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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
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