80,000 = 5,60,000
Total number of Literate graduates of City Q = 22500 = 45%
x = Literate population of City Q
45% of x = 22500 => x = 50000
Total number of Literate graduates of City S = 52000 = 65%
y = Literate population of City S
65% of y = 52000 => x = 80000
% = (690000/1820000) * 100 = 37.9%
Data Interpretation MCQ -5 - Question 14
(Q.no: 15-20). Refer to the table and answer the given questions.
Total Population in Five Cities = 15,00,000
The difference between the total population of City R and T is 30,000. The ratio of total population of City R and T is 44:41. Total % of Literate graduates of City R and T is 90%. If the % Literate graduates of City T is equalled to two times of Literate graduates of City R then what is the difference between the number of graduates from City R and City T?
Detailed Solution for Data Interpretation MCQ -5 - Question 14
3/85 * x = 30000 => x = 850000
44/85*850000 = 440000 = City R Population
7/11*440000 = 280000
Number of graduates from City R = 30% of 280000 = 84000
41/85*85000 = 410000 = City T Population
2/5*410000 = 164000
Number of graduates from City T = 60% of 1,64,000 = 98400
Difference = 14400
Data Interpretation MCQ -5 - Question 15
(Q.no: 15-20). Refer to the table and answer the given questions.
Total Population in Five Cities = 15,00,000
What is the sum of female population in City Q and male population in City S together ?
Detailed Solution for Data Interpretation MCQ -5 - Question 15
Sum = 3/10 * 415000 + 11/15 *525000 = 124500 + 385000 = 509500
Data Interpretation MCQ -5 - Question 16
(Q.no: 16 -20). Refer to the table and answer the given questions.
In 2013, if the number of valid votes of female was 100, what was the respective ratio of number of valid votes of male and number of valid votes of female same year?
Detailed Solution for Data Interpretation MCQ -5 - Question 16
38% of 1000 = 380
Valid votes of Female = 100
Valid votes of Male = 380 – 100 = 280
=> 280:100
=> 14:5
Data Interpretation MCQ -5 - Question 17
(Q.no: 16 -20). Refer to the table and answer the given questions.
The total number of votes increased by 50% from 2012 to 2016. If 16% of the votes valid in 2016, what was the number of valid votes in 2016?
Detailed Solution for Data Interpretation MCQ -5 - Question 17
50% of 500 = 250
= 500 + 250 = 750
16% of 750 = 120
Data Interpretation MCQ -5 - Question 18
(Q.no: 16 -20). Refer to the table and answer the given questions.
If the average number of valid votes in 2012 and 2015 was 635, approximately, what percent of total votes valid in 2012?
Detailed Solution for Data Interpretation MCQ -5 - Question 18
Number of valid votes in 2012 and 2015 = 1270
2015: 40% of 2500 = 1000
1270 – 1000 = 270
(270/500)*100 = 54%
Data Interpretation MCQ -5 - Question 19
(Q.no: 16 -20). Refer to the table and answer the given questions.
In 2014, if the difference between number of valid votes of male and number of valid votes of female was 150, what was the total number of votes in 2014?
Detailed Solution for Data Interpretation MCQ -5 - Question 19
(1/6)*x = 15
x = 900
60% of y = 900
y = 1500
Data Interpretation MCQ -5 - Question 20
(Q.no: 16 -20). Refer to the table and answer the given questions.
In 2011, the respective ratio of total number of votes to valid votes was 5:4. Number of valid votes of female in 2011 constitutes what percent of the total number of votes in the same year?
Detailed Solution for Data Interpretation MCQ -5 - Question 20
[4x * (3/8) / 5] * 100 = 30%
## Quantitative Techniques for CLAT
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## Quantitative Techniques for CLAT
55 videos|35 docs|98 tests | 3,153 | 11,447 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.703125 | 4 | CC-MAIN-2023-23 | latest | en | 0.9011 |
http://cs.nyu.edu/~icpc/wiki/index.php?title=DNA_Sorting&oldid=6179 | 1,444,162,477,000,000,000 | text/html | crawl-data/CC-MAIN-2015-40/segments/1443736679145.29/warc/CC-MAIN-20151001215759-00243-ip-10-137-6-227.ec2.internal.warc.gz | 73,929,529 | 5,978 | # DNA Sorting
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)
This problem has been solved by hjfreyer.
Sorter: Uncategorized East Central North America 1998 http://acm.pku.edu.cn/JudgeOnline/problem?id=1007
DNA Sorting is problem number 1007 on the Peking University ACM site. The idea of the problem is to find the "unsortedness" of a given string, in this particular example, it happens to be a string of DNA, but the solution applies to any string.
The unsortedness of a string is defined to be the number of pairs of entries that are out of order with respect to each other. More concretely: for every letter in the string, how many letters are there to the right of it, which should be to the left. After finding this figure, sort the strings by their disorder, and print them out.
(Note: I believe the same figure is achieved when you reverse the words "left" and "right" but I have not proven this)
## Algorithm
The obvious n^2 solution is in fact the correct one (n^2 is fine because each string has at most 50 letters). For each letter in the string, count through all the letters to it's right. Keep a count of all such letters that come before it lexicographically. After looping through each letter, the total count you have represents the disorder of the string. Find out the disorder for each entry, sort them though however means you like, and print them.
## Solution
```import java.util.*;
public class Main{
public static Scanner in;
public static StringBuilder out;
public static int STRLEN;
public static Entry[] ents;
public static void main(String[] args){
in=new Scanner(System.in);
out=new StringBuilder();
doStuff();
System.out.println(out);
}
public static void doStuff(){
//Read in the length of the string and how many there are
STRLEN=in.nextInt();
int N=in.nextInt();
//ents is an array of pairs of Strings and Disorder
ents=new Entry[N];
for(int i=0;i<N;i++){
ents[i]=new Entry();
//Solve the problem
solve(i);
}
//They want the output sorted
Arrays.sort(ents);
//Prinnit!
for(Entry e:ents){
System.out.println(e);
}
}
public static void solve(int k){
ents[k].str=in.next();
ents[k].disorder=0;
String tmp=ents[k].str;
//n^2 is fine because the number of letters is small
//For each letter in the string
for(int i=0;i<tmp.length();i++)
//Check each letter to its right
for(int j=i+1;j<tmp.length();j++)
//If it belongs on its left
if(tmp.charAt(j)<tmp.charAt(i))
//Increase the counter
++ents[k].disorder;
}
}
class Entry implements Comparable{
String str;
int disorder;
public int compareTo(Object o){
if(!(o instanceof Entry))
throw new AssertionError();
Entry e=(Entry) o;
if(disorder==e.disorder)
return 0;
return (disorder>e.disorder)?1:-1;
}
public String toString(){
return str;
}
}
``` | 672 | 2,785 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.703125 | 4 | CC-MAIN-2015-40 | latest | en | 0.875256 |
http://www.nabla.hr/IA-CircleAndLine3.htm | 1,716,126,313,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971057788.73/warc/CC-MAIN-20240519132049-20240519162049-00821.warc.gz | 49,550,516 | 7,110 | Conic Sections
Equation of a tangent at a point of a translated circle
Circle and Line
Line circle intersection
A line and a circle in a plane can have one of the three positions in relation to each other, depending on the distance d of the center S(p, q) of the circle (x - p)2 + (y - q)2 = rfrom the line Ax + By + C = 0, where the formula for the distance:
If the distance of the center of a circle from a line is such that: d < r, then the line intersects the circle in two points, d = r, the line touches the circle at only one point D, d > r, the line does not intersect the circle, and they have no common points.
Equation of a tangent at a point of a circle with the center at the origin
The direction vector of the tangent at the point P1 of a circle and the radius vector of P1 are perpendicular to each other so their scalar product is zero.
Points, O, P1 and P in the right figure, determine vectors,
or written in components x1x + y1y = r2
This is equation of a tangent at the point P1(x1, y1) of a circle with the center at the origin
Equation of a tangent at a point of a translated circle (x - p)2 + (y - q)2 = r2
The direction vector of the tangent at the point P1(x1, y1), of a circle whose center is at the point S(p, q), and the direction vector of the normal, are perpendicular, so their scalar product is zero.
Points, O, S, P1 and P in the right figure, determine vectors,
Since their scalar product is zero, that is
Therefore, is vector equation of a tangent at the point of a translated circle,
or, when this scalar product is written in component form, (x1 - p) · (x - p) + (y1 - q) · (y - q) = r2
it represents the equation of the tangent at the point P1 (x1, y1), of a circle whose center is at S(p, q).
Example: Find the angle formed by tangents drawn at points of intersection of a line x - y + 2 = 0 and
the circle
x2 + y2 = 10.
Solution: Solution of the system of equations gives coordinates of the intersection points,
Plug coordinates of A and B into equation of the tangent:
Example: At intersections of a line x - 5y + 6 = 0 and the circle x2 + y2 - 4x + 2 - 8 = 0 drown are tangents, find the area of the triangle formed by the line and the tangents.
Solution: Intersections of the line and the circle are also tangency points. Solutions of the system of equations are coordinates of the tangency points,
(1) x - 5y + 6 = 0 => x = 5y - 6 => (2) (2) x2 + y2 - 4x + 2y - 8 = 0 (5y - 6)2 + y2 - 4(5y - 6) + 2y - 8 = 0 y2 - 3y + 2 = 0, Þ y1 = 1 and y2 = 2 x1 = 5 · 1 - 6 = -1, => A(-1, 1), x2 = 5 · 2 - 6 = 4, => B(4, 2). Rewrite the equation of the circle into standard form, (x - p)2 + (y - q)2 = r2
x2 + y2 - 4x + 2 - 8 = 0 => (x - 2)2 + (y + 1)2 = 13, thus S(2, -1) and r = Ö13.
Plug tangency points A and B into equation of the tangent,
A( -1, 1) and B(4, 2) => (x1 - p) · (x - p) + (y1 - q) · (y - q) = r2
(-1 - 2) · (x - 2) + (1 + 1) · (y + 1) = 13 => t1 :: - 3x + 2y - 5 = 0,
(4 - 2) · (x - 2) + (2 + 1) · (y + 1) = 13 => t2 :: 2x + 3y - 14 = 0.
Solution of the system of equations of tangents determines the third vertex C of the triangle,
- 3x + 2y - 5 = 0 Tangents are perpendicular since their slopes satisfy the condition, m1 = - 1/m2.
2x + 3y - 14 = 0, C(1, 4)
The triangle ABC is right isosceles, whose area A = 1/2 · AC 2 = 1/2 · Ö(22 + 32)2 = 13/2 square units.
Intermediate algebra contents | 1,177 | 3,435 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.5625 | 5 | CC-MAIN-2024-22 | latest | en | 0.889955 |
https://nrich.maths.org/public/topic.php?code=-203&cl=4&cldcmpid=6412 | 1,571,744,692,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570987817685.87/warc/CC-MAIN-20191022104415-20191022131915-00376.warc.gz | 634,220,200 | 5,803 | # Search by Topic
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### Slide
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This function involves absolute values. To find the slope on the slide use different equations to define the function in different parts of its domain.
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Generalise the sum of a GP by using derivatives to make the coefficients into powers of the natural numbers.
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##### Age 16 to 18 Challenge Level:
What is the quickest route across a ploughed field when your speed around the edge is greater? | 670 | 3,070 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.53125 | 4 | CC-MAIN-2019-43 | longest | en | 0.787875 |
null | null | null | null | null | null | Groupon tour slide
Google is reportedly interested in buying Groupon, for good reason: Groupon is growing insanely fast and seems to have cracked local online advertising, which could be hugely valuable to Google.
But here's how Google could destroy the deal.
If history is any indication, Google will probably try to change Groupon, to make it more Googly.
But the two companies have vastly different structures and cultures -- like oil and water. And based on our understanding of Groupon, Google could potentially ruin Groupon very easily.
Why? Because Groupon is primarily a sales and writing organization, built on people-power and smiles. It employs some 3,000 people around the world, most in sales, customer support, and editorial.
"I'm not sure this is a tech company that gets built in San Francisco or Palo Alto," Groupon cofounder and CEO Andrew Mason told us when we visited his office last week. "It's so heavily people dependent -- not really consistent with the way people think about building startups in the Valley." (Mason refused to discuss anything Google-related, and did not deny the talks.)
In Groupon's bustling 1,000-person Chicago headquarters, heroes include comedy writers, improv actors, and buzzing rooms of salespeople. There is an office elaborately decorated as a bedroom for an imaginary, deranged tenant. It's hilarious, but also the kind of thing that the brains at Google probably wouldn't find funny, or support.
Google, meanwhile, is the master algorithm. Yes, there are lots of people at Google, too, including many sales and support staff. But Google's culture is dominated by engineers, the sort of high-GPA geniuses who can answer all of the brain-teaser Google interview questions, and by the image of being clean and non-evil.
Sure, it's possible that Google could promise Groupon complete autonomy, with the freedom to maintain and grow its counter-culture, and to hire its staff the way it always has, with Google's assistance mainly for infrastructure and investment. YouTube is sort of like this, even 4 years after being acquired.
DoubleClick, on the other hand, was swallowed whole, and quickly Google-fied. Same with pretty much every other Google acquisition, large and small. And that's the sort of thing that could totally screw up Groupon.
| null | null | null | null | null | null | null | null | null |
null | null | null | null | null | null | This is the coolest ALS Ice Bucket Challenge video, period.
Check out famous race and aerobatic pilot Bruce Bohannon taking the ALS Ice Bucket Challenge inside his plane. He placed the ice water in a little bucket between his seat harness and his shirt, took off, and then did a loop. The only thing cooler than this would be if an astronaut did it in the ISS.* [Updated] » 8/27/14 7:22pm 8/27/14 7:22pm | null | null | null | null | null | null | null | null | null |
null | null | null | null | null | null | Rock's Backpages Library Rock's Backpages
Rock's Backpages
Prince: Chaos and Disorder
Charles Shaar Murray, MOJO, August 1996
ALTERNATIVE TITLE: GROINHEAD GOES TO FLORIDA. A sort of musical honeymoon, one presumes: flushed with post-nuptial bliss after snatching mighty Mayte, the abdominal show-woman herself, right from under the nose of the editor of this magazine, Squiggle gathers up the NPG (now once again featuring the deeply fabulous Rosie Gaines), hauls ‘em off to a studio in Carl Hiaasen country, and cuts the kind of album he only seems to create when jaded commentators and connoisseurs start to suspect that he’s lost touch with the known universe and vanished into warp space.
Total word count of piece: 896 | null | null | null | null | null | null | null | null | null |
https://knilt.arcc.albany.edu/index.php?title=Unit_1:_Understanding_the_Benefits_of_the_Nspire_in_Mathematics_Instruction&diff=prev&oldid=26712 | 1,670,310,350,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446711074.68/warc/CC-MAIN-20221206060908-20221206090908-00461.warc.gz | 370,080,958 | 8,900 | # Difference between revisions of "Unit 1: Understanding the Benefits of the Nspire in Mathematics Instruction"
Back to the course outline: Better Understanding of Mathematical Concepts through the Graphing Calculator
## Objective: By the end of Unit 1, you should be able to explain and understand why utilizing the graphing calculator for mathematics instruction is beneficial for student understanding.
#### Introduction
Darling-Hammond (2008) explains their view of the learning of math in school:
“. . [Math] can be experienced in two different ways. For mathematicians, and for those students lucky enough to experience it this way in classrooms. Mathematics is a form of sense making, a result of which is that patterns, rules and procedures, and results all cohere in a meaningful way. It is a discipline in which one can explore, and make and verify discoveries. For most, however, mathematics has been experienced as a set of somewhat arbitrary rules and procedures to be memorized and applied” (p. 113 - 114).
The graphing calculator is one way in which students can better understand mathematics and not simply view it as a set of rules.
### Sample Activity
Here is an example of an activity where students learn using the TI (Texas Instruments) Nspire graphing calculator. The images you see are snap shots of the different screens students see as they are completing the activity to learn about equations in point-slope form:
The above image just gives you a preview to what the Nspire is capable of demonstrating for your students. Check out the complete file at the website below. It contains links to download the actual tns file for the Nspire and activity sheet (Word or PDF): [[1]]
### Article
Below is a link to an article “Graphing Calculators in the Mathematics Classroom.” It presents some general ideas of how graphing calculators are beneficial in the classroom for students’ understanding of mathematics. It also provides references and summaries to further studies and informative articles. Please read the article:[[2]]
### Reflection of Research & Sample Activity
After reading the article and looking at the Shark Attack activity, discuss the following with a partner:
1) Create a list of reasons why utilizing the graphing calculator would be beneficial to use in the mathematics classroom.
2) Do you agree with everything the research says? Why or why not?
3) Did you find the Shark Attack activity to be engaging and beneficial for students' learning of point-slope form? Why or why not?
4) After thinking about the ways the graphing calculator can be beneficial in the mathematics classroom, reflect how you might try to use it in your own classroom. A unit your students have traditionally struggle with might be a great place to start!
Back to the course outline: Better Understanding of Mathematical Concepts through the Graphing Calculator | 570 | 2,896 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5625 | 4 | CC-MAIN-2022-49 | latest | en | 0.932103 |
https://www.coursehero.com/file/6962092/Again-I-need-to-point-out-that-this-is-NOT-the-most-technically/ | 1,498,259,407,000,000,000 | text/html | crawl-data/CC-MAIN-2017-26/segments/1498128320201.43/warc/CC-MAIN-20170623220935-20170624000935-00703.warc.gz | 872,637,000 | 23,505 | Alg_Complete
# Again i need to point out that this is not the most
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Unformatted text preview: 10 y - 1 = 0 [Solution] Solution Neither of these equations are in standard form and so to determine the center and radius well need to put it into standard form. We actually already know how to do this. Back when we were solving quadratic equations we saw a way to turn a quadratic polynomial into a perfect square. The process was called completing the square. This is exactly what we want to do here, although in this case we aren’t solving anything and we’re going to have to deal with the fact that we’ve got both x and y in the equation. Let’s step through the process with the first part. (a) x 2 + y 2 + 8 x + 7 = 0 We’ll go through the process in a step by step fashion with this one. Step 1 : First get the constant on one side by itself and at the same time group the x terms together and the y terms together. x 2 + 8 x + y 2 = -7 In this case there was only one term with a y in it and two with x’s in them. Step 2 : For each variable with two terms complete the square on those terms. © 2007 Paul Dawkins 172 http://tutorial.math.lamar.edu/...
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## This note was uploaded on 06/06/2012 for the course ICT 4 taught by Professor Mrvinh during the Spring '12 term at Hanoi University of Technology.
Ask a homework question - tutors are online | 371 | 1,506 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.921875 | 4 | CC-MAIN-2017-26 | latest | en | 0.954029 |
https://schoollearningcommons.info/question/4-by-which-number-175-is-to-be-multiplied-so-that-the-product-is-a-perfect-square-also-find-the-19839684-92/ | 1,632,563,042,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780057615.3/warc/CC-MAIN-20210925082018-20210925112018-00499.warc.gz | 558,837,119 | 13,554 | ## 4. By which number 175 is to be multiplied so that the product is a perfect square? Also, find the lsquare root of the product.
Question
4. By which number 175 is to be multiplied so that the product is a perfect square? Also, find the lsquare
root of the product.
in progress 0
1 month 2021-08-16T13:36:20+00:00 2 Answers 0 views 0
Perfect square= 1225
Its root = 35
Step-by-step explanation:
Let’s do the prime factorization of 175
175 = 5² x 7
So to make it a perfect square we need to multiply 175 and 7.
Therefore 175 x 7 = 1225.
To find its root:
1225 = (5² x 7) x7
√1225 = √(5² x 7²)
Therefore √1225 = 5 x 7 = 35 | 211 | 636 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.34375 | 4 | CC-MAIN-2021-39 | latest | en | 0.853892 |
https://www.math-only-math.com/opposite-sides-of-a-parallelogram-are-equal.html | 1,720,886,988,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514510.7/warc/CC-MAIN-20240713150314-20240713180314-00116.warc.gz | 720,713,772 | 13,538 | # Opposite Sides of a Parallelogram are Equal
Here we will discuss about the opposite sides of a parallelogram are equal in length.
In a parallelogram, each pair of opposite sides are of equal length.
Given: PQRS is a parallelogram in which PQ ∥ SR and QR ∥ PS.
To prove: PQ = SR and PS = QR
Construction: Join PR
Proof:
Statement In ∆PQR and ∆RSP;1. ∠QPR = ∠SRP2. ∠QRP = ∠RPS3. PR = PR4. ∆PQR ≅ ∆RSP 5. PQ = SR and PS = QR. (Proved) Reason 1. PQ ∥ RS and RP is a transversal.2. PS ∥ QR and RP is a transversal.3. Common side4. By ASA criterion of congruency. 5. CPCTC
Converse of the above given theorem
A quadrilateral is a parallelogram if each pair of opposite sides are equal.
Given: PQRS is a quadrilateral in which PQ = SR and PS = QR
To prove: PQRS is a parallelogram
Proof: In ∆PQR and ∆RSP, PQ = SR, QR = SP (given) and PR is the common side.
Therefore, by SSS criterion of congruency, ∆PQR ≅ ∆RSP
Therefore, ∠QPR = ∠PRS, ∠QRP = ∠RPS (CPCTC)
Therefore, PQ ∥ SR, QR ∥ PS
Hence, PQRS is a parallelogram.
Solved examples based on the theorem of opposite sides of a parallelogram are equal in length:
1. In the parallelogram PQRS, Pq = 6 cm and SR : RQ = 2 : 1. Find the perimeter of the parallelogram.
Solution:
In the parallelogram PQRS, PQ ∥ SR and SP ∥ RQ.
The opposite sides of a parallelogram are equal. So, SR + PQ = 6 cm.
AS SR : RQ = 23 : 1, $$\frac{6 cm}{RQ}$$ = $$\frac{2}{1}$$
⟹ RQ = 3 cm
Also, RQ = SP = 3 cm.
Therefore, perimeter = PQ + QR + RS + SP
= 6 cm + 3 cm + 6 cm + 3 cm
= 18 cm.
2. In the parallelogram ABCD, ∠ABC = 50°. Find the measures of ∠BCD, ∠CBA and ∠DAB.
Solution:
AS AB ∥ DC, ∠ABC + ∠BCD = 180°
Therefore, ∠BCD = 180° - ∠ABC
= 180° - 50°
= 130°
As opposite angles in a parallelogram are equal,
∠CDA = ∠ABC = 50° and
∠DAB = ∠BCD = 130°
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Jul 12, 24 12:59 AM
To convert an improper fraction into a mixed number, divide the numerator of the given improper fraction by its denominator. The quotient will represent the whole number and the remainder so obtained… | 980 | 3,175 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.875 | 5 | CC-MAIN-2024-30 | latest | en | 0.862769 |
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kieran saunders
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Remove the nth column from input matrix A and return the resulting matrix in output B. So if A = [1 2 3; 4 5 6]; ...
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Times 2 - START HERE
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A recipe for one large cake calls for 3 2/3 cups of flour. Anya has decided to increase each ingredient in the recipe by 2 1/2. How many cups of flour does she need?
• math - ,
what is 3 2/3+2 1/2?
3 4/6 + 2 3/6=5 7/6= 6 1/6 cups
• math - ,
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Saturday, December 20, 2014
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Enron’s Skilling gets more than 10 years cut off sentence
Mike Tolso, Houston Chronicle
Updated 1:56 pm, Friday, June 21, 2013
• Former Enron CEO Jeff Skilling arrives to Bob Casey Federal Courthouse Friday, June 21, 2013, for a new sentencing hearing. Photo: Melissa Phillip, Houston Chronicle
Photo: Melissa Phillip, Houston Chronicle
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The long-running battle between federal prosecutors and former Enron CEO Jeff Skilling formally came to an end Friday when U.S. District Judge Sim Lake signed off on a negotiated sentence agreement that will end Skilling's incarceration in about four years.
In agreeing to the minimum sentence within the range of 168 and 210 months, Lake said the interest of justice would not be served by having Skilling serve longer. He noted that the architect of Enron's rise from an obscure pipeline company to an innovative and diverse energy giant will end up spending more time in prison than anyone else connected to the company's stunning 2001 collapse into bankruptcy.
"This is not an easy decision to make," Lake said about the number of months to choose.
"The two most significant factors are the need for the sentence to deter others from similar action and to reflect the seriousness of the offense — 168 months adequately reflects both concerns. The court is not persuaded a longer sentence is necessary."
Appellate reversals of part of the government's case against Skilling, as well as one of grounds used by Lake in setting the original sentence of 24 years, meant that Skilling already was entitled to a reduced sentence. The agreement knocked about 20 months off the lower end of the revised sentence range, in exchange for which Skilling agreed to forgo any further appeals.
That finality, which brings the Enron saga to a close, frees up about $40 million in Skilling's assets to be divided among investors, protects the government from the possibility of more appeals reversals and assures Skilling of at least the possibility of having an extended period of time as a free man before he dies.
He has been in federal custody since December 2006, about two months after he and Enron Chairman Ken Lay were convicted of various financial crimes. Lay died of a heart attack before being sentence.
"This is absolutely a resonable resolution," said attorney Philip Hilder, a onetime federal prosecutor who represented so-called Enron whistleblower Sheron Watkins, an accountant who became famous when her memo to Lay warning of the consequences of potential accounting scandals became public. "It frees up about $41 million and puts and end to the appeals. And Skilling will have served a significant amount of time."
In accordance with the terms of the agreement, government attorneys did not recommend a particular sentence within the negotiated range. However, prosecutor Patrick Stokes emphasized that Skilling, while not a criminal in the sense of corporate looter, used deceptive accounting to keep the company from having to face the consequences of bad business decisions.
"He had a tremendous work ethic and was very talented," Stokes said. "That's where the contradiction comes in. He (eventually) used those talents to lie and cheat. Why he did it, we don't know. But he used lies to hide from the public and the people of Enron what was going on." | null | null | null | null | null | null | null | null | null |
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I seek metrics on complete manifolds whose scalar curvature represents a singularity of the form $\frac{h}{\rho^2}$ where $h$ is a continuous function and $\rho$ vanishes on the boundary of some compact set of the manifold.
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A simple example might be to form a surface of revolution by rotating the superellipse $x^m+y^m=1$ about the $y$ axis, and using the induced intrinsic metric on that surface. The $m=2/3$ case, called an astroid, gives a surface of revolution whose Gaussian curvature blows up like $\rho^{-1/2}$, where $\rho$ is the distance from the cusp. I would expect that some $m<2/3$ would give a singularity with the exponent you want. (You didn't define $\rho$, but I assume you intend it to be the distance from the boundary.) – Ben Crowell Jul 31 '12 at 15:43
3 Answers 3
Take the graph $t=\sqrt{x^2+y^2+z^2}$ with induced intrinsic metric. In $(x,y,z)$-coordinates, the scalar curvature is $$\frac C{x^2+y^2+z^2}.$$
Are you happy?
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Did you mean some other function, perhaps something like $z=1/|x+y|$, rather than $z=|x+y|$? – Ben Crowell Jul 31 '12 at 12:47
I think, Anton answered question which OP intended to ask, namely, construct a singular Riemannian metric whose curvature behaves like $h/\rho^2$, where $\rho(x)=d(x_0, x)$ and $h(x_0)\ne 0$. Instead, OP asked a question which does not make whole lot of sense, since in his question scalar curvature becomes infinite outside of a bounded region $C$ (assuming that he does not want $h$ to vanish outside of $C$, otherwise one could take, say, $h=-\rho$ and have curvature equal to $-1$ everywhere). – Misha Jul 31 '12 at 13:39
Anton's formula for the scalar curvature doesn't match up correctly with the equation he's given for the surface. The surface $z=|x+y|$ consists of two half-planes joined at the line $x=-y$. The scalar curvature of this surface does not equal $C/|x+y|^2$. The induced intrinsic curvature of this surface is in fact zero everywhere, even on the line $x=-y$. (One way to see that it's zero even on the line is to consider the join between the two planes as the limiting case of a piece of a cylinder with radius $r$ approaching zero; the intrinsic curvature is zero for all $r$.) – Ben Crowell Jul 31 '12 at 14:34
@Anton: Assuming you intended $z=|x+y|^p$ with some other exponent $p\ne 1$, there is still a problem with the answer, because the OP wanted the singularity to occur on the boundary of a compact set, and the line $x=-y$ is not the boundary of a compact set. @Misha: I assume the OP intended $\lim_{\rho\rightarrow0}h\ne 0$. – Ben Crowell Jul 31 '12 at 15:49
Ben, isn't the one point the boundary of the compact set consisting of that same point? – Deane Yang Jul 31 '12 at 18:45
There is a description a plane curve called its Cesàro equation, in which the curvature is given as a function $k$ of arc length $s$. Suppose that $k$'s domain is $(0,1]$. Then it should be clear that, given one point and a tangent vector at that point, the corresponding curve $\mathbf{r}(s)$ exists and is unique for $s\in[0,1]$. (Even if the function $k$ misbehaves at $s=0$, the limit $\mathbf{r}(0)$ exists.)
Define the curve C whose Cesàro equation for $s\in(0,1]$ is $k=1/s^2$, with (arbitrarily, but for concreteness) $\mathbf{r}(0)=(1,0)$, and its tangent vector upward at $s=0$. Form a surface of revolution S by revolving C about the $y$ axis. S is compact, and the locus $s=0$ is a unit circle forming a boundary of S.
The Gaussian curvature $\kappa$ is equal in magnitude to the product of the curvatures along the two principal axes. The curvature along the azimuthal axis is 1 for $s=0$, so $|\kappa|\sim 1/s^2$ as $s\rightarrow 0$.
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You can use the warped product of two Riemannian manifolds, with a warping function which vanishes on a compact region.
Let $(B,g_B)$ and $(F,g_F)$ be two Riemannian manifolds, and $f$ a smooth function on $B$. The warped product of $B$ and $F$ with warping function $f$ is the manifold $$B\times_f F:=\big(B\times F, \pi^*_B(g_B) + (f\circ \pi_B)\pi^*_F(g_F)\big),$$ where $\pi_B:B\times F \to B$ and $\pi_F: B \times F \to F$ are the canonical projections. It is customary to call $B$ the base and $F$ the fiber of the warped product $B\times_f F$.
Let $B \times_f F$ be a warped product, with $\dim F>1$. Then, the scalar curvature $s$ of $B \times_f F$ is related to the scalar curvatures $s_B$ and $s_F$ of $B$ and $F$ by $$s = s_B + \frac {s_F}{f^2} + 2\dim F\frac{\Delta f}{f} + \dim F(\dim F - 1)\frac{\langle grad f, grad f\rangle_B}{f^2}.$$
You can play with this formula to obtain what you are looking for - by trying to make $s_B + 2\dim F\frac{\Delta f}{f}$ vanish, or at least to be of the form $\frac{h}{f^2}$. To make it vanish, you may solve the equation $$\Delta f+\frac{s_B}{2\dim F}f=0.$$
More about warped products in O'Neill's "Semi-Riemannian geometry: with applications to relativity", and in this paper.
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1 GS/ECON 5010 Answers to Assignment November 2007 Q1. What is the equation of the supply curve of a firm which has a total cost function with the equation T C(q) = (q 6) + q where q is the quantity of output produced by the firm? A1. Given the total cost function, the marginal cost for the firm is MC(q) = (q 6) 2 + (1 1) Equation (1 1) defines a U shaped curve, which reaches a minimum (of MC = ) at q = 6. The firm s supply curve is the upward sloping part of its marginal cost curve, above its intersection with the average cost curve. So the profit maximizing relation p = MC, and equation (1 1), imply a supply curve with the equation or p = (q 6) 2 + p q = + 6 (1 2) Equation (1 2) defines the firm s supply curve, but only for values of the price which are above the minimum average cost. To find the minimum average cost, divide the total cost by q to get AC(q) = (q 6) q + (1 ) From equations (1 1) and (1 ), the marginal cost equals the average cost when or (expanding both sides of (1 4)) (q 6) 2 q = (q 5) (1 4) q 6q q = q 18q q which is equivalent to 2q(q 9) = 0 (1 5) so that MC(q) = AC(q) at q = 9. Checking, from equation (1 ), AC (q) = 1 q 2 [q(q 6)2 (q 6) 216] = 2(q + )(q 6)2 216 q 2 (1 6) 1
2 From (1 6), AC (q) < 0 for small values of q, and AC (q) = 0 at q = 9. At q = 9, MC = AC = 0. p So the firm s supply curve is q = 0 if p 0 amd q = + 6 if p 0. [Note that this is a long run supply curve ; the firm s total cost equals 0 when q = 0 here so that there are no fixed costs.] Q2. What is the equation of the long run supply curve for an industry, if it has one firm of type t, for each value of t = 1, 2,,,, and if a firm of type t has a long run total cost function T C(q) = (q 6) + tq where q is the quantity of output produced by the firm? A2. The total cost function in the first question is just a special case of the total cost function in this question, with t =. So, from the answer to the first question, a firm of type t has a U shaped average cost curve, with a minimum at q = 9, and a minimum average cost (at q = 9) of 27 + t. p t Also, the supply curve for a type t firm is q = 0 for p 27 + t, and q = + 6 if p 27 + t. So a type 1 firm will produce a positive level of output only if the price is 28 or more, a type 2 firm will produce only if the price is 29 or more, and so on. This means that if the price is less than 28, no firms will produce anything ; the price is below each firm s minimum average cost, and no output will be supplied. If the price is between 28 and p 1 29, 1 firm (the type 1 firm) will produce, and it will produce + 6 units. If the price is p 1 p 2 between 29 and 0, 2 firms will produce, and their total output will be In general, if the price is between k and k + 1, where k is an integer greater than 27, then k 27 firms will produce, and the industry s total quantity supplied will be 6(k 27) + k 27 i=1 p i Q. Suppose that a (single price) monopoly faced a market demand with inverse demand curve p(q) (with p (Q) < 0), and could produce under constant returns to scale at a marginal cost of c per unit produced. However, the marginal cost c depends on the firm s fixed investment F in technology ; c (F ) < 0, since more investment in technology leads to lower cost production techniques. (i) Derive an expression for the firm s profit maximizing level of technology investment. (ii) Would consumers, in aggregate, be able to bribe the monopoly to invest in further technology investment, above the level which maximizes profits? Explain briefly. 2
3 A. (i) Let q (c) denote the profit maximizing output level for a monopoly with a constant marginal cost of c, that is the solution to the equation p(q [c]) + p (q [c])q [c] = c ( 1) The monopoly s profit, ignoring its fixed cost of technology, would be π(c) = p(q [c])q (c) cq [c] ( 2) Differentiating equation ( 2), π (c) = [ p (q [c]q [c] + p(q [c]) c ] q c q [c] ( ) But equation ( 1) implies that the term in large square brackets in ( ) is zero that s the envelope theorem again so that π = q. Now consider the monopoly s choice of technology. It wants to choose an investment F in technology to maximize its overall profit π(c(f )) F ( 4) implying a first order condition c (F )q (c(f )) = 1 ( 5) That is, one dollar more invested in technology should raise its profits (net of fixed costs) π by one dollar. (ii) The aggreate benefit to consumers can be measured as the area underneath the aggregate demand curve, and above the price. This consumers surplus is CS = q 0 p(q)dq p(q )q (that is, the difference between total consumer surplus, and the total amount paid to the monopoly) if the firm produces q units of output. So (since the derivative of an integral is the function itself) CS q = p (q )q > 0 ( 6) At the monopoly s profit maximizing level of technology investment the solution to equation ( 5) consumers could get the monopoly to invest more in cost saving technology if they could get together and coordinate a campaign to bribe the firm. At the margin (that is, at the level of F for which c (F )q (F ) = 1) a slight increase in technology investment has no effect on the monopoly s profit, since the cost savings are just offset by the expense of the added technology
4 ivestment. But since the lower marginal cost would result in greater output and a lower price, consumers would benefit. So the monopoly invests too little in technology, in that a further technology improvement would benefit consumers more than it would harm the firm. Q4. Find every Cournot Nash equilibrium in a duopoly, in which the demand function for the homogeneous product has the equation Q = 14 p and in which both firms have the same total cost function, T C(q) = q q > 0 = 0 q = 0 A4. The fixed cost (of 12) does not matter for each firm s reaction function provided that the firm does decide to produce a positive level of output. So formula (4.15) of Jehle & Reny can just be plugged in here, implying an output level for each firm of since here a = 14, b = 1, c = 2, J = 2. q 1 = q 2 = 14 2 = 4 (4 1) But q 1 = q 2 = 4 will be an equilibrium only if each firm is making non negative profits ; otherwise the firm would prefer to produce nothing at all (at a total cost of 0). When q 1 = q 2 = 4, then p = 14 (q 1 + q 2 ) = 6, so that each firm s revenue is 6 4 = 24. The firm s total costs of production are 2(4)+15 = 2, so that each firm is making positive profits. The output combination q 1 = q 2 = 4 is a Cournot Nash equilibrium. But here, becasue of the positive fixed costs, q 1 = q 2 = 4 is not the only Cournot Nash equilibrium. If q 1 is high enough, then firm 2 would be unable to cover its fixed costs, no matter what it does. For example, if q 1 = 8, equation (4.14) of Jehle & Reny shows that firm 2 s best output would be q 2 = 2, If q 1 = 8 and q 2 = 2, then p = 4, and firm 2 s net profit would be 2(4) 2(2) 15 = 11 < 0. So a high output level, such as q 1 = 8 would deter entry by the other firm : if q 1 = 8, then firm 2 s best reaction would be to produce nothing. However (8, 0) is not a Cournot Nash equilibrium here, becasue firm 1 would not choose to produce 8 units of output. If firm 2 produces nothing, firm 1 s output must be its own best reaction to q 2 = 0. Here that best reaction is q 1 = 6. You can see that q 1 = 6 and q 2 = 0 satisfies (4.14) for j = 1. In fact, q 1 = 6 is exactly the output firm 1 would choose if it were a single price monopoly which it is, if firm 2 chooses not to produce. So is (6, 0) a Cournot Nash equilibrium here? Certainly firm 1 will want to produce q 1 = 6 if q 2 = 0. What will firm 2 want to do if q 1 = 6? Equation (4.14) shows that firm 2 s best choice of 4
5 output (other than 0) when q 1 = 6 is q 2 =. If q 1 = 6 and q 2 =, then p = 5, and firm 2 would earn profits of 5() 2() 15 = 6. Firm 2 makes a loss, and would rather produce 0. So q 1 = 6, q 2 = 0 is also a Cournot Nash equilibrium. Firm 1 wants to produce 6 when firm 2 produces 0, and firm 2 wants to produce 0 when firm 1 produces 6. Why should firm 1 be the lucky one? There is also another Cournot Nash equilibrium, in which q 1 = 0 and q 2 = 6. Therefore, there are Cournot Nash equilibria with this cost function and demand function : (4, 4), (6, 0), (0, 6). Q5. What would be the equilibrium in a duopoly, if the demand and cost functions were those of question #4 above, but in which the firms moved sequentially (as in the Stackelberg duopoly presented in question 4.9 of the text)? That is, firm 1 moves first, committing to a quantity of output which it will produce. Firm 2 then observes firm 1 s quantity, and, after doing so, chooses its own quantity. (Here firm #1 is aware that firm #2 will be choosing its output level last, and will be reacting to firm #1 s own choice of quantity.) A5. When firm 1 moves first, in can anticipate how its rival will react to its choice of output. In this case, firm 2 will choose to produce q r 2 = 6 q 1 2 (5 1) if it chooses to produce anything at all. Equation (5 1) is firm 2 s reaction function, and can be derived from (4.14) of Jehle & Reny with j = 2. But firm 2 will only choose the output level defined by (5 1) if it will make non negative profit ; otherwise it will choose to produce nothing. So if it were not for the fixed cost, firm 1 would want to choose its output q 1 so as to maximize its own profits [a b(q 1 + q r 2(q 1 )]q 1 C(q 1 ) (5 2) an expression which takes into account the fact that more output by firm 1 will lead to less output by firm 2. [When C(q 1 ) = cq 1, then maximizing (5 2) actually leads to an optimal choice for firm 1 of q 1 = a c 2b which would be the answer to the question if there were no fixed costs.] (5 ) But here, the problem is actually easy for firm 1 to solve. From the solution to question #4 above, it knows that if it chooses an output of q 1 = 6, that firm 2 would then react by choosing to produce nothing. That is, an initial choice of q 1 = 6 will deter entry by firm 2. But q 1 = 6 is also the profit maximizing output for firm 1 to choose if it does not need to worry about another firm. So the equilibrium here, when firm 1 has a first mover advantage is for firm 1 to choose q 1 = 6, and for firm 2 to react be choosing not to produce at all. 5
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### Universidad Carlos III de Madrid Game Theory Problem Set - Dynamic Games
Universidad Carlos III de Madrid Game Theory Problem Set - Dynamic Games Session Problems 1, 2, 3, 4, 5, 6 1 (no SPNE) 2 7, 8, 9, 10, 11 3 12, 13, 14, 15, 16 4 17, 18, 19 5 Test 1. The next figure shows | 9,547 | 37,922 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.890625 | 4 | CC-MAIN-2018-51 | latest | en | 0.915728 |
http://mathhelpforum.com/advanced-algebra/212216-normal-extensions.html | 1,481,158,490,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698542288.7/warc/CC-MAIN-20161202170902-00395-ip-10-31-129-80.ec2.internal.warc.gz | 187,492,236 | 10,254 | 1. ## Normal extensions
Obtain a normal extension of $\mathbb{Q}$ containing $3+\sqrt{3}$ and $\sqrt{15}$ and express this extension in the form $\mathbb{Q}(\alpha)$
2. ## Re: Normal extensions
clearly Q(√3) contains 3+√3. if we then take Q(√3,√5) this contains 3+√3 and √15 = √3√5.
note that Q(√3) is a normal extension of Q, since all roots of x2 - 3 are contained in Q(√3) (it is a splitting field of that polynomial).
in much the same vein, Q(√3,√5) = (Q(√3)(√5) is a splitting field of x2 - 5 over Q(√3), so Q(√3,√5) is normal over Q(√3). we will see later than Q(√3,√5) is actually normal over Q.
but first, let's find an element u in Q(√3,√5) such that Q(√3,√5) = Q(u). i claim we can take u = √3+√5.
it is obvious that Q(√3+√5) is a subfield of Q(√3,√5). so to show that √3+√5 actually generates Q(√3,√5) it suffices to show that √3 and √5 are contained in Q(√3+√5).
if we compute (√3+√5)3 = 18√3 + 14√5, it becomes obvious that (1/4)((√3+√5)3 - 14(√3+√5)) = √3, which shows that √3 is in Q(√3+√5).
therefore √5 = (√3+√5) - √3 is likewise in Q(√3+√5), so Q(√3,√5) = Q(√3+√5).
as promised we will now show Q(√3+√5) is normal by exhibiting an irreducible polynomial in Q[x] that splits completely in Q(√3+√5):
f(x) = (x + √3+√5)(x - √3+√5)(x + √3-√5)(x - √3-√5)
= (x2 - (8+2√15))(x2 - (8-2√15))
= x4 - ((8+2√15)+(8-2√15))x2 + (8+2√15)(8-2√15)
= x4 - 16x2 + (64-60)
= x4 - 16x2 + 4
convince yourself that if f(x) = (x2+ax+b)(x2+cx+d) that we must have c = -a, and a(d-b) = 0, and that a = 0 leads to b2 - 16b +4 = 0, while b = d leads to a2 = 12, neither of which can occur. note as well that f(x) is a quadratic in x2, and there is no x2 in Q that is a root, since the discriminant 256 - 4(4) = 240 is not a perfect square. thus this has no rational roots x (since x2 is surely rational if x is), and so conclude that f(x) is irreducible over Q.
but this means Q(u) is a splitting field of an irreducible polynomial in Q[x], and is thus a normal extension. in fact, since f(x) is separable, it is a galois extension. | 797 | 2,039 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 4, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.0625 | 4 | CC-MAIN-2016-50 | longest | en | 0.897067 |
null | null | null | null | null | null | TPM Livewire
Blitzer Calls Out Gingrich On Sebelius 'Nixon' Comment
AP Photo / Rex Features
Blitzer asked Gingrich Wednesday afternoon if he would like to revise his accusation.
"I want to know if you want to revise making a comparison to Nixon and the only president forced out of office because he broke the law," Blitzer said to Gingrich.
At first, Gingrich declined.
"No, he was forced out of office in part, he resigned in part because he had said a series of things that weren't true," Gingrich said. "How can you have been through a month of the website breaking down? If you want to make the argument, she's out of touch with reality, it wasn't a lie because she hasn't noticed."
After more back and forth, Gingrich said that Sebelius was just as bad as Nixon.
"I'll modify it. Equals anything. How is that?" Gingrich asked Blitzer. | null | null | null | null | null | null | null | null | null |
null | null | null | null | null | null | Yushchenko to overhaul security agency - Taipei Times
Sun, Jan 23, 2005 - Page 7 News List
Yushchenko to overhaul security agency
One of the toughest tasks facing Viktor Yushchenko as Ukraine's new president likely will be overhauling the State Security Service, the KGB successor agency alleged to have been involved in an array of devious and deadly activities.
The cases of other Soviet-bloc countries show that success or failure could determine how well democracy takes root in Ukraine -- but the issue is complicated by indications that security forces played a role in averting a crackdown on Yushchenko's "Orange Revolution."
Since the 38,000-strong SBU was formed in 1991 after independence from the Soviet Union, it is alleged to have been connected to organized crime, shady weapons deals and the deaths of several prominent opposition politicians and journalists.
It is suspected of involvement in the September dioxin poisoning of Yushchenko, which took him off the campaign trail for weeks and left his face badly disfigured. He fell ill from the poisoning within hours after having dinner with top SBU officials.
Throughout the region, security agency reform has been a mixed bag of results.
The Czech Republic -- now a successful market-driven democracy -- dissolved its former communist secret police, the STB, after the 1989 Velvet Revolution and barred former high-ranking communists and secret police agents from holding public office.
By contrast, Belarus, whose President Alexander Lukashenko is widely considered Europe's last dictator, has retained unchanged the structure -- and the name -- of the feared Soviet-era security agency, the KGB.
In Russia, the KGB was broken up into several agencies, but the main successor organization, the FSB, retains much of the clout of its predecessor. President Vladimir Putin, himself a former KGB officer, has brought into government many high-ranking former security officials.
Ukraine's security agency is highly factionalized, with internal clans loyal to different political camps. Which one was dominant came under question the day after the Nov. 21 election, when demonstrators jammed Kiev to protest voting fraud after Yushchenko was declared the loser.
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null | null | null | null | null | null | Singing the Dogstar Blues
Singing the Dogstar Blues
Goodman, Alison
Seventeen-year-old Joss is a rebel--the daughter of a famous newscaster and a sperm donor; a wild girl who can play a mean harmonica; a student of time travel at the prestigious Centre for Neo-Historical Studies. This year, for the first time, the Centre has an alien student. Mavkel, from the planet Choria, has chosen Joss--of all people--as his study partner. Can this very different pair manage to live together, much less learn from each other? Singing the Dogstar Blues is a genre-breaking mix of humor, science fiction, and adventure that will keep readers riveted.
In this novel set in futuristic Australia, Joss gets to learn about the past firsthand--through time travel!
This item appears in the following Booklists:
Call Number:
YP F Goodman
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https://www.educationquizzes.com/gcse/maths/distance-speed-and-time-problems-f/print/ | 1,590,861,764,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347410284.51/warc/CC-MAIN-20200530165307-20200530195307-00485.warc.gz | 687,078,179 | 3,772 | # GCSE Maths Quiz - Distance, Speed and Time Problems (F) (Questions)
Want to go further, faster? In that case you had better have a good understanding of speed calculations! See if you can speed through this GCSE Maths quiz.
We all know that speed is something to do with distance and time, but can you remember how they are related? One way to remember it is to think of road speed limits. For instance, in towns the speed limit is 30 miles per hour. Speed = miles (Distance) per hour (Time). ‘Per’ means divide, so we have S = D/T. Once you have that it is a case of filling in what you know, to work out what you need.
Always remember to apply a common sense check to your answer – for instance, if you are travelling at 80km/h, it doesn’t make sense that you will cover 160km in half an hour!
Another area that catches people out is converting between m/s and km/h. It is best to work through this in stages – work up from seconds to minutes, then to an hour. Then, scale that answer from metres to kilometres. If you try to do it all in one step you are likely to trip up!
A speed of 60 units per hour is always quite nice, as this means you cover one unit every minute. Remember that next time you are on a car journey, and want to know how long until you get there!
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GCSE Maths Quiz - Distance, Speed and Time Problems (F) (Answers)
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http://combine.se/blog/page/2/ | 1,547,941,920,000,000,000 | text/html | crawl-data/CC-MAIN-2019-04/segments/1547583684033.26/warc/CC-MAIN-20190119221320-20190120003320-00045.warc.gz | 49,381,702 | 36,803 | Blog | Combine
# Blog
## Introduction
At Combine, we play board games during so-called “Game Nights”. On several occasions, there have been discussions regarding how to shuffle cards efficiently (i.e. having an unpredictable order of the cards).
A deck of ordinary playing cards consists of four suites of 13 cards each giving a total of 52 cards. The total number of permutations of the card deck is given by:
$$P^n_r = \frac{n!}{(n-r)!}$$
where $$n = 52$$ is the total number of cards and $$r = 52$$ is the length of the sequence we want to generate from the $$n$$ cards. Since $$n = r$$ we obtain $$n! = 52!$$, which is approximately $$8 \cdot 10^{67}$$ combinations.
How to shuffle cards have been studied before and also discussed in other ways, and these sources have been used as a foundation for this text.
Given a deck of 52 cards, each card is numbered from 0 to 51 in order ($$F_i = i$$). After shuffling the deck we then know the id of each card in the new order. The Shannon Entropy is then calculated by first estimating the distribution of distances between each card in order:
$$\Delta F_j = F_{j+1} – F_j$$
The Shannon Entropy is then calculated using
$$E = \sum_{j=0}^{51} -p_j \log_2(p_j)$$
The variable $$p_j$$ is a normalized histogram of distances between cards.
The maximum possible entropy is $$\log_2(52) = 5.7$$ (measured in the unit “bits”), which is useful as a reference.
According to literature, it is enough to cut the deck and riffle shuffle seven times to obtain an unpredictable order.
## Overhand Shuffle
Not everyone is able to perform the riffle shuffle and might instead use the overhand shuffle. We experimented with the overhand shuffle, wrote down the order of the cards for each shuffle, and ended up with the following increase in entropy (the red line is the maximum possible entropy).
The first iteration does not increase the entropy at all since the deck was only cut once without any shuffling taking place.
## Hash Shuffle
One idea which was discussed at one Game Night was to shuffle cards using a method similar to how hash values are calculated in computer science. This is a deterministic shuffling method, but by adding some random elements, like cutting the deck, we get some interesting results.
The idea is to choose a number of piles and divide the cards between them. This would force cards to interleave with each other, introducing a distance between them. Just doing this once without any random elements gives the following entropies for different numbers of piles:
If we apply the hash shuffle twice and try all combinations between 2 and 10 piles we find that using 5 piles to start with and then 5 piles a second time again gives the highest entropy. In practice, you should cut the deck between each operation as well.
If you want to repeat the hash shuffle for a fixed number of piles you should obviously avoid 2 and 4 piles.
## Riffle Shuffle
The riffle shuffle is claimed to be one of the best ways to shuffle a deck of cards. And, indeed, given the Shannon Entropy measure the riffle shuffle is by far the best way to shuffle according to the following figure:
The entropy rises very fast. Using other measures it is claimed that 7 riffle shuffles should be enough, and more than $$2 \log_2(52) = 11.4$$ shuffles is not necessary.
## Conclusion
When shuffling you should use the riffle shuffle. Just make sure that you cut the deck between each shuffle since the top and bottom cards tend to get stuck otherwise.
In recent years, the hype around artificial intelligence (AI) has grown a lot. AI is not a new concept and the term has been around since the 1950’s when computer scientist John McCarthy coined the term. But the concept of creating artificial beings has been around much longer and can be found in greek mythology or in Mary Shelley’s Frankenstein.
AI has also been portrayed in numerous books and movies and some are more interesting than others. There are great classics like Stanley Kubrick’s 2001: A Space Odyssey (1968) inspired by a short story by Arthur C. Clarke. In the movie the ship’s computer HAL (Heuristically programmed ALgorithmic computer) 9000 is the main antagonist. This is one of the first movies that portrayed the idea of a human made AI to the masses.
Another classic movie is Blade Runner (1982) based on Philip K. Dick’s novel Do Androids Dream of Electric Sheep?. In the movie humans have designed intelligent androids, called replicants, that the main protagonist, Rick Deckard, must hunt down and retire (terminate).
One common problem in movies, and sci-fi movies in general is how to balance an intricate and interesting story with special effects and action scenes. The sci-fi genre has so much potential when it comes to explore concepts about science, technology, existentialism, human evolution and the possibility of extraterrestrial life. When the boundaries of our current paradigm is not a hard limit, imagination and science can blend in the most interesting ways. The movies and TV shows below are not based around special effects or action but rather trying to explore what it is to be human.
#### Her (2013)
(image from themoviedb)
Director: Spike Jonze
Starring: Joaquin Phoenix, Amy Adams, Rooney Mara, Olivia Wilde, Scarlett Johansson
Genres: Sci-fi, Romance, Drama
Synopsis: Joaquin Phoenix plays a man who installs a new operating system with artificial intelligence to help him with various tasks and over time their relationship becomes more and more romantic.
Being able to convey a romantic relationship without the need of a materialized body, using only a voice, is beautifully executed. Most other movies that explore romantic human/machine involvement are doing so almost by cheating. Using an android, more or less indistinguishable from a human, makes it much easier to relate to. Human/computer interaction by voice is not something new and has been around for a quite some time, mostly as an accessibility tool for people not able to use standard equipment. In October 2011, Siri was launched (by Apple) as the first smart phone integrated voice controlled virtual assistant. At the time, the actions it could perform was rather limited, problems understanding voice input and could be seen as more of a gimmick than a valuable tool. Since then other companies have released voice controlled virtual assistants and the biggest competitors are Amazon Alexa released in 2014 and the Google Assistant released in 2016. These products are getting better and better and voice controlled virtual assistants are probably here to stay although they are not perfect.. yet.
Her, artificial intelligence and the concept of time
Humans, like animals are evolutionary equipped to experience some basic aspects of time. What happens when an artificial intelligence emerge with the computational power of living years, decades or millenniums every day, hour or second? Their concept of time will be something completely different which could have huge implications. For some perspective, OpenAI Five recently competed with five bots against five former professional gamers in the competitive game Dota 2. Everyday the AI played 180 years worth of games against itself running on 256 GPUs and 128 000 CPU cores. When summing up each character (a total of five) it amounts to 900 years every day. Time is relative but even more so for an AI running on a giant cluster and there is a great scene about this in the movie.
#### Ex Machina (2014)
(image from themoviedb)
Director: Alex Garland
Starring: Domhnall Gleeson, Alicia Vikander, Oscar Isaac
Genres: Sci-fi, Drama, Mystery, Thriller
Synopsis: Meet Caleb Smith, a talented programmer at a large tech company, who wins an office competition to stay a week at the CEO’s remotely located house. When he arrives he is introduced to an android who according to Nathan Bateman, the CEO, has passed the Turing test but he wants more validation before going public with the news. Surprise, surprise.. Caleb starts to develop feelings for the android.
How close is humanity to develop an human-like conscious AI and what happens then?
What happens then?, is a question people have been debating for a long time. Some view this as the inevitable future and hope for the human race and others see it as humanity’s demise. This has also been explored in sci-fi books and movies over the years and The Terminator (1984) portrays one of the darker scenarios for mankind. But how close are we to develop an AI able to pass the Turing test? Probably not very close. Even though great progress has been made in the last few years, with everything from AlphaGo being able to beat the top players at Go to Libratus a poker AI able to beat top players in heads up no-limit Texas hold ’em (and of course the OpenAI Dota 2 bots mentioned earlier), the Turing test is something else to beat. The AI’s created today are heavily specialized and contextual but can’t do much outside their specialization. And in order to pass the Turing test an AI would have to behave like a human when exposed to a multitude of different questions that could range from anything to everything.
The idea of sitting down with an AI, like Caleb, having a conversation and slowly become more and more amazed how well it performs, is thrilling. This is something else than the Turing test because it should not be known beforehand whether it is a human or not, but that doesn’t make it less interesting. Problems will arise when the AI’s starting to become too humanlike and at the same time have their own agenda. Sooner or later it would become really difficult to be able to control all the possible outcomes when dealing with an AI.
#### Black Mirror (2011)
(image from themoviedb)
Creator: Charlie Brooker
Genres: Sci-fi, Drama, Mystery, Thriller
Black Mirror is a TV show exploring new technology, society and possible scenarios from today to a more distant future. It has been called a modern day The Twilight Zone (1959), also a great show but many episodes can feel rather outdated. Every episode is standalone and explore a new theme. This means that viewers don’t have to see the episodes in a chronological order and can cherry pick the ones that seems most interesting, and even skip the first episode in the first season. This episode is mostly built on shocking the audience and is not representative for the rest of the show. Some favourites:
The Entire History of You (2011)
Synopsis: Some people have an implant recording everything they see and hear.
Think Google Glass but built into the body and it is not hard that this could create all kinds of problems. At the same time a lot of people were really excited about the idea of wearing glasses recording everything and honestly it is really not that far from our current reality.
San Junipero (2016)
Synopsis: Two women meet in a California-esque small town named San Junipero in 1987.
This is not a story about relationship with an AI. This is about human relationships, love, consciousness and how future technology could change our lives for the better.
Hang the DJ (2017)
Synopsis: In a maybe dystopian future two people, Amy and Frank, try out a dating/matching service that always puts an expiration date on all relationships. The reason for this is that each short relationship will give “the system” more data to eventually find the optimal match for each individual using the service.
So what if you could see the expiration date for a relationship, is it something you really want to know. And how would this knowledge change your actions? It is not that far from the classic theme of knowing the day you will die.
You have been working with electrical systems and software in vehicles your entire career. Why?
Actually, I’m not into cars or any other vehicles specifically. What drives me is mainly to solve problems in a challenging technical environment. It might as well have been airplanes or something else entirely. It just so happened that cars and buses had the right combination of technology and challenges to attract me.
How has your career developed at Combine?
The first step was into the telematics area. From there I moved on to Infotainment, and currently I am helping a customer with IT and processes relating to software development and software management.
When Combine sends you to help a customer, what can the customer expect?
I have realized that I have a knack for understanding how the processes, support systems and organization in a company are meant to facilitate technical development. Once I understand this, I bring out my broomstick and begin clean-up operations so that things work the way they are supposed to. Consequently, my CV is full of activities such as ”responsible for project documentation”, ”task force leader”, ”team leader” and similar. In my current assignment I also have the opportunity to develop improvements and IT that raise the quality of the customers processes.
If you could choose a completely different assignment, what would it be?
I believe that we have the technology, or the ability to develop it, needed to help solve some of the big issues facing us globally, issues like our impact on the climate. Creating the right incentives and mechanisms, as well as developing the solutions themselves, would be really stimulating and interesting.
It certainly does. I have been brewing beer for many years and I finally started a microbrewery called Sad Robot Brewing. Being who I am I tried to learn as much as possible about the engineering side of all the steps in brewing processes, such as chemistry and thermodynamics. Just like at an assignment I like things to be clean and controlled, so the only solution was to team up with some friends and do it ourselves. It was just like a second job. I spend less time on brewing nowadays, but one of the things I have done lately is to mentor a thesis project at Combine aimed at controlling and monitoring the brewing process (Editor’s note: you can read about this in the Combine blog here).
I also do some acting in theatre and movies, so not everything is technical. And yes, you can probably figure out what kind of books and movies I like from the name of the brewery.
#### Image processing in Sympathy for Data
This is the second blog post in a series of posts on image processing using Sympathy for Data, an Open-Source tool for graphically programming data-flows. See the previous entry for an example of how you can read the time from an analog clock using only basic image processing building blocks. No programming required.
When it comes to object recognition today most people think about deep learning and throw vast datasets onto deep machine learning algorithms — hoping that something will stick. One thing that all such algorithms have in common is that they all have a large number of parameters, requiring an even larger number of examples to be trained. There are two major costs associated with this approach: firstly the computational cost in training the datasets, usually using a single or a cluster of high-end graphic cards; and secondly the difficulty in acquiring large enough datasets to do the training with. Sure, there exists techniques for artificially extending existing datasets into larger ones in order to help against over fitting, but even these cannot handle the case of datasets with only a hand full of examples. With all the hype of deep learning it is easy to forget that earlier approaches to object recognition, while much more limited in what they could solve, did not suffer from these difficulties and can sometimes still be favourable to be used.
If we look back at when image recognition was first considered as a problem to be solved with computers we see that the problem was at-first greatly underestimated. Back in the summer of 1966 a very optimistic project was started at MIT using only the student summer workers that year and with the aim of solving the computer vision problem. As you can read in the PDF the final goal was, in hindsight, a quite ambitious one indeed:
“The final goal is OBJECT IDENTIFICATION which will actually name objects by matching them with a vocabulary of known objects”.
Needless to say, this task proved more complex that what was first imagined, and have since led the the creation of a whole field of research. It is not until recently, more than 50 years after that summer project that we can say that general purpose object recognition is a more or less solved or solvable problem.
In my previous image processing post we looked at a simple image processing task in reading the time from an analog clock, and showed how this could be solved using the image processing tools available in Sympathy for Data, all without having to write a single line of code. A major factor in this solution was by limiting ourselves only to images acquired in a very specific way. This solution generalizes more to industrial image processing such as eg. reading a pressure valve rather than doing general purpose like reading like a random clock you find on the side of a building.
In this and the upcoming image processing post I will show how we can use the image processing tools and the machine learning tools of Sympathy to similarly solve an object recognition task under well defined circumstances. These circumstances generalizes again more to an industrial setting, such as analysing objects on a conveyor belt, where we can have a clearly defined environment and camera setup.
For this purpose we will have a camera mounted straight above the incoming objects. The objects are photographed against a neutral background (white) clearly distinguishable from the objects themselves (metallic grey). Furthermore we ensure that the lighting is smooth and even over the whole area and that no sharp shadows are cast by the objects themselves or anything else. In the example dataset used here we use pictures of a mix of fasteners, with the target of identifying the screws. Furthermore we ensure that objects are overlapping since it would require more advanced techniques to separate overlapping objects, a problem almost as hard as object recognition itself. If we would like to do this in an industrial setting we could use a mechanical solution to ensure this before the objects enter the belt, eg. using a suitable hopper.
#### Segmenting the image
We will start by solving the problem of segmenting and labelling an input image, with the task of deciding which areas of the image correspond to different objects. The intention here is to pick out individual objects and to classify each found object whether it matches the target object.
Thus our workflow will contain the following steps:
1. Separate the image into pixels that belong to objects or to the background
2. Cleanup this image to remove noise and to completely close all objects
3. Create labels for each pixel
4. Extract a list of binary image masks, one per found label.
A typical step in many image segmentation tasks is to use a simple thresholding algorithm. We can use simple thresholding and the fact that the metallic grey objects all are darker than the background paper in order to create a binary representation of the pixels that belong to objects. We start by attempting to use a simple basic threshold at the value 0.5.
Note that we added a filtering step that inverts the image by scaling it by a factor of -1 and adding an offset 1 to it before we do the thresholding. Thus we can ensure that a completely dark pixel (value 0) becomes 1.0 before thresholding and is classified as a “true” boolean after the thresholding.
We can also note that the result of the basic thresholding is quite poor, We incorrectly classify the bottom half of the image as belonging to an object. If we raise the threshold until no background is classified as an object, then we instead start losing pixels from the objects that are classified as background. You can see this effect in the images below, where we have a higher threshold on the right side than on the left side.
Furthermore, just using a simple scalar value as a hard-coded threshold will not work very well if there is even the slightest change in global illumination from picture to picture.
We can use one of the automatic thresholding algorithms that automatically finds a scalar suitable for thresholding. The simplest automatic thresholding algorithm is the mean or median which sets the threshold such that half the image will be True and half the image False. This is however seldom good, and most definitively not good for our application since we are almost guaranteed that background (which is more than 50% of the image) is classified as part of the objects.
Other alternatives to automatic thresholding include a number of algorithms that consider the overall distribution of pixel values and tries to find a suitable threshold. For example the Otsu algorithm assumes that the pixel values follows a bi-modal distribution and find a global threshold that minimises the variance within each found class.
The results of Otsu is surprisingly good for most images, as you can see in the image above. However we note that this algorithm still misses some parts of the objects (see the upper edge of the circular washers in the image above). Sometimes, it is impossible to get a good enough result by just setting a single global threshold value.
Other alternatives exists that perform an adaptive threshold that considers a window around each pixel and calculates a threshold value for that pixel based on this window. With this technique we for instance can easily compensate for any unevenness in the overall lighting.
One example of this is an adaptive gaussian thresholding method. Here we first perform a low pass filtering with a gaussian kernel of size 21 and sigma 11. We take the lowpass filtered value and apply an offset (-0.01) before testing if it is higher or lower than the pixel that is being thresholded. We picked the value for the kernel size based on the overall size of the objects (the circular ones are approximately 20 pixels wide). The offset compensates for small irregularities in the background itself.
The noise on the background can be removed in a later stage using morphological opening. Before we progress to this however we consider one more approach which is to instead extract all the edges in the image and to perform morphological operations to close the objects based on the edge data. We do this by applying a Canny edge detector to the raw input image (no pre-scaling step needed anymore). As we can see below this method generates no false positives and does capture all sides of the objects.
The interior of the objects can filled in using morphological closing after the Canny edge detector. What this does is to perform to perform a dilation operation followed by an erosion operation where the dilation makes all objects “thicker” by a given radius and the erosion makes them correspondingly “thinner”. Each of these operations are done by checking a neighbourhood around each pixel and taking the MAX or MIN value in the neighbourhood, respectively.
Consider the image on the left side below. In this image if we perform dilation then we get a white pixel in the areas marked red and green and only the area marked in blue would get a black pixel. If we instead perform erosion then we get black pixels in the red and blue areas and only the green area stays white. In the right side of the example below we can see the result of performing the erosion operation followed by a dilation operation. It has first made the white objects significantly thinner, followed by thicker.
For many objects making them thicker followed by thinner would not change the overall shape of the object. However, if two edges both become thick enough to touch each other then there is no black areas in the middle that can make them thinner again. Thus the end-result is that the objects have been closed as can be seen in the images below:
One problem that we can spot with the morphologically closed image is that some objects are now touching each other due to the thickening radius being larger that the distance between the objects which have created small bridges between some of the objects. To compensate for this we can perform a morphological opening that removes the small bridges between the objects. This step also removes all the small dots of false positives given by the thresholding algorithm if that one is used instead of the edge detection.
For the final step before we can start working with the objects it to use labeling to create a unique ID for each object. The labeling algorithm takes a binary image as input and creates an image with integers for each pixel. The integer values of a pixel correspond to a unique value for each object. If there were even a single pixel linking two objects to each other then both objects would be assigned the same integer value. We can visualise the result of this step by clicking on the object, this gives a pseudo-colour for each object based on a default colour map.
Note that since objects that are close to each other have similar ID’s then they are mapped to almost the same color. The ID values assigned differs even when not evident in the image below:
One final node that is useful is to create a list of all the found objects. The node Image to List can be used to convert the labeled image into a list of images. Use the configure menu to select “from labels” to do this conversion.
As we can see in the preview window below we have a list that contains many images. Each entry in the list is an image mask that is true only for one single object (as defined by the unique ID’s given by the labeling operation). We will use these images as the inputs to our classification algorithm to detect the individual objects.
#### Summary
In this post we have looked at the segmentation problem and shown how simple thresholding or edge detection algorithms can be used together with morphological operations and labeling to create a list of objects in an input image. This list of consists of a mask singling out each individual object in the image, one at a time. In part 2 we will continue to perform the classification of each found object.
Märta, why did you choose engineering?
I have always been interested in technology, wanted to know how stuff works. I also liked math and physics and thought it was kind of easy. In gymnasium I first planned to study natural science but ended up choosing more technology-oriented classes since the combination of math and reality was tempting. I think that might also have been a reason to why I focused so much on control theory.
What was the best part of your engineering studies?
Without a doubt my time spent as an exchange student at University of California Santa Cruz!
That sound like a great experience!
Yes, it was fun to take other courses than what was available at Lund University. I also got the opportunity to work in the Autonomous Systems Lab, playing around with robots and drones. This was very valuable since it was like a mix of working, studying and doing research. California is also such a great place so besides studying I spent a lot of time surfing and skateboarding.
Autonomous drones sound like the optimal way to apply math in reality. How did you move on from that?
Well, after California I returned to Sweden in time for my master thesis. Since I had spent quite some time working with autonomous systems and drones I wanted to do my thesis in that area. With that said, I was thrilled when the perfect project was available at SAAB.
It was about controlling a swarm of autonomous flying drones. Having multiple drones in a swarm leads to many interesting problems ranging from internal distance estimation between the drones to the high-level behavior of the swarm.
So now you work as an engineer, is it all you thought it would be?
Well, I never really had any clear picture of exactly what it means to “be an engineer”. It wasn’t until the final years at the university I started to get a better picture of what it means. But yes, I work with applied mathematics every day so in that sense it is what I envisioned.
How does a typical day at work look like?
I work in an agile environment, kind of like scrum-ish… The day starts with a daily scrum meeting where we go through what we work on and potential issues. After the meeting it’s time to start work on my current tasks. Right now, my main focus is on PLC programming, coding new features and testing them out at the machine or in a virtual environment. Some time is also spent on developing the virtual test rigg, bug fixes etc. My days are very flexible, and I control a lot of the time myself and that suits me perfect.
Why did you choose Combine?
I started my career at a larger consulting company. I liked the role of a consultant, but I felt that I wanted to work for a company more focused on the technologies I’m interested in. I had also heard good things about Combine from friends.
Do you also want to work with applied mathematics and control systems development as a consultant at Combine? See if we have any available positions, or just give us a call and see if we have something coming up soon.
For todays post we have a guest author, Lia Silva, that works in Data Science and have her own blog https://statsletters.com/ on mathematics, statistics and other fun stuff. Without any further presentation, see what Lia have to write about studying our relationship with consumerism through graph theory and statistics:
One important difference since a decade ago is what can be measured about how a user consumes a product. Lately, it seems like every little thing that we do can be used to build a projection of ourselves from our habits. And in the end, that projection can be used to poke the reptilian parts of your reward circuitry so they release the right cocktail of hormones. A cocktail that makes you choose bright red over dull gray, reach for your wallet or click “I accept the terms and conditions”.
Through the years, recipes for such cocktails have been perfected by different disciplines. As an educated consumer, actively tasting those recipes in modern products can be as interesting as wine tasting, minus the inebriation. This is the first of a series of posts intended to help you be more aware of your own reward circuitry by using interpretations that different algorithms build from observing your measurable actions.
Another intention with this series of blog posts is to show that the methods are not necessarily:
• Absolute
• Inherently objective
• Infallible
And definitely NOT suitable to use blindly e.g. “press the Analytics Button and have the neural network tell me everything”*. If anyone promises that without disclosing any assumptions, make sure to ask LOTS of questions.
The “Serpent people” series will present some textbook representations suitable for modeling this problem, aspects that are better reflected on each one of them, and trying out different open-source libraries on different artificially generated models of “people according to what we know about them”.
Some of that material is already in a very fluid shape in this notebook if you can’t wait to play by yourself :). The representation in there is simply what I considered to be natural for the problem itself. I plan on elaborating that representation with classics such as Frequent Itemset Mining and Associative Classification. For those, you can start by checking out Chapter 10 of “Data Mining” by Mehmed Kantardzic.
And that’s the teaser for what will come. For now, I will leave you with this David Bowie Song. Granted, it’s “Cat People” instead of “Serpent People”, but pretty cool still.
## The basic idea
The basic idea we have for reading the time (or any other analog device!) is to first capture an image of the device using a web camera and use the image processing tools in Sympathy for Data to read the hour and minute hands from the clock. Our goal is to do this with only the open-source image processing tools that are included in Sympathy for Data, and no custom code.
Our aim is not to create a solution that works for any image of any clock, but rather to show how we can reliably read the values of the given clock given a very specific camera angle and lighting situation. To do this for any clock and lighting setup requires more advanced techniques, often involving machine learning, and which harder to guarantee that it works in all of the target situations.
For the purpose of this blog post, we will not be using machine learning or any other advanced algorithms but rather rely on basic (traditional) image processing techniques. This problem resembles very much that of designing image processing algorithms that are used every day in industrial production — where you have a highly controlled environment and want simple and fast algorithms. By controlling the environment we can make this otherwise complex task very simple and easy to design an algorithm for. When doing image processing for industrial purposes it is commonly found that you can simplify the problem and increase robustness by heavily controlling the environment and the situation in which your images are acquired.
For the impatient, you can download the dataset with images that we prepare in the first part as well as the finished Sympathy for Data workflow.
## Acquiring the data
Step one is to acquire some images of the clock and to try to analyse them in Sympathy. If you don’t want to repeat these steps yourself you can simply download the full dataset from here.
We start with a naive approach and just place the web camera in front of the clock and record images once per minute over a full day. Some example of these images are below:
What we can see in the images above is two main problems: the lighting varies widely depending on the time of day, and the lighting in the image itself varies widely due to specular reflections and makes part of the hour-hand invisible second image above. Whatever algorithm we come up with will have a hard time to deal with an invisible hour-hand.
We can also note that with a stronger and directional light we would have shadows cast by the hour and minute arms of the clock is projected at different part of the face of the clock depending on the incoming light direction. If we where to directly try to analyse these images we would need to compensate for the shifting position of the shadows, and we could not use any simple thresholding steps since the overall light changes widely. Any algorithms we create for these kind of images will inherently be more complex and possibly more prone to failures when the light conditions change. We would need to test if over a wide range of conditions (day, night, sunset, rainy weather, sunny weather, summer, winter) to be sure that it works correctly under all conditions.
An attempted first fix for the specular reflections was to remove the cover-glass of the clock, the idea being that the face paint of the clock wouldn’t be reflective enough to give these issues. This however wasn’t enough to solve the problem with disappearing arms for all lighting conditions, and are we to apply this idea to an industrial setting it may oftentimes not be possible to make such a modification. A better solution is therefore to remove all direct light in favour of a setup with only diffuse lights.
To solve both these problems we choose to make a controlled environment with only diffuse light and where we know that the only thing that changes are the positions of the clock’s arms. We force a constant light level by enclosing the clock and camera in a box with a lightsource. This way we can also eliminate the the shadows of the clocks arms by placing the lightsource from the same direction as the camera.
## Building a camera friendly light source
In order to eliminate the shadows from the clock arms and to provide a even and diffuse lighting conditions we place a ring of white LED’s around the camera. Usually this is done using professional solutions such a light ring for photography, but we’ll manage with a simple 3d-printed design and some hobby electronics. You can find the downloads for these over a Thingiverse.
The design for this light is simple ring where we can add the lights plus a diffuser on top of it to avoid any sharp reflections.
After printing the parts above we place a number of white LED’s in the small holes in the middle part above. By twisting the pins together with each other on the underside (take care with the orientation of anode and cathode!) we can easily keep them in place while at the same time connecting it all up. See if you can spot the mistake I did below in the wiring. Fortunately it was salvageable.
Next step is to place the diffuser over the LED’s and to attach it onto your web camera. Power it with approximately 3V per LED used. Point it straight at the clock and put an enclosure over it. We can used a simple carton box as a simple enclosure that removes all external light.
Congratulations, you can now get images of the clock with perfect lighting conditions regardless of sunlight, people walking bye, or any other factors that would complicate the readings.
## Pre-processing the images
When doing image processing it is common to operate on grayscale images unless the colour information is an important part of the recognition task. For this purpose we first run a pre-processing step on the whole dataset where we convert the images to greyscale and downsample since we don’t need the full resolution of the camera for the rest of the calculations. This can easily be done using Sympathy for Data.
First create a new flow and point a Datasources directory to a copy of the dataset (we will overwrite the files in place). Add a lambda node by right clicking anywhere in the flow and select it. Connect a map node to apply the lambda for every datasource found.
Before you can run it, add the nodes below into the lambda node to do the actual image conversion. You need to select greyscale in the configuration menu of the colour space conversion node, and rescale X/Y by 0.5 in the transform image node. Note that the “save image” node here overwrites the images in place, so try not to run the node until everything is ok. Another option would be to compute a new filename to be used instead using eg. a datasource to tables node and a calculator node.
In the dataset that you can download we have already done these conversions (downscaling to 800×500 pixels) to save on bandwidth.
## Analysing the images
Our goal in this section is to create a Sympathy workflow that allows us to take any image of the clock and convert it into an hour and minute representation of the time.
### Creating a template
For the first step we want to extract only the arms of the clock that should be analysed. For this purpose we will use a practical trick to easily detect only the moving parts of the image. We do this by first calculating a template image that show how the images would look if all the moving parts were removed. This trick only works when the camera is fixed and there are no major changes in overall lighting.
Start by calculating the median (or in our case max since we know the arm’s are black) of a few different images from the dataset. This will give an image where the arms of the clock is removed.
Select max as operator in the Overlay Images node below. This works since we know that the arms are darker than the background, and whichever pixel has a light colour in any of the images will have a light colour in the final image.
The results look surprisingly good given that we only used a few images (where the arms where all in different positions):
We save this template in a separate file so that we don’t have to redo the calculation for each image that should be processed.
### Extracting the minute and hour arms
Continue by creating a new workflow and load one of the images to be analysed. We start by making a subtraction of the image to be analysed from the template image.
This will give an image where only the arms are visible, you will need to select “subtract” as the operation for the “overlay images” node.
The next step will be to perform a threshold to pick out only the arms of the clock as a binary image. To do so we use a Threshold node and set it to basic threshold. We can figure out a good threshold level by looking at the histogram above of the image after subtraction. We see that the maximum value of the image is 0.55 and that something significant seem to happen around the 0.4 mark (note that the graph is logarithmic!). We set the threshold to 0.35 and get the results shown above.
Since there can be some small smudges and missed spots on the binary image we apply morphological closing on the image using a structuring element of size 20 which should be more than enough to compensate for any missed pixels caused by noise, scratches on the object/lens or the otherwise black areas of the image.
Finally, we can note that we only actually need to see the tips of the hour and minute hand in order to read the time. If we sample to check for the the minute arm at every point in a circle with a radius closer to the edge of the picture we can know which pixels belong to the minute arm. Similarly, if we sample every point in in a smaller circle we get one or two sets of pixels corresponding to the hour arm when it is below the minute arm or both the hour arm and minute arm when they are not overlapping.
We can do this sampling by first creating two new templates that we use to select a subset of the pixels. This is done by drawing a white ring on an otherwise empty image for each of the two selections. We can do this using the Draw on image node and a Manually create table node that gives the XY coordinates (416, 256) and radius 200 for a circle with colour 1.0 and 170 for a circle with colour 0.0 — corresponding to the ring selecting the minute arm below:
After multiplying these two templates with the thresholded image (using again the overlay images node) we get two new images with blobs corresponding to the tips of the hour and minute hand:
All that is left now is to extract the coordinates of these blobs and to apply some math to convert them into hours and minutes.
### Computing the minute
We can compute the position of the minute hand by using a Image statistics node with the algorithm “blob, DoG” with a threshold of 0.1. This algorithm finds “blobs”, or light area on a dark background, in an image by subtracting two low-pass (gaussian) filtered version of the image filtered at different scales.
All other parameters can be default, but the default value for threshold of the difference-of-gaussian algorithm is too high for our inputs.
Now all we have to do is to convert the XY values 443, 426 of the tip of the minute hand into an actual value in the range 0 – 60. We can do this by calculating the vector from the center of the clock determined from the raw image as (416, 256) to the point of the detected blob. This gives us a vector (187, 10). By taking the arctan of this vector in a calculator node we can get the angle to this point and convert it into minutes. Note that we invert the y-component of this vector to compensate for the difference in coordinate systems (y-axis in images point down):
### Computing the hours
In order to compute the hour we need to eliminate one of two possible candidates for the hour. Consider the blobs shown below, from just this data it is hard to know which hour it is:
However, what we can do is to take the position of the minutes that we calculate above and clear out one of the two blobs above. Since we know the radius that we used for the circle multiplied with the data, we can easily draw a black area on top of the location where the minute arm is located and at the given distance from the centre:
For this purpose we use another calculator node to compute the X/Y coordinate above and to draw a black circle onto the image at that location. For clarity it has been drawn as a brown circle in the example above to see what area of the image is deleted.
We extract the hour value from the remaining blob, if there is one, similarly as to how the minute value was calculated. Note that if there is no blob in the image containing the tip of the hour arm then the expression belows gives a NaN value. This happens when it is under the minute arm.
Finally, we can finish the flow by adding in a special case calculation that checks if the ‘hour’ column has the NaN value and if so instead derive the hour position from the minute position. In this step we also round the minutes to even number and round the hours down to nearest smaller integer. Note that we subtract a fraction of the minutes from the hours before rounding due to how the hour arm moves closer and closer to the next hour as the minutes raise.
We also subtract 1 (modulo 60) from the minute position since the captured images where all slightly rotated clockwise. We could have compensated this in the original pre-processing if we had noticed it earlier.
## Time to check the results
Before we are happy with the flow, let’s check how well it performs versus the ground truth. Since the timestamp was saved when each image was captured we can easily compare these values with the results of the flow. Due to the sampling process and since the seconds of this clock wasn’t synchronized with the seconds of the computer sampling them — we should expect to be off by one minute in some of the readings.
As we can see in the table below we successfully read the time, with a difference of at most one minute, for first 100 images.
By only allowing ourselves to use the first 50 images when we developed the flow, and then validating it by running on a larger dataset we gain confidence that the algorithm works for all the situations it will encounter. We have run it the full dataset of 700+ hours without any other errors.
Sympathy for Data is a visual software tool that helps to link scripts and data between different systems and enables analysis.
The tool targets all domains where data analysis is carried out off line and is challenging and repetitive. Our focus is on automating the tasks of importing, preparing, analyzing and reporting data. In some respects, it is also a visual programming environment.
Sympathy for Data is used by a number of leading companies in different industries, such as the automotive, process and automation industries. At Combine we believe that automation is the key to long-term success, yet a large proportion of current analysis is done manually. This is no longer sustainable as data continues to grow in every respect. Sympathy for Data supports subscription services to databases, network disks, etc. It can find, filter, sort and analyze data as well as create automatically formatted reports.
Applications
Among other things the tool allows you to:
– Transfer data from and to all sources (DB, web & remote, local).
– Move your favourite scripts to nodes and turn them into an analytical flow.
– Easily share data processes across an entire organization.
– Run data processes in batch mode to enable scheduled reporting.
– Easily share data from a data process with web reports or other chosen formats.
One of the aims of Sympathy for Data is to allow users to manage data from many different sources and in different formats. You can connect the tool to databases, select files from different network devices, or analyze thousands of Excel files automatically. We have developed the tool to manage large amounts of data along with other forms of useful data, such as metadata (units, dates, etc.), results and other custom data needed for analysis.
A tool for the entire organization
Sympathy for Data can easily be used throughout an organization, thanks to the flexible configuration options. The tool can be deployed as a standalone application, integrated in a server environment or combined with other enterprise solutions such as SQL (Server Integration Services), SharePoint, etc. The purpose is not to lock users into the platform, but instead to serve as an interface between different systems, data formats, etc.
### 1. Grabbing that cup of coffee
In the figure below, a two link robot arm is depicted. We say this depicts your arm. The upper
arm has the length r0 and the forearm the length r1. The position (x; y) of your wrist s described
by the two angles q0 and q1. Your wrist is currently located at the red dot. Say we want to the
grab a cup of coffee that is located at the blue dot. Pretend that you are a robot being controlled
by a computer giving commands to your upper- and forearm. What should the angles q1 and q2
be to achieve this? This type of problem is usually called the inverse kinematics problem. First,
we note that there should be two valid solutions, one with the “elbow” up and one with it down,
right? However, assuming this is us grabbing a cup of coffee, our arm is imposed to kinematical
constraints meaning that the elbow can only be down. I skip the equations for this, but they
can be found in for instance Robotics by Bruno Siciliano et.al.. The solution for this problem is
analytical, but however not completely trivial I would say.
Let’s just say we found the angles q0 and q1 and full of confidence we are to grab our cup of coffee.
Preferably we should move along a straight line, because we have a computer screen in front of
us. However, these angles say nothing about how we reach our cup of coffee. If we just first set q1
to our calculated target value, and then q0 the robot arm end point might move along the dashed
line or something similar. We would knock down our screen and maybe even that cup of coffee in
the process, which is all but a good start of the day.
So, we don’t just want to move to a certain point of interest, we usually want to reach that point
while being constrained to a specific trajectory. A common way in the digital world to do this
is to solve the inverse kinematics in very small steps, where all small steps adds up to a straight
line. This implies that the angular velocities q_1 and q_2 should be coordinated so we get a smooth
movement. A challenge here is that the dynamics of your arm is highly non-linear. Even if your
arm were actuated by ideal electrical motors, the equations of motion could well fill up a whole
page. In this case your arm are actuated by various muscles, which further adds complexity.
Then, we probably want the Cartesian velocity (x_ ; y_)T along the straight line to fulfill some
criterion. Probably we want to move a bit faster at first and slower when we reach the cup so we
can slow down in time and not spill coffee all over the place. This implies that we should want to
employ some trajectory planning. In the end we have a trajectory involving joint space variables
q1; q2; q_1; q_2 as well as Cartesian space variables x; y; x_ ; y_. So this starts to sound at least like a bit
of a hairy problem, eh? Then, add other challenges as that maybe we cannot even reach the blue
dot given that our arm is too short? Or that there is a bowl of cereals, a computer screen or some
other constraint in the way that we need to circumvent.
Then let’s say we add another link to your arm. Now we have three joint variables q1; q2; q3. Having
three input variables and two output variables (x; y) we can suddenly have infinite solutions to the
inverse kinematics problem. This is called a kinematic redundant manipulator. Off course, your
arm probably doesn’t have three links (would be quiet cool though?). But, your arm has far more
degrees of freedom than depicted here in 2D. You can tilt your forearm, upper arm and what not.
Each and every of your fingers are even more complex than the planar manipulator depicted. For
this 2D problem, your arm is kinematically redundant and we couldn’t find a closed form solution
for the inverse kinematics problem. We have to select some of the infinite solutions according to
criterion. A way to do this is to formulate some form of optimization problem. Machine learning
is also being employed sometimes.
Figure 1: A figure used to derive some basic ideas related to inverse kinematics
### 2 Kinematics and the evolution
Are you convinced now that grabbing a cup of coffee is maybe not as trivial as you might have
thought? Back to the philosophical question, what makes us human? For one, we are an animal
with relatively high intelligence compared to others creatures we know in this world. Without
dwelling into the subjective definition of what intelligence is, I think we can all agree on this. This
is what enabled us to develop tools helping us to gather, store and process foods among other
things. It encompasses all from weapons such as spears and archery, agriculture, taming fire and
so on. But what would these ideas concepts be without our ability to physically manipulate the
world? Agriculture and fire are merely theoretical pipe dreams when lacking some sort of ability to
achieve this things in practice. Surely, thinking might be existing. But you couldn’t think without
eating. Not in the way humankind exists today anyway.
This is where our advanced kinematics come into play. Advanced kinematics control is as crucial
to define humanity as we know it, as our intelligence I would say. You as a reader, might
not be dependent on your ability to make fire in order to survive in this world. Perhaps clacking
the keyboard of a computer is quiet enough. Which is applying kinematics control. But Stephen
Hawking could do without it you say then. Sure, he personally, yes. But those who built the first
computers? Those who scrabbled down the theorems necessary to build the first computer? Nope.
Then, say we all would communicate and operate like Hawking’s did. You would still at least
need to eat, right? Even if you could communicate with merely your mind, you would still need
to somehow run agriculture. Which again, means physically manipulating the world. It seems
inevitable that our existence is depending completely on our ability to physically manipulate the
world.
Let’s assume that the evolution made us interested in things that were beneficial for our survival
and reproduction in various ways. This is probably where sports come in. Apart form competition,
it is a way of refining our kinematic abilities. Spear-throwing, wrestling and boxing can probably
be directly related to the need of defense and hunting. A tribe being good at handball or a precursor
to it, would probably also be better at throwing stones at both prey and attacking enemies.
Those playing football would probably catch a hare better. Most culture employ dancing. While
dance can fulfill a variety of purposes, one is certainly displaying reproductional benefits. It could
be a way showing off genes being able to handle kinematics well. Do we have complex robots?
Yes. Does anybody of them dance well yet? Well, no. A robot that could dance well would thus
arguably be more complex and all robots so far known and would be able to perform many other
complex tasks then dancing. A person showing off some smooth moves at the dance floor basically
communicates “hey, my gene-pool is probably super-good for a variety of tasks in this world that
is beneficial for our existence. Good from all to hunting rabbits to climb trees”.
2
So, smooth and advanced kinematics is not a of interest for engineers and such only. The interest
for it is probably even coded into the DNA of each and every human being. It is fascinating
how we as a humanity, consciously or not, many times happen to mimic features and stages of the
evolution.
Controlling the temperature during beer brewing is essential for the quality of the end product, as well as ensuring an efficient production. By eliminating manual control in the production of beer a more consistent product can be produced. During this thesis project the temperature control strategies for a heat exchanger and the fermentation process has been modelled and developed.
During fermentation of beer, a brewmaster wants to control the temperature at which the fermentation occurs with great precision. Not only does the temperature need to be steady, different temperatures are needed during different stages of the fermentation process. With a precise controller capable of ensuring a unique temperature profile the brewmaster can repeatably create the best tasting beer possible. This is due to the temperature during fermentation affects the flavour profile of the finished product. With the additional benefits of being able to monitor the progress of the fermentation through the online monitoring system, being present at the brewery is no longer needed for routine checks of the beer.
By modelling the biological behaviour when yeast fermenting a variety of different sugars into ethanol and carbon dioxide, precise control of the temperature is achieved. Implementation of a mathematical model based on a modified version of the equations derived by Engrasser was used in Simulink.
During the beer brewing process, the last step before the fermentation starts is the cooling of the wort. This is done by pumping the boiling hot wort through a heat exchanger. By modelling the thermodynamical system a controller could be developed. It is of importance to ensure that the wort is at a precise temperature when the fermentation starts, in order to give the yeast the best possible environment. By implementing our controller the flow used in production can be increased as much as 65%, with the added benefit of a consistent and predictable temperature in the fermentation vessel, eliminating the guesswork from manual control. | 11,835 | 58,043 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 2, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.125 | 4 | CC-MAIN-2019-04 | longest | en | 0.954789 |
https://mathematica.stackexchange.com/questions/128475/how-can-i-find-the-roots-of-the-maximizer-of-a-function | 1,579,738,288,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579250607596.34/warc/CC-MAIN-20200122221541-20200123010541-00008.warc.gz | 550,198,037 | 27,676 | # How can I find the roots of the maximizer of a function? [duplicate]
Suppose I define $x(b) = \arg \max (-x^2/2+bx)$. I'd like to find the values of $b$ such that $x(b) = 0$. I tried running the code
x[b_] := x /. FindMaximum[-x^2/2 + b x, x][[2]]
FindRoot[x[b] == 0, {b, 0}]
but I get an error, even though the function x seems to work. I know this is a somewhat trivial example (since $x(b) = b$), but in the application I have in mind the maximizer may not have a closed form expression.
Thanks!
• Use _?NumericQ: x[b_?NumericQ] := x /. FindMaximum[-x^2/2 + b x, x][[2]]; – Michael E2 Oct 11 '16 at 19:41
• Specifically, see this answer – Michael E2 Oct 11 '16 at 19:42
Use NMaximize instead of FindMaximum, with _?NumericQ as pointed out by Michael E2 in the comments (I changed the name of the function from x[b] to max[b] because I simply dislike naming different things with the same symbol):
max[b_?NumericQ] := x /. NMaximize[-x^2/2 + b x, x][[2]]
max[1]
1
Plot[max[b], {b, -1, 1}]
FindRoot[max[b] == 0, {b, 0}]
{b -> 0.}
With a different example:
max[b_?NumericQ] := x /. NMaximize[Sin[b x] + Cos[x] - x^2 + (b - 1)/b x, x][[2]];
max[0.5]
-0.167512
Plot[max[b], {b, 0.1, 1}]
FindRoot[max[b] == 0, {b, 0.5}]
{b -> 0.618034}
• What determines whether Maximize, NMaximize, or FindMaximum work in this context? – David Oct 11 '16 at 19:40 | 490 | 1,373 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.703125 | 4 | CC-MAIN-2020-05 | latest | en | 0.776425 |
https://www.coursehero.com/file/6770098/Hw4Fl05Alg/ | 1,519,143,450,000,000,000 | text/html | crawl-data/CC-MAIN-2018-09/segments/1518891812978.31/warc/CC-MAIN-20180220145713-20180220165713-00086.warc.gz | 870,392,787 | 57,753 | {[ promptMessage ]}
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Hw4Fl05Alg
# Hw4Fl05Alg - Input(w p =(3 2(8 12(5 7 the knapsack-limit by...
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CSE 5211/4081 Algorithm Analysis HomeWork 4 Due: 11/3/05 Points: UG: 20/ Grad:20 .1. Write a recursive algorithm for the following formula. Input: a matrix of integers pij, 1<=i<=n, 1<=j<=n, for problem size n. C(i, j) = 0, for all 1<= i>j <=n. C(i, j) = min{ C(i+k1, j) + pij, C(i, j-k2) – pij | for all k1, k2 with 1<=k1<=n-i, 1<=k2<=j}, for all 1<= i<=j <=n .2. Write a dynamic programming algorithm for the above formula. Analyze the complexity for this algorithm. .3. Apply the dynamic programming algorithm for solving the following 0-1 Knapsack problem. Show the full matrix.
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Unformatted text preview: Input: (w, p) = ((3, 2), (8, 12), (5, 7)), the knapsack-limit by weight is M=7. GRAD 3b. Show how you would find the knapsack content from the matrix. [2] .4. Apply the dynamic programming algorithm for solving the following matrix chain-product problem. Show the full matrix and details of a sample calculation for each of the subproblem sizes >2. Input sequence of matrix-dimensions: (4x1)(1x5)(5x2)(2x3)(3x1) GRAD 4b. Keep track of the breakpoint data and recover the actual optimal order of matrix-chain multiplication in the above problem. [2]...
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Ask a homework question - tutors are online | 469 | 1,519 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5 | 4 | CC-MAIN-2018-09 | latest | en | 0.746699 |
https://thealmarty.com/category/lambda-calculus/ | 1,725,808,142,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651013.49/warc/CC-MAIN-20240908150334-20240908180334-00051.warc.gz | 564,033,490 | 12,359 | # Category: Lambda Calculus
• ## Lambda Calculus in OCaml: “fun” and “function”
Lambda is fun! Lambda is certainly fun, but what I mean here is that the λ in lambda calculus is similar to the expression fun in OCaml. Recall that in lambda calculus, we have function expressions and function applications: λx.λy.x+y (*A function expression*) λx.λy.x+y 3 4 (*A function application*) In OCaml, you can express the same…
• ## Lazy or Eager? Order of Evaluation in Lambda Calculus and OCaml
Recall in lambda calculus, two items side by side is an application. One applies the left item (the function) to the right item (the input). E.g.: f x is read as “apply f to x”, in which f and x can be any lambda expressions. Therefore, f and/or x may be expressions that can be evaluated…
• ## Currying in Lambda Calculus and OCaml
Currying Recall that in lambda calculus, a function can have more than one input, each preceded by a λ symbol. Another way of thinking about more than one input is currying. Currying a function of two inputs turns that function into a function with one input by passing one of the inputs into it. In other…
• ## Encoding Recursion with the Y Combinator
In this post I’ll go through some exercises and encode some recursive functions with the Y combinator. Encoding with rec Continuing on from my last post, Professor Hutton gave us two exercises in the Y combinator video: Encode loop (the function that just calls itself) with rec. I.e., loop = rec (?) Encode the factorial function…
• ## Recursion in Lambda Calculus: The Y Combinator
In the last post I talked about how powerful lambda calculus is. In this post I further proves the point by encoding recursion in it. This enables you to do recursion in any languages! If you haven’t read my last post already, please do so! It’d be easier for you to follow this post, especially…
• ## Simple Yet Powerful: Lambda Calculus
I’ve long since heard of “Lambda Calculus” but I didn’t really know what it is about until I saw this video. It got me super excited! What I love about it is that it’s built on almost nothing! Only the concept of functions. It’s so simple and elegant! Professor Graham Hutton also listed good reasons to… | 532 | 2,227 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.796875 | 4 | CC-MAIN-2024-38 | latest | en | 0.818949 |
null | null | null | null | null | null | Odds and ends from Wednesday:
• Columbia Pictures has pre-emptively snatched up the spec script Jack and Jill as a possible starring vehicle for Adam Sandler. As of right now, the actor is only producing the pic through his Sony-based Happy Madison banner (appropriately named after the two films that made him). However, since he is involved in some capacity, it's only right to hype his possible involvement in front of the camera. The story is being under double secret probation right now, but is said to have something to do with fraternal twins.
• Call me crazy but I've never been a big Todd Phillips fan. While Road Trip and Old School definitely had their high points, I've heard too many bad stories about the guy, so much so that they have tainted my view of his films. Nevertheless, the guy does know how to direct a successful commercial comedy and, according to Variety, Warner Brothers has tapped him to helm Men. Pic, which will be an Americanized version of the German 80's comedy, will tell of a man who becomes roommates with the guy his wife is having an affair with. Seeing as Phillips already has two films in pre-production (one of them being Old School 2), I'm curious to watch him pull it all off.
• After shoving out a direct-to-dvd sequel, the folks behind Starship Troopers are now looking to produce a third installment of the cult hit. However, this time they want to go back to where it all began and, not only create a film much like the first, but also use some of the original talent. Yes folks, Casper Van Dien said he's involved and will be reprising the role of Johnny Rico. Killer awesome! Not for nothing Casper, but upon your return to the alien planet, perhaps you should try and find the acting career you accidentally left there the last time. | null | null | null | null | null | null | null | null | null |
http://schools-wikipedia.org/wp/p/Pascal%2527s_triangle.htm | 1,500,893,447,000,000,000 | text/html | crawl-data/CC-MAIN-2017-30/segments/1500549424846.81/warc/CC-MAIN-20170724102308-20170724122308-00068.warc.gz | 292,746,704 | 12,308 | # Pascal's triangle
#### Background to the schools Wikipedia
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$\begin{matrix} &&&&&1\\ &&&&1&&1\\ &&&1&&2&&1\\ &&1&&3&&3&&1\\ &1&&4&&6&&4&&1 \end{matrix}$
The first five rows of Pascal's triangle
In mathematics, Pascal's triangle is a geometric arrangement of the binomial coefficients in a triangle. It is named after Blaise Pascal in much of the western world, although other mathematicians studied it centuries before him in India, Persia, China, and Italy. The rows of Pascal's triangle are conventionally enumerated starting with row zero, and the numbers in odd rows are usually staggered relative to the numbers in even rows. A simple construction of the triangle proceeds in the following manner. On the zeroth row, write only the number 1. Then, to construct the elements of following rows, add the number directly above and to the left with the number directly above and to the right to find the new value. If either the number to the right or left is not present, substitute a zero in its place. For example, the first number in the first row is 0 + 1 = 1, whereas the numbers 1 and 3 in the third row are added to produce the number 4 in the fourth row.
Each number in the triangle is the sum of the two directly above it.
This construction is related to the binomial coefficients by Pascal's rule, which states that if
${n \choose k} = \frac{n!}{k! (n-k)!}$
is the kth binomial coefficient in the binomial expansion of (x + y)n, where n! is the factorial of n, then
${n \choose k} = {n-1 \choose k-1} + {n-1 \choose k}$
for any nonnegative integer n and any integer k between 0 and n.
Pascal's triangle has higher dimensional generalizations. The three-dimensional version is called Pascal's pyramid or Pascal's tetrahedron, while the general versions are called Pascal's simplices — see also pyramid, tetrahedron, and simplex.
## The triangle
Below are rows zero to sixteen of Pascal's triangle:
1
1 1
1 2 1
1 3 3 1
1 4 6 4 1
1 5 10 10 5 1
1 6 15 20 15 6 1
1 7 21 35 35 21 7 1
1 8 28 56 70 56 28 8 1
1 9 36 84 126 126 84 36 9 1
1 10 45 120 210 252 210 120 45 10 1
1 11 55 165 330 462 462 330 165 55 11 1
1 12 66 220 495 792 924 792 495 220 66 12 1
1 13 78 286 715 1287 1716 1716 1287 715 286 78 13 1
1 14 91 364 1001 2002 3003 3432 3003 2002 1001 364 91 14 1
1 15 105 455 1365 3003 5005 6435 6435 5005 3003 1365 455 105 15 1
1 16 120 560 1820 4368 8008 11440 12870 11440 8008 4368 1820 560 120 16 1
## Pascal's triangle and binomial expansions
Pascal's triangle determines the coefficients which arise in binomial expansions. For an example, consider the expansion
(x + y)2 = x2 + 2xy + y2 = 1x2y0 + 2x1y1 + 1x0y2.
Notice the coefficients are the numbers in row two of Pascal's triangle: 1, 2, 1. In general, when a binomial like x + y is raised to a positive integer power we have:
(x + y)n = a0xn + a1xn−1y + a2xn−2y2 + … + an−1xyn−1 + anyn,
where the coefficients ai in this expansion are precisely the numbers on row n of Pascal's triangle. In other words,
$a_i = {n \choose i}.$
This is the binomial theorem.
Notice that entire right diagonal of Pascal's triangle corresponds to the coefficient of yn in these binomial expansions, while the next diagonal corresponds to the coefficient of xyn-1 and so on.
To see how the binomial theorem relates to the simple construction of Pascal's triangle, consider the problem of calculating the coefficients of the expansion of (x + 1)n+1 in terms of the corresponding coefficients of (x + 1)n (setting y = 1 for simplicity). Suppose then that
$(x+1)^n=\sum_{i=0}^n a_i x^i.$
Now
$(x+1)^{n+1} = (x+1)(x+1)^n = x(x+1)^n + (x+1)^n = \sum_{i=0}^n a_i x^{i+1} + \sum_{i=0}^n a_i x^i.$
The two summations can be reorganized as follows:
\begin{align} & \sum_{i=0}^{n } a_{i } x^{i+1} + \sum_{i=0}^n a_i x^i \\ & {} = \sum_{i=1}^{n+1} a_{i-1} x^{i } + \sum_{i=0}^n a_i x^i \\ & {} = \sum_{i=1}^{n } a_{i-1} x^{i } + \sum_{i=1}^n a_i x^i + a_0x^0 + a_{n}x^{n+1} \\ & {} = \sum_{i=1}^{n } (a_{i-1} + a_i)x^{i } + a_0x^0 + a_{n}x^{n+1} \\ & {} = \sum_{i=1}^{n } (a_{i-1} + a_i)x^{i } + x^0 + x^{n+1} \end{align}
(because of how raising a polynomial to a power works, a0 = an = 1).
We now have an expression for the polynomial (x + 1)n+1 in terms of the coefficients of (x + 1)n (these are the ais), which is what we need if we want to express a line in terms of the line above it. Recall that all the terms in a diagonal going from the upper-left to the lower-right correspond to the same power of x, and that the a-terms are the coefficients of the polynomial (x + 1)n, and we are determining the coefficients of (x + 1)n+1. Now, for any given i not 0 or n + 1, the coefficient of the xi term in the polynomial (x + 1)n+1 is equal to ai (the figure above and to the left of the figure to be determined, since it is on the same diagonal) + ai−1 (the figure to the immediate right of the first figure). This is indeed the simple rule for constructing Pascal's triangle row-by-row.
It is not difficult to turn this argument into a proof (by mathematical induction) of the binomial theorem.
An interesting consequence of the binomial theorem is obtained by setting both variables x and y equal to one. In this case, we know that $(1+1)^n = 2^n$, and so
${n \choose 0} + {n \choose 1} + \cdots +{n \choose n-1} + {n \choose n} = 2^n.$
In other words, the sum of the entries in the nth row of Pascal's triangle is the nth power of 2.
## Patterns and properties
Pascal's triangle has many properties and contains many patterns of numbers.
### The diagonals
Some simple patterns are immediately apparent in the diagonals of Pascal's triangle:
• The diagonals going along the left and right edges contain only 1's.
• The diagonals next to the edge diagonals contain the natural numbers in order.
• Moving inwards, the next pair of diagonals contain the triangular numbers in order.
• The next pair of diagonals contain the tetrahedral numbers in order, and the next pair give pentatope numbers. In general, each next pair of diagonals contains the next higher dimensional "d- simplex" numbers, which can be defined as
$\textrm{tri}_1(n) = n \quad\mbox{and}\quad \textrm{tri}_{d}(n) = \sum_{i=1}^n \mathrm{tri}_{d-1}(i).$
An alternative formula is as follows:
$\textrm{tri}_d(n)=\begin{cases} 1 & \mbox{if } d=0 \\ n & \mbox{if } d=1 \\ \displaystyle \frac{1}{d!}\prod_{k=0}^{d-1} (n+k) & \mbox{if } d\ge 2.\end{cases}$
The geometric meaning of a function trid is: trid(1) = 1 for all d. Construct a d- dimensional triangle (a 3-dimensional triangle is a tetrahedron) by placing additional dots below an initial dot, corresponding to trid(1) = 1. Place these dots in a manner analogous to the placement of numbers in Pascal's triangle. To find trid(x), have a total of x dots composing the target shape. trid(x) then equals the total number of dots in the shape. A 1-dimensional triangle is simply a line, and therefore tri1(x) = x, which is the sequence of natural numbers. The number of dots in each layer corresponds to trid − 1(x).
Sierpinski triangle
### Other patterns and properties
• The pattern obtained by coloring only the odd numbers in Pascal's triangle closely resembles the fractal called Sierpinski triangle, and this resemblance becomes more and more accurate as more rows are considered; in the limit, as the number of rows approaches infinity, the resulting pattern is the Sierpinski triangle. More generally, numbers could be colored differently according to whether or not they are multiples of 3, 4, etc.; this results in other patterns and combinations.
• Imagine each number in the triangle is a node in a grid which is connected to the adjacent numbers above and below it. Now for any node in the grid, count the number of paths there in the grid (without backtracking) which connect this node to the top node (1) of the triangle. The answer is the Pascal number associated to that node. The interpretation of the number in Pascal's Triangle as the number of paths to that number from the tip means that on a Plinko game board shaped like a triangle, the probability of winning prizes nearer the centre will be higher than winning prizes on the edges.
• The value of each row, if each number in it is considered as a decimal place and numbers larger than 9 are carried over accordingly, is a power of 11 (specifically, 11n, where n is the number of the row). For example, row two reads '1, 2, 1', which is 112 (121). In row five, '1, 5, 10, 10, 5, 1' is translated to 161051 after carrying the values over, which is 115. This property is easily explained by setting x = 10 in the binomial expansion of (x + 1)row number, and adjusting the values to fit in the decimal number system.
### More subtle patterns
There are also more surprising, subtle patterns. From a single element of the triangle, a more shallow diagonal line can be formed by continually moving one element to the right, then one element to the bottom-right, or by going in the opposite direction. An example is the line with elements 1, 6, 5, 1, which starts from the row 1, 3, 3, 1 and ends three rows down. Such a "diagonal" has a sum that is a Fibonacci number. In the case of the example, the Fibonacci number is 13:
1
1 1
1 2 1
1 → 3 ↓ 3 1
1 4 →6 → 4 ↓ 1
1 5 10 10 →5 → 1 ↓
1 → 6 ↓ 15 20 15 6 →1
1 7 →21 35 35 21 7 1
1 8 28 56 70 56 28 8 1
1 9 36 84 126 126 84 36 9 1
1 10 45 120 210 252 210 120 45 10 1
1 11 55 165 330 462 462 330 165 55 11 1
1 12 66 220 495 792 924 792 495 220 66 12 1
1 13 78 286 715 1287 1716 1716 1287 715 286 78 13 1
1 14 91 364 1001 2002 3003 3432 3003 2002 1001 364 91 14 1
1 15 105 455 1365 3003 5005 6435 6435 5005 3003 1365 455 105 15 1
1 16 120 560 1820 4368 8008 11440 12870 11440 8008 4368 1820 560 120 16 1
The second highlighted diagonal has a sum of 233. The numbers 'skipped over' between the move right and the move down-right also sum to Fibonacci numbers, being the numbers 'between' the sums formed by the first construction. For example, the numbers skipped over in the first highlighted diagonal are 3, 4 and 1, making 8.
In addition, if row m is taken to indicate row $(n+1)$, the sum of the squares of the elements of row m equals the middle element of row $(2m-1)$. For example, $1^2 + 4^2 + 6^2 + 4 ^2 + 1^2 = 70$. In general form:
$\sum_{k=0}^n {n \choose k}^2 = {2n \choose n}.$
Another interesting pattern is that on any row m, where m is odd, the middle term minus the term two spots to the left equals a Catalan number, specifically the (m + 1)/2 Catalan number. For example: on row 5, 6 − 1 = 5, which is the 3rd Catalan number, and (5 + 1)/2 = 3.
Also, the sum of the elements of row m is equal to 2m−1. For example, the sum of the elements of row 5 is $1 + 4 + 6 + 4 + 1 = 16$, which is equal to $2^4 = 16$. This follows from the binomial theorem proved above, applied to (1 + 1)m−1.
Some of the numbers in Pascal's triangle correlate to numbers in Lozanić's triangle.
Another interesting property of Pascal's triangle is that in rows where the second number (the 1st number following 1) is prime, all the terms in that row except the 1s are multiples of that prime.
Binomial matrix as matrix exponential (illustration for 5×5 matrices). All the dots represent 0.
### The matrix exponential
Due to its simple construction by factorials, a very basic representation of Pascal's triangle in terms of the matrix exponential can be given: Pascal's triangle is the exponential of the matrix which has the sequence 1, 2, 3, 4, … on its subdiagonal and zero everywhere else.
### Geometric properties
Pascal's triangle can be used as a lookup table for the number of arbitrarily dimensioned elements within a single arbitrarily dimensioned version of a triangle (known as a simplex). For example, consider the 3rd line of the triangle, with values 1, 3, 3, 1. A 2-dimensional triangle has one 2-dimensional element (itself), three 1-dimensional elements (lines, or edges), and three 0-dimensional elements ( vertices, or corners). The meaning of the final number (1) is more difficult to explain (but see below). Continuing with our example, a tetrahedron has one 3-dimensional element (itself), four 2-dimensional elements (faces), six 1-dimensional elements (edges), and four 0-dimensional elements (vertices). Adding the final 1 again, these values correspond to the 4th row of the triangle (1, 4, 6, 4, 1). Line 1 corresponds to a point, and Line 2 corresponds to a line segment (dyad). This pattern continues to arbitrarily high-dimensioned hyper-tetrahedrons (simplices).
To understand why this pattern exists, one must first understand that the process of building an n-simplex from an (n − 1)-simplex consists of simply adding a new vertex to the latter, positioned such that this new vertex lies outside of the space of the original simplex, and connecting it to all original vertices. As an example, consider the case of building a tetrahedron from a triangle, the latter of whose elements are enumerated by row 3 of Pascal's triangle: 1 face, 3 edges, and 3 vertices (the meaning of the final 1 will be explained shortly). To build a tetrahedron from a triangle, we position a new vertex above the plane of the triangle and connect this vertex to all three vertices of the original triangle.
The number of a given dimensional element in the tetrahedron is now the sum of two numbers: first the number of that element found in the original triangle, plus the number of new elements, each of which is built upon elements of one fewer dimension from the original triangle. Thus, in the tetrahedron, the number of cells (polyhedral elements) is 0 (the original triangle possesses none) + 1 (built upon the single face of the original triangle) = 1; the number of faces is 1 (the original triangle itself) + 3 (the new faces, each built upon an edge of the original triangle) = 4; the number of edges is 3 (from the original triangle) + 3 (the new edges, each built upon a vertex of the original triangle) = 6; the number of new vertices is 3 (from the original triangle) + 1 (the new vertex that was added to create the tetrahedron from the triangle) = 4. This process of summing the number of elements of a given dimension to those of one fewer dimension to arrive at the number of the former found in the next higher simplex is equivalent to the process of summing two adjacent numbers in a row of Pascal's triangle to yield the number below. Thus, the meaning of the final number (1) in a row of Pascal's triangle becomes understood as representing the new vertex that is to be added to the simplex represented by that row to yield the next higher simplex represented by the next row. This new vertex is joined to every element in the original simplex to yield a new element of one higher dimension in the new simplex, and this is the origin of the pattern found to be identical to that seen in Pascal's triangle.
A similar pattern is observed relating to squares, as opposed to triangles. To find the pattern, one must construct an analog to Pascal's triangle, whose entries are the coefficients of (x + 2)Row Number, instead of (x + 1)Row Number. There are a couple ways to do this. The simpler is to begin with Row 0 = 1 and Row 1 = 1, 2. Proceed to construct the analog triangles according to the following rule:
${n \choose k} = 2\times{n-1 \choose k-1} + {n-1 \choose k}.$
That is, choose a pair of numbers according to the rules of Pascal's triangle, but double the one on the left before adding. This results in:
1
1 2
1 4 4
1 6 12 8
1 8 24 32 16
1 10 40 80 80 32
1 12 60 160 240 192 64
1 14 84 280 560 672 448 128
The other way of manufacturing this triangle is to start with Pascal's triangle and multiply each entry by 2k, where k is the position in the row of the given number. For example, the 2nd value in row 4 of Pascal's triangle is 6 (the slope of 1s corresponds to the zeroth entry in each row). To get the value that resides in the corresponding position in the analog triangle, multiply 6 by 2Position Number = 6 × 22 = 6 × 4 = 24. Now that the analog triangle has been constructed, the number of elements of any dimension that compose an arbitrarily dimensioned cube (called a hypercube) can be read from the table in a way analogous to Pascal's triangle. For example, the number of 2-dimensional elements in a 2-dimensional cube (a square) is one, the number of 1-dimensional elements (sides, or lines) is 4, and the number of 0-dimensional elements (points, or vertices) is 4. This matches the 2nd row of the table (1, 4, 4). A cube has 1 cube, 6 faces, 12 edges, and 8 vertices, which corresponds to the next line of the analog triangle (1, 6, 12, 8). This pattern continues indefinitely.
To understand why this pattern exists, first recognize that the construction of an n-cube from an (n − 1)-cube is done by simply duplicating the original figure and displacing it some distance (for a regular n-cube, the edge length) orthogonal to the space of the original figure, then connecting each vertex of the new figure to its corresponding vertex of the original. This initial duplication process is the reason why, to enumerate the dimensional elements of an n-cube, one must double the first of a pair of numbers in a row of this analog of Pascal's triangle before summing to yield the number below. The initial doubling thus yields the number of "original" elements to be found in the next higher n-cube and, as before, new elements are built upon those of one fewer dimension (edges upon vertices, faces upon edges, etc.). Again, the last number of a row represents the number of new vertices to be added to generate the next higher n-cube.
In this triangle, the sum of the elements of row m is equal to 3m − 1. Again, to use the elements of row 5 as an example: $1 + 8 + 24 + 32 + 16 = 81$, which is equal to $3^4 = 81$.
Yang Hui (Pascal's) triangle, as depicted by the Chinese using rod numerals.
### Calculating an individual row
This algorithm is an alternative to the standard method of calculating individual cells with factorials. Starting at the left, the first cell's value is 1. For each cell after, the value is determined by multiplying the value to the left by a slowly changing fraction:
$v(c) = \frac{r-c}{c}$
Where r = row + 1, starting with 0 at the top, and c = the column, starting with 0 on the left. For example, to calculate row 5, r=6. The first value is 1. The next value is 1 x 5/1 = 5. The numerator decreases by one, and the denominator increases by one with each step. So 5 x 4/2 = 10. Then 10 x 3/3 = 10. Then 10 x 2/4 = 5. Then 5 x 1/5 = 1. Notice that the last cell always equals 1, the final multiplication is included for completeness of the series.
A similar pattern exists on a downward diagonal. Starting with the one and the natural number in the next cell, form a fraction. To determine the next cell, increase the numerator and denominator each by one, and then multiply the previous result by the fraction. For example, the row starting with 1 and 7 form a fraction of 7/1. The next cell is 7 x 8/2 = 28. The next cell is 28 x 9/3 = 84.
Note that for any individual row you only need to calculate half (rounded up) the number of values in the row. This is because the row is symmetrical.
## History
The earliest explicit depictions of a triangle of binomial coefficients occur in the 10th century in commentaries on the Chandas Shastra, an ancient Indian book on Sanskrit prosody written by Pingala between the 5th– 2nd centuries BC. While Pingala's work only survives in fragments, the commentator Halayudha, around 975, used the triangle to explain obscure references to Meru-prastaara, the "Staircase of Mount Meru". It was also realised that the shallow diagonals of the triangle sum to the Fibonacci numbers. The Indian mathematician Bhattotpala (c. 1068) later gives rows 0-16 of the triangle.
At around the same time, it was discussed in Persia (Iran) by the mathematician Al-Karaji (953–1029) and the poet- astronomer-mathematician Omar Khayyám (1048-1131); thus the triangle is referred to as the "Khayyam triangle" in Iran. Several theorems related to the triangle were known, including the binomial theorem. In fact we can be fairly sure that Khayyam used a method of finding nth roots based on the binomial expansion, and therefore on the binomial coefficients.
In 13th century, Yang Hui (1238-1298) presented the arithmetic triangle, which was the same as Pascal's Triangle. Today Pascal's triangle is called " Yang Hui's triangle" in China.
Finally, in Italy, it is referred to as "Tartaglia's triangle", named for the Italian algebraist Niccolò Fontana Tartaglia who lived a century before Pascal (1500-1577); Tartaglia is credited with the general formula for solving cubic polynomials (which may be really from Scipione del Ferro but was published by Gerolamo Cardano 1545).
Petrus Apianus ( 1495 -1552 ) published the Triangle on the frontispiece of his book on business calculations 1531/32 and an earlier version in 1527 the first record of it in Europe.
In 1655, Blaise Pascal wrote a Traité du triangle arithmétique (Treatise on arithmetical triangle), wherein he collected several results then known about the triangle, and employed them to solve problems in probability theory. The triangle was later named after Pascal by Pierre Raymond de Montmort (1708) and Abraham de Moivre (1730). | 6,370 | 22,427 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 21, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.78125 | 5 | CC-MAIN-2017-30 | longest | en | 0.891955 |
https://metanumbers.com/15813 | 1,624,630,033,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623487630175.17/warc/CC-MAIN-20210625115905-20210625145905-00335.warc.gz | 347,298,706 | 10,901 | ## 15813
15,813 (fifteen thousand eight hundred thirteen) is an odd five-digits composite number following 15812 and preceding 15814. In scientific notation, it is written as 1.5813 × 104. The sum of its digits is 18. It has a total of 4 prime factors and 12 positive divisors. There are 9,000 positive integers (up to 15813) that are relatively prime to 15813.
## Basic properties
• Is Prime? No
• Number parity Odd
• Number length 5
• Sum of Digits 18
• Digital Root 9
## Name
Short name 15 thousand 813 fifteen thousand eight hundred thirteen
## Notation
Scientific notation 1.5813 × 104 15.813 × 103
## Prime Factorization of 15813
Prime Factorization 32 × 7 × 251
Composite number
Distinct Factors Total Factors Radical ω(n) 3 Total number of distinct prime factors Ω(n) 4 Total number of prime factors rad(n) 5271 Product of the distinct prime numbers λ(n) 1 Returns the parity of Ω(n), such that λ(n) = (-1)Ω(n) μ(n) 0 Returns: 1, if n has an even number of prime factors (and is square free) −1, if n has an odd number of prime factors (and is square free) 0, if n has a squared prime factor Λ(n) 0 Returns log(p) if n is a power pk of any prime p (for any k >= 1), else returns 0
The prime factorization of 15,813 is 32 × 7 × 251. Since it has a total of 4 prime factors, 15,813 is a composite number.
## Divisors of 15813
1, 3, 7, 9, 21, 63, 251, 753, 1757, 2259, 5271, 15813
12 divisors
Even divisors 0 12 6 6
Total Divisors Sum of Divisors Aliquot Sum τ(n) 12 Total number of the positive divisors of n σ(n) 26208 Sum of all the positive divisors of n s(n) 10395 Sum of the proper positive divisors of n A(n) 2184 Returns the sum of divisors (σ(n)) divided by the total number of divisors (τ(n)) G(n) 125.75 Returns the nth root of the product of n divisors H(n) 7.24038 Returns the total number of divisors (τ(n)) divided by the sum of the reciprocal of each divisors
The number 15,813 can be divided by 12 positive divisors (out of which 0 are even, and 12 are odd). The sum of these divisors (counting 15,813) is 26,208, the average is 2,184.
## Other Arithmetic Functions (n = 15813)
1 φ(n) n
Euler Totient Carmichael Lambda Prime Pi φ(n) 9000 Total number of positive integers not greater than n that are coprime to n λ(n) 750 Smallest positive number such that aλ(n) ≡ 1 (mod n) for all a coprime to n π(n) ≈ 1841 Total number of primes less than or equal to n r2(n) 0 The number of ways n can be represented as the sum of 2 squares
There are 9,000 positive integers (less than 15,813) that are coprime with 15,813. And there are approximately 1,841 prime numbers less than or equal to 15,813.
## Divisibility of 15813
m n mod m 2 3 4 5 6 7 8 9 1 0 1 3 3 0 5 0
The number 15,813 is divisible by 3, 7 and 9.
## Classification of 15813
• Arithmetic
• Deficient
### Expressible via specific sums
• Polite
• Non-hypotenuse
## Base conversion (15813)
Base System Value
2 Binary 11110111000101
3 Ternary 210200200
4 Quaternary 3313011
5 Quinary 1001223
6 Senary 201113
8 Octal 36705
10 Decimal 15813
12 Duodecimal 9199
36 Base36 c79
## Basic calculations (n = 15813)
### Multiplication
n×i
n×2 31626 47439 63252 79065
### Division
ni
n⁄2 7906.5 5271 3953.25 3162.6
### Exponentiation
ni
n2 250050969 3954055972797 62525487097838961 988715527478127490293
### Nth Root
i√n
2√n 125.75 25.0999 11.2138 6.91517
## 15813 as geometric shapes
### Circle
Diameter 31626 99356 7.85558e+08
### Sphere
Volume 1.65627e+13 3.14223e+09 99356
### Square
Length = n
Perimeter 63252 2.50051e+08 22363
### Cube
Length = n
Surface area 1.50031e+09 3.95406e+12 27388.9
### Equilateral Triangle
Length = n
Perimeter 47439 1.08275e+08 13694.5
### Triangular Pyramid
Length = n
Surface area 4.33101e+08 4.6599e+11 12911.3
## Cryptographic Hash Functions
md5 931bbcc1be76e466aea6eb35b6717730 9fa2383c9fbafb8a2166e9aa0c1cae5182505112 b8dcb3ef8bd34d41e27be529d987228a744af7d4fedc416be81fed3ae9ea806d 69ad0a0702079e70947ffb62886fb691af098ccb7898cd2f757fc6d3334363a3342f94fe8af523bc1c36786b2845d905c6b1ab079ced560b5bd4b06b162428c9 fcebbe54b5442db583c17c7cd5c2bc538b87ea2a | 1,454 | 4,118 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.859375 | 4 | CC-MAIN-2021-25 | latest | en | 0.805754 |
null | null | null | null | null | null | « | Main | »
Michelle Rhee: Superstar!
Michelle Rhee got star treatment when she came to Tallahassee, literally. She was introduced as a movie star to legislators.
I think Rhee could be a valuable front person for reform.
But come on. Jeb Bush has been preaching these reforms since Rhee was snapping her gum in high school. Florida Education Commissioner Eric Smith may well be the best education commissioner in the country. He’s been at this for a couple decades and has produced remarkable gains. Rhee was in Washington DC for a couple years and barely nudged the test scores.
Another problem is Rhee’s penchant for self-promotion, in which reform becomes so closely linked to Rhee that the credibility of reform becomes too closely linked to the credibility of Rhee. As we saw with Al Gore’s ownership of global warming, this can be a two-edged sword.
Rhee testified about abolishing tenure, saying it is bad for students. She is right. The purpose of tenure is to protect teacher jobs. The purpose of schools is to educate kids. Keeping an ineffective teacher in the classroom puts the needs of up to 150 kids behind the needs of one teacher.
Rhee also is right about not paying teachers for masters degrees. The studies are pretty conclusive about this. Advanced degrees do not increase a teacher’s effectiveness. However, I have seen data that links learning gains with teachers that have national certifications. So that should be rewarded with higher pay.
Of course Rhee supports merit pay, which is coming. Paying all teachers the same, using a formula based on longevity, hardly makes sense.
Rhee correctly criticizes the last-one-in, first-one-out firing policies that come into play during teacher layoffs. As a result, more teachers must be laid off because beginning teachers make less money. We also are losing the next generation of teachers. However, there must be some objective measure in place when it comes time for layoffs so excellent, veteran teachers aren’t targeted simply because they have the biggest paychecks.
I have no idea why Rhee felt that Florida’s pension reform and education budget fell within her area of expertise. She said teachers should pay 5 percent of their paychecks into their pension fund. I don’t disagree. I just don’t understand why this is a reform issue or why she felt compelled to weigh in on it other than zinging teachers. Nor do I understand how someone with no knowledge of our education budget felt it could be slashed without harming classrooms.
This was simply gratuitous sucking up to Rick Scott and playing into her image at a teacher basher, an image she professes to regret.
Lastly, and I raised this in a past column, Rhee’s message, like the message of many reformers, is mainly about weeding out bad teachers. If you look at the data, this would indeed have a huge impact on improving learning gains. HOWEVER. The flip side is attracting a new generation of high-achieving teachers from the top ranks of college classes. And I’ve seen nothing from the reformers about putting serious money on the table to do that. If 90 percent of your message is firing, and 10 percent is rewarding, we are going to face a serious teacher shortage down the road because nobody is going to sign up for this gig.
Ya’ll might want to check out what’s going on over at Jay Matthews’ blog (@ the Washington Post). http://voices.washingtonpost.com/class-struggle/2011/02/michelle_rhees_early_test_scor.html
Seems a blogger has uncovered a few facts about Michelle Rhee’s claims on her resume regarding miraculous test scores.
Matthews has four separate posts discussing the “miracle.”
Then, check out how Rhee mishandled the firing of 75 D.C. teachers and the mistake is costing the district 7.5 million dollars in back pay.
While it is true that our education system has many ineefective teachers and they should be gotten rid of long ago, we hear no legislative proposals to get rid of ineffective administrators and principals, which are the real key to improving our schools as they stand on the frontline of hiring and firing ineffective teachers, so it makes no sense that ineffective adminstrators are left along in any educational reform package! Also, the other “shoe to drop” is what of ineffective parents? If they have no discipline in the home, and in fact never appear at teacher-parent conferences, what exactly is the school room teacher supposed to do? The administrators look the other way and do everything possible to avoid enforceing reasonable disciplinary standards. Finally, how about our neo-con tax-cutting legislators and governor? They specialize in warm fuzzy sound-bite reforms but in reality have enacted no meaningful reforms requiring more vigorous teacher preparation standards in at least 50-years, as they want easy access to a cheap labor supply. For example, an elementary school science teacher can qualify by having taken only one- 100-level general science course in undergrad school, which is unchanged since 1956. So much for placing a high priority on math and science education to enable us to compete in the global marketplace. Which leads me to the last question, why aren’t our so-called “educational journalists” fully profiling all of these areas of concern with meaningful exposes instead of siting on their butts and pulling stories off the internet? Yes, it is easy to beat-up on the teachers but if real academic advancement is to be attained, then our half-vast legislators need to have an adult conversation with their constitutents.
As a rule–if Scott’s behind it it’s either immoral, illegal, nonsensical or all three.
Michelle Rhee is a Michelle Malkin wannabe.
Mike, The last paragraph of your article is EXACTLY what veteran teachers have been saying for years. It is about time you’ve seen the light. And do you know what? Dean Cannon and Mike Haridopolos don’t care. Neither does Jeb Bush or John Thrasher. This is not about providing money for good teachers. This is an ideological struggle. This is simply about retribution against teachers, most of whom are ideologically different. Florida’s teachers have been poorly paid sense we achieved statehood. There is no reason for Florida’s teachers to ever think they have a chance for a good paycheck. This crowd of Republican ideologues proves just that. You’re going to get your wet dream – teacher pay, an end to tenure, pension reform. But the joke is on you because they will never, ever provide the money for meaningful merit pay reform. In 5-10 years you will be calling for job protections for teachers as a way to attract people to the classroom since the state won’t pay us. And by the way, teachers in Georgia pay into their pension, but they retire at 60% of their salary – not 48% like in Florida.
Sorry for the typos. There really needs to be an edit feature on this blog.
My wife is from Asia with a masters degree in Engineering. She became a public school teacher three years ago because of tenure and stablity for family. Now she is pissed off that politicians are making the life of teachers more difficult. Most teachers do not become teachers because of pay. The irony is that by eliminating tenure and reforming pensions, you are wiping out the very things that draws people in(other than a desire to teach). My wife is ready to give up on teaching in the US as here teachers are treated like dirt while in Asia they are well respected. The current pay system is the same idea as in the military. There are some careers that you can not measure sucess easily.
As for Rhee, her speaking about education is like General Westmoreland speaking about how to win the current wars. She failed and yet is railing about failing teachers.
Looks like I can move to Tennessee when I get fired from Florida. Let’s make sure all of Florida’s teachers know that a solid majority of Tennesseeans don’t like the idea of judging teachers based on test scores.Do the link. http://www.tennessean.com/article/20110208/NEWS04/102080338/Poll-Most-Tennesseans-don-t-favor-paying
It makes a lot of sense to have someone who couldn’t make it as a teacher be a consultant on education (teacher) reform. This lady has no credentials or experience that qualifies her for such a position. She doesn’t care about children or their education, she is after power and money.
Can I ask a really dumb question? Who is Rhee? It says she’s a movie star. I never heard of her. What movie was she in and why is she in Tallahassee talking about education?
@America, Michelle Rhee used to be in charge of Wash DC schools. She was in a movie called Waiting for Superman. The movie is stirring, but it’s more propaganda than fact. One review, by Diane Ravitch (who by the way is a conservative appointed to the Bush Sr. administration then reappointed by Clinton) does some debunking: http://www.nybooks.com/articles/archives/2010/nov/11/myth-charter-schools/?page=1
In September of 2010, this alleged educational miracle worker was speaking to a gathering at the Columbia Heights Education Campus in Washington DC. The topic was missteps made by beginning teachers. (Apparently, Rhee actually was a teacher, once.) She explained her misstep as being this: She put duct tape on the lips of students who were not behaving as she wished, and their lips bled.
I’m neither a teacher nor a parent. But this is an example, to me, of a bad teacher, someone to be weeded out. How did she become, instead, the so-called savior of education?
How does paying teachers due to test scores help Special Need kids? If it wasn’t for a teacher believing in me and supporting me, I would not be graduating with a standard diploma. The teachers that have to mentor students like me need to be paid more money not less because they help kids with disorders. What is going to be done for them?
I love this sentence in the article…”The flip side is attracting a new generation of high-achieving teachers from the top ranks of college classes.” Well, that’s seems a little obvious and rhetorically-without any substance. How do you throw out that sentence with zero procedures laid out to get there? Why don’t you just throw in “world peace” while you’re at it. The whole issue is tough, that’s the point of suggesting solutions.
I would rather have a voucher.
Great post.
Sounds like we disagree on a lot (like on Eric Smith), but we can disagree agreeably.
Check out:
Disagree agreeably?
What country do you live in!!!
Go to the first post and follow the links. Based on your past positions, you will probably love her.
go to the first post and follow the link
based on your past positions, you will probably love her.
It appears Julian covered all that needed to be said!
Look at the blog post that shows a breakdown of Michelle Rhee’s teaching ability and claims of success. Also she shouldn’t be saying anything about pensions. In DC she continually overspent and over-hired.
Spending and budget:
Student Achievement:
Her resume and claims:
Now- Do you really want her around your children?
Forget about the allegations of her fiancee’s alleged sexual indiscretions with teenagers.
Sadly, through great PR and a complacent media. The info is out there, but she’s too busy pimping or being pimped. Our kids will pay and we’ll rue the day.
Unfortunately what is occurring above is exactly what Mike Thomas said would occur. Throwing out the ideas because you discredit the source.
Rhee is too linked to the word “reform” and for some that comes with a bad taste (and one would think bitter feelings from some in
Washington DC too).
The ideas are sound whether you like the source or not.
Remove the ideas from the individual please and explore them on their own worth – because as Thomas shows – they make a great deal of sense. All of them. And they culminate in that last paragraph. You cannot have that one, without the others.
As for the poster above who said “As a rule–if Scott’s behind it it’s either immoral, illegal, nonsensical or all three.” – while I may or may not agree with your stance on our Governor, having that attitude is the exact problem plaguing the country right now. The feeling that “if the other guy from the other team says it, then we must not listen” – A very American disease right now.
Mike Rhee is just doing what you are…spouting off at the mouth with NO expertise in the area whatsoever. Teachers in Florida don’t have tenure…not since 1976…a little research and you would have found this out for yourself. Not paying for master’s degrees and forcing a 5% pay cut is just changing the rules that were in place when we signed up for this gig, as you call it. If Lying politicians, Past Governors with their own personal agendas and legislators saving their corporate buddies while hurting the hard working low paid teachers, police and firemen are what you support…so be it…Keep your head in the sand if you want, but some of us will choose to live with our eyes wide open. Rhee has done nothing, but cause problems for public education(see the DCPS settlement by an arbitrator), yet these moronic legislators welcome her with open arms and call her a Rock Star!?! Another sad commentary and another ridiculous article by Mike!
Back iln the early 50’s schools were underunded and teachers were underpaid. Yet they produced a generation of high achieving graduates who today are successful business people, doctors, lawyers, newspaper columnists, engineers, etc. So what has changed? The Federal government started handing out money to states, with strings attached. You had to follow the Federal guidelines to get the money. The Feds (read that Dept. of Education} began to control and micromanage all aspects of local level education.
Then we were inundated by immigrants who could not speak English, and the Feds told us, incorrectly, how to teach those immigrants. Then we started rewarding people like Michelle Rhee to come in and tell us what we already know: we have good teachers and bad teachers. Surprisingly, you can find this phenomena in all walks of life. The difference is that in private industry they weed out the bad eggs. In education, our administrators who would like to do their job are impeded by a bureaucratic morass of union rules and regulations, including tenure.
Tenure does not get in the way in Florida and if there were “Just cause” to weed out the bad eggs Nothing would get in your way!
@julian- what evaluation tool did you use to come up with this statement
When making a statement like this, one should check spelling first.
We have qualified and dedicated teachers in Florida who have been doing a reasonably good job under sometimes trying conditions. Now that their pay has caught up with their contemporaries, they still have one insurmountabled obstacle they have faced the past several years but which they are loathe to mention. The politically correct word for this problem is “diversity.” If you want to compare results in various states, just look at Minnesota, North Dakota, South Dakota, Montana, Idaho, etc., where they very little “diversity,” and compare that with Florida, California, Mississippi, Alabama, Georgia, New York City, and Wash. DC, where they have serious “diversity” problems. A “diversity” problem exists where the parents (or parent) are not interested in the education of the child, or is incompetent in raising the child. Measuring these teachers against a national yardstick is patently unfair.
Please excuse the typos & spelling mistakes. I failed to proofread.
Do away with tenure?! Then the top paid teachers will lose their jobs because you can hire two new teachers for the price of one. Additionally, maybe you are too young to remember teachers getting fired for teaching evolution or other controversial subjects. Why go into teaching with such low salaries and no job security? One reason teachers accept low salaries is because pensions were considered as a compensation.
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# 14 a 40 a 61 cm c 6 cm 20 a 60 a 508 cm c
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Unformatted text preview: a2 or cos A = 2 bc 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. Given a c A b C Given b = 2 cm a = 4 cm b = 7 cm a = 5 cm a = 6 cm a = 3 cm a = 4 cm a = 12 cm b = 3.5 cm a = 7.8 cm a = 10.2 cm a = 7.4 cm a = 8 cm c = 3 cm b = 6 cm c = 9 cm c = 8 cm b = 7 cm b = 5 cm b = 2 cm b = 5 cm c = 4.7 cm c = 5.3 cm b = 6.4 cm b = 9.6 cm b = 10 cm 14. Given a = 4.8 cm b = 7.1 cm c = 5.5 cm 15. 16. Given a = 8.5 cm a = 17.1 cm b = 13.2 cm b = 28.6 cm 17. The goal-posts in football are 7.32 metres apart. A ball is placed on the ground 8 metres from one goal-post and 6 metres from the other. Within what angle must the ball be kicked along the ground in order to score? 18. Two people start walking from the same place at the same time. They are walking on level ground. One walks at 4 kilometres an hour going due North, and the other walks at a speed of 3.5 kilometres an hour going North-east. After 3 hours, how far apart will they be? 19. The principal 'legs' of a step-ladder are usually of two different length...
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Ask a homework question - tutors are online | 489 | 1,333 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.78125 | 4 | CC-MAIN-2018-05 | latest | en | 0.871572 |
http://mathhelpforum.com/number-theory/99840-perfect-square-test.html | 1,524,472,573,000,000,000 | text/html | crawl-data/CC-MAIN-2018-17/segments/1524125945855.61/warc/CC-MAIN-20180423070455-20180423090455-00203.warc.gz | 199,971,001 | 11,044 | 1. ## Perfect Square Test
Are there any tests you can perform to see whether an integer is a perfect square? The number I want to test is odd. The only test I know is: If $\displaystyle k$ is odd, then $\displaystyle k^2 \equiv 1 \mod 8$. Unfortunately, the number I have passes this test. Are there any others I can do?
2. Um, from a previously answered question, you can also check the number of divisors it has . If it's odd, then it's a perfect square, if it's even then it's not.
Not sure if that's the kind of test you're looking for, but it works
3. I don't know the number, though. I'll describe the problem below, but I DO NOT WANT THE ANSWER OR EVEN A SOLUTION; I just want some tests that I can perform and maybe a nudge in the right direction.
Problem:
Can a number of the form $\displaystyle 200...009$ (i.e. $\displaystyle 2\cdot 10^n + 9, n\in\mathbb{N}$) be a perfect square?
Any potential square root has to be of the form $\displaystyle 10j+3$ or $\displaystyle 10k+7$. Squaring the first option gives $\displaystyle 100j^2+60j+9=20..09 \implies j(5j+3) = 10^m, m\in\mathbb{N}$. I think with a bit of work I could show that there are no integer solutions to that equation; however I'm not so sure I can do the same for $\displaystyle (10k+7)^2=20..09$.
Does this seem like a good way to go about this problem, or should I try something completely different?
4. Originally Posted by redsoxfan325
Are there any tests you can perform to see whether an integer is a perfect square? The number I want to test is odd. The only test I know is: If $\displaystyle k$ is odd, then $\displaystyle k^2 \equiv 1 \mod 8$. Unfortunately, the number I have passes this test. Are there any others I can do?
I do not understand the question.
What is wrong with extracting the Square Root?
If the square root is an integer, it is a perfect square.
That just seems to be the most efficient way to get the result.
5. I can't do that because I don't know the number. (I explain the problem in detail in my second post.) I know that it starts with 2, ends with 9, and has an arbitrary number of zeroes in between. The question: can a number like that ever be a perfect square (for any number of zeroes).
6. You're looking too far.
Hint : any integer is congruent mod $\displaystyle 3$ to the sum of its decimal digits.
Now what are the squares, mod $\displaystyle 3$?
7. Originally Posted by Bruno J.
You're looking too far.
Hint : any integer is congruent mod $\displaystyle 3$ to the sum of its decimal digits.
Now what are the squares, mod $\displaystyle 3$?
Brilliant! | 691 | 2,582 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.59375 | 4 | CC-MAIN-2018-17 | latest | en | 0.927991 |
http://www.jiskha.com/display.cgi?id=1190346881 | 1,498,597,408,000,000,000 | text/html | crawl-data/CC-MAIN-2017-26/segments/1498128321553.70/warc/CC-MAIN-20170627203405-20170627223405-00715.warc.gz | 573,578,520 | 4,408 | posted by .
I have the solutions but I don't understand how to get them!
1. Perform the indicated operations and simplify:
[(2/x)-1]/(x^2 -4)
2. Subtract and simplify:
[12/(x^2 -4)] - [(3-x)/(x^2 + 2x)]
3. Perform the indicated operations and simplify:
[(x/y)-(y/x)]^-1
4. Divide and simplify:
[(a^2 b)/(a-b)]/[(a+b)/(a^2-b^2)]
5. Perform the indicated operations and simplify
[(x/x-3)-(2x/x^2-2X-3)]/[(2/x+1)-(1/x)
Solutions:
1. -1/(x^2 + 2x)
2. [x^2 + 7x + 6]/[x(x+z)(x-z)]
--typo in book?
3. xy/(x^2 - y^2)
4. a^2 b
5. x^2/(x-3)
1. Rewrite (2/x) -1 in the numerator as -(x-2)/x
Factor (x^2/4) in the denominator as (x-2)(x+2)
Then cancel out the (x-2) terms in numerator and denominator. You wil be left with -1/[x(x+2)]
3. Rewrite x/y and y/x with a common denominator, xy.
[(x^2- y^2)/xy]^-1 = xy/(x^2- y^2)
You try the others. They are all exercises in factoring and canceling tems
8/9a^2 divide 4a^2-4a-24/a^2-6a+9 Perform the indicated operation
(2/3-1/6)divide by(1/4+4/5) reduce to lowes terms
idk edsgfjhdsh
### Related Questions
More Related Questions
Post a New Question | 425 | 1,104 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.859375 | 4 | CC-MAIN-2017-26 | latest | en | 0.83293 |
https://csitcollege.com/floor-and-ceiling-function/ | 1,701,326,964,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100172.28/warc/CC-MAIN-20231130062948-20231130092948-00028.warc.gz | 232,628,969 | 27,922 | # Floor and ceiling function with example, program
In computer science or discrete structure, there are two important function are use i.e. Floor and ceiling function . In this tutorial we discuss definition, example using concept computer science and program of these function. Notation: ⌈x⌉ is ceiling function, ⌊x⌋ is floor function where x is variable.
Let x be a real number then ⌊x⌋ called the floor function of x, assigns to the real number x the largest integer that is less than or equal to x. this function rounds x down to the closest integer less than or equal to x. The floor function is also called the greatest integer function.
Let x be a real number then ⌈x⌉ called the ceiling function of x, assigns to the real number x the smallest integer that is greater than or equal to x. this function rounds x up to the closest integer less than or equal to x.
#### Example of Floor and ceiling function
• 5.3 Floor value ⌊x⌋=5 & Ceiling value ⌈x⌉=6
• 4.5 Floor value ⌊x⌋=4 & Ceiling value ⌈x⌉=5
• -4 Floor value ⌊x⌋=-4 & Ceiling value ⌈x⌉=-4
• 1/2 Floor value ⌊x⌋=0 & Ceiling value ⌈x⌉=1
Prove or disprove that ⌈x+y⌉=⌈x⌉+⌈y⌉ for all real number x and y.
solution, A counter example is supplied by x=1/2 and y=1/2. with these values we found that ⌈x+y⌉=⌈1/2+1/2⌉=⌈1⌉=1 but ⌈x+y⌉=⌈1/2+1/2⌉=1+1=2
Another example: Data stored on a computer or transmitted over a data network are usually represent as a string of bytes. Each byte is made up of 8 bits. How many bytes are required to encode 300 bits of data?
solution: To determine the number of bytes needed, we determine the smallest integer that is at least as large as the quotient when 300 is divided by 8. Therefore, the number of bytes required = ⌈300/8⌉=⌈37.5⌉=38 bytes.
#### Program to find Floor and ceiling function
#include <stdio.h>
#include <math.h>
int main()
{
float val;
float fVal,cVal;
printf("Enter a float value: ");
scanf("%f",&val);
fVal=floor(val);
cVal =ceil(val);
printf("floor value:%f \nceil value:%f\n",fVal,cVal);
return 0;
} | 641 | 2,022 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.1875 | 4 | CC-MAIN-2023-50 | latest | en | 0.790049 |
https://www.esaral.com/q/the-function-41352 | 1,726,478,966,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651682.69/warc/CC-MAIN-20240916080220-20240916110220-00507.warc.gz | 695,948,706 | 11,597 | # The function
Question:
The function $f(x)=\frac{x^{3}+x^{2}-16 x+20}{x-2}$ is not defined for $x=2$. In order to make $f(x)$ continuous at $x=2$, Here $f(2)$ should be defined as
(a) 0
(b) 1
(C) 2
(d) 3
Solution:
Here,
$x^{3}+x^{2}-16 x+20$
$=x^{3}-2 x^{2}+3 x^{2}-6 x-10 x+20$
$=x^{2}(x-2)+3 x(x-2)-10(x-2)$
$=(x-2)\left(x^{2}+3 x-10\right)$
$=(x-2)(x-2)(x+5)$
$=(x-2)^{2}(x+5)$
So, the given function can be rewritten as
$f(x)=\frac{(x-2)^{2}(x+5)}{x-2}$
$\Rightarrow f(x)=(x-2)(x+5)$
If $f(x)$ is continuous at $x=2$, then
$\lim _{x \rightarrow 2} f(x)=f(2)$
$\Rightarrow \lim _{x \rightarrow 2}(x-2)(x+5)=f(2)$
$\Rightarrow f(2)=0$
Hence, in order to make $f(x)$ continuous at $x=2, f(2)$ should be defined as $0 .$
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Programming Technology
Comments are More Important than Code 1021
Posted by timothy
from the if-you-disagree-provide-comments dept.
CowboyRobot writes "I was going through some code from 2002, frustrated at the lack of comments, cursing the moron who put this spaghetti together, only to realize later that I was the moron who had written it. An essay titled Comments Are More Important Than Code goes through the arguments that seem obvious only in hindsight - that 'self-documenting' code is good but not enough, that we should be able to write code based on good documentation, not the other way around, and that the thing that separates human-written code from computer-generated code is that our stuff is readable to future programmers. But I go through this argument with my colleagues, who say that using short, descriptive variable names 'should' be enough as long as the code is well-organized. Who's right?"
Comments are More Important than Code
Comments Filter:
• by Anonymous Coward on Tuesday April 26, 2005 @09:12PM (#12354434)
I used to grade student's code as a TA at my university, and I'll tell you what is more annoying than NO comments, this:
printf("Encrypt message..."); /* print "Encrypt message..." to the console */
and then followed by about 150 lines of uncommented spaghetti code
• by eobanb (823187)
Well then, just employ a code comment moderation system.
• by Klar (522420) < minus physicist> on Tuesday April 26, 2005 @09:22PM (#12354521) Homepage Journal
hah I also grade as a university TA, and let me tell you what is more annoying than that... people who don't know how to indent their code, and want you to help debug it! Ahhhhh!!!!!
• by Anonymous Coward on Tuesday April 26, 2005 @09:46PM (#12354709)
You know whose fault that is? The morons that mark you down if you don't comment trivial programs like Hello World.
I actually ended up doing three introductory programming courses at college and university (overlap in curricula), and I've been marked down for not including this type of comment in all three courses.
printf("Hello, world!\n"); /* send "Hello, world!" to the console and move to the next line. */
The problem is that they try and teach you to write comments prematurely. When all most people on the course can write are programs like Hello World, they really can't see the use of comments and there are no good examples to give. The use of comments should come after things like functions etc IMHO, so that students can actually see how comments can be useful.
• by T-Ranger (10520) <jeffw@chebuct o . n s .ca> on Tuesday April 26, 2005 @10:56PM (#12355211) Homepage
Ive been in enough classes this year (well below my level.. dammit, I need the "letters") that Ive come to the conclusion that introductory programming courses are taught all wrong.
Write 5 lines. Write 20 lines. Write 100 lines. Of useless and pointless code.
What should be done is: Take this 1000 line programme. Add on 5 lines. Add on 20 lines. Add on 100 lines. Beacuse that would require being able to read code, too. Being able to understand what is already there is frequently more then half the battle. Saying "What the fuck does this mean?" a few dozen times is the only way to get to write comments. Being a sysadmin, rather then a full time programmer, this took me litteraly years to learn. And it was usualy "What the fuck were you doing here, brain? Dammit, one of these days were going to have to start commenting our code."
About 6 months ago, I read through a project request on one of these ebay-for-coders sites. Language optional; 1 comment per 5 LOC, but if using Perl, 1 comment per 2 LOC.. With requrements like that you are only asking to get BS comments like "Print X to the console".
• by tomhudson (43916) < ... <nosduh.arabrab>> on Tuesday April 26, 2005 @11:07PM (#12355317) Journal
Better yet - remove 5 lines. remove 20 lines. remove half the lines ... and it better still work!
• by Moraelin (679338) on Wednesday April 27, 2005 @02:45AM (#12356530) Journal
It's not just that the programs to write are small, it's that they're write-only. You write them once, get graded, that's it. We churn generation after generation of students who are taught that code is written once, then never ever maintained.
Sure, you learn lots of things about design, software engineering, etc, in university, but they're pure theory. And seemingly useless theory at the moment. There is _nothing_ to illustrate there why some code organization is good, and why spagetti code is bad. All those lessons about maintenance as wasted when you never have to maintain anything, nor ever write anything big enough.
So while I'll say your idea does have merit, I think it can be done better. Don't just give them 1000 lines of someone else's code. Make them keep building and expanding the same program until the last year.
E.g., ok, in introductory programming they had to write some 100 line trivial program. But don't throw it away. When the next course comes along, give them the assignment to change or expand that original program.
E.g., if at some point you also get a computer graphics course, make them add a graphics module to that program. GUI programming? Sure, add a GUI to it. Database programming? Sure, make it save the data in a database. YACC? Ok, make them add a small scripting language to it. Different language? Make them port it to that language. Etc.
Make it a part of the grade to explain _what_ had to be changed and _why_.
Eventually it _will_ grow to be 1000 lines, and then it will grow even larger. And more importantly it'll be example of why code has to be readable and maintainable.
• by JohnsonWax (195390) on Tuesday April 26, 2005 @11:07PM (#12355314)
No, the problem is that they try and teach you to write code prematurely.
Document your application, requirements, constraints, and system interactions (what the engineer does). Then write the code (what the coder does).
What you will quickly learn is that it's better to be the engineer than the coder. What you'll realize is that the engineer is the client-side guy that figures out how to solve the challenge presented and the coder is the guy who can live in Manipal.
The computer scientist is another guy altogether, that sets the boundaries within which the engineer must work and provides many of the tools.
Somehow CS education has gotten horribly derailed, and asks students to combine the equivalents of electromagnetic theory, power system design, and basic home wiring in one curriculum. No wonder enrollments are plummeting - nobody knows what a CS major is or should do.
• by Fubari (196373) on Wednesday April 27, 2005 @01:01AM (#12356023)
Here is an example of some Pretty Evil Shit:
I want to laugh when I read it.
Some of it is funny.
Some of it is just scary.
The human mind can be a freakishly messed up thing. []
It is part of a larger essay about writting crappy code.
Anybody that even comes close to software development
should check it out.
--- begin excerpts ---
Avoid Documenting the "Obvious" : : If, for example, you were
writing an airline reservation system, make sure there are at
come after you have no business modifying your code without
thoroughly understanding every line of it.
Units of Measure : : Never document the units of measure of any
in engineering work.
constants, or how the values were derived.
incorrect units of measure in the comments.
measure; name it after yourself or some obscure person and never
you could use integer rather than floating point arithmetic.
On the Proper Use of Design Documents : : When implementing a very
complicated algorithm, use the classic software engineering principles
of doing a sound design before beginning coding. Write an extremely
detailed design document that describes each step in a very complicated
algorithm. The more detailed this document is, the better.
there are over 500 such auto-numbered paragraphs.
For example, one paragraph might be: (this is a real example) - Display all impacts for activity where selected
mitigations can apply (short pseudocode omitted).
you write a corresponding global function named:
Do not document these functions. After all, that's what the
design document is for!
contradictory, early drafts of the design document hidden on some
dusty shelving in the back room near the dead 286 computers.
--- end excerpts ---
• Don't forget about students who use assert()'s for validating user input!
• by el-spectre (668104) on Tuesday April 26, 2005 @10:18PM (#12354924) Journal
As a freshman I had a professor who kept giving me shit about not having enough comments, so my next program looked something like this: ...
a=5; //assign integer variable 'a' a value of 5
b=2; //assign integer variable 'b' a value of 2
print(a+b); //print to the console the sum of integer variable 'a' and integer variable 'b' ...
and so on, for about 200 lines. Worthless commenting? Sure, and childish too... but it was amusing and I never got docked for insufficient commenting again :)
• by Ratbert42 (452340) on Tuesday April 26, 2005 @11:03PM (#12355269)
I just ripped this comment out of our code today: // close the file
• by rve (4436) on Wednesday April 27, 2005 @12:41AM (#12355943)
At work, we are not allowed to use comments in the code.
Allowing comments would "encourage coders to use clever tricks" according to the technical director.
• by TripMaster Monkey (862126) * on Tuesday April 26, 2005 @09:12PM (#12354435)
From the Summary:
You are. No question.
I have a saying I like to use when people (usually managers) try to persuade me to take the quick/cheap way out:
"Any time/money you save by cheaping out now, you'll wind up having to pay back twice over on the back end."
I've yet to see this maxim disproved, and it's just as applicable to coding as it is to anything else. Your colleagues are certainly correct when they state that the code must be well organized, but this simply isn't enough. If you don't put in the necessary time (minimal, really) to properly document your code, you'll wind up spending a lot of time trying to figure out just what you did and why. Also, even if you can remember exactly what your code is all about, the guy that comes after you probably won't...proper documentation is professional courtesy. I suppose they'll learn after they spend a few hours puzzling over a piece of old code (that's how I learned...:P ). Sooner or later, not documenting properly will bite them in the ass.
• by Fjornir (516960) on Tuesday April 26, 2005 @09:14PM (#12354446)
The canonical form of your maxim is "If you don't have time to do it right the first time, when will you find time to do it again?". The correlary is: "The problem with quick and dirty is that dirty remains long after quick is forgotten."
• And don't forget the maxim that seems to be prevalent everywhere, even though everyone "knows" this shouldn't be....
"There's never enough time to do it right, but always enough money to do it twice."
• Judging from your indenting, I'm sure glad I don't have to read YOUR code.
• I agree. Giving your variables descriptive names is a good practice, but it's not nearly enough "commenting". If you have a piece of code that manipulates the values of those descriptive variables with bit masks to spit out a value, how is someone supposed to know what you just did? I think the adage "you don't document your code, but you code your document" is a little extreme, but it gets across the importance of code documentation. I'm working on an open source project of my own right now. I went back to modify a piece of code that I hadn't touched for nearly 4 months, only to make absolutely no sense of what the hell I was doing there. Luckly I commented it extremely well so after reading the comments I was able to get that "Oh yeah..." feeling and make the appropriate modifications without breaking the code. Lesson: commenting code may take you some more time now, but it's going to save you a hell of a lot of time later.
On that topic, what are some good examples of well-commented code? Rather than just see words, I'd rather see real-life applications of these practices. For starters, here's typically how much I comment my code: Allacrost source code [] (note: not all files were written by me, by I try to stress heavy commenting on the rest of my team)
• by gregmac (629064) on Tuesday April 26, 2005 @11:05PM (#12355292) Homepage
tmp = last_update; // tmp = last update time
last_update = SDL_GetTicks(); // Set the last update time to now
tmp = last_update - tmp; // tmp = difference between now and last update
fps_timer += tmp; // Increase our fps millisecond timer
fps_counter++; // Increase our frame count
There's also something as far as too much commenting. Don't get me wrong, I can't stand uncommented code. I learnt, like most people, the hard way -- go back to edit something I did 6 months ago, and have no idea what's happening.
But I don't think this is useful: // tmp = last update time. You've just spelled out in english what tmp = last_update; says in code. Comments don't add anything if you can just read the code and figure it out. If anything, it's a waste of time.
Here's my version of your same section
// find the time since our last update
tmp = last_update;
last_update = SDL_GetTicks(); // now
tmp = last_update - tmp;
// add time and increment counter
fps_timer += tmp;
This isn't really the greatest example, but I like to comment sections of code. The details can be read by looking through the code, with "hints" as needed (for example, reminding me that SDL_GetTicks() is the current time).
Well commented blocks mean you can skim through the code, getting a quick overview of what's happening, without having to read in much detail.
// find matching entry in hash table
while (....
You don't have to care about the details of the look, you just know by the time you hit the } you've found what you wanted.
Don't take this personally or anything. Your code is very readable.. there's just lots of extra comments (granted, I've seen worse) that are really just not worth putting in.
(also, proper indenting is as, if not more, important.. of course, /. is butchering my indenting.. so please don't harass me about that because i couldn't agree with you more ;) )
• by ComputerSlicer23 (516509) on Wednesday April 27, 2005 @12:38AM (#12355927)
The concept I remember most from "Code Complete" is comments should tell you why or what are you doing something, but it shouldn't tell you how you are doing. The how should be obvious from the code.
Never tell me you are adding shifting the varaible by 4 bits, instead tell me, you are converting from bytes to 16 byte blocks due to a chunks size conversion because that's the size the output device expects blocks to be written in.
Don't tell me you are swapping variables, tell me why the values should be exchanged.
Don't tell me you searching for the max value, tell me what the max value is used for when you find it. That sort of concept.
One easy way to do that, is to look at code and the comments. If the code and the comments could get out of sync because you changed the implementation, the comment is wrong. You documented the how not the what.
There are cases where the how is important, especially when there are things where the "how" doesn't behave naturally. Oh, this sorts numbers, but it is intentionally sorting the ASCII values of the numbers when represented as a string. That'd be very useful to know. I'd expect a sort to work based on the binary values.
The other rule I remember from code complete, and from "Writting Solid Code", was the concept of laying out the pseudo code that explained what you wanted to do at a high level. Then filling in between the comments with the implementation. It was a very good way to end up with documented code.
• by (321932) on Wednesday April 27, 2005 @12:42AM (#12355945) Homepage Journal
Here's what my version would look like:
// TODO: gracefully handle wrap-around of SDL_GetTicks()
// not implemented yet because it only happens every 40 years
tmp = last_update;
last_update = SDL_GetTicks();
tmp = last_update - tmp;
fps_millis_timer += tmp;
My philosophy is that:
1) production code is not a tutorial of the C(++) language. It's okay to assume the reader knows their stuff. A bad example:
tmp = last_update; // tmp = last update time
2) production code is not documentation on the APIs that it uses. APIs have their own documentation, the reader either is familiar with it, or can find documentation. A bad example:
last_update = SDL_GetTicks(); // now
If the API is only used in one file, then you could point at the documentation right where the header is included. If it's used on a project level, then it should be pointed to at the project documentation.
3) comments are just as much about why you didn't do something a certain way as it is about why you did do something a certain way.
4) the obvious; name your variables correctly. Instead of:
fps_timer += tmp; // Increase our fps millisecond timer
fps_millis_timer += tmp;
5) if you are implementing using a spec, somewhere in the top of the file describe the revision of the spec. Additional comments can describe the section of the spec that they relate to.
For example:
// Init GDT, see section 6.3.1
virtual_machine.gdt.base = base;
virtual_machine.gdt.limit = limit;
6) NEVER put actual values in comments. For example:
com1.baudrate = 9600; //set baudrate to 9600
The code already shows the actual value, and all too often the comment doesn't get updated.
Sometimes it's okay to list the possible values:
out_p16_d8( base + CSR, 0x66 ); //0x61 = 1x, 0x62 = 2x, 0x66 = 4x, other val illegal
but I only prefer that when this is the _only_ place where it is used. Otherwise you'd use enums and/or refer to the documentation section/page
7) write top level documentation to explain what the software is actually supposed to do. It's amazing how often this is missing. Write one bloody page that describes to a programmer that doesn't know ANYTHING about the software what the hell it's supposed to do, and how it's accomplishing this. It also helps tremendously if every file has a little description in the header. And I'm not talking about what license the code is distributed under.
• Comments are essential to maintaining any project.
It's annoying when I go to the person who wrote a chunk of code fom a couple months ago to ask them some questions, they look at the code and have no idea why they were doing what they were doing.
I've always followed the philosphy that I should write and comment code so that someone who wasn't a programmer could read it and understand what was going on. This makes it very easy to go back even years and know exactly what the code was doing and why it was do
• by snorklewacker (836663) on Tuesday April 26, 2005 @09:36PM (#12354641)
You are. No question.
Hello. I question. Look, if you need comments to explain the code flow because your method spans four screens and has six levels of conditionals in two levels of loops, all the comments in the world won't help you.
Especially if you change the code and now the comments are wrong
I have had the joy of maintaining lovingly commented code with all these huge blocks at the start showing what args get passed and what happens, and I can't understand a god damned thing about what it's doing, because the code all looks like
(void *((ebuf->qs) > VRT_FBUF) ? etranf(&q) ...
All generalisms are wrong.
• The code needs to be well documented, where "well" means both "correctly" and "frequently." Lots and lots of poor documentation is just as bad as little or no documentation. And documentation, no matter how well done, will not fix lousy code.
So I don't question. The story poster is correct. Code needs to be well documented, and neither good organization nor descriptive variable names is a sufficient replacement for documentation. But you're absolutely correct too that documentation is only a small pa
That's not true. One line comments won't help, but a large descriptive comments will. If you can explain the code by talking about it, then you can explain the code by writing about it.
Especially if you change the code and now the comments are wrong
You're incompetent if you don't change th
• by mosel-saar-ruwer (732341) on Tuesday April 26, 2005 @10:58PM (#12355228)
GRANDPARENT: Especially if you change the code and now the comments are wrong
PARENT: You're incompetent if you don't change the comments to match the code. You're equally incompetent if you come across incorrect comments and leave them in. You're supposed to the job, so do it...
PARENT: As Fred Brooks said, "There is no silver bullet."
A database backend would go a long ways towards providing a silver bullet, i.e. if instead of writing your code to an ASCII text file, you were writing to a document management system that kept doubly linked associations between the lines of code and the comments associated with those lines of code, and if code/comment pairs had dirty bits, so that if you changed one [e.g. the code], then the dirty bit wouldn't get changed to clean until you verified that the other [e.g. the comment] was correct, then that would go a long way towards solving the problem.
I think we are still in the infancy of code/documentation/database integration, however.
• by spammacus (805242) on Tuesday April 26, 2005 @11:03PM (#12355267)
All generalisms are wrong.
Isn't that a paradox? :)
• by rkcallaghan (858110) on Tuesday April 26, 2005 @09:47PM (#12354721)
Actually I have found that more often than not, not documenting properly only bites someone else in the ass.
This is likely the source of the problem, and the least likely to change. I suppose it could be part of the QA process to check for notation on code, but I somehow suspect that with programming jobs on a one way trip to Bangalore that readability it secondary to "works cheap."
• If you don't put in the necessary time [...] to properly document your code, you'll wind up spending a lot of time trying to figure out just what you did and why
The corollary to this is the need to maintain existing comments when the code is changed. Misleading/outdated comments can sometimes be worse than none at all.
proper documentation is professional courtesy
I couldn't agree more. Good documentation gives you context and insight into the motivation of a past developer, which can be priceless when
• The only time comments should be in code, is when the effect of the code is different than it appears to be - different than what a reasonably skilled programmer would expect. That is the only reasonable use for comments. For reference, here is a quick list of arguments and my canned responses:
I need comments because I can't understand the code otherwise.
You should be writing simpler code. If the code is too hard to understand by itself, then it is your job to make it simpler.
It's not the code t
• by jesup (8690) * <> on Tuesday April 26, 2005 @11:08PM (#12355325) Homepage
comments can lie
I worked with a programmer who disliked comments so much he'd remove them before looking at a function. Ok. So I wrote some code and he came to me and said "why do you have an empty else case?" I was puzzled, then realized that I'd written something like this:
/* we don't have to do anything in the else case because of x y and z */
where x y and z would be non-obvious to anyone who wasn't fully immersed in this code. He's run it through a filter that removed all comments. He was a genius programmer - but wrote code that almost no one else could ever maintain. Tons of reliances on edge conditions without comment, reuse of generically-named variables (1 and 2 character names), tricky (but efficient) algorithms. So far as I know, I was the only one there ever to manage to really grok his code, and that required days of immersing myself.
x = x++; // add one to x
is obviously not useful.
// Test FU_E (End bit) after FU_A/FU_B test! If there's a gap, do not consider
// hitting the End bit a marker to stop - continue until we see another
// packet/timestamp (in which case we return TRUE), or until we are
// at the end of the buffer (in which case we return FALSE and keep
// hoping to assemble it).
if (((*curr)->GetPayload()[1] & FU_E_BIT) && !gap)
break; // no error in fragment
// if there's a gap we still won't return true unless
// we find a non-fragment packet (or one from another fragment!)
This is an example of a useful comment - and yes, it has to be maintained if the line of source were to change. I chose that at random; there are better examples - such as explaining what the edge cases are (especially if not handled), and under what circumstances they would become relevant, and how they could be dealt with then.
Please excuse the incorrent indenting above; "<ecode>" doesn't work exactly like 'pre'
• by skraps (650379) on Tuesday April 26, 2005 @11:50PM (#12355609)
That comment is a good match for the code it is with. I never write comments like that unless dealing with a 3rd party interface, or something else that is a brick wall I can't refactor across, and there is something very quirky or unusual about it. The 'if' in the last snippet you quoted is very complex, and I would probably try to decompose it into a few functions and/or objects as necessary to make it clear what is happening. For example, you could start by replacing the 'if' with:
This clearly expresses the intent of the code you quoted (assuming I understood it correctly). Here, I've created a 'message' object to encapsulate some of the logic. The entire loop may be changed to look something like this:
packet_t packet=connection.get_next_packet();
Now, the "is complete" decision making is abstracted to the "message", which seems to me the right place for that responsibility. Another thought is that the [1] should be changed. I'm guessing that you are referring to the second byte of the received packet, and using it as a flags field of some kind. If that is the case, why not go ahead and make a struct that represents the header layout, and reference the flags through a bitfield in the struct? If you do that,
...which seems clearer to me.
You could continue along these lines, and get to a point where it would be obvious what the code does. The comment in that snippet also addressed "why", in addition to "how". I would move the "why" explanation into a document that defines the protocol.
• by Jeremi (14640) on Tuesday April 26, 2005 @11:51PM (#12355617) Homepage
x = x++; // add one to x
is obviously not useful
Not only is the comment not useful, it's factually incorrect... the behaviour of that line of code is undefined [].
• by Joey7F (307495) on Tuesday April 26, 2005 @09:13PM (#12354439) Homepage Journal
Just go ahead and use the long names, that alone will reduce documentation/commenting
• Nah (Score:5, Funny)
by Anonymous Coward on Tuesday April 26, 2005 @09:23PM (#12354531)
Variable names should be proportional to the size of their scope within the code.
• by p3d0 (42270) on Tuesday April 26, 2005 @09:45PM (#12354701)
This same point arises every frigging time someone talks about comments. As I say every time: A variable name can only tell you what the variable is, not why you chose to make it that way.
Now with that out of the way, here's my philosophy on variable names... Every variable name should be as long as necessary to describe what the variable is. Having said that, the shorter, the better. If you have a lot of long variable names, then you probably have not found the most elegant solution to your problem.
• by Rollie Hawk (831376) on Tuesday April 26, 2005 @09:14PM (#12354441) Homepage
Has he read the ones here?
• Haven't you ever heard the maxim, "Debug code, not comments". It reflects the fundamental problem with comments.
The problem with comments is there is no constraint to keep them...
• Current
• Correct
• Meaningful
• Unambiguous.
Code Reviews help, but not enough.
There is a better way. It is called Test Driven Development. Tests are the least ambiguous, most correct and current commentary on the code. They are also always at the correct (higher) semantic level than just paraphrasing the code.
Given a choi
• by eobanb (823187) on Tuesday April 26, 2005 @09:14PM (#12354442) Homepage
...where comments are more important than articles.
• by LogicX (8327) * <[su.xcigol] [ta] [todhsals]> on Tuesday April 26, 2005 @09:14PM (#12354444) Homepage Journal
How about Perl's POD Documentation? I do a lot of hacking of Matt Simerson []'s Mail::Toaster [] and Nictool [] projects, and I find that the Perl POD Documentation system, combined with well-named variables is easy on the eyes, and leads to it being well interpreted by an outsider.
• Comment. (Score:3, Insightful)
by stealth.c (724419) on Tuesday April 26, 2005 @09:16PM (#12354454)
Always comment. Always. I'm not a career coder, but I've done enough to know that if it's a project of any noticeable size at all, and you intend to read the code later, take the extra few seconds and COMMENT THE STUPID THING. It makes life a lot easier. The only thing that probably doesn't need commenting is a simple BAT file or a shell script. Comment!
• by John Seminal (698722) on Tuesday April 26, 2005 @09:18PM (#12354477) Journal
This works for code I write that nobody else will ever maintain. Even then I can get tripped up, I'll have to lean back in my chair and try to remember what I was thinking when I wrote the code.
But if you write code you're getting paid for, or code for an organization, anything but personal stuff, write good comments. Variable names might give a good idea about what data the variable holds, but it does not tell us much about how it is used.
When I took my first programming class, the most frustrating part was the documentation. I thought it was retarded and stupid and a waste of time. But now I realize it is very important once you write something more significant than "Hello World".
• ..and try to remember what I was thinking when I wrote the code
People sometimes forget that you are not only documenting WHAT the code is doing, but WHY you are doing it that way. You can usually figure out the what from the code itself, but trying to figure out the why can be next to impossible.
I wrote some code years ago to calculate (Australian) Capital Gains Tax liabilities. I knew it would be a hassle to update the code when the tax laws were inevitably reviewed and updated, so I went out of my wa
• by kwoo (641864) <kjwcode&gmail,com> on Tuesday April 26, 2005 @09:19PM (#12354488) Homepage Journal
In my opinion, comments are useful -- but literate programming is where it's at if you're looking for the best way to document your code.
Knuth did a lot of work in the area -- if I remember correctly, all of the sources to TeX are written in a style understood by the "web" literate programming tool.
There was also a good article by one of the Perl folks (Nathan Torkington? M.J. Dominus? Chromatic? I can't remember.) on POD, and how although POD wasn't literate programming, it was still useful. That article was great in that it showed a middle ground that may be more palatable to your non-LP-fanatic programmer.
That being said, I prefer full-on LP for large projects.
• Something i notice (Score:5, Insightful)
by bmajik (96670) <> on Tuesday April 26, 2005 @09:19PM (#12354490) Homepage Journal
i do a lot of code reviews at work and nothing makes me happier than good comments.
but just putting a bunch of blocks of comments that are like "get customer", "build record", etc are basically useless. If you use programming by intent then its more or less obvious that the code
is going to do something with a database and get a customer. a comment telling me that is useless.
what i like to do is write a few paragraphs of text at the top of a function. it explains my general approach, why im doing certain things the way i am, why im not doing other things, and why the function even exists.
essentially the comments should be enough that anyone that knew the problem space ought to be able to read them and come up with more or less a similar implementation.
then, in the body of the method anytime i do something that i feel isn't completely obvious, i put a 1 or 2 liner infront of, i.e. "we have to do this because zergs are sensitive"
the end result of all of this is that code reviews can see what you were thinking at the time the code was written.. and you're documenting assumptions about the problem, the apis, your understanding of the language, etc, all right in the code. it makes finding errors pretty easy.. someone that can't even read source code can read the comments and get an idea of the correctness of your approach.
• Joel on Software (Score:5, Interesting)
by TrueJim (107565) on Tuesday April 26, 2005 @09:24PM (#12354535) Homepage
It's harder to read code than to write it."
From Joel on Software 69.html []
Always comment.
• by X-rated Ouroboros (526150) on Tuesday April 26, 2005 @09:24PM (#12354543) Homepage
If the comments are clear, a better programmer than you can come along later and say "What the hell was this guy doing?" and then replace your lines of fumbling crap with much cleaner/clearer code.
It's the difference between being able to see what you were trying to do vs. figuring out what you actually did.
Call it "Intent Oriented Programming" if you want.
• They are (Score:4, Insightful)
by MarkusQ (450076) on Tuesday April 26, 2005 @09:25PM (#12354549) Journal
Comments are a maintenence nightmare. They get out of sync with the code, and (especially when the code is bad to begin with) people read them instead of the code.
That means over time the human's understanding of what the program does starts to diverge from the computer's understanding.
This is not good.
If something is too hard to understand the way it is written without comments, it should be rewritten. You will save time in the long run, trust me.
Remember the old adage: Don't get suckered in by the comments--the bug is in the code.
• Uhh... (Score:5, Funny)
by PrimeWaveZ (513534) on Tuesday April 26, 2005 @09:26PM (#12354557)
UsngShrtCmtsIsOftNotEnghAsOneMyNdToReWrtShtInTheFt r.
• by jjoyce (4103) on Tuesday April 26, 2005 @09:26PM (#12354559)
The problem is usually functions that are too long and are not orthogonal. Write short, orthogonal functions and you'll see your need for heavy commenting go away along with the need for long variable names.
• Raskin (Score:5, Informative)
by kebes (861706) on Tuesday April 26, 2005 @09:27PM (#12354575) Journal
For those who don't know (which apparently includes whoever is in charge of the linked article), Jef Raskin passed away this february. You can view the official press release, [] or read more about [] his contributions to computer science. I don't know when the article was written, but it seems it should mention that Raskin has passed away. In any case, his advice about commenting is good, just as his advice on user-interface design has always been lucid and helpful.
• Indeed (Score:5, Funny)
by screwballicus (313964) on Tuesday April 26, 2005 @09:29PM (#12354591)
Imagine my relief when I came upon a helpful comment:
What the hell was I thinking when I wrote this?
Comments save the day once again.
• by Bob9113 (14996) on Tuesday April 26, 2005 @09:38PM (#12354659) Homepage
I found the following in some production code, which quickly and concisely demonstrates why many comments are highly questionable:
/** Always returns true. */
public boolean isMagilla() {
return false;
That's the core problem with many comments, but it can be avoided. Comments are good when they state the intent or business case for a block of code, acting as a guide to the meaning of a block for subsequent developers. They are bad when they profess to know the actual outcome or implementation of a block; only the code itself can accurately reflect the state of the code.
The above would be far more useful like this:
/** Tells whether this instance meets the magilla criteria. */
public boolean isMagilla() {
return false; // currently, no MyClass meets the magilla criteria.
Now the intent of the method is clear, and anyone coming along who wonders why it's hard coded will know under what circumstances they should change it to a formula (namely, if MyClass becomes capable of meeting the Magilla criteria).
Comments can be good, but they should always be a guidepost to the intent of a block of code, and not attempt to explain how the code achieves that goal.
• by bunratty (545641) on Tuesday April 26, 2005 @09:43PM (#12354692)
My basic rule of thumb about comments is to comment the interface. In C/C++ this would mean writing the comments at the beginning of all function declarations in .h files instead of .c files, and in Java would mean writing Javadoc comments at the beginning of all methods. The idea is that if you know what a function/method does, you should be able to understand how it does it. The inside of functions/methods should have minimal comments, usually just a one-line comment for each major section explaining what that section does. Sometimes how the function/method works needs to be commented, as in:
void sort(int arr[]); // sort arr, ascending
// Uses quicksort
void sort(int arr[]) { ... }
Of course, there are always exceptions. When I was writing low-level code that manipulated hardware registers, I wrote a multi-line comment before each line of bit-fiddling code, complete with what the code did and a cross-reference into the hardware manual. Something like:
// Set the serial port to big-endian mode
// See SlurpSCC manual page 3-5
bitset(slurp->serial, bit(13));
• by iplayfast (166447) on Tuesday April 26, 2005 @09:46PM (#12354717)
I find that for simple code that requires little thought, little in the way of comments is required. Sometimes comments just get in the way as the code progresses over time, and the comments don't. It's better to code with long descriptive variable names, and structures then depend on comments.
For example which would you rather read.
#define AgentNameLen 41
#define CustomerNameLen 51
struct CustomerID
int CustomerID;
char AgentName[AgentNameLen];
char CustomerName[CustomerNameLen];
struct CoID //Customer ID structure
int ID; // ID number for the customer
char AName[41]; // Agent Name
char CName[51]; // Customer Name
To me the first is much more clear, and throughout the code will be obvious as to it's use. The second is mildly clear, but will degrade as new things get added to the structure.
The first also allows things like
for(int i=0;iAgentNameLen;i++)
which makes it very clear that you are iterating throught the agent name.
Algorithm's should be documented, as much as possible, at the top of the function, and any function that takes more then a screen should be looked at to see if it can be broken up into smaller functions.
(Almost) Always code towards maintainability, never speed. Usually it pays off within the year.
So to sum up. In order of importance:
80% code clearly with an eye towards maintainablity
20% comment
• by wowbagger (69688) on Tuesday April 26, 2005 @09:59PM (#12354794) Homepage Journal
There are three things the poor maintainance programmer who has to maintain your code needs to know: The What, the Why, and the How.
The What: What is this code trying to do. This is where your design documents come in, as they tell what the overall goal of the program is.
The Why: Why are you doing what you are doing? Why are you writing to this hardware register twice? Why do you divide this value by 1800 here? This is where comments are needed - to explain that the hardware sometimes doesn't take the value on the read, or that the nominal deviation of this signal is 1800 Hz.
The How: How are you getting the work done. This is where well written code with well chosen variable names comes in.
• Quality not quantity (Score:5, Interesting)
by MagikSlinger (259969) on Tuesday April 26, 2005 @10:03PM (#12354828) Homepage Journal
Having RTFA, I can see what he's trying to get at, but as someone who has (unfortunately) found himself spending most of my 10 year career in programming cleaning up other people's poop. At first I thought it was because I must have done something wrong that I kept ending up being assigned this work, but as I came to realise, it was because I make the code better than I found it and I have a knack for fixing stuff other people give up on. I also had silly managers who assign work to the people least qualified to do it.
At any rate, some observations:
1. 20 lines of comments "documenting" your code before you write it (or even after you write it) is far less useful than writing the code COHERENTLY and CORRECTLY in the first place.
Last month, I had a 1200 (yes 1,200) line method with huge blocks of documentation before big pieces of code. I still can't quite tell you what it thought it was doing. The code was a for loop wrapping around code to handle 3 different and mutually exclusive situations. Instead of identifying which of the 3 situations it was and creating a method for each situation, the person just stuffed it all in with lots of comments documenting everything the article's author said. The code was still unmaintainable.
2. Comments are useless unless they are kept up to date
Part of the reason that code was so difficult to figure out was because most of those big verbose documentation comments referred to a completely different implementation. After the programmer had written the first case, she encountered some other bad cases and eventually had to completely change a block of code embedded in this 1200 line for-loop. The code was now correct, but the comments no longer had anything to do with that block of code.
3. Don't be clever when you can be clear
I have made a solemn vow to hunt down and hurt anyone who puts "clever" code in my project. I am so sick of trying to figure out what some obfuscated piece of code in C, C++ or even SmallTalk is doing. And find out it was just a "clever" way of doing something pretty straight forward like iterating over a list. There was no speed gain from the clever trick, and the code wasn't even a bottleneck to begin with. *sigh*
4. If you don't know how to solve the problem, write some experiment code in a separate app to figure it out. Then take the time to do the "right" thing in the production code.
3 days from final for a video game. The CD streaming library for the Sony Playstation was making this strange "hic-up" sound at rare moments. By this time, the original author of this code has long since gone to another company. So I plunge into the code and found that the original programmer didn't know how to write streaming code so he created this hack of a hack of a hack of a test (ad nauseum). The code was programmed by accident, not design. No amount of comment before coding could help this. If the author had dumped the code, wrote documentation describing everything he learned then wrote the code, things would have been a lot better.
5. Unrelated to comments, but use variable names that make sense. Don't name them arbitrarily or to amuse yourself!
That CD sound streamer code I mentioned above used quirky names for variables. Can you tell what "little_ninja" is supposed to be just from the name? When I confronted the coder about this quirk of his (in another library he wrote), he got all huffy and didn't understand why people didn't appreciate his little puzzles or his sense of humor. It galls me he still earns a paycheck in the industry.
• by francisew (611090) on Tuesday April 26, 2005 @10:03PM (#12354838) Homepage
I kid you not, this is real code my supervisor writes.
if(preproct(7) >=thres), for j=size(at,2),at(:,j)=at(:,j)-maa;,for i=1:size(at,1),at(i,j)=(at(i,j))/(stda(i)+.0000000 01);,end,end,end
Note that this is matlab code, where commas are both an end-of-statement indicator (it's also possible to use just a semicolon), and an array index separator. The nice thing about this code, is that I can at least guess where most of the variable names come from. Oh, and there was *no* line break in the original code. Hooray for the avoidance of those wasteful '/n' characters?!?!
To answer the original poster: yes, comments are of vital importance.
Then again, if a program is structured right, things can be organized into sections, each of which is then commented, as opposed to a bunch of seemingly random lines with comments spaced throughout. Sometimes the layout of code, in conjunction with good variable names, is the best possible method of commenting it. The one thing comments are good for is to assure that someone not familiar with the particular language being used, will still understand the purpose of the code.
• Speak, Memory! (Score:3, Interesting)
by guet (525509) on Tuesday April 26, 2005 @10:07PM (#12354857)
program.files.each do | class |
class.methods.sort.each do | method |
5.times { method.refactor! unless method.elegant? }
if problem_domain > current_language
choices << comp.lang.each
current_language =
comments << intentions.remove(implimentation_details)
puts comments
def refactor!
method.split! unless method.size < Too_Big
method.rename! unless
def evangelise(lang)
puts "#{lang} is the only real language"
goto (1.0/0.0) and $beyond
• by Univac_1004 (643570) on Tuesday April 26, 2005 @10:14PM (#12354905) Journal
Raskin (who did not write code himself, but was more of a essayist) was a devote of "Literate Programming", first promulgated by Knuth: []
In "Literate Programming" the comments are all important and the code itself is trivialized. The code, as Jef told me, is like "raisons in the muffin of the comments." There are paragraphs of verbiage which might go on about the history of the project, why certain features were discarded, etc., etc, and might not even explain what the following line or two of code was concerned with.
It's really very difficult to deal with code that has been written in this style ("literatized?" ) since the actual structure of the program is severly obscured. It serves as an example of how overdoing a good thing is usually a bad thing.
Jef was a nice guy, and I recommend his book, "The Humane Interface" for its many interesting ideas. His attempt to put them into practice in the Archy project [], was not completed before his death. Even for that, Archy is very close to his vision.
But since Jef was in many ways an extremist, one can demonstrate in Archy his pushing of his concepts to the limit resulted in the end fully maching his goals. Somewhat like "Literate Programming", in fact.
• by menace3society (768451) on Tuesday April 26, 2005 @10:18PM (#12354929)
There will always be loser "programmers" who write code without comments, or write code without useful comments, or modify usefully-commented code without modifying the comments. Everyone I've seen who's ever put up an example of how code is self-documenting and that comments are just extra text saying the same thing have fallen into the second category. Of course things like "prints a message" or "check to see if i==0" are stupid comments. But that's just a straw man if you want to say that all comments are useless (I dare these people to read uncommented assembly of more than about 40 lines, and tell me the code documents itself).
The true useful skill lies in reading sloppy and/or wizardly code. Some people think that they have job security if they write impenetrable code, but then they can just be fired and all their code rewritten. If you can read others' "unmaintainable" code, you enable your employer to save money by not having to rewrite everything the guy they just fired wrote. So they'll want to keep you around as they fire/downsize everyone else. I It doesn't really matter what kind of code you write, since you can read whatever. advise everyone to start reading up on the Obfuscated C Contest, and practice figuring out what evertyhing does. Then you can handle any kind of code thrown at you, and the code you actually write becomes of secondary importance.
• by localman (111171) on Tuesday April 26, 2005 @10:21PM (#12354947) Homepage
Definitely not. No matter how clear your code is, it only gives you an idea of what is going on the small scale. You often need comments to describe your big-picture motivation. But the comments have to be good too.
Here's a classic bad comment:
// add one to the total
That's useless. But a comment can help:
// the total doesn't count the root node of the tree
// (since there's no actual entry in the DB for node "0"),
// so we just add that in here...
No matter how clear you write the code you're not going to get that much understanding without a comment.
It's just a simple example off the top of my head, but that's how I use comments. The code clarity is all that's needed to tell what you are doing, but comments can tell you why you are doing it. You need both for truly maintainable code.
• by amigabill (146897) on Tuesday April 26, 2005 @10:24PM (#12354975)
My learning experience about commenting code was a difficult one. Like many, while in college I wrote the code and then went back to comment it so the profs were happy.
Then I did a co-op with an automated storage/retreival systems company in their software department. One of the processes involved in a communications system needed some work. The code was licensed from another company in another country. There was no documentation for this communications system. There was very little commenting in the code. Luckily it wasn't in a foreign language. Unluckily it was wrong, apparently the structure of this program was similar to that of another, which was mostly gutted and rewritten, but a few old-program comments survived to be the ONLY comments in the new program.
Sure, the sources could be reverse engineered to provide the documentation required. I did it. It took a few weeks.
After that, I didn't leave comments for last anymore. It's been a good thing. I now work for a semiconductor design company and often write perl scripts or skill-language scripts to automate tedious tasks. I think I'm abou thte only one in the office that comments such scripts in any way. It's nice to read what stuff does when I have to revisit code many months or years later. I hate having to revisit someone else's code because it's nearly guaranteed to be completely barren of anything human-readable.
Listen up kids! Commenting is GOOD! Your professors aren't just being jerks. Learn the easy way and hopefully save yourself a great deal of trouble with your own code. Other people's code will always suck, but your own shouldn't have to.
• by os2fan (254461) on Tuesday April 26, 2005 @11:25PM (#12355452) Homepage
I have written programs in both raw and literate programming style, i prefer the latter. In fact, i wrote a literate-program pre-processor to write programs, and it made the program easier to write and more bug-free.
In literate programming, you rely on a pre-processor to make the output production, so you are free to put things together as you want. What this means, is that bracketing code (eg open, close files), can be written in the same block, which are invoked separately.
The main program then ends up looking like a rough scetch, full of commands to include other bits. With wing comments, it is easy to see what is going on.
/* A rexx script */
!inc rexxsets ; standard settings
!inc cmdopts ; process command line options
!inc fileopen ; open files
!inc mainprog; main program
!inc closefile ; close files
!inc subs ; subroutines
One uses a folding editor to search for strings like "!topic". This will not show you a consolidated index, but you can use it to also show where you're are, and any missing bugs.
On the main, Jon Bentley's comments on Literate Programming are fair (that is, it creates a good environment for writing single-purpose code), but one needs to consider the context the program is written for.
The form i use was specifically designed to allow all sorts of text-output, so the same file can make as output, eg .CMD, .REX, .TXT and .HTM output, which means that when you run the script you get a perfectly matched set of files, all correctly pointing to each other.
• Actual code block (Score:4, Interesting)
by GrouchoMarx (153170) on Wednesday April 27, 2005 @12:08AM (#12355754) Homepage
I love telling this story...
Last year I had a brief stint at a small software company that had just taken a project in-house that was developed by an outside contractor. My job was to take the code they'd just inherited (which no one there knew anything about) and add some features to it on a tight schedule. Documentation? What documentation?
The extent of all of the code comments it had was the following (and no, I'm not making this up):
if(...) {
break; // break
If that wasn't bad enough, I knew the original developer personally. She was a former professor of mine and I'd worked for her company only a few months before she had taken that contract.
As someone who has had to deal with code with descriptive names and no comments or docs to go with them: If you write such code, may you rot in the lowest level of hell along with traitors, used car salesmen, and people who answer cell phones during movies.
• by Todd Knarr (15451) on Wednesday April 27, 2005 @12:42AM (#12355944) Homepage
I'd say you're right, comments are more important. Clearly-written code should make how it's doing things obvious, yes. Comments, though, should say what is being done and why it's being done the way it is.
| null | null | null | null | null | null | null | null | null |
https://icsemath.com/2016/12/24/class-8-chapter-30-triangles-exercise-30-b/ | 1,548,273,929,000,000,000 | text/html | crawl-data/CC-MAIN-2019-04/segments/1547584350539.86/warc/CC-MAIN-20190123193004-20190123215004-00029.warc.gz | 520,881,285 | 30,744 | Question 1: Find the value of $x$ in each of the following figures:
i) $\angle ABC=\angle BAC=35^{\circ} \ (since \ BD=DA)$
$\angle BDA = 180 - 35 - 35 = 110^{\circ}$
Similarly, $\angle ECA=\angle AEC=28^{\circ} (since EC=EA)$
$\angle AEC=180-28-28=124^{\circ}$
$x+ \angle ADE+ \angle AED=180^o$
$x+(180-110)+(180-124)=180^o$
$\Rightarrow x=54^{\circ}$
ii) $\angle OBC= \angle OCB=40^{\circ} (Since \ OB=OC)$
$\angle BOC=180-40-40=100^o$
$\angle OCA= \angle OAC=30^{\circ} (Since \ OA=OC)$
$\angle AOC=180-30-30=120^{\circ}$
We know $\angle BOC+ \angle COA+ \angle AOB=360^o$
$\Rightarrow \angle AOB=(360-100-120)=140^{\circ}$
$\Rightarrow 2x+140=180^o$
$\Rightarrow x=20^{\circ}$
$\\$
Question 2: State giving reasons, whether it is possible to construct a triangle or not with sides of lengths:
(i) $3 \ cm, \ 4 \ cm,\ 7 \ cm$
(ii) $9 \ cm, \ 8 \ cm,\ 16 \ cm$
(iii) $7 \ cm, \ 9.2 \ cm, \ 6.7 \ cm$
(iv) $3 \ cm, \ 6.2 \ cm, \ 10.8 \ cm$
Answer: (Note: The sum of any two sides of a triangle is always greater than the third side)
i) Not Possible to construct the triangle because (3+4 not greater than 7
ii) Possible to construct the triangle as sum of any two sides is greater then the third side.
iii) Possible to construct the triangle as sum of any two sides is greater then the third side.
iv) Not possible to construct a triangle as sum of $4.3+6.2<10.8$
$\\$
Question 3: In $\Delta ABC, \angle B=60^{\circ} \ and\ \angle C=45^{\circ}$. Find $\angle A$. Name i) the largest side of $\Delta ABC$, ii) the smallest side of $\Delta ABC$ iii) write the sides of $\Delta ABC$ in ascending order of their lengths.
$\angle A=180-60-45=75^{\circ}$ . Therefore
i) the largest side of $\Delta ABC=BC$
ii) the smallest side of $\Delta ABC=AB$
iii) $AB < AC < AB$
$\\$
Question 4: In $\Delta XYZ, \angle X=53^{\circ} \ and\ \angle Y=67^{\circ}$ . Name i) the smallest side of $\Delta XYZ$, ii) the largest side of $\Delta XYZ$ iii) write the sides of $\Delta XYZ$ in ascending order of their lengths.
$\angle Z=180-53-67=60^{\circ}$ . Therefore
i) the largest side of $\Delta XYZ=XZ$
ii) the smallest side of $\Delta XYZ=YZ$
iii) $YZ < XY
$\\$
Question 5: In $\Delta PQR, \angle P= \angle R=64^{\circ}$ . Name the smallest side and the equal sides.
$\angle Q=180-64-64=52^{\circ}$
Smallest side $=PR$
Equal Sides $=QR \ \&\ PQ$
$\\$
Question 6: In $\Delta ABC, \angle A= \angle B \ and\ \angle C=100^{\circ}$. Name the largest side and the equal sides.
$x+x+100=180^o (let \ \angle A= \angle B=x)$
$\Rightarrow x=40^{\circ}$
Largest side $=AB$, Equal Sides $=AC \ \&\ BC$
$\\$
Question 7: In $\Delta DEF, \angle D \colon \angle E \colon \angle F=7 \colon 3\colon 5$. Name the smallest side and the largest sides.
$7x+3x+5x=180 \Rightarrow x=12$
$\angle D=84^{\circ} , \angle E=36^{\circ} \ and\ \angle F=60^{\circ}$
Largest side $=EF$, Smallest Side $=DF$
$\\$
Question 8: In the adjoining figure, $AC=DC, \angle CAD=50^{\circ} \ and\ \angle BAD=23^{\circ}$. Find $\angle ADC, \angle ACD, \angle ADB \ and\ \angle ABD$. Also show that $AD>AC, \ AD>BD \ and\ AB>BC$
$AD=AD \Rightarrow \angle ADC= \angle DAC=50^{\circ}$
$\angle ACD=180-50-50=80^{\circ}$
Therefore $\angle ABD=180-50=130^{\circ}$
$\Rightarrow \angle ABD=180-130-23=27^{\circ}$
Since $\angle ACD> \angle ADC \Rightarrow AD>AC$
Since $\angle ABD> \angle BAD \Rightarrow AD>BD$
Since $\angle ACB> \angle BAC\Rightarrow AB>BC$
$\\$
Question 9: In the adjoining figure, $BA \parallel CD \ and\ AB=BC$. If $\angle BAC=56^{\circ} \ and \ \angle COD=85^{\circ}$, find the values of $x, \ y \ and \ z$.
Since $AB=BC, \angle BAC= \angle ACB=56^{\circ}$
$x+y=180-112=68^o ...i)$
$\angle AOB= \angle DOC=85^{\circ}$ (opposite angles)
Therefore $x+56+85=180^o \Rightarrow x=39^o$
Which implies that $y=68-39=29^o$
Becasue $AB \parallel CD, \angle ABC=\angle DCX=x+y=39+29=68^{\circ}$
$\\$
Question 10: In the adjoining figure, $BA \parallel CD, BA=BC and CE=DE$. If $\angle ABC=66^{\circ} , \angle DCE=48^{\circ} \ and\ \angle DAC=63^{\circ}$ , find the value of $x, \ y \ and\ z$.
$\angle BAC= \angle BCA=y$
$2y+66=180 \Rightarrow y=57^{\circ}$
$\angle ECD= \angle EDC=48^o$
$x+48+48=180 \Rightarrow x=84^{\circ}$
Since $BA \parallel CD, \angle BAC= \angle ACD=y=57^o$
Therefore $63+57+z=180^o \Rightarrow z=60^{\circ}$
$\\$
Question 11: In the adjoining figure, $AD=BD=CD$ and $\angle CAD=40^{\circ}$. Show that $AC>AD, AB>AD, AC>AB$.
Since $AD=CD, \angle DAC= \angle DCA=40^{\circ}$
$\angle ADB=180-(180-40-40)=80^{\circ}$
Since $AD=BD, \angle DAB= \angle DBA$
$\angle ADC=100 \Rightarrow \angle ADB=80^{\circ}$
$\angle DAB= \angle DBA=50^{\circ}$
Since$\angle ADC> \angle ACD \Rightarrow AC>AD$
Since $\angle ADB> \angle ABD \Rightarrow AB>AD$
Since $\angle ABD> \angle ACB\Rightarrow AC>AB$
$\\$
Question 12: If $D$ is a point on side $BC of \Delta ABC$, prove that: $i) \ AB+BD>AD \ ii)\ AC+CD>AD \ iii)\ AB+BC+AC>2AD$
In $\Delta ABD$, because it is a valid triangle, $AB+BD>AD$
In $\Delta ADC$, because it is a valid triangle, $AC+CD>AD$
In $\Delta ABC$, because it is a valid triangle,
Add the above two relations
$AB+(BD+CD)+AC>2 AD$
$\Rightarrow AB+BC+AC>2AD$
$\\$
Question 13: In the adjoing figure, $ABCD$ is a quadrilateral. Prove that: $i)AD+DC>AC \\ ii)AB+BC>AC \\ iii) AB+BC+AD+DC>2AC \\ iv) AD+AB+BC>CD$
In $\Delta ADC, AD+DC>AC$
In $\Delta ABC AB+BC>AC$
Adding the above two expresions we get
$AB+BC+AD+DC>2AC$
$AB+BC>AC and AD+AC>CD$. Add these two expressions we get
$AD+AB+BC>CD$ | 2,047 | 5,577 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 120, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.5 | 4 | CC-MAIN-2019-04 | latest | en | 0.595048 |
https://api-project-1022638073839.appspot.com/questions/how-do-you-find-the-derivative-of-x-2-1 | 1,723,698,718,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641151918.94/warc/CC-MAIN-20240815044119-20240815074119-00870.warc.gz | 79,757,925 | 5,902 | How do you find the derivative of x^2+1?
May 28, 2015
Simply using the power rule $\left({x}^{a}\right) ' = a {x}^{a - 1}$:
$\left({x}^{2} - 1\right) ' = \left({x}^{2}\right) ' + 1 ' = 2 x + 0 = 2 x$
Apr 1, 2018
$2 x$
Explanation:
Using the power rule:
$f \left(x\right) = {x}^{n} \to f ' \left(x\right) = n \times {x}^{n - 1}$
$\frac{d}{\mathrm{dx}} \left({x}^{2} + 1\right)$
$\to 2 {x}^{1}$
$\to 2 x$
Anything without an $x$ will always give an answer of $0$ | 207 | 471 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 9, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.28125 | 4 | CC-MAIN-2024-33 | latest | en | 0.59938 |
http://math.stackexchange.com/questions/147199/a-maximal-ideal-among-those-avoiding-a-multiplicative-set-is-prime | 1,469,682,220,000,000,000 | text/html | crawl-data/CC-MAIN-2016-30/segments/1469257827791.21/warc/CC-MAIN-20160723071027-00159-ip-10-185-27-174.ec2.internal.warc.gz | 164,780,902 | 19,608 | # A maximal ideal among those avoiding a multiplicative set is prime
1. Let $S$ be a multiplicatively closed subset of a ring $R$, and let $I$ be an ideal of $R$ which is maximal among ideals disjoint from $S$. Show that $I$ is prime.
2. If $R$ is an integral domain, explain briefly how one may construct a field $F$ together with a ring homomorphism $R\to F$.
3. Deduce that if $R$ is an arbitrary ring, $I$ an ideal of $R$, and $S$ a multiplicatively closed subset disjoint from $I$, then there exists a ring homomorphism $f\colon R\to F$, where $F$ is a field, such that $f(x)=0$ for all $x\in I$ and $f(y)\neq 0$ for all $y\in S$.
[You may assume that if $T$ is a multiplicatively closed subset of a ring, and $0\notin T$, then there exists an ideal which is maximal among ideals disjoint from $T$.]
Here is a question I need to answer. For the first part I can show if $x, y$ are not in $I$ then nor does $xy$. The second part is just the field of fractions.
For the third part, I think I need to find an ideal $J$ containing $I$ such $J$ is prime, so that $R/J$ is an integral domain, and use the second part. To find $J$ prime, I think as suggested by the hint, I should go for a maximal ideal disjoint from $S$ containing $I$, but how can I do that?
-
Wait, please slow down. You just asked a question that turned out to be trivially false, and some people there (including me!) have been putting some effort into trying to figure out what you meant to ask. Could you please address this before moving on to a new question? – Pete L. Clark May 20 '12 at 0:12
In fact, the first question you ask is answered in a link I gave in a comment to your last question. This link was provided based on educated guesswork by @Benjamin Lim and myself on what you might have meant. So again, please slow down. Also, you have reproduced verbatim what looks to be a portion of a problem set. Is this homework of some sort? What is the source you are quoting from? – Pete L. Clark May 20 '12 at 0:16
@PeteL.Clark This is from a set of practice questions given to us for exam preparation. – Montez May 20 '12 at 0:18
@Zhang: thanks for the clarification. I don't really feel comfortable helping you solve practice problems for an exam. I don't find it ethically problematic per se, but I strongly recommend that you discuss them with your instructor and/or your classmates first. If you cannot do the practice problems, that's useful information for the instructor to have. Besides, getting other people to solve your practice problems may not be such good practice. – Pete L. Clark May 20 '12 at 0:23
@BenjaminLim Yes. If $x, y$ are outside $I$ then by maximality, $(x, I), (y, I)$ both intersect $S$. So for some $r, s\in R$, $i, j\in I$, $xr+i, ys+j\in S$. So $(xr+i)( ys+j)\in S$, and on expanding one sees $(xy, I)$ intersects $S$, so $xy$ is outside $I$. Hence $I$ is prime. – Montez May 20 '12 at 0:24
Let us slow down and first do $(1)$. By a standard application of Zorn's Lemma one can show that there is an ideal $P \supset I$ maximal subject to the condition that $P \cap S = \emptyset$.
Now the following steps lead to a solution: You want to show that for all $f,g \in R$ such that $f \notin P$, $g \notin P$ then $fg \notin P$.
1) If $f,g \notin P$ what can you say about the ideals $P + (f)$ and $P + (g)$? (Look at the maximality condition on $P$)
2) Recall that $S$ is a multiplicative set.
Conclude your result from here (also called Krull's Lemma).
Supplementary problem: Using Krull's Lemma prove that the set of zero - divisors in a ring is a union of prime ideals.
Now we come to actually proving $(3)$. By Krull's Lemma you know that you can find a prime ideal $P$ containing $I$ maximal with respect to the property that $P \cap S = \emptyset$. Now consider the following diagram
$$R \stackrel{h}{\longrightarrow} R/P \stackrel{g}{\longrightarrow} \textrm{Frac}(R/P)$$
where $h$ is a surjective map from $R$ onto $R/ P$ and $g$ is the canonical morphism from the integral domain $R/P$ into its fraction field. The canonical morphism is now injective because $R/P$ is an integral domain.
You should now be able to complete your problem by asking:
What is the kernel of the map $g \circ h$?
Is $I$ contained in the kernel of the map $g \circ h$?
Can you complete your problem from here?
-
Thank you. But I have not learnt any set theory yet, is there a way to tackle the question without Zorn's lemma? – Montez May 20 '12 at 0:34
@Zhang No I don't believe so. By the way you CANNOT do commutative algebra without knowing Zorn's Lemma. – user38268 May 20 '12 at 0:36
@Zhang Can you complete your problem based on what I've put out above? – user38268 May 20 '12 at 0:41 | 1,320 | 4,703 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.59375 | 4 | CC-MAIN-2016-30 | latest | en | 0.958668 |
https://www.jiskha.com/search?query=Critical+Thinking%2FMath&page=90 | 1,537,778,221,000,000,000 | text/html | crawl-data/CC-MAIN-2018-39/segments/1537267160233.82/warc/CC-MAIN-20180924070508-20180924090908-00203.warc.gz | 779,869,227 | 6,202 | # Critical Thinking/Math
146,623 results, page 90
1. ## math
7 yd - 3 yd 1 ft
-3(4-2g)-6
6-4 3/10=
4. ## math
what is 1 plus 1
5. ## math
23 + 5 × 10 ÷ 2 – 33 + (11 – 8)
40-n=28
7. ## Math
Does 5/8 + 5/6 = 1 11/24
8. ## Math
(15 - 2x)0, for x = -3
9. ## math
8-(1/x)-(6/x^2)
-3 + n/2 = 9
11. ## Math
Does 5/8 + 5/6 = 1 11/24
6+y=17
13. ## math
2( a + b ) x 4 1/2 (a + b ) =
14. ## Math
(X+1)/2 = 4 - (X+2)/7
41 3/4
16. ## Math
96 * .5 + 30 =
17. ## math
(16+ )+(14-7)=18+22
18. ## Math
If m2 = m1*d1/d2 what is d1 and d2 when m1 is 50 and m2 is 84
19. ## math
-(-4)-5=-1????? Or.......
20. ## Math
If -2x+3=7 and 3x+1=5+y, the value of y is: A. 1 B. 0 C.-10 D.10 ????
21. ## math
-3(x-1)+7+5(x-2)
22. ## math
x + 3 + 2X =X+3 can someone help with this?
23. ## Math
4 x (2 +7 +6 )
y = -x +b
70% of 72=
5-16/3=
7x/7x-21
3x+6
1/4 + 1/5
f(x)=2x-3
31. ## math
(X+2Y)+6X(X-2Y)+8X
32. ## math
ya but where did you get the 4x? SORRY!
33. ## math
(16+ )+(14-7)=18+22
1/2÷5/8
35. ## Math
What does ( ) mean
x^2+6x=24
7^x-9=4^2x
11 -3 1/2
39. ## math
3/4 k-1/6=1/2
40. ## math
-2x(-6)= - x - = + 12
41. ## math
((r+10)/r)+(r/(r+10))
2x+3y=9
43. ## math
[7-(-8)]|-6|=
44. ## math
7.8*10^5+3.9*10^6
–0.4 n ≤ 6.8
46. ## math
6y+12+(-8y+60)=30
15=4x+7
48. ## MATH 2
WHAT IS 2X+8=X+7
49. ## math
6-2x=5x-9x+14
50. ## Math
x + 3x^2 + x^2
8 < 5 + y
52. ## math
1 1/3 x 2 1/2
I need help
54. ## math
16 : x :: x : 25 what is the value of x.
X/1/3=1/2 X=
56. ## math
If 2p-4p-2r is what
57. ## math
what is the value of x if 4:x=24:28
(1 - 5 ) + 9
14=5x-6
-6=4x/7+2
61. ## Math
( 9 / 3 x 6 )
62. ## math
3y+6y^3-2y^2+5y^3+5y^2-4y+2
63. ## math
If xy+2x+3y+4=5, then y= A. 1-2x/3 B. 1-2x/x C. 1-2x/x+3 D. 1-3x/xy+3
64. ## Math
e(x + 1) + e(x - 4) = ? e(2x - 3) or 2e(2x - 3)?
65. ## math
3y+6y^3-2y^2+5y^3+5y^2-4y+2
x/2-x/3=5
67. ## math
7/10 = h - 4/5
-3/4-1/5
69. ## Math
( 3 - 4 - 1 )
3x+5y
( 6-6 ) +7
5/6- ?=1/3
73. ## math
( (-1)^5/(-2)^-3 )^2 THANKS!
74. ## math
y-7-1/3=y-7, 9 2 3
75. ## Math
If -T = 5, what is T ?
8-7b=-6b+9
12-x=5x+16
78. ## Math
(√2÷5)^8÷(√2÷5)^1/3
1/2x=2
-6+5b=6b-7
81. ## math
1(1/6) x 3/8=
82. ## Math
Does 1.3*10^0=1.3??
(-16)^-2/3
84. ## Math
y=3x+2
4 2/5x=12 1/10 A. 7 7/10 B.2 3/4 c.2 1/2 D.4/11
5m = ? km
n+12=4
5k+2=6
89. ## Math
need help on this one... 8^3 * 0^-1 Thanks...
90. ## Math
what is "a" in 2(a-5)-5=3
91. ## math
2 over 5m-14=9+3 over 5m
92. ## math
x + y = 4 2x - 3y = -2
93. ## math
f f + (-20) > -7 f = ?
9! = ?
20 - 0
96. ## Math
(2/3)^-2 + (-4)^-2
2+2
6-4(3/10)=
99. ## math
What is 1/3 of 12
100. ## math
20%is less than 32 | 1,551 | 2,659 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.859375 | 4 | CC-MAIN-2018-39 | latest | en | 0.288865 |
null | null | null | null | null | null | YOU ARE HERE: LAT HomeCollectionsBridges
View From the Bridge Helps Her Earn a Living
Jobs: D.J. Galloway may be the only bowling ball lookout in the world. It's a post she treasures.
PORTLAND, Ore. — Another 24 hours have rolled off into the sunset without a single bowling ball crashing from the Vista Avenue Viaduct. For yet another day, and by now about six whole months free of bowling ball droppings, the people of Portland owe their thanks to D.J. Galloway, brave and vigilant bowling ball patrolwoman.
Galloway spends eight hours a day, six days a week perched on the concrete bridge, keeping a lookout for errant bowlers. This is her job, a job for which she gets union wages, a job she calls "the best job in the whole damn state."
Below the bridge, along Southwest Jefferson Street, construction crews now work without worry of bridge-to-street bowling ball strikes. They're building the new light-rail tracks. Galloway used to work down below as a spotter, holding a stop sign when heavy equipment interrupted traffic.
But after the third bowling ball crashed through the complacency of the construction site, she got reassigned. She's been guarding the bridge since April; for at least another year, through winter and another hot summer, she'll continue her patrol.
Only the perpetrators of the first bowling ball dropping, two high school students, have been caught and punished for the destruction of a truck's cab. They were easy to catch: the ball had a name tag.
The next two droppings, which occurred after a newspaper article appeared about the first one, were clearly copycats. Fortunately, they were, as bowling ball droppings go, gutter balls and caused no significant damage.
Galloway, a 54-year-old divorced grandmother of five, arrives at the bridge by about 6:15 a.m.--sometimes earlier, just to keep would-be bowlers on their toes. She settles in to one of the four pedestrian alcoves built into the 1925 viaduct, arranging her cooler, coffee thermos, ashtray, radio and cushioned footstool on the bench. Short and plump, with a hard hat resting on her long, auburn hair and a walkie-talkie clipped to her belt, she hoists herself atop the stool, lights a cigarette, and goes to work.
A car passes and the driver honks and waves. Galloway stretches out her arm and waves vigorously. "Good morning!" A school bus goes by. Galloway waves and shouts, "Hi, kids!" She sees a woman walking her dog. She calls out to the dog, "Cookies! I've got cookies for you!" As the dog trots eagerly toward her, she pulls a huge handful of dog biscuits from a bag.
"The first week I hated it," Galloway recalls. "There wasn't nothing to do. But I just kept waving and talking to people. The second week I wouldn't give it up for anything."
By the second week she was a fixture of the posh West Hills neighborhood. A woman who lives a block away began bringing her hot lunch every day. She got acquainted with many of the elderly ladies who live in a nearby apartment building. And soon she knew all the walkers (and their dogs) by name.
The gift of gab and natural empathy she developed during 23 years as a bartender have served her well in her job as bowling ball patrolwoman. On the bridge that has long been known as Suicide Bridge and Lovers Leap, she has averted at least one suicide.
One day last spring she spotted a teenage boy nervously smoking and looking over the edge of the bridge. She tried to talk to him and he moved away. When she saw him starting to climb onto the side, she went into action.
"I said, 'I've got two sons. You can talk to me.' " Slowly she won him over and the boy was soon sitting at Galloway's side, pouring his heart out, telling how his girlfriend broke up with him on prom night, and how his teacher wouldn't let him graduate.
"I said, 'If you promise not to do anything, I'll call the superintendent of schools and get this straightened out.' He said, 'Ah, you don't know him.' I said, 'Sure I do. He and I are old drinking buddies.' "
Galloway did call the boy's principal and said, "What the hell kind of school have you got there!" Through her dogged efforts, she learned that a teacher had misplaced one of the boy's final exams. He was allowed to graduate and is now in college. Galloway calls his parents occasionally to learn how he's doing.
"That was my only close call," she says. "I'm glad I was here."
Galloway takes her job very seriously and would never consider reading a magazine or book while she's on duty. Her only distraction is the radio that plays old standards, such as "Hello Young Lovers," while she sings along. She insists she is never bored.
"I do my job and I make everybody happy," she says. "That's what counts."
Over the months she has observed the comings and goings of all the neighbors. She knows who's having landlord problems and who is entertaining out-of-town guests. She also knows who's getting married because she's been invited to the wedding.
By now, all the neighbors know why the pranks stopped as soon as Galloway arrived on the scene. Obviously, it was her beautiful smile that did the trick. It's enough to make anyone leave their bowling ball at home.
Los Angeles Times Articles | null | null | null | null | null | null | null | null | null |
null | null | null | null | null | null | Law & Disorder / Civilization & Discontents
The government leaves Google alone—what it means for users
Google's win is a win for consumers, too, despite its critics' complaints.
The biggest tech news this week is: Google is safe.
Of course, from a business perspective, Google never looked endangered in the first place. The company has close to 70 percent of the US search market, and pushes 90 percent in some parts of Europe; in mobile search, it utterly dominates, with around 98 percent. Google's market share in the US would surely be even higher but for all of the computers pre-installed with Microsoft software.
When a company gets that big, its biggest threat starts to come not from its competitors—who, at least in the Web search business, are few—but from governments, or big public-relations slip-ups. "We’ve always accepted that with success comes regulatory scrutiny" was Google's very diplomatic way of describing that predicament yesterday.
Now the FTC's antitrust investigation is over, with no charges brought, and some quite minor changes to Google's business practices. On the patent front, Google will make a sacrifice that's more substantial—but still doesn't outweigh the enormous benefits of being "in the clear" as far as the feds are concerned.
Is it a good thing that regulators won't be messing with Google? On balance, yes. Corporations that amass too much power are worth worrying about, and need to be watched—by governments, by non-profits, by journalists and by citizens at large. But at the end of the day, the insistence that Google wasn't playing fair wasn't coming from an enraged citizenry. It was coming from Google's rivals.
Google didn't make the Internet, and it doesn't own the Internet. But it has become a dominant force in navigating the net. Today, those tools of net-navigation touch everyone, like it or not. It's hard to imagine what the net would look like without them. Much of the war against Google that's getting pushed in courts and in Congress—by newspapers, by the entertainment industry, by traditional software makers—is, in reality, a proxy war. The real battle is an impossible fight against the disruption and change wrought by the Internet itself. That's not a fight worth cheering for, even for those deeply concerned with Google's power.
The antitrust attack that rivals pined for
Microsoft and other rivals had hopes for an assault from government regulators against Google—ideally, the same kind of massive antitrust litigation Microsoft had to endure back in the 1990s. Now, it's quite clear that such action isn't forthcoming in the US, and it's less likely to happen in Europe now, either.
As the regulatory probes against Google dragged on, it has become clear that in the antitrust area, they were sponsored by the company's chief corporate rivals—Microsoft especially. Almost no consumer or public interest groups expressed concerns about "search bias." The competitors' criticisms reached a new volume when Google completed its acquisition of ITA a few years ago, and a coalition called FairSearch was formed, which ultimately included Oracle and Microsoft.
When FTC Chairman Jon Leibowitz announced the end of his investigation yesterday, he seemed to be speaking to some of Google's competitors/critics directly. And he sounded ready to head them off at the pass.
"Some may believe that we should have done more," said Leibowitz yesterday. That may come out of a "mistaken belief that criticizing us will influence other jurisdictions," he added.
In other words, he knew the criticism of Google would keep coming, in hopes to get the EU to take a tougher line against the company. And that's exactly what happened.
FairSearch quickly called the FTC's decision to not charge Google "premature," and today put up a second blog post saying that it's the EU and state attorneys general who will get the final word.
Microsoft took a similar position."[T]here appears to be no reason, despite the FTC’s optimistic statements this morning, to believe that Google recognizes its responsibilities as an industry leader," wrote Deputy GC Dave Heiner in a blog post.
The only silver lining in the end of the antitrust investigation against Google, in his view? That the other investigations still aren't over. "In Europe Vice President Almunia has made clear that he will close his investigation of Google only with a formal, binding order that addresses search bias and other issues," wrote Heiner.
A patent "loss" that isn't too bad
On the patent front, Google did suffer a meaningful setback. The company has agreed it won't use its big stash of standards-essential patents to get injunctions or "exclusion orders" in most cases. A hefty chunk of the $12.5 billion purchase price of Motorola was understood to be for its 17,000 patents, which it could use to counter-sue against attacks that Microsoft and Apple brought against the company, or its Android partners.
So was that purchase a bad deal now? Hardly.
First of all, no one should think the Moto purchase was all about patents. Google got much more than that: Motorola Mobility had manufacturing facilities, research facilities, more than 20,000 employees.
But it also got a giant stash of useful patents, the best of which are "standards essential" patents that have become controversial. Google did need those patents; and they're now somewhat encumbered because they can't be used to ask for injunctions. Microsoft and Apple, meanwhile, are flush with "feature" patents that aren't related to the international standards, so they're still in a position to try to kick products off the market.
That's not as big a setback as it may seem, though. First of all, Google and its partners can still use those patents to fight back in court. They can ask for significant damages based on such patents. Injunctions aren't going to be the end-all of the patents wars; Apple didn't get one in its court battle against Samsung, but it's still fighting and Samsung has the potential to suffer significant setbacks from that litigation. And the Moto patent portfolio is a big moneymaker in its own right, which has already been licensed to parties that aren't interested in fighting all the time.
The clearest sign that Google didn't get hit hard here? It's that arch-rival Microsoft is so unhappy about it. "The FTC’s proposed consent decree on patents runs for 13 pages, most of which spell out exceptions," the company notes in its responsive blog post. And Google can still threaten an injunction if it believes the patents being used against it are standards-essential themselves, even if its opponents disagree. That's often a murky line.
An unfortunate focus
Google's competitors have been lobbying for regulatory intervention for years now. In hindsight that focus is unfortunate, and probably misguided, as former FTC policy director David Balto noted in a recent article in The Hill. (Balto has done work for Google in the past.) Google's critics have switched up some of their theories and have tried turning to the DOJ, the other main antitrust regulatory agency; but at the end of the day, it really was an exhaustive investigation that, in terms of the evidence, never got past the first step.
"This came down to a simple proposition—are consumers harmed?" said Balto in an interview with Ars. "They never got past that first initial question." Even if consumer harm had been found, Google may have had defenses. It could have pushed evidence supporting its line that its industry is competitive, with the competition just "a click away," as the company has said before. Or it could have argued that search results are a kind of editorial product, protected under the First Amendment just like the front page article selection on the New York Times (which obeys its own mysterious, if more human, algorithms.) But it never got that far.
The government's investigation of search was exhaustive. It went on for 19 months, and involved millions of documents and countless discussions and interviews with Google executives. The FTC also spoke to Google's rivals, as well as small businesses that were concerned they were getting a raw deal from a leading search that favored its own services. (Some of those businesses continue to be highlighted by FairSearch.)
Maybe the government will have to take a close look at Google's algorithm again one day. But for now, the right regulators have taken a very close look—it has been one of the most comprehensive investigations in the history of the FTC—and the company has come up clean. Critics' continuing hope that Google will get snared in a brawl with the government is misguided.
"They'd rather try to win by hobbling Google through regulatory action rather than through inventing better products," said Balto. "Now Google will be able to innovate as much as they want, and price as aggressively as they want, without leading to an over-aggressive regulator." And that will be good for consumers.
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Science News
The Paradox of Choice 537
Posted by Hemos
from the so-many-options dept.
sproketboy writes "Psychology professor Barry Schwartz has written a book which is a must read by those wanting to get Linux on the Desktop. Dr. Schwartz examines the problem of too much choice in our society. Maybe Microsoft has it right after all? Here's a video interview with Dr. Schwartz, a review of the book from the New Yorker and more info from PBS." Of course, the choice issue applies to far more than desktop computers, but is still instructive in that area. Thanks to Stefan Hudson for a SciAm story that has more information.
The Paradox of Choice
Comments Filter:
• Freedom of Choice (Score:5, Insightful)
by The Queen (56621) on Monday March 29, 2004 @11:04AM (#8703770) Homepage
is what you got...
Freedom From Choice
Is what you want.
(Are we not men?)
• Re:Freedom of Choice (Score:5, Interesting)
by bobcat7677 (561727) on Monday March 29, 2004 @12:18PM (#8704634) Homepage
There are alot of people that seem to think that choice = freedom. While this is true at some levels, there are deeper issues to be considered. For an American example being that America was founded on the idea of freedom: Is a choice between Busy Kerry Nader really freedom? Sure you have a choice but you are limited to only 3 options. Is that really pure freedom or just a little bit of choice allowed you by a governmental system so you feel like you have freedom.
Going back to the current topic however, it seems like everyone is making this desktop choice issue way too political IMHO. It should be about what the needs are of the users. Isn't that what we as technology professionals are supposed to look to? (tech hippocratic oath if you will?) For some users, we want to limit their options because they don't have the knowledge/experience/brain capacity? to choose the correct option for what they are trying to do. Thank you M$ for aknowledging that. However, there are increasingly more people who DO have the knowledge/experience/... (especially with a whole generation of kids being brought up to use this stuff) that need to have the choices. If the future inventors, artists, and innovators have their tools dictated to them in nice neat little "luser" packages, then how much will that limit their ability to invent, create, and innovate? And how much will the corporation that controls all the tools become in control of the society in dictating who has the tools to do things and who doesn't. [Maybe a caste system of technology dictated by M$ on the horizon?]
-- my random thoughts
• Re:Freedom of Choice (Score:4, Interesting)
by cygonik (679205) <> on Monday March 29, 2004 @01:26PM (#8705525)
It seems relatively simple to me: Have the first choice be "How much choice do you want?". We have verbosity levels in logs, why not have a 'busyness' or 'user level' option for ui and configuration? With a few clicks I could change my interface from Default to Clean or Busy, and change my configuration screens from Beginner to Intermediate or Advanced. One could even have a default that is login, desktop, or system-wide. ----- If only Ben Franklin had known people would be using Linux, the choice between Freedom and Security would never have been an issue.
• by groomed (202061)
Another choice to choose the level of choice? What a nightmare. Can you imagine the documentation? "Select Tools->Options, unless your user level is Clean, in which case you must go to Control Panels->User level, except when you're in Beginner mode, in which case you have to log out and ..."
That's not making things easier. It's deferring responsibility.
• Re:Freedom of Choice (Score:5, Interesting)
by SillyNickName4me (760022) <> on Monday March 29, 2004 @01:48PM (#8705879) Homepage
It is limited freedom.
When the USA was founded, freedom was an important argument, but it should be seen in the settings of the late 18th century.
I have been reading a lot about the early days of the USA and the following is my recolelction of what I read about the discussions regarding the exact form of government that the USA got at the time.
In that time, there has been a lot of discussion in the USA and France about the different models of government without monarch.
There is a choice between a few systems there, and 2 of them were discussed a lot in detail:
- The republic of Sparta
- The democracy of Athens.
What they ended with is somethign that looks a lot more like the republic of Sparta then the democracy of Athens.
Bottomline, an elite is in charge of the country, however, this elte is elected.
This means that people cannot make direct choices in matters that concern the country as a whole, but they can appoint those who can make those choices.
At the time, people were afraid that the purely democratic way would result in chaos and unlimited individualism. The Spartan system didn't provide for the freedom that people demanded and was too much of a tirany.
In the end, it did end up folowing the Spartan model, but with an elected elite.
What this tells me is that the founders were actually looking for a way to limit individualism at least to the point where peopel would not act against the common good, and in the hope that peopel would contribute to the common good, while at the same time trying to maintain as much freedom as possible.
I believe it is a bit simplistic to say that Freedom is THE thing the USA was founded on, it was an important aspect, but in the end, balance to get a state that worked for as big a part of its citizens as possible, and finding the right balance between individualism and the common good were at least as important if not more important.
It seems to me that the way political parties function in the USA is pretty much a continuation of English tradition. A rather substantial part of the representative democracies in the world have more then 2 major parties, and do indeed need coalition governments. Few of those have the problems that we have seen for decades in Italy where a government wouldn't last more then a few months, in fact, Germany, The Netherlands and Belgium are 3 examples of countries with very stable governments while having many political parties.
A basicly 2 party system (with all respect for the man, I'll skip Nadar, untill some major change happens to how the US population percieves politics, I am afraid he has little chance whatsoever) makes life easy.
Political views can be put into a black/white perspective, and there is no need for cooperation since one side will end up beign in power while the other side will have to wait and watch untill the next elections (yeah yeah, I know it is a bit more complex then that due to the way the senate and congress work in the USA where you can have a republican president with a democratic congress for example).
The black/white choice makes it easy because people don't have to think too much about things of which they often don't see the direct relation to their daily life.
Most people want simple choices if any at all for things that they are not really interested in but want huge variety of choice once they are interested.
• by IronTek (153138) * on Monday March 29, 2004 @11:06AM (#8703783) Homepage
This would have been an informed post, but there was a link to a video of the guy discussing the paradox of choice, a link to the article about the book, and a link to an interview with the guy in the video who wrote the book that the article was about... I couldn't decide.
• by QEDog (610238) on Monday March 29, 2004 @11:06AM (#8703791)
Should I post in this story or in the other stories? What to do? what to do? argh! I'm going crazy!!!
• by -Surak- (31268) * on Monday March 29, 2004 @11:06AM (#8703795)
Looks like link this was going to be included in the article, but something got messed up. Sciam digital subscription required for the full article, unfortunately...
Scientific American: The Tyranny of Choice [ PSYCHOLOGY ] []
Logic suggests that having options allows people
to select precisely what makes them happiest. But, as studies show, abundant choice often makes for misery
• Like you, I didn't read the entire article since it requires a digital subscription.
There was a time when I would have chosen to read Scientific American, but now I choose not to.
Gee - choice. And I chose. And I'm happy with my choice.
That wasn't too difficult. I don't need a massage. OK, I do, but that's because I did a 60 mile bike ride this weekend and my legs are a bit cranky.
Personally, I think Professor Schwartz misses the cause and effect. These people who stress over too many choices could b
• by Chris_Stankowitz (612232) on Monday March 29, 2004 @11:07AM (#8703799)
Thats like saying ther is such a thing as getting laid too much.
Linux (and Pizza) is like a Blowjob, no matter how bad it is, its still pretty good!
• by Penguinisto (415985) on Monday March 29, 2004 @11:07AM (#8703805) Journal
Last Monday, Miguel de Icaza (at Novell's BRainshare here in Salt Lake City) mentioned Novell's push for the Linux desktop, and covered a lot of the same ground, but he presented it quite intelligently...
You can have a simple desktop that Joe Sixpack can play with, and at the same time set up a dialogue that allows the tweaker in some of us to have free reign over what each little widget and bit of desktop does.
I just don't get why it has to be such an "either or" choice here...
• I feel so justified. I was talking about this long ago. I think for (any distribution of) Linux to be really successful on the desktop market, they'll need to deliver at least two versions of the product. This can be chosen at installation time or literally be different products, but desktop success NEEDS to have a wizard/automatic driven desktop for people who just want the thing to work, and linux NEEDS to have the geeky engine exposed to be accepted by the experts.
The company that does this well and
• by Valar (167606)
I think it is even easier than that. For every configuration item, come up with a good guess. Then in the dialog to change it, present a simple list of the options which joe-user can be expected to understand (hm, do i want big icon or small icons?). Then, down in the corner, put a little checkbox labeled 'advanced'. If they click it, they get the 'tweaker' box.
• As for the "either/or" mentality, I guess it doesn't matter much. The important thing, I think, is to draw a line and decide which is part of the "easy to use" desktop and which is part of the "experts only" desktop. Then you can support one or the other without telling Grandma to open up her config file in vi and edit the daemon options to support more client connections.
There are plenty of "easy to use" packages for linux, but having 10 easy to use programs with 30 hard to use programs makes for a desk
• by Nurgled (63197)
As much as I hate to say it, the problem isn't "hiding" the choice, it's making the "desktop" of every linux system the same, so that users can transfer knowledge. This is true of MacOS, and it's true of Windows within a given version. (MS are less good at this.)
If a user sits down at a high school linux system and learns how to start his or her favourite word processor, then sits down at another system elsewhere with a different Window manager and desktop environment... what now?
Sure, they'll learn not t
• by Angst Badger (8636) on Monday March 29, 2004 @03:23PM (#8707009)
Though it really, really grieves me to say this, Apple got this right first, and Microsoft eventually learned from Apple and their own customer feedback. The only people who want a whole bunch of choice and configurability from computers are geeks like you and me who enjoy the computer as a thing-in-itself. Everyone else is just trying to use the computer to accomplish their jobs, a particular hobby, or something else where the machine is just a means to an end. And those people are where the money is. (There certainly isn't much money in folks like me who rejected the Macintosh because they preferred the joys of the Applesoft command prompt and 6502 machine language programming!)
It also means less expensive support if you don't have to train your support drones to answer questions about a million conceivable configuration possibilities.
This is no doubt what Novell is thinking. For all I know, the executives at Novell think free and open software is a great thing in and of itself, but at the end of the day, their jobs depend on making money, so reducing interface choice is an eminently rational route for them to choose.
Novell's efforts will go to whatever they decide is the "best" interface, period. If geeks like you and me want special feature X, we'll have to code it ourselves, because only we care. There is a sliding scale of preference for complexity in users, starting with zero for the general public and sliding all the way to infinity for Java development toolchains, in inverse proportion to the likelihood of profit.
This is, however, nothing new.
• annoying... (Score:5, Funny)
by spangineer (764167) on Monday March 29, 2004 @11:08AM (#8703815) Homepage
Tons of choices can be annoying - going to a restaurant and being forced to select from a huge list of foods can be overwhelming. Usually, all I end up doing is finding one thing I like and then ordering that all the time, without checking out other stuff. It's too much of a hassle to try out every choice that exists in the world.
Then again, if we didn't have as many choices, I might not be able to find one thing I like in the first place, and thus probably wouldn't go back to eat there - I'll choose to go somewhere else.
But if that choice was taken away, I'd have to eat something I didn't particularly like, which never killed anyone.
Morale of the story? Having too many choices is the real reason I'm a picky eater.
• Re:annoying... (Score:3, Insightful)
by EnderWiggnz (39214)
but at a restaurant, you get a menu of complete dishes, and not a list of ingredients.
imagine if the waiter came out and said:
well, the chef has some chicken, some salmon, a couple a nice cuts of beef... some vegetables, this list of spices, and some potatoes...
what would you like?
• Wow... (Score:2, Interesting)
by Anonymous Coward
Somebody really fubared their links...
But anyway, I know I enjoy my choices. I can choose linux or windows. I much prefer choice to no choice. Does anyone really believe that we are better off when we can't make decisions for ourselves?
Sure, it might be nice to be a little drone in the big hive... You don't have to put any effort into thinking for yourself, or expanding your mind, since the hive could really care less about your individuality. In fact, indivduality is discouraged.
I dunno, I think lin
• Re:Wow... (Score:3, Informative)
I actually saw this on TV yesterday (on the show "Daily Planet"), and it's clearly apparent you didn't see it at all.
Basically, the idea is not to take away all choices from you, but to limit the choices to a respectable number. This is NOT a new idea. Your choice of Windows or Linux is not a problem. The problem, as was explained, is like when you try to choose a cereal. You have over 200 different brands. It becomes overwhelming the number of choices. It's also like wine. Walk into a SAQ (Canadian
• by toupsie (88295) on Monday March 29, 2004 @11:08AM (#8703819) Homepage
Apple used to have a massive product line with dizzying list of model numbers. Not only did it confuse customers but it also brought down quality and delayed shipping of many of the models. Now you can just buy a notebook (iBook and PowerBook) and a desktop (eMac, iMac & PowerMac) from Apple. Sure you can supe up the basic model they sell but you are still buying a standardized item.
• I got a 20g iPod a couple of months ago and in some ways it's made walking with music a worse experience for me than before.
My first Walkmen (dating back to 1984) were cassettes, and while you could carry extra tapes, you were largely stuck with one or two and even then skipping around wasn't much of an option (no music search on a Walkman I could afford until the mid 1990s). So you listened to bands you really liked or spent a lot of time making a few mix tapes to guarantee you'd like most of the songs.
• by glassware (195317) on Monday March 29, 2004 @12:33PM (#8704800) Homepage Journal
On the other hand, Apple's software philosophy is to have only one way to do something, and to have that work well and be obvious. Check out the Macintosh Human Interface Design documents.
Even more importantly, this philosophy extended to the Macintosh API. Even Microsoft moved in this direction. Bill Gates once said, "Why should everyone in the world have to write a File-Open dialog?" The Microsoft Common Controls API was the best thing that happened to Win16 programmers back in the early '90s.
Yet, after a few years, Microsoft started putting together OLE, DDE, ActiveX, and a bunch more stuff - there were tons of choices. Consider Microsoft's media player: there was a text-based API, a procedure call API, and an object oriented API. Microsoft programming has been getting harder, thus they introduce .NET and standardize everything again.
I'm all for choice when it works. For example, KDE offers you tons of choices; by default there's this multiple-virtual-desktop thing with all sorts of options and shortcut keys and soforth. But the one choice I want - the ability to stop files and folders on my local harddrive from acting like hyperlinks - isn't available. I suppose that, given a few months of practice, I could get used to treating my hard drive like a website, but it isn't working out for me at the moment.
I dunno if I have a real point here. But I think Extreme Programming has at least one useful idea: customer stories. Try writing down all the things a user wants to do - "Map a Network Drive", "Change double-click behavior", "Organize My Documents" - and then put together an obvious way for the user to do it, or (if it's too hard to make it obvious) at least a straightforward help page that explains the task.
Am I rambling? Feel free to call me redundant.
• Nothing new (Score:5, Interesting)
by Anonymous Coward on Monday March 29, 2004 @11:11AM (#8703846)
The observations are a direct consequence of a well known usability heuristic called Hick's Law []. Hick's Law states (roughly) that the time an individual requires to make a decision increases with the number of alternatives available.
• Re:Nothing new (Score:5, Insightful)
by rev063 (591509) on Monday March 29, 2004 @01:21PM (#8705442) Homepage
Put another way, choice is an abdication of responsibility on the behalf of the programmer.
When an interface gives you dozens or hundreds of different choices, it's because the programmer (or designer) was lazy. Instead of trying to figure out -- in advance, or by context -- what options would be best for the user, the programmer throws his hands up in the air and says: "YOU figure it out, loser!".
There are SO MANY instances where programs insist on making you make irrelevant, useless choices. Just look at the typical installation program, for example. Like 95% of users, I don't CARE where the program is installed, what the application is named, or what skin I'd like the interface to use. I just want the damn thing installed -- and stop bothering me, dammit!
An interface with fewer options is easier to use, no doubt about it. An interface with fewer, well-selected options also makes a BETTER program.
• Good Title (Score:5, Interesting)
by Doesn't_Comment_Code (692510) on Monday March 29, 2004 @11:11AM (#8703850)
The Paradox of Choice
From the title, I thought this was going to be a deep mathematical or philosophical piece that I would have to give a lot of thought to.
I do agree with concept that we have too much choice in our society, or rather, we are deep in information overload. Too much choice is not a problem if you can quickly whittle down what you want and what you don't want. The problem is when the choices become confusing and ambiguous - and I think that has happened for the average individual. For instance, go into an applience store and say you want a tv, then hold on to your butts, because you're going to be there for a while. Then pretend you didn't know what all the fancy jargon stuff means (like the average consumer). If that wasn't bad enough, I think marketers actually inflate the problem on purpose, making it seem that there is more choice than there actually is - since that boosts the chances that a consumer will buy your product.
• Re:Good Title (Score:5, Insightful)
by BiggerIsBetter (682164) on Monday March 29, 2004 @11:49AM (#8704312)
They do this, and it's really bad in small countries; in some markets, you can go to every store in a town and find the same range of products, which have all come through the same two importers (two, so there's no monopoly, I guess).
But there's another insidious problem with "choice" - Most of the time, you aren't making one. I went to a local supermarket which is the only one open at 4 am, and there are signs saying "Thank you for choosing to shop at [supermarket name here]". The only choice I made ws to get food now or later, not to shop there. Or you get a Dell and it has Windows on it, and there's a little note saying "Thank you for choosing Microsoft" or similar. You didn't choose it, it just came with the computer whether you liked it or not.
People are so used to being told they are making choices when they plainly are not. When confronted with a real decision, it overwhelms them and they freak out and run back to their comfort zone. The paradox of choice is not that we have too much choice, it's that when given a real one most people don't want it anyway.
• by G4from128k (686170) on Monday March 29, 2004 @12:14PM (#8704582)
Very good points. I see the issue in terms of 4 factors:
Rising Cost of Decision Making: Excessive options and excessive information on each option drive up the cost of choice. The cost of decision making can easily exceed the marginal benefit of making the decison.
Psychological Risk of Decision Making: Some people are more comfortable without choice because it absolves them of responsibility. If you have only one choice, you get to bitch about it. If you have multiple choices and you chose incorrectly, you have only yourself to blame.
Cost of Competition: We seem to live in a competative, judgemental socitey in which people are judged by the choices they make. This increases the importance of every minor decision. Faced with a number of reasonably good options, people often spend too much time deciding. They feel compelled to do this because of the perceived social penalty of making the wrong choice. Nobody wants to pick the second-best option even if it is nearly as good as the #1 option.
Scale of Society: The bigger problem is the increasing scale of society. Many might think that have umpteen types of mustard, text editors, or cars is too much. But there is no unanimous agreement on which alternatives to remove.
This problem will only get worse. I would wager that in most industries, the number of economically viable choices scales with the log of the market-accessible population. With global trade and rising standards of living, we will only see more choices.
• "Too many choices, not enough voices" which I interpreted as too many people willing to settle for the norm and not enough people who will demand more, go out on a limb and challenge the status quo, which I think is the main reason that microsoft has its hold.
• by ajs (35943) <> on Monday March 29, 2004 @11:13AM (#8703887) Homepage Journal
Choice is a fine thing for now. Most of the world is still being introduced to desktop computing. It is not yet time to select the best technologies for any given application because we don't understand the application well enough yet.
Even something as "basic" as word processing has changed radically in the last 10 years as a wider variety of people have gained access to computers. The "outliers" in the sample set have, in some cases, become the majority of users.
Open source OSes are especially subject to this. Our systems are designed by those who have a combination of real-world-need and ability to implement. As time goes on that will be a broader and broader segment, and others will be brought in to implement for those who have the need, but not the ability (certainly already happened in some areas).
Give computing 20 or so more years to find its feet and it will be time to make hard decisions, but for now I think choice is a good thing.
Now, moving on to the officeplace (which is where most people think of desktop computing in terms of adoption strategies), I think it's key that OS vendors such as Red Hat, Mandrakesoft, SuSE/Novell and others produce a desktop with clear defaults and clear ways for admins to limit choices. This is important for large scale systems admin where you are maintaining 2,000 systems on people's desks. You need some uniformity in order to scale that support reasonably. This does NOT meant that choice should not be available, but that it should be available to the admins who install the systems and the system should behave well once those choices are made.
I think Red Hat and Mandrake do a decent job here. I'm not as familliar with SuSE, so I can't say. But, that is clearly one of the jobs of a vendor: to establish best practices and ease compatibility.
What? Sorry, no. More than ten years ago, I was doing word-processing using Mac Write II on a Macintosh classic, now I'm using Word X on a Dual G4. Would you please tell me what radical changes happened in word processing? The interface is basically the same, the text window, the ruler above with the tab stops and some buttons for getting bold etc. Many as
• by burgburgburg (574866) <> on Monday March 29, 2004 @11:14AM (#8703900)
while not forcing them to constantly make them. Having a simple, functional default desktop but with the adaptability/personalization we've come to expect is the best way. For those willing/desiring to modify, their options are open. For those who have better things to do (like work), the default is there for them.
• by Joe Tie. (567096) on Monday March 29, 2004 @11:18AM (#8703946)
Q: Choice is bad? A: Yes Q: Can anyone understand the issues? A: Think of how many letters there are in one word. Now multiply that by how many words are in a page, and then the book. Then by how many books there are. That's so much information! You shouldn't even try. Q: I like choice A: No you don't. You'd be happier if you didn't have them. Q: No, really, I like choice A: Well, here's some proof for you. People with cancer like having doctors treat them instead of creating their own chemo routines. Do you think you're better than people with cancer or something! THE END
• There's only a lot of choice in areas where there is still a lot of experimentation into the possible solutions. In areas where a suitable and economic solution has been found, choice is really rather limited.
It's a standard aspect of evolution: early forms show extraordinary variation and complexity; as time goes on the simplest and most economical solutions get standardized and the bizarre varieties get killed off.
The same happens in technology, which is why we converge on mature standards such as TCP/IP and (dare I say it) Linux.
• by stateofmind (756903) on Monday March 29, 2004 @11:20AM (#8703990)
It's just like when I'm trying to find some good porn, I've overwhelmed!
So many fetishes, so little kleenex.
• by Anonymous Coward
Psychology professor Barry "FUD" Schwartz receives $50 million from a mysterious donor...
• by Perl-Pusher (555592) on Monday March 29, 2004 @11:21AM (#8704002)
I can concede that 50+ operating systems with no data exchange compatibility would be a bad thing. But that is not the same as having no choice at all. The old Soviet Union had one choice state owned monopolies. Look where it got them. The addition of choice becomes less of a problem when they all follow standards. Take a look at cars, they all have a steering wheel, brakes, etc. They all use similiar motor oil, the same gas etc. Having a choice in cars is good. Being locked into one supplier or manufacturer is bad. It's the same with computers. Open standards, choice, competition spurring innovation, all good things. One supplier, added features and imcompatibilties just to force an upgrade and maintain monopoly, bad!
• by gurps_npc (621217) on Monday March 29, 2004 @11:22AM (#8704015) Homepage
In many ways he is correct - with so many choices it takes a lot of work to figure out what is worth what.
The problem is made worse by the rapid improvement. Rules that apply last year do not apply this year.
But on the other side of that if the manufactures were not scum, that problem can easily be dealt with.
All it takes is a classification system, similar to what we do with cars.
People know what you mean when you say:
station waggon
What we need are some similar terms for the newer technologies to become more common.
We need categories like: game-system (high end video/audio), word-system (low-end MS word,Excel,presentationsm with low memory, low speed etc.), net-server (designed to host a web site or other network), etc. etc. to be come common terms that everyone knows and uses.
• by heironymouscoward (683461) <> on Monday March 29, 2004 @11:25AM (#8704041) Journal
It's a common mistake to confuse choices with decisions. Decisions are what confuse and annoy people, not choices.
Some simple illustrations of this. Choice: "these are the desktop themes you can play with". Decision: "please choose a desktop theme to continue installation.
Choice: "tired of your wife? Here are ten more girls to choose from." Decision: "you gonna marry me or what?!"
Choice: "choose from fifty different fabric colors for your car interior". Decision: "what color interior do you want your next subway car to have?"
Basically a good designer maximises choice but minimizes the decisions needed to get started.
I believe the article has made the error of confusing the two.
• by globalar (669767) on Monday March 29, 2004 @12:05PM (#8704487) Homepage
It is easy to adopt a new program, it is not so easy to adopt a new OS and everything else.
The barrier to Linux adoption is mostly entry. There are not only so many choices (some required, most just clouding the decision) to make the first step, but a new way of thinking about software ("How good can free software be?"), new applications, (not really) new security dynamics, new names, new acronyms, new conventions, etc.
The way to mediate all these is to make a common, extremely well documented and supported, simple, and well-crafted base design. Introduce the design (maybe through big corporate rollouts, preinstallations on PC's) and then let people play with it. But there can only be so many designs to fit the market. The average users does not need to consider over a dozen Linux flavors (let alone two desktop environments).
I think Linux could be a little bit more like OS X in these regards.
• Even Windows has choice, but minimizes decision. For example, there are multiple screen savers available, but one is chosen for you by default upon installation. Same for themes. Windows has one standard text editor (notepad), but you can easily install others if you wish. Same for browsers, etc. Choice is alive and well.
It seems that most people posting WRT this issue think that it is a question between choice or no choice. Of course, I want choice. Just keep my decisions to a minimum, and provide
• by bperkins (12056) * on Monday March 29, 2004 @11:26AM (#8704055) Homepage Journal
I saw this on a silly cable TV show and have been thinking a lot about it. Choice is nothing new, it's just that the types of choices we all have are changing. If you think about what career you should taken or where exactly you should live, the choices are absolutely staggering. These, for the most part aren't new developments, though more people have the ability to make a wider array of them.
What's interesting to me is that things that people have had to choose from for many number of years have special agents who specialize in making these choices; travel agents, real estate agents and career counselors. I expect that we'll see more and more of these agents in the future, though it's hard for me to imagine how a breakfast cereal agent would work exactly.
I understand that some people may feel overwhelmed by the breadth of choices presented to the average person, but it seems rather condescending to imply that you ought to give up your choices. The underlying attitude seems to be choice is bad for _you_, and I'll go ahead and keep reading the Economist and drinking my reserve cognac.
Concluding that choice is bad because it causes indecision is like concluding that the sun is bad because it causes sunburn.
After all, is freedom really slavery; ignorance, strength?
• OS Winner by TKO (Score:5, Insightful)
by Mulletproof (513805) on Monday March 29, 2004 @11:28AM (#8704087) Homepage Journal
Um, Microsoft being right or wrong doesn't really factor in here. It's the lack of effective competition that's creating a lack of choice. Apple OS has more or less limited themselves to their own platform, which is generally more expenisive than the average computer user is willing to pay, while Linux is still too obscure for the average user to screw around with. It's not that Windows is a spectacular product that by nature crushes all competition in it's path, it's the fact that what competition exists has been limiting itself in one form or another, giving MS free reign on the PC. As such, most products now cater to it, which makes it more popular.
Too much competition doesn't even begin to enter into the PC OS market, because there never has been that amount of competition. MS won by default, which has nothing to do with them being right or wrong.
• by sheldon (2322) on Monday March 29, 2004 @03:26PM (#8707043)
There was that much competition in the 1980's, before Microsoft became dominant.
You must be too young to remember this, but there was a day when you could go to a store like Computerland and be faced with a choice of 8 different computers. None of which interoperated with one another.
Over time people got tired of this, they got tired of seeing something and finding out it wouldn't work on their computer. So they started making purchasing decisions based upon compatibility. This led to the final decision to standardize on Microsoft. This has also been going on with hardware, the advent of CPU sockets and SIMMs, IDE ports, USB ports, and so forth is all about making the computing easy to understand and hook up and make work.
You're right in that this doesn't make MS right or wrong. Microsoft never made that decision, it was instead the Consumers.
But it does help to explain the dominance of Microsoft software.
• by MattRog (527508) on Monday March 29, 2004 @11:34AM (#8704164)
This is a well-known phenomenon in people management. If you're trying to persuade someone to make a choice and give them 50 options they are most likely to not choose any of them (or in this case, stick with Windows). When you have so many options they get worried and confused - did I pick the right one? What if I had picked XYZ? What makes option XYZ better than option ABC?
Now, I'm not suggesting that choice is bad - but if you want someone to decide you must initially present them with a small number of options - A or B - not A or B or C or D or .. N, etc..
• The choices also must mean something. "Gnome or KDE?" or "SMB or IPP?" probably don't mean much to your average user, at least when they're getting started. If they need to make those choices before they can get any work done, it'll be considerably off-putting.
• with voting there is a rule/problem (I don't know its name - Condorcet is all that comes to mind) that any vote with more than two participants may yield cycles (for example, in some cases, A is prefered to B is preferred to C is preferred to A....) or other nonoptimalities; thus voting methods may not give results consistent with what everyone would want of independent of how the vote is done. The problem with runoff systems (do you want A or B? do you want B or C?...) is that they depend on the order or o
• Makes sense to me: (Score:3, Interesting)
by airrage (514164) on Monday March 29, 2004 @11:35AM (#8704173) Homepage Journal
I have often found that cell-phone (mobile) carriers have the vast array of plans which overlap and seem to not really give you feeling that you have a well-fitting plan.
Also, it's that Coke in a 1-liter bottle versus 6-cans versus 6 glass-bottles versus...
I tend to re-buy crap for this very reason: the first purchase I realize now why it was so cheap, the second purchase while more expensive I realize it's just over-hyped. The third typically is a good cost-to-quality ratio.
• Joel on Software (Score:5, Insightful)
by Boing (111813) on Monday March 29, 2004 @11:41AM (#8704228)
Chapter Three [] of Joel Spolsky's User Interface Design for Programmers has an excellent, clear presentation of this problem.
The summary (as I read it)? People like choice when it's related to what they want to do. If they're making a greeting card, they want to choose what font it uses and what overused clip-art to use. They don't want to choose its orientation as it comes out of the printer, or whether it's saved in MS Word or PDF or RTF or HTML or BMP.
So when I install a linux distribution, and I want to compose a word processing document? I don't care all that much whether I'm using KOffice or StarOffice or or AbiWord or whatever, because the point is not what program I'm using. The point is to write a document, and I shouldn't have to make a needless choice just to get to that point.
That's why modularity (versus "yes" or "no" to compiling it in) in the linux kernel is such a good idea, for example. It allows me to say, "make this choice for me if I need it, and don't hassle me about it."
• by Junks Jerzey (54586) on Monday March 29, 2004 @11:49AM (#8704304)
This is a simple, fundamental principle. Every option you give the user means that you dodged a design decision. Sometimes this is fine, but it can be tremendously overdone. In a great many cases, what you're doing is forcing the *user* to make design decisions: which fonts look good together, which icons are the clearest, which keys work best for various features, and so on. Have some spine! Keep things simple and clean!
With Linux things are worse, because the decisions forced on the user run much deeper. Now you don't just wade through pages of configuration settings in KDE, you have to choose which window manager to use in the first place. Bleah. I'm a techie, a programmer, and I don't want to mess with this stuff. Just give me something reliable and WELL THOUGHT OUT, and I'll use it.
• by Neil Watson (60859) on Monday March 29, 2004 @11:51AM (#8704338) Homepage
I just installed Linux on a friend's laptop. He does not have much experience with Linux. To help in through the transition I installed Openbox as his window manager and gave him a simply menu and hot key list:
• Web Browser: F1
• Chat: F2
• Email: F3
• Word Processor F4
• Speadsheet F5
• Presentation F6
• Xterm F7
• Run Command F8
• Exit
Less choice, less questions, less confusion. So far I have had no complaints. Obviously, as he gets comfortable he will want more choices later. At the beginning, I think the overwhelming amount of choice is what turns new users away from Linux.
• by Quebec (35169) on Monday March 29, 2004 @12:04PM (#8704470) Homepage
I've seen one professor talking about the problem
of having too much choices (I think it was the author of this book) and he was clear about
something, it only is a problem for one type of people; those who are not satisfied with their choice until they absolutly know for sure they made the best choice. Those with a "good is good enough for me" mentality do not have any problem with too many choices.
• by Guardian452 (761937) on Monday March 29, 2004 @12:06PM (#8704499)
This concept was driven home for me in elementary school with the "Choose Your Own Adventure" books. I could NOT read one of those without jamming my fingers between pages to mark interesting divergences in case my choice didn't work out! It drove me nuts to think that I might be missing out on something interesting somewhere else.
"Do you want to repair the damaged robot? Turn to p. 42"
"Or you want to flee with the princess? Turn to p.22"
Choices? Bah! I just gave up and went with the old "one narrative only" books. Much more satisfying.
• Apple, not Microsoft (Score:4, Interesting)
by hak1du (761835) on Monday March 29, 2004 @12:38PM (#8704858) Journal
That's Apple's secret: Apple picks a lot of things for you. They don't always make the best choices, but they usually make workable choices, and even when their choices are technically bad (as they are from time to time), at least they still make them look good.
Microsoft, on the other hand, is all about choice (within well-defined, money-making parameters): you get zillions of audio and video CODECs, lots of configuration options for the UI, preference panels with sub-panels until your eyes glaze over, dozens of classes that all do the same thing, and let's not forget an ever expanding list of third-party utilities and add-ons to make up for the choices Microsoft didn't give you and the problems Microsoft created while creating all that choice. Microsoft isn't kidding when they are saying that they are giving you choices.
UNIX, like Apple, traditionally has made choices and stuck by them. For example, the UNIX folks at Bell Labs understood that the use of "tab" in Makefiles probably was a mistake, but it wasn't a big enough mistake to create another "make" utility (at least not for a couple of decades). And, yes, the file system may not be the ideal IPC or database mechanism, but it worked well enough and provided a good, simple answer.
Linux has inherited some of the UNIX simplicity and philosophy, although, sadly, there has been a lot of uncertainty and waffling come into it, mostly from people who are trying to turn Linux into Windows.
• by buzzoff (744687) on Monday March 29, 2004 @12:43PM (#8704936)
I recently bought two sets of tires for my SUVs. The set of brands and models were overwhelming. I got through the process by looking at reviews for the tires. Within 20 minutes I had narrowed the list to four models of tires. I checked prices locally and made purchases within the week from two different vendors.
Not only is choice of tires good, but choice of vendors. The qualification is, you have to be smart about it. I can see how choice could be bad for people with low comprehension skills. For those who negotiate prices and want the best quality, the more choice the better.
Reviews, either formal or informal, are key for high involvement purchases (choice in low involvement purchases don't matter as much, because the product is inexpensive, not critical, etc.).
• Biased... (Score:3, Interesting)
by mercuryresearch (680293) * on Monday March 29, 2004 @01:00PM (#8705155) Journal
I thought the name of the professor sounded familiar, and sure enough, it was who I thought it was. His position is not agenda-free (not that anyone's is.)
Schwartz wrote a paper for the January, 2000 edition of the Journal of the America Psychological Association, American Psychologist, titled "Self-Determination, the Tyranny of Freedom."
The artical basically lays the groundwork for restricting freedom for people's own good, and to force beliefs on people for their own good. Coming from a libertarian viewpoint myself, the entire article was disturbing in a very subtle way -- and it was clear that a political or social agenda was a subtext.
It appears he is simply continuing on this theme.
• by SmallFurryCreature (593017) on Monday March 29, 2004 @02:10PM (#8706166) Journal
Don't want to confuse the lower classes with all these options on the ballot paper you know. Might scare them.
The old MS joke comes to mind:
One world, One web, One program - Microsoft Ad
Ein Volk, Ein Reich, Ein Fuhrer
Either this guy think this is a good idea or he doesn't understand choice.
The current choice is simply:
• Apple
Easy oneway to do things. (well if you don't make use of its unix background wich you never need to touch if you don't want to)
• Windows
Oneway for the OS. (easy until you become an admin, the old change network settings example comes to mind)
Choice for everything else. Just check how many email programs and office suits there really are for windows.
• Linux
Roll your own. Choice in everything except hardware. And even there you got choice. Just write your own.
• BSD
For the necrophiliacs.
But how is this any different from choosing a car? Choosing a house? Choosing a meal?
Do I want a car I can fix myself if needed (handy if you drive in remote places) or just a little town cruiser how about no car at all? Do I want my own house I can rebuild however I want or do I want a nice rented apartment fully furnitured with a maid? Do I want a meal I can microwave and be ready or do I want to spend hours in the kitchen to create a feast fit a king?
Now some people here seem to want everyone to make the same choice they have. This applies equally to all computer OSes. These people accuse other OS users of being zealots and never realise they are a very black pot.
I for one am not scared of choice. I am scared one day I will not have a choice. Choose whatever OS you want. But let it be YOUR choice.
• by Trolling4Dollars (627073) on Monday March 29, 2004 @02:13PM (#8706205) Journal
Recently I was discussing what I see to be a central conflict in human beings. It is the conflict of two desires that every human being has:
1. The desire to be an individual. To be unique.
2. The desire to be part of a group. To be accepted as being the same as others.
I'm not sure of the forces driving these two desires, but it certainly has an effect on making choices. Which leads to another issue which I think this article was focusing on: who has the control?
In situations where one entity has all the control (a centralized system), there are fewer choices to make and therefore the system tends to work better. However the downside is that the system will entity will naturally impose restrictions. This is the point where the desire for individuality comes into direct conflict with a centralized system.
In a situation where the individual has control, the system they are working within must allow for a variety of choices to be made since no individual is the same. With all this choice, the system has a tendency to be very complex and break down frequently (witness Macs vs. PCs, with PCs being more complex). It also has a tendency to lead to situaitons where there is no continuity. On the PC you have a gazillion choices no matter what OS you choose. Expand that to the hardware, and you have even more choices. With a Mac, you only have a handful of choices, but they are the "best" choices based on experience. The user gives up a certain level of control for a simpler experience.
So... what's the answer? There really isn't one. It's a flaw in human design. We would be largely better off without the desire for individuality and centralized control, but we would also be a lot less interesting. However, the trains would probably run on time... ;P
• by spun (1352) <`moc.oohay' `ta' `yranoituloverevol'> on Monday March 29, 2004 @03:24PM (#8707017) Journal
Maybe there are people genetically predisposed to want to make choices, to want to choose from the widest array of options. These people would sometimes make the wrong choices and die. Then maybe there are others, people genetically predisposed to do what everyone else is doing. They would become unhappy when forced to make choices on their own (knowing, genetically speaking, that making the wrong choice may kill them.) Sheep and goats, you see. Maybe geeks tend towards the latter, and perhaps that's why most people here have a hard time grasping the fact that many people don't like to have too many choices.
Or maybe I'm just full of it. I don't know. Maybe the moderators will decide for me.
• by FreshFunk510 (526493) on Monday March 29, 2004 @03:26PM (#8707042)
This week's Newsweek has an op-ed article called Afflictions of Affluence that speaks on this very topic. According to it, there are 3 consequences of our rich society: obesity, time crunch and buyer's remorse.
In short the article goes on to say that because we're so rich and food is so cheap our portion sizes have been getting bigger. And that's why we're becoming fat.
We're facing a constant time crunch because we constantly view our time as more and more valuable (time is money in our capitalistic culture) ergo there's this need to cram all our activities into shorter time periods.
Lastly, ther'es buyer's remorse simply because we havfe so many choices out there. You buy one mp3 player but have time to research all 100. You're likely to find a feature in another mp3 player you wish you had.
| null | null | null | null | null | null | null | null | null |
null | null | null | null | null | null | A Cunt Fagot carries every connotation of a fagot except homosexual. In a man a Cunt Fagot indicates a person who thinks they are superior because of their artsy taste in obscure music, movies, books etc... In a woman it attaches the same associations except the connection between an actual homosexual is even more detached. A Cunt Fagot is significantly more concerned with whether you are aware of their bizarre taste in obscure media than whether or not they actually enjoy said media.
Kanye West is a "Cunt Fagot".
That "Cunt Fagot" just tried to sell Boards of Canada to me.
That "Cunt Fagot" with the wolf on his shirt only likes Lil Wayne because its ironic.
by The Prophet El Matador November 02, 2009
7 Words related to Cunt Fagot
Free Daily Email
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null | null | null | null | null | null | living wage
: an amount of money you are paid for a job that is large enough to provide you with the basic things (such as food and shelter) needed to live an acceptable life
Full Definition of LIVING WAGE
: a subsistence wage
: a wage sufficient to provide the necessities and comforts essential to an acceptable standard of living
Examples of LIVING WAGE
1. He was barely earning a living wage.
First Known Use of LIVING WAGE
Next Word in the Dictionary: living willPrevious Word in the Dictionary: living unitAll Words Near: living wage
July 04, 2015
stringent Hear it
rigorous, strict, or severe
Take a 3-minute break and test your skills!
Test your vocab with our fun, fast game
Ailurophobia, and 9 other unusual fears | null | null | null | null | null | null | null | null | null |
https://socratic.org/questions/586ecddcb72cff5f5aa0426c | 1,638,617,426,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964362969.51/warc/CC-MAIN-20211204094103-20211204124103-00002.warc.gz | 595,353,998 | 5,907 | # Question #0426c
Jan 5, 2017
$0.8 \overline{7} = \frac{79}{90}$
#### Explanation:
Let $x = 0.8 \overline{7}$
$\implies 10 x = 8. \overline{7}$
$\implies 10 x - x = 8. \overline{7} - 0.8 \overline{7}$
$\implies 9 x = 8.7 - 0.8$
$\implies 9 x = 7.9$
$\implies 9 x = \frac{79}{10}$
$\implies x = \frac{79}{10} \cdot \frac{1}{9}$
$\therefore x = \frac{79}{90}$
This technique of multiplying by $10$ (or a power of $10$) and subtracting can be used to find the fractional representation of any number with infinitely repeating decimal digits. This answer contains a detailed explanation. | 223 | 595 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 11, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.34375 | 4 | CC-MAIN-2021-49 | latest | en | 0.70874 |
http://mathhelpforum.com/algebra/49630-square-root-simplification.html | 1,503,043,350,000,000,000 | text/html | crawl-data/CC-MAIN-2017-34/segments/1502886104612.83/warc/CC-MAIN-20170818063421-20170818083421-00408.warc.gz | 269,544,017 | 10,513 | 1. ## Square Root Simplification
(sqroot(x^3+x^2))/x
Anyone know how to simplify this?
2. Originally Posted by gearshifter
(sqroot(x^3+x^2))/x
Anyone know how to simplify this?
If this is what you wrote:
$\frac{\sqrt{x^3+x^2}}{x}$, then
$\frac{\sqrt{x^3+x^2}}{x}=\frac{\sqrt{x^2(x+1)}}{x} =\frac{x\sqrt{x+1}}{x}=\sqrt{x+1}$
3. how does $\frac{\sqrt{x^2(x+1)}}{x}$ equal $\frac{x \sqrt {x+1}}{x}$
?
4. Originally Posted by hana_102
how does $\frac{\sqrt{x^2(x+1)}}{x}$ equal $\frac{x \sqrt {x+1}}{x}$
?
Think of it like this
$[(x^2)(x+1)]^{\frac{1}{2}}$
Distribute the power of $\frac{1}{2}$
$x\sqrt{x+1}$
5. what if it was "For x0"?
6. Originally Posted by gearshifter
what if it was "For x0"?
If x=0 or x=-1, the radicand is 0.
If x<-1, the radicand is negative thus producing imaginary solutions. | 311 | 814 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 9, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.84375 | 4 | CC-MAIN-2017-34 | longest | en | 0.728625 |
https://azdikamal.com/500-divided-by-5/ | 1,716,171,018,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058147.77/warc/CC-MAIN-20240520015105-20240520045105-00558.warc.gz | 100,954,672 | 10,513 | # 500 Divided By 5: An Easy Math Solution
## Introduction
Mathematics is an essential subject that every student must learn. It helps to improve our logical and analytical skills, which are useful in our daily lives. One of the fundamental concepts in mathematics is division. In this article, we will discuss how to divide 500 by 5.
## The Basics of Division
Before we dive into the problem, let us first review the basics of division. Division is a mathematical operation that involves splitting a number into equal parts. The number being divided is called the dividend, while the number of parts it is being split into is called the divisor. The result of division is called the quotient.
## How to Divide 500 by 5
Dividing 500 by 5 is a straightforward process. We need to divide the dividend (500) by the divisor (5) to get the quotient. To do this, we follow these steps:
### Step 1: Write the Dividend and Divisor
Write the dividend (500) and divisor (5) on a piece of paper.
### Step 2: Divide the First Digit
Start by dividing the first digit of the dividend (5) by the divisor (5). The result is 1.
### Step 3: Write the Quotient
Write down the quotient (1) above the dividend (500).
### Step 4: Multiply the Divisor by the Quotient
Multiply the divisor (5) by the quotient (1). The result is 5.
### Step 5: Subtract the Product from the Dividend
Subtract the product (5) from the dividend (500). The result is 495.
### Step 6: Bring Down the Next Digit
Bring down the next digit (0) from the dividend and write it next to the remainder (495).
### Step 7: Repeat the Process
Repeat the process by dividing the new dividend (4950 by the divisor (5). The result is 99.
### Step 8: Write the Quotient and Remainder
Write down the quotient (99) above the dividend (4950). The final answer is 100.
## Why is Dividing Important?
Dividing is a fundamental operation in mathematics that has many real-life applications. We use division when dividing money between friends, calculating time and distance, and even when sharing food. It is an essential skill that we use in our daily lives.
## Conclusion
In conclusion, dividing 500 by 5 is a simple process that involves following a few steps. It is a fundamental concept in mathematics that has many practical applications. By learning how to divide numbers, we can improve our logical and analytical skills, which are useful in our daily lives. | 561 | 2,424 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 5.03125 | 5 | CC-MAIN-2024-22 | latest | en | 0.886996 |
https://www.physicsforums.com/threads/deriving-the-duffing-equation.668393/ | 1,719,091,655,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198862420.91/warc/CC-MAIN-20240622210521-20240623000521-00769.warc.gz | 814,916,881 | 15,245 | # Deriving the duffing equation?
• great_sushi
In summary, the conversation discussed a mass on a non-linear spring with damping. The restoring force of the spring was given by F=-kx+x^3, and this was set equal to Newton's second law F=mx'' = -kx+x^3. The damping, which is dependent on velocity, was then added to the equation. The system is also driven by a periodic force dependent on time. The final equation is x'' + bx' + w0*x + x^3 = Fcos(wt), and there was some confusion regarding the signs in the equation.
great_sushi
Say I have a mass m on a non linear spring k with some damping b.
I start with the restoring force of the spring F=-kx+x^3... the x^3 is the non linearity.
Set that equal to Newtons second law F=mx'' = -kx+x^3
Add in the damping which is dependent of velocity bx'
......mx'' = -kx + x^3 + bx'
......= x'' + k/m*x + x^3 + bx'
......= x'' + bx' + w0*x + x^3... Now because my system is driven I add in a periodic force dependent on t.
Fcos(wt) = x'' + bx' + w0*x + x^3
Is this the correct method? I may have gotten mixed up with my signs I tend to do that :(
We kindly ask that you not post the same question twice. Thank you.
## 1. What is the duffing equation?
The duffing equation is a nonlinear differential equation used to model a damped, driven harmonic oscillator. It is named after Georg Duffing, a German mathematician who studied this type of equation in the early 20th century.
## 2. How is the duffing equation derived?
The duffing equation can be derived from the equation of motion for a damped, driven harmonic oscillator by adding a nonlinear term that accounts for the restoring force becoming nonlinear at large amplitudes.
## 3. What are the applications of the duffing equation?
The duffing equation has applications in various fields, including physics, engineering, and mathematics. It can be used to model the behavior of many physical systems, such as pendulums, electrical circuits, and biological systems.
## 4. What are the challenges in solving the duffing equation?
One of the main challenges in solving the duffing equation is that it is a nonlinear equation, which means that there is no general solution that works for all cases. Instead, numerical methods or approximations must be used to solve the equation.
## 5. What are the implications of the duffing equation in chaos theory?
The duffing equation is often used as an example of a chaotic system, meaning that small changes in the initial conditions can lead to drastically different outcomes. This has implications for predicting the long-term behavior of certain systems and has been studied extensively in the field of chaos theory.
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1K | 741 | 2,937 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.75 | 4 | CC-MAIN-2024-26 | latest | en | 0.946775 |
http://slideplayer.com/slide/2474978/ | 1,524,491,865,000,000,000 | text/html | crawl-data/CC-MAIN-2018-17/segments/1524125946011.29/warc/CC-MAIN-20180423125457-20180423145457-00365.warc.gz | 281,844,891 | 18,980 | # Time to Teach Presents Year 5 (National Numeracy Strategy) (Based on DFEE Sample Lessons) www.timetoteach.co.uk.
## Presentation on theme: "Time to Teach Presents Year 5 (National Numeracy Strategy) (Based on DFEE Sample Lessons) www.timetoteach.co.uk."— Presentation transcript:
Time to Teach Presents Year 5 (National Numeracy Strategy) (Based on DFEE Sample Lessons) www.timetoteach.co.uk
Resources Demonstration number line
Mental Learning Objective I can count forwards and backwards in nines and twelves.
Mental Learning Task We are going to start by practising counting with nines and twelves.
Mental Learning Task Count in nines with your teacher to over 100 and count back to your start number… starting with…
Mental Learning Task Count in nines with your teacher to over 120 and count back to your start number… starting with…
Mental Learning Task Count back in nines to 0, starting from…..
Mental Learning Task Why was counting in nines easy? You can quickly add 10 and take off 1.
Mental Learning Task Counting in twelves is also easy. You can quickly add 10 and 2. The numbers are always even.
Mental Learning Objective I can count forwards and backwards in nines and twelves.
Main Learning Objective I can understand division uses repeated subtraction. I know two methods of repeated subtraction.
Key idea
Main Learning Task This lesson will look at the way division can be seen as repeated subtraction. You will be developing an efficient method of division of three-digit numbers by one-digit numbers.
Main Learning Task 360 8 = We can think of 360 divided by 8 as ‘how many 8s in 360?’ We could find the answer by repeatedly subtracting 8. This would take a long time… How could we make it faster?
Main Learning Task 360 8 = We could work with ‘chunks’ of eight… such as 8, ten times 8, five times 8, or twenty times 8. Lets first first roughly estimate what the answer might be…
Main Learning Task There are two ways we could solve this problem… 360 8 = 360 - 80 (10 lots of 8) 280 - 80 (10 lots of 8) 200 - 80 (10 lots of 8) 120 -80 (10 lots of 8) 40 -40 (5 lots of 8) 0 4 x 10 lots of 8 and five lots of 8 together make 45.
Main Learning Task There is a second faster way to solve this…. 360 8 = 360 - 160 (20 lots of 8) 200 -160 (20 lots of 8) 40 - 40 (5 lots of 8) 0 20 + 20 + 5 lots of 8 This makes 45 lots of 8 360 8 = 45
Main Learning Task Now try and solve these sums using repeated subtractions. 351 9 360 8 693 3 168 8 168 7 168 12 Challenge 256 6 1452 5 482 15
Main Learning Task Simplification:- Some children will feel happier with smaller steps and a longer calculation. Encourage children to work at the level they understand but encourage discussion about the efficiency of different chunks.
Main Learning Objective I can understand division uses repeated subtraction. I know two methods of repeated subtraction.
Plenary How did you solve 351 9? Compare short and long methods.
Review of Key Idea I can discuss methods using repeated subtraction. Did you learn this today?
Download ppt "Time to Teach Presents Year 5 (National Numeracy Strategy) (Based on DFEE Sample Lessons) www.timetoteach.co.uk."
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##### Would you like to try a portion of a dynamic, interactive MathLinks: Essentials lesson with your students? Just follow these 3 easy steps!
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#### Description
NUMBER
BASE-TEN
(NBT1) Whole Number Multiplication and Division
Lesson excerpt:
Division Strategies
Students use repeated subtraction and a chunking strategy to make sense of division prior to tackling the standard long division algorithm.TE: 9, 9ab, 10
SP: 9, 10
SD: NBT1.2a
FRACTIONS
(FR1) Fraction Concepts
Lesson excerpt:
Fraction Strips 1
Students create fraction strips (a linear model) for halves, fourths, and eighths, and they explore relationships between these fractions.TE: 3, 3a, 4
SP: 3, 4
SD: FR1.1a
Reproducible: R1
(FR 2) Fraction Addition and Subtraction
Lesson excerpt:
Students begin the study of fraction operations by using contexts, pictures, and mental math to perform fraction addition.TE: 4, 4a
SP: 4
SD: FR2.1a
INTEGERS
(IN1) Introduction to Integers
Lesson excerpt:
Temperature Change
Students use a temperature context and number lines to make sense of temperature changes involving signed numbers.TE: 4, 4ab, 5
SP: 4,5
SD: IN1.1
(IN2) Integers 2
Lesson excerpt:
A Counter Model
Students use counters to represent positive and negative numbers, an important step in leading up to using counters for operations on signed numbers. TE: 3, 3ab, 4
SP: 3, 4
SD: INT2.1
Reproducible: R7
EXPRESSIONS
AND EQUATIONS
(EE1) Variables and Balance
Lesson excerpt:
Variables and Expressions
Students define and use variables to write and evaluate expressions in a menu context.TE: 9, 9a, 10
SP: 9, 10
SD: EE1.2a
(EE2) Expressions and Balance
Lesson excerpt:
The Hundred Chart Puzzle
Students use 7th grade level equation solving skills to solve number problems based on the hundred chart.TE: 18, 19, 19ab
SP: 18, 19
SD: EE2.3
Reproducible: R4
PROPORTIONAL
REASONING
(PR1) Ratio Representations
Lesson excerpt:
Paint Mixtures
In this introduction to tape diagrams, students use cards depicting paint mixtures to reason intuitively in a proportional context.TE: 3, 4, 5
SP: 3, 4, 5
Reproducible: R1
(PR3) Proportional Reasoning Applications
Lesson excerpt:
Art Supplies
Students represent costs and quantities using tables and double number lines.TE 3, 3a, 4
SP 3, 4
SD: RP3.1
FUNCTIONS
(FUN1) Introduction to Linear Functions
Lesson excerpt:
Saving for a Bicycle
Students represent input-output rules using equations, tables and graphs. TE 3, 3ab, 4, 5
SP: 3, 4, 5
SD: FUN1.1 | 902 | 3,329 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.34375 | 4 | CC-MAIN-2024-38 | latest | en | 0.8483 |
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Please hel me to solve the following applications.
Business and finance. A store has T-shirts on sale at 2 for \$5.50. At this rate, what
do five T-shirts cost?
Thank you for using the Jiskha Homework Help Forum. Why not find out what one shirt costs first? Divide \$5.50 by 2 and you get \$2.75, which is the price for one shirt. Now Take \$2.75 times 5 = \$13.75.
There is also another way to do this. Take \$5.50 times 4 and then add the \$2.75 which is half the \$5.50. The price comes out the same = \$13.75.
## Similar Questions
1. ### Math
Help me- Proportions A store has T-shirts on slae at 2 for \$5.50. At this rate, what do five T-shirts cost?
2. ### Math
Business and finance. A store has T-shirts on sale at 2 for \$5.50. At this rate, what do five T-shirts cost?
3. ### math
a store has 3 small childrens shirts, 3 medium and 1 large childrens shirts. they have 5 large adults shirts for sale. what is the mathematical probability of someone buying a large shirt?
4. ### math
Traditions Clothing Store is having a sale. Shirts that were regularly priced at \$20 are on sale for \$17. What is the percentage of decrease in the price of the shirts?
5. ### algebra
the student store sells schools supplies,hats,and t-shirts with NAMS printed on them.The person who counted the hats and t-shirts can't remember exactly how many of each there were but he does remeber these 2 facts: -there were four …
6. ### algebra
Speciality t shirts are being sold online for \$30 plus one time handling fee of \$2.75. The total cost is a function of the number of t shirts bought. What function rule models the cost of the t shirts?
7. ### math
teemu's t-shirts designs and manufactures t-shirts for a \$15 flat design rate and \$7.50 for each t-shirt.he wants to sell his t-shirts to the senior class for their graduation party. he is going to charge them \$10 for each t-shirt. … | 492 | 1,904 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.71875 | 4 | CC-MAIN-2018-05 | latest | en | 0.957069 |
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## Recommended Posts
1) Out of 90 tins, 24 contained corns, 12 contained beans, 36 contained peas and the rest conatined mushrooms. Find the probability that Sean will have to open more than 2 tins before he finds one which does not contain baby corns.
I am not very sure how to tackle this question. I thought of using complement, but I don't know what is the eact opposite of this situation. Actually I don't really understand what the question means.
2) A box contains 6 black balls, 4 red balls and 2 yellow balls. 3 are drawn with replacement. Calculate the probability that all 3 balls are of the same colour.
Why isn't it (0.5)^3 x (1/3)^3?
3) What is the probability of getting all three cards being red out of a ordinary pack of 52 playing cards?
Why isn't it (26/52)x(25/51)x(24/50)?
Please help me understand where I am wrong! I have been constantly making mistakes like that , but I don't know what is my error in thinking. Thank you!
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1) Out of 90 tins, 24 contained corns, 12 contained beans, 36 contained peas and the rest conatined mushrooms. Find the probability that Sean will have to open more than 2 tins before he finds one which does not contain baby corns.
I am not very sure how to tackle this question. I thought of using complement, but I don't know what is the eact opposite of this situation. Actually I don't really understand what the question means.
2) A box contains 6 black balls, 4 red balls and 2 yellow balls. 3 are drawn with replacement. Calculate the probability that all 3 balls are of the same colour.
Why isn't it (0.5)^3 x (1/3)^3?
3) What is the probability of getting all three cards being red out of a ordinary pack of 52 playing cards?
Why isn't it (26/52)x(25/51)x(24/50)?
Please help me understand where I am wrong! I have been constantly making mistakes like that , but I don't know what is my error in thinking. Thank you!
I think I can answer these bast in reverse order:
your answer the #3 is correct in most typical situations, that is, being dealt 3 cards in a row and looking at all three in your hand and the chances of them all being red. But, the answer obviously implies that it was sampling with replacement, because the chance of a red card out of a normal 52 card deck is one half and the answer given is (1/2)^3. So, there is probably a mistake in the question statement of #3
#2 can probably be easiest seen if you break the question statement down into its component parts:
Turn "what is the probability of drawing the same color three times in a row" into "what is the prob. of drawing black 3 times in a row + what is the prob. of drawing red 3 times in a row + what is the prob. of drawing yellow 3 times in a row"
Calculate each of the three in the right hand side, and then add them up and see if that doesn't get your answer.
#1 is actually going to be just like your answer (not the book's answer) to #3. Do the same thing -- find what the chances are of not opening corn the first time, then what the chances are of not opening corn the second time (which changes because it is sampling without replacement).
I hope this helps.
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1. its best to do a tree diagram, find the p of 2 then find the p of opening 3 and thats one way. Other way is finding it indirectly like bignose said
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Thanks a lot! Your pieces of advice have helped me tremendously =)
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https://www.nagwa.com/en/explainers/862138531945/ | 1,721,294,505,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514826.9/warc/CC-MAIN-20240718064851-20240718094851-00743.warc.gz | 799,535,401 | 38,911 | Lesson Explainer: Scientific Notation | Nagwa Lesson Explainer: Scientific Notation | Nagwa
# Lesson Explainer: Scientific Notation Mathematics • First Year of Preparatory School
## Join Nagwa Classes
In this explainer, we will learn how to express numbers in scientific notation and how to convert numbers between their normal and scientific forms.
When numbers have a very large or very small absolute value, writing them implies writing a lot of digits, for instance, 23 410 000 000 or 0.0000016482. This is not very convenient. Therefore, we use another way of writing numbers called scientific notation (also referred to as standard form). It is a compact way of writing numbers that proves particularly useful for numbers with a very large or very small absolute value.
To explain scientific notation, let us consider a big number: 300 000 000. We note that this number is a 3 followed by 8 zeros, and ideally, we want to be able to express this without having to write all of the zeros down. We note that one way to write a number with 8 zeros succinctly is to write it in powers of 10:
Since the number we are interested in is three times this number, we can write it as
In this way, we have written the number purely in terms of an exponent of 10 and a number multiplying it, meaning it is much quicker to write.
We can extend this approach to very small values in a similar way. For instance, 0.0000002 can be written in scientific notation as follows:
We can see that the main difference with this form compared with the larger number is that the power of 10 is now negative. Generally speaking, larger numbers can be written in terms of positive powers of 10, while smaller numbers can be written using negative powers.
Let us consider how this approach can be generalized with the following definition.
### Definition: Scientific Notation
A number written in scientific notation (also called standard form) is of the form where .
The exponent is positive for numbers with a large absolute value. It is negative for numbers with a very small absolute value.
Let us quickly check our understanding of this definition with the first example.
### Example 1: Identifying When a Number Is Correctly Written in Scientific Notation
Which of the following numbers is not in standard form?
Recall that a number written in scientific notation (also called standard form) is of the form , where .
The four options listed here are all in the form of a decimal or integer number multiplied by , with either positive or negative. However, in option A, this number is the decimal , where , which is smaller than 1.
Therefore, the number is not written in standard form.
When converting a number from its normal form to scientific notation and vice versa, it is important to understand what happens when a number is divided or multiplied by 10 or powers of 10.
Recall that, in our decimal system, each digit has a value that can be written in the form of a power of ten. In a place value table, ones can indeed be replaced with , tens with , hundreds with , and so on, and as for the digits after the decimal point, tenths are , hundredths are , thousandths are , and so forth.
Multiplying a number by 10 is therefore equivalent to moving all its digits one place left (a one becomes a ten, a ten becomes a hundred, a tenth becomes a one, etc.), while dividing by 10 is equivalent to moving all the digits one place right (a one becomes a tenth, a ten becomes a one, a tenth becomes a hundredth, etc.). Look, for instance, at 347.58, , and .
It follows that multiplying by 100, or , is equivalent to moving all the digits two places left, while dividing by is equivalent to moving them two places right, and so forth. Remember that dividing by is the same as multiplying by since .
These translations of the digits to the left or to the right when multiplying by a power of ten can also be tracked by the movement of the decimal point. When a number is multiplied by 10, its decimal point moves one place right. When a number is divided by 10, it moves one place left.
When a number is written in scientific notation, we can easily express it in normal form by carrying out multiplication by the power of 10 that appears in its scientific notation form. Let us look at a couple of examples to see how this works; the first involves a large number.
### Example 2: Expressing a Large Number in Normal Form
Express in normal form.
Recall that a number written in scientific notation is of the form , where .
Note that the given number is already written in scientific notation. To express in normal form, we need to multiply 3.06707 by .
Multiplying 3.06707 by 10 moves the decimal point 1 place right to give 30.6707, multiplying it by 100 moves the decimal point 2 places right to give 306.707, multiplying it by 1 000 gives 3 067.07, and multiplying it by 10 000 gives 30 670.7.
However, once we get to multiplying 3.06707 by 100 000, moving the decimal point 5 places right means that it effectively reaches the end of the digits of our original number, giving 306 707. The way to go beyond this, enabling us to multiply by further powers of 10, is to add as many trailing zeros to our original number as we need. Note that we can do this without changing its value, because 3.06707 has the same numerical value as 3.067070, 3.0670700, or (as needed here) 3.0670700000.
Therefore, multiplying 3.06707 by moves the decimal point 10 places to the right, so we get 30 670 700 000.
Now, we work through the same method for a small number.
### Example 3: Expressing a Small Number in Normal Form
Write in normal form.
Recall that a number written in scientific notation is of the form , where .
The given number is already written in scientific notation; to express in normal form, we need to multiply 3.01 by .
Multiplying 3.01 by is equivalent to dividing by 10, which moves the decimal point 1 place left. However, this means the decimal point effectively reaches the start of the digits of our original number, so we need to add a leading zero, giving the number 0.301.
The way to go beyond this, enabling us to divide by further powers of 10, is to add as many leading zeros to our original number as we need. Note that we can do this without changing its value, because 3.01 has the same numerical value as 03.01, 003.01, or (as needed here) 000 003.01.
Therefore, multiplying 3.01 by moves the decimal point 5 places to the left, so we get 0.0000301.
Next, we will learn what to do when going in the opposite direction. We want to be able to write any number in scientific notation.
Let us start with the very large number 23 410 000 000. For this, we are going to write 23 410 000 000 in a place value table.
10 10 10 10 10 10 10 10 10 10 10 2 3 4 1 0 0 0 0 0 0 0
The number 23 410 000 000 can be decomposed as the sum of and
We see that the digit with the highest value is the first digit on the left, here 2, and its value is . Therefore, .
Note that can be decomposed as , and we have indeed , , and .
Now, we look at a very small number. Choosing 0.0000016482, again we start by writing it in a place value table.
10 10 10 10 10 10 10 10 10 10 10 0 0 0 0 0 0 1 6 4 8 2
We can decompose here as well 0.0000016482 as the sum of and
The digit with the highest value in 0.0000016482 is the first nonzero digit after the decimal place, here 1, and its value is . Therefore, .
Note that we have indeed , , , and .
We see that writing a number in a place value table is a very effective method to write the number in scientific notation. However, it is a little bit long. A faster method is to compare the position of the decimal point in the original, normal number with that in the number from its scientific notation version . Note that when given any normal number, we can always write down the corresponding value of . This is because there will be only one placement of the decimal point that results in a number with an absolute value that is at least 1 and less than 10. The value of is then the number of places the decimal point needs to move from its position in to its position in the original, normal number. If the decimal point needs to move right, then is positive because is smaller than the number, and so must be multiplied by a power of 10. If the decimal point needs to move left, then is negative because is larger than the number and so must be divided by a power of 10, that is, multiplied by a power of 10 with a negative exponent.
To test our knowledge of this method, let us go through a few examples. The first one concerns finding a missing value of when a normal number is expressed in its scientific notation form .
### Example 4: Finding an Unknown by Comparing Numbers in Scientific Notation
Given that , find the value of .
Recall that a number written in scientific notation is of the form , where .
Here, we have a number written in its normal form, together with the value of when it is written in scientific notation. We need to work out the value of , the missing exponent in the scientific notation form .
The number tells us how many times 4.16 needs to be multiplied or divided by 10 to give 4 160 000. Here, we see that 4.16 is smaller than 4 160 000, so 4.16 must be multiplied by 10 several times.
To find the value of , we count how many places the decimal point in 4.16 would need to move right to its position in 4 160 000. We find that it is six times.
Therefore, , so .
Next, we will practice writing a large number in scientific notation form.
### Example 5: Expressing a Large Number in Scientific Notation
Express 874 527 893 in scientific notation.
Recall that a number written in scientific notation is of the form , where .
To express 874 527 893 in scientific notation, we need first to determine the value of .
The number is made with exactly the same digits of 874 527 893 and in the same order, but its absolute value must be greater than or equal to 1 and less than 10. Hence, .
Then, we need to find the value of . The number tells us how many times needs to be multiplied or divided by 10 to give 874 527 893. Here, we see that 8.74527893 is smaller than 874 527 893. Therefore, needs to be multiplied several times by 10.
To easily find how many times the multiplication by 10 needs to be repeated, we find by how many places the decimal point in 8.74527893 would need to move right to its position in 874 527 893 upon this repeated multiplication by 10. We find that it is eight times.
Therefore, , and so if we express 874 527 893 in scientific notation, we get .
We follow this with an example of what happens for a small number.
### Example 6: Expressing a Small Number in Scientific Notation
Express 0.00447 in scientific notation.
Recall that a number written in scientific notation is of the form , where .
To express 0.00447 in scientific notation, we need first to determine the value of .
The number is made with exactly the same digits of 0.00447 and in the same order, but its absolute value must be greater than or equal to 1 and less than 10. Hence, . The zeros to the left of the first 4 do not need to be written.
Then, we need to find the value of . The number tells us how many times needs to be multiplied or divided by 10 to give 0.00447. Here, we see that 4.47 is larger than 0.00447. Therefore, needs to be divided several times by 10. We are going to write this repeated division by 10 in the form of a repeated multiplication by . In other words, is negative and its absolute value tells us how many times the multiplication by needs to be repeated.
To easily find the absolute value of , we find by how many places the decimal point in 4.47 would need to move left to its position in 0.00447 upon this repeated multiplication by . We find that it is three times.
Therefore, , and so if we express 0.00447 in scientific notation, we get .
Since many quantities that feature in real-world situations can be either very large (e.g., the area of a country in square kilometres) or very small (e.g., the mass of an atom in grams), then scientific notation is extremely useful in helping us to describe and compare the numbers involved. Let us finish with an example of this type, which features large numbers.
### Example 7: Ordering Numbers from Least to Greatest Involving Scientific Notation in a Real-World Context
The table shows the populations of five countries in a given year. List the countries from the smallest to the largest population.
Country Population A B C D E 5.3×10 6.1×10 5.8×10 6.5×10 4.2×10
Here, we are given the population numbers of five countries written in scientific notation. We need to compare these numbers so that we can list the countries from the smallest to the largest population.
Recall that a number written in scientific notation is of the form , where . To answer this question, it is important to understand the following two facts about numbers written in scientific notation:
1. For any two numbers that have different exponents of 10, the larger number will be the one with the larger exponent.
2. For any two numbers that have the same exponent of 10, the larger number will be the one with the larger value of .
In this question, we have five population numbers with five different exponents of 10, namely 4, 5, 6, 7, and 8. Point I above tells us that the larger the exponent, the larger the number. This means that we can rank the populations from smallest to largest without doing any detailed calculations.
The country with the smallest population is C with a population of . Next comes country B with a population of , followed by D with a population of , E with a population of , and A with a population of .
Therefore, listing the countries from the smallest to the largest population gives C, B, D, E, and A.
Let us finish by recapping some key concepts from this explainer.
### Key Points
• A number written in scientific notation (also referred to as standard form) is of the form where . The exponent is positive for numbers with a large absolute value; it is negative for numbers with a very small absolute value.
• To write a number in scientific notation, we first find , that is, the number with exactly the same digits as our number, but with . Then, we find the value of , that is, the number of places the decimal point needs to move from its position in to its position in the original number.
If (i.e., if the decimal point needs to move to the right from to the number), then is positive. If (i.e., if the decimal point needs to move to the left from to the number), then is negative.
• When a number is written in scientific notation, we can express it in its normal form by carrying out the multiplication by the appropriate power of 10.
• For any two numbers written in scientific notation, if they have different exponents of 10, then the larger number will be the one with the larger exponent. If they have the same exponent of 10, then the larger number will be the one with the larger value of .
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• Realistic Exam Questions | 3,769 | 15,376 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.9375 | 5 | CC-MAIN-2024-30 | latest | en | 0.925635 |
http://puzzle.queryhome.com/1824/i-am-formed-by-6-letters-letters-1-2-6-spell-out-a-pet | 1,501,144,772,000,000,000 | text/html | crawl-data/CC-MAIN-2017-30/segments/1500549427750.52/warc/CC-MAIN-20170727082427-20170727102427-00387.warc.gz | 258,745,008 | 30,994 | # I am formed by 6 letters Letters 1-2-6 spell out a pet................
615 views
I am formed by 6 letters
Letters 1-2-6 spell out a pet,
Letters 6-5-2 spell out a drink,
Letters 4-5-2-3 spell out a fruit,
Letters 3-2-6 spell out a pest,which often gets
eaten by 1-2-6
What am I ?
posted Jul 7, 2014
## 1 Solution
Its `Carpet`
1-2-6 = Cat (pet)
6-5-2= tea (drink)
4-5-2-3=pear (fruit)
3-2-6= rat (often eaten by cat)
So, 1=c, 2=a,3=r, 4=p, 5=e & 6=t.
solution Jul 9, 2014
Similar Puzzles
+1 vote
I am a 6 letter word.
Letters 6-5-2 spell out a drink.
Letters 4-5-2-3 spell out a fruit.
Letters 1-2-6 spell out a pet.
Letters 3-2-6 spell out a pest, which often gets eaten by 1-2-6.
What am I?
I have 9 letters,
4,5,6 is an animal,
7 is me,
3 is you,
2+8+9+1 is completed,
Who am I ?
–1 vote
Guess this Indian Name
A) I am a 6 letter word (name)
B) 1,4 letter means yourself.
C) 2,3 letters is a country name.
D) 2,5 is a country name.
E) Remove first two letters means hair.
F) 6,2,1 letters means "we" in hindi. | 399 | 1,028 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.890625 | 4 | CC-MAIN-2017-30 | latest | en | 0.791551 |
https://math.stackexchange.com/questions/3986888/air-fan-problem-fluid-dynamic | 1,653,361,464,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662562410.53/warc/CC-MAIN-20220524014636-20220524044636-00621.warc.gz | 448,539,998 | 66,685 | # Air fan problem - fluid dynamic
For preparing an exam me and my classmates are unable to solve this question. Can anybody help us. Thanks.
At a conference my colleagues had a room without air conditioning. At the reception they could get a fan to cool down the warm room.
(a) Assume that the open part of the window is a vertical opening not much wider than the fan, but much higher in height. Where would it be better, for my colleagues to put the fan, at the bottom of the opening, in the middle or at the top? Why? What physical limitation do you have to make on the blowing speed of the fan to get to physically relevant results?
(b) The flow rate of the inflow caused by the fan is Q and the inflow air has a temperature Ti , and the volume of the chamber is Vk. Assume (as opposed to question a) that the outflow from the window at any time is the average temperature of the chamber T(t) (and thus there is a very efficient mixing of the air). Set up a differential equation for T(t) and calculate how long it takes for the room to cool from a high initial temperature $$T_0$$ to $$T = (T_0 + T_i)/2$$\
(c) The cooling effect of a fan is caused by the layer of air around the skin (heated by thermal conduction) being replaced by cool air of the passing air flow of the fan. We can write the influence of the thermal conduction with the help of the P´eclet number, which takes the ratio of the advection transport of thermal energy and the conductive transport. Calculate this dimensionless number (analog of the Reynolds number) using the equation for temperature transport: $$\frac{dT}{dt}=k \nabla^2T$$ where κ is the thermal conduction coefficient
• I think this is a question for physics.stackexchange.com Jan 15, 2021 at 21:36
• There is a typo in part (c) : Peclet number is mistakenly written P' eclet number. This makes the problem harder to understand for non-physicists. Jan 18, 2021 at 12:30
## 1 Answer
Welcome to MSE. Please note that at least the first part of your question may be more suitable for a physics forum. Please also note that in here we expect you to show your efforts towards solving your own problems. Otherwise it may seem that you are having others do your homework for you! That said, I will be happy to give you a boost with part (b). You can use the hints in this comment and try to solve the problem yourselves. And if you still find it difficult, you can share your work with us, and hopefully some of us will help you again. Alright, let's do this:
We are assuming air flow to be incompressible in this example. Therefore, the flow rates of inlet and exit are equal ($$Q$$). In the rest of this post we assume that $$Q$$ is volume flow rate (and not mass flow rate). In an infinitesimally short duration, $$dt$$ , volume of in-coming air is $$Qdt$$ . This air, with temperature $$T_i$$ mixes with the air in the rest of the room (with volume $$V-Qdt$$) at temperature $$T(t)$$ . As a result, the temperature of the room changes by $$dT$$. The temperature of the mixture is a weighted average of the temperatures of the two volumes of air, with weights being proportional to their respective volumes. $$T(t)+dT = \frac{Qdt}{V}T_i + \frac{V-Qdt}{V}T(t)$$ Rearranging the terms, we will have: $$\frac{dT}{dt} + \frac{Q}{V}T(t) = \frac{Q}{V}T_i \qquad(1)$$ This is a first-order linear ordinary differential equation (ODE) that is not homogeneous. From the text of the problem I assume you are familiar with ODEs, so I will suffice to a hint for solving equation (1). You must find the general solution of (1) and then apply the initial condition $$T(0) = T_0$$ . The general solution is the sum of the solution of the homogeneous equation (in which the right hand side is zero) and a particular solution of the non-homogenous equation. The solution of the homogeneous equation $$y'+ry=0$$ is $$y_h=e^{-rx}$$. To find a particular solution to the non-homogeneous equation $$y'+ry=A$$ , try $$y_p=y_h+B$$ and use it in the equation to find $$B$$. The general solution will be $$y_g=cy_h+y_p$$ , wherein $$c$$ is a constant that should satisfy the initial condition.
There, you now have the equation and you know how to solve it. Good luck!
• how do you become (1)? Because: $T(t)+dT=(Qdt T_{i}/V)+(V-Qdt)T(t)/V)$ gives $dT/dt+1=QT_{i}/V+T/dt-QT/V$ Jan 16, 2021 at 13:05
• @questmath I checked my calculations again: equation (1) is correct. I suggest you check your calculations too. If you would like, you can share your calculations in a comment or as another answer, and I will be able to identify the mistake in them. Jan 17, 2021 at 1:20
• Yes I divided everything trough dt so , T/dt+dT/dt=QT_{i}/V+T/dt-QT/V =>dT/dt=QT_{i}/V-QT/V, no you're right, I made a mistake :) Jan 18, 2021 at 7:02 | 1,238 | 4,747 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 21, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.640625 | 4 | CC-MAIN-2022-21 | latest | en | 0.941081 |
https://indianpythonista.wordpress.com/2017/03/14/linear-regression/ | 1,597,436,500,000,000,000 | text/html | crawl-data/CC-MAIN-2020-34/segments/1596439739370.8/warc/CC-MAIN-20200814190500-20200814220500-00554.warc.gz | 344,119,796 | 26,761 | # Linear Regression
This article discusses the basics of linear regression and its implementation in Python programming language.
Linear regression is a statistical approach for modelling relationship between a dependent variable with a given set of independent variables.
Note: In this article, we refer dependent variables as response and independent variables as features for simplicity.
In order to provide a basic understanding of linear regression, we start with the most basic version of linear regression, i.e. Simple linear regression.
## Simple Linear Regression
Simple linear regression is an approach for predicting a response using a single feature.
It is assumed that the two variables are linearly related. Hence, we try to find a linear function that predicts the response value(y) as accurately as possible as a function of the feature or independent variable(x).
Let us consider a dataset where we have a value of response y for every feature x:
For generality, we define:
x as feature vector, i.e $x = [x_1, x_2, ...., x_n]^T$,
y as response vector, i.e $y = [y_1, y_2, ...., y_n]^T$
for n observations (in above example, n=10).
A scatter plot of above dataset looks like:-
Now, the task is to find a line which fits best in above scatter plot so that we can predict the response for any new feature values. (i.e a value of x not present in dataset)
This line is called regression line.
The equation of regression line is represented as:
$h(x_i) = \beta_0 + \beta_1x_i$
Here,
• $h(x_i)$ represents the predicted response value for $i^{th}$ observation.
• $\beta_0$ and $\beta_1$ are regression coefficients and represent y-intercept and slope of regression line respectively.
To create our model, we must “learn” or estimate the values of regression coefficients $\beta_0$ and $\beta_1$. And once we’ve estimated these coefficients, we can use the model to predict responses!
In this article, we are going to use the Least Squares technique.
Now consider:
$y_i = \beta_{0} + \beta_{1}x_i + \epsilon_i = h(x_i) + \epsilon_i \Rightarrow \epsilon_i = y_i - h(x_i)$
Here, $\epsilon_i$ is residual error in $i^{th}$ observation.
So, our aim is to minimize $\sum_{i=1}^{n}\epsilon_i^2$.
We define the squared error or cost function, J as:
$J(\beta_0,\beta_1) = \frac{1}{2n}\sum_{i=1}^{n}\epsilon_{i}^2$
and our task is to find the value of $\beta_0$ and $\beta_1$ for which $J(\beta_0,\beta_1)$ is minimum!
Without going into the mathematical details, we present the result here:
$\newline \beta_{1} = \frac{SS_{xy}}{SS_{xx}} \newline \newline \beta_0 = \bar{y} - \beta_{1}\bar{x}$
where $SS_{xy}$ is the sum of cross-deviations of y and x:
$SS_{xy} = \sum_{i=1}^{n}(x_{i} - \bar{x})(y_{i}-\bar{y}) = \sum_{i=1}^{n}y_{i}x_{i}-n\bar{x}\bar{y}$
and $SS_{xx}$ is the sum of squared deviations of x:
$SS_{xx} = \sum_{i=1}^{n}(x_{i} - \bar{x})^2 = \sum_{i=1}^{n}x_{i}^2-n(\bar{x})^2$
Note: The complete derivation for finding least squares estimates in simple linear regression can be found here.
Given below is the python implementation of above technique on our small dataset:
import numpy as np
import matplotlib.pyplot as plt
def estimate_coef(x, y):
# number of observations/points
n = np.size(x)
# mean of x and y vector
m_x, m_y = np.mean(x), np.mean(y)
# calculating cross-deviation and deviation about x
SS_xy = np.sum(y*x - n*m_y*m_x)
SS_xx = np.sum(x*x - n*m_x*m_x)
# calculating regression coefficients
b_1 = SS_xy / SS_xx
b_0 = m_y - b_1*m_x
return(b_0, b_1)
def plot_regression_line(x, y, b):
# plotting the actual points as scatter plot
plt.scatter(x, y, color = "m",
marker = "o", s = 30)
# predicted response vector
y_pred = b[0] + b[1]*x
# plotting the regression line
plt.plot(x, y_pred, color = "g")
# putting labels
plt.xlabel('x')
plt.ylabel('y')
# function to show plot
plt.show()
def main():
# observations
x = np.array([0, 1, 2, 3, 4, 5, 6, 7, 8, 9])
y = np.array([1, 3, 2, 5, 7, 8, 8, 9, 10, 12])
# estimating coefficients
b = estimate_coef(x, y)
print("Estimated coefficients:\nb_0 = {} \
\nb_1 = {}".format(b[0], b[1]))
# plotting regression line
plot_regression_line(x, y, b)
if __name__ == "__main__":
main()
Output of above piece of code is:
Estimated coefficients:
b_0 = -0.0586206896552
b_1 = 1.45747126437
And graph obtained looks like this:
## Multiple linear regression
Multiple linear regression attempts to model the relationship between two or more features and a response by fitting a linear equation to observed data.
Clearly, it is nothing but an extension of Simple linear regression.
Consider a dataset with p features(or independent variables) and one response(or dependent variable).
Also, the dataset contains n rows/observations.
We define:
X (feature matrix) = a matrix of size n X p where $x_{ij}$ denotes the values of $j^{th}$ feature for $i^{th}$ observation.
So,
$\mathbf{X} =\begin{pmatrix} x_{11} & \cdots & x_{1p} \\ x_{21} & \cdots & x_{2p} \\ \vdots & \ddots & \vdots \\ x_{n1} & \cdots & x_{np} \end{pmatrix}$
and
y (response vector) = a vector of size n where $y_{i}$ denotes the value of response for $i^{th}$ observation.
$y = \begin{bmatrix} y_{1}\\ y_{2}\\ .\\ .\\ y_{n}\\ \end{bmatrix}$
The regression line for p features is represented as:
$h(x_i) = \beta_0 + \beta_1x_{i1} + \beta_2x_{i2} + ..... + \beta_px_{ip}$
where $h(x_i)$ is predicted response value for $i^{th}$ observation and $\beta_0, \beta_1, ..., \beta_p$ are the regression coefficients.
Also, we can write:
$\newline y_i = \beta_0 + \beta_1x_{i1} + \beta_2x_{i2} + ..... + \beta_px_{ip} + \epsilon_i \newline or \newline y_i = h(x_i) + \epsilon_i \Rightarrow \epsilon_i = y_i - h(x_i)$
where $\epsilon_i$ represents residual error in $i^{th}$ observation.
We can generalize our linear model a little bit more by representing feature matrix X as:
$\mathbf{X} =\begin{pmatrix} 1 & x_{11} & \cdots & x_{1p} \\ 1 & x_{21} & \cdots & x_{2p} \\ \vdots & \vdots & \ddots & \vdots \\ 1 & x_{n1} & \cdots & x_{np} \end{pmatrix}$
So now, the linear model can be expressed in terms of matrices as:
$y = X\beta + \epsilon$
where,
$\beta = \begin{bmatrix} \beta_0\\ \beta_1\\ .\\ .\\ .\\ \beta_p\\ \end{bmatrix}$
and
$\epsilon = \begin{bmatrix} \epsilon_1\\ \epsilon_2\\ .\\ .\\ \epsilon_n\\ \end{bmatrix}$
Now, we determine estimate of $\beta$, i.e. $\hat{\beta}$ using Least Squares method.
As already explained, Least Squares method tends to determine $\hat{\beta}$ for which $\sum_{i=1}^{n}\epsilon_i^2$ is minimized.
We present the result directly here:
$\hat{\beta} = (X'X)^{-1}X'y$
where $'$ represents the transpose of the matrix while $-1$ represents the matrix inverse.
Knowing the least square estimates, $\hat{\beta}$, the multiple linear regression model can now be estimated as:
$\hat{y} = X\hat{\beta}$
where $\hat{y}$ is estimated response vector.
Note: The complete derivation for obtaining least square estimates in multiple linear regression can be found here.
Given below is the implementation of multiple linear regression technique on the Boston house pricing dataset dataset using Scikit-learn.
import matplotlib.pyplot as plt
import numpy as np
from sklearn import datasets, linear_model, metrics
# defining feature matrix(X) and response vector(y)
X = boston.data
y = boston.target
# splitting X and y into training and testing sets
from sklearn.model_selection import train_test_split
X_train, X_test, y_train, y_test = train_test_split(X, y, test_size=0.4,
random_state=1)
# create linear regression object
reg = linear_model.LinearRegression()
# train the model using the training sets
reg.fit(X_train, y_train)
# regression coefficients
print('Coefficients: \n', reg.coef_)
# variance score: 1 means perfect prediction
print('Variance score: {}'.format(reg.score(X_test, y_test)))
# plot for residual error
## setting plot style
plt.style.use('fivethirtyeight')
## plotting residual errors in training data
plt.scatter(reg.predict(X_train), reg.predict(X_train) - y_train,
color = "green", s = 10, label = 'Train data')
## plotting residual errors in test data
plt.scatter(reg.predict(X_test), reg.predict(X_test) - y_test,
color = "blue", s = 10, label = 'Test data')
## plotting line for zero residual error
plt.hlines(y = 0, xmin = 0, xmax = 50, linewidth = 2)
## plotting legend
plt.legend(loc = 'upper right')
## plot title
plt.title("Residual errors")
## function to show plot
plt.show()
Output of above program looks like this:
Coefficients:
[ -8.80740828e-02 6.72507352e-02 5.10280463e-02 2.18879172e+00
-1.72283734e+01 3.62985243e+00 2.13933641e-03 -1.36531300e+00
2.88788067e-01 -1.22618657e-02 -8.36014969e-01 9.53058061e-03
-5.05036163e-01]
Variance score: 0.720898784611
and Residual Error plot looks like this:
In above example, we determine accuracy score using Explained Variance Score.
We define:
$explained\_variance\_score = 1 - \frac{Var\{y - \hat{y}\}}{Var\{y\}}$
where $\hat{y}$ is the estimated target output, $y$ the corresponding (correct) target output, and $Var$ is Variance, the square of the standard deviation.
The best possible score is 1.0, lower values are worse.
## Assumptions
Given below are the basic assumptions that a linear regression model makes regarding a dataset on which it is applied:
• Linear relationship: Relationship between response and feature variables should be linear. The linearity assumption can be tested using scatter plots. As shown below, 1st figure represents linearly related variables where as variables in 2nd and 3rd figure are most likely non-linear. So, 1st figure will give better predictions using linear regression.
• Little or no multi-collinearity: It is assumed that there is little or no multicollinearity in the data. Multicollinearity occurs when the features (or independent variables) are not independent from each other.
• Little or no auto-correlation: Another assumption is that there is little or no autocorrelation in the data. Autocorrelation occurs when the residual errors are not independent from each other. You can refer here for more insight into this topic.
• Homoscedasticity: Homoscedasticity describes a situation in which the error term (that is, the “noise” or random disturbance in the relationship between the independent variables and the dependent variable) is the same across all values of the independent variables. As shown below, figure 1 has homoscedasticity while figure 2 has heteroscedasticity.
As we reach to the end of this article, we discuss some applications of linear regression below.
## Applications:
1. Trend lines: A trend line represents the variation in some quantitative data with passage of time (like GDP, oil prices, etc.). These trends usually follow a linear relationship. Hence, linear regression can be applied to predict future values. However, this method suffers from a lack of scientific validity in cases where other potential changes can affect the data.
2. Economics: Linear regression is the predominant empirical tool in economics. For example, it is used to predict consumption spending, fixed investment spending, inventory investment, purchases of a country’s exports, spending on imports, the demand to hold liquid assets, labor demand, and labor supply.
3. Finance: Capital price asset model uses linear regression to analyze and quantify the systematic risks of an investment.
4. Biology: Linear regression is used to model causal relationships between parameters in biological systems. | 3,146 | 11,517 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 38, "codecogs_latex": 16, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.0625 | 4 | CC-MAIN-2020-34 | latest | en | 0.877088 |
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#### Resources tagged with Comparing and Ordering numbers similar to Light the Lights:
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### There are 45 results
Broad Topics > Numbers and the Number System > Comparing and Ordering numbers
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Place this "worm" on the 100 square and find the total of the four squares it covers. Keeping its head in the same place, what other totals can you make?
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Pick two rods of different colours. Given an unlimited supply of rods of each of the two colours, how can we work out what fraction the shorter rod is of the longer one?
### Farey Sequences
##### Age 11 to 14 Challenge Level:
There are lots of ideas to explore in these sequences of ordered fractions. | 1,927 | 8,200 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.984375 | 4 | CC-MAIN-2019-43 | latest | en | 0.799895 |
https://wikivisually.com/wiki/3D_rotation_group | 1,568,763,250,000,000,000 | text/html | crawl-data/CC-MAIN-2019-39/segments/1568514573124.40/warc/CC-MAIN-20190917223332-20190918005332-00156.warc.gz | 740,023,985 | 88,118 | # 3D rotation group
In mechanics and geometry, the 3D rotation group, often denoted SO(3), is the group of all rotations about the origin of three-dimensional Euclidean space R3 under the operation of composition.[1] By definition, a rotation about the origin is a transformation that preserves the origin, Euclidean distance (so it is an isometry), and orientation (i.e. handedness of space). Every non-trivial rotation is determined by its axis of rotation (a line through the origin) and its angle of rotation. Composing two rotations results in another rotation; every rotation has a unique inverse rotation; and the identity map satisfies the definition of a rotation. Owing to the above properties (along composite rotations' associative property), the set of all rotations is a group under composition. Rotations are not commutative (for example, rotating R 90° in the x-y plane followed by S 90° in the y-z plane is not the same as S followed by R), making it a nonabelian group. Moreover, the rotation group has a natural structure as a manifold for which the group operations are smoothly differentiable; so it is in fact a Lie group, it is compact and has dimension 3.
Rotations are linear transformations of R3 and can therefore be represented by matrices once a basis of R3 has been chosen. Specifically, if we choose an orthonormal basis of R3, every rotation is described by an orthogonal 3×3 matrix (i.e. a 3×3 matrix with real entries which, when multiplied by its transpose, results in the identity matrix) with determinant 1. The group SO(3) can therefore be identified with the group of these matrices under matrix multiplication; these matrices are known as "special orthogonal matrices", explaining the notation SO(3).
The group SO(3) is used to describe the possible rotational symmetries of an object, as well as the possible orientations of an object in space, its representations are important in physics, where they give rise to the elementary particles of integer spin.
## Length and angle
Besides just preserving length, rotations also preserve the angles between vectors; this follows from the fact that the standard dot product between two vectors u and v can be written purely in terms of length:
${\displaystyle \mathbf {u} \cdot \mathbf {v} ={\tfrac {1}{2}}\left(\|\mathbf {u} +\mathbf {v} \|^{2}-\|\mathbf {u} \|^{2}-\|\mathbf {v} \|^{2}\right).}$
It follows that any length-preserving transformation in R3 preserves the dot product, and thus the angle between vectors. Rotations are often defined as linear transformations that preserve the inner product on R3, which is equivalent to requiring them to preserve length. See classical group for a treatment of this more general approach, where SO(3) appears as a special case.
## Orthogonal and rotation matrices
Every rotation maps an orthonormal basis of R3 to another orthonormal basis. Like any linear transformation of finite-dimensional vector spaces, a rotation can always be represented by a matrix. Let R be a given rotation. With respect to the standard basis e1,e2,e3 of R3 the columns of R are given by (Re1, Re2, Re3). Since the standard basis is orthonormal, and since R preserves angles and length, the columns of R form another orthonormal basis. This orthonormality condition can be expressed in the form
${\displaystyle R^{\mathsf {T}}R=RR^{\mathsf {T}}=I,}$
where RT denotes the transpose of R and I is the 3 × 3 identity matrix. Matrices for which this property holds are called orthogonal matrices; the group of all 3 × 3 orthogonal matrices is denoted O(3), and consists of all proper and improper rotations.
In addition to preserving length, proper rotations must also preserve orientation. A matrix will preserve or reverse orientation according to whether the determinant of the matrix is positive or negative. For an orthogonal matrix R, note that det RT = det R implies (det R)2 = 1, so that det R = ±1. The subgroup of orthogonal matrices with determinant +1 is called the special orthogonal group, denoted SO(3).
Thus every rotation can be represented uniquely by an orthogonal matrix with unit determinant. Moreover, since composition of rotations corresponds to matrix multiplication, the rotation group is isomorphic to the special orthogonal group SO(3).
Improper rotations correspond to orthogonal matrices with determinant −1, and they do not form a group because the product of two improper rotations is a proper rotation.
## Group structure
The rotation group is a group under function composition (or equivalently the product of linear transformations), it is a subgroup of the general linear group consisting of all invertible linear transformations of the real 3-space R3.[2]
Furthermore, the rotation group is nonabelian; that is, the order in which rotations are composed makes a difference. For example, a quarter turn around the positive x-axis followed by a quarter turn around the positive y-axis is a different rotation than the one obtained by first rotating around y and then x.
The orthogonal group, consisting of all proper and improper rotations, is generated by reflections; every proper rotation is the composition of two reflections, a special case of the Cartan–Dieudonné theorem.
## Axis of rotation
Every nontrivial proper rotation in 3 dimensions fixes a unique 1-dimensional linear subspace of R3 which is called the axis of rotation (this is Euler's rotation theorem). Each such rotation acts as an ordinary 2-dimensional rotation in the plane orthogonal to this axis. Since every 2-dimensional rotation can be represented by an angle φ, an arbitrary 3-dimensional rotation can be specified by an axis of rotation together with an angle of rotation about this axis. (Technically, one needs to specify an orientation for the axis and whether the rotation is taken to be clockwise or counterclockwise with respect to this orientation).
For example, counterclockwise rotation about the positive z-axis by angle φ is given by
${\displaystyle R_{z}(\varphi )={\begin{bmatrix}\cos \varphi &-\sin \varphi &0\\\sin \varphi &\cos \varphi &0\\0&0&1\end{bmatrix}}.}$
Given a unit vector n in R3 and an angle φ, let R(φ, n) represent a counterclockwise rotation about the axis through n (with orientation determined by n). Then
• R(0,n) is the identity transformation for any n
• R(φ,n) = R(−φ,−n)
• R(π + φ,n) = R(π − φ,−n).
Using these properties one can show that any rotation can be represented by a unique angle φ in the range 0 ≤ φ ≤ π and a unit vector n such that
• n is arbitrary if φ = 0
• n is unique if 0 < φ < π
• n is unique up to a sign if φ = π (that is, the rotations R(π, ±n) are identical).
In the next section, this representation of rotations is used to identify SO(3) topologically with three-dimensional real projective space.
## Topology
The Lie group SO(3) is diffeomorphic to the real projective space RP3.[3]
Consider the solid ball in R3 of radius π (that is, all points of R3 of distance π or less from the origin). Given the above, for every point in this ball there is a rotation, with axis through the point and the origin, and rotation angle equal to the distance of the point from the origin; the identity rotation corresponds to the point at the center of the ball. Rotation through angles between 0 and −π correspond to the point on the same axis and distance from the origin but on the opposite side of the origin. The one remaining issue is that the two rotations through π and through −π are the same. So we identify (or "glue together") antipodal points on the surface of the ball. After this identification, we arrive at a topological space homeomorphic to the rotation group.
Indeed, the ball with antipodal surface points identified is a smooth manifold, and this manifold is diffeomorphic to the rotation group, it is also diffeomorphic to the real 3-dimensional projective space RP3, so the latter can also serve as a topological model for the rotation group.
These identifications illustrate that SO(3) is connected but not simply connected; as to the latter, in the ball with antipodal surface points identified, consider the path running from the "north pole" straight through the interior down to the south pole. This is a closed loop, since the north pole and the south pole are identified; this loop cannot be shrunk to a point, since no matter how you deform the loop, the start and end point have to remain antipodal, or else the loop will "break open". In terms of rotations, this loop represents a continuous sequence of rotations about the z-axis starting and ending at the identity rotation (i.e. a series of rotation through an angle φ where φ runs from 0 to 2π).
Surprisingly, if you run through the path twice, i.e., run from north pole down to south pole, jump back to the north pole (using the fact that north and south poles are identified), and then again run from north pole down to south pole, so that φ runs from 0 to 4π, you get a closed loop which can be shrunk to a single point: first move the paths continuously to the ball's surface, still connecting north pole to south pole twice. The second half of the path can then be mirrored over to the antipodal side without changing the path at all. Now we have an ordinary closed loop on the surface of the ball, connecting the north pole to itself along a great circle; this circle can be shrunk to the north pole without problems. The plate trick and similar tricks demonstrate this practically.
The same argument can be performed in general, and it shows that the fundamental group of SO(3) is cyclic group of order 2. In physics applications, the non-triviality of the fundamental group allows for the existence of objects known as spinors, and is an important tool in the development of the spin-statistics theorem.
The universal cover of SO(3) is a Lie group called Spin(3); the group Spin(3) is isomorphic to the special unitary group SU(2); it is also diffeomorphic to the unit 3-sphere S3 and can be understood as the group of versors (quaternions with absolute value 1). The connection between quaternions and rotations, commonly exploited in computer graphics, is explained in quaternions and spatial rotations; the map from S3 onto SO(3) that identifies antipodal points of S3 is a surjective homomorphism of Lie groups, with kernel {±1}. Topologically, this map is a two-to-one covering map. (See the plate trick.)
## Connection between SO(3) and SU(2)
In this section, we give two different constructions of a two-to-one and onto homomorphism of SU(2) onto SO(3).
### Using quaternions of unit norm
The group SU(2) is isomorphic to the quaternions of unit norm via a map given by
${\displaystyle q=a\mathbf {1} +b\mathbf {i} +c\mathbf {j} +d\mathbf {k} =\alpha +j\beta \leftrightarrow {\begin{bmatrix}\alpha &-{\overline {\beta }}\\\beta &{\overline {\alpha }}\end{bmatrix}}=U,\quad q\in \mathbb {H} ,\quad a,b,c,d\in \mathbb {R} ,\quad \alpha ,\beta \in \mathbb {C} ,\quad U\in \mathrm {SU} (2).}$[4]
Let us now identify ${\displaystyle \mathbb {R} ^{3}}$ with the span of ${\displaystyle \mathbf {i} ,\mathbf {j} ,\mathbf {k} }$. One can then verify that if ${\displaystyle v}$ is in ${\displaystyle \mathbb {R} ^{3}}$ and ${\displaystyle q}$ is a unit quaternion, then
${\displaystyle qvq^{-1}\in \mathbb {R} ^{3}}$.
Furthermore, the map ${\displaystyle v\mapsto qvq^{-1}}$ is a rotation of ${\displaystyle \mathbb {R} ^{3}}$. Moreover, ${\displaystyle (-q)v(-q)^{-1}}$ is the same as ${\displaystyle qvq^{-1}}$. This means that there is a 2:1 homomorphism from quaternions of unit norm to SO(3).
One can work this homomorphism out explicitly: the unit quaternion, q, with
{\displaystyle {\begin{aligned}q&{}=w+\mathbf {i} x+\mathbf {j} y+\mathbf {k} z,\\1&{}=w^{2}+x^{2}+y^{2}+z^{2},\end{aligned}}}
is mapped to the rotation matrix
${\displaystyle Q={\begin{bmatrix}1-2y^{2}-2z^{2}&2xy-2zw&2xz+2yw\\2xy+2zw&1-2x^{2}-2z^{2}&2yz-2xw\\2xz-2yw&2yz+2xw&1-2x^{2}-2y^{2}\end{bmatrix}}.}$
This is a rotation around the vector (x,y,z) by an angle 2θ, where cos θ = w and |sin θ| = ||(x,y,z)||. The proper sign for sin θ is implied, once the signs of the axis components are fixed. The 2:1-nature is apparent since both q and q map to the same Q.
### Using Möbius transformations
Stereographic projection from the sphere of radius 1/2 from the north pole (x, y, z) = (0, 0, 1/2) onto the plane M given by z = −1/2 coordinatized by (ξ, η), here shown in cross section.
The general reference for this section is Gelfand, Minlos & Shapiro (1963); the points P on the sphere S = {(x, y, z) ∈ ℝ3: x2 + y2 + z2 = 1/4} can, barring the north pole N, be put into one-to-one bijection with points S(P) = P´ on the plane M defined by z = −1/2, see figure. The map S is called stereographic projection.
Let the coordinates on M be (ξ, η). The line L passing through N and P can be parametrized as
${\displaystyle L(t)=N+t(N-P)=(0,0,1/2)+t((0,0,1/2)-(x,y,z)),\quad t\in \mathbb {R} .}$
Demanding that the z-coordinate of ${\displaystyle L(t_{0})}$ equals 1/2, one finds ${\displaystyle t_{0}={\frac {1}{z-{\frac {1}{2}}}}}$. We have ${\displaystyle L(t_{0})=(\xi ,\eta ,-1/2)}$. Hence the map
${\displaystyle S:\mathbf {S} \rightarrow M;\qquad P\mapsto P'}$
is given by
${\displaystyle (x,y,z)\mapsto (\xi ,\eta )=\left({\frac {x}{{\frac {1}{2}}-z}},{\frac {y}{{\frac {1}{2}}-z}}\right)\equiv \zeta =\xi +i\eta ,}$
where, for later convenience, the plane M is identified with the complex plane .
For the inverse, write L as
${\displaystyle L=N+s(P'-N)=\left(0,0,{\frac {1}{2}}\right)+s\left(\left(\xi ,\eta ,-{\frac {1}{2}}\right)-\left(0,0,{\frac {1}{2}}\right)\right),}$
and demand x2 + y2 + z2 = 1/4 to find s = 1/1 + ξ2 + η2 and thus
${\displaystyle S^{-1}:M\rightarrow \mathbf {S} ;\qquad P'\mapsto P;\qquad (\xi ,\eta )\mapsto (x,y,z)=\left({\frac {\xi }{1+\xi ^{2}+\eta ^{2}}},{\frac {\eta }{1+\xi ^{2}+\eta ^{2}}},{\frac {-1+\xi ^{2}+\eta ^{2}}{2+2\xi ^{2}+2\eta ^{2}}}\right).}$
If g ∈ SO(3) is a rotation, then it will take points on S to points on S by its standard action Πs(g) on the embedding space 3. By composing this action with S one obtains a transformation S ∘ Πs(g) ∘ S−1 of M,
${\displaystyle \zeta =P'\quad \mapsto \quad P\quad \mapsto \quad \Pi _{s}(g)P=gP\quad \mapsto \quad S(gP)\equiv \Pi _{u}(g)\zeta =\zeta '.}$
Thus Πu(g) is a transformation of associated to the transformation Πs(g) of 3.
It turns out that g ∈ SO(3) represented in this way by Πu(g) can be expressed as a matrix Πu(g) ∈ SU(2) (where the notation is recycled to use the same name for the matrix as for the transformation of it represents). To identify this matrix, consider first a rotation gφ about the z-axis through an angle φ,
{\displaystyle {\begin{aligned}x'&=x\cos \varphi -y\sin \varphi ,\\y'&=x\sin \varphi +y\cos \varphi ,\\z'&=z.\end{aligned}}}
Hence
${\displaystyle \zeta '={\frac {x'+iy'}{{\frac {1}{2}}-z'}}={\frac {e^{i\varphi }(x+iy)}{{\frac {1}{2}}-z}}=e^{i\varphi }\zeta ={\frac {e^{\frac {i\varphi }{2}}\zeta +0}{0\zeta +e^{-{\frac {i\varphi }{2}}}}},}$
which, unsurprisingly, is a rotation in the complex plane. In an analogous way, if gθ is a rotation about the x-axis through and angle θ, then
${\displaystyle w'=e^{i\theta }w,\quad w={\frac {y+iz}{{\frac {1}{2}}-x}},}$
which, after a little algebra, becomes
${\displaystyle \zeta '={\frac {\cos {\frac {\theta }{2}}\zeta +i\sin {\frac {\theta }{2}}}{i\sin {\frac {\theta }{2}}\zeta +\cos {\frac {\theta }{2}}}}.}$
These two rotations, gφ, gθ, thus correspond to bilinear transforms of 2 ≃ ℂ ≃ M, namely, they are examples of Möbius transformations.
A general Möbius transformation is given by
${\displaystyle \zeta '={\frac {\alpha \zeta +\beta }{\gamma \zeta +\delta }},\quad \alpha \delta -\beta \gamma \neq 0.}$.
The rotations, gφ, gθ generate all of SO(3) and the composition rules of the Möbius transformations show that any composition of gφ, gθ translates to the corresponding composition of Möbius transformations. The Möbius transformations can be represented by matrices
${\displaystyle \left({\begin{matrix}\alpha &\beta \\\gamma &\delta \end{matrix}}\right),\quad \quad \alpha \delta -\beta \gamma =1,}$
since a common factor of α, β, γ, δ cancels.
For the same reason, the matrix is not uniquely defined since multiplication by I has no effect on either the determinant or the Möbius transformation. The composition law of Möbius transformations follow that of the corresponding matrices; the conclusion is that each Möbius transformation corresponds to two matrices g, −g ∈ SL(2, ℂ).
Using this correspondence one may write
{\displaystyle {\begin{aligned}\Pi _{u}(g_{\varphi })&=\Pi _{u}\left[\left({\begin{matrix}\cos \varphi &-\sin \varphi &0\\\sin \varphi &\cos \varphi &0\\0&0&1\end{matrix}}\right)\right]=\pm \left({\begin{matrix}e^{i{\frac {\varphi }{2}}}&0\\0&e^{-i{\frac {\varphi }{2}}}\end{matrix}}\right),\\\Pi _{u}(g_{\theta })&=\Pi _{u}\left[\left({\begin{matrix}1&0&0\\0&\cos \theta &-\sin \theta \\0&\sin \theta &\cos \theta \end{matrix}}\right)\right]=\pm \left({\begin{matrix}\cos {\frac {\theta }{2}}&i\sin {\frac {\theta }{2}}\\i\sin {\frac {\theta }{2}}&\cos {\frac {\theta }{2}}\end{matrix}}\right).\end{aligned}}}
These matrices are unitary and thus Πu(SO(3)) ⊂ SU(2) ⊂ SL(2, ℂ). In terms of Euler angles[nb 1] one finds for a general rotation
{\displaystyle {\begin{aligned}g(\varphi ,\theta ,\psi )&=g_{\varphi }g_{\theta }g_{\psi }=\left({\begin{matrix}\cos \varphi &-\sin \varphi &0\\\sin \varphi &\cos \varphi &0\\0&0&1\end{matrix}}\right)\left({\begin{matrix}1&0&0\\0&\cos \theta &-\sin \theta \\0&\sin \theta &\cos \theta \end{matrix}}\right)\left({\begin{matrix}\cos \psi &-\sin \psi &0\\\sin \psi &\cos \psi &0\\0&0&1\end{matrix}}\right)\\&=\left({\begin{matrix}\cos \varphi \cos \psi -\cos \theta \sin \varphi \sin \psi &-\cos \varphi \sin \psi -\cos \theta \sin \varphi \cos \psi &\sin \varphi \sin \theta \\\sin \varphi \cos \psi +\cos \theta \cos \varphi \sin \psi &-\sin \varphi \sin \psi +\cos \theta \cos \varphi \cos \psi &-\cos \varphi \sin \theta \\\sin \psi \sin \theta &\cos \psi \sin \theta &\cos \theta \end{matrix}}\right),\end{aligned}}}
(1)
one has[5]
{\displaystyle {\begin{aligned}\Pi _{u}(g(\varphi ,\theta ,\psi ))&=\pm \left({\begin{matrix}e^{i{\frac {\varphi }{2}}}&0\\0&e^{-i{\frac {\varphi }{2}}}\end{matrix}}\right)\left({\begin{matrix}\cos {\frac {\theta }{2}}&i\sin {\frac {\theta }{2}}\\i\sin {\frac {\theta }{2}}&\cos {\frac {\theta }{2}}\end{matrix}}\right)\left({\begin{matrix}e^{i{\frac {\psi }{2}}}&0\\0&e^{-i{\frac {\psi }{2}}}\end{matrix}}\right)\\&=\pm \left({\begin{matrix}\cos {\frac {\theta }{2}}e^{i{\frac {\varphi +\psi }{2}}}&i\sin {\frac {\theta }{2}}e^{i{\frac {\varphi -\psi }{2}}}\\i\sin {\frac {\theta }{2}}e^{-i{\frac {\varphi -\psi }{2}}}&\cos {\frac {\theta }{2}}e^{-i{\frac {\varphi +\psi }{2}}}\end{matrix}}\right).\end{aligned}}}
(2)
For the converse, consider a general matrix
${\displaystyle \pm \Pi _{u}(g_{\alpha ,\beta })=\pm \left({\begin{matrix}\alpha &\beta \\-{\overline {\beta }}&{\overline {\alpha }}\end{matrix}}\right)\in \mathrm {SU} (2).}$
Make the substitutions
{\displaystyle {\begin{aligned}\cos {\frac {\theta }{2}}&=|\alpha |,\quad \sin {\frac {\theta }{2}}=|\beta |,\quad (0\leq \theta \leq \pi ),\\{\frac {\varphi +\psi }{2}}&=\arg \alpha ,\quad {\frac {\psi -\varphi }{2}}=\arg \beta .\end{aligned}}}
With the substitutions, Π(gα, β) assumes the form of the right hand side (RHS) of (2), which corresponds under Πu to a matrix on the form of the RHS of (1) with the same φ, θ, ψ. In terms of the complex parameters α, β,
${\displaystyle g_{\alpha ,\beta }=\left({\begin{matrix}{\frac {1}{2}}(\alpha ^{2}-\beta ^{2}+{\overline {\alpha ^{2}}}-{\overline {\beta ^{2}}})&{\frac {i}{2}}(-\alpha ^{2}-\beta ^{2}+{\overline {\alpha ^{2}}}+{\overline {\beta ^{2}}})&-\alpha \beta -{\overline {\alpha }}{\overline {\beta }}\\{\frac {i}{2}}(\alpha ^{2}-\beta ^{2}-{\overline {\alpha ^{2}}}+{\overline {\beta ^{2}}})&{\frac {1}{2}}(\alpha ^{2}+\beta ^{2}+{\overline {\alpha ^{2}}}+{\overline {\beta ^{2}}})&-i(+\alpha \beta -{\overline {\alpha }}{\overline {\beta }})\\\alpha {\overline {\beta }}+{\overline {\alpha }}\beta &i(-\alpha {\overline {\beta }}+{\overline {\alpha }}\beta )&\alpha {\overline {\alpha }}-\beta {\overline {\beta }}\end{matrix}}\right).}$
To verify this, substitute for α. β the elements of the matrix on the RHS of (2). After some manipulation, the matrix assumes the form of the RHS of (1).
It is clear from the explicit form in terms of Euler angles that the map p:SU(2) → SO(3);Πu(±gαβ) ↦ gαβ just described is a smooth, 2:1 and onto group homomorphism. It is hence an explicit description of the universal covering map of SO(3) from the universal covering group SU(2).
## Lie algebra
Associated with every Lie group is its Lie algebra, a linear space of the same dimension as the Lie group, closed under a bilinear alternating product called the Lie bracket; the Lie algebra of SO(3) is denoted by so(3) and consists of all skew-symmetric 3 × 3 matrices.[6] This may be seen by differentiating the orthogonality condition, ATA = I, A ∈ SO(3).[nb 2] The Lie bracket of two elements of so(3) is, as for the Lie algebra of every matrix group, given by the matrix commutator, [A1, A2] = A1A2A2A1, which is again a skew-symmetric matrix. The Lie algebra bracket captures the essence of the Lie group product in a sense made precise by the Baker–Campbell–Hausdorff formula.
The elements of so(3) are the "infinitesimal generators" of rotations, i.e. they are the elements of the tangent space of the manifold SO(3) at the identity element. If R(φ, n) denotes a counterclockwise rotation with angle φ about the axis specified by the unit vector n, then
${\displaystyle \left.{\operatorname {d} \over \operatorname {d} \varphi }\right|_{\varphi =0}R(\varphi ,{\boldsymbol {n}}){\boldsymbol {x}}={\boldsymbol {n}}\times {\boldsymbol {x}}}$
for every vector x in R3.
This can be used to show that the Lie algebra so(3) (with commutator) is isomorphic to the Lie algebra R3 (with cross product). Under this isomorphism, an Euler vector ${\displaystyle {\boldsymbol {\omega }}\in \mathbb {R} ^{3}}$ corresponds to the linear map ${\displaystyle \mathbf {\tilde {\omega }} }$ defined by ${\displaystyle \mathbf {\tilde {\omega }} ({\boldsymbol {x}})={\boldsymbol {\omega }}\times {\boldsymbol {x}}}$.
In more detail, a most often suitable basis for so(3) as a 3-dimensional vector space is
${\displaystyle L_{\mathbf {x} }={\begin{bmatrix}0&0&0\\0&0&-1\\0&1&0\end{bmatrix}},\quad L_{\mathbf {y} }={\begin{bmatrix}0&0&1\\0&0&0\\-1&0&0\end{bmatrix}},\quad L_{\mathbf {z} }={\begin{bmatrix}0&-1&0\\1&0&0\\0&0&0\end{bmatrix}}.}$
The commutation relations of these basis elements are,
${\displaystyle [L_{\mathbf {x} },L_{\mathbf {y} }]=L_{\mathbf {z} },\quad [L_{\mathbf {z} },L_{\mathbf {x} }]=L_{\mathbf {y} },\quad [L_{\mathbf {y} },L_{\mathbf {z} }]=L_{\mathbf {x} }}$
which agree with the relations of the three standard unit vectors of R3 under the cross product.
As announced above, one can identify any matrix in this Lie algebra with an Euler vector in ℝ3,[7]
{\displaystyle {\begin{aligned}{\boldsymbol {\omega }}&=(x,y,z)\in \mathbb {R} ^{3},\\{\boldsymbol {\tilde {\omega }}}&={\boldsymbol {\omega \cdot L}}=xL_{\mathbf {x} }+yL_{\mathbf {y} }+zL_{\mathbf {z} }={\begin{bmatrix}0&-z&y\\z&0&-x\\-y&x&0\end{bmatrix}}\in {\mathfrak {so}}(3).\end{aligned}}}
This identification is sometimes called the hat-map.[8] Under this identification, the so(3) bracket corresponds in 3 to the cross product,
${\displaystyle [{\tilde {\mathbf {u} }},{\tilde {\mathbf {v} }}]={\widetilde {\mathbf {u} \!\times \!\mathbf {v} }}.}$
The matrix identified with a vector u has the property that
${\displaystyle {\tilde {\mathbf {u} }}\mathbf {v} =\mathbf {u} \times \mathbf {v} ,}$
where ordinary matrix multiplication is implied on the left hand side. This implies that u is in the null space of the skew-symmetric matrix with which it is identified, because u × u = 0.
### A note on Lie algebra
In Lie algebra representation, the group SO(3) is compact and simple of rank 1, and so it has a single independent Casimir element, a quadratic invariant function of the three generators which commutes with all of them; the Killing form for the rotation group is just the Kronecker delta, and so this Casimir invariant is simply the sum of the squares of the generators, ${\displaystyle J_{x},\,J_{y},\,J_{z}}$, of the algebra
${\displaystyle [J_{\mathbf {x} },J_{\mathbf {y} }]=J_{\mathbf {z} },\quad [J_{\mathbf {z} },J_{\mathbf {x} }]=J_{\mathbf {y} },\quad [J_{\mathbf {y} },J_{\mathbf {z} }]=J_{\mathbf {x} }.}$
That is, the Casimir invariant is given by
${\displaystyle J^{2}\equiv {\boldsymbol {J\cdot J}}=J_{x}^{2}+J_{y}^{2}+J_{z}^{2}\propto I~.}$
For unitary irreducible representations Dj, the eigenvalues of this invariant are real and discrete, and characterize each representation, which is finite dimensional, of dimensionality 2j+1. That is, the eigenvalues of this Casimir operator are
${\displaystyle J^{2}=-j(j+1)~I_{2j+1}~,}$
where j is integer or half-integer, and referred to as the spin or angular momentum.
So, above, the 3×3 generators L displayed act on the triplet (spin 1) representation, while the 2×2 ones (t) act on the doublet (spin-½) representation. By taking Kronecker products of D1/2 with itself repeatedly, one may construct all higher irreducible representations Dj. That is, the resulting generators for higher spin systems in three spatial dimensions, for arbitrarily large j, can be calculated using these spin operators and ladder operators.
For every unitary irreducible representations Dj there is an equivalent one, Dj−1. All infinite-dimensional irreducible representations must be non-unitary, since the group is compact.
In quantum mechanics, the Casimir invariant is the "angular-momentum-squared" operator; integer values of spin j characterize bosonic representations, while half-integer values fermionic representations, respectively. The antihermitian matrices used above are utilized as spin operators, after they are multiplied by i, so they are now hermitian (like the Pauli matrices). Thus, in this language,
${\displaystyle [J_{\mathbf {x} },J_{\mathbf {y} }]=iJ_{\mathbf {z} },\quad [J_{\mathbf {z} },J_{\mathbf {x} }]=iJ_{\mathbf {y} },\quad [J_{\mathbf {y} },J_{\mathbf {z} }]=iJ_{\mathbf {x} }.}$
and hence
${\displaystyle J^{2}=j(j+1)~I_{2j+1}~.}$
Explicit expressions for these Dj are,
{\displaystyle {\begin{aligned}\left(J_{z}^{(j)}\right)_{ba}&=(j+1-a)~\delta _{b,a}\\\left(J_{x}^{(j)}\right)_{ba}&={\frac {1}{2}}(\delta _{b,a+1}+\delta _{b+1,a}){\sqrt {(j+1)(a+b-1)-ab}}\\\left(J_{y}^{(j)}\right)_{ba}&={\frac {1}{2i}}(\delta _{b,a+1}-\delta _{b+1,a}){\sqrt {(j+1)(a+b-1)-ab}}\\&1\leq a,b\leq 2j+1~,\end{aligned}}}
for arbitrary j.
For example, the resulting spin matrices for spin 1, spin 3/2, and 5/2 are:
For ${\displaystyle j=1}$
{\displaystyle {\begin{aligned}J_{x}&={\frac {1}{\sqrt {2}}}{\begin{pmatrix}0&1&0\\1&0&1\\0&1&0\end{pmatrix}}\\J_{y}&={\frac {1}{\sqrt {2}}}{\begin{pmatrix}0&-i&0\\i&0&-i\\0&i&0\end{pmatrix}}\\J_{z}&={\begin{pmatrix}1&0&0\\0&0&0\\0&0&-1\end{pmatrix}}\end{aligned}}}
(Note, however, how these are in an equivalent, but different basis, the spherical basis, than the above i Ls in the Cartesian basis.[nb 3])
For ${\displaystyle j=\textstyle {\frac {3}{2}}}$:
{\displaystyle {\begin{aligned}J_{x}&={\frac {1}{2}}{\begin{pmatrix}0&{\sqrt {3}}&0&0\\{\sqrt {3}}&0&2&0\\0&2&0&{\sqrt {3}}\\0&0&{\sqrt {3}}&0\end{pmatrix}}\\J_{y}&={\frac {1}{2}}{\begin{pmatrix}0&-i{\sqrt {3}}&0&0\\i{\sqrt {3}}&0&-2i&0\\0&2i&0&-i{\sqrt {3}}\\0&0&i{\sqrt {3}}&0\end{pmatrix}}\\J_{z}&={\frac {1}{2}}{\begin{pmatrix}3&0&0&0\\0&1&0&0\\0&0&-1&0\\0&0&0&-3\end{pmatrix}}.\end{aligned}}}
For ${\displaystyle j=\textstyle {\frac {5}{2}}}$:
{\displaystyle {\begin{aligned}J_{x}&={\frac {1}{2}}{\begin{pmatrix}0&{\sqrt {5}}&0&0&0&0\\{\sqrt {5}}&0&2{\sqrt {2}}&0&0&0\\0&2{\sqrt {2}}&0&3&0&0\\0&0&3&0&2{\sqrt {2}}&0\\0&0&0&2{\sqrt {2}}&0&{\sqrt {5}}\\0&0&0&0&{\sqrt {5}}&0\end{pmatrix}}\\J_{y}&={\frac {1}{2}}{\begin{pmatrix}0&-i{\sqrt {5}}&0&0&0&0\\i{\sqrt {5}}&0&-2i{\sqrt {2}}&0&0&0\\0&2i{\sqrt {2}}&0&-3i&0&0\\0&0&3i&0&-2i{\sqrt {2}}&0\\0&0&0&2i{\sqrt {2}}&0&-i{\sqrt {5}}\\0&0&0&0&i{\sqrt {5}}&0\end{pmatrix}}\\J_{z}&={\frac {1}{2}}{\begin{pmatrix}5&0&0&0&0&0\\0&3&0&0&0&0\\0&0&1&0&0&0\\0&0&0&-1&0&0\\0&0&0&0&-3&0\\0&0&0&0&0&-5\end{pmatrix}}.\end{aligned}}}
and so on.
### Isomorphism with su(2)
The Lie algebras so(3) and su(2) are isomorphic. One basis for su(2) is given by[9]
${\displaystyle t_{1}={\frac {1}{2}}{\begin{bmatrix}0&-i\\-i&0\end{bmatrix}},\quad t_{2}={\frac {1}{2}}{\begin{bmatrix}0&-1\\1&0\end{bmatrix}},\quad t_{3}={\frac {1}{2}}{\begin{bmatrix}-i&0\\0&i\end{bmatrix}}.}$
These are related to the Pauli matrices by ti1/2iσi. The Pauli matrices abide by the physicists' convention for Lie algebras. In that convention, Lie algebra elements are multiplied by i, the exponential map (below) is defined with an extra factor of i in the exponent and the structure constants remain the same, but the definition of them acquires a factor of i. Likewise, commutation relations acquire a factor of i. The commutation relations for the ti are
${\displaystyle [t_{i},t_{j}]=\varepsilon _{ijk}t_{k},}$
where εijk is the totally anti-symmetric symbol with ε123 = 1. The isomorphism between so(3) and su(2) can be set up in several ways. For later convenience, so(3) and su(2) are identified by mapping
${\displaystyle L_{x}\leftrightarrow t_{1},\quad L_{y}\leftrightarrow t_{2},\quad L_{z}\leftrightarrow t_{3},}$
and extending by linearity.
## Exponential map
The exponential map for SO(3), is, since SO(3) is a matrix Lie group, defined using the standard matrix exponential series,
${\displaystyle \exp \colon {\mathfrak {so}}(3)\to SO(3);\quad A\mapsto e^{A}=\sum _{k=0}^{\infty }{\frac {1}{k!}}A^{k}=I+A+{\tfrac {1}{2}}A^{2}+\cdots .}$
For any skew-symmetric matrix Aso(3), eA is always in SO(3). The proof uses the elementary properties of the matrix exponential
${\displaystyle (e^{A})^{T}e^{A}=e^{A^{T}}e^{A}=e^{A^{T}+A}=e^{-A+A}=e^{A-A}=e^{A}(e^{A})^{T}=e^{0}=I.}$
since the matrices A and AT commute, this can be easily proven with the skew-symmetric matrix condition. This is not enough to show that so(3) is the corresponding Lie algebra for SO(3), and shall be proven separately.
The level of difficulty of proof depends on how a matrix group Lie algebra is defined. Hall (2003) defines the Lie algebra as the set of matrices AMn(ℝ)| etA ∈ SO(3) ∀t, in which case it is trivial. Rossmann (2002) uses for a definition derivatives of smooth curve segments in SO(3) through the identity taken at the identity, in which case it is harder.[10]
For a fixed A ≠ 0, etA, −∞ < t < ∞ is a one-parameter subgroup along a geodesic in SO(3). That this gives a one-parameter subgroup follows directly from properties of the exponential map.[11]
The exponential map provides a diffeomorphism between a neighborhood of the origin in the so(3) and a neighborhood of the identity in the SO(3).[12] For a proof, see Closed subgroup theorem.
The exponential map is surjective; this follows from the fact that every R ∈ SO(3), since every rotation leaves an axis fixed (Euler's rotation theorem), and is conjugate to a block diagonal matrix of the form
${\displaystyle D=\left({\begin{matrix}\cos \theta &-\sin \theta &0\\\sin \theta &\cos \theta &0\\0&0&1\end{matrix}}\right)=e^{\theta L_{z}},}$
such that A = BDB−1, and that
${\displaystyle Be^{\theta L_{z}}B^{-1}=e^{B\theta L_{z}B^{-1}},}$
together with the fact that so(3) is closed under the adjoint action of SO(3), meaning that BθLzB−1so(3).
Thus, e.g., it is easy to check the popular identity
${\displaystyle e^{-\pi L_{x}/2}~e^{\theta L_{z}}~e^{\pi L_{x}/2}=e^{\theta L_{y}}~.}$
As shown above, every element Aso(3) is associated with a vector ω = θ u, where u = (x,y,z) is a unit magnitude vector. Since u is in the null space of A, if one now rotates to a new basis, through some other orthogonal matrix O, with u as the z axis, the final column and row of the rotation matrix in the new basis will be zero.
Thus, we know in advance from the formula for the exponential that exp(OAOT) must leave u fixed. It is mathematically impossible to supply a straightforward formula for such a basis as a function of u, because its existence would violate the hairy ball theorem; but direct exponentiation is possible, and yields
{\displaystyle {\begin{aligned}\exp({\tilde {\boldsymbol {\omega }}})&{}=\exp(\theta ~({\boldsymbol {u\cdot L}}))=\exp \left(\theta {\begin{bmatrix}0&-z&y\\z&0&-x\\-y&x&0\end{bmatrix}}\right)\\[4pt]&{}={\boldsymbol {I}}+2cs~({\boldsymbol {u\cdot L}})+2s^{2}~({\boldsymbol {u\cdot L}})^{2}\\[4pt]&{}={\begin{bmatrix}2(x^{2}-1)s^{2}+1&2xys^{2}-2zcs&2xzs^{2}+2ycs\\2xys^{2}+2zcs&2(y^{2}-1)s^{2}+1&2yzs^{2}-2xcs\\2xzs^{2}-2ycs&2yzs^{2}+2xcs&2(z^{2}-1)s^{2}+1\end{bmatrix}},\end{aligned}}}
where c = cosθ2, s = sinθ2. This is recognized as a matrix for a rotation around axis u by the angle θ: cf. Rodrigues' rotation formula.
## Logarithm map
Given R ∈ SO(3), let
${\displaystyle A={\frac {R-R^{\mathrm {T} }}{2}}}$
denote the antisymmetric part and let ${\displaystyle \|A\|={\sqrt {-{\text{Tr}}(A^{2})/2}}}$.
Then, the logarithm of A is given by[8]
${\displaystyle \log R={\frac {\sin ^{-1}\|A\|}{\|A\|}}A.}$
This is manifest by inspection of the mixed symmetry form of Rodrigues' formula,
${\displaystyle e^{X}=I+{\frac {\sin \theta }{\theta }}X+2{\frac {\sin ^{2}{\frac {\theta }{2}}}{\theta ^{2}}}X^{2},\quad \theta =\|X\|,}$
where the first and last term on the right-hand side are symmetric.
## Baker–Campbell–Hausdorff formula
Suppose X and Y in the Lie algebra are given. Their exponentials, exp(X) and exp(Y), are rotation matrices, which can be multiplied. Since the exponential map is a surjection, for some Z in the Lie algebra, exp(Z) = exp(X) exp(Y), and one may tentatively write
${\displaystyle Z=C(X,Y),}$
for C some expression in X and Y. When exp(X) and exp(Y) commute, then Z = X + Y, mimicking the behavior of complex exponentiation.
The general case is given by the more elaborate BCH formula, a series expansion of nested Lie brackets.[13] For matrices, the Lie bracket is the same operation as the commutator, which monitors lack of commutativity in multiplication; this general expansion unfolds as follows,[nb 4]
${\displaystyle Z=C(X,Y)=X+Y+{\tfrac {1}{2}}[X,Y]+{\tfrac {1}{12}}[X,[X,Y]]-{\tfrac {1}{12}}[Y,[X,Y]]+\cdots ~.}$
The infinite expansion in the BCH formula for SO(3) reduces to a compact form,
${\displaystyle Z=\alpha X+\beta Y+\gamma [X,Y],}$
for suitable trigonometric function coefficients (α, β, γ).
The trigonometric coefficients
The (α, β, γ) are given by
${\displaystyle \alpha =\varphi \cot(\varphi /2)~\gamma ,\qquad \beta =\theta \cot(\theta /2)~\gamma ,\qquad \gamma ={\frac {\sin ^{-1}d}{d}}{\frac {c}{\theta \varphi }}~~,}$
where
{\displaystyle {\begin{aligned}c&={\frac {1}{2}}\sin \theta \sin \varphi -2\sin ^{2}{\frac {\theta }{2}}\sin ^{2}{\frac {\varphi }{2}}\cos(\angle (u,v)),\quad a=c\cot(\varphi /2),\quad b=c\cot(\theta /2),\\d&={\sqrt {a^{2}+b^{2}+2ab\cos(\angle (u,v))+c^{2}\sin ^{2}(\angle (u,v))}}~~,\end{aligned}}}
for
${\displaystyle \theta ={\frac {1}{\sqrt {2}}}\|X\|~,\quad \varphi ={\frac {1}{\sqrt {2}}}\|Y\|~,\quad \angle (u,v)=\cos ^{-1}{\frac {\langle X,Y\rangle }{\|X\|\|Y\|}}~.}$
The inner product is the Hilbert–Schmidt inner product and the norm is the associated norm. Under the hat-isomorphism,
${\displaystyle \langle u,v\rangle ={\frac {1}{2}}\operatorname {Tr} X^{\mathrm {T} }Y,}$
which explains the factors for θ and φ. This drops out in the expression for the angle.
It is worthwhile to write this composite rotation generator as
${\displaystyle \alpha X+\beta Y+\gamma [X,Y]~{\underset {{\mathfrak {so}}(3)}{=}}~X+Y+{\tfrac {1}{2}}[X,Y]+{\tfrac {1}{12}}[X,[X,Y]]-{\tfrac {1}{12}}[Y,[X,Y]]+\cdots ,}$
to emphasize that this is a Lie algebra identity.
The above identity holds for all faithful representations of so(3). The kernel of a Lie algebra homomorphism is an ideal, but so(3), being simple, has no nontrivial ideals and all nontrivial representations are hence faithful. It holds in particular in the doublet or spinor representation; the same explicit formula thus follows in a simpler way through Pauli matrices, cf. the 2×2 derivation for SU(2).
The SU(2) case
The Pauli vector version of the same BCH formula is the somewhat simpler group composition law of SU(2),
${\displaystyle e^{ia'({\hat {u}}\cdot {\vec {\sigma }})}e^{ib'({\hat {v}}\cdot {\vec {\sigma }})}=\exp \left({\frac {c'}{\sin c'}}\sin a'\sin b'~\left((i\cot b'{\hat {u}}+i\cot a'{\hat {v}})\cdot {\vec {\sigma }}+{\frac {1}{2}}[i{\hat {u}}\cdot {\vec {\sigma }},i{\hat {v}}\cdot {\vec {\sigma }}]\right)\right)~,}$
where
${\displaystyle \cos c'=\cos a'\cos b'-{\hat {u}}\cdot {\hat {v}}\sin a'\sin b'~,}$
the spherical law of cosines. (Note a', b' ,c' are angles, not the a,b,c above.)
This is manifestly of the same format as above,
${\displaystyle Z=\alpha 'X+\beta 'Y+\gamma '[X,Y],}$
with
${\displaystyle X=ia'{\hat {u}}\cdot \mathbf {\sigma } ,\quad Y=ib'{\hat {v}}\cdot \mathbf {\sigma } ~\in {\mathfrak {su}}(2),}$
so that
{\displaystyle {\begin{aligned}\alpha '&={\frac {c'}{\sin c'}}{\frac {\sin a'}{a'}}\cos b'\\\beta '&={\frac {c'}{\sin c'}}{\frac {\sin b'}{b'}}\cos a'\\\gamma '&={\frac {1}{2}}{\frac {c'}{\sin c'}}{\frac {\sin a'}{a'}}{\frac {\sin b'}{b'}}~.\end{aligned}}}
For uniform normalization of the generators in the Lie algebra involved, express the Pauli matrices in terms of t-matrices, σ →2i t, so that
${\displaystyle a'\mapsto -{\frac {\theta }{2}},\quad b'\mapsto -{\frac {\varphi }{2}}.}$
To verify then these are the same coefficients as above, compute the ratios of the coefficients,
{\displaystyle {\begin{aligned}{\frac {\alpha '}{\gamma '}}&={\theta }\cot {\frac {\theta }{2}}&={\frac {\alpha }{\gamma }}\\{\frac {\beta '}{\gamma '}}&=\varphi \cot {\frac {\varphi }{2}}&={\frac {\beta }{\gamma }}~.\end{aligned}}}
Finally, γ = γ' given the identity d = sin 2c'.
For the general n × n case, one might use Ref.[14]
The quaternion case
The quaternion formulation of the composition of two rotations RB and RA also yields directly the rotation axis and angle of the composite rotation RC=RBRA.
Let the quaternion associated with a spatial rotation R is constructed from its rotation axis S and the rotation angle φ this axis. The associated quaternion is given by,
${\displaystyle S=\cos {\frac {\varphi }{2}}+\sin {\frac {\varphi }{2}}\mathbf {S} .}$
Then the composition of the rotation RR with RA is the rotation RC=RBRA with rotation axis and angle defined by the product of the quaternions
${\displaystyle A=\cos {\frac {\alpha }{2}}+\sin {\frac {\alpha }{2}}\mathbf {A} \quad {\text{and}}\quad B=\cos {\frac {\beta }{2}}+\sin {\frac {\beta }{2}}\mathbf {B} ,}$
that is
${\displaystyle C=\cos {\frac {\gamma }{2}}+\sin {\frac {\gamma }{2}}\mathbf {C} ={\Big (}\cos {\frac {\beta }{2}}+\sin {\frac {\beta }{2}}\mathbf {B} {\Big )}{\Big (}\cos {\frac {\alpha }{2}}+\sin {\frac {\alpha }{2}}\mathbf {A} {\Big )}.}$
Expand this product to obtain
${\displaystyle \cos {\frac {\gamma }{2}}+\sin {\frac {\gamma }{2}}\mathbf {C} ={\Big (}\cos {\frac {\beta }{2}}\cos {\frac {\alpha }{2}}-\sin {\frac {\beta }{2}}\sin {\frac {\alpha }{2}}\mathbf {B} \cdot \mathbf {A} {\Big )}+{\Big (}\sin {\frac {\beta }{2}}\cos {\frac {\alpha }{2}}\mathbf {B} +\sin {\frac {\alpha }{2}}\cos {\frac {\beta }{2}}\mathbf {A} +\sin {\frac {\beta }{2}}\sin {\frac {\alpha }{2}}\mathbf {B} \times \mathbf {A} {\Big )}.}$
Divide both sides of this equation by the identity, which is the law of cosines on a sphere,
${\displaystyle \cos {\frac {\gamma }{2}}=\cos {\frac {\beta }{2}}\cos {\frac {\alpha }{2}}-\sin {\frac {\beta }{2}}\sin {\frac {\alpha }{2}}\mathbf {B} \cdot \mathbf {A} ,}$
and compute
${\displaystyle \tan {\frac {\gamma }{2}}\mathbf {C} ={\frac {\tan {\frac {\beta }{2}}\mathbf {B} +\tan {\frac {\alpha }{2}}\mathbf {A} +\tan {\frac {\beta }{2}}\tan {\frac {\alpha }{2}}\mathbf {B} \times \mathbf {A} }{1-\tan {\frac {\beta }{2}}\tan {\frac {\alpha }{2}}\mathbf {B} \cdot \mathbf {A} }}.}$
This is Rodrigues' formula for the axis of a composite rotation defined in terms of the axes of the two rotations, he derived this formula in 1840 (see page 408).[15]
The three rotation axes A, B, and C form a spherical triangle and the dihedral angles between the planes formed by the sides of this triangle are defined by the rotation angles.
## Infinitesimal rotations
The matrices in the Lie algebra are not themselves rotations; the skew-symmetric matrices are derivatives. An actual "differential rotation", or infinitesimal rotation matrix has the form
${\displaystyle I+A\,d\theta ~,}$
where is vanishingly small and Aso(3).
These matrices do not satisfy all the same properties as ordinary finite rotation matrices under the usual treatment of infinitesimals .[16] To understand what this means, one considers
${\displaystyle dA_{\mathbf {x} }={\begin{bmatrix}1&0&0\\0&1&-d\theta \\0&d\theta &1\end{bmatrix}}~.}$
First, test the orthogonality condition, QTQ = I. The product is
${\displaystyle dA_{\mathbf {x} }^{T}\,dA_{\mathbf {x} }={\begin{bmatrix}1&0&0\\0&1+d\theta ^{2}&0\\0&0&1+d\theta ^{2}\end{bmatrix}},}$
differing from an identity matrix by second order infinitesimals, discarded here. So, to first order, an infinitesimal rotation matrix is an orthogonal matrix.
Next, examine the square of the matrix,
${\displaystyle dA_{\mathbf {x} }^{2}={\begin{bmatrix}1&0&0\\0&1-d\theta ^{2}&-2d\theta \\0&2\,d\theta &1-d\theta ^{2}\end{bmatrix}}~.}$
Again discarding second order effects, note that the angle simply doubles; this hints at the most essential difference in behavior, which we can exhibit with the assistance of a second infinitesimal rotation,
${\displaystyle dA_{\mathbf {y} }={\begin{bmatrix}1&0&d\varphi \\0&1&0\\-d\varphi &0&1\end{bmatrix}}.}$
Compare the products dAx dAy to dAydAx,
{\displaystyle {\begin{aligned}dA_{\mathbf {x} }\,dA_{\mathbf {y} }&{}={\begin{bmatrix}1&0&d\varphi \\d\theta \,d\varphi &1&-d\theta \\-d\varphi &d\theta &1\end{bmatrix}}\\dA_{\mathbf {y} }\,dA_{\mathbf {x} }&{}={\begin{bmatrix}1&d\theta \,d\varphi &d\varphi \\0&1&-d\theta \\-d\varphi &d\theta &1\end{bmatrix}}.\\\end{aligned}}}
Since ${\displaystyle d\theta \,d\varphi }$ is second-order, we discard it: thus, to first order, multiplication of infinitesimal rotation matrices is commutative. In fact,
${\displaystyle dA_{\mathbf {x} }\,dA_{\mathbf {y} }=dA_{\mathbf {y} }\,dA_{\mathbf {x} },\,\!}$
again to first order. In other words, the order in which infinitesimal rotations are applied is irrelevant.
This useful fact makes, for example, derivation of rigid body rotation relatively simple, but one must always be careful to distinguish (the first order treatment of) these infinitesimal rotation matrices from both finite rotation matrices and from Lie algebra elements. When contrasting the behavior of finite rotation matrices in the BCH formula above with that of infinitesimal rotation matrices, where all the commutator terms will be second order infinitesimals one finds a bona fide vector space. Technically, this dismissal of any second order terms amounts to Group contraction.
## Realizations of rotations
We have seen that there are a variety of ways to represent rotations:
## Spherical harmonics
The group SO(3) of three-dimensional Euclidean rotations has an infinite-dimensional representation on the Hilbert space
${\displaystyle L^{2}(\mathbf {S} ^{2})=\operatorname {span} \left\{Y_{m}^{\ell },\ell \in \mathbf {N} ^{+},-\ell \leqslant m\leqslant \ell \right\},}$
where ${\displaystyle Y_{m}^{\ell }}$ are spherical harmonics. Its elements are square integrable complex-valued functions[nb 5] on the sphere; the inner product on this space is given by
${\displaystyle \langle f,g\rangle =\int _{\mathbf {S} ^{2}}{\overline {f}}g\,d\Omega =\int _{0}^{2\pi }\int _{0}^{\pi }{\overline {f}}g\sin \theta \,d\theta \,d\varphi .}$
(H1)
If f is an arbitrary square integrable function defined on the unit sphere S2, then it can be expressed as[17]
${\displaystyle |f\rangle =\sum _{\ell =1}^{\infty }\sum _{m=-\ell }^{m=\ell }|Y_{m}^{\ell }\rangle \langle Y_{m}^{\ell }|f\rangle ,\qquad f(\theta ,\varphi )=\sum _{\ell =1}^{\infty }\sum _{m=-\ell }^{m=\ell }f_{\ell m}Y_{m}^{\ell }(\theta ,\varphi ),}$
(H2)
where the expansion coefficients are given by
${\displaystyle f_{\ell m}=\langle Y_{m}^{\ell },f\rangle =\int _{\mathbf {S} ^{2}}{\overline {Y_{m}^{\ell }}}f\,d\Omega =\int _{0}^{2\pi }\int _{0}^{\pi }{\overline {Y_{m}^{\ell }}}(\theta ,\varphi )f(\theta ,\varphi )\sin \theta \,d\theta \,d\varphi .}$
(H3)
The Lorentz group action restricts to that of SO(3) and is expressed as
${\displaystyle (\Pi (R)f)(\theta (x),\varphi (x))=\sum _{\ell =1}^{\infty }\sum _{m=-\ell }^{m=\ell }\sum _{m'=-\ell }^{m'=\ell }D_{mm'}^{(\ell )}(R)f_{\ell m'}Y_{m}^{\ell }\left(\theta (R^{-1}x),\varphi (R^{-1}x)\right),\qquad R\in \mathrm {SO} (3),\quad x\in \mathbf {S} ^{2}.}$
(H4)
This action is unitary, meaning that
${\displaystyle \langle \Pi (R)f,\Pi (R)g\rangle =\langle f,g\rangle \qquad \forall f,g\in \mathbf {S} ^{2},\quad \forall R\in \mathrm {SO} (3).}$
(H5)
The D() can be obtained from the D(m, n) of above using Clebsch–Gordan decomposition, but they are more easily directly expressed as an exponential of an odd-dimensional su(2)-representation (the 3-dimensional one is exactly so(3)).[18][19] In this case the space L2(S2) decomposes neatly into an infinite direct sum of irreducible odd finite-dimensional representations V2i + 1, i = 0, 1, … according to[20]
${\displaystyle L^{2}(\mathbf {S} ^{2})=\sum _{i=0}^{\infty }V_{2i+1}\equiv \bigoplus _{i=0}^{\infty }\operatorname {span} \left\{Y_{m}^{2i+1}\right\}.}$
(H6)
This is characteristic of infinite-dimensional unitary representations of SO(3). If Π is an infinite-dimensional unitary representation on a separable[nb 6] Hilbert space, then it decomposes as a direct sum of finite-dimensional unitary representations;[17] such a representation is thus never irreducible. All irreducible finite-dimensional representations (Π, V) can be made unitary by an appropriate choice of inner product,[17]
${\displaystyle \langle f,g\rangle _{U}\equiv \int _{\mathrm {SO} (3)}\langle \Pi (R)f,\Pi (R)g\rangle \,dg={\frac {1}{8\pi ^{2}}}\int _{0}^{2\pi }\int _{0}^{\pi }\int _{0}^{2\pi }\langle \Pi (R)f,\Pi (R)g\rangle \sin \theta \,d\varphi \,d\theta \,d\psi ,\quad f,g\in V,}$
where the integral is the unique invariant integral over SO(3) normalized to 1, here expressed using the Euler angles parametrization. The inner product inside the integral is any inner product on V.
## Generalizations
The rotation group generalizes quite naturally to n-dimensional Euclidean space, Rn with its standard Euclidean structure. The group of all proper and improper rotations in n dimensions is called the orthogonal group O(n), and the subgroup of proper rotations is called the special orthogonal group SO(n), which is a Lie group of dimension n(n − 1)/2.
In special relativity, one works in a 4-dimensional vector space, known as Minkowski space rather than 3-dimensional Euclidean space. Unlike Euclidean space, Minkowski space has an inner product with an indefinite signature. However, one can still define generalized rotations which preserve this inner product; such generalized rotations are known as Lorentz transformations and the group of all such transformations is called the Lorentz group.
The rotation group SO(3) can be described as a subgroup of E+(3), the Euclidean group of direct isometries of Euclidean R3. This larger group is the group of all motions of a rigid body: each of these is a combination of a rotation about an arbitrary axis and a translation along the axis, or put differently, a combination of an element of SO(3) and an arbitrary translation.
In general, the rotation group of an object is the symmetry group within the group of direct isometries; in other words, the intersection of the full symmetry group and the group of direct isometries. For chiral objects it is the same as the full symmetry group.
## Remarks
1. ^ This is effected by first applying a rotation gφ through φ about the z-axis to take the x-axis to the line L, the intersection between the planes xy and x´y´, the latter being the rotated xy-plane. Then rotate with gθ through θ about L to obtain the new z-axis from the old one, and finally rotate by gψ through an angle ψ about the new z-axis, where ψ is the angle between L and the new x-axis. In the equation, gθ and gψ are expressed in a temporary rotated basis at each step, which is seen from their simple form. To transform these back to the original basis, observe that gθ =gφgθgφ−1. Here boldface means that the rotation is expressed in the original basis. Likewise, gψ =gφgθgφ−1gφgψ[gφgθgφ−1gφ]−1. Thus gψgθgφ = gφgθgφ−1gφgψ[gφgθgφ−1gφ]−1*gφgθgφ−1*gφ = gφgθgψ.
2. ^ For an alternative derivation of so(3), see Classical group.
3. ^ Specifically, ${\displaystyle UJ_{\alpha }U^{\dagger }=iL_{\alpha }}$ for ${\displaystyle U={\frac {1}{\sqrt {2}}}\left({\begin{matrix}-1&0&1\\-i&0&-i\\0&{\sqrt {2}}&0\end{matrix}}\right)}$.
4. ^ For a full proof, see Derivative of the exponential map. Issues of convergence of this series to the correct element of the Lie algebra are here swept under the carpet. Convergence is guaranteed when ||X|| + ||Y|| < log 2 and ||Z|| < log 2. The series may still converge even if these conditions aren't fulfilled. A solution always exists since exp is onto in the cases under consideration.
5. ^ The elements of L2(S2) are actually equivalence classes of functions. two functions are declared equivalent if they differ merely on a set of measure zero. The integral is the Lebesgue integral in order to obtain a complete inner product space.
6. ^ A Hilbert space is separable if and only if it has a countable basis. All separable Hilbert spaces are isomorphic.
## Notes
1. ^ Jacobson (2009), p. 34, Ex. 14.
2. ^ n × n real matrices are identical to linear transformations of Rn expressed in its standard basis.
3. ^ Hall 2015 Proposition 1.17
4. ^ Rossmann 2002 p. 95.
5. ^ These expressions were, in fact, seminal in the development of quantum mechanics in the 1930s, cf. Ch III, § 16, B.L. van der Waerden, 1932/1932
6. ^ Hall 2015 Proposition 3.24
7. ^ Rossmann 2002
8. ^ a b Engø 2001
9. ^ Hall 2015 Example 3.27
10. ^ See Rossmann 2002, theorem 3, section 2.2.
11. ^ Rossmann 2002 Section 1.1.
12. ^ Hall 2003 Theorem 2.27.
13. ^ Hall 2003, Ch. 3; Varadarajan 1984, §2.15
14. ^ Curtright, Fairlie & Zachos 2014 Group elements of SU(2) are expressed in closed form as finite polynomials of the Lie algebra generators, for all definite spin representations of the rotation group.
15. ^ Rodrigues, O. (1840), Des lois géométriques qui régissent les déplacements d’un système solide dans l’espace, et la variation des coordonnées provenant de ses déplacements con- sidérés indépendamment des causes qui peuvent les produire, Journal de Mathématiques Pures et Appliquées de Liouville 5, 380–440.
16. ^ (Goldstein, Poole & Safko 2002, §4.8)
17. ^ a b c Gelfand, Minlos & Shapiro 1963
18. ^ In Quantum Mechanics – non-relativistic theory by Landau and Lifshitz the lowest order D are calculated analytically.
19. ^ Curtright, Fairlie & Zachos 2014 A formula for D() valid for all is given.
20. ^ Hall 2003 Section 4.3.5.
## References
• Boas, Mary L. (2006), Mathematical Methods in the Physical Sciences (3rd ed.), John Wiley & sons, pp. 120, 127, 129, 155ff and 535, ISBN 978-0471198260
• Curtright, T. L.; Fairlie, D. B.; Zachos, C. K. (2014), "A compact formula for rotations as spin matrix polynomials", SIGMA, 10: 084, arXiv:1402.3541, Bibcode:2014SIGMA..10..084C, doi:10.3842/SIGMA.2014.084
• Engø, Kenth (2001), "On the BCH-formula in so(3)", BIT Numerical Mathematics, 41 (3): 629–632, doi:10.1023/A:1021979515229, ISSN 0006-3835 [1]
• Gelfand, I.M.; Minlos, R.A.; Shapiro, Z.Ya. (1963), Representations of the Rotation and Lorentz Groups and their Applications, New York: Pergamon Press
• Hall, Brian C. (2015), Lie Groups, Lie Algebras, and Representations: An Elementary Introduction, Graduate Texts in Mathematics, 222 (2nd ed.), Springer, ISBN 978-3319134666
• Jacobson, Nathan (2009), Basic algebra, 1 (2nd ed.), Dover Publications, ISBN 978-0-486-47189-1
• Joshi, A. W. (2007), Elements of Group Theory for Physicists, New Age International, pp. 111ff, ISBN 81-224-0975-X
• Rossmann, Wulf (2002), Lie Groups – An Introduction Through Linear Groups, Oxford Graduate Texts in Mathematics, Oxford Science Publications, ISBN 0 19 859683 9
• van der Waerden, B. L. (1952), Group Theory and Quantum Mechanics, Springer Publishing, ISBN 978-3642658624 (translation of the original 1932 edition, Die Gruppentheoretische Methode in Der Quantenmechanik).
• Veltman, M.; 't Hooft, G.; de Wit, B. (2007). "Lie Groups in Physics (online lecture)" (PDF). Retrieved 2016-10-24.. | 16,838 | 53,405 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 112, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.6875 | 4 | CC-MAIN-2019-39 | longest | en | 0.934767 |
null | null | null | null | null | null | News Article
New Phoenix Wright: Dual Destinies Trailers Reveal Mature Rating
Posted by Tim Latshaw
It’s “the nature of the various crimes and storylines”
From the web
Game Screenshots
User Comments (113)
Jaz007 said:
I'm certainly not getting this now. This seriously has to be an idiotic move by capcom though, they have nothing to gain by an M rating. People aren't going to want to buy the game more because it's rated M. People are only going to not buy the game because of it. An M rating doesn't even fit Phoenix Wright, it's too lighthearted and fun. I'm not even sure I really believe this, I really can't understand why it would be M. This can't possibly do the series any favors. I honestly think this will hurt the games sales.
Dark-Luigi said:
M rated...But the others were rated T...Hm. I don't know, but I guess I'll still consider it... :(
psycoticdev1l said:
@Jaz007 I don't understand why you won't buy it all because of the rating, when you seem to have liked the series before. I personally think this could be a good thing, it's possible they are trying something new in the series or getting into better detail that they could not have done before with a T rating.
ToastyYogurt said:
@psycoticdev1l: Considering the inclusion of sentences like "Nobody is going to buy the game becuase of it. People are only going to not buy the game because of it." in his comment, I believe he is joking. :)
Jaz007 said:
@psycoticdev1l I've played Trials and Tribulations, I still need to finish. This game isn't that high on my radar. I think it's a legitamittly bad business move. Most people who want a game to be M rated don't want to play Phoenix Wright, it's 3DS which has a large amount of customers that don't play or want M games. An M rating doesn't represent the franchise very well either. I don't see the franchise going where a lot of fans of it want it to with an M rating either. I think they will be alienating the existing without really gaining any new ones here. The series doesn't really need either. Most poeple don't play PW for detailed rape cases or detailed gory murder scenes.
psuboy172 said:
Still a 1 day release for me...but this is what happened to Mirror of Fate...I don't like when games try to be M. The first 5 AA games were all great! Why change rating?
Jaz007 said:
@ToastyYogurt I didn't mean it that way, I meant that I don't think the M rating is going to increase sales. I edited it to try to clarify what I meant. I don't think nobody will buy it due an M rating. As you can see, the reaction to it here hasn't been very positive even form people who buy M games and will still buy it too.
Kyloctopus said:
@Jaz007 I see your view of things, but isn't most of the AA fans at the age of 17 now? Or will buy it anyways. Afterall, that doesn't seem to stop Activision going to the bank every year for Call of Duty.
CanisWolfred said:
Nor can anybody be stopped from buying an M Rated game from the eShop. Who exactly is gonna card them, huh?
haniwa said:
Ha, maybe they just squeezed in some swear word or two to get the M rating.
CAM290 said:
I'm not sure how things are nowadays for younger gamers, but I remember when i was around 10 (now 24) i had quite a few M rated titles on my N64. My mom or grandma could always acknowledge & confirm that they were ok with it to the lady (or gentlemen) over the counter when i would purchase these games with allowance money. Funny enough, I was just ID carded at Best Buy when i purchased Ninja Gaiden 3 for Wii U, I showed my ID card and laughed. "How old do you think I am?" I asked the lady over the counter. "hmm 16 or 17?" I laughed and said "wow now your just trying to sweet talk me!"
Jukilum said:
@Kyloctapus: I am old enough to buy an M rated game at retail should I desire, but I have moral qualms with the content in such games. I have been a fan of the series and have played all of the games that are out so far in the West, but Capcom has lost me with this.
Caryslan said:
Ok, before everyone blames Capcom, keep in mind one thing. The ESRB has become alot more strict over the years in how they rate games. What might have flown 10 years ago as a Teen rated game might get an M rating now. The content might not be any worse then the other games overall. It might have been the crime scene photos that hurt the game.
No, I don't think we are going to get F-bombs or sex scenes in this game. I think the rating was due to the graphic nature of the crime scenes and photos that have been in the series. SImply put, virtually every case in the series has been a murder case. Maybe that and the increased power of the 3DS played a role in causing the shift in rating.
Or maybe its the ESRB playing it safe since video games have been targeted in recent months. What might have flown in the previous games as a T rating might now be an M rating now. Keep in mind, the standards keep changing over time.
So unless Capcom really crossed the line, I don't think the blame lies with them alone. I think the content is the same as its always been. Maybe a bit more detail on crime scene photos might have been enough to push it over the edge. The AA games always danced close to the line several times in the past.
Gregor said:
@Jaz007 number one, that's EXTREMELY brash and shallow minded of you and number two, that humor might have earned it the M rating have you ever thought of that? The humor of the series can be quite crude at times.
psycoticdev1l said:
You're implying almost no one over the age of 18 has a 3DS/doesn't want to play M rated games, which is so far from true. And since you have played these games, you should know that they have pushed that T rating as far a it will go, from murders/suicides, mild sexual themes and the like, very 'mature' stuff. My theory for why the rating is now an M, is because the crime scenes and murders have become much more detailed and have raised the rating because of that. Or it could be for some other reason, who knows, maybe they wanted to try something new or different with the series instead of resting on their laurels and playing it safe. But honestly, if you are not going to play a game that is in a series you like all because of the rating change, that's really sad you think like that.
Jaz007 said:
@UgliestSoup You don't need to make fun of me. I may be a bit unhappy about this, but I think it a bad idea for Capcom business wise. Can you tell me how the sales will increase from this? I'm not just saying all this because I don't personally like the M rating. Ethics aside I think it's a bad idea. I get unhappy that a number of games get M rated, but I can understand why they would do it, but I can't make sense of this.
Gregor said:
@Caryslan I like this person. He/she has a point. Besides, do you HONESTLY think that Capcom would purposefully net the game a mature rating? C'mon people.
Gregor said:
@Jaz007 you think Capcom purposefully got the game an M rating? Geez. How daft can one be. I can garuntee you that almost no company would go after an M rating just to increase sales. It is what it is, Pheonix wright. If you were an actual fan you would know that.
Gregor said:
@Jukilum does this look like Call of Duty to you? It it an FPS with zombies and war and dull textures? No. It's a point and click adventure.
Jaz007 said:
@psycoticdev1l I'm not implying that, I'm implying that there is a good porition of the market this could hurt. A lot of kids own 3DSs who can't play M because of their parents, not all 3DS owners, but the market is definitely there. I haven't personally noticed that much extreme stuff from my time with trials and tribulations, it could have just been me with that. I'm sorry if you don't like that I can't play M games or don't like the content they have. Can you some of you guys stop jumping down my throat too? I'm the only person that has gotten blasted for personnally not being so sure about the game now because it's M.
JesseMcCloud said:
This is yet another reason the ESRB needs to look at PEGI's example and start making more ratings, with more detailed descriptors. At LEAST a T-15 rating for games like Uncharted, Halo, and this: Stuff a little too edgy for a T, but not adult enough to get lumped in with Resident Evil and God of War.
Objection said:
Stop being butthurt, Community. Geez.
Anyway, I'm surprised but it's probably the ESRB being over cautious. I've long thought we need a Teen+ rating since E10 largely replaced the old Teen without having a Text Adventure Mystery game get an Adult rating. (I'm aware that this also has a C CERO and 16 PEGI fyi.)
I do hope that this is part of the game trying new things instead of being needlessly or style-breakingly violent/whatever.
I don't think the rating will effect sales in the same way it would if it was retail. You can enter any age on the eshop in my experience. (I'm too lazy to add my exact age each time.)
ThetaRobo said:
@Jukilum Yes, because boycotting a video game that decides to appeal to different (decidedly more "adult") sensibilities totally reflects on one's morality.
@Jaz007 Learn how the ratings system works, please. The usual trend with game companies is they tone-down a game to avoid higher ratings (e.g. OOT). Ace Attorney has been (and always will be) a franchise that revolves around courtroom scenarios that usually stem from murder (gruesome ones at that, depending on the case). Bravo to Capcom for planning on releasing the game as intended. There's a big difference between "hardcore" blood and "realistic" blood; one is there for a reason. You seem to be speaking out of context a lot, plus you've proven your experience with the franchise to be very little, so I assume that would serve to detach yourself further from the topic at hand.
SMT Games are historically M Rated, not as "hardcore" as you describe M Rated games to be.
TimLatshaw said:
I of course can only speculate without official word from Capcom or the ESRB, but I find it highly unlikely that Capcom would intentionally make Dual Destinies an M-rated title as a means of boosting sales. It doesn't seem to make any business sense to do so--especially with an established franchise that's largely considered Nintendo-centric--and it doesn't seem they've been pushing the game to potential audiences as more "mature," either. I would have to guess now that the ESRB has either become more strict as noted in above comments or there is a bit more of something than in other games just as part of the story--but probably not for pure shock value. Sometimes just a bit more blood could be enough to change a rating.
My opinion does not reflect the opinions of Nintendo Life or any of its subsidiaries, yada yada.
ThetaRobo said:
This whole thing brings back memories of the Metroid Prime 3 M Rating controversy. Of course, it never was rated M to begin with, but people sure did freak out when those fake GameStop cases went on display before release.
Jukilum said:
@ThetaRobo I'm not entirely sure whether you're rrplying to my second or third comment. I'll assume the latter. I was mainly referring to how most of the replies to @Jaz007 include personal insults because of his views instead of relying on the merits of his arguments. There are some people replying to his actual arguments, but most of those still include insults.
AVahne said:
Meh, don't care. Ratings like these mean absolutely nothing to me as long as the game is fun. Since I'm interested by the fact that'll be eShop only in the states, I might pick this up next year. Hoping Senran Kagura Burst and Project Mirai 1 and 2 can make it here through the eShop.
SCAR392 said:
Ya, I'm betting it wasn't intentional. Besides, anyone who played Pheonix Wright when it first came out has gotten at least 9 years older.
ThetaRobo said:
@Jukilum Fair enough, but where are the morals in shooting down a game based on the rating instead of the content? People are WAY too quick to judge a game by its packaging these days. From what we can see, there aren't any big changes in the content themes between the GBA/DS games to this one. Simply turning down a game that you've considered buying based simply on the rating and not how much you like the content isn't "ethical", more like flawed judgement.
Morpheel said:
Shut up about the rating already... Does that second video imply Wright is dead?!
Objection said:
@morphtorok- It does seem to imply that. And also simultaneously imply that he is again the defendant.
Dude, if they actually killed him off, that would be the wackest s---, I don't know if I'd be angry or impressed.
theblackdragon said:
so first everyone flips out because some people have said they don't want to buy digital, now you're all flipping out because someone isn't interested in an M-rated PW game. can't we all just respect that other people have their own reasons for doing things and leave it at that without jumping down their throats? :/
Senario said:
@Jaz007 Agreed, maybe some small subset of people think that an M rating actually means the people who play the game are mature. After all, isn't that what most young people want?
I wasn't getting this game because of the digital only thing, this rating doesn't really affect me.
psycoticdev1l said:
@Jaz007 I'm not trying to jump down your throat, however I just don't think the game being rated M will change it dramatically. If you don't like that, fine, I'm not gonna force you to, but it seems silly that you won't buy it because of the rating, especially since we don't know what caused it yet. If I were you, I'd wait to see what made it get a M rating and judge it then instead of immediately dismissing it now.
naut said:
I think the negative reaction is silly. I for one am excited that Phoenix Wright may be dealing with more mature subjects. If you're going to automatically write a game off because of its rating then I disagree with you.
benjamines said:
Doesn't make a bit of difference to me. I'll be getting this as soon as it's released.
RR529 said:
I don't think Capcom was going for an "M" rating, and I don't think the tone of the game is going to be much, if any, darker than the previous entries.
Despite the lighthearted tone of the series, murder cases (including blood) are nothing new to it either. It's probably that the images appear to be more graphic simply due to the 3DS's power, and that jump in detail was just enough for the ESRB to list it as "M", instead of "T".
Gregor said:
the question was rhetorical. please play nicely or you may wind up banned — TBD
BigDaddysPizza said:
Wow. I'm shocked. I guess this fits with the dark tone.
coolvw93 said:
RevolverLink said:
I've been greatly anticipating this game for 5 and a half years now, so an "M" next to the title or even a digital-only release won't deter me in the slightest, but I will admit that this seems a bit of an odd decision by the ESRB at the moment.
I have yet to see anything from Dual Destinies that would indicate a significant change in mood, theme or tone from all of the previous Ace Attorney entries, but I also haven't played the game and I'm basically on media blackout until release, so for all I know, the violence could be less implied than it previously was in the series.
Stark_Nebula said:
This could spring up a pretty good discussion on the rating of games versus the subject matter of books. In many ways, Ace Attorney is like a novel, filled with stories of crimes and murder. The irony is that the educational system often promotes classics with such subject matter for 8th and 9th graders but is never frowned upon, while the opposite may be true for this. I'm not saying 10 year old should play it, nor am I saying this game is in the same way a "classic" with superb writing, but I'm sure a lot of book-loving teens would enjoy the story this game has to offer because it did not limit itself as much/in anyway.
SonataAndante said:
In case nobody considered it the original games were on the GameBoy Advance, then ported to DS. As a result, the crime scenes and whatnot weren't exactly bursting with graphical fidelity. It was all just static pictures. Now everything is done in 3D, making it a bit more "realistic," or as realistic as PW will ever look probably. That alone could jump the rating.
Magikarp3 said:
Gosh, this is a strange move.
Let's just hope it's rated M for Maya XD
apologies for making the worst jokes ever
DaemonSword said:
Let's hope one of the cases doesn't involve a neighborhood watch guy shooting a kid in hoodie, with a bag of skittles in his hand. lol
DualWielding said:
I think the M writing is because of the terrorism bombing thing for some reason censors are more sensitive to stuff that deal with terrorism and I can't believe anyone would complain about the M rating is something Capcom has no control over and doesn't affect the quality of the game at alll
Shambo said:
I hope it simply gets grittier and more serious, while keeping it's humor. I also -perhaps even more- hope it will get a boxed European release. Together with the Layton crossover and Time Travellers.
Dormouse said:
Wow this sounds interesting. Ive been curious about these games cuz I hear good tings but now im really curious about this one.
Einherjar said:
I cant understand the fuzz about it. Did the ratings ever stop from kids getting their hands on any Grand theft modern battlefield 2000 ?
Its an attorney game about murder cases, i think its ok that it gets a slightly higher rating. That shows me that its cases are probably a bit more mature instead of just goofy. I cant get it into my head why this would keep people from buyng it...
Tasuki said:
Well its not the rating thats keeping me from getting this its the digital only release thats keeping me from getting this. Unless that petition that Capcom started gets enough signatures that they will release a physical copy. Oh wait silly me the guy that started that no longer works for Capcom.
Zombie_Barioth said:
I doubt Capcom has anything to do with this, sometimes games just end up with a higher rating than normal. Even a few Nintendo have earned themselves higher ratings despite usually being rated E. Capcom made their game and it Just happened to earn an M rating.
I don't see this affecting sales much either. Many Parents still buy their kids M-rated games no matter what they're told and anyone familar with crime shows or murder mysteries should know what they're getting into. You can't sugar-coat crime and murder while keeping things engaging and true to the formula. Past games were as tame as they were due to hardware limitations, but the 3DS is a lot more capable so they can do a lot more with the series on it. Its only natural the series would catch up to the nature of the genre eventually.
Csaw said:
This is a bad move on capcom's part because if previous experiences have taught us anything, it's that M rated games sell like crap on nintendo systems. With this news and the going digital only I'm kinda wondering if capcom wants this to flop so they have an excuse not to localize anymore AA games.
grimbldoo said:
I haven't played enough Phoenix Wright to know, but don't they usually not show the entire bloodied body? Like the bodies in the picture? That is probably what gave it the M rating, that or there are rape cases because even the cases in the first game were pretty brutal if you thought about them.
Dizzard said:
I'll buy it anyway, the M rating doesn't exclude me.
It's unfortunate though, the rating can't help get more purchases. If any parents are reading this....if this game is anything like previous games in the series I wouldn't be concerned about teens playing it.
JakobG said:
Does anyone in the comments realize Capcom has NOTHING, NOTHING, I repeat, NOTHING to do with the rating? The ESRB has the only say in how the game is to be rated, and there's no indication that Dual Destinies is designed to be a darker entry; if it was, you could clearly tell.
I'll say it once again, in case you forgot it: COMPANIES CAN NOT PICK RATINGS.
Stop citing your "Mature rating to boost sales" garbage, it's a lie on so many levels.
BigBluePanda said:
Probably because I am of age, but this is no big deal.
Only worry I would have if it hurt it's chances of coming over. I remember Trials and Tribulations nearly didn't make Europe because of the nature of the cases.
Peach64 said:
I can't believe so many people will let an age rating dictate their purchases. Once I was 18 and didn't have to worry about wether or not I'd get sold the game, I don't think I've ever looked at an age rating. I think it might be mostly a US thing where I understand a lot of stores won't carry anything rated adult? I don't get that. To me that says this is only suitable for adults, but in the US it seems to be taken as something else.
jackaroo said:
CoAndy said:
@Caryslan i agree that the rating is probably due to the murders, in the third game when terry dies in court i was shocked that it was put in (i was 17 at the time) so yeah i can see the game now getting a more mature theme oh and those who want a fun AA story watch the MLP cross over :P
DreamOn said:
I refuse to buy games with M rated content. I don't enjoy it and the world is brutal enough and beyond my control. Entertainment is within my control.
Zombie_Barioth said:
The only places that do that here are stores like Toys R Us that are specifically for kids. All the fuss seems to be because either people are upset that a lot of kids won't be aloud to play it or don't understand how the ratings work and are upset that its being bumped up a notch. I really don't get that, not all games need to be suitable for everyone so why do they need to be watered down to reach a wider audience? You never hear people tell authors or film makers to water down their work.
To me Ace Attorney is just like any other crime show, like Bones or Castle and should be treated accordingly.
DualWielding said:
Moreover ratings are very subjective, the same game can be either T or M depending on which of the ESRB guys got the task of rating it
Rargon said:
World said:
@iMii I didn't exactly realize that this was also why I don't play M-rated games, but I think it is.
Also, an entire Phoenix Wright comment string, with a guy saying things people didn't agree with at that, and not one person said "Objection!"
Perhaps it is the fanbase that is M for Mature.
BakaKnight said:
I don't care for the Rating, I know it exist for a good reason, but the rating systems are always too strict and exagerated, making hard to really judge a game's contenent from their opinion >.>; (Still being this a game with murders and blood around I could understand their reasons, even crime movies and films are rated "adult only" usually.)
Anyway PW is PW and those trailers make me feel so hyped!!! Must Buy Game!!! :D
Kyloctopus said:
@Jukilum A guys thought this game couldn't be for ages 14, 15, and 16. That doesn't mean it is such an awful and disgusting game. I have seen M-rated games have not such an awful story.
roryscott said:
Ace Attorney is my favourite video game series and I'm only 14. I will still get it, whatever my parents say, as I have been waiting for it for ages. I'M SO EXCITED! :D I can not wait to play it!
AJWolfTill said:
I'm shocked that fans of the series would consider parting because of an altered age rating, at the very least wait for the reviews to try and determine why this has happened. It might simply be that the ESRB have treated similar scenes in the new art style more harshly. It could also be sensitivity about the blowing up of public buildings which we know to be a part of the plot. Don't assume the series has drastically changed in tone or direction simply because of this.
TheXboxHero said:
I guess I can't get it. Oh well, I was really looking forward to this as My first Ace Attorney....
BigDaddysPizza said:
You can buy points on the eShop and then buy Dual Destinies at a later time. The eShop won't stop you from buying an M rated game.
Plus, M is an extreme stretch. It's going to be nearly the same as the past games. Which means it'll be fantastic. :)
AlexSora89 said:
There, I said it.
Jonny said:
Clayfrd said:
@Jaz007 Anyone that prefers games solely based on ESRB rating should reconsider. Buy games because they're good or bad, not because of what they're rated. I don't even look at ratings when I buy games.
Also, people seem to be bringing a lot of baseless conjecture into this argument. I don't think this game is rated M because Capcom was trying to attract the FPS demographic or anything like that. I think it's rated M because the 3DS's hardware is advanced such that truly representing the scene of a murder has now become more realistically graphic. Nothing wrong with that. I don't see the reason to get in a fuss over an ESRB stamp.
tenshinohone said:
Does that mean the characters can finally drink wine instead of 'grape juice'? That was an embarrassment! Brutal murders were A-OK but alcohol drinking by responsible adults is not? Seriously that was always so annoying.
Bass_X0 said:
Maybe Phoenix is actually teetotal? It wouldn't be out of character for him to choose not to drink alcohol.
Assuming they are old enough to play Mature games.
tenshinohone said:
@Bass_X0 As is Edgeworth, Kristophe Gavin and other characters? Possible but unlikely. It's been done with some anime localisations too - conversion of all alcohol into fruit juice to make it child friendly, but violence all over the place... I really don't get it
sillygostly said:
Meanwhile, in Australia, it's been rated PG for "Mild violence and themes" (akin to PEGI's 7+ rating and ESRB's E10+ rating), the same rating as all of the other Ace Attorney games. :P
It is also said to contain "very mild" (i.e. able to be accommodated under the G rating) coarse language, drug references and sexual references.
ToadFan said:
Zombie_Barioth said:
Not to mention having a drink or two before going to court (or during court for that matter) isn't exactly a wise decision.
I doubt the series will gain a "darker tone". I think the rating is due to the 3D art style and not using static images. What they've shown so far isn't that bad its just more detailed than past games. I imagine the other games would have been rated M if they were made with the same amount of detail as this one.
MagicEmperor said:
An M rating? That was surprising... but not a deal-breaker at all.
@theblackdragon I have specific reasons for believing why this is the wrong game to boycott (since the future of the series in America pretty much depends on this game), but ultimately it's everyone's choice. With all due respect, since I know we're on opposing sides of the issue (at least, from what I recall in previous articles).
Also, I think some people flipped out over his/her opinion because the M rating isn't Capcom's call--it's the ESRB's. So to boycott the game for that rating, to certain people, isn't a good stance. Of course, we won't know for sure until the game comes out.
Afrodudex said:
@Jaz007 I think one of the main reasons why it's rated M is because of the first case where a courtroom is blown up. With all the terrorist acts happening nowadays, it's obvious that the ERSB has stricter policies with terrorism in video games
Jaz007 said:
@Afrodudex I'm pretty sure the ESRB is not supposed to operate that way though. It would be a breach of policy to operate like that.
B3ND3R said:
This doesn't really affect me in any way, so I am perfectly okay with the M-rating, more Ace Attorney is never a bad thing.
VoiceOfReason said:
Next thing you know, Professor Layton, Luke, and Flora'll be droppin F-Bombs
and swearing all over the place. Especially Flora.
jmr908 said:
I think that this is a HUGE mistake on Capcom's part. After reading some of the comments on other websites, I have to say that I think that it's M only in america because of the Bombing case at the beginning. Gee, the government is getting really jumpy of bombing related things. Especially in video games. I wonder why...
EpicSkye said:
@Jaz007 I have to say that I agree with you. Nintendo definitely the gaming company with the most young audience members, and rating this m will really hurt sales.
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| null | null | null | null | null | null | null | null | null |
https://gauss.vaniercollege.qc.ca/gwikis/pwiki/index.php/Newton%27s_Law_of_Universal_Gravitation | 1,708,775,251,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474533.12/warc/CC-MAIN-20240224112548-20240224142548-00027.warc.gz | 269,439,796 | 14,145 | # Newton's Law of Universal Gravitation
Kreshnik Angoni
• The law of Universal Gravitation is formulated by Newton for two point particles with masses ${\displaystyle m_{1}}$ and ${\displaystyle m_{2}}$ at a distance r between them. The magnitude of force exerted on one particle by the other one is given by:
${\displaystyle F_{12}=F_{21}=G{\frac {m_{1}m_{2}}{r^{2}}}}$
The measurements show that the universal constant of gravitation is ${\displaystyle G=6.67*10^{-11}Nm^{2}/kg^{2}\,\!}$
• The gravitational force is a vector directed versus the source that exerts this force. So, its vector form is:
${\displaystyle F_{12}=-G{\frac {m_{1}m_{2}}{r^{2}}}{\frac {\vec {r_{21}}}{r_{21}}}}$
where ${\displaystyle {\frac {\vec {r_{21}}}{r_{21}}}}$ (or, ${\displaystyle {\hat {r}}_{21}}$) is the unit vector with tail at mass ${\displaystyle m_{2}}$ – the source of ${\displaystyle {\vec {F}}_{12}}$.
${\displaystyle F_{21}=-G{\frac {m_{1}m_{2}}{r^{2}}}{\frac {\vec {r_{12}}}{r_{12}}}}$
${\displaystyle {\frac {\vec {r_{12}}}{r_{12}}}}$ (or, ${\displaystyle {\hat {r}}_{12}}$) is the unit vector with tail at mass ${\displaystyle m_{1}}$ – the source of ${\displaystyle {\vec {F}}_{21}}$.
• To apply the gravitational law for two bodies close to each other one must use integration techniques and
the difficulty of calculi depends on the form of the two bodies. But, if the bodied are far enough to each other, one may model them as point particles and apply the law in its original form.
• In particular, with some approximations, we are able to model the interaction of earth with an object on its surface as if the whole mass of the earth is concentrated at its center and the objects are at distance ${\displaystyle r=R_{earth}}$.
• Many experiments have shown that; when several particles interact gravitationally between them the
principle of linear superposition applies. So, inside a system of particles ${\displaystyle m_{1},m_{2},m_{3}}$,…${\displaystyle m_{n}}$ , the force exerted on mass ${\displaystyle m_{1}}$ is:
${\displaystyle {\vec {F_{1}}}=\sum _{i=2}^{n}{\vec {F}}_{1i}}$
### THE GRAVITATIONAL AND THE INERTIAL MASS
• When expressing the second law of Newton we use the inertial mass:
${\displaystyle {\vec {F}}_{NET}=m_{in}{\vec {a}}}$
When formulating the gravitation law, Newton was not sure that the mass of particles in this law is the same as their inertial mass. Let’s verify this issue.
We start by supposing that the mass in the gravitational law may be different from ${\displaystyle m_{in}}$. So, we call it ${\displaystyle m_{gr}}$.
Let’s consider now a body in free fall close to earth surface. The earth will exert on it the gravitational force with magnitude:
${\displaystyle F_{gr}=G{\frac {m_{gr}M_{Earth}}{R_{Earth}^{2}}}}$
Here we assume that the body is close to the surface so that its distance form earth center is Earth ≅ ${\displaystyle R_{Earth}}$
As this is the net force exerted on the body we apply the second law of Newton
${\displaystyle F_{NET}=m_{in}a=F_{gr}=F_{gr}=G{\frac {m_{gr}M_{Earth}}{R_{Earth}^{2}}}}$
So, we get
${\displaystyle m_{in}a=m_{gr}({\frac {GM_{Earth}}{R_{Earth}^{2}}})=m_{gr}g}$, where ${\displaystyle {\frac {GM_{Earth}}{R_{Earth}^{2}}}=g}$
Then,
${\displaystyle a={\frac {m_{gr}}{m_{in}}}g}$
A big number of measurements show that the acceleration of free bodies is equal to ${\displaystyle g=9.8m/s^{2}}$. This means that ${\displaystyle m_{gr}/m_{in}=1}$, and ${\displaystyle m_{in}=m_{gr}}$.
So, the experiments confirm that the gravitational mass is the same as the inertial mass.
• Let us apply the gravitation law for the force exerted by earth over a mass 1kg close to earth.
${\displaystyle {\vec {F}}=-G{\frac {M_{Earth}}{(R_{Earth}+h)^{2}}}*{\frac {\vec {R_{Earth}}}{R_{Earth}}}\equiv {\vec {g}}}$
So, the g-vector is equal to the gravitation force exerted on a mass 1kg.
By measuring the force exerted on the mass 1kg in different locations on the earth one gets a whole system of g-vectors (fig.2). The totality of these vectors forms the gravitational field of the earth.
Note that the g-vector magnitude decreases with the increase of distance “h” from the earth but it is always directed versus the center of the earth. The gravitational field of the earth has a spherical symmetry.
In fact, it is not exactly spherical, because the model of earth as a uniform density sphere is not very precise. Now, the object weight is equal to gravitational force exerted by this field
${\displaystyle {\vec {W}}=m{\vec {g}}}$
So, the weight of the same object is a vector that is different in different points of gravitational field of the earth.
### KEPLER’S LAWS ON PLANETARY MOTION
• First law: The planets move on elliptic orbits around the sun that is located at one of its focuses.
The minor axis is long 2b and the major axis is long 2a. The closest distance to sun is called perihelion and the biggest distance to sun is called aphelion (fig.3).
• Second law: The line sun- planet sweeps out equal areas for equal interval of times.
• Third law: The square of the period of planet motion is proportional to the cube of average distance from the sun. Calculations show that the average of distance sun-planet is equal to half of major axis a. Then, satellite motion tells that
${\displaystyle T^{2}=ka^{3}}$ where
${\displaystyle k_{sun}={\frac {4\pi ^{2}}{GM_{sun}}}}$
Note : Kepler’s laws are valid for elliptical paths of any planet around a central body; for example the moon moving around the earth but in this case
${\displaystyle k_{Earth}={\frac {4\pi ^{2}}{GM_{Earth}}}}$
### THE ENERGY OF PLANETS
• As the mass of other planets is much smaller than the mass of sun
we neglect their action on the motion of the studied planet. We consider that the system sun-planet is a conservative system, i.e. the forces originated from outside it are zero.
In these circumstances:
a) The torque of exterior force is zero and we can apply the principle for conservation of angular momentum.
b) The work done by Net exterior force is zero and we can apply the principle of energy conservation for the system sun-planet.
• The principle of angular momentum conservation tells that ${\displaystyle {\vec {L_{A}}}={\vec {L_{P}}}}$
or ${\displaystyle {\vec {r_{A}}}\times {\vec {p_{A}}}={\vec {r_{P}}}\times {\vec {p_{P}}}}$
The equality of magnitudes brings to condition
${\displaystyle r_{A}p_{A}Sin90^{o}=r_{P}p_{P}Sin90^{o}\Rightarrow r_{A}m_{planet}v_{A}=r_{P}m_{planet}v_{P}}$
So, we get
${\displaystyle r_{A}v_{A}=r_{P}v_{P}\,\!}$
• The principle of energy conservation tells that that ${\displaystyle E_{A}=E_{P}\,\!}$
As the mechanical energy is E = K + U where
${\displaystyle K={\frac {m_{pl}v_{pl}^{2}}{2}}}$ and
${\displaystyle U=-G{\frac {m_{pl}M_{Sun}}{r_{pl-sun}}}}$
we get
${\displaystyle {\frac {m_{pl}v_{A}^{2}}{2}}-G{\frac {m_{pl}M_{Sun}}{r_{A}}}={\frac {m_{pl}v_{P}^{2}}{2}}-G{\frac {m_{pl}M_{Sun}}{r_{P}}}\,\!}$
This gives, on cancelling ${\displaystyle m_{pl}}$ from the equation:
${\displaystyle {\frac {v_{A}^{2}}{2}}-G{\frac {M_{Sun}}{r_{A}}}={\frac {v_{P}^{2}}{2}}-G{\frac {M_{Sun}}{r_{P}}}\,\!}$
${\displaystyle 2GM_{Sun}({\frac {1}{r_{P}}}-{\frac {1}{r_{A}}})=v_{P}^{2}-v_{A}^{2}}$
Now, we know ${\displaystyle r_{A}v_{A}=r_{P}v_{P}\,\!}$
and the fact that:
${\displaystyle r_{A}+r_{P}=2a\,\!}$
Thus,
${\displaystyle v_{A}^{2}={\frac {GM}{a}}{\frac {r_{P}}{r_{A}}}}$ and,
${\displaystyle v_{P}^{2}={\frac {GM}{a}}{\frac {r_{A}}{r_{P}}}}$
Finally, by substituting one of this expressions at the expression of total energy (at perihelion or aphelion) we get
${\displaystyle E=-G{\frac {m_{pl}M_{Sun}}{2a}}}$ | 2,355 | 7,695 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 52, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.0625 | 4 | CC-MAIN-2024-10 | latest | en | 0.880569 |
http://www.talkstats.com/showthread.php/59155-Regression-cor(X1-Yhat)-cor(X1-Y)-R-How-to-show?p=169301 | 1,508,640,782,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187825057.91/warc/CC-MAIN-20171022022540-20171022042540-00597.warc.gz | 557,941,547 | 12,620 | # Thread: Regression: cor(X1, Yhat) = cor(X1, Y)/R How to show?
1. ## Regression: cor(X1, Yhat) = cor(X1, Y)/R How to show?
Hello everyone. I am taking a class in regression that's way over my head. We are working with the sample model Y = b0 + b1X1 + b2X2 + e. In it I am asked to show that the correlation between the predictor X1 and the predicted values Yhat can be obtained by dividing the simple correlation cor(X1,Y) by the square root of R-squared. I have verified this with sample datasets and I can see the relationship holds. So it is true that:
This is what I have tried so far but I think I'm not getting anywhere. Does anyone have a hint on how I can proceed?
I am starting off by noticing:
Now, I know that . The intercept is a constant so I can drop it and using the rules of covariance algebra I can expand this to be:
But when I expand the denominator (so the product of the square roots of and )
I end up with where the big first term is .
I don't see how the denominator is going to become R (the square root of R2), and I definitely not see where I am going to get from
Any help, please? I'm so stuck!
2. ## Re: Regression: cor(X1, Yhat) = cor(X1, Y)/R How to show?
Try to solve the problem by working with matrices. It becomes easier that way.
3. ## Re: Regression: cor(X1, Yhat) = cor(X1, Y)/R How to show?
Thank you, but this course assumes no knowledge of matrix algebra. My guess is that the solution can be obtained purely from covariance algebra and some properties of regression.
Do you have any insights?
4. ## Re: Regression: cor(X1, Yhat) = cor(X1, Y)/R How to show?
hi,
maybe you could plug in the formula for b1 into the equation?
regards
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null | null | null | null | null | null | Back when Rudy Sandoval was about thirteen years old, a good friend of his had a brother that was in Amigos Car Club. "He and I used to take off in his brother's Caprice Classic, and we used to cruise Arroyo Seco with him," recalls Rudy. While Lowriders surrounded him when he was coming up, it wasn't until three years ago that Rudy took his plunge to build one himself. "I own a variety of cars including a Barracuda, a Porsche, a '56 Boyd Coddington and few others, but I grew up in the Lowrider scene."
"One of my good friends, Carlos, has been a member of Groupe Car Club for a long time and he helped push me, so I ended up getting in the club," says the proud plaque-flier. When the time came to purchase his first Lowrider, there was no question about the make and model car that Rudy was going to get. "I knew I wanted to build a traditional Lowrider, so I went and got an LTD from my friend Saul, who had already done the '76 front clip swap on it." Once the car was purchased, it was taken apart and basically everything on it was redone. "I didn't want a show car; I wanted something I can drive, slam it on the ground, scrape, and just enjoy it without having to worry too much about it." Rudy also owns a '58 convertible and has also owned '61 and '62 convertibles, but his favorite car by far is the Ford LTD. "To me there is nothing like it, they lay real low and they look like they are a mile long," Rudy says with a smile. Rudy currently has a '76 Ford LTD in the works and if everything goes well, it will be out this summer.
Model: MeeMee B
Makeup: Shanella Genice (
Tech Specs
"True Blue"
Owner: Rudy Sandoval
Vehicle: 1973 Ford LTD
City/State: Los Angeles, CA
Club: Groupe SoCal
Body/Paint: After shaving all the emblems and adding a '76 front clip and '74 trim, Fabian/Burrito, at Carlos Customs, in East L.A., finished the bodywork and Carlos, of Carlos Customs, candied the LTD by using true blue pearl on the car and oriental candy blue on the top. Danny D. came in to finish off the paintjob by adding his fine lines.
Sound System: Beto, of Los Angeles, installed the Alpine head unit and amp with four 6x9-inch Sony speakers.
Chrome: Yolanda, of Yolanda's in East L.A.
Tires: Coker 5.20's with a 5/8 white wall.
Wheels: 72-spoke 14x7s. | null | null | null | null | null | null | null | null | null |
http://mymathforum.com/algebra/42296-one-lengthiest.html | 1,571,467,285,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570986692126.27/warc/CC-MAIN-20191019063516-20191019091016-00204.warc.gz | 133,850,180 | 8,799 | My Math Forum which one is the lengthiest
Algebra Pre-Algebra and Basic Algebra Math Forum
March 22nd, 2014, 07:26 PM #1 Senior Member Joined: Apr 2012 Posts: 799 Thanks: 1 which one is the lengthiest $\triangle ABC ,\,\,$ with side length : $a,b,\,\, and \,\, c$ we have : $(1) a^2-a-2b-2c=0$ $(2)a+2b-2c+3=0$ now determine among : $a,b,c$ which one is the lengthiest
March 22nd, 2014, 08:13 PM #2 Math Team Joined: Dec 2013 From: Colombia Posts: 7,690 Thanks: 2669 Math Focus: Mainly analysis and algebra Re: which one is the lengthiest You want to express b in terms of a only and c in terms of a only. You can do this by eliminating b and c in turn from the equations in the same manner that you would to solve a system of simultaneous equations. You should then be able to order them appropriately. I made $c > b$ for all values of $a$ and $a > c$ when $1 \lt a \lt 3$.
March 22nd, 2014, 08:23 PM #3 Math Team Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 14,597 Thanks: 1039 Re: which one is the lengthiest c. Also, if a,b,c are integers, then b>a except for one case: (a,b,c) = (5,3,7)
March 22nd, 2014, 09:26 PM #4 Senior Member Joined: Mar 2014 Posts: 112 Thanks: 8 0 = aČ - 4c + 3 i.e. 4c = aČ + 3. Hence c = (aČ + 3)/4. 0 = aČ - 2a - 4b - 3 i.e. 4b = aČ - 2a - 3 = (a - 3)(a + 1). Hence b = (a - 3)(a + 1)/4. Since b > 0 is correct, a > 3 must also be correct. If we use the solution Denis gave we are done.
March 24th, 2014, 09:14 PM #5
Math Team
Joined: Oct 2011
Posts: 14,597
Thanks: 1039
Re: which one is the lengthiest
Quote:
Originally Posted by Albert.Teng $(2)a+2b-2c+3=0$
Was diddling with this "Albert Special" during the hockey game...
Above equation can be rewritten as:
2c - 2b = a + 3 [1]
So evident that c > b (since a,b,c are all > 0, being triangle sides)
Can't find "as simple" a way to show c > a...
March 25th, 2014, 11:24 AM #6 Math Team Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 14,597 Thanks: 1039 Re: which one is the lengthiest Albert, something for your "useless information(!)" file; here's the first 8 integer solutions: Code: n a b c 1 5 3 7 2 7 8 13 3 9 15 21 4 11 24 31 5 13 35 43 6 15 48 57 7 17 63 73 8 19 80 91 a = 2(n - 1) + 5 b = n(n + 2) c = n^2 + 3n + 3 You're welcome; no charge :P
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null | null | null | null | null | null | "The cache folder has no files or folders relating to iTunes."<br><br>If you're running Leopard and have no specific iTunes cache folder within the Caches folder of your user Library, something is very amiss. Have you at any time run a third party cache cleaning utility? This is the only way I can imagine that cache database was eliminated.<br><br>- a.k.a. Mississauga -
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http://logika.sireum.org/dschmidt/08-predicate-logic/index.html | 1,532,194,615,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676592650.53/warc/CC-MAIN-20180721164755-20180721184755-00154.warc.gz | 216,090,008 | 19,596 | Logika: Programming Logics
8. The Predicate-Logic Quantifiers
# 8. The Predicate-Logic Quantifiers¶
In the previous chapter, we studied how to combine primitive propositions with the operators, ^ (math: ∧), v (math: ∨), -> (math: →), and ~ (math: ¬). When we wrote propositions like (p ^ q) -> r, we pretended that p, q, and r stood for complete, primitive statements like “It is raining” or “x + 1 > 0”. We did not try to disassemble p, q, and r.
Now it is time to decompose and analyze primitive propositions in terms of their “verbs” (called predicates) and their “nouns” (called individuals). This leads to predicate logic.
First, some background: When we study a particular “universe” or “domain” consisting of “individuals”, we make assertions (propositions) about the individuals in the domain. Example domains are: the domain of all animals, the domain of U.S. Presidents, the domain of days-of-the-week, the domain of crayon colors, the domain of integers, the domain of strings, etc. We assemble propositions by using the individuals in the domain along with some predicates. For example, for the domain of integers, we use predicates like == and >, like this: 3 > 5, 2 * x == y + 1, etc. (Here, 3 and 5 are individuals, and x and y are names of individuals.) As these examples show, we might also use functions, like * and +, to compute new individuals.
For nonnumeric domains like humans, animals, and objects, predicates are written in a function-call style, like this: hasFourLegs(_), isTheMotherOf(_,_), isHuman(_), isOlderThan(_,_), etc. So, if Lassie is an individual animal, we write hasFourLegs(Lassie) to make the proposition, “Lassie has four legs”. Another example is isOlderThan(GeorgeWashington, AbrahamLincoln), which uses the individuals GeorgeWashington and AbrahamLincoln.
Predicate logic has two important new operators that let us write stronger propositions than what we can do with mere predicates. These operators are called quantifiers. The quantifers are “for all” (math: ∀), and “exists” (math: ∃). In this chapter, we will learn to use the quantifiers to reason about data structures.
The ∀ quantifier helps us write propositions about all the individuals in a domain. Say we consider the domain of animals. The sentence, “All humans are mortal” is written like this:
∀x (isHuman(x) -> isMortal(x))
That is, if an individual, x, is human, then x is mortal also. (Notice that dogs like Lassie are individuals in the domain, but the above proposition cannot be used to show that Lassie is mortal, since dogs aren’t human. Sadly, dogs are nonetheless mortal.)
An arithmetic example looks like this: for the domain of ints, “every value is less-than-or-equal to its square”:
∀n (n <= n * n)
A data-structure example looks like this: For array, r, we can assert that every element of r is positive:
∀i ((i >= 0 ^ i < len(r)) --> r[i] > 0)
That is, for every index int, i, in the range of 0 up to (but not including) len(r) (the length of r), the indexed element r[i] is greater than 0. The previous statement is often written in a “shorthand” like this:
∀ 0 <= i < len(r), r[i] > 0
which we later use in many of our programming examples.
The ∃ quantifier helps us write propositions about specific individuals in a domain, where the name of the individual is unimportant or unknown. For example, we can say that Lassie has a mother like this:
∃x isMotherOf(x, Lassie)
(“There exists some x such that x is the mother of Lassie”.) Here is how we write that every individual in the domain has a mother:
∀x∃y isMotherOf(y, x)
For the domain of integers, we can make assertions like these:
∃x (x * x = x)
∃y (y + 2 = 9)
∀x (x > 1) -> (∃y (y > 0 and y + 1 = x))
For array r, we can say that r holds at least one negative int like this:
∃i (i >= 0 ^ i < len(r) ^ r[i] < 0)
(The shorthand version is ∃ 0 <= i < len(r), r[i] < 0.) Lots more examples will follow.
With the new format of primitive propositions, we can write proofs like before:
isHuman(Socrates) -> isMortal(Socrates), isHuman(Socrates) |- isMortal(Socrates) ^ isHuman(Socrates)
{
1. isHuman(Socrates) -> isMortal(Socrates) premise
2. isHuman(Socrates) premise
3. isMortal(Socrates) ->e 1 2
4. isMortal(Socrates) ^ isHuman(Socrates) ^i 3 2
}
But more importantly, we will learn to prove claims like this:
∀x(isHuman(x) -> isMortal(x)), isHuman(Socrates) |- isMortal(Socrates)
## 8.1. The Universal quantifier and Its Deduction Rules¶
Like the other logical operators, ∀ has an introduction rule and an elimination rule. It works best to introduce the rules via examples. First, here is the most famous claim in logic:
All humans are mortal.
Socrates is human.
Therefore, Socrates is mortal.
We express this ancient claim like this:
∀x (isHuman(x) -> isMortal(x)), isHuman(Socrates) |- isMortal(Socrates)
Clearly, we require a kind of matching/binding rule to prove that the human individual, Socrates, is mortal. The rule is ∀e (“all elimination”):
∀x (isHuman(x) -> isMortal(x)), isHuman(Socrates) |- isMortal(Socrates)
{
1. ∀x (isHuman(x) -> isMortal(x)) premise
2. isHuman(Socrates) premise
3. isHuman(Socrates) -> isMortal(Socrates) ∀e 1 Socrates
4. isMortal(Socrates) ->e 3 2
}
Line 3 shows that the claim on Line 1, which holds for all individuals in the domain, can apply specifically to Socrates, an individual member of the domain. We use the new knowledge on Line 3 to reach the conclusion on Line 4.
∀e tailors a general claim, prefixed by ∀x, to any individual element (who replaces the x). We see this in Line 3 above. Here is the rule’s schematic:
∀x P(x)
∀e: -----------
P(v) that is, [v/x]P(x), where v is an individual in the domain
(Here, P(x) stands for a proposition that contains some occurrences of x. Recall that [v/x]P(x) is “substitution notation”: substitute v for occurrences of x in P(x).) For example, from the premise, ∀i (i + 1 > i), we apply ∀e to deduce [3/i](i + 1 > i), that is, 3 + 1 > 3.
The other deduction rule, ∀i (“all-introduction”), deduces propositions that are prefixed by ∀. Here is a motivating example, in the domain of integers:
∀x((x + 1) > x), ∀x(n > (x - 1)) |- ∀x((x + 1) > x ^ x > (x - 1))
Alternatively this can be written using uninterpreted functions:
∀x gt(inc(x), x), ∀x gt(x, dec(x)) |- ∀x gt(inc(x), x) ∧ gt(x, dec(x))
That is, we wish to prove that for every possible int, the int is smaller than its successor and larger than its predecessor. How do we do this?
Clearly, we will not inspect all of ..., -2, -1, 0, 1, 2, ... and verify that (-2 + 1) > -2 ^ -2 < (-2 - 1), (-1 + 1) > -1 ^ -1 < (-1 - 1), (0 + 1) > 0 ^ 0 < (0 - 1), etc.! Instead, we write a single, generic, general-purpose argument — a “case analysis” — that applies to whichever, arbitrary int we would ever consider. Let a stand for the arbitrary int we will discuss. The case analysis appears in the proof like this:
∀x gt(inc(x), x), ∀x gt(x, dec(x)) |- ∀x gt(inc(x), x) ∧ gt(x, dec(x))
{
1. ∀x gt(inc(x), x) premise
2. ∀x gt(x, dec(x)) premise
3. {
4. a
5. gt(inc(a), a) ∀e 1 a
6. gt(a, dec(a)) ∀e 2 a
7. gt(inc(a), a) ∧ gt(a, dec(a)) ∧i 5 6
}
8. ∀x gt(inc(x), x) ∧ gt(x, dec(x)) ∀i 3
}
Lines 3-7 are the generic argument: let a be the arbitrary/anybody integer we discuss. By Lines 1 and 2, we must have that (a + 1) > a and that a > (a - 1). Line 6 uses ^i to show a has the property (a + 1) > a ^ a > (a - 1).
Since the argument in Lines 3-7 is not specific to any specific integer, we can use the argument on all the individual integers – that is, we can substitute -2 for a and the argument holds; we can substitute -1 for a and the argument holds; we can substitute 0 for a and the argument holds; and so on!
Line 8 is justified by the new deduction rule, ∀i, which asserts that the generic case analysis in Lines 3-7 applies to all the individual integers. Here is the rule’s schematic:
{ a (a is fresh)
. . . P(a) }
∀i: ---------------
∀x P(x) (That is, P(x) is [x/a]P(a).
Thus, a _does not appear_ in P(x), and
every premise and assumption visible
to ∀x P(x) _does not mention_ a)
To repeat this important idea: The rule says, to prove a claim of form, ∀x P(x), we undertake a case analysis: we prove property P(a) for an arbitrary member, a, of domain D. (Call the element, “Mister a” — Mister arbitrary — Mister anybody — Mister anonymous). Since Mister a is a complete unknown, it stands for “everyone” in doman D. We know that we can substitute whichever domain element, d, from domain D we want into the proof and we get a proof of P(d). In this way, we have proofs of P for all elements of domain D.
Here is the same idea, used in a proof about a domain of people: “Everyone is healthy; everyone is happy. Therefore, everyone is both healthy and happy”:
∀x isHealthy(x), ∀y isHappy(y) |- ∀z(isHealthy(z) ^ isHappy(z))
{
1. ∀x isHealthy(x) premise
2. ∀y isHappy(y) premise
3. {
4. a
5. isHealthy(a) ∀e 1 a
6. isHappy(a) ∀e 2 a
7. isHealthy(a) ^ isHappy(a) ^i 5 6
}
8. ∀z(isHealthy(z) ^ isHappy(z)) ∀i 3
}
Say that we have a domain of living beings. This next example requires nested cases
All humans are mortal
All mortals have soul
Therefore, all humans have soul
∀x (isHuman(x) -> isMortal(x)), ∀y (isMortal(y) -> hasSoul(y)) |- ∀x (isHuman(x) -> hasSoul(x))
{
1. ∀x (isHuman(x) -> isMortal(x)) premise
2. ∀y (isMortal(y) -> hasSoul(y)) premise
3. {
4. a
5. {
6. isHuman(a) assume
7. isHuman(a) -> isMortal(a) ∀e 1 a
8. isMortal(a) ->e 7 6
9. isMortal(a) -> hasSoul(a) ∀e 2 a
10. hasSoul(a) ->e 9 8
}
11. isHuman(a) -> hasSoul(a) ->i 5
}
12. ∀x (isHuman(x) -> hasSoul(x)) ∀i 3
}
Line 4 states that we use a to stand for an arbitrary individual of the domain. Line 6 starts a nested case, which assumes a is human. Then we can prove that a has a soul, hence by ->i, isHuman(a) -> hasSoul(a). Since the outer case is stated in terms of the arbitrary, anonymous individual, a, we can finish the proof on Line 12 by ∀i.
Here is a last, important example. Let the domain be the members of one family. We can prove this truism
Every (individual) family member who is healthy is also happy.
Therefore, if all the family members are healthy, then all the members are happy.
∀x (healthy(x) -> happy(x)) |- (∀y healthy(y)) -> (∀x happy(x))
{
1. ∀x healthy(x) -> happy(x) premise
2. {
3. ∀y healthy(y) assume
4. {
5. a
6. healthy(a) ∀e 3 a
7. healthy(a) -> happy(a) ∀e 1 a
8. happy(a) ->e 7 6
}
9. ∀ x happy(x) ∀i 4
}
10. (∀y healthy(y)) -> (∀x happy(x)) ->i 2
}
We commence by assuming all the family is healthy (Line 3). Then, we consider an arbitrary/anonymous family member, a, and show that healthy(a) is a fact (from Line 3). Then we deduce happy(a). Since a stands for anyone/everyone in the family, we use ∀i to conclude on Line 7 that all family members are happy. Line 10 finishes.
Consider the converse claim; is it valid?
If all the family members are healthy, then all are happy.
Therefore, for every (individual) family member, if (s)he is healthy then
(s)he is also happy.
Well, no – perhaps the family is so close-knit that, if one one family member is unhealthy; then other, healthy, family members might well be unhappy with worry. This is a subtle point, so take a moment and think about it!
Let’s try to prove the dubious claim and see where we get stuck:
(∀y healthy(y)) -> (∀x happy(x)) |- ∀x (healthy(x) -> happy(x))
{
1. (∀y healthy(y)) -> (∀x happy(x)) premise
2. {
3. a assume
4. {
5. healthy(a) assume WE ARE TRYING TO PROVE happy(a)?!
}
}
6. ∀y healthy(y) ∀i 2?? NO --- WE ARE TRYING TO FINISH THE OUTER
BLOCK BEFORE THE INNER ONE IS FINISHED!
No matter how you might try, you will see that the “block structure” of the proofs warns us when we are making invalid deductions. It is impossible to prove this claim.
### 8.1.1. More Examples¶
We state some standard exercises with ∀, where the domains and predicates are unimportant:
∀x f(x) |- ∀y f(y)
{
1. ∀x f(x) premise
2. {
3. a
4. f(a) ∀e 1 a
}
5. ∀y f(y) ∀i 2
}
∀z (f(z) ^ g(z)) |- (∀x f(x)) ^ (∀y g(y))
{
1. ∀z (f(z) ^ g(z)) premise
2. {
3. a
4. f(a) ^ g(a) ∀e 1 a
5. f(a) ^e1 4
}
6. ∀x f(x) ∀i 2
7. {
8. b
9. f(b) ^ g(b) ∀e 1 b
10. g(b) ^e2 9
}
11. ∀y g(y) ∀i 7
12. (∀x f(x)) ^ (∀y g(y)) ^i 6 11
}
The earlier example about healthy and happy families illustrates an important structural relationship between ∀ and ->:
∀x (F(x) --> G(x)) |- (∀x F(x)) --> (∀x G(x))
can be proved, but the converse cannot.
This last one is reasonable but the proof is a bit tricky because of the nested subproofs:
∀x∀y f(x,y) |- ∀y∀x f(x,y)
{
1. ∀x∀y f(x,y) premise
2. {
3. b
4. {
5. a
6. ∀y f(a,y) ∀e 1 a
7. f(a,b) ∀e 6 b
}
8. ∀x f(x,b) ∀i 4
}
9. ∀y∀x f(x,y) ∀i 2
}
### 8.1.2. Tactics for the ∀-rules¶
As in the previous chapter, we now give advice as to when to use the ∀i and ∀e rules.
• (***) ∀i-tactic: To prove Premises |- ∀x P(x),
1. assume a, for a new, anonymous “Mister a
2. prove Premises |- P(a)
3. finish with ∀i.
The proof structure looks like this:
1. Premises premise
i. {
j. a
(fill in)
k. P(a)
}
l. ∀x P(x) ∀i i
This tactic was applied in Lines 2-9 of the previous (correct) example
proof.
• (*) ∀e-tactic: To prove Premises, ∀x P(x) |- Q, then for an individual, i, that appears in the proof so far, use the ∀e rule to deduce the new fact, P(i):
1. Premises premise
2. ∀x P(x) premise
. . .
j. P(i) ∀e 2
(fill in)
k. Q
This tactic should be used only when it is clear that the new fact makes a significant step forwards to finishing the proof. Steps 4 and 5 of the previous (correct) example proof used this tactic.
### 8.1.3. Other Ways of Proving Propositions with The Universal Quantifier¶
How do we prove an assertion of the form, ∀x P(x)? We just saw that ∀i can do this for any domain whatsoever. But there are, in fact, three different approaches, depending on the form of domain we use:
• Approach 1: use conjunctions for a finite domain
Say that the domain we study is a finite set, D = {e0, e1, ..., ek}. (An example domain is the days of the week, {sun, mon, tues, weds, thurs, fri, sat}.)
This makes ∀x P(x) just an abbreviation itself of this much-longer assertion:
P(e0) ^ P(e1) ^ ... ^ P(ek)
For example, when the domain is the days of the week, the assertion, ∀d isBurgerKingDay(d), abbreviates:
isBurgerKingDay(sun) ^ isBurgerKingDay(mon) ^ isBurgerKingDay(tues) ^ ... ^ isBurgerKingDay(sat)
To prove such a ∀x P(x) for a finite domain D, we must prove P(ei), for each and every ei in D.
We can use this approach when we are analyzing all the elements of a finite-length array. Say that array r has length 4. We can say that the domain of its indexes is {0, 1, 2, 3}. So, if we wish to prove that ∀ 0 <= i < 4, r[i] > 0, we need only prove that r[0] > 0 ^ r[1] > 0 ^ r[2] > 0 ^ r[3] > 0.
• Approach 2: for the domain of nonnegative ints, use mathematical induction
The domain, Nat = {0, 1, 2, ... } is infinite, so we cannot use the previous technique to prove properties like ∀ n > 0, (n + 1) > n – we would have to write separate proofs that 0 + 1 > 0, 1 + 1 > 1, 2 + 1 > 2, ..., forever. But we can use mathematical induction. Remember how it works: we write two proofs:
• basis case: a proof of P(0)
• induction case: a proof of P(k) --> P(k+1), where k is a brand new variable name
(At this point, it would be a very good idea for you to review the Section on An Introduction to Mathematical Induction.)
• Approach 3: for any domain, finite or infinite whatsoever, use the ∀i-law
Finally, we might be using a large domain that is not as organized as the nonnegatives, 0,1,2,.... Maybe the domain is the domain of all humans or all the citizens of Peru or the members of the Republican party or all the objects on Planet Earth. How can we prove ∀ x P(x) for such huge collections?
To prove a claim of form, ∀x P(x), for an arbitrary domain, we undertake a kind of case analysis: we prove property P(a) for an arbitrary member, a, of domain D. (Call the element, “Mister a” — Mister arbitrary — Mister anybody — Mister anonymous). Since Mister a is a complete unknown, it stands for “everyone” in doman D. We know that we can substitute whichever domain element, d from domain D, we want into the proof and we get a proof of P(d). In this way, we have proofs of P for all elements of domain D.
This is the idea behind the ∀i-rule.
## 8.2. The Existential Quantifier¶
The existential quantifier, ∃ (math: ∃), means “there exists” or “there is”. We use this phrase when we do not care about the name of the individual involved in our claim. Here are examples:
There is a mouse in the house: ∃m (isMouse(m) ^ inHouse(m))
(We don't care about the mouse's name.)
There is a number that equals its own square: ∃n n == n*n
For every int, there is an int that is smaller: ∀x ∃y y < x
If we have a fact about an individual in a domain, we can use the fact to deduce a fact that begins with an existential quantifier. For example, if we know that:
isHuman(Socrates) ^ isMortal(Socrates)
surely we can conclude that:
∃h (isHuman(h) ^ isMortal(h))
that is, “there is someone who is human and mortal”. The identity of the human is no longer important to us. In the next section, we see that the ∃i-rule makes such deductions.
### 8.2.1. The Existential-Introduction Rule¶
Often ∃ is used to “hide” secret information. Consider these Pat Sajack musings from a typical game of Wheel of Fortune:
• Pat thinks: “There is an ‘E’ covered over on Square 14 of the game board”. In predicate logic, this can be written:
isCovered(Square14) ^ holds(Square14,'E').
• Pat thinks: “Wait – I can’t say that on TV! Perhaps I can say, There is a vowel covered on Square 14 of the game board”. In predicate logic, this is written:
isCovered(Square14) ^ (∃c isVowel(c) ^ holds(Square14,c)).
In this way, Pat does not reveal the letter to the game players and TV viewers.
• Because it isn’t fair to tell the players which squares hold vowels, Pat announces on the air, “There is a vowel that is still covered on the game board”:
∃s (isSquare(s) ^ isCovered(s) ^ (∃c isVowel(c) ^ holds(s,c)))
This statement hides the specific square and letter that Pat is thinking about.
Pat’s announcement was deduced from its predecessors by means of the ∃i-rule, which we see in a moment.
What can a game player do with Pat’s uttered statement? A player might deduce these useful facts:
• There is a square still covered: ∃s isSquare(s) ^ isCovered(s)
• There is a vowel: ∃c isVowel(c)
• There is a covered letter, A, E, I, O, U (assuming the vowels are exactly A, E, I, O, U):
∃s isSquare(s) ^ isCovered(s) ^ (holds(s,'A') v holds(s,'E') v holds(s,'I') v holds(s,'O') v holds(s,'U'))
Although the game player does not know the letter and square that Pat Sajak “hid” with his statement, the player can still make useful deductions. We will use the ∃e rule to deduce these style of propositions.
#### ∃-Introduction Rule¶
The rule for ∃i has this format:
P(d) where d is an individual in the domain D
∃i: -----------
∃x P(x)
The ∃i rule says, if we locate an individual d (a “witness”, as it is called by logicians) that makes P true, then surely we can say there exists someone that has P and hide the identity of the individual/witness.
The rule was used in the previous section in a tiny example:
isHuman(Socrates), isMortal(Socrates) |- ∃h (isHuman(h) ^ isMortal(h))
{
1. isHuman(Socrates) premise
2. isMortal(Socrates) premise
3. isHuman(Socrates) ^ isMortal(Socrates) ^i 1 2
4. ∃h (isHuman(h) ^ isMortal(h)) ∃i 3 Socrates
}
Since Socrates is an individual that is both human and mortal, we deduce Line 3. Line 4 “hides” Socrates’ name.
Let’s do a Wheel-Of-Fortune example: Pat Sajak uses two premises and the ∃i rule to deduce a new conclusion:
isVowel(e), holds(Square14,e) |- ∃c(isVowel(c) ^ ∃s holds(s,c))
{
1. isVowel(e) premise
2. holds(Square14,e) premise
3. ∃s holds(s,e) ∃i 2 Square14
4. isVowel(e) ^ ∃s holds(s,e) ^i 1 3
5. ∃c(isVowel(c) ^ ∃s holds(s,c)) ∃i 4 e
}
Line 3 hides the number of the square (“there is a square that holds ‘e’”), and Line 5 hides the ‘e’ (“there is a letter that is a vowel and there is a square that holds the letter”).
From the same two premises we can also prove this:
isVowel(e), holds(Square14,e) |- ∃c∃s(isVowel(c) ^ holds(s,c))
{
1. isVowel(e) premise
2. holds(Square14,e) premise
3. isVowel(e) ^ holds(Square14,e) ^i 1 2
4. ∃s(isVowel(e) ^ holds(s,e)) ∃i 3 Square14
5. ∃c∃s(isVowel(c) ^ holds(s,c)) ∃i 4 e
}
This reads, “there are a letter and square such that the letter is a vowel and the square holds the letter”. The proposition differs slightly from the previous one, but the two seem to have identical information content. (When we learn the ∃e-rule, we can prove the two conclusions have identical content.)
#### The ∃-Elimination Rule¶
Since the ∃i-rule constructs propositions that begin with ∃, the ∃e-rule disassembles propositions that begin with ∃. The new rule employs a subtle case analysis.
Here is a quick example (in the universe of things on planet Earth), to get our bearings:
All humans are mortal
Someone is human
Therefore, someone is mortal
We don’t know the name of the individual human, but it does not matter – we can still conclude someone is mortal. The steps we will take go like this
• Since “someone is human” and since we do not know his/her name, we will just make up our own name for them – “Mister A”. So, we assume that “Mr. A is human”
• We use the logic rules we already know to prove that “Mr. A is mortal”.
• Therefore “someone is mortal” and their name does not matter.
This approach is coded into the last logic law, ∃e (exists-elimination).
Say we have a premise of the form, ∃x P(x). Since we do not know the name of the individual “hidden” behind the ∃x, we make up a name for it, say a, and discuss what must follow from the assumption that P(a) holds true:
{a P(a) assume // where a is a new, fresh name
∃x P(x) ... Q }
∃e: ----------------------------------- // a MUST NOT appear in Q
Q
That is, if we can deduce Q from P(a), and we do not mention a within Q, then it means Q can be deduced no matter what name the hidden individual has. So, Q follows from ∃x P(x).
We can work the previous example, with ∃e:
All humans are mortal
Someone is human
Therefore, someone is mortal
We make up the name, a, for the individual whose name we do not know, and do a case analysis:
∀h(isHuman(h) -> isMortal(h)), ∃x isHuman(x) |- ∃y isMortal(y)
{
1. ∀h(isHuman(h) -> isMortal(h)) premise
2. ∃x isHuman(x) premise
3. {
4. a isHuman(a) assume
5. isHuman(a) -> isMortal(a) ∀e 1 a
6. isMortal(a) ->e 5 4
7. ∃y isMortal(y) ∃i 6 a
}
8. ∃y isMortal(y) ∃e 2 3
}
Line 4 proposes the name a and the assumption that isHuman(a). The case analysis leads to Line 7, which says that someone is mortal. (We never learned the individual’s name!) Since Line 7 does not explicitly mention the made-up name, a, we use Line 8 to repeat Line 7 – without knowing the name of the individual “hiding” inside Line 2, we made a case analysis in Lines 4-7 that prove the result, anyway. This is how ∃e works.
To repeat: The ∃e rule describes how to discuss an anonymous individual (a witness) without knowing/revealing its identity: Assume the witness’s name is Mister a (“Mister Anonymous”) and that Mister a makes P true. Then, we deduce some fact, Q, that holds even though we don’t know who is Mister a. The restriction on the ∃e rule (Q cannot mention a) enforces that we have no extra information about the identity of Mister a – the name a must not leave the subproof.
Here is a Wheel-of-Fortune example that uses ∃e:
∃c (isVowel(c) ^ ∃s holds(s,c)) |- ∃y isVowel(y)
{
1. ∃c (isVowel(c) ^ ∃s holds(s,c)) premise
2. {
3. a isVowel(a) ^ ∃s holds(s,a) assume
4. isVowel(a) ^e1 3
5. ∃y isVowel(y) ∃i 4 a
}
6. ∃y isVowel(y) ∃e 1 2
}
We do not know the identity of the vowel held in an unknown square, but this does not prevent us from concluding that there is a vowel.
#### Standard Examples¶
For practice, we do some standard examples:
∃x p(x) |- ∃y p(y)
{
1. ∃x p(x) premise
2. {
3. a p(a) assume
4. ∃y p(y) ∃i 3 a
}
5. ∃y p(y) ∃e 1 2
}
∃x(f(x) ^ g(x)) |- (∃y f(y)) ^ (∃z g(z))
{
1. ∃x(f(x) ^ g(x)) premise
2. {
3. a f(a) ^ g(a) assume
4. f(a) ^e1 3
5. ∃y f(y) ∃i 4 a
6. g(a) ^e2 3
7. ∃z g(z) ∃i 6 a
8. (∃y f(y)) ^ (∃z g(z)) ^i 5 7
}
9. (∃y f(y)) ^ (∃z g(z)) ∃e 1 2
}
Notice that you cannot prove the converse:
(∃y f(y)) ^ (∃z g(z)) |- ∃x(f(x) ^ g(x)).
For example, say that the universe of discussion is the cast of Wheel of Fortune, and f == isMale and g == isFemale. Clearly, Pat Sajak is male and Vanna White is female, so (∃y f(y)) ^ (∃z g(z)) is a true premise. But ∃x(f(x) ^ g(x)) does not follow.
The following proof uses the ve-tactic – a cases analysis. See the assumptions at lines 3 and 6, based on Line 2:
∃x (p(x) V q(x)) |- (∃x p(x)) V (∃x q(x))
{
1. ∃x (p(x) V q(x)) premise
2. {
3. a p(a) V q(a) assume
4. {
5. p(a) assume
6. ∃x p(x) ∃i 5 a
7. (∃x p(x)) V (∃x q(x)) Vi1 6
}
8. {
9. q(a) assume
10. ∃x q(x) ∃i 9 a
11. (∃x p(x)) V (∃x q(x)) Vi2 10
}
12. (∃x p(x)) V (∃x q(x)) Ve 3 4 8
}
13. (∃x p(x)) V (∃x q(x)) ∃e 1 2
}
As an exercise, prove the converse of the above:
(∃x p(x)) V (∃x q(x)) |- ∃x (p(x) V q(x)).
#### An Important Example¶
We finish with this crucial example. We use the domain of people:
∃x ∀y isBossOf(x,y)
Read this as, “there is someone who is the boss of everyone”. From this strong fact we can prove that everyone has a boss, that is, ∀u∃v isBossOf(v,u):
∃x∀y isBossOf(x,y) |- ∀u∃v isBossOf(v,u)
{
1. ∃x∀y isBossOf(x,y) premise
2. {
3. b ∀y isBossOf(b,y) assume
4. {
5. a
6. isBossOf(b,a) ∀e 3 a
7. ∃v isBossOf(v,a) ∃i 6 b
}
8. ∀u∃v isBossOf(v,u) ∀i 4
}
9. ∀u∃v isBossOf(v,u) ∃e 1 2
}
In the above proof, we let b be our made-up name for the boss-of-everyone. So, we have the assumption that ∀y isBossOf(b,y). Next, we let a be “anybody at all” who we might examine in the domain of people. The proof exposes that the boss of “anybody at all” in the domain must always be b. ∀i and then ∃i finish the proof.
Here is the proof worked again, with the subproofs swapped:
∃x∀y isBossOf(x,y) |- ∀u∃v isBossOf(v,u)
{
1. ∃x∀y isBossOf(x,y) premise
2. {
3. a
4. {
5. b ∀y isBossOf(b,y) assume
6. isBossOf(b,a) ∀e 5 a
7. ∃v isBossOf(v,a) ∃i 6 b
}
8. ∃v isBossOf(v,a) ∃e 1 4
}
9. ∀u∃v isBossOf(v,u) ∀i 2
}
Can we prove the converse? That is, if everyone has a boss, then there is one boss who is the boss of everyone?
∀u∃v isBossOf(v,u) |- ∃x∀y isBossOf(x,y) ???
No – we can try, but we get stuck:
∀u∃v isBossOf(v,u) |- ∃x∀y isBossOf(x,y)
{
1. ∀u∃v isBossOf(v,u) premise
2. {
3. a
4. ∃v isBossOf(v,a) ∀e 1 a
5. {
6. b isBossOf(b,a) assume
}
6. ∀y isBoss(b,y) ∀i 2 NO --- THIS PROOF IS TRYING TO FINISH THE OUTER
SUBPROOF WITHOUT FINISHING THE INNER ONE FIRST.
We see that the “block structure” of the proofs warns us when we are making invalid deductions.
It is interesting that we can prove the following:
∃x∀y isBossOf(x,y) |- ∃z isBossOf(z,z)
(“if someone is the boss of everyone, then someone is their own boss”):
∃x∀y isBossOf(x,y) |- ∃z isBossOf(z,z)
{
1. ∃x∀y isBossOf(x,y) premise
2. {
3. b ∀y isBossOf(b,y) assume
4. isBossOf(b,b) ∀e 3 b
5. ∃z isBossOf(z,z) ∃i 4 b
}
6. ∃z isBossOf(z,z) ∃e 1 2
}
Line 4 exposes that the “big boss”, b, must be its own boss.
#### Domains and Models¶
The examples of bosses and workers illustrate these points:
1. You must state the domain of individuals when you state premises. In the bosses-workers examples, the domain is a collection of people. Both the bosses and the workers belong to that domain. Here are three drawings of possible different domains, where an arrow, person1 —> person2, means that person1 is the boss of person2:
Notice that ∀u∃v isBossOf(v,u) (“everyone has a boss”) holds true for the first two domains but not the third. ∃x∀y isBossOf(x,y) holds true for only the second domain.
2. When we make a proof of P |- Q and P holds true for a domain, then Q must also hold true for that same domain. We proved that ∃x∀y isBossOf(x,y) |- ∃z isBossOf(z,z), and sure enough, in the second example domain, ∃z isBossOf(z,z) holds true.
Our logic system is designed to work in this way! When we do a logic proof, we are generating new facts that must hold true for any domain for which the premises hold true. This property is called soundness of the logic, and we will examine it more closely in a later section in this chapter.
3. A domain can have infinitely many individuals. Here is a drawing of a domain of infinitely many people, where each person bosses the person to their right:
In this domain, ∀u∃v isBossOf(v,u) holds true as does ∀u∃v isBossOf(u,v) (“everyone bosses someone”), but ∃z isBossOf(z,z) does not hold true.
The third example domain is famous – it is just the integer domain, where isBossOf is actually >:
. . . < -3 < -2 < -1 < 0 < 1 < 2 < 3 < . . .
Indeed, one of the main applications of logic is proving properties of numbers. This leads to a famous question: Is it possible to write a collection of premises from which we can deduce (make proofs of) all the logical properties that hold true for the domain of integers?
The answer is NO. In the 1920s, Kurt Goedel, a German PhD student, proved that the integers, along with +, -, *, /, are so complex that it is impossible to ever formulate a finite set (or even an algorithmically defined infinite set) of premises that generate all the true properties of the integers. Goedel’s result, known as the First Incompleteness Theorem, set mathematics back on its heels and directly led to the formulation of theoretical computer science (of which this course is one small part). There is more material about Goedel’s work at the end of this chapter.
#### Tactics for The ∃-Rules¶
There are two tactics; neither is easy to master:
• (***) ∃e-tactic: To prove Premises, ∃x P(x) |- Q,
1. assume a and P(a), where a is a brand new anonymous name
2. prove Premises, P(a) |- Q
3. apply ∃e
The proof looks like this:
1. Premises premise
2. ∃x P(x) premise
i. {
j. a P(a) assume
(fill in)
k. Q // Q does not mention a!
}
l. Q ∃e 2 i
• (*) ∃i-tactic: To prove Premises |- ∃x P(x), try to prove P(e) for some e that already appears in the partially completed proof. Finish with ∃i:
1. Premises premise
. . .
i. ...e...
(fill in)
j. P(e)
k. ∃x P(x) ∃i j e
Look at the Wheel-of-Fortune proofs for instances where these tactics were applied.
## 8.3. Equivalences in Predicate Logic¶
Here are some important equivalences in predicate logic. (We include the pbc-rule to prove the third and fourth ones.)
• ∀x ∀y P(x,y) -||- ∀y ∀x P(x,y)
• ∃x ∃y P(x,y) -||- ∃y ∃x P(x,y)
• ~(∀x P(x)) -||- ∃x ~P(x)
• ~(∃x P(x)) -||- ∀x ~P(x)
• Q ^ (∀x P(x)) -||- ∀x (Q ^ P(x)) (where x does not appear in Q)
• Q v (∀x P(x)) -||- ∀x (Q V P(x)) (where x does not appear in Q)
• Q ^ (∃x P(x)) -||- ∃x (Q ^ P(x)) (where x does not appear in Q)
• Q V (∃x P(x)) -||- ∃x (Q V P(x)) (where x does not appear in Q)
## 8.4. Soundness and Completeness of Deduction Rules¶
Once again, it is time to consider what propositions mean and how it is that ∀i, ∀e, ∃i, ∃e preserve meaning.
At this point, it would be good to review the section on models for Applications of Propositional Logic to Program Proving. There, we saw that the connectives, ^, V, ~, -> were understood in terms of truth tables. Also, the primitive propositions were just letters like P, Q, and R, which were interpreted as either True or False.
Within predicate logic, we use predicates, like isMortal() and >, to build propositions, and we might also use functions, like +, within the predicates. We must give meanings to all predicates and functions so that we can decide whether propositions like isMortal(God) and (3+1)>x are True or False. The act of giving meanings to the predicates and functions is called an interpretation.
### 8.4.1. Interpretations¶
When we write propositions in a logic, we use predicates and function symbols (e.g., ∀i (i*2)>i). An interpretation gives the meaning of
1. the underlying domain – what set of elements it names;
2. each function symbol – what answers it computes from its arguments from the domain; and
3. each predicate – which combinations of arguments from the domain lead to True answers and False answers.
Here is an example. Say we have the function symbols, +, -, *, /, and predicate symbols, >, =. What do these names and symbols mean? We must interpret them.
• The standard interpretation of arithmetic is that int names the set of all integers; +, -, *, / name integer addition, subtraction, multiplication, and division, and = and > name integer equality comparison and integer less-than comparison. With this interpretation of arithmetic, we can interpret propositions. For example, ∀i (i*2)>i interprets to False, and ∃j (j*j)=j interprets to True. If we use any extra, “constant names”, (e.g., pi), we must give meanings to the constants, also.
• Now, given function names +, -, *, /, and predicates, =, >, we can choose to interpret them in another way. For example, we might interpret the underlying domain as just the nonnegative integers. We can interpret +, *, /, >, = as the usual operations on ints, but we must give a different meaning to -. We might define m - n == 0, whenever n > m.
• Yet another interpretation is to say that the domain is just {0,1}; the functions are the usual arithmetic operations on 0,1, modulo 2; and > is defined 1 > 0 (and that’s it).
These three examples show that the symbols in a logic can be interpreted in multiple different ways. (In the Chapter on Applications of Propositional Logic to Program Proving, we called an interpretation a “context”. In this chapter, we see that a “context” is quite complex – domain, functions, and predicates.)
Here is a second example. There are no functions, and the predicates are isMortal(_), isLeftHanded(_), isMarriedTo(_,_). An interpretation might make all (living) members of the human race as the domain; make isMortal(h) True for every human, h; make isLeftHanded(j) True for exactly those humans, j, who are left handed; and set isMarriedTo(m,f) True for all pairs of humans m, f, who have their marriage document in hand.
You get the idea….
We can ask whether a proposition is True within one specific interpretation, and we can ask whether a proposition is True within all possible interpretations. This leads to the notions of soundness and completeness for predicate logic:
A sequent, P_1, P_2, ..., P_n |- Q is valid in an interpretation, I, provided that when all of P_1, P_2, ..., P_n are True in interpretation I, so is Q. The sequent is valid exactly when it is valid in all possible interpretations. We have these results for the rules of propositional logic plus ∀i, ∀e, ∃i, ∃e:
1. soundness: When we use the deduction rules to prove that P_1, P_2, ..., P_n |- Q, then the sequent is valid (in all possible interpretations).
2. completeness: When P_1, P_2, ..., P_n |- Q is valid (in all possible interpretations), then we can use the deduction rules to prove the sequent.
Note that, if P_1, P_2, ..., P_n |- Q is valid in just one specific interpretation, we are not guaranteed that our rules will prove it. This is a famous trouble spot: For centuries, mathematicians were searching for a set of deduction rules that could be used to build logic proofs of all the True propositions of arithmetic, that is, the language of int, +, -, *, /, >, =. No appropriate rule set was devised.
In the early 20th century, Kurt Gödel showed that it is impossible to formulate a sound set of rules customized for arithmetic that will prove exactly the True facts of arithmetic. Gödel showed this by formulating True propositions in arithmetic notation that talked about the computational power of the proof rules themselves, making it impossible for the proof rules to reason completely about themselves. The form of proposition he coded in logic+arithmetic stated “I cannot be proved”. If this proposition is False, it means the proposition can be proved. But this would make the rule set unsound, because it proved a False claim. The only possibility is that the proposition is True (and it cannot be proved). Hence, the proof rules remain sound but are incomplete.
Gödel’s construction, called diagonalization, opened the door to the modern theory of computer science, called computability theory, where techniques from logic are used to analyze computer programs. Computability theory tells us what problems computers cannot solve, and why, and so we shouldn’t try. (For example, it is impossible to build a program-termination checker that works on all programs – the checker won’t work on itself!) There is also an offshoot of computability theory, called computational complexity theory, that studies what can be solved and how fast an algorithm can solve it.
Given an interpretation of a predicate logic, we can say that the “meaning” of a proposition is exactly the set of interpretations (cf. Chapter on Applications of Propositional Logic to Program Proving – “contexts”) in which the proposition is True. This returns us to the Boolean-algebra model of logic. Or, we can organize the interpretations so that an interpretation grows in its domain and knowledge over time. This returns us to the Kripke models. Or, we can introduce two new programming constructs, the abstract data type and the parametric polymorphic function and extend the Heyting interpretation.
All of these are possible and are studied in a typical second course on logic.
## 8.5. Summary¶
Here are the rules for the quantifiers, stated in terms of their tactics:
• (***) ∀i: use to prove Premises |- ∀x P(x):
1. Premises premise
i. {
j. a
(fill in)
k. P(a)
}
l. ∀x P(x) ∀i i
• (***) ∃e-tactic: use to prove Premises, ∃x P(x) |- Q:
1. Premises premise
2. ∃x P(x) premise
i. {
j. a P(a) assume
(fill in)
k. Q // Q does not mention a!
}
l. Q ∃e 2 i
• (*) ∀e-tactic: use to prove Premises, ∀x P(x) |- Q:
1. Premises premise
2. ∀x P(x) premise
. . .
j. P(i) ∀e 2 // where individual i appears earlier in the proof
(fill in)
k. Q
• (*) ∃i-tactic: use to prove Premises |- ∃x P(x):
1. Premises premise
. . .
i. ...e...
(fill in)
j. P(e)
k. ∃x P(x) ∃i j e
This note was adapted from David Schmidt's CIS 301, 2008, Chapter 6 course note. | 11,730 | 40,037 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.9375 | 4 | CC-MAIN-2018-30 | latest | en | 0.842592 |
https://www.numbersaplenty.com/3220 | 1,656,801,828,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656104205534.63/warc/CC-MAIN-20220702222819-20220703012819-00465.warc.gz | 978,362,119 | 3,557 | Search a number
3220 = 225723
BaseRepresentation
bin110010010100
311102021
4302110
5100340
622524
712250
oct6224
94367
103220
112468
121a44
131609
141260
15e4a
hexc94
3220 has 24 divisors (see below), whose sum is σ = 8064. Its totient is φ = 1056.
The previous prime is 3217. The next prime is 3221. The reversal of 3220 is 223.
3220 = T7 + T8 + ... + T26.
3220 = 72 + 82 + ... + 212.
It is a Harshad number since it is a multiple of its sum of digits (7).
It is a plaindrome in base 11.
It is a nialpdrome in base 10 and base 16.
It is a congruent number.
It is not an unprimeable number, because it can be changed into a prime (3221) by changing a digit.
It is a pernicious number, because its binary representation contains a prime number (5) of ones.
It is a polite number, since it can be written in 7 ways as a sum of consecutive naturals, for example, 129 + ... + 151.
It is an arithmetic number, because the mean of its divisors is an integer number (336).
It is an amenable number.
It is a practical number, because each smaller number is the sum of distinct divisors of 3220, and also a Zumkeller number, because its divisors can be partitioned in two sets with the same sum (4032).
3220 is an abundant number, since it is smaller than the sum of its proper divisors (4844).
It is a pseudoperfect number, because it is the sum of a subset of its proper divisors.
3220 is a wasteful number, since it uses less digits than its factorization.
3220 is an odious number, because the sum of its binary digits is odd.
The sum of its prime factors is 39 (or 37 counting only the distinct ones).
The product of its (nonzero) digits is 12, while the sum is 7.
The square root of 3220 is about 56.7450438364. The cubic root of 3220 is about 14.7667625193.
Adding to 3220 its reverse (223), we get a palindrome (3443).
The spelling of 3220 in words is "three thousand, two hundred twenty", and thus it is an iban number. | 565 | 1,942 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.703125 | 4 | CC-MAIN-2022-27 | latest | en | 0.902711 |
https://byjus.com/question-answer/the-dynamic-mass-of-a-photon-having-a-wavelength-of-10-nm-is-2-21/ | 1,638,568,547,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964362919.65/warc/CC-MAIN-20211203212721-20211204002721-00213.warc.gz | 214,553,805 | 26,450 | Question
The dynamic mass of a photon having a wavelength of 10 nm is:2.21×10−35 kg2.21×10−34 kg2.21×10−33 kg2.21×10−32 kg
Solution
The correct option is B 2.21×10−34 kgAccording to de-Broglie hypothesis, λ=hp where, h = Planck's constant p = momentum of photon So p = hλ = 6.626×10−3410×10−9 = 6.626×10−26 kg.ms Since photon travels at the speed of light, v=c Hence, p = mc where, c=3×108 m/s So, dynamic mass m = 6.626×10−26kg.ms3×108 m/s ≈2.21×10−34 kg
Suggest corrections | 194 | 494 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.703125 | 4 | CC-MAIN-2021-49 | latest | en | 0.521987 |
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