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# A2 2 linear fxns notes
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### A2 2 linear fxns notes
1. 1. Algebra 2<br />Slope<br />EX1:Find the intercepts of the line y=3x-9. Then graph the line. <br />32004004191000<br />Slope<br />How is slope defined? Slope = ----------------------- = ---------------------------------------<br />EX2: What is the slope of the following graph?<br />What is the y-intercept of this graph?<br />EX3:What is the slope of the line going through the ordered pairs (4, 2) and (-3, 16)?<br />The most common form for a linear equation is called __________________________________.<br />The equation for this form is: ________________________________<br /><ul><li>x represents the _________________________
2. 2. y represents the _________________________
3. 3. m is the _____________________ of the line
4. 4. b is the ___________________________________</li></ul>To graph a line in this form:<br /><ul><li>Plot the _______________________________ first. (It’s the one point on the line that you already know.)
5. 5. _____________________________________________, use the slope to rise and run to the next point.
6. 6. _______________________________ with a line.</li></ul>320040022161500EX4:Graph the line y=23x-1.<br />HW “Line graph intuition” and “Slope of a line” exercises on Khan Academy<br /> | 443 | 1,643 | {"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} | 4.25 | 4 | CC-MAIN-2017-39 | latest | en | 0.714691 |
https://math.stackexchange.com/questions/1227353/a-graph-that-all-its-vertices-are-vertices-cut | 1,582,536,851,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875145910.53/warc/CC-MAIN-20200224071540-20200224101540-00268.warc.gz | 463,463,764 | 27,598 | # A graph that all its vertices are vertices cut [duplicate]
Is there any graph that all its vertices are cut vertices? I couldn't find a graph with this property? and if there is no such graph how can i prove that it does not exist.
• The earlier question is about connected graphs, but it’s clear that if very connected graph has a non-cut vertex, then every graph has a non-cut vertex. – Brian M. Scott Apr 9 '15 at 18:25
Suppose such a graph $G$ existed. We may assume WLOG it is connected. For each vertex $v$, let $f(v)$ be the minimum number of vertices in the connected components of the graph $G - v$ obtained by deleting $v$. Take $v$ that minimizes $f$, and let $C$ be a connected component of $G - v$ with $f(v)$ vertices. At least one vertex $w$ of $C$ is adjacent to $v$. Now in $G - w$, one connected component must contain $v$, and this will also contain all other vertices not in $C$. Any other connected component of $G - w$ will therefore be a proper subset of $C$, so $f(w) < f(v)$, contradiction. | 266 | 1,020 | {"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} | 3.578125 | 4 | CC-MAIN-2020-10 | latest | en | 0.936458 |
https://rebeccaprinting.com/gszes/855a23-warshall-algorithm-transitive-closure | 1,627,546,474,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046153854.42/warc/CC-MAIN-20210729074313-20210729104313-00086.warc.gz | 497,007,582 | 24,321 | # warshall algorithm transitive closure
In mathematics, the transitive closure of a binary relation R on a set X is the smallest relation on X that contains R and is transitive. The main advantage of Floyd-Warshall Algorithm is that it is extremely simple and easy to implement. Blog. Randomized Dictionary Structures:Structural Properties of Skip Lists. I am writing a program that uses Warshall's algorithm for to find a transitive closure of a matrix that represents a relation. ⢠Let A denote the initial boolean matrix. Example: Apply Floyd-Warshall algorithm for constructing the shortest path. Later it recognized form by Robert Floyd in 1962 and also by Stephen Warshall in 1962 for finding the transitive closure of a graph. Here is a link to the algorithm in psuedocode: http://people.cs.pitt.edu/~adamlee/courses/cs0441/lectures/lecture27-closures.pdf (page … Apply Warshall's algorithm to find the transitive closure of the digraph defined by the following adjacency matrix. Warshall's Algorithm The transitive closure of a directed graph with n vertices can be defined as the nxn boolean matrix T = {tij}, in which the element in the ith row and the jth column is 1 if there exists a nontrivial path (i.e., directed path of a positive length) from … Once we get the matrix of transitive closure, each query can be answered in O(1) time eg: query = (x,y) , answer will be m[x][y] To compute the matrix of transitive closure we use Floyd Warshall's algorithm which takes O(n^3) time and O(n^2) space. For calculating transitive closure it uses Warshall's algorithm. The algorithm thus runs in time θ(n 3). The running time of the Floyd-Warshall algorithm is determined by the triply nested for loops of lines 3-6. QUESTION 5 1. Transitive Closure it the reachability matrix to reach from vertex u to vertex v of a graph. The transitive closure of a binary relation R on a set X is the minimal transitive relation R^' on X that contains R. Thus aR^'b for any elements a and b of X provided that there exist c_0, c_1, ..., c_n with c_0=a, c_n=b, and c_rRc_(r+1) for all 0<=r | 513 | 2,086 | {"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-2021-31 | latest | en | 0.908825 |
https://mathhelpboards.com/threads/exponential-equations.3561/#post-15747 | 1,638,539,803,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964362879.45/warc/CC-MAIN-20211203121459-20211203151459-00261.warc.gz | 459,511,646 | 15,049 | # Exponential equations
#### linapril
##### New member
I need help with understanding two questions:
1. (2x+2)2
The answer I got was: 22x+2x++4, but according to the answers that is wrong. Can any explain why?
2. 84x/32x
I simplified this to become (84/32)x, but I don't know how to go from there... The answer is apparently 27x, but I don't understand how that can be...
Would appreciate the help enormously,
//APRIL
#### linapril
##### New member
Nevermind, I understood question 2 now, but question 1 is still a mystery to me...
#### Jameson
Staff member
I need help with understanding two questions:
1. (2x+2)2
The answer I got was: 22x+2x++4, but according to the answers that is wrong. Can any explain why?
I think the middle term is wrong. The first and the last terms look good though. Check your work and if you still don't see it then post your attempt and we'll help you sort it out
#### Prove It
##### Well-known member
MHB Math Helper
I need help with understanding two questions:
1. (2x+2)2
The answer I got was: 22x+2x++4, but according to the answers that is wrong. Can any explain why?
2. 84x/32x
I simplified this to become (84/32)x, but I don't know how to go from there... The answer is apparently 27x, but I don't understand how that can be...
Would appreciate the help enormously,
//APRIL
\displaystyle \begin{align*} \left( 2^x + 2 \right)^2 &= \left( 2^x \right)^2 + 2\cdot 2\cdot 2^x + 2^2 \\ &= 2^{2x} + 2^{x + 2} + 2^2 \end{align*} | 451 | 1,470 | {"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} | 3.984375 | 4 | CC-MAIN-2021-49 | latest | en | 0.952914 |
https://www.physicsforums.com/threads/rank-the-cases-from-greatest-to-smallest-in-order-of-magnitude.923394/ | 1,601,285,876,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600401598891.71/warc/CC-MAIN-20200928073028-20200928103028-00714.warc.gz | 962,858,824 | 15,372 | # Rank the cases from greatest to smallest in order of magnitude
## Homework Statement
http://imgur.com/PUrHBaa
Question 13 in the middle of the page.
Each case in the figure shows an example of force vectors exerted on an object. These forces are all of the same magnitude F_o. Assume the forces lie in the plane of the paper. Rank the cases from greatest to smallest in order of magnitude of the total force. Note: the total force is the vector sum of the individual forces exerted on the objct.
## The Attempt at a Solution
Okay so they say they all have the same magnitude, so I chose value 1 for F_o just to make it simple.
That gives me a magnitude of 1 for case C.
For case B, you have F_y = 1 + sin (5deg) and F_x = cos (5 deg). using (a^2 + b^2 )^(1/2) you get 1.47 N
For case A you have F_y = 1 - sin (5 deg) and F_x = cos (5 deg). Using (a^2 + b^2 )^(1/2) you get 1.35 N
but my book tells me the answer is B>C>A
Is my book just wrong? Or am I doing this wrong? This is simple pythagorean theorem..
Related Introductory Physics Homework Help News on Phys.org
haruspex | 303 | 1,088 | {"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} | 3.640625 | 4 | CC-MAIN-2020-40 | latest | en | 0.847038 |
https://www.jiskha.com/questions/763748/noah-and-brianna-want-to-calculate-the-distance-between-their-houses-which-are-opposite | 1,623,581,774,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623487608702.10/warc/CC-MAIN-20210613100830-20210613130830-00352.warc.gz | 766,586,479 | 4,601 | # Math
Noah and Brianna want to calculate the distance between their houses which are opposite sides of a water park. They mark a point, A, 120m long the edge of the water park from Brianna's house. The measure <NBA as 75degree and <BAN as 70degree. Determine the distance between their houses.
1. 👍
2. 👎
3. 👁
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log (1/2)^-1/2 = log2^1/2
Thanks.
2. Standard rules of exponents - the log term is a red herring for this part
$a^{-b} = \dfrac{1}{a^b} = \left(\dfrac{1}{a}\right)^b$
In your case $a= \dfrac{1}{2}$ and $b = \dfrac{1}{2}$
$\left(\dfrac{1}{2}\right)^{-1/2} = \left(\dfrac{1}{\frac{1}{2}}\right)^{1/2} = 2^{1/2}$
3. Originally Posted by SWEngineer
In the "Foundation Maths" book, chapter (20), "The Logarithm Function", I'm just having an issue of understanding how we got the result in the following (I have used "^" for power): log (1/2)^-1/2 = log2^1/2
$\log \left( {2^{\frac{{ - 1}}{2}} } \right) = \frac{1}{2}\left[ {\log (2) - \log (1)} \right] = \log \left( {2^{\frac{1}{2}} } \right)$
4. Thanks a lot for your replies. | 339 | 913 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 5, "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-2016-50 | longest | en | 0.849683 |
https://www.bankersadda.com/data-sufficiency-for-ibps-rrb-prelims | 1,571,534,275,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570986700560.62/warc/CC-MAIN-20191020001515-20191020025015-00025.warc.gz | 802,231,091 | 31,481 | # Data Sufficiency for IBPS RRB Prelims: 27th July 2018
Dear Aspirants,
Reasoning Questions for IBPS RRB 2018
Reasoning Ability is an onerous section. With the increasing complexity of questions, it becomes hard for one to give it the cold shoulder. The only way to make the grade in this particular section in the forthcoming banking exams like IBPS RRB is to practice continuously with all your heart and soul. And, to let you practice with the best of the latest pattern questions, here is the Adda247 Reasoning Quiz based on the study plan and the exact same pattern of questions that are being asked in the exams.
Directions (1-5): Each of the questions below consists of a question and two statements numbered I and II given below it. You have to decide whether the data provided in the statements are sufficient to answer the question. Read both the statements. Give answer-
(a) If the data in statement I alone are sufficient to answer the question, while the data in statement II alone are not sufficient to answer the question.
(b) If the data in statement II alone are sufficient to answer the question, while the data in statement I alone are not sufficient to answer the question.
(c) If the data either in statement I alone or in statement II alone are sufficient to answer the question.
(d) If the data in both statements I and II together are not sufficient to answer the question.
(e) If the data in both statements I and II together are necessary to answer the question.
Q1. How is Ritika related to Ritesh?
I. Ritika who has two children is daughter of Radhika. Radhika has only one grandson named Ritesh.
II. Paras who is brother of Ritika has only one child named Ridhi.
Q2. Who stays at 4th floor of the building having 5 floors (1st is the bottommost floor and 5th is the topmost floor, also ground floor is empty)?
I. Madhuri stays on an odd-numbered floor.
II. Akshay stays three floors above Madhuri.
Q3. What is the distance between points X and Y?
I. Point P is 10 m west of point C. Point D is 5 m north of point X which is 5 m west of point C.
II. Point D is equidistant from points X and Y.
Q4. How is ‘great’ written in that code?
I. In a certain code, ‘great are those days’ is written as ‘ki vo tu mpi’ and ‘those days were good’ is written as ‘ki fo mpi ta’
II. In a certain code, ‘many days passed’ is written as ‘ti mpi dis’ and ‘those good years’ is written as ‘ko ki ka’.
Q5. On which day is Nikita’s birthday?
I. Shruti remembers that Nikita’s birthday falls in February. Preeti remembers that Nikita’s birthday falls on either 29 or 30.
II. Krish remembers that Nikita’s birthday occurs in the last week of a month.
Directions (6-10): Each of the questions below consists of a question and two statements numbered I and II given below it. You have to decide whether the data provided in the statements are sufficient to answer the question.
(a) If statements I alone is sufficient to answer the question, but statement II alone is not sufficient to answer the question.
(b) If statement II alone is sufficient to answer the question, but statement I alone is not sufficient to answer the question.
(c) If statement either I or II is sufficient to answer the question.
(d) If both the statements I and II taken together are not sufficient to answer the questions.
(e) If both the statements I and II taken together are sufficient to answer the questions.
Q6. What is Sonali’s position from the left end of a row?
I. There are 4 students between Gauri and Sonali. Bhavna is 6th to the right of Gauri.
II. Gini is 6th to the left of Sonali and is 2nd from the left end.
Q7. How many students are sitting between P and Q?
I. P is 5th to the right of R and 6th to the left of S. Q is 6th to the right of R.
II. In a row of 25 students, P is 5th from left end and Q is 20th from the right end.
Q8. Among G, H, I, J and K who has got the marks in middle position?
I. G has got 1 mark less than H and has got greater marks than K.
II. I and J have got greater marks than G.
Q9. How is N related to Q?
I. C is daughter of Q, who is also the mother of D.
II. B is brother of C and N is the wife of D.
Q10. Who has got highest marks among A, B, C, D and E?
I. D has got greater marks than A and less than C. Also, B has got greater marks than A and less than E.
II. E has got less marks than C and greater than B and D. A has got the lowest marks.
Directions (11-15): Each of the questions below consists of a question and two statements numbered I and II given below it. You have to decide whether the data provided in the statements are sufficient to answer the question. Read both the statements.
(a) If the data in statement I alone are sufficient to answer the question, while the data in statement II alone are not sufficient to answer the question
(b) If the data in statement II alone are sufficient to answer the question, while the data in statement I alone are not sufficient to answer the question
(c) If the data either in statement I alone or in statement II alone are sufficient to answer the question
(d) If the data in both statements I and II together are not sufficient to answer the question
(e) If the data in both statements I and II together are necessary to answer the question
Q11. How is ‘duster’ written in a code language?
I. ‘Buy three dusters get one free’ is written as ‘dd ee jj oo tt yy’ in that code language.
II. ‘Purchase one sharpener get three free’ is written as ‘dd ee ll oo ww yy’ in that code language.
Q12. L is in which direction of M?
I. C is towards East of L and towards North-East of M.
II. C is towards North of D, which is towards South-East of L, which is towards west of C.
Q13. Amongst G, H and I, who scored the highest number of runs in the series XYZ?
I. Strike rate of G was less than H but higher than I in the series.
II. Number of total centuries and fifties made by H were higher than G and I both, in the series.
Q14. P, Q, R, S, T and V are sitting in a circle, facing towards the center of the circle. Who is sitting on the immediate right of Q?
I. P is facing S. Only R is between P and Q. V is between T and P.
II. V is on the immediate left of T. Only S is between Q and T. R is not adjacent to V.
Q15. Among A, E, I, O and U, which book is the costliest?
I. Price of the book E is more than those of A and U only and the prices of the books I and O are not equal.
II. Number of pages in the books A, E, I, O and U are 800, 700, 600, 500 and 400, respectively.
You may also like to read: | 1,641 | 6,493 | {"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} | 3.578125 | 4 | CC-MAIN-2019-43 | latest | en | 0.942859 |
http://www.algebra.com/algebra/homework/word/geometry/Geometry_Word_Problems.faq.question.159620.html | 1,371,701,941,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368710274484/warc/CC-MAIN-20130516131754-00092-ip-10-60-113-184.ec2.internal.warc.gz | 314,400,022 | 4,729 | # SOLUTION: The area of a square is a 1/4 as large as the area of a triangle. The triangle has a 16 inch side and altitude H the same as the square. Draw the diagrams. Find the length of the
Algebra -> Algebra -> Customizable Word Problem Solvers -> Geometry -> SOLUTION: The area of a square is a 1/4 as large as the area of a triangle. The triangle has a 16 inch side and altitude H the same as the square. Draw the diagrams. Find the length of the Log On
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Word Problems: Geometry Solvers Lessons Answers archive Quiz In Depth
Question 159620: The area of a square is a 1/4 as large as the area of a triangle. The triangle has a 16 inch side and altitude H the same as the square. Draw the diagrams. Find the length of the side of the square.Answer by checkley77(12569) (Show Source): You can put this solution on YOUR website!(16h/2)1/4=h^2 16h/8=h^2 8h^2=16h 8h^2-16h=0 8h(h-2)=0 h=2 answer. area of the square=2*2=4 Proof. (16*2/2)1/4=2*2 32/2*1/4=4 32/8=4 4=4 | 364 | 1,214 | {"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} | 4.03125 | 4 | CC-MAIN-2013-20 | latest | en | 0.86121 |
https://www.thestudentroom.co.uk/showthread.php?t=38378 | 1,590,529,794,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347391309.4/warc/CC-MAIN-20200526191453-20200526221453-00089.warc.gz | 939,641,673 | 40,064 | # for those who want P2 practice
Watch
This discussion is closed.
#1
i've just done this beast of a P2 volumes of revolution question.
try it, it's good practice.
The diagram shows the region R bounded by the curve y^2=x-1, the x-axis and y-axis, and the line y=2
Find the volume when R is rotated 360 degrees around:
a) the x axis
b) y axis
a)12pi b)206pi/15
0
15 years ago
#2
(Original post by hihihihi)
i've just done this beast of a P2 volumes of revolution question.
try it, it's good practice.
The diagram shows the region R bounded by the curve y^2=x-1, the x-axis and y-axis, and the line y=2
Find the volume when R is rotated 360 degrees around:
a) the x axis
b) y axis
a)12pi b)206pi/15
a. volume= 20pi-pi*int(x-1)[5 and 1]
=20pi-pi(x^2/2-x)
=20pi-pi(12.5-5-0.5+1)
=20pi-8pi
=12pi
0
15 years ago
#3
0 and 1 i think
0
15 years ago
#4
(Original post by lgs98jonee)
wot r the limits when u integrate around the xaxis??
well you know y = 2...
cmon dude, dont slip now!
0
15 years ago
#5
oh yeh lol i fell in the same trap i didnt see y=2! deary me
0
#6
(Original post by lgs98jonee)
a. area= 6-pi*int(x-1)[3 and 1]
=6-pi(x^2/2-x)
=6-pi(4.5-0.5-3+1)
=6-2pi
area?? it's volumes!!
0
15 years ago
#7
yeh i just noticed the y=2! oops
0
15 years ago
#8
(Original post by hihihihi)
you also need to find another limit..clue: when y=2, x=...
oh thnx for that clue.. .give us a chance to do the question....seems like a really patronisng tone..ok then children wot if y=1+x and x=1, y=...
0
15 years ago
#9
r u sure that it is not 12.5pi?
0
15 years ago
#10
I got for a) (25pi-5pi)-(1pi-1pi) = 20pi and for b) 404/15pi...
When y = 2... 4 = x-1 so x=5... You seem to have integrated the upper value as being 4... Why?
For a) y was squared already so I just integrated x^2 - x between 5 and 1...
Have I done something horrible wrong?
0
15 years ago
#11
why do u do 20pi-what u get? where are u minusing it from?
0
15 years ago
#12
I am such a retard... I forgot to find the intersection points... I will fail this bloody exam!
0
15 years ago
#13
(Original post by hihihihi)
i've just done this beast of a P2 volumes of revolution question.
try it, it's good practice.
The diagram shows the region R bounded by the curve y^2=x-1, the x-axis and y-axis, and the line y=2
Find the volume when R is rotated 360 degrees around:
a) the x axis
b) y axis
a)12pi b)206pi/15
b. volume=piINT(y^2+1)^2 dx
=pi INT(y^4+2y^2+1)
=pi(y^5/5+2y^3/3+x) [between 2 and 0]
=pi(32/5+16/3+2)
=96/15+80/15+30/15
=206pi/15
0
15 years ago
#14
(Original post by Sapphira)
why do u do 20pi-what u get? where are u minusing it from?
well i did the big cylinder (pi *2^2*5)-the area under the curve between x=0 and x=5
0
#15
(Original post by lgs98jonee)
a. volume= 20pi-pi*int(x-1)[5 and 0]
=20pi-pi(x^2/2-x)
=20pi-pi(12.5-5)
=20pi-7.5pi
=12.5pi
you need to do between 5 and 1??
0
#16
(Original post by lgs98jonee)
b. volume=piINT(y^2+1)^2 dx
=pi INT(y^4+2y^2+1)
=pi(y^5/5+2y^3/3+x) [between 2 and 0]
=pi(32/5+16/3+2)
=96/15+80/15+30/15
=206pi/15
high five!!
0
15 years ago
#17
(Original post by hihihihi)
high five!!
they werent easy
0
15 years ago
#18
The diagram shows the region R bounded by the curve y^2=x-1, the x-axis and y-axis, and the line y=2
Find the volume when R is rotated 360 degrees around:
a) the x axis
b) y axis
a)
2^2 = x-1
x = 5
Vol. = (5/0){ pi(y^2) dx
= (5/0){ pi x - pi) dx
= [pi.x^2/2 - pi.x] (5/0)
= [pi.25/2 - 5pi] - 0
= 12.5pi
b)
y^2=x-1
x = (y^2 + 1)
x^2 = y^4 + 2y^2 + 1
Vol. = (2/0){ pi(x^2) dy
= (2/0){ (pi.y^4 + 2pi.y^2 + pi) dy
= [pi.y^5/5 + 2pi.y^3/3 + pi.y](2/0)
= [(32/5)pi + (16/3)pi + (2)pi]
= (206/15)pi
0
15 years ago
#19
(Original post by mik1a)
Vol. = (5/0){ pi(y^2) dx
= (5/0){ pi x - pi) dx
= [pi.x^2/2 - pi.x] (5/0)
= [pi.25/2 - 5pi] - 0
= 12.5pi
i got that but apparently it was 12pi :-(
0
15 years ago
#20
I don't get why between 5 and 1.
0
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http://threadspodcast.com/margin-of/marginal-error-statistics.html | 1,518,985,426,000,000,000 | text/html | crawl-data/CC-MAIN-2018-09/segments/1518891812259.18/warc/CC-MAIN-20180218192636-20180218212636-00166.warc.gz | 351,572,472 | 4,420 | Home > Margin Of > Marginal Error Statistics
# Marginal Error Statistics
## Contents
Most surveys you come across are based on hundreds or even thousands of people, so meeting these two conditions is usually a piece of cake (unless the sample proportion is very Linearization and resampling are widely used techniques for data from complex sample designs. The new employees appear to be giving out too much ice cream (although the customers probably aren't too offended). Swinburne University of Technology. navigate here
MathWorld. According to sampling theory, this assumption is reasonable when the sampling fraction is small. Definition The margin of error for a particular statistic of interest is usually defined as the radius (or half the width) of the confidence interval for that statistic.[6][7] The term can In astronomy, for example, the convention is to report the margin of error as, for example, 4.2421(16) light-years (the distance to Proxima Centauri), with the number in parentheses indicating the expected https://en.wikipedia.org/wiki/Margin_of_error
## Margin Of Error Definition Statistics
How to Find the Critical Value The critical value is a factor used to compute the margin of error. What is a Survey?. Find the degrees of freedom (DF). Different confidence levels For a simple random sample from a large population, the maximum margin of error, Em, is a simple re-expression of the sample size n.
For example, suppose the true value is 50 people, and the statistic has a confidence interval radius of 5 people. Here's an example: Suppose that the Gallup Organization's latest poll sampled 1,000 people from the United States, and the results show that 520 people (52%) think the president is doing a One example is the percent of people who prefer product A versus product B. Margin Of Error Synonym That means that in order to have a poll with a margin of error of five percent among many different subgroups, a survey will need to include many more than the
For more complex survey designs, different formulas for calculating the standard error of difference must be used. Margin Of Error Calculator The math behind it is much like the math behind the standard deviation. When the sampling distribution is nearly normal, the critical value can be expressed as a t score or as a z score. https://en.wikipedia.org/wiki/Margin_of_error The Math Gods just don't care.
When estimating a mean score or a proportion from a single sample, DF is equal to the sample size minus one. Margin Of Error Excel Political Animal, Washington Monthly, August 19, 2004. Let's say you picked a specific number of people in the United States at random. Retrieved February 15, 2007. ^ Braiker, Brian. "The Race is On: With voters widely viewing Kerry as the debate’s winner, Bush’s lead in the NEWSWEEK poll has evaporated".
## Margin Of Error Calculator
In astronomy, for example, the convention is to report the margin of error as, for example, 4.2421(16) light-years (the distance to Proxima Centauri), with the number in parentheses indicating the expected http://www.dummies.com/education/math/statistics/what-the-margin-of-error-tells-you-about-a-statistical-sample/ Multiply by the appropriate z*-value (refer to the above table). Margin Of Error Definition Statistics Comparing percentages In a plurality voting system, where the winner is the candidate with the most votes, it is important to know who is ahead. Acceptable Margin Of Error Survey Research Methods Section, American Statistical Association.
Retrieved from "https://en.wikipedia.org/w/index.php?title=Margin_of_error&oldid=744908785" Categories: Statistical deviation and dispersionErrorMeasurementSampling (statistics)Hidden categories: Articles with Wayback Machine links Navigation menu Personal tools Not logged inTalkContributionsCreate accountLog in Namespaces Article Talk Variants Views Read Edit check over here MSNBC, October 2, 2004. Now, if it's 29, don't panic -- 30 is not a magic number, it's just a general rule of thumb. (The population standard deviation must be known either way.) Here's an The margin of error is a measure of how close the results are likely to be. Margin Of Error In Polls
The chart shows only the confidence percentages most commonly used. Z Score 5. The margin of error of an estimate is the half-width of the confidence interval ... ^ Stokes, Lynne; Tom Belin (2004). "What is a Margin of Error?" (PDF). http://threadspodcast.com/margin-of/marginal-error-wiki.html This may not be a tenable assumption when there are more than two possible poll responses.
For simplicity, the calculations here assume the poll was based on a simple random sample from a large population. Margin Of Error Sample Size Refer to the above table for the appropriate z*-value. It can be calculated as a multiple of the standard error, with the factor depending of the level of confidence desired; a margin of one standard error gives a 68% confidence
## Suppose you know that 51% of people sampled say that they plan to vote for Ms.
This maximum only applies when the observed percentage is 50%, and the margin of error shrinks as the percentage approaches the extremes of 0% or 100%. If we use the "relative" definition, then we express this absolute margin of error as a percent of the true value. You need to include the margin of error (in this case, 3%) in your results. Margin Of Error Confidence Interval Calculator Introductory Statistics (5th ed.).
Typically, you want to be about 95% confident, so the basic rule is to add or subtract about 2 standard errors (1.96, to be exact) to get the MOE (you get One example is the percent of people who prefer product A versus product B. The likelihood of a result being "within the margin of error" is itself a probability, commonly 95%, though other values are sometimes used. weblink Post a comment and I'll do my best to help!
For example, a poll might state that there is a 98% confidence interval of 4.88 and 5.26. | 1,288 | 5,987 | {"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-2018-09 | latest | en | 0.910723 |
https://jeopardylabs.com/play/addition-and-subtraction-without-regrouping-5 | 1,638,857,508,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964363336.93/warc/CC-MAIN-20211207045002-20211207075002-00268.warc.gz | 417,017,465 | 9,230 | Subtraction- no regrouping
Simple Subtraction word problems
Simple Multiplication
100
73+6=
79
100
98-0=
98
100
I have 5 apples and my mom gave me 7 more apples. How many apples do I have all together?
7+5 = 12
100
If I had 30 pencils and 10 of them broke, How many pencils are left?
30-10=20
100
5x5=
25
200
51+7=
58
200
85-35=
50
200
Lisa's friend gave her 22 stickers. She had 15 stickers before. How many stickers are there now?
22+15=37
200
The temperature was 25 degrees. It went down by 4 degrees at night. What is the temperature now?
25-4=21
200
100x1=
100
300
38+20=
58
300
91-30=
61
300
Bill had 36 rocks. He collected 4 more from the beach. How many rocks does he have now?
36+4=40
300
How much is 55-12=?
43
300
3x11=
33
400
27+42=
69
400
84-11=
73
400
There were 15 birds on the fence. 4 more birds join them. How many birds are there total?
15+4=19
400
If we ordered 20 pizza boxes and finished 5 boxes before the party and 5 boxes after the party. How many boxes will be left?
20-5-5= 10
400
5x10=
50
500
63+34=
97
500
30-20=
10
500
What does 8+8?=
16
500
Tony had 14 candies and he ate 7 of them. How many candies does he still have?
14-7=7
500
9x9=
81
Click to zoom | 421 | 1,257 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.984375 | 4 | CC-MAIN-2021-49 | latest | en | 0.951285 |
https://www.thoughtco.com/archimedes-volume-and-density-3976031 | 1,685,693,769,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224648465.70/warc/CC-MAIN-20230602072202-20230602102202-00547.warc.gz | 1,112,121,833 | 40,014 | # How to Measure Volume and Density
The Tale of Archimedes and the Gold Crown
Archimedes needed to determine if a goldsmith had embezzled gold during the manufacture of the royal crown for King Hiero I of Syracuse. How would you find out if a crown was made of gold or a cheaper alloy? How would you know if the crown was a base metal with a golden exterior? Gold is a very heavy metal (even heavier than lead, though lead has a higher atomic weight), so one way to test the crown would be to determine its density (mass per unit volume). Archimedes could use scales to find the mass of the crown, but how would he find the volume? Melting the crown down to cast it into a cube or sphere would make for an easy calculation and an angry king.
After pondering the problem, it occurred to Archimedes that he could calculate volume based on how much water the crown displaced. Technically, he didn't even need to weigh the crown, if he had access to the royal treasury since he could just compare the displacement of water by the crown with the displacement of water by an equal volume of the gold the smith was given to use. According to the story, once Archimedes hit upon the solution to his problem, he burst outside, naked, and ran through the streets yelling, "Eureka! Eureka!"
Some of this might be fiction, but Archimedes' idea to calculate the volume of an object and its density if you know the object's weight was fact. For a small object, in the lab, the easiest way to do this is to partly fill a graduated cylinder large enough to contain the object with water (or some liquid in which the object won't dissolve). Record the volume of water. Add the object, being careful to eliminate air bubbles. Record the new volume. The volume of the object is the initial volume in the cylinder subtracted from the final volume. If you have the object's mass, its density is the mass divided by its volume.
## How to Do It at Home
Most people don't keep graduated cylinders in their homes. The closest thing to it would be a liquid measuring cup, which will accomplish the same task, but with a lot less accuracy. There is another way to calculate volume using Archimede's displacement method.
1. Partially fill a box or cylindrical container with liquid.
2. Mark the initial liquid level on the outside of the container with a marker. | 509 | 2,341 | {"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} | 4.09375 | 4 | CC-MAIN-2023-23 | latest | en | 0.970734 |
https://www.coursehero.com/file/3034815/Class-3-Review/ | 1,490,867,241,000,000,000 | text/html | crawl-data/CC-MAIN-2017-13/segments/1490218193288.61/warc/CC-MAIN-20170322212953-00534-ip-10-233-31-227.ec2.internal.warc.gz | 880,765,696 | 34,452 | Class 3 Review
# Class 3 Review - Evaluate the appropriate expression(2-D or 3-D truss For the 2-D case 2*Number of Joints – Number of Links – 3 = 2*J – L –
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ENGR 111 - Track A 1 Truss Class 03
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ENGR 111 - Track A 2 How are trusses built? Links and Joints Trusses are built by joining bars (links) to form a lattice-like structure. The joints transmit forces from one link to another but are incapable of transmitting torques. This is one link! This is one link! Links (a.k.a. bars) Joints
ENGR 111 - Track A 3 Calculating minimum number of links Q: How do I find out how many links are necessary to create stable trusses? 2-D: No. of Links Needed = 2*No. of Joints -3 3-D: No. of Links Needed = 3*No. of Joints -6
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ENGR 111 - Track A 4 How to determine stability of a truss Now you can determine whether your truss is unstable (underconstrained, links missing) trusses: a truss that can slide and rotate as a whole. stable (correct number of links) overconstrained (too many links).
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Unformatted text preview: Evaluate the appropriate expression (2-D or 3-D truss). For the 2-D case: 2*Number of Joints – Number of Links – 3 = ? 2*J – L – 3 = ? For the 3-D case: 3*Number of Joints – Number of Links – 6 = ? 3*J – L – 6 = ? If positive, the truss is unstable, you need to add links. If zero, the truss is stable, good job! If negative, the truss is over constrained, you could remove some links. ENGR 111 - Track A 5 Cross-bracing Q: How do I stabilize an unstable truss? Stabilize an “underconstrained” or wobbly truss by adding diagonals or “cross-braces” Convert all quadrilaterals to triangles: Design guideline: “add short cross-braces” (B) is better than (C) (A) Unstable (B) Short brace (C) Long brace...
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## This note was uploaded on 04/27/2009 for the course ENGR 111 taught by Professor Walker during the Spring '07 term at Texas A&M.
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Class 3 Review - Evaluate the appropriate expression(2-D or 3-D truss For the 2-D case 2*Number of Joints – Number of Links – 3 = 2*J – L –
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Ask a homework question - tutors are online | 819 | 2,518 | {"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} | 4.09375 | 4 | CC-MAIN-2017-13 | longest | en | 0.345662 |
https://discuss.leetcode.com/topic/34729/different-idea-incomplete-help | 1,513,264,981,000,000,000 | text/html | crawl-data/CC-MAIN-2017-51/segments/1512948544677.45/warc/CC-MAIN-20171214144324-20171214164324-00265.warc.gz | 532,879,338 | 13,660 | # Different idea, incomplete... help?
• Before I came up with my merge sort solution, I tried another idea: Attach the index to each number, then sort the numbers (with their indexes), then see how much each number moved. Take for example input `[5, 2, 6, 1, 3]` and sort it so it's `[1, 2, 3, 5, 6]`.
The 5 moved three steps to the right, and that's because there were three smaller numbers on its right initially. So in the result, write down "3" for the 5. Well, it doesn't quite work, but that's the basic idea.
You can see that it doesn't quite work by checking the 2: It moves zero steps, even though there was a smaller number on its right, namely the 1. But while the 1 did push the 2 one step to the right, the 5 pushed it one step to the left, so that overall the 2 didn't move.
So maybe the index differences of the basic idea can somehow be adjusted by the number of rightmovers? Here are the numbers of rightmovers for each index:
``````index: [0, 1, 2, 3, 4]
before: [5, 2, 6, 1, 3]
after: [1, 2, 3, 5, 6]
rightmover changes: [1, 0, 1, -1, -1]
1 numbers moved right from index 0
1 numbers moved right from index 1
2 numbers moved right from index 2
1 numbers moved right from index 3
0 numbers moved right from index 4
``````
That was computed by this code in O(nlogn) for sorting plus O(n) for the rest:
``````nums = [5, 2, 6, 1, 3]
rightmover_changes = [0] * len(nums)
indexed_and_sorted = sorted(enumerate(nums), key=lambda x: x[::-1])
for i, (orig_i, num) in enumerate(indexed_and_sorted):
if orig_i < i:
rightmover_changes[orig_i] += 1
rightmover_changes[i] -= 1
print 'index: ', range(len(nums))
print 'before:', nums
print 'after: ', sorted(nums)
print 'rightmover changes:', rightmover_changes
rightmovers = 0
for i, change in enumerate(rightmover_changes):
rightmovers += change
print '{} numbers moved right from index {}'.format(rightmovers, i)
``````
I might try again to make this work, but maybe I won't. For now I just wanted to put this out here.
• what if you track not the right movers, but the leftmovers?
so for 1 you have that it moved 3 positions left, hence it's "area of influence" in the sorted array would be from 1 to 3.
at each point of the array you track how many leftmovers "influence" the current number:
for 2, you calculate 1-1 (index difference, move) +1 (number of symbols that were initially after 2)
and for 3, that would be 2-4 (move) + 1 (for the 1 that moved to the left) => total s to -1, so
I didn't think through how to track the number of "influencers" - with stack maybe ?
would this work?
Looks like your connection to LeetCode Discuss was lost, please wait while we try to reconnect. | 778 | 2,661 | {"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} | 3.6875 | 4 | CC-MAIN-2017-51 | latest | en | 0.89381 |
https://quant.stackexchange.com/questions/9729/unique-equivalent-martingale-measure-in-incomplete-markets | 1,709,474,587,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947476374.40/warc/CC-MAIN-20240303111005-20240303141005-00509.warc.gz | 464,519,109 | 39,449 | # unique equivalent martingale measure in incomplete markets
Do you have any idea about how we can prove, and under which conditions, that an equivalent martingale measure (EMM) in an incomplete market is unique? The assumptions we have made are:
1) that the stochastic process St of the asset is a semi martingale (continuous) and
2) that this EMM exists.
In other words, that the variance optimal measure is unique.
Thanks.
• Can you clarify your question a little? A market being incomplete is equivalent to there being multiple equivalent martingale measures. Are you asking how to show that the variance optimal EMM is unique? The answer to that question is basically strict convexity, I can elaborate if that's what you're after. Dec 15, 2013 at 19:31
• yes, this is exactly what I am trying to show.
– jim
Dec 15, 2013 at 20:09
Suppose that there are multiple martingale measures $Q_1$ and $Q_2$ that attain the minimal variance. Then the convex combination $Q_* := \frac{1}{2}Q_1 + \frac{1}{2}Q_2$ is also a martingale measure. Due to the strict convexity of $f(x) = x^2$, it can be shown that $$E_P \left[\frac{dQ_*}{dP}^2 \right] < \frac{1}{2} E_P \left[ \frac{dQ_1}{dP}^2 \right] + \frac{1}{2} E_P \left[ \frac{dQ_2}{dP}^2 \right].$$
To make this completely airtight, you need to use the notion of uniform convexity, i.e. $f \left(\frac{x+y}{2} \right) < \frac{1}{2} \left( f(x) + f(y) \right) -\epsilon | x - y|$, for some $\epsilon$. The details are tedious but not hard.
In any case, you have a contradiction, and the minimizer must be unique.
• Are you missing come expectations in the right-hand side?
– SBF
Apr 8, 2014 at 15:34
• yeah, thanks. this was a while ago, i have to try and understand what i wrote. Apr 8, 2014 at 19:22
• Sure :) also, what does these square mean? That you take an expectation/integral of squared R-N derivative?
– SBF
Apr 9, 2014 at 7:32
• yeah. you apply $f(x)$ pointwise. Apr 9, 2014 at 8:01 | 594 | 1,948 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.625 | 4 | CC-MAIN-2024-10 | latest | en | 0.906692 |
https://q4interview.com/aptitude-ques-ans-discussion.php?qid=5284&t=6&qnum=5&cat=10 | 1,542,273,860,000,000,000 | text/html | crawl-data/CC-MAIN-2018-47/segments/1542039742569.45/warc/CC-MAIN-20181115075207-20181115101207-00510.warc.gz | 694,147,860 | 17,844 | Note 1
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# Arithmetic Aptitude :: Profit and Loss
## 5. An article is sold at a certain price. By selling it at 2/3 of that price one loses 10%. Find the gain percent at original price.
Answer: Option B
Explanation :
Let the original S.P be Rs. X.
Then new S.P = Rs. (2/3)X, Loss =10%
So, C.P = Rs. [100/90 * (2/3)X ] = 20X/27.
Now C.P = Rs. 20X/27, S.P =Rs. X.
Gain = Rs. [ X -20X/27 ] =Rs.7X/27.
Hence, Gain % = [ 7X/27 *27/20X *100 ]% = 35%.
#### Asked In :: iGate
Post Your Answer Here:
No Discussion on this question yet! | 328 | 1,109 | {"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} | 4.09375 | 4 | CC-MAIN-2018-47 | latest | en | 0.794448 |
http://goatleaps.xyz/euler/maths/Project-Euler-616.html | 1,566,333,865,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027315618.73/warc/CC-MAIN-20190820200701-20190820222701-00357.warc.gz | 83,020,648 | 7,001 | # WARNING: The text below provides a guidance for a Project Euler problem.
• If you pride yourself in tackling problems entirely on your own, you are encouraged to exit this post right now.
• If you are simply looking for a hint for the problem, please visit the »Project Euler’s official forum«. You may find some hints here, but if you don’t want your problem spoiled, scroll cautiously.
• If you are looking for a readily available copy/paste answer, you will not find it here. All of the code snippets in the post have been stripped of some crucial lines (these are clearly marked). It is my aim that anyone able to replicate the answer based on such snippets will have to understand the solution conceptually. At that point, by virtue of learning something new, I believe he deserves it anyway.
• Guidance for future problems will not be published before 100 people have solved the problem and there are at least two more recent problems.
### Creative numbers
The problem asks us to sum up so-called creative numbers. Number $n$ is creative if, starting from list $L = \{n\}$, any integer greater than one is reachable via sequence of steps of two types:
• Remove $a, b$ from $L$ and substitute them by $a^b$. (For example $\{2,3\} \rightarrow \{9\}$)
• Remove any integer of the form $a^b; a, b > 1$ from the list and substitute it by $a, b$. (For example $\{16\} \rightarrow \{2, 4\}$)
We need to get a better grasp on which numbers are creative. Well, for starters, any number $n$ that is not on the form $a^b; a, b > 1$ cannot be creative, as we will get stuck in the beginning. So we consider only numbers of this form. What if both $a$ and $b$ are prime? In such case the only course of action available to us is:
and we are stuck again. What if the exponent $b$ is composite? Then we find ourselves able to reach:
What if the number we are taking power of is composite? Then we can reach:
After these simple observations we arrive at the claim that is the backbone of the whole problem.
Claim. Any number $m > 1$ is reachable from $\{a, b, c\};\ a,b,c > 1$, unless $a = b = c = 2$.
Proof. In case $a = b = c = 2$, we are stuck as the only reachable lists are $\{2,2,2\}, \{2, 4\}, \{16\}$. So let from now on $c \geq 3$. We may obtain:
By the following chain of steps we can grow our number in magnitude:
Eventually, we can reach, for any $m > 2$, some number $b^{b^x}$ where $x > m + 1$. Then we simply do:
whereupon we have reached $m$. $\square$
Finally, the problem tasks us with finding $\sum_{i = 1}^{k} i \cdot \mathbb 1_{i \text{ is creative}}$ (Where $\mathbb 1_{i \text{ is creative}}$ is $1$ if $i$ is creative and $0$ otherwise.) which is by the previous discussion equal to (Any $a^b$ with at least one of $a, b$ composite can be written as $b^p$ where $b$ is composite and $p$ prime.):
A straightforward implementation seems to be fast enough with PyPy3 runtime just below one second. | 783 | 2,929 | {"found_math": true, "script_math_tex": 36, "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} | 4.125 | 4 | CC-MAIN-2019-35 | longest | en | 0.926962 |
https://www.aimsedu.org/tag/visual/?full-site=true | 1,620,697,257,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243991553.4/warc/CC-MAIN-20210510235021-20210511025021-00003.warc.gz | 621,250,471 | 12,597 | # Tag Archives: Visual
### There’s More Than Meets The Eye
This week’s Puzzle Corner activity is a modification of a puzzle that has been around for many years. All the versions of the puzzle I have seen use two identical arcs that are placed one above the other. While the two arcs have the same length, the one on top seems shorter. This illusion persists even when the arcs are switched.
The illusion in There’s More Than Meets the Eye is much more dramatic than the illusion in the puzzle described above. There’s More Than Meets the Eye consists of two arcs placed in such a way that they appear to have the same length. In reality, the top arc is about an inch longer than the bottom arc. This difference in length is not apparent until the arcs are cut out and reversed. This reversal produces the opposite effect and greatly exaggerates the differences in length. To “see” the true size relationship between the two arcs, they need to be placed directly on top of the other. The challenge in this puzzle is to come up with a reasonable theory to explain the illusion.
I have used this puzzle for a number of years and it is always one of the student’s favorites. I use two arcs (similar in size to the arcs on the student sheet) cut from the rim of a paper plate. I introduce the puzzle by holding the longer arc above the shorter arc and asking students to tell me how they compare in size. The students invariably tell me that the two arcs are the same length. I then tease the students and tell them that the arcs are made of a special material that can be stretched. I pretend to stretch the top arc and move it under the other arc. This causes great consternation since the this arc now appears to be much longer than the other arc.
Next, I “stretch” the arc now in the top position and place it below the other. The two arcs again appear to be the same size. After “stretching” and switching the arcs several more times, I leave them at a center so that students have ample time to “play” with them. I challenge students to discover as much as they can about the arcs and about what happens when they are placed in various positions. I also ask them to come up with some theories to explain the illusions. At the end of the week we have a class discussion on the arcs. Students share their discoveries and theories. After this discussion, I pass out paper plates and each student makes a set of arcs to take home and share with their families.
There’s More Than Meets the Eye can be done as describe above or it can be done as a whole-class activity by giving each student a copy of the sheet on the next page. In either approach, students should be given ample time to “play” with the arcs. If using the whole-class approach, pairs of students can share arcs so that one student has both short arcs and the other has both long arcs. In this way, students can see how the illusion works for two identical arcs.
Look at the two arcs. What is their apparent size relationship?
Cut out the arcs and reverse their positions.
What happens?
Experiment with the arcs and see what else you discover. Write about your discoveries. | 660 | 3,142 | {"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-2021-21 | latest | en | 0.949682 |
https://classroom.synonym.com/science-projects-shooting-hoops-7968563.html | 1,529,486,717,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267863516.21/warc/CC-MAIN-20180620085406-20180620105406-00217.warc.gz | 596,848,965 | 20,324 | For some students, coming up with a science project idea and executing that idea can be difficult. But if you think about activities that you love to do and the science involved in some of those activities, a science project idea might present itself -- and you might even have fun doing it! Basketball lovers, for example, can find ample material in shooting hoops.
## Shooting Percentage
One statistic basketball analysts use to judge players' performance is shooting percentage -- the percentage of baskets a player makes out of all of the shots he takes. You can design a science project based on this principle by comparing novice and advanced players. Begin by making a hypothesis about what percentage of shots 10 novice players will make and 10 advanced players will make. Then have the 20 players shoot 10 free throws, noting how many shots the two groups made and how many the two groups missed. The math is easy -- because each group will have tried 100 shots, the number they sink will add up to the shooting percentage. Was your hypothesis confirmed?
## Technique
Other science projects might focus on the best and worst techniques for shooting hoops. A specific experiment might have you testing whether it is best to shoot from the chest, from the chin or over the head. To complete the experiment, ask about 10 subjects who have similar basketball capabilities to shoot 10 free throws from the chest, 10 free throws from the chin and 10 free throws from over the head. Record how many baskets your subjects made from each position and analyze the data to decipher which technique is the best for shooting hoops. | 317 | 1,631 | {"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-2018-26 | latest | en | 0.953452 |
https://sciencedocbox.com/Astrology/86106450-Lesson-1-7-circles-notebook-september-19-geometry-agenda.html | 1,628,103,678,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046154897.82/warc/CC-MAIN-20210804174229-20210804204229-00115.warc.gz | 476,215,740 | 24,025 | # Lesson 1.7 circles.notebook. September 19, Geometry Agenda:
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Start display at page:
Transcription
1 Geometry genda: Warm-up 1.6(need to print of and make a word document) ircle Notes 1.7 Take Quiz if you were not in class on Friday Remember we are on 1.7 p.72 not lesson 1.8 1
2 Warm up 1.6 For Exercises 1 3, use the graph at right. 1. Locate D so that D is a rectangle 2. Locate E so that E is a trapezoid. 3. Locate G so that points,,, and G determine a parallelogram that is not a rectangle. 2
3 Sketch the following 4. arallelogram GR 5. Square SQRE 6. Rhombus RHM with acute H. 7. Trapezoid TR with TR, RE,and, E, and collinear. 3
4 4
5 5
6 Lesson 1.7 ircles Note Sheet ircle: set of points in a plane that are the distance (radiu from a given point (center). If =5 and =5, Then and are the radii of the circle. Radius: a segment from the of a circle to a on the edge of the circle. Diameter: a line segment containing the center of the circle, with its on the circle. The length of this segment is also called the diameter. Draw the diameter D D ongruent ircles: two circles are congruent if they have the same radius. If = or, then oncentric ircles: two circles are concentric if they are and have the same. rc of a ircle: the two points on the circle (endpoints) and the continuous (unbroken) part of the circle the two points. Semicircle: an arc whose endpoints are the endpoints of a. Use three points to name a semicircle. Major rc: an arc that is than a semicircle. Use points to name a major arc. an you name a semicircle, a major arc, and a minor arc?? D Minor rc: an arc that is smaller than a. Use two letters to name a minor arc. 6
7 entral ngle: an angle whose vertex is located at the center of a circle. an you identify a central angle? How do central angles relate to major and minor arcs? The of a minor arc is the as the measure of the central angle. hord: a line segment whose lie on the circle. an a chord be a diameter? Name the chords: Tangent: line that a circle at only one point. D M T Where the line intersects the circle is called the. an you identify the points of tangency? Name the tangent lines. H Use the ordered pair rule to relocate the four points of the given circle. an the four new points be connected to create a new circle? Is the new circle congruent to the original circle? 7
8 8
9 Lesson 1.7 ircles Note Sheet ircle: set of points in a plane that are the distance (radius) from a given point (center). If =5 and =5, Then and are the radii of the circle. Radius: a segment from the of a circle to a on the edge of the circle. Diameter: a line segment containing the center of the circle, with its on the circle. The length of this segment is also called the diameter. Draw the diameter D D ongruent ircles: two circles are congruent if they have the same radius. If = or, then oncentric ircles: two circles are concentric if they are and have the same. rc of a ircle: the two points on the circle (endpoints) and the continuous (unbroken) part of the circle the two points. Semicircle: an arc whose endpoints are the endpoints of a. Use three points to name a semicircle. Major rc: an arc that is than a semicircle. Use points to name a major arc. an you name a semicircle, a major arc, and a minor arc?? D Minor rc: an arc that is smaller than a. Use two letters to name a minor arc. 9
10 entral ngle: an angle whose vertex is located at the center of a circle. an you identify a central angle? How do central angles relate to major and minor arcs? The of a minor arc is the as the measure of the central angle. hord: a line segment whose lie on the circle. an a chord be a diameter? Name the chords: Tangent: line that a circle at only one point. D M T Where the line intersects the circle is called the. an you identify the points of tangency? Name the tangent lines. H Use the ordered pair rule to relocate the four points of the given circle. an the four new points be connected to create a new circle? Is the new circle congruent to the original circle? 10
### What is the longest chord?.
Section: 7-6 Topic: ircles and rcs Standard: 7 & 21 ircle Naming a ircle Name: lass: Geometry 1 Period: Date: In a plane, a circle is equidistant from a given point called the. circle is named by its.
### radii: AP, PR, PB diameter: AB chords: AB, CD, AF secant: AG or AG tangent: semicircles: ACB, ARB minor arcs: AC, AR, RD, BC,
h 6 Note Sheets L Shortened Key Note Sheets hapter 6: iscovering and roving ircle roperties eview: ircles Vocabulary If you are having problems recalling the vocabulary, look back at your notes for Lesson
### DO NOW #1. Please: Get a circle packet
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### Mth 076: Applied Geometry (Individualized Sections) MODULE FOUR STUDY GUIDE
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KEY STANDARDS ADDRESSED:. Students will understand the properties of circles. a. Understand and use properties of chords, tangents, and secants an application of triangle similarity. b. Understand and | 8,386 | 31,256 | {"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} | 4.1875 | 4 | CC-MAIN-2021-31 | latest | en | 0.897422 |
https://raptid.com/discussion/question/159 | 1,656,327,494,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103331729.20/warc/CC-MAIN-20220627103810-20220627133810-00597.warc.gz | 534,813,606 | 6,855 | ## Discussion
#### Q.
A shopkeeper buys toffees at 4 for ₹ 3 and sells all the toffees at 3 for ₹ 4. Find his profit percentage.
A. 77.25% B. 77.5% C. 77.55% D. 77.77%
#### Explanation:
Quantity Price
CP 4 3 9 (CP)
SP 3 4 16 (SP)
Profit = 16 - 9 = 7
Profit% = | 124 | 317 | {"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} | 3.90625 | 4 | CC-MAIN-2022-27 | latest | en | 0.441243 |
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### Magical Multiplication
1. 1. Magical Multiplication<br />Created By: Kristin Genung<br />Math 105<br />
2. 2. Learning To Multiply:<br />A FEW MYSTIFYING TIPS<br />
3. 3.
4. 4.
5. 5. It’s as simple as that!!<br />0 x 1 = 0<br />0 x 2 = 0<br />0 x 3 = 0<br />
6. 6.
7. 7.
8. 8. 1’s<br />1 x 4 = 4<br />1 x 5 = 5<br />1 x 6 = 6<br />
9. 9.
10. 10. 2 times any number is that number doubled<br />=<br />+<br />+<br />2<br />2<br />2<br />6<br />2 x 3 = 6<br />
11. 11. 2’s<br />2’s<br />
13. 13.
14. 14.
15. 15. 4’s<br />
16. 16. 4’s<br />
17. 17. 4 x 1 = 4 ( 1 doubled is 2 and 2 doubled is 4 )<br />4 x 2 = 8 ( 2 doubled is 4 and 4 doubled is 8 )<br />4 x 3 = 12 ( 3 doubled is 6 and 6 doubled is 12 )<br />
18. 18.
19. 19. If you are multiplying by an odd number..<br />Cut the number you are multiplying by in half then drop the decimal <br />
20. 20. 5 x 3 = 15 ( half of 3 is 1.5, then drop decimal)<br />5 x 7 = 35 ( half of 7 is 3.5, then drop the decimal)<br />5 x 9 = 45 ( half of 9 is 4.5, then drop the decimal)<br />
21. 21. 5<br />If it is an even number…<br />- Cut the number in half and add a 0<br />
22. 22.
23. 23. 6’s<br />
24. 24. <ul><li>Take the number your multiplying by and divide it in half.
25. 25. Then multiply that number by ten.
26. 26. Then add the original number you are multiplying by. </li></li></ul><li>6 x 6 = 36 <br />Half of 6 = 3 multiplied by 10 equals 30 then add 6 and you get 36. <br />6 x 7 = 42 <br />Half of 7 is 3.5 multiplied by 10 equals 35 then add 7 and you get 42.<br />6 x 8 = 48<br />Half of 8 is 4 multiplied by 10 equals 40 then add 8 and you get 48.<br />
27. 27. 7’s<br />
28. 28. No need to be slick there isn’t a trick…<br />7 x 7 = 49<br />7 x 8 = 56<br />
29. 29. 8’s<br />
30. 30. Once again use the tricks.. <br />8 x 8 = 64<br />
31. 31. 9’s<br />
32. 32. Subtract 1 from the number you are multiplying by then add a second number that adds up to nine..<br />
33. 33. 9 x 7 = ( 7 – 1 = 6, 6 +3 = 9 so the answer is 63)<br />9 x 8 = ( 8 – 1 = 7, 7 +2 = 9 so the answer is 72)<br />9 x 9 = ( 9 – 1 = 8, 8 +1 = 9 so the answer is 81)<br />
34. 34. 10’s<br />
35. 35. Just add a 0 to the number <br />you are multiplying by<br />10 x 2 = 20 <br />10 x 3 = 30<br />10 x 4 = 40<br />
36. 36. 11’s<br />
37. 37. Any number 1-10 multiplied by 11 is just duplicated.<br />
38. 38. 11 x 3 = 33<br />(3 is multiplied by 11 so you just duplicate 3 side by side) <br />11 x 4 = 44<br />11 x 5 = 55<br />
39. 39. 12 were done…Now let’s have fun…<br /> Use your tricks 1-11 …<br /> And 12 x 12 = 144 <br /> | 1,223 | 3,087 | {"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-2016-40 | latest | en | 0.681152 |
https://www.beatthegmat.com/og-13-problem-t288364.html | 1,518,917,337,000,000,000 | text/html | crawl-data/CC-MAIN-2018-09/segments/1518891811243.29/warc/CC-MAIN-20180218003946-20180218023946-00281.warc.gz | 879,749,394 | 34,991 | • 1 Hour Free
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## OG 13 Problem
oquiella Master | Next Rank: 500 Posts
Joined
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#### OG 13 Problem
Tue Dec 22, 2015 3:08 pm
If K is an integer and 2
A. One
B. Two
C. Three
D. Four
E. Five
### GMAT/MBA Expert
[email protected] Elite Legendary Member
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Thu Feb 15, 2018 7:05 pm
Hi All,
We're told that K is an integer and 2 < K < 7. We're asked for the number of different values of K that would create a triangle with sides of lengths 2, 7 and K.
This question is based on a math rule called the Triangle Inequality Theorem. In simple terms, it means that if you're dealing with a triangle, then the sum of ANY 2 sides will be greater than the 3rd side. For example, with a 3/4/5 right triangle....
3 + 4 > 5
3 + 5 > 4
4 + 5 > 3
So these 3 lengths (and these 3 inequalities) PROVE that we're dealing with an actual triangle.
In that same way, we can use the rule to determine when we're NOT dealing with an actual triangle... For example, you CANNOT have a triangle with sides of 1, 1 and 100 (because 1+1 is NOT greater than 100).
With this question, we're given two sides: 2 and 7 and we're asked to determine what the third side (K) could be. We're given a range of values for K and we're told that K must be an INTEGER. Since 2 + K must be GREATER than 7, the only possible value that fits all of the given information is K=6. Thus, there's only one value for K.
GMAT assassins aren't born, they're made,
Rich
_________________
Contact Rich at [email protected]
Marty Murray Legendary Member
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Tue Dec 22, 2015 10:47 pm
Quote:
If K is an integer and 2 < k < 7, for how many different values of k is there a triangle with sides of lengths 2,7, and k?
A. One
B. Two
C. Three
D. Four
E. Five
For three line segments to work as a triangle, the lengths of any two of the segments have to add up to a sum greater than the length of the third. Otherwise two sides will not be long enough to connect with each other and with the ends of the third side.
The integers k such that 2 < k < 7 are 3, 4, 5 and 6.
See how many of those could form a triangle with sides 2, 7, and k.
If the lengths of the three segments were 2, 7 and 3, the segments 2 and 3 long would not be long enough to connect with each other and with the ends of the segment 7 long to form a triangle.
2 + 4 < 7 Doesn't work.
2 + 5 = 7 The only way to connect all the ends is to put them together flat. That's not a triangle.
2 + 6 > 7 -- 7 + 2 > 6 -- 7 + 6 > 2 That works.
So only K = 6 works and the correct answer is A.
_________________
Marty Murray
GMAT Coach
[email protected]
http://infinitemindprep.com/
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https://idealcalculator.com/coefficient-of-variation-calculator-with-steps/ | 1,716,984,450,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971059239.72/warc/CC-MAIN-20240529103929-20240529133929-00103.warc.gz | 262,533,167 | 25,684 | # Coefficient of variation calculator with steps (Sample, Population and Grouped)
## Coefficient of variation calculator
Example 1 Example 2
The online Coefficient of Variation calculator is a useful tool to easily calculate the variability of a given data set, including samples, populations, or grouped data. This calculator is especially helpful for statistics students as it provides step-by-step explanations for each calculation. The solution has three parts: the first calculates the mean, the second explains the standard deviation calculation in detail, and the last part shows the process to obtain the Coefficient of Variation.
To use the Coefficient of Variation calculator, follow these steps:
Step 1: Select the type of data you will be working with from the dropdown menu above the “Calculate” button. You can choose from population, sample, or grouped data.
Step 2: Enter your data. If you selected population or sample in Step 1, enter your data separated by commas in the yellow input box. If you selected grouped data, the interface will change and you will see fields to enter the class intervals and their respective frequencies.
Step 3: Once you have entered your data, click the “Calculate” button. A new box will appear with the solution explained step by step.
## What is coefficient of variation?
The coefficient of variation, also known by the term relative dispersion, is a statistical measure used to assess the variability of a data set in relation to its standard deviation.
Formally, the coefficient of variation is defined as the relationship between the standard deviation and the arithmetic mean of a distribution expressed as a percentage. It is commonly represented with the letters CV.
The formula for calculating the coefficient of variation is:
CV = (standard deviation / mean) x 100%
## How to find coefficient of variation
To calculate this coefficient you need to perform the following steps:
1. Find the arithmetic mean of the data set
2. Calculate the standard deviation
3. Apply the coefficient of variation formula, which consists of dividing the standard deviation by the arithmetic mean and all of this multiplied by 100.
When you use our calculator you will be able to see step by step the procedure to find CV.
## Coefficient of variation interpretation
The CV is used to determine the consistency of the data. When we talk about consistency we refer to the uniformity in the data values. Therefore, when comparing two data sets, the one with the lowest CV will be the one with the highest consistency.
In finance, the coefficient of variation allows investors to estimate volatility compared to the expected return on investment.
It is also often used to compare results of different tests or surveys. For example, if we assume that the CV of two surveys, A and B, is 3% and 8%, respectively, we could say that the data from Survey B are less consistent than those from Survey A, given its higher level of variability.
So, we can say that the smaller the CV value, or the smaller the ratio of the standard deviation to the mean, the better it is. A lower ratio suggests a higher tradeoff between risk and return.
## Solved examples
Example 01: Find the sample coefficient of variance of the given data set (12.3, 22.5, 35.8, 44.8)
$\overline{x}=\frac{\sum x}{n}=\frac{115.39999999999999}{4}=28.85$
$x$ $x-\overline{x}$ $\left(x-\overline{x}{\right)}^{2}$ $12.3$ $-16.55$ $273.9$ $22.5$ $-6.35$ $40.32$ $35.8$ $6.95$ $48.3$ $44.8$ $15.95$ $254.4$ $\mathrm{\Sigma }$ $0$ $616.93$
${s}^{2}=\frac{\sum \left(x-\overline{x}{\right)}^{2}}{n-1}=\frac{616.93}{4-1}=205.64$ $s=\sqrt{{s}^{2}}=\sqrt{205.64}=\overline{)14.34}$
$\begin{array}{rl}\text{CV}& =\left(\frac{s}{\overline{x}}×100\right)\mathrm{%}\\ \\ \text{CV}& =\left(\frac{14.34}{28.85}×100\right)\mathrm{%}=\overline{)49.71\mathrm{%}}\end{array}$ | 947 | 3,878 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 22, "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} | 4.25 | 4 | CC-MAIN-2024-22 | longest | en | 0.892208 |
https://byjusexamprep.com/csir-net/the-half-life-of-a-radioactive-nuclide-is-100-hours-the-fraction-of-original-activity-that-will-remain-after-150-hours-would-be | 1,712,942,719,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296816045.47/warc/CC-MAIN-20240412163227-20240412193227-00876.warc.gz | 130,034,006 | 28,663 | # The half-life of a radioactive nuclide is 100 hours. The fraction of original activity that will remain after 150 hours would be a. 2/3√2 b. ½ c. 1/2√2 d. ⅔
By BYJU'S Exam Prep
Updated on: September 13th, 2023
The half-life of a radioactive nuclide is defined as the time in which half of the original number of radioactive atoms has decayed.
Given
TH = 150 hours
t/2 = 100 hours
Table of content
We know that the activity of a radioactive substance is written as
A = Ao (1/2)TH/t/2
Where TH is the fraction of the original activity
t/2 is the half-life of a radioactive nuclide
Substituting the values
A/Ao = (1/2)150/100
By further simplification
A/Ao = (1/2)3/2
So we get
A/Ao = 1/2√2
Therefore, the fraction of original activity that will remain after 150 hours would be 1/2√2.
Summary:
4. ## ⅔
The half-life of a radioactive nuclide is 100 hours. The fraction of original activity that will remain after 150 hours would be 1/2√2.
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GradeStack Learning Pvt. Ltd.Windsor IT Park, Tower - A, 2nd Floor, Sector 125, Noida, Uttar Pradesh 201303 [email protected] | 350 | 1,149 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.703125 | 4 | CC-MAIN-2024-18 | latest | en | 0.884999 |
http://www.onlinemath4all.com/Mode-in-statistics.html | 1,511,185,866,000,000,000 | text/html | crawl-data/CC-MAIN-2017-47/segments/1510934806066.5/warc/CC-MAIN-20171120130647-20171120150647-00684.warc.gz | 468,921,825 | 11,820 | # MODE IN STATISTICS
Mode in statistics :
Mode is one of the measures of central tendency which can be defined as follows.
For a given set of observations, mode may be defined as the value that occurs the maximum number of times.
Thus, mode is that value which has the maximum concentration of the observations around it. This can also be described as the most common value with which, even, a layman may be familiar with.
Thus, if the observations are 5, 3, 8, 9, 5 and 6, then Mode (Mo) = 5 as it occurs twice and all the other observations occur just once.
The definition for mode also leaves scope for more than one mode. Thus sometimes we may come across a distribution having more than one mode. Such a distribution is known as a multi-modal distribution. Bi-modal distribution is one having two modes.
Furthermore, it also appears from the definition that mode is not always defined. As an example,
If the marks of 5 students are 50, 60, 35, 40, 56, there is no modal mark as all the observations occur once i.e. the same number of times.
We may consider the following formula for computing mode from a grouped frequency distribution:
Where,
l₁ = LCB of the modal class
f₀ = frequency of the modal class
f = frequency of the pre modal class
f = frequency of the post modal class
C = class length of the modal class
## Mode in statistics- Practice problem
Compute the mode for the following distribution
Solution :
Computation of mode
For the given data, the formula to find mode is given by
Going through the frequency column, we note that the highest frequency i.e. f₀ is 82.
So we have,
f₋₁ = 58
f₁ = 65
Also, the modal class i.e. the class against the highest frequency is 410 - 419
Thus,
l = LCB = 409.50
C = 429.50 - 409.50 = 20
C = 429.50 – 409.50 = 20
Plugging these values in the above formula, we get
Mode = 409.50 + [ (82-58) / (2x52 - 58 - 65) ] x 20
Mode = 409.50 + 11.71
Mode = 421.21
## Relation among mean median and mode
When it is difficult to compute mode from a grouped frequency distribution, we may consider the following empirical relationship between mean, median and mode:
Mean - Mode = 3(Mean - Median)
or
Mode = 3 x median - 2 x mean
The above result holds for holds for a moderately skewed distribution.
We also note that if y = a + bx, then we have
Mode of "y" = a + b (Mode of "x")
For example, if y = 2 + 1.50x and mode of x is 15, then the mode of y is given by
Mode of "y" = 2 + 1.50(Mode of "x")
Mode of "y" = 2 + 1.50x15
Mode of "y" = 2 + 22.50
Mode of "y" = 24.50
## Properties of mode
1) Although mode is the most popular measure of central tendency, there are cases when mode remains undefined.
2) Unlike mean, it has no mathematical property.
3) Mode is affected by sampling fluctuations.
After having gone through the stuff given above, we hope that the students would have understood "Mode in statistics".
Apart from the stuff given on this web page, if you need any other stuff in math, please use our google custom search here.
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Sum of all three four digit numbers formed using 1, 2, 5, 6 | 1,354 | 5,312 | {"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-2017-47 | latest | en | 0.882997 |
http://oeis.org/A063436 | 1,556,217,607,000,000,000 | text/html | crawl-data/CC-MAIN-2019-18/segments/1555578732961.78/warc/CC-MAIN-20190425173951-20190425195951-00387.warc.gz | 137,457,529 | 4,287 | This site is supported by donations to The OEIS Foundation.
Hints (Greetings from The On-Line Encyclopedia of Integer Sequences!)
A063436 Write 1,2,3,4,... counterclockwise in a hexagonal spiral around 0 starting left down, then a(n) is the sequence found by reading from 0 in the vertical upward direction. 2
0, 15, 54, 117, 204, 315, 450, 609, 792, 999, 1230, 1485, 1764, 2067, 2394, 2745, 3120, 3519, 3942, 4389, 4860, 5355, 5874, 6417, 6984, 7575, 8190, 8829, 9492, 10179, 10890, 11625, 12384, 13167, 13974, 14805, 15660, 16539, 17442, 18369 (list; graph; refs; listen; history; text; internal format)
OFFSET 0,2 COMMENTS Related to parity of Beatty sequences for exp(-(1/2)/n). Let f(k,n)=-sum(i=1,n,sum(j=1,i,(-1)^floor(j*exp(-(1/2)/n)))), then a(n)=Max{f(k,n) : 1<=k<=4*a(n)-2} and for 0<=i<=4*a(n)-3, f(i,n)=f(4*a(n)-2-i,n). - Benoit Cloitre, May 26 2004 Or, sum of multiples of 2 and 3 from 0 to 6n. - Zak Seidov, Aug 06 2016 REFERENCES B. Cloitre, On parity properties of certain Beatty sequences, in preparation 2004 LINKS Harry J. Smith, Table of n, a(n) for n = 0..1000 Index entries for linear recurrences with constant coefficients, signature (3,-3,1). FORMULA a(n) = 3*n*(4*n+1) = 3*A007742(n). a(n) = 24*n + a(n-1) - 9 (with a(0)=0). - Vincenzo Librandi, Aug 07 2010 From Colin Barker, Jul 07 2012: (Start) a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3). G.f.: 3*x*(5+3*x)/(1-x)^3. (End) EXAMPLE The spiral begins: . 16--15--14 / \ 17 5---4 13 / / \ \ 18 6 0 3 12 / / / / / 19 7 1---2 11 26 \ \ / / 20 8---9--10 25 \ / 21--22--23--24 PROG (PARI) { for (n=0, 1000, write("b063436.txt", n, " ", n*(12*n + 3)) ) } \\ Harry J. Smith, Aug 21 2009 CROSSREFS Cf. A062783, A000567. Sequence in context: A194454 A219384 A198955 * A010004 A172073 A059145 Adjacent sequences: A063433 A063434 A063435 * A063437 A063438 A063439 KEYWORD nonn,easy AUTHOR Floor van Lamoen, Jul 21 2001 STATUS approved
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Last modified April 25 13:31 EDT 2019. Contains 322461 sequences. (Running on oeis4.) | 925 | 2,353 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.53125 | 4 | CC-MAIN-2019-18 | latest | en | 0.449421 |
http://www.targetmathematics.com/2018/12/area-between-curves-igcse-o-level.html | 1,606,395,483,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141188146.22/warc/CC-MAIN-20201126113736-20201126143736-00144.warc.gz | 164,983,844 | 29,131 | # Area between Curves (IGCSE O LEVEL) - Problems and Solutions
If the fumction $\displaystyle f(x)$ and $\displaystyle g(x)$ intersect at $\displaystyle x_1=a$ and $\displaystyle x_2=b$, then the bounded area between $\displaystyle f(x)$ and $\displaystyle g(x)$ is
$\displaystyle \int_{a}^{b}{{\left[ {f(x)-g(x)} \right]}}\ dx$
Where $\displaystyle f(x)$ is the curve of upper boundry and $\displaystyle g(x)$ is the curve of lower boundry.
ဆရာႀကီး ေဒါက္တာ ေရႊေက်ာ္Drill For Exam Blog မွ Area under curve ပုစာၦမ်ား ျဖစ္ပါသည္။
Problem (1)
The diagram shows part of the curve $\displaystyle y=9x^2−x^3,$ which meets the $\displaystyle x$-axis at the origin $\displaystyle O$ and at the point $\displaystyle A$. The line $\displaystyle y−2x+18=0$ passes through $\displaystyle A$ and meets the $\displaystyle y$-axis at the point $\displaystyle B.$
(i) Show that, for $\displaystyle x≥0, 9x^2−x^3≤108.$
(ii) Find the area of the shaded region bounded by the curve, the line $\displaystyle AB$ and the $\displaystyle y$-axis.
Show/Hide Solution
$\displaystyle \begin{array}{l}\ \ \ \ \text{Line : }y-2x+18\Rightarrow y=2x-18\\\\\ \ \ \ \text{Curve : }y=9{{x}^{2}}-{{x}^{3}}\end{array}$
$\displaystyle \ \ \ \ \frac{{dy}}{{dx}}=18x-3{{x}^{2}}=3x(6-x)$
$\displaystyle \ \ \ \ \frac{{dy}}{{dx}}=0\ \text{when}\ 3x(6-x)=0$
$\displaystyle {\therefore \ \ x=0\ \text{or }x=6}$
$\displaystyle \therefore \ \ \text{When }x=0, y=0$
$\displaystyle {\therefore \ \ \text{When }x=6,\ }$
$\displaystyle {\ \ \ \ y=9({{6}^{2}})-{{{(6)}}^{3}}=108}$
$\displaystyle \ \ \ \ \frac{{{{d}^{2}}y}}{{d{{x}^{2}}}}=-6x$
$\displaystyle \therefore \ \ \text{When }x=6,\frac{{{{d}^{2}}y}}{{d{{x}^{2}}}}=-36<0$
$\displaystyle \begin{array}{l}\therefore \ \ y=108\ \text{is a maximum value}\text{.}\\\\\therefore \ \ \text{For}\ x\ge 0,\ y\le 0.\\\\\ \ \ \ \text{According to the diagram, }\\\ \ \ \ \text{the line and curve intersect when }y=0.\\\\\therefore \ \ 2x-18=0\Rightarrow x=9.\\\\\ \ \ \ \text{Let the curve of upper boundry be }{{y}_{1}}\text{ and }\\\ \ \ \ \text{that of the lower boundry be }{{y}_{2}}\text{, and }\\\ \ \ \ \text{let the area of the shaded region be }A.\\\\\therefore \ \ {{y}_{1}}=9{{x}^{2}}-{{x}^{3}},\ {{y}_{2}}=2x-18\ \operatorname{and}\end{array}$
$\displaystyle \ \ \ A=\int_{0}^{9}{{\left( {{{y}_{1}}-{{y}_{2}}} \right)}}\ dx$
$\displaystyle \ \ \ \ \ \ \ =\int_{0}^{9}{{\left( {9{{x}^{2}}-{{x}^{3}}-2x+18} \right)}}\ dx$
$\displaystyle \ \ \ \ \ \ \ =\left. {\frac{9}{3}{{x}^{3}}-\frac{{{{x}^{4}}}}{4}-\frac{{2{{x}^{2}}}}{2}+18x} \right]_{0}^{9}$
$\displaystyle \ \ \ \ \ \ \ =\left. {3{{x}^{3}}-\frac{{{{x}^{4}}}}{4}-{{x}^{2}}+18x} \right]_{0}^{9}$
$\displaystyle \ \ \ \ \ \ \ =3{{(9)}^{3}}-\frac{{{{{(9)}}^{4}}}}{4}-{{(9)}^{2}}+18(9)$
$\displaystyle \ \ \ \ \ \ \ =627.75$
Problem (2)
The diagram shows part of the curve $\displaystyle y = 2\sin 3x.$ The normal to the curve $\displaystyle y = 2\sin 3x$ at the point where $\displaystyle x =\frac{\pi}{9}$ meets the $\displaystyle y$-axis at the point $\displaystyle P.$
(i) Find the coordinates of $\displaystyle P.$
(ii) Find the area of the shaded region bounded by the curve, the normal and the $\displaystyle y$-axis.
Show/Hide Solution
$\displaystyle \ \ \ \text{Curve : }y=2\sin 3x,$
$\displaystyle \ \ \ \text{When }x=\frac{\pi }{9},$
$\displaystyle \ \ \ y=2\sin 3\left( {\frac{\pi }{9}} \right)$
$\displaystyle \ \ \ y=2\sin 3\left( {\frac{\pi }{9}} \right)$
$\displaystyle \ \ \ \ \ =\sqrt{3}$
$\displaystyle \ \ \ \frac{{dy}}{{dx}}=6\cos 3x$
$\displaystyle \ \ \ {{\left. {\frac{{dy}}{{dx}}} \right|}_{{x=\frac{\pi }{9}}}}=6\cos 3\left( {\frac{\pi }{9}} \right)=3$
$\displaystyle \ \ \ {{\left. {\frac{{dy}}{{dx}}} \right|}_{{x=\frac{\pi }{9}}}}=6\cos 3\left( {\frac{\pi }{9}} \right)=3$
$\displaystyle \ \ \ y=\sqrt{3}-\frac{1}{3}\left( {x-\frac{\pi }{9}} \right)$
$\displaystyle \ \ \ y=\sqrt{3}+\frac{\pi }{{27}}-\frac{x}{3}$
$\displaystyle \ \ \ \text{When the normal cuts the }y\text{-axis,}\ x=0.$
$\displaystyle \therefore \ y=\sqrt{3}+\frac{\pi }{{27}}$
$\displaystyle \therefore \ P=(0,\sqrt{3}+\frac{\pi }{{27}})=(0,1.85)$
$\displaystyle \ \ \ \text{Let the area of the shaded region be }A.$
$\displaystyle \therefore A=\left. {\left( {\sqrt{3}+\frac{\pi }{{27}}} \right)x-\frac{{{{x}^{2}}}}{6}+\frac{2}{3}\cos 3x} \right]_{0}^{{\frac{\pi }{9}}}$
$\displaystyle \therefore A=\left( {\sqrt{3}+\frac{\pi }{{27}}} \right)\frac{\pi }{9}-\frac{{{{\pi }^{2}}}}{{486}}+\frac{2}{3}\cos 3\left( {\frac{\pi }{9}} \right)-\frac{2}{3}\cos (0)$
$\displaystyle \therefore A=\frac{{\sqrt{3}\pi }}{9}+\frac{{{{\pi }^{2}}}}{{243}}-\frac{{{{\pi }^{2}}}}{{486}}+\frac{1}{3}-\frac{2}{3}=2.912$
Problem (3)
The diagram shows part of the curve $\displaystyle y=\sin \frac{1}{2}x$. The tangent to the curve at the point $\displaystyle P\left( {\frac{{3\pi }}{2},\frac{{\sqrt{2}}}{2}} \right)$ cuts the $\displaystyle x$-axis at the point $\displaystyle Q$. (i) Find the coordinates of $\displaystyle Q$. (ii) Find the area of the shaded region bounded by the curve, the tangent and the $\displaystyle x$-axis.
Show/Hide Solution
$\displaystyle \ \ \ \ y=\sin \frac{1}{2}x$
$\displaystyle \ \ \ \ y=0\Rightarrow \sin \frac{1}{2}x=0$
$\displaystyle \therefore \ \ \frac{1}{2}x=0\ \text{or}\ \frac{1}{2}x=\pi \ \ (\text{for first half cycle)}$
$\displaystyle \therefore \ \ x=0\ \text{or}\ x=2\pi$
$\displaystyle \ \ \ \ \frac{{dy}}{{dx}}=\frac{1}{2}\cos \frac{1}{2}x$
$\displaystyle \ \ \ \ \text{At }P\left( {\frac{{3\pi }}{2},\frac{{\sqrt{2}}}{2}} \right),\frac{{dy}}{{dx}}=\frac{1}{2}\cos \frac{{3\pi }}{4}=-\frac{{\sqrt{2}}}{4}$
$\displaystyle \therefore \ \ \text{Equation of tangent at }P\ \text{is}$
$\displaystyle \ \ \ \ y=\frac{{\sqrt{2}}}{2}-\frac{{\sqrt{2}}}{4}\left( {x-\frac{{3\pi }}{2}} \right)$
$\displaystyle \therefore \ \ y=\frac{{\sqrt{2}}}{2}+\frac{{3\sqrt{2}\pi }}{8}-\frac{{\sqrt{2}}}{4}x$
$\displaystyle \ \ \ \ \text{When the tangent cuts the }x\text{-axis, }y=0.$
$\displaystyle \therefore \ \ \frac{{\sqrt{2}}}{2}+\frac{{3\sqrt{2}\pi }}{8}-\frac{{\sqrt{2}}}{4}x=0$
$\displaystyle \therefore \ \ x=2+\frac{{3\pi }}{2}$
$\displaystyle \therefore \ \ Q=\left( {2+\frac{{3\pi }}{2},0} \right)=(6.71,0)$
$\displaystyle \ \ \ \ \text{Area of }\Delta PQR=\frac{1}{2}\cdot PQ\cdot PR$
$\displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\frac{1}{2}(2)\left( {\frac{{\sqrt{2}}}{2}} \right)$
$\displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\frac{{\sqrt{2}}}{2}$
$\displaystyle \ \ \ \ \text{Area of yellow region}=\int_{{\frac{{3\pi }}{2}}}^{{2\pi }}{{\left( {\sin \frac{1}{2}x} \right)dx}}$
$\displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\left. {-2\cos \frac{1}{2}x} \right]_{{\frac{{3\pi }}{2}}}^{{2\pi }}$
$\displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =-2\left( {\cos \pi -\cos \frac{{3\pi }}{4}} \right)$
$\displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =-2\left( {-1+\frac{{\sqrt{2}}}{2}} \right)$
$\displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =2-\sqrt{2}$
$\displaystyle \ \ \ \ \text{Let the required area be }A.$
$\displaystyle \therefore \ \ A=\text{Area of }\Delta PQR-\text{Area of yellow region}$
$\displaystyle \therefore \ \ A=\frac{{\sqrt{2}}}{2}-2+\sqrt{2}=\frac{{3\sqrt{2}}}{2}-2$
ဆက္လက္ ေဖၚျပပါမည္ | 3,078 | 7,463 | {"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} | 4.46875 | 4 | CC-MAIN-2020-50 | latest | en | 0.571967 |
https://mathoverflow.net/questions/149633/using-topological-pressure-to-determine-a-subshift-of-finite-type | 1,550,629,161,000,000,000 | text/html | crawl-data/CC-MAIN-2019-09/segments/1550247494125.62/warc/CC-MAIN-20190220003821-20190220025821-00559.warc.gz | 603,746,868 | 29,794 | # Using topological pressure to determine a subshift of finite type
I am interested in recognising graphs (or matrices, or subshifts of finite type) using topological pressure. Suppose that we play the following game:
${\bf Step 1:}$ I write down an irreducible n x n matrix A of zeroes and ones, which I ask you to try and work out. I tell you what n is.
Let $\sigma$ be the left shift on $\Sigma=\{1,2,..,n\}^{\mathbb N}$ and $\sigma_A$ be the left shift on $\Sigma_A$, which is the shift space corresponding to incidence matrix A.
${\bf Step 2:}$ You think of a function $f:\Sigma\to\mathbb R$ and ask me what the value of the topological pressure $P_{\sigma_A}(f|_{\Sigma_A})$ is. I tell you the answer. Repeat step 2 until you can tell me what A is.
${\bf Question:}$ How many times do you have to repeat step 2 before you know the matrix A?
It's easy to see that $n^2$ is enough. Define $f_{i,j}(\underline a) = 1$ if $a_1=i$, $a_2=j,$ 0 otherwise. Then $P_{\sigma_A}(f_{ij}|_{\Sigma_A})$ is different from $h_{top}(\sigma_A)$ if and only if $A_{ij}=1$. You don't know what $h_{top}(\sigma_A)$ is, but you can work this out from the topological pressure information.
Can we do any better? I'm particularly interested in the situation where you are only allowed to pick locally constant functions f, but the general case is also interesting. One could make minor improvements to the above idea to reduce $n^2$, but I'm really interested in whether there is a constant k independent of n such that we can always determine an n x n matrix by knowing k pressure functions.
• If we were to insist that $f$ should be of the form $f_{i,j}$ above then this would correspond to a matrix analysis question: but what question, exactly? How hard is that question, and does it give us any information? (I'm just thinking out loud here.) – Ian Morris Nov 22 '13 at 15:50
• First, you should allow me to assume that $A$ is irreducible. Otherwise there are some words you only see at most once, and you can never tell what $f$ is like on places you only see once. Given this, I think you must be able to do this with one $f$. Here's how I'd try: Choose $x_1\ll x_2 \ll x_3 \ll x_4\ll\ldots\ll x_{n^2}$, order the edges and let the function be the sum of the $x_i$'s corresponding to the edges. By $\ll$ here I mean there's a really really massive additive difference between the $x$'s. – Anthony Quas Nov 22 '13 at 17:24
• Thanks for your answers. Ian, I'm constantly finding questions that I think should be easy if only I knew more matrix analysis... Anthony, yes sorry I should have written irreducible rather than aperiodic, and after a bit of thought I think I believe that your approach ought to work. Thanks! – Tom Kempton Nov 25 '13 at 8:28 | 753 | 2,746 | {"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} | 3.625 | 4 | CC-MAIN-2019-09 | latest | en | 0.90349 |
https://www.encyclopediaofmath.org/index.php/Cauchy%E2%80%93Riemann_conditions | 1,569,314,514,000,000,000 | text/html | crawl-data/CC-MAIN-2019-39/segments/1568514572896.15/warc/CC-MAIN-20190924083200-20190924105200-00230.warc.gz | 806,956,216 | 8,823 | # Cauchy-Riemann equations
(Redirected from Cauchy–Riemann conditions)
2010 Mathematics Subject Classification: Primary: 30-XX Secondary: 32-XX [MSN][ZBL]
Also known as Cauchy-Riemann conditions and D'Alembert-Euler conditions, they are the partial differential equations that must be satisfied by the real and imaginary parts of a complex-valued function $f$ of one (or several) complex variable so that $f$ is holomorphic.
### One complex variable
More precisely, assume $D\subset \mathbb C$ is some open set and $f: D \to \mathbb C$ a map with real and imaginary parts given by $u$ and $v$ (i.e. $f(z) = u (z) + i v (z)$). If we introduce the real variables $x,y$ so that $z= x+iy$, we can consider $u$ and $v$ as real functions on a domain in $\mathbb R^2$. If $u$ and $v$ are differentiable (in the real-variable sense), then they solve the Cauchy-Riemann equations if $$\label{e:CR} \frac{\partial u}{\partial x} = \frac{\partial v}{\partial y} \qquad \frac{\partial u}{\partial y} = - \frac{\partial v}{\partial x}\, .$$ Such equations are equivalent to the fact that $f$ is complex-differentiable at every point $z_0\in D$, where its complex derivative is defined by $f' (z_0) = \lim_{w\in \mathbb C, w\to 0} \frac{f(z_0+w)-f(z_0)}{w}\, .$ It then turns out that $$\label{e:f'} f' = \frac{\partial u}{\partial x} + i \frac{\partial v}{\partial x}$$ (and evidently we can use \eqref{e:CR} to derive three other similar formulas).
#### Harmonicity and conjugacy
Any pair of solutions of the Cauchy-Riemann equations turn out to be infinitely differentiable and in fact analytic. Moreover they are also conjugate harmonic functions. Viceversa if $u$ (or $v$) is a given harmonic function on a simply connected open $D$, then there is a conjugate harmonic function to $u$, that is a $v$ satisfying \eqref{e:CR}. Such $v$ is unique up to addition of a constant.
#### Different systems of coordinates
The conditions \eqref{e:CR} can be written, equivalently, for any two orthogonal directions $s$ and $n$, with the same mutual orientation as the $x$- and $y$-axes, in the form: $\frac{\partial u}{\partial s} = \frac{\partial v}{\partial n} \qquad \frac{\partial u}{\partial n} = - \frac{\partial v}{\partial s}\, .$ For example, in polar coordinates $r, \phi$, for $r\neq 0$ the Cauchy-Riemann equations read: $\frac{\partial u}{\partial r} = \frac{1}{r} \frac{\partial v}{\partial \phi} \qquad \frac{1}{r} \frac{\partial u}{\partial \phi} = - \frac{\partial v}{\partial r}\, .$
#### $\frac{\partial}{\partial \bar{z}}$ and $\frac{\partial}{\partial z}$ operators
Defining the complex differential operators by $\frac{\partial}{\partial \bar{z}}=\frac{1}{2} \left(\frac{\partial}{\partial x} + i \frac{\partial}{\partial y}\right)\qquad\mbox{and}\qquad \frac{\partial}{\partial z}=\frac{1}{2} \left(\frac{\partial}{\partial x} - i \frac{\partial}{\partial y}\right)\, ,$ one can rewrite the Cauchy–Riemann equations \eqref{e:CR} as $\frac{\partial f}{\partial \bar{z}} = 0$ and the identity \eqref{e:f'} as $f' = \frac{\partial f}{\partial z}\, .$
#### Conformality
The Cauchy-Riemann equations are equivalent to the fact that the map $(u,v): D \to \mathbb R^2$ is conformal, i.e. it preserves the angles and (locally where it is injective) the orientation. Such condition can indeed be expressed at the differential level with the property that at each point $(x_0, y_0)$ the Jacobian matrix of $(u,v)$ is a multiple of a rotation. In turn this can be easily seen to be equivalent to \eqref{e:CR}.
### Several complex variables
For analytic functions of several complex variables $z = (z_1, \ldots, z_n)$, with $z_k = x_k + iy_k$ , the Cauchy–Riemann equations is given by the following system of partial differential equations (overdetermined when $n>1$) for the functions \begin{align*} & u (x_1, \ldots, x_n, y_1, \ldots , y_n) = {\rm Re}\, f (x_1+iy_1, \ldots, x_n + iy_n)\\ & v (x_1, \ldots, x_n, y_1, \ldots , y_n) = {\rm Im}\, f (x_1+iy_1, \ldots, x_n + iy_n): \end{align*} $$\label{e:CR-sys} \frac{\partial u}{\partial x_k} = \frac{\partial v}{\partial y_k} \qquad \frac{\partial u}{\partial y_k} = - \frac{\partial v}{\partial x_k}\, \qquad k = 1, \ldots, n\, ,$$ or, in terms of the complex differential operators: $\frac{\partial f}{\partial \bar{z}_k} = 0\, .$ Each of the two functions $u$ and $v$ satisfying the conditions \eqref{e:CR-sys} (which as in the one-variable case, turn out to be infinitely differentiable and analytic) is a pluriharmonic function of the variables $x_k$ and $y_k$. When $n>1$ the pluriharmonic functions constitute a proper subclass of the class of harmonic functions. The conditions \eqref{e:CR-sys} are conjugacy conditions for two pluriharmonic functions $u$ and $v$: knowing one of them, one can determine the other by integration (up to addition of a constant in each connected component of the domain of definition).
### Historical remarks
The conditions \eqref{e:CR} apparently occurred for the first time in the works of J. d'Alembert [DA]. Their first appearance as a criterion for analyticity was in a paper of L. Euler, delivered at the Petersburg Academy of Sciences in 1777 [Eu]. A.L. Cauchy utilized the conditions \eqref{e:CR} to construct the theory of functions, beginning with a memoir presented to the Paris Academy in 1814 (see [Ca]). The celebrated dissertation of B. Riemann on the fundamentals of function theory dates to 1851 (see [Ri]).
### References
[Al] L.V. Ahlfors, "Complex analysis" , McGraw-Hill (1966) MR0188405 Zbl 0154.31904 [Ca] A.L. Cauchy, "Mémoire sur les intégrales définies" , Oeuvres complètes Ser. 1 , 1 , Paris (1882) pp. 319–506 [DA] J. d'Alembert, "Essai d'une nouvelle théorie de la résistance des fluides" , Paris (1752) [Eu] L. Euler, Nova Acta Acad. Sci. Petrop. , 10 (1797) pp. 3–19 [Ma] A.I. Markushevich, "Theory of functions of a complex variable" , 1–3 , Chelsea (1977) (Translated from Russian) MR0444912 Zbl 0357.30002 [Ri] B. Riemann, "Grundlagen für eine allgemeine Theorie der Funktionen einer veränderlichen komplexen Grösse" H. Weber (ed.) , Riemann's gesammelte math. Werke , Dover, reprint (1953) pp. 3–48 [Sh] B.V. Shabat, "Introduction of complex analysis" , 1–2 , Moscow (1976) (In Russian) Zbl 0799.32001 Zbl 0732.32001 Zbl 0732.30001 Zbl 0578.32001 Zbl 0574.30001
How to Cite This Entry:
Cauchy–Riemann conditions. Encyclopedia of Mathematics. URL: http://www.encyclopediaofmath.org/index.php?title=Cauchy%E2%80%93Riemann_conditions&oldid=31186 | 2,040 | 6,480 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 3, "x-ck12": 0, "texerror": 0} | 3.671875 | 4 | CC-MAIN-2019-39 | latest | en | 0.826097 |
http://www.jiskha.com/display.cgi?id=1292300420 | 1,495,544,721,000,000,000 | text/html | crawl-data/CC-MAIN-2017-22/segments/1495463607636.66/warc/CC-MAIN-20170523122457-20170523142457-00591.warc.gz | 542,757,539 | 3,722 | posted by on .
40/n = 28/22
what is n?
• Math 7th grade - ,
cross multiply
so 28 x n= 28n
and
40x22= 880
28n=880
divide by n on both sides
n= 31.42 | 65 | 154 | {"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-2017-22 | latest | en | 0.899491 |
https://www.itcsdaixie.com/java%E4%BB%A3%E5%86%99%EF%BD%9Cma117-project-3-determinants-of-matrices/ | 1,686,180,389,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224654016.91/warc/CC-MAIN-20230607211505-20230608001505-00173.warc.gz | 856,722,267 | 18,408 | # Java代写｜MA117 Project 3: Determinants of Matrices
## 1 Formulation of the Problem
Matrices are one of the most important mathematical concepts to be modelled by computer, being used in many problems from solving simple linear systems to modelling complex partial differential equations.
Whilst a matrix (in our formulation) is simply an element of the vector space ℝ!×#, it usually possesses some structure which we can exploit to gain computational speed. For example, a matrix-matrix multiplication generally requires of the order of 𝑛\$ floating-point operations. If the matrix has some special structure which we can exploit using a clever method, then we might be able to reduce this to 𝑛 operations. For large values of 𝑛, this significantly improves the performance of our code.
In this project, you will write two classes representing matrices of the form:
𝐴 is a dense 𝑚 × 𝑛 matrix which, in general, has no special structure and no zero entries. 𝐵 is a tri-diagonal matrix, where all entries are zero apart from along the diagonal and upper and lower diagonals. Note that although 𝐵 is only a 5 × 5 matrix, your classes should represent a general 𝑛 × 𝑛 tri-diagonal matrix. Also, the tri-diagonal matrices you need to represent will always be square.
In a similar fashion to Fraction, you will then write functions to perform various matrix operations:
2. scalar and matrix-matrix multiplication;
3. calculating the determinant of the matrix.
Clearly calculating the determinant is the trickiest task here. Probably you will already have seen expansion by minors as a possible method. Whilst this is an excellent method for calculating determinants by hand, you should not use it for this task. The reason is that calculating the determinant of a 𝑛 × 𝑛 matrix requires 𝑂(𝑛!) operations, since for each 𝑛 × 𝑛 matrix, we must calculate the values of the 𝑛 − 1 sub-determinants. This is extremely slow.
A much better method is called LU decomposition. In this, we write a matrix 𝐴 as product of two matrices 𝐿 and 𝑈 which are lower- and upper- triangular respectively. For example, for a 4×4 matrix, we would find matrices so that
Such a factorisation is not guaranteed to exist (and indeed is not unique), but typically it does. In this project,you don’t really need to worry about this – your code will be tested with matrices for which the LU decomposition exists. It is up to you to figure out how to calculate the determinant from the LU decomposition!
Throughout the formulation, matrices will be represented by indices running between 1 ≤ 𝑖, 𝑗 ≤ 𝑚, 𝑛.
However, in your code, you should stay consistent with Java notation and indices should start at 0
(i.e. 0 ≤ 𝑖, 𝑗 ≤ 𝑚 − 1, 𝑛 − 1).
## 2 Programming Instructions
On the course web page for the project, you will find files for the following classes. As with the previous projects, the files have some predefined methods that are either complete or come with predefined names and parameters. You must keep all names, parameter types and return types of public objects and methods as they are in the templates. Other methods must be filled in and it is up to you to design them properly.
There are five classes in this project:
• Matrix: a general class defining the basic properties and operations on matrices.
• MatrixException: a subclass of the RuntimeException class which you should use to throw matrix-related exceptions. This class is complete – you do not need to alter it.
• GeneralMatrix: a subclass of Matrix which describes a general 𝑚 × 𝑛 real matrix.
• TriMatrix: another subclass of Matrix which describes a 𝑛 × 𝑛 real tri-diagonal matrix.
• Project3: a separate class which will use Matrix and its subclasses to collect some basic statistics involving random matrices.
• Please note that unlike other projects, you may not assume that the data you receive will be valid.
Therefore, you will need to check, amongst other things, that matrix multiplications are done using matrices of valid sizes, the user is not trying to access matrix elements which are out of bounds, etc. If something goes wrong, you are expected to throw a MatrixException.
The classes you need to work on are briefly described below.
### 2.1 The Matrix class
This is the base class from which you will build your specialised subclasses. Matrix is abstract – as described in the lectures, this means that some of the methods are not defined, and they need to be implemented in the subclasses. The general idea is that each subclass of Matrix can implement its own storage schemes, whilst still maintaining various common methods inherent in all matrices.
In particular, the following functions are not abstract, and need to be filled in inside Matrix:
• the protected constructor function;
• toString, which should return a String representation of the matrix.
Additionally, the following abstract methods will be implemented by the subclasses of Matrix:
• getIJ and setIJ: accessor and mutator methods to get/set the 𝑖𝑗th entry of the matrix.
• add: returns a new Matrix containing the sum of the current matrix with another.
• multiply(double scaler): multiply the matrix by a constant 𝑠𝑐𝑎𝑙𝑎𝑟 ∈ ℝ.
• multiply(Matrix B): multiply the matrix by another matrix. Note that this is intended to be aleft multiplication; i.e. A.multiply(B) corresponds to the multiplication 𝐴𝐵.
• random(): fills current the matrix with random numbers, uniformly distributed between 0 and
1. For a tri-diagonal matrix, this should fill the three diagonals with random numbers.
In subclasses, you should pay attention to what type of matrix needs to be returned from each of the functions. For example, when adding two GeneralMatrix objects the result should be a GeneralMatrix (which is then typecast to a Matrix).
E-mail: [email protected] 微信:itcsdx | 1,292 | 5,817 | {"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-2023-23 | longest | en | 0.927141 |
https://www.zigya.com/study/book?class=12&board=nbse&subject=Chemistry&book=Chemistry+I&chapter=Solutions&q_type=&q_topic=Expressing+Concentration+of+Solutions&q_category=&question_id=CHEN12043920 | 1,545,006,458,000,000,000 | text/html | crawl-data/CC-MAIN-2018-51/segments/1544376828018.77/warc/CC-MAIN-20181216234902-20181217020902-00054.warc.gz | 1,135,050,121 | 16,729 | What is the sum of the mole fraction of all the components in a three component system? from Chemistry Solutions Class 12 Nagaland Board
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What is the sum of the mole fraction of all the components in a three component system?
Some of the mole fractions of all the components is unity.
x1 + x2 + x3 = 1.
144 Views
Calculate the mole fraction of benzene in solution containing 30% by mass in carbon tetrachloride.
Let the total mass of the solution be 100g and mass of benzene be 30 g
therefore mass of tetrachloride= (100-30)g = 70g
Molar mass of benzene,
897 Views
Calculate the molarity of each of the following solution (a) 30 g of Co(NO3)2.6H2O in 4.3 L solution (b) 30 mL of 0.5 MH2SO4 diluted to 500 mL.
solution;
Molarity (M) is defined as number of moles of solute dissolved in one litre (or one cubic decimetre) of solution.
(a) Mol. mass of
Moles of $\mathrm{Co}\left(\mathrm{NO}{\right)}_{3}.6{\mathrm{H}}_{2}\mathrm{O}$
Volume of solution = 4.3 L
Molarity,
(b) Number of moles present in 1000 ml of 0.5M H2SO4= 0.5 mol
therefore number of moles present in 30ml of 0.5M H2SO4=$\frac{0.5×30}{1000}$mol =0.015mol
therefore molarity =0.015/0.5L
thus molarity is 0.03M
844 Views
Calculate (a) molality (b) molarity and (c) mole fraction of KI if the density of 20% (mass/mass) aqueous KI is 1.202 g mL-1.
(a) 20% (mass/mass) means that 20 g of KI is present in 80 g of water.
Therefore, Moles of KI in solution
moles of KI = 20/166 =0.12mol
moles of water =80/18 =4.44mol
therefore, mole fraction of KI
=
1010 Views
Calculate the mass percentage of benzene (C6H6) and carbon tetrachloride (CCl4) if 22 g of benzene is dissolved in 122 g of carbon tetrachloride.
Mass % of benzene
Mass% of carbon tetrachloride = 100 - 15.28
= 84.72%
1703 Views
Calculate the mass of urea (NH2CONH2) required in making 2.5 kg 0.25 of molal aqueous solution.
Solution:
Molality (m) is defined as the number of moles of the solute per kilogram (kg) of the solvent and is expressed as:
Mol. mass of urea ${\mathrm{NH}}_{2}{\mathrm{CONH}}_{2}$
= 14 + 2 + 12 + 16 + 14 + 2
=
Molality (m) =
or Moles of solute
= 0.25 x 0.25 = 0.625
Mass of urea
= Moles of solute x Molar mass
= 0.625 x 60 = 37.5 g
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Definition A topological space is called noetherian if every descending chain of closed subsets terminates.
Remark If R is anoetherian ring, then X = Spec(R) is a noetherian topological space, since the closed subsets correspond to ideals.
Proposition Every closed subset in a noetherian topological space can be uniquely decomposed as an irredundant union of irreducible subspaces.
Proof: Use the same proof that showed the corresponding result for algebraic subsets of affine space over a field.
Definition Let X be a noetherian topological space. The dimension of X is defined to be
dim(X) = sup { n : there exists a chain of nonempty irreducible distinct closed subsets Z0 ⊂ Z1 ⊂ … ⊂ Zn in X},
provided that this supremum exists.
### Examples
1. dim(*) = 0.
2. If k is a field, then dim(A1k) = 1, since the irreducible sets are either points or all of A1.
3. If k is a field, then dim(A2k) = 2, since the irreducible sets are either all of A2, an irreducible curve, or a point.
4. With the usual topology on the real line R, one has dim(R)=0. For R is not irreducible, and the only irreducible closed subsets of R are single points.
5. dim(Spec(Z)) = 1, since every prime ideal is either zero or maximal.
Definition Let P be a prime ideal in a ring R (as always, commutative with 1). The height of P is defined to be
ht(P) = sup { n : there exists a chain of distinct prime ideals P0⊂ P1⊂…⊂ Pn = P⊂ R}
Definition The Krull dimension of a ring R is defined to be
dim(R) = sup{ht(P) : P R is prime}.
Lemma Let X = Spec(R) be an affine scheme. Then
dim(X) = dim(R).
Proof: There is a one-to-one correspondence between the irreducible closed subsets of X and the prime ideals of R.
Proposition Let X be a noetherian topological space. Then
1. dim(X) = sup(dim(Xi)), where the Xi range through the irreducible components of X.
2. If U X is a dense open set and if dim(U) is finite, then dim(X) = dim(U).
3. If Y- X, then dim(Y) dim(X). Moreover, if X is irreducible and the inclusion is proper, then the inequality is strict.
Proof: (i) Let Z0 Zn be a proper chain of irreducible closed subsets of X. Since
Zn = i(Xi Zn)
and Zn is irreducible, there exists an i such that Zn Xi. But then the entire chain is contained inside Xi.
(ii) By (i), we may assume that X is irreducible. If Z0 Zn is a proper chain of closed irreducible subsets of U, then their closures Z-0Z-n form a proper chain of closed irreducibles of X. (Properness follows because Zi = UZ-i, and closures always stay irreducible.) So dim(U) dim(X).
Since dim(U) is finite, we can choose a chain Z0 Zn in U of maximal length. Now let W be any irreducible closed subset of X such that Z-i WZ-i+1. Intersecting back with U yields
Zi W U Zi+1.
Since the original chain had maximal length, one of these inclusions must be an equality. But W U is dense in the irreducible space W, hence one of the earlier inclusions in X was already an equality. It follows that Z-0Z-n is a maximal chain in X, and the dimensions are equal.
(iii) Without loss of generality, we may assume that Y=Y-. Let Z0 Zn be a proper chain in Y. Then it is also a proper chain in X. If the containment is proper, then we can add X to the top of the chain to lengthen it.
Proposition Let X be a scheme that is a noetherian topological space. Then dim(X)=0 if and only if X is a finite set with the discrete topology.
Proof: It is clear that any set with the discrete topology has dimension zero, since the only irreducible subsets are singeltons. On the other hand, we know that every scheme contains at least one closed point (coming from a maximal ideal inside some ring defining an open affine subset). If X contained a non-closed point, then its closure would be irreducible, forcing the dimension of X to be greater than zero. So, any zero-dimensional scheme has the property that every point is closed. Thus, in any affine open subset Spec(R) of X, the ring R must have the property that every prime ideal is maximal. This forces Spec(R), and hence X to have the discrete topology. Now use the finite decomposition into irreducibles for the noetherian topological space X to obtain the desired conclusion.
Theorem (Dimension Theorem) Let X be an algebraic variety over a field k. Then
dim(X) = tr.degk(k(X)).
Proof: Since every variety contains an open, dense, affine variety, the result will follow if we can establish the following three facts.
1. The coordinate ring A(X) is an integral extension of a polynomial ring.
2. If B A is an integral extension of noetherian rings, then dim(B)=dim(A) and tr.deg(B)=tr.deg(A).
3. If k is a field, then dim(Ank) = n.
Proposition If k is a field, then dim(Ank) = n.
Proof: Since (0) (x1) (x1, x2) (x1,, xn) is a proper chain of prime ideals, we have dim(An) n.
On the other hand, we know that n = tr.deg(k[x1,,xn]). Moreover, if A is an integral domain and if P is any nonzero prime ideal in A, then tr.deg(A/P)<tr.deg(A). For, take any algebraically independent set x-1, , x-r A/P. Lift them arbitrarily to elements x1, , xr A, and choose a nonzero element x P. Then {x1, , xr, x} is an algebraically independent set in A. To see this, suppose F(t1,, tr, tr+1) k[t1,,tr+1] is any polynomial such that F(x1,,xr,x) = 0 in A. Since A is an integral domain, we can assume that F is an irreducible polynomial. Then F(x-1,,x-r,0) = 0 in A/P. Since there are no nonzero algebraic relations between these elements in A/P, the polynomial F(t1,,tr,0) must be identically zero. But then F is divisible by tr+1; by irreducibility, F=tr+1. Since x0 in A; this is a contradiction.
Now let P0 Pt be a chain of prime ideals in k[x1, , xn]. Then
n > tr.deg(A/P0) > tr.deg(A/P1) >> tr.deg(A/Pt),
so n t. Thus, dim(An) n, and the result follows.
In order to establish the other two parts, we need some preliminaries from commutative algebra; in particular, we need to know how prime ideals behave in integral extensions of rings. To understand that, we introduce one of the essential tools of commutative algebra: localization.
Definition If A is a ring and if P is a prime ideal in A, we define the localization of A at P to be the ring AP obtained by inverting all elements of A\ P.
Remark The localization AP is a local ring, with maximal ideal PAP.
Definition Let f:B A be an extension of rings. If P A is a prime ideal, then Q=f-1(P) = P B is a prime ideal in B. In this circumstance, we say that P lies over Q.
Lemma Let B be a local ring with maximal ideal Q. Let B A be an integral extension. Then the set of prime ideals of A lying over Q is just the set of maximal ideals of A.
Proof: We show first that every maximal ideal M of A lies over Q. Define N=M B. Then A- = A/M is a field that is integral over the subring B- = B/N. Now, let 0 x B-. Since 1/xA-, we have a monic polynomial equation
(1/x)n + b1(1/x)n-1++bn = 0
with coefficients in B-. Now multiply by xn-1 to get
(1/x) = - (b1 + b2 x ++ bnxn-1) B-.
So, B- is a field and N=Q is the unique maximal ideal in B.
Next, we must show that every ideal P in A that lies over Q is maximal. This time, A- = A/P is an integral domain that is integral (thus algebraic) over the field B-=B/Q. Take 0 y A-. There is an algebraic dependence relation
b0yn + b1yn-1 ++bn = 0
of minimal degree with coefficients in B-. By minimality, bn0. Since B- is a field, we can divide the polynomial by bn, and assume that bn=1. Now we have
[1 / y] = -(b0yn-1 + b1yn-2 ++bn-1) A-.
Hence, A- is a field and P is maximal.
Proposition (The Lying-Over Theorem) Let B A be an integral extension of rings. If Q B is any prime ideal, then there exists a prime ideal P A lying over Q.
Proof: Let BQ be the localization of B at Q. Then AQ = (B\ Q)-1A is an integral extension of BQ, and contains it as a subring. The prime ideals of A lying over Q correspond to the prime ideals of AQ lying over QBQ, which are necessarily the maximal ideals of AQ. Since BQ0, we know that AQ is nonzero, so it has maximal ideals.
Proposition (The Going-Up Theorem) Let B A be an integral extension of rings. Let Q Q' B be prime ideals and let P A be a prime ideal lying over Q. Then there exists an ideal P' lying over Q' such that P P'.
Proof: The quotient A/P contains and is integral over B/Q. By the Lying Over Theorem, there exists a prime P'/P in A/P lying over the prime ideal Q'/Q. By the isomorphism theorems, P' is a prime ideal of A lying over Q'.
Lemma Let B be an integral domain that is integrally closed in its field of fractions L. Let K/L be a normal extension with Galois group G, and let A be the integral closure of B in K. If Q is any prime ideal of B, then G acts transitively on the set of prime ideals lying over Q.
Proof: One can assume (with some work that I'm omitting) that K/L is a finite Galois extension. Suppose now that P' and P are two primes lying over Q, and that P' is not contained in any of the conjugates Pi= σi(P) of P for σi G. Then there is an element x P' that is contained in no Pi. But y=NK/L(x) B is not in Q (because all σi(x) P) and is in P' A. This is a contradiction.
Proposition (The Going-Down Theorem) Let B A be an integral extension of integral domains, and assume that B is integrally closed in its field of fractions. Let Q Q' B be prime ideals and let P' A be a prime ideal lying over Q'. Then there exists an ideal P lying over Q such that P P'.
Proof: Let K be the field of fractions of A, let L be the field of fractions of B, and let F be the Galois closure of the field extension K/L. Inside F, let C denote the integral closure of B (and hence also of A). By the Lying Over Theorem, there exists a prime ideal Q0 in C lying over Q. By the Going Up Theorem, there exists a prime ideal Q'0 Q0 lying over Q'. Also by the Lying Over Theorem, there exists a prime ideal Q'1C lying over P'. Now the ideals Q'0 and Q'1 lie over the same ideal Q' in B. Because the extension is Galois, there exists an automorphism σ Gal(F/L) such that σ(Q'0) = Q'1. Now the result follows by taking P= σ(Q0) A.
Now we can carry out part (ii) of the proof of the dimension theorem.
Proposition Let B A be an integral extension of noetherian rings. Then dim(B)=dim(A) and tr.deg(B) = tr.deg(A).
Proof: Since integral extensions are algebraic, the equality for transcendence degrees is trivial. The Lying Over and Going Up Theorems show that any chain of prime ideals in B can be lifted to a chain in A of the same length; thus, dim(A) dim(B). Conversely, if P P' are distinct prime ideals in A, then the ideals P B P' B must also be distinct. (Localize at the prime ideal P B; since the ideals lying over its maximal ideal are precisely the maximal ideals of the localization of A, there cannot be any proper containments between them.) Thus, any chain of primes in A produces a chain of the same length in B, and dim(A) dim(B).
Finally, we can carry out step (i).
Theorem Let A=k[x1, , xn] be a polynomial ring over a field k, and let P be an ideal of A of height h. Then there exist elements v1, , vn A such that
1. A is integral over k[v], and
2. P k[v] = (v1, , vh).
Proof: The proof is by induction on the height h=ht(P). The case h=0 is trivial, for then P=0 and we can take vi=xi. A better base for the induction, however, is provided by the case h=1. Choose a nonzero polynomial v1=f(x) P. Write f(x) = i=1s ci Mi(x) where ci k and each Mi(x) is a monomial. Given any positive integers d1,,dn, we define the d-weight of a monomial M(x)= xiai to be aidi. Now choose weights d1=1, d2, , dn so that all the monomials appearing in f(x) have distinct weights. (We can achieve this by using large prime powers for the weights.) Put vi=xi-x1di for i=2,,n. Then
v1 = f(x) = f(x1, v2+x1d2, …, vn + x1dn) = ajx1e + g(x1, v2, …, vn)
where g is a polynomial whose degree in x1 is strictly less than e, and aj is the coefficient of the term of largest weight in f. It follows that x1 is integral over the ring k[v1, , vn], and hence so are the elements xi= vi+x1di.
We're not done yet, however, since we only know that k[v] k[x] is an integral extension. We still need to mod out the prime ideal P and its pullback P k[v]. We clearly have a containment (v1) P k[v]. In fact, however, both these ideals are primes of height 1, so they must be equal. (Note: that P k[v] is height 1 uses the fact that k[v] is integrally closed, so the Going Down Theorem applies.)
Now suppose h>1. Choose an ideal Q P of height h-1. By induction, there exist w1, , wn such that k[x] is integral over k[w] and Q k[w] = (w1, , wh-1). Define P'=P k[w]. Then P' also has height h, so it contains (w1,,wh-1) properly. Choose a nonzero polynomial of the form f(wh, , wn) P' and repeat the weight argument of the case h=1.
Theorem (Noether Normalization Theorem) Let A be an integral domain that is finitely generated as an algebra over a field k. Then there exist elements y1, , yr A that are algebraically independent over k such that A is integral over k[y1, , yr].
Proof: Write A=k[x1, , xn]/P where P is a prime ideal of height n-r. By the previous theorem, there are elements v1, , vn in k[x1, , xn] such that k[v] k[x] is integral and such that P k[v] = (vr+1, , vn). The result follows by taking yi vi mod P.
Corollary Let A be an integral domain that is finitely generated as a k-algebra. Then for any prime P in A, we have
ht(P) + dim(A/P) = dim(A).
Proof: Reduce to the case A=k[x1,,xn] and use the proof of the Noether Normalization Theorem.
Corollary If X and Y are algebraic varieties, then dim(X× Y) = dim(X) + dim(Y).
Proof: The transcendence degree of a tensor product satisfies the corresponding relation.
Proposition Let X be a hypersurface in Ank. Then every irreducible component of X has dimension n-1.
Proof: We may assume that X is irreducible, and hence of the form Z(f) for a nonconstant irreducible polynomial f k[x1,,xn]. By renumbering the variables, we can assume that xn actually occurs in f. Now let ti A(X) be the image of xi. We claim that {t1, , tn-1} is an algebraically independent set, and therefore dim(X) n-1. For, suppose that G(t1, , tn-1) =0 is an algebraic relation. Then G I(X)= (f), so f divides G. But this is impossible, since xn occurs in f but not in G. On the other hand, the proof that An has dimension n shows that dim(X) n-1.
Theorem Let XAnk be an algebraic set all of whose irreducible components have dimension n-1. Then X is a hypersurface.
Proof: Without loss of generality, we may assume that X is irreducible. Let f I(X) be a nonzero polynomial. By irreducibility, some irreducible factor h of f must vanish on X. So, X Z(h). But this is an inclusion of irreducible sets of the same dimension, so it must be an equality. | 4,037 | 14,659 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.65625 | 4 | CC-MAIN-2021-49 | longest | en | 0.888707 |
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Coin Word Problem
Jenny received \$6.10 in tips one afternoon. All of her tips were in quarters, dimes, and nickels. there five less dimes than quarters and seven less nickels than dimes. How many of each kind of coin was there?
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6.10 = .25Q + .10(Q-5) + .05 (Q-5-7)
6.10 = .25 Q +.10Q-.50 + .05Q -.60
6.10 = .40Q -1.10
7.20 = .40Q
18=Q
so 18 quarters (4.50)
18-5=13 dimes (1.30)
13-7=6 nickels (.30)
4.50+1.30+.30= 6.10
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This coin word problem is worked out in the following manner: \$6.10 = 0.25Q + 0.10(Q-5) + 0.05 (Q-50 -7); 6.10 = .25 Q +.10Q-.50 + .05Q -.60; 6.10 = .40Q -1.10. You will get 7.20 = .40Q, which translates to 18=Q, meaning there were 18 quarters, or about \$4.50. The dimes were 13 (18-5=13) dimes and the nickels were 6 (13-7=6).
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We'll provide some tips to help you choose the best Diamond problem solver calculator for your needs. Math can be a challenging subject for many students.
## The Best Diamond problem solver calculator
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An arithmetic sequence solver is a tool that can be used to find the next number in a sequence of numbers that follow a specific pattern. For example, if you know that the first two numbers in a sequence are 1 and 3, you can use an arithmetic sequence solver to find the next number in the sequence.
There are a number of ways to solve ln equations, depending on the complexity of the equation. For simple equations, taking the natural log of both sides of the equation will usually suffice. However, for more complex equations, more sophisticated methods may be necessary. In any case, it is always important to carefully consider the equation before attempting to solve it, in order to ensure that the chosen method is appropriate. | 440 | 2,174 | {"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} | 4.375 | 4 | CC-MAIN-2023-06 | latest | en | 0.954874 |
http://mypages.iit.edu/~smile/ma9003.html | 1,558,885,744,000,000,000 | text/html | crawl-data/CC-MAIN-2019-22/segments/1558232259316.74/warc/CC-MAIN-20190526145334-20190526171334-00544.warc.gz | 142,060,650 | 2,103 | `Tower of HanoiPaul Eubanks Washington High School 3535 E. 114th Street Chicago, Illinois 60617 312-646-4860Objectives:(For grades 6 through 12)1. To discover a pattern.2. To evaluate an exponential expression.Materials:Tower of Hanoi puzzle and circular discs of various sizes.Procedure:Pass out five circular discs of different sizes to each student. Have each student draw three large circles on a sheet of paper. Then have them place the largest disc in one of the circles as you place the largest disc of the puzzle on one of the three pegs (towers). Explain the rules of the puzzle as follows. The object is to move all the discs from one tower (one circle) to another tower (another circle). Only one disc can be moved at a time and a larger disc can never be placed on top of a smaller one. The discs should be moved from one tower to another in the least number of moves. With one disc it only takes one move to place that disc on another tower (circle). Next have the students place the two largest discs in any circle. Have the students try to move the two discs to another circle. Ask them to tell how many moves it took. It will take three moves. Place the smaller disc in either open circle. Then move the larger disc to the other open circle. Finally place the smaller disc on top of the larger disc. Now have the students place the three largest discs in any circle. Allow the students some time to try to move the three discs to another circle. This will take at least seven moves. Move the top two discs to another circle in three moves as explained above. Next move the largest disc to the open circle. Then use three moves to get the two smaller discs on top of the largest disc. Now have the students place the four largest discs in any circle and try to move them to another circle. This will take at least 15 moves. It will take seven moves to move the top three discs to another circle as we just did with three discs in a circle. After moving the top three discs, the largest disc is moved to the open circle.Next the other three discs must be moved on top of this largest disc. This will take seven more moves, however, the first move is crucial. When moving an odd number of discs, you must move the first disc to the circle (tower) that you want all the discs to end up on. When moving an even number of discs you must move the first disc to the circle that you do not want the discs to end up on. At this point you should have the students notice how the number of moves increased, i.e., 1, 3, 7 and 15. Have the students look for a pattern and predict the number of moves for five discs. The number is 31. Give the students a few minutes to see if anyone can come up with the formula - which is M = 2n - 1 where M is the number of moves and n is the number of discs. If no one comes up with the formula give it to them and have them verify it for n equal to 1, 2, 3, 4, and 5. Next have them use the formula to find out the number of moves for 6, 7, 8, and 10 discs. Finally tell them about the legend that a monk has 64 discs on a tower and he moves one every second. When he has moved all the discs the world will end. Have them estimate how long this would be. Use a calculator to evaluate 264 - 1. The answer is approximately 589,942,417,400 years. ` | 806 | 3,455 | {"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} | 4.0625 | 4 | CC-MAIN-2019-22 | latest | en | 0.931249 |
https://scribesoftimbuktu.com/solve-for-q-5q62qq2/ | 1,669,819,303,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710764.12/warc/CC-MAIN-20221130124353-20221130154353-00462.warc.gz | 546,142,150 | 14,450 | # Solve for q 5q+6=2q+q+2
5q+6=2q+q+2
Since q is on the right side of the equation, switch the sides so it is on the left side of the equation.
2q+q+2=5q+6
Add 2q and q.
3q+2=5q+6
Move all terms containing q to the left side of the equation.
Subtract 5q from both sides of the equation.
3q+2-5q=6
Subtract 5q from 3q.
-2q+2=6
-2q+2=6
Move all terms not containing q to the right side of the equation.
Subtract 2 from both sides of the equation.
-2q=6-2
Subtract 2 from 6.
-2q=4
-2q=4
Divide each term by -2 and simplify.
Divide each term in -2q=4 by -2.
-2q-2=4-2
Cancel the common factor of -2.
Cancel the common factor.
-2q-2=4-2
Divide q by 1.
q=4-2
q=4-2
Divide 4 by -2.
q=-2
q=-2
Solve for q 5q+6=2q+q+2
### Solving MATH problems
We can solve all math problems. Get help on the web or with our math app
Scroll to top | 337 | 825 | {"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} | 4.0625 | 4 | CC-MAIN-2022-49 | latest | en | 0.798755 |
https://jovian.ai/siddhantramteke369/m-coloring-problem | 1,632,372,767,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780057417.10/warc/CC-MAIN-20210923044248-20210923074248-00560.warc.gz | 370,230,873 | 40,368 | Learn data science and machine learning by building real-world projects on Jovian
siddhantramteke369/m-coloring-problem
5 months ago
m Coloring Problem
In [1]:
``project_name = "m_coloring_problem"``
In [2]:
``!pip install jovian --upgrade --quiet``
In [3]:
``import jovian``
In [5]:
``jovian.commit(project=project_name)``
```[jovian] Attempting to save notebook.. [jovian] Updating notebook "siddhantramteke369/m-coloring-problem" on https://jovian.ai [jovian] Uploading notebook.. [jovian] Uploading additional files... [jovian] Committed successfully! https://jovian.ai/siddhantramteke369/m-coloring-problem ```
Out[5]:
``'https://jovian.ai/siddhantramteke369/m-coloring-problem'``
Problem Statement
Given an undirected graph and a number m, determine if the graph can be coloured with at most m colours such that no two adjacent vertices of the graph are colored with the same color. Here coloring of a graph means the assignment of colors to all vertices.
The Method
Here's the systematic strategy we'll apply for solving problems:
1. State the problem clearly. Identify the input & output formats.
2. Come up with some example inputs & outputs. Try to cover all edge cases.
3. Come up with a correct solution for the problem. State it in plain English.
4. Implement the solution and test it using example inputs. Fix bugs, if any.
5. Analyze the algorithm's complexity and identify inefficiencies, if any.
6. Apply the right technique to overcome the inefficiency. Repeat steps 3 to 6.
Let's apply this approach step-by-step.
Solution
1. State the problem clearly. Identify the input & output formats.
Problem
Given a 2D array having 1 at position (i, j) if there is an edge between vertex i and j or 0 if there isn't and an integer m, we have to find the array of colors which represent the color given to ith vertex so that no two neighbour vertices have the same color or output false if the graph cannot be colored with just m colors.
Input
1. A 2D array graph[V][V] where V is the number of vertices in graph and graph[V][V] is adjacency matrix representation of the graph. A value graph[i][j] is 1 if there is a direct edge from i to j, otherwise graph[i][j] is 0.
2. An integer m which is the maximum number of colors that can be used.
Output
1. An array color[V] that should have numbers from 1 to m. color[i] should represent the color assigned to the ith vertex. The code should also return false if the graph cannot be colored with m colors.
Based on the above, we can now create a signature of our function:
In [4]:
``````def graphColoring():
pass``````
In [8]:
``import jovian``
In [10]:
``jovian.commit()``
```[jovian] Attempting to save notebook.. [jovian] Updating notebook "siddhantramteke369/m-coloring-problem" on https://jovian.ai [jovian] Uploading notebook.. [jovian] Uploading additional files... [jovian] Committed successfully! https://jovian.ai/siddhantramteke369/m-coloring-problem ```
Out[10]:
``'https://jovian.ai/siddhantramteke369/m-coloring-problem'``
2. Come up with some example inputs & outputs. Try to cover all edge cases.
Our function should be able to handle any set of valid inputs we pass into it. Here's a list of some possible variations we might encounter:
1. General case
2. Graph with all zeros
3. Graph with all 1s
4. m with 1
5. m with 0
We'll express our test cases as dictionaries, to test them easily. Each dictionary will contain 2 keys: `input` (a dictionary itself containing one key for each argument to the function and `output` (the expected result from the function).
In [5]:
``````test = {
'input': {
'graph': [[ 0, 1, 1, 1 ],
[ 1, 0, 1, 0 ],
[ 1, 1, 0, 1 ],
[ 1, 0, 1, 0 ]],
'm': 3,
'color': [0 for i in range(4)]
},
'output': [1, 2, 3, 2]
}
``````
Create one test case for each of the scenarios listed above. We'll store our test cases in an array called `tests`.
In [6]:
``tests = []``
In [7]:
``tests.append(test)``
In [8]:
``````tests.append({
'input': {
'graph': [[ 0, 0, 0, 0 ],
[ 0, 0, 0, 0 ],
[ 0, 0, 0, 0 ],
[ 0, 0, 0, 0 ]],
'm': 3,
'color': [0 for i in range(4)]
},
'output': [1, 1, 1, 1]
})
``````
In [9]:
``````tests.append({
'input': {
'graph': [[ 1, 1, 1, 1 ],
[ 1, 1, 1, 1 ],
[ 1, 1, 1, 1 ],
[ 1, 1, 1, 1 ]],
'm': 3,
'color': [0 for i in range(4)]
},
'output': None
})
``````
In [10]:
``````tests.append({
'input': {
'graph': [[ 0, 1, 0, 1 ],
[ 1, 0, 1, 0 ],
[ 0, 1, 0, 0 ],
[ 1, 0, 0, 0 ]],
'm': 1,
'color': [0 for i in range(4)]
},
'output': None
})
``````
In [11]:
``````tests.append({
'input': {
'graph': [[ 0, 1, 1, 1 ],
[ 1, 0, 1, 0 ],
[ 1, 1, 0, 0 ],
[ 1, 0, 0, 0 ]],
'm': 0,
'color': [0 for i in range(4)]
},
'output': None
})
``````
In [12]:
``````tests.append({
'input': {
'graph': [[ 0, 1, 0 ],
[ 1, 0, 1 ],
[ 0, 1, 0 ]],
'm': 3,
'color': [0 for i in range(3)]
},
'output': [1, 2, 1]
})
``````
3. Come up with a correct solution for the problem. State it in plain English.
Our first goal should always be to come up with a correct solution to the problem, which may not necessarily be the most efficient solution. Come with a correct solution and explain it in simple words below:
Bruteforce technique
We will first try the brute force technique where we will try all the possible configurations to find correct solution.
Steps:
1. We will create a recursive function that takes graph, current index, no. of vertices and output color array
2. If current index is equal to number of vertices. Check if the output color configuration is safe i.e. check if the adjacent vertices does not have same color. If the condition is true print the configurations and break.
3. Assign color to a vertex
4. For every assigned color recursively call the function with next index and number of vertices.
5. If any recursive function returns true break the loop and return true.
6. If the current index equals 0 and all the recursive functions returned true return color array.
Let's save and upload our work before continuing.
In [17]:
``jovian.commit()``
```[jovian] Attempting to save notebook.. [jovian] Updating notebook "siddhantramteke369/m-coloring-problem" on https://jovian.ai [jovian] Uploading notebook.. [jovian] Uploading additional files... [jovian] Committed successfully! https://jovian.ai/siddhantramteke369/m-coloring-problem ```
Out[17]:
``'https://jovian.ai/siddhantramteke369/m-coloring-problem'``
4. Implement the solution and test it using example inputs. Fix bugs, if any.
In [13]:
``````# Checks if the graph is safe or not (if its acceptable or not)
def isSafe(graph, color):
l = len(graph)
for i in range(l):
for j in range(i + 1, l):
# if both are adjacent vertices and their color is same
if (graph[i][j] and color[j] == color[i]):
return False
return True
# This function solves the m Coloring problem using recursion.
def graphColoring(graph, m, color, i=0):
l = len(graph)
if (i==l):
if (isSafe(graph, color)):
#print(color)
return True
return False
for j in range(1, m + 1):
color[i] = j
if (graphColoring(graph, m, color, i + 1)):
if (i == 0):
return color
return True
color[i] = 0
return None``````
In [14]:
``````graph1 = test['input']['graph']
m = test['input']['m']
color = test['input']['color']
print('output:', graphColoring(graph1, m, color, 0))
print('actual: ', test['output'])``````
```output: [1, 2, 3, 2] actual: [1, 2, 3, 2] ```
In [15]:
``tests``
Out[15]:
``````[{'input': {'graph': [[0, 1, 1, 1], [1, 0, 1, 0], [1, 1, 0, 1], [1, 0, 1, 0]],
'm': 3,
'color': [1, 2, 3, 2]},
'output': [1, 2, 3, 2]},
{'input': {'graph': [[0, 0, 0, 0], [0, 0, 0, 0], [0, 0, 0, 0], [0, 0, 0, 0]],
'm': 3,
'color': [0, 0, 0, 0]},
'output': [1, 1, 1, 1]},
{'input': {'graph': [[1, 1, 1, 1], [1, 1, 1, 1], [1, 1, 1, 1], [1, 1, 1, 1]],
'm': 3,
'color': [0, 0, 0, 0]},
'output': None},
{'input': {'graph': [[0, 1, 0, 1], [1, 0, 1, 0], [0, 1, 0, 0], [1, 0, 0, 0]],
'm': 1,
'color': [0, 0, 0, 0]},
'output': None},
{'input': {'graph': [[0, 1, 1, 1], [1, 0, 1, 0], [1, 1, 0, 0], [1, 0, 0, 0]],
'm': 0,
'color': [0, 0, 0, 0]},
'output': None},
{'input': {'graph': [[0, 1, 0], [1, 0, 1], [0, 1, 0]],
'm': 3,
'color': [0, 0, 0]},
'output': [1, 2, 1]}]``````
Evaluate your function against all the test cases together using the `evaluate_test_cases` (plural) function from `jovian`.
In [16]:
``from jovian.pythondsa import evaluate_test_cases``
In [17]:
``evaluate_test_cases(graphColoring, tests)``
``` TEST CASE #0 Input: {'graph': [[0, 1, 1, 1], [1, 0, 1, 0], [1, 1, 0, 1], [1, 0, 1, 0]], 'm': 3, 'color': [1, 2, 3, 2]} Expected Output: [1, 2, 3, 2] Actual Output: [1, 2, 3, 2] Execution Time: 0.064 ms Test Result: PASSED TEST CASE #1 Input: {'graph': [[0, 0, 0, 0], [0, 0, 0, 0], [0, 0, 0, 0], [0, 0, 0, 0]], 'm': 3, 'color': [1, 1, 1, 1]} Expected Output: [1, 1, 1, 1] Actual Output: [1, 1, 1, 1] Execution Time: 0.018 ms Test Result: PASSED TEST CASE #2 Input: {'graph': [[1, 1, 1, 1], [1, 1, 1, 1], [1, 1, 1, 1], [1, 1, 1, 1]], 'm': 3, 'color': [0, 0, 0, 0]} Expected Output: None Actual Output: None Execution Time: 0.262 ms Test Result: PASSED TEST CASE #3 Input: {'graph': [[0, 1, 0, 1], [1, 0, 1, 0], [0, 1, 0, 0], [1, 0, 0, 0]], 'm': 1, 'color': [0, 0, 0, 0]} Expected Output: None Actual Output: None Execution Time: 0.011 ms Test Result: PASSED TEST CASE #4 Input: {'graph': [[0, 1, 1, 1], [1, 0, 1, 0], [1, 1, 0, 0], [1, 0, 0, 0]], 'm': 0, 'color': [0, 0, 0, 0]} Expected Output: None Actual Output: None Execution Time: 0.005 ms Test Result: PASSED TEST CASE #5 Input: {'graph': [[0, 1, 0], [1, 0, 1], [0, 1, 0]], 'm': 3, 'color': [1, 2, 1]} Expected Output: [1, 2, 1] Actual Output: [1, 2, 1] Execution Time: 0.015 ms Test Result: PASSED SUMMARY TOTAL: 6, PASSED: 6, FAILED: 0 ```
Out[17]:
``````[([1, 2, 3, 2], True, 0.064),
([1, 1, 1, 1], True, 0.018),
(None, True, 0.262),
(None, True, 0.011),
(None, True, 0.005),
([1, 2, 1], True, 0.015)]``````
In [104]:
``jovian.commit()``
```[jovian] Attempting to save notebook.. [jovian] Updating notebook "siddhantramteke369/m-coloring-problem" on https://jovian.ai [jovian] Uploading notebook.. [jovian] Uploading additional files... [jovian] Committed successfully! https://jovian.ai/siddhantramteke369/m-coloring-problem ```
Out[104]:
``'https://jovian.ai/siddhantramteke369/m-coloring-problem'``
5. Analyze the algorithm's complexity and identify inefficiencies, if any.
Time Complexity:
Since each node can be coloured using any of the m available colours, the total number of colour configurations possible are m^V.
So the time complexity is O(m^V).
In [18]:
``time_complexity_brute = 'O(m^V)'``
Space Complexity:
Recursive Stack of graphColoring(…) function will require O(V) space.
In [19]:
``space_complexity_brute = 'O(V)'``
Backtracking Algorithms
Backtracking is an algorithmic-technique for solving problems recursively by trying to build a solution incrementally, one piece at a time, removing those solutions that fail to satisfy the constraints of the problem at any point of time.
Backtracking Method:
The idea is to assign colors one by one to different vertices, starting from the vertex 0. Before assigning a color, check for safety by considering already assigned colors to the adjacent vertices i.e check if the adjacent vertices have the same color or not. If there is any color assignment that does not violate the conditions, mark the color assignment as part of the solution. If no assignment of color is possible then backtrack and return false.
7. Come up with a optimized solution for the problem. State it in plain English.
Steps:
1. Create a recursive function that takes the graph, current index, number of vertices and output color array.
2. If the current index is equal to number of vertices. Print the color configuration in output array.
3. Assign color to a vertex (1 to m).
4. For every assigned color, check if the configuration is safe, (i.e. check if the adjacent vertices do not have the same color) recursively call the function with next index and number of vertices
5. If any recursive function returns true break the loop and return true.
6. If no recusive function returns true then return None.
Let's save and upload our work before continuing.
In [24]:
``jovian.commit()``
```[jovian] Attempting to save notebook.. [jovian] Updating notebook "siddhantramteke369/m-coloring-problem" on https://jovian.ai [jovian] Uploading notebook.. [jovian] Uploading additional files... [jovian] Committed successfully! https://jovian.ai/siddhantramteke369/m-coloring-problem ```
Out[24]:
``'https://jovian.ai/siddhantramteke369/m-coloring-problem'``
8. Implement the solution and test it using example inputs. Fix bugs, if any.
In [28]:
``````class Graph():
def __init__(self, vertices):
self.V = vertices
self.graph = [[0 for column in range(vertices)] for row in range(vertices)]
def isSafe(self, v, color, c):
for i in range(self.V):
if self.graph[v][i] == 1 and color[i]==c:
return False
return True
def graphColorUtil(self, m, color, v):
if v == self.V:
return True
for c in range(1, m + 1):
if self.isSafe(v, color, c) == True:
color[v] = c
if self.graphColorUtil(m, color, v + 1) == True:
return True
color[v] = 0
def graphColoring(self, m):
color = [0] * self.V
if self.graphColorUtil(m, color, 0) == None:
return None
#print(color)
return color``````
In [21]:
``````g = Graph(4)
g.graph = test['input']['graph']
m = test['input']['m']
color = test['input']['color']
print('Output:', g.graphColoring(m))
print('Actual:', test['output'])``````
```Output: [1, 2, 3, 2] Actual: [1, 2, 3, 2] ```
Evaluate on all the test cases
In [22]:
``````passed, failed = 0, 0
for i in range(len(tests)):
gt = Graph(len(tests[i]['input']['color']))
gt.graph = tests[i]['input']['graph']
m = tests[i]['input']['m']
color = tests[i]['input']['color']
print('Output:', gt.graphColoring(m))
print('Actual:', tests[i]['output'])
if (gt.graphColoring(m) == tests[i]['output']):
print(gt.graphColoring(m) == tests[i]['output'], '\n')
passed += 1
else:
print(gt.graphColoring(m) == tests[i]['output'], '\n')
failed += 1
if (i == len(tests)-1):
print('Total:', passed+failed, 'Passed:', passed, 'Failed:', failed)``````
```Output: [1, 2, 3, 2] Actual: [1, 2, 3, 2] True Output: [1, 1, 1, 1] Actual: [1, 1, 1, 1] True Output: None Actual: None True Output: None Actual: None True Output: None Actual: None True Output: [1, 2, 1] Actual: [1, 2, 1] True Total: 6 Passed: 6 Failed: 0 ```
9. Analyze the algorithm's complexity and identify inefficiencies, if any.
Time Complexity:
Since each node can be coloured using any of the m available colours, the total number of colour configurations possible are m^V.
So time complexity is O(m^V).
The upperbound time complexity remains the same but the average time taken will be less.
In [23]:
``time_complexity_bt = 'O(m^V)'``
Space Complexity:
Recursive Stack of graphColoring(…) function will require O(V) space.
In [24]:
``space_complexity_bt = 'O(V)'``
In [ ]:
``jovian.commit()``
```[jovian] Attempting to save notebook.. ``` | 4,916 | 14,924 | {"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} | 3.734375 | 4 | CC-MAIN-2021-39 | latest | en | 0.789914 |
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4x-2y-8z+8w+7q=5
2x+4y-3z+6w+8q=9
6x-6y-6z+1w+2q=1
2x+4y-4z+9w+4q=5
Note, you must have n number of equations and n number of unknowns (in this case 5 equations and 5 unknowns)
Say you have 4 equations and 5 unknowns.. then you are under defined. Say you have 5 equations and 4 unknowns.. then you are over defined. So you want and equal number of equations and unknowns.
Anyway, to solve this problem throw the coefficients into a mattrix and put them into Reduced Row Echelon Form! This will solve it for any number of unknowns as long as you have enough equations to solve for them. The best part is that it involves simple addition and multiplication! | 355 | 1,215 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.53125 | 4 | CC-MAIN-2019-51 | longest | en | 0.916497 |
https://cs.stackexchange.com/questions/72224/perform-dijkstra-on-graph-with-negative-edges-by-adding-a-big-enough-constant-to | 1,713,868,319,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296818474.95/warc/CC-MAIN-20240423095619-20240423125619-00795.warc.gz | 166,689,405 | 40,428 | # Perform Dijkstra on graph with negative edges by adding a big enough constant to every edges
I know that this doesn't work because shortest path with a lot of edges may have bigger weight than a longer path with less edges. But what if, you keep track of the edges that our current weight path on each node and when we compare we subtract the number of edges on this path * the constant so that we get the original weight. I understand there is also the negative cycles problem, but is there an efficient algorithm that tell if a graph contains a negative cycle ?
• Right, I guess we could preprocessed it using Bellman-ford then perform this Dijkstra on the graph if it actually work. Mar 29, 2017 at 14:59
It's called Johnson's algorithm.
Instead of trying to keep track of the number of edges, it searches for a function $\delta:V\to \mathbb R$ so that for all edge $(v,v')$, $w(v,v')+\delta(v)-\delta(v')\ge 0$ and set that as the new weight $w'(e,e')$ of the edge $(e,e')$.
Then, for any path $p=v_1\cdot \dots\cdot v_n$ the $\delta$ values of all intermediate vertices cancel so that $w'(p)=w(p)+\delta(v_1)-\delta(v_n)$. And the shortest paths for $w$ and $w'$ therefore coincide.
Then remains the problem of finding such a $\delta$. The magic is that you can just add a new vertex $v_*$ and edges $(v_*,v)$ of weight $0$ for every vertex $v$, and then set $\delta (v)$ to be the weight of a shortest path $p_v$ from $v_*$ to $v$. Then the positiveness condition simply becomes $w(v,v')+w(p_v)-w(p_{v'})\ge 0$ which can be rewritten as $w(p_v)+w(v,v')\ge w(p_{v'})$ which is true because $p_v\cdot v'$ is a path from $v_*$ to $v'$ and $p_{v'}$ is a path from $v_*$ to $v'$ of minimal weight.
• But does the algorithm I mentioned above correct? Since Johnson a lot more complicated, I highly doubted that the algorithm above can do the same things as Johnson with less complexity. Mar 29, 2017 at 18:11
• @TienanhNguyen No. If you get the shortest path using $w'$, you may not get a shortest path for $w$ because of the length problem. And if you always keep track of the length and then use $w'(p)-|p|W$ whenever you compare, then you're just using $w$ which can have negative edges so Dijkstra may not work. Mar 30, 2017 at 11:04 | 619 | 2,246 | {"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} | 3.8125 | 4 | CC-MAIN-2024-18 | latest | en | 0.906114 |
http://commoncoregeometry.blogspot.com/2015/01/section-11-4-midpoint-formula-day-97.html | 1,526,844,186,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794863684.0/warc/CC-MAIN-20180520190018-20180520210018-00580.warc.gz | 62,030,708 | 14,631 | ## Thursday, January 22, 2015
### Section 11-4: The Midpoint Formula (Day 97)
Last night I tutored my geometry student. Section 4-5 of the Glencoe text is on ASA and AAS Congruence, so this clearly corresponds to Chapter 7 of the U of Chicago text.
I wanted to show him my worksheet from this blog -- the one where I proved the ASA Congruence Theorem using Common Core principles. At first he only took one glance at my proof outline and was intimidated when he saw that it was a ten-step proof. But then he had to finish his homework assignment, which required what turned out to be a nine-step proof.
Notice that the congruence chapters in the two texts are organized differently. In the U of Chicago, SSS, SAS, ASA, and AAS are all theorems that need to be proved. And so Section 7-2 contains these proofs, so that Section 7-3 has students prove the higher-level theorems that use these medium-level Congruence Theorems, and then Section 7-4 contains more complex proofs, the ones with overlapping triangles.
But Glencoe, like most pre-Common Core texts, gives SSS, SAS, and ASA as postulates. Because of this, the lessons go straight into having the students make higher-level proofs. And some of these proofs had overlapping triangles. So unlike the U of Chicago, with separate sections for the simpler and harder proofs, Sections 4-4 (SSS, SAS) and 4-5 (ASA, AAS) contained both simpler and harder proofs.
We look at Section 7-4 of the U of Chicago, and see Example 1 of a proof with overlapping triangles. Here, B lies on AD and C lies on AE.
Given: AC = ABAD = AE
Prove: Angle D = E
Statements Reasons
1. AC = AB, AD = AE 1. Given
2. Angle A = A 2. Reflexive Property of Congruence
3. ABE, ACD cong. tri. 3. SAS Congruence (steps 1, 2, 1)
4. Angle D = E 4. CPCTC
Now Glencoe contains a similar problem, except that the given is different. F is where BE and CD meet.
Given: AB, AC cong. seg., BDF, CDE cong. tri.
Prove: D, E cong. angles
Statements Reasons
1. BDF, CDE cong. tri. 1. Given
2. BD cong. CE 2. CPCTC
3. AB cong. AC 3. Given
4. BD = CE, AB = AC 4. Segment Congruence Theorem
5. AB + BD = AC + CE 5. Addition Property of Equality
6. AD = AE 6. Betweenness Theorem
7. Angle A cong. A 7. Reflexive Property of Congruence
8. ABE, ACD cong. tri. 8. SAS Congruence (steps 3, 7, 6)
9. Angle D cong. E 9. CPCTC
The reason that this proof is so long is that in many texts, including Glencoe, the statements "segment AB is congruent to AC" and "length AB is equal to AC," while equivalent, are not identical. In the U of Chicago, the Segment Congruence Theorem of Chapter 6 tells us that segments of equal lengths are congruent -- that is, that there is an isometry mapping one segment to the other. But after introducing the theorem, the book freely interchanges the congruence of the segments and the equality of the lengths, without mentioning it as a separate step in the proof.
Here the problem is that we need the Betweenness Theorem -- what Glencoe calls the Segment Addition Postulate -- and that only numbers, not figures, can be added. A length is a number, but a segment is a figure -- so we must convert the given segment congruence statement into an equal length statement before we can add anything, and then convert it back to segment congruence before we can invoke SAS. Most geometry teachers just handwave over the distinction between segment congruence and length equality -- as does the U of Chicago text, as do I on this blog, since the congruence symbol is hard to type in ASCII.
But the Glencoe text painfully includes every distinction between segment congruence and length equality -- as became apparent when I looked at one of the answers (a flow proof) in the back of the text. Moreover, my student's math teacher has been known to deduct half-points for leaving out such steps!
Even in the proof I posted above, notice that Step 6 is AD = AE, but I failed to convert this back to the congruence of segments AD and AE. Not only that, but technically speaking, the Segment Addition Postulate tells us that AB + BD = AD and AC + CE = AE. This should be the step that immediately follows step 5, and then the next step should derive AD = AE by substituting the above two equations into step 5.
So a fully complete proof would actually have eleven steps, not nine. The student and I came to the agreement that these nine steps should be sufficient for the teacher not to deduct any points -- but we won't know until today.
To me, having to make the painful distinction between segment congruence and length equality, as well as adding the substitution step, changes this problem to an exercise in formality, rather than help the student to see how Segment Addition can be used in an overlapping triangles proof. Even Dr. Wu admits that he skips steps in proofs when they don't enhance student understanding -- which is the primary goal of teaching proofs.
After this proof, I hardly had time to show my student the full proof of the ASA Congruence Theorem, as planned. What I decided to do was cut out two triangles satisfying the ASA condition, and then showed how I could slide, flip, and turn one triangle so that it coincided with the other. In particular, I showed him exactly how the Side-Switching Theorem was used to show that the reflection image of C' is F.
Afterward, my student noticed another proof in the Glencoe text and wondered whether he could use reflections, as I had to in the proof of the ASA Congruence Theorem, to prove the triangles in that problem congruent. Actually, he had a point there. Notice that in general, in a congruence proof, the first given triangle isn't necessarily a mere reflection image of the second. But in overlapping triangles, the triangles actually are mirror images of each other. Indeed, we look at the diagrams in U of Chicago Lesson 7-4, and notice that every single diagram in the lesson has reflectional symmetry!
In the diagram for this problem, point E lies on AB.
Given: Angles BCE, DCE congruent, Angles BEC, DEC congruent
Let's compare the traditional Glencoe proof with a Common Core/U of Chicago proof.
Glencoe Proof:
Statements Reasons
1. Angles BCE, bla, bla 1. Given
2. CE cong. CE 2. Reflexive Property of Congruence
3. BCE, DCE cong. tri. 3. ASA Congruence (steps 1, 2, 1)
4. BE cong. DE 4. CPCTC
5. AEB, CEB linear pair, 5. Definition of linear pair
AED, CED linear pair
6. Angle AEB cong. AED 6. Congruent Supplements Theorem
7. AE cong. AE 7. Reflexive Property of Congruence
8. AEB, AED cong. tri. 8. SAS Congruence (steps 4, 5, 6)
9. AB cong. AD 9. CPCTC
For the Common Core/U of Chicago proof, we'll start out just like the ASA Congruence Theorem proof and proceed from there:
Common Core/U of Chicago Proof:
Statements Reasons
1. Angles BCE, bla, bla 1. Given
2. Ray EB refl. over line AC is ED 2. Side-Switching Theorem
3. Ray CB refl. over line AC is CD 3. Side-Switching Theorem
4. The image of B lies on ED, CD 4. Figure Reflection Theorem
5. The image of B is D 5. Line Intersection Theorem
6. The image of A is A 6. Definition of reflection (point on mirror refl. to itself)
7. AB refl. over AC is AD 7. Figure Reflection Theorem
8. AB cong. AD 8. Definition of congruence (since the refl. is the isometry)
The Common Core proof is slightly shorter -- mainly because of the wasted step in the Glencoe about showing that the angles form a linear pair. And I forget whether Glencoe would make the student add yet another step, that the angles in a linear pair are supplementary! It's because of these extra formalistic steps that make the reflectional proof shorter.
Also, this week I've been subbing at the same continuation school for the second week in a row. One student is preparing to take the national test known as the Armed Services Vocational National Battery -- or ASVAB, for short. He is having trouble passing the ASVAB because it covers much material from the second semester of Algebra I, whereas he has just barely finished the first semester of algebra.
Notice that the ASVAB has two test formats -- computerized and written -- and it is given by the federal government -- specifically the military.
Section 11-4 of the U of Chicago text covers the other important formula of coordinate geometry -- the Midpoint Formula. As the text states, this is one of the more difficult theorems to prove.
In fact, the way we prove the Midpoint Formula is to use the Distance Formula to prove that, if M is the proposed midpoint of PQ, then both PM and MQ are equal to half of PQ. The rest of the proof is just messy algebra to find the three distances. The U of Chicago proof uses slope to prove that M actually lies on PQ. Since we don't cover slope until next week, instead I just use the Distance Formula again, to show that PM + MQ = PQ, so that M is between P and Q. The algebraic manipulation here is one that's not usually used -- notice that instead of taking out the four in the square root of 4x^2 to get 2x (as is done in the last exercise, the review question), but instead we take the 2 backwards inside the radical to get 4, and then distribute that 4 so that it cancels the 2 squared in the denominator.
I don't have nearly as much to say about the Midpoint Formula as the Pythagorean Theorem and its corollary, the Distance Formula. To me, it's a shame that I had to bury the Pythagorean Theorem in the middle of this Coordinate Geometry unit. The main theorem named for a mathematician really deserves its own lesson, but due to time constraints I had to combine it with the Distance Formula the way I just did it in yesterday's lesson. | 2,461 | 9,995 | {"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} | 4.125 | 4 | CC-MAIN-2018-22 | latest | en | 0.918823 |
https://www.wiziq.com/online-tests/44663-speed-time-distance-for-gmat-gre-bat | 1,660,949,784,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882573849.97/warc/CC-MAIN-20220819222115-20220820012115-00479.warc.gz | 896,245,128 | 35,520 | # Speed, Time & Distance for GMAT/GRE/BAT Online Test
A motorboat, whose speed in 15 km/hr in still water goes 30 km downstream and comes back in a total of 4 hours 30 minutes. The speed of the stream (in km/hr) is:
A boat running upstream takes 8 hours 48 minutes to cover a certain distance, while it takes 4 hours to cover the same distance running downstream. What is the ratio between the speed of the boat and speed of the water current respectively?
A man's speed with the current is 15 km/hr and the speed of the current is 2.5 km/hr. The man's speed against the current is:
A boat can travel with a speed of 13 km/hr in still water. If the speed of the stream is 4 km/hr, find the time taken by the boat to go 68 km downstream.
In a 500 m race, the ratio of the speeds of two contestants A and B is 3 : 4. A has a start of 140 m. Then, A wins by:
A and B take part in 100 m race. A runs at 5 kmph. A gives B a start of 8 m and still beats him by 8 seconds. The speed of B is:
In a 100 m race, A can give B 10 m and C 28 m. In the same race B can give C:
The length of the bridge, which a train 130 metres long and travelling at 45 km/hr can cross in 30 seconds, is:
A train 125 m long passes a man, running at 5 km/hr in the same direction in which the train is going, in 10 seconds. The speed of the train is:
A train running at the speed of 60 km/hr crosses a pole in 9 seconds. What is the length of the train?
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This Test consists of 10 questions on Speed, Time & Distance for BAT/GRE/GMAT or any other entrance exam. All questions consists of 4 options & there is a time limit of 15 minutes to complete these 10 questions. There will be more upcoming tests so that students can prepare well for there targeted exams. Best of Luck.
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Price:\$3.77 | 575 | 2,154 | {"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} | 3.59375 | 4 | CC-MAIN-2022-33 | latest | en | 0.925131 |
http://crispexcel.com/excel-rand-and-randbetween-functions/ | 1,527,462,342,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794870497.66/warc/CC-MAIN-20180527225404-20180528005404-00219.warc.gz | 72,527,956 | 9,873 | # Excel RAND() and RANDBETWEEN Functions
Human being cannot consciously generate random numbers sequences as they tend to avoid/favor picking specific numbers yet there is so much application of randomness in life.
Random numbers are useful in a number of ways, such as generating data encryption keys,GUIDS, Strong Passwords, selecting random samples or creating survey groups, generating lottery numbers e.t.c
Scientist, auditors, cryptographers, betting firms, polling firms, lottery players endeavor very hard to ensure the numbers they use are really random numbers as shown below by Scott Adams’s Dilbert cartoon.
If you want to be sure the numbers you generate are random, use excel RAND and RANDBETWEEN functions.
►RAND function returns a random number between 0 and 0.99999999999999999. The function can generate upto 17 decimal places precision.
`Syntax = RAND()`
►RANDBETWEEN function returns values in a range you specify.
```Syntax = RANDBETWEEN(bottom, top)→→bottom(smallest integer) & top (largest integer)
```
### NOTES:
►RAND function can also be used to generate random numbers between zero and any k number.
`= RAND() * k+1`
For example to generate random numbers between 0 and 13,
`=RAND() * 13+1`
►RAND can also be used to generate random numbers in a given floor and a ceiling value just like RANDBETWEEN.
For example to generate random numbers between 34 and 64
`=RAND() * (64-34)+34`
►RANDBETWEEN is prone to generate duplicates, therefore if you want to generate unique numbers in given range use below;
• Create a helper row with random numbers use RAND() then use MATCH and LARGE to fetch the ROW numbers
`=MATCH(LARGE(\$A\$2:\$A\$20,ROWS(\$C\$2:C12)),\$A\$2:\$A\$20,0)`
In this formula, MATCH is used to return the row which contains the Largest k number.
• If you do not want to use a helper row, then you can use array formula shown below
```{=IFERROR(LARGE(ROW(INDIRECT(\$J\$3&":"&\$J\$4))*NOT(COUNTIF(\$F\$5:F5,ROW(INDIRECT(\$J\$3&":"&\$J\$4))))
,RANDBETWEEN(1,\$J\$4-\$J\$3-ROW(A1)+2)),"")}``` | 530 | 2,048 | {"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} | 3.828125 | 4 | CC-MAIN-2018-22 | latest | en | 0.722706 |
http://polymathematics.typepad.com/polymath/2006/06/and_finally.html | 1,534,447,673,000,000,000 | text/html | crawl-data/CC-MAIN-2018-34/segments/1534221211167.1/warc/CC-MAIN-20180816191550-20180816211550-00367.warc.gz | 318,231,191 | 17,895 | We write that as 1/8=.125 with no problem (although it clearly could also be written as .125000...).
Or .124999...
:)
Dave S.: And in fact, I've definitely seen my fair share of proofs involving decimal expansions that start with "assume that the decimal expansion of x doesn't end with infinitely many zeros"!
Here, here polymath. Now move on to some other interesting math, please.
Well the thing I cannot understand is how someone can admit that they don't have an extensive education in mathematics and are very much a math novice but assert the fact that they know what they are talking about and that all well-educated mathematicians are wrong. I guess it goes along with the same ignorance that people use towards doctors when they believe that the doctor doesn't know what he/she is talking about.
Sigh...oh well
The finding a number between 0.99|9 and 1.00|0 really is the simplest way to argue the case.
I'll wait around to see if there are still objectors...chuckle.
We should clearly teach the controversy in our schools here. Who needs experts when we have democracy?
"but even I know you're waay wrong"
That's hilarious! Good job btw, polymath. You tell 'em.
I entered comments way back. Just for the record - for you as a math teacher -from me as a former math teacher: I still think that the problem is caused because they never learned the difference between number and numeral. Making that distinction explicit makes a lot of arithmetic and algebra much easier.
And, by the way, since the standard way to change a fraction into a decimal is with the standard long division algorithm, I point out the following: it is just an algorithm, it can be modified. Consider calculating the decimal for 9/9. As you did above: 9 goes into 9 one time - put a 1 above the line, then 1 goes inot zero zero times, put a zero after the decimal point - that's standard. But consider this, suppose you don't notice that 9 goes into 9 one time, so you look at how many times 9 goes into 90. Since you can only use a single digit, it goes 9 times. Put a 9 in the first place to the right of the decimal point. 9 times 9 is 81. 90 - 81 = 9. Bring down the next zero. 9 goes into 90 9 times again, and again and again. What do you get as the quotient - .999...
Cute?
Good one Karl, I had thought about that too. In fact, that would work with any long division problem that comes out to a whole number. You could always end up with one number down, followed by .999...
Karl,
I had noticed that too. The reason I didn't post it is that it's not the "standard" version of long division, so I suspect it would draw criticism from people who don't see that it's correct. But it is an excellent point.
I came over from Mathforge to see the brouhaha. It reminds me of a humbling experience I had a while ago when I said to one of my engineering friends that mathematics was largely free of pseudoscience and crackpots. Unfortunately I had spoken much too soon and was immediately lead to a round-trip tour of the kookiest mathematics sites on the internet. The most amusing was definitely the site where the claim was made that "mathematicians are biased to the positive" and therefore -1 * -1 should be equal to -1. After rolling around laughing for several minutes I was forced to write an (quick) axiomatic development of the real number system. The strange thing that I keep encountering is that while to us it seems completely obvious that -1 * -1 = 1 (it is really forced on us by the axioms chosen for the real number system - and besides, could you imagine what it would do to analysis if it were suddenly changed ;-)) it really isn't very obvious to the lay person. In explaining some of what I do to other people I frequently encounter people who are quite literally still stuck in Zeno's paradoxes. The limit process is an utterly incomprehensible concept to many people. The old example of walking across a room half the remaining distance at a time still stumps people. I've found that people simply cannot disconnect what they believe to the rigour of mathematics. To them the mathematical concept of infinity seems wrong and obscure. I really applaud your blog for taking this often controversial topic on and bringing some clarity to people. For the insult slinging people out there who believe this person to be lying about the topic, let me make a recommendation. Pick up "A Course in Pure Mathematics" by Hardy, "Foundations of Analysis" by Landau and then when you've read both in detail, come back in a few years and try to argue your point again. Until then, trust the mathematicians.
Hey, that's a good refutation. I would really like to know more about the hyperreal numbers, but I'm currently just researching infinite set theory... so I suppose I'll get to that in due time. This whole .999... = 1 "debate" has really gotten me interested in math again. Thanks polymath!
Just a quick note about hyperreals -- they are a field extension of the real numbers, which in particular means that if .999...=1 is true in the reals, it's also true in the hyperreals. In particular, the algebraic proof (and I think the infinite series proof) holds independent of what field extension of the reals you consider.
Of course, I'm not well-versed in hyperreals, as they're far outside my research area; I just know a fair bit about field extensions.
Egad. Wow. Some people's ability to convince themselves of things beyond all evidence to the contrary is truly stunning. It's hard for me to believe that that many people posted to argue against the fact that 0.999.... equals 1--I guess there are things about infinity that are just too counterintuitive for some people to be willing to accept them. (And, of course, the irony is that some of those commenters accused you of not understanding infinity...wow.)
Anyway, it's my first time visiting this blog--came here from your recent reply to a post in Good Math, Bad Math--but I'll probably be by again. And you have my sympathies regarding your having to deal with so many nutcases making bad arguments against basic mathematics...
Looks like I've kicked the debate off again on the Yahoo answers website . Link....
Also, the best answer so far (from the YAHOO site):
Question:
"Does 0.99999... recurring = 1????"
Yes.
Try finding a number between the two!
This is an interesting discussion, but is, at its heart, irrelevant. Let me opine on the reason why.
The basis for argument that .999...=1 is sound, if you ignore the basic flaw in any argument that tries to express fractions in decimal equivalents. A fraction is essentially an external reference into some form of unit. Such as 2/3 of a block of cheese. The size of the resulting fraction is only rational when viewed in relation to the assigned unit.
Yet still, we persist in trying to assign decimal meaning to a fractional representation. This is erroneous. Yet, for as long as we do so, we must then accept that .999...=1 for all of the well reasoned approaches outlined by Mr. Polymath.
But it still "sits wrongly" with those who understand that there is an underlying flaw ... but just can't put their finger on it. They understand that .333... does not equal 1/3, nor do any of the other fractions equate to their most common decimal expressions.
This problem reares its ugly head in other ways. When I was teaching network engineering to young EE majors, this problem frequently appeared in computations designed to originate in one type of unit and conclude in another. The students frequently forgot to manipulate the units as an entity in their own right. Which underscores my previous point on fractions being an external reference, not something at absolute as a real number on a number line. (Note that this becomes even more complex when working with electrical power logarithms, deciBels, and half power points.)
A final example would be the common representation of Internet Protocol addresses. A simple 32 bit binary number used in a logical AND function with a 32 bit binary mask to form the truth table resultant of the ANDed function, which is then applied to the primary address for the purpose of "local reading" of the delivery address. While logically this is a simple solution, and equates very well to an analogy of the U.S. Postal Mail system ... we just couldn't leave it alone! We had to first break these 32 bits up into four sections of 8 bits each (since we already had the "Byte" concept going ... but that could be another post) and called them Octets (even though they were not Octal). We then further compounded this confusion by representing each of these octets as a decimal expression! Now, try to teach this "simplification" to a room full of supposedly intelligent college seniors and post-grads. If we teach our mathematicians, scientists, and engineers to think in a logical and empirical manner ... why do we continue to try to make math and science "neater" by tying up perceived "loose ends". There are no loose ends ... merely theorms without (as yet) sufficient proofs.
As the saying goes, "God doesn't throw dice with the Universe." But I think this is because the Universe would clean God out of folding money!
Until we get the neat-niks out of math and science, we will continually be faced with issues such as these.
Irrelevant as they may be.
"They understand that .333... does not equal 1/3, nor do any of the other fractions equate to their most common decimal expressions."
Guh? In what sense do you mean that?
A fraction is a rational number. The rational numbers are a (dense) subset of the real numbers. All real numbers have (not necessarily finite) decimal representations. And representations in any other base, for that matter. Where's the "flaw"?
Here's my opinion on this.
0,999.... is not equal 1. Why? Because it defies common sence. No matter how many nines you put after that zero, it's not gonna equal 1.
However, the difference between 0,999... and 1 is infinitely small, which is mathematically equal with 0.
Therefore, 0,999... = 1
"Yet still, we persist in trying to assign decimal meaning to a fractional representation. This is erroneous."
A decimal IS a fraction! That's what it is. 0.5 is 5/10. 0.125 is 125/1000. They are one and the same.
By the way in other number systems, you have the same kind of phenominon. In octal, .777... = 1. In hexadecimal, .FFF... = 1. In binary, .111.. = 1.
Any argument that opens with the position from "common sense" is not by any stretch of the imagination a proof. A proof of something one way or the other will trump any common sense argument 100% of the time.
I. Will. Never. Ever. Again. Make the first comment on one of your math posts.
Ever.
I can't comment on Robinson's non-standard analysis as I have never studied it but at least in the non-standard analysis of Detlef Laugwitz 0.99999... does not equal 1! There is a very readable explanation as to why not as an apendix to the doctoral thesis of Detlef Spalt "Vom Mythos der Mathematischen Vernunft". Unfortunately only available in German. Laugwitz defines infinitesimals simple by introducing the arithmetic operations for use with them in analogy to the introduction of 'i' and the complex numbers thereby creating a perfectly valid calculus of infintesimal numbers in which 0.999... does not equal 1.
i think he is right. the evidence given that .999repeating = 1 is much more convincing than the arguments that its not.
The comments to this entry are closed.
## Other blogs I like
• EvolutionBlog
He writes mostly about evolution, but he's a math guy.
Scienceblogs finally has a math guy!
• Kung Fu Monkey
A very smart, high-profile screen writer and comic with sensible politics and an amazing ability to rant
• Math Spectrometer
My ideas about life, teaching, and politics
• Pharyngula
Biology, lefty politics, and strident anti-Intelligent Design | 2,699 | 11,896 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.96875 | 4 | CC-MAIN-2018-34 | latest | en | 0.973662 |
https://www.gradesaver.com/textbooks/math/trigonometry/CLONE-68cac39a-c5ec-4c26-8565-a44738e90952/chapter-6-inverse-circular-functions-and-trigonometric-equations-section-6-4-equations-involving-inverse-trigonometric-functions-6-4-exercises-page-286/34 | 1,709,614,621,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707948217723.97/warc/CC-MAIN-20240305024700-20240305054700-00023.warc.gz | 790,208,169 | 12,531 | # Chapter 6 - Inverse Circular Functions and Trigonometric Equations - Section 6.4 Equations Involving Inverse Trigonometric Functions - 6.4 Exercises - Page 286: 34
$\frac{12}{5}$
#### Work Step by Step
Given: $arctanx=arccos\frac{5}{13}$ Consider $arccos\frac{5}{13}=P$ $cos P=\frac{5}{13}$ Apply trigonometric identity for cosine. $cosx=\frac{adj}{hyp}=\frac{5}{13}$ Thus, $opp=\sqrt {13^{2}-5^{2}}=\sqrt {169-25}=12$ Likewise, $tanP=\frac{opp}{adj}=\frac{12}{5}$ $arctanx=P$ gives $x=tanP$ Hence, $x=\frac{12}{5}$
After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback. | 230 | 681 | {"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} | 4.125 | 4 | CC-MAIN-2024-10 | latest | en | 0.573308 |
http://www.docstoc.com/docs/141193469/Introduction-to-Statistics%E2%80%A6 | 1,371,629,242,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368708144156/warc/CC-MAIN-20130516124224-00027-ip-10-60-113-184.ec2.internal.warc.gz | 410,932,387 | 13,743 | # Introduction to Statistics…
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``` Introduction to Statistics
Chapter 1
Introduction to Statistics
Lecture 2 1
Introduction to Statistics… Chapter 1
Overview
A common goal of studies and surveys and other
data collecting tools is to collect data from a small
part of a larger group so we can learn something
In this section we will look at some of the ways to
describe data.
Lecture 2 2
Introduction to Statistics… Chapter 1
Definitions
Data
Observations (such as measurements, genders,
survey responses) that have been collected
Statistics
a collection of methods for planning studies and
experiments, obtaining data, and then organizing,
summarizing, presenting, analyzing, interpreting, and
drawing conclusions based on the data
Lecture 2 3
Introduction to Statistics… Chapter 1
Definitions
Population
the complete collection of all elements (scores, people,
measurements, and so on) to be studied; the collection
is complete in the sense that it includes all subjects to
be studied
Census
Collection of data from every member of a population
Sample
Sub collection of members selected from a population
Lecture 2 4
Introduction to Statistics… Chapter 1
Lecture 2 5
Introduction to Statistics… Chapter 1
Parameter
a numerical measurement describing some
characteristic of a population.
Population
Parameter
Lecture 2 6
Introduction to Statistics… Chapter 1
Statistic
a numerical measurement describing some
characteristic of a sample.
Sample
Statistic
Lecture 2 7
Introduction to Statistics… Chapter 1
Common Summary Measures
Sample Statistic Population Parameter
Mean X
Standard
Deviation S
Variance S2 2
Lecture 2 8
Introduction to Statistics… Chapter 1
Types of Data
Data
Lecture 2 9
Introduction to Statistics… Chapter 1
Quantitative data
Numbers representing counts or measurements.
Example: The income of college graduates
Qualitative (or attribute) data
can be separated into different categories that are distinguished
by some nonnumeric characteristic
Example: The genders (male/female)
Lecture 2 10
Introduction to Statistics… Chapter 1
Working with Quantitative Data
Quantitative data can further be described by
distinguishing between discrete and continuous
types.
Lecture 2 11
Introduction to Statistics… Chapter 1
Discrete data
Result when the number of possible values is
either a finite number or a ‘countable’
number
(i.e. the number of possible values is
0, 1, 2, 3, . .)
Example: The number of Lumps that a
factory can produce
Lecture 2 12
Introduction to Statistics… Chapter 1
Continuous (numerical) data
Result from infinitely many possible values that
correspond to some continuous scale that covers
a range of values without gaps, interruptions, or
jumps
Lecture 2 13
Introduction to Statistics… Chapter 1
Uses & Abuses of Statistics
Lecture 2 14
Introduction to Statistics… Chapter 1
Voluntary response sample
(or self-selected sample)
one in which the respondents themselves
decide whether to be included
In this case, valid conclusions can be made only
about the specific group of people who agree to
participate.
Lecture 2 15
Introduction to Statistics… Chapter 1
Misuse # 2- Small Samples
Conclusions should not be based on samples that
are far too small.
Example: Basing a school suspension rate on a
sample of only three students
Lecture 2 16
Introduction to Statistics… Chapter 1
Misuse # 3- Graphs
To correctly
interpret a graph,
you must analyze
the numerical
information given
in the graph, so as
not to be misled by
the graph’s shape.
Lecture 2 17
Introduction to Statistics… Chapter 1
Misuse # 4- Pictographs
Part (b) is
designed to
exaggerate
the
difference by
increasing
each
dimension in
proportion to
the actual
amounts of
oil
Lecture 2 18
consumption
Introduction to Statistics… Chapter 1
Misuse # 5- Percentages
sometimes used. For example, if you take
100% of a quantity, you take it all. 110% of
an effort does not make sense.
Lecture 2 19
Introduction to Statistics… Chapter 1
Sample Size
Lecture 2 20
Introduction to Statistics… Chapter 1
Sample Size
use a sample size that is large enough to see the
true nature of any effects and obtain that sample
using an appropriate method, such as one based
on randomness
Lecture 2 21
Introduction to Statistics… Chapter 1
Random Sample
Members of the population are selected in
such a way that each individual member has an
equal chance of being selected
Lecture 2 22
Introduction to Statistics… Chapter 1
Methods of Sampling
Random
Sampling
selection so that
each
individual
member has an
equal chance of
being selected
Lecture 2 23
Introduction to Statistics… Chapter 1
Systematic Sampling
Select some starting point and then
select every k th element in the population
Lecture 2 24
Introduction to Statistics… Chapter 1
Convenience Sampling
use results that are easy to get
Lecture 2 25
Introduction to Statistics… Chapter 1
Stratified Sampling
subdivide the population into at
least two different subgroups that share the same
characteristics, then draw a sample from each subgroup
Lecture 2 26
Introduction to Statistics… Chapter 1
Cluster Sampling
divide the population into sections
(or clusters); randomly select some of those clusters; choose all
members from selected clusters
Lecture 2 27
Introduction to Statistics… Chapter 1
Methods of Sampling - Summary
Random
Systematic
Convenience
Stratified
Cluster
Lecture 2 28
```
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Other docs by xiangpeng | 1,391 | 6,077 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.984375 | 4 | CC-MAIN-2013-20 | longest | en | 0.800885 |
https://tomrocksmaths.com/2020/02/11/funbers-31-32-and-33/?replytocom=31073 | 1,639,055,513,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964364169.99/warc/CC-MAIN-20211209122503-20211209152503-00084.warc.gz | 635,020,318 | 53,603 | # Funbers 31, 32 and 33
Fun facts about numbers that you didn’t realise you’ve secretly always wanted to know…
## 31 – Thirty-one
Hands up if you like ice cream? And your favourite brand? I’m not sure I could pick a favourite myself, but Baskin-Robbins is certainly up there. They have a total of 31 flavours of ice cream which means the name of their shops in Japan literally translates as ’31 Ice Cream’. In theory a great idea, but what if they discover a magical new thirty-second flavour…
Aside from frozen goods, thirty-one is also the number of teams in the National Hockey League, with 24 coming from the US and 7 from their Northern neighbours Canada. Each season, the teams battle it out to win the Stanley Cup — the oldest trophy to be awarded in professional sport in North America, and also the one with the infamously large base (see below). Originally, key members of the winning team were engraved on the base, which means its grown a fair few inches over the past 126 years. However, these days the oldest band is removed and replaced with a new one to prevent the trophy from getting any bigger. Less fun no doubt, but perhaps sensible given its already considerable size…
Credit: Michael Miller
Thirty-one is also a Mersenne Prime — the third such one in fact. A Mersenne Prime is a prime number that can be expressed as exactly one less than a power of two: 2^n — 1 for some positive whole number n. To get 31, we take n=5: 2⁵ = 2 x 2 x 2 x 2 x 2 = 32 minus 1 gives 31. A perhaps surprisingly large (and possibly infinite) amount of prime numbers take this form, with the current largest known prime number also being a Mersenne Prime: 2⁸²⁵⁸⁹⁹³³–1 = a number with 24,862,048 digits, aka too many for me to write out here!
## 32 — Thirty-two
Sticking with maths, thirty-two has the very nice property that it can be written as 1 to the power 1 plus 2 to the power 2 plus three to the power three, or in its neatest form: 1¹ + 2² + 3³ = 32. Here’s a challenge for you: can you work out the next largest number that follows the same pattern?
32 can also be written as 2⁴ + 4² = 32 which makes it a Leyland Number. Any number that can be written using two other numbers x and y, in the pattern x to the power y plus y to the power x, is classified as a Leyland Number. Here, we take x = 2 and y = 4 to get: 2⁴ = 2 x 2 x 2 x 2 = 16, plus 4² = 4 x 4 = 16, giving a total of 16 + 16 = 32. Other Leyland Numbers include: 8, 17, 54, 57 and 100 — I’ll leave it to you to figure out the specific values of x and y needed to satisfy the formula x^y + y^x = Leyland Number for each of the cases above.
Outside of the mathematical world, thirty-two is the number of completed piano sonatas by Ludwig van Beethoven; the number of black (or white) squares, and total number of pieces on a chessboard; the number of teeth generally found in an adult human; and the number of described physical characteristics of the historical Buddha, according to the text of the Pāli Canon in the Theravada Buddhist tradition. It’s a pretty long list, but I think it is best enjoyed in its entirety, so here you go:
1. Level feet
2. Thousand-spoked wheel sign on feet
3. Long, slender fingers
4. Pliant hands and feet
5. Toes and fingers finely webbed
6. Full-sized heels
7. Arched insteps
8. Thighs like a royal stag
9. Hands reaching below the knees
10. Well-retracted male organ
11. Height and stretch of arms equal
12. Every hair-root dark coloured
13. Body hair graceful and curly
14. Golden-hued body
15. Ten-foot aura around him
16. Soft, smooth skin
17. Soles, palms, shoulders, and crown of head well-rounded
18. Area below armpits well-filled
19. Lion-shaped body
20. Body erect and upright
21. Full, round shoulders
22. Forty teeth
23. Teeth white, even, and close
24. Four canine teeth pure white
25. Jaw like a lion
26. Saliva that improves the taste of all food
28. Voice deep and resonant
29. Eyes deep blue
30. Eyelashes like a royal bull
31. White ūrṇā curl that emits light between eyebrows
32. Fleshy protuberance on the crown of the head
## 33 — Thirty-three
Let’s start with the bad. Thirty-three is one of the symbols of the Ku Klux Klan, with K being the 11th letter of the alphabet and 3 x 11 or 3 K’s giving 33. It is also believed to be the age of Jesus when he was crucified by the Romans.
On a more positive note, 33 is the longest winning streak ever recorded in NBA history, which was achieved by the Los Angeles Lakers in the 1971–72 season. We also find 33 vertebrae in a normal human spine when the bones that form the coccyx (the tail-like part at the bottom) are counted individually.
Now I don’t say this often — mainly because I think I’m supposed to be impartial when writing these — but this next fun fact is one of my all-time favourites. Long playing records, or LPs as they are more commonly known, are referred to as 33’s in the record industry, because they rotate 33 and a third times per minute when playing on a gramophone. So, next time you see a record player you know what to do… | 1,385 | 5,064 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.75 | 4 | CC-MAIN-2021-49 | latest | en | 0.964601 |
https://www.enotes.com/homework-help/please-make-sure-that-distance-value-1-7mm-not-245669 | 1,674,857,470,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764495012.84/warc/CC-MAIN-20230127195946-20230127225946-00242.warc.gz | 745,428,284 | 19,126 | # What is the area of the plates if capacitance is 1 F in the following case: Consider a parallel plate capacitor with a distance 1.7 mm . What would be the area A of each plate if the capacitance is C = 1 F? Use: ε = 8.85 x 10^-12 F/m . Let the area be in square kilometers accurate to 2 significant places.
The capacitance of a parallel plate capacitor which has plates of area A, a distance of separation of the plates equal to d and the permittivity epsilon is equal to e, is given by the formula
C = e*A / d
Here we are given that the capacitance of the capacitor is 1 F. The plates are separated by a distance equal to 1.7 mm and e = 8.85*10^-12. Substituting the given values in the formula we get
1 = 8.85*10^-12 * A / 1.7*10 ^-3
=> A = 1.7* 10^-3 / 8.85* 10^-12
=> A = 192.09 * 10^6 m^2
=> 192.09 km^2
The area of the plates in square kilometers is 192.09.
Approved by eNotes Editorial Team | 280 | 908 | {"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} | 4.125 | 4 | CC-MAIN-2023-06 | latest | en | 0.862888 |
https://www.univerkov.com/an-excess-of-hydrochloric-acid-was-added-to-300-00-g-of-potassium-carbonate/ | 1,716,590,137,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058751.45/warc/CC-MAIN-20240524214158-20240525004158-00159.warc.gz | 922,455,149 | 6,381 | # An excess of hydrochloric acid was added to 300.00 g of potassium carbonate.
An excess of hydrochloric acid was added to 300.00 g of potassium carbonate. Determine the mass of the salt obtained and the volume of gas evolved.
To solve the problem, let’s compose an equation, arrange the coefficients:
CaCO3 + 2HCl = CaCl2 + H2O + CO2 – exchange reaction, obtained salt calcium chloride, carbon dioxide;
Let’s make calculations using the formulas:
M (CaCO3) = 100 g / mol;
M (CaCl2) = 111 g / mol;
M (CO2) = 44 g / mol.
Determine the amount of moles of calcium carbide:
Y (CaCO3) = m / M = 300/100 = 3 mol.
According to the reaction equation, the amount of moles of calcium carbonate and calcium chloride is 1 mole, which means that Y (CaCl2) = 3 moles.
We calculate the mass of the CaCl2 salt by the formula:
m (CaCl2) = Y * M = 3 * 111 = 333 g.
We find the volume of carbon dioxide, taking into account the data of the reaction equation:
Y (CO2) = 3 mol;
V (CO2) = 3 * 22.4 = 67.2 liters.
Answer: the mass of calcium chloride is 333 g, the volume of carbon dioxide is 67.2 liters.
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https://edurev.in/course/quiz/attempt/1643_Test-Alternating-Current--Competition-Level-1-/030fab52-2649-4bea-be89-92cff1961869 | 1,601,284,342,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600401598891.71/warc/CC-MAIN-20200928073028-20200928103028-00636.warc.gz | 365,242,572 | 39,610 | Courses
# Test: Alternating Current (Competition Level 1)
## 30 Questions MCQ Test Physics Class 12 | Test: Alternating Current (Competition Level 1)
Description
This mock test of Test: Alternating Current (Competition Level 1) for Class 12 helps you for every Class 12 entrance exam. This contains 30 Multiple Choice Questions for Class 12 Test: Alternating Current (Competition Level 1) (mcq) to study with solutions a complete question bank. The solved questions answers in this Test: Alternating Current (Competition Level 1) quiz give you a good mix of easy questions and tough questions. Class 12 students definitely take this Test: Alternating Current (Competition Level 1) exercise for a better result in the exam. You can find other Test: Alternating Current (Competition Level 1) extra questions, long questions & short questions for Class 12 on EduRev as well by searching above.
QUESTION: 1
Solution:
QUESTION: 2
Solution:
QUESTION: 3
### The instantaneous value of current in an ac circuit is I = 2 sin (100pt + p/3) A. The current at the beginning (t = 0) will be
Solution:
QUESTION: 4
If instantaneous value of current is I = 10 sin (314 t) A, then the average current for the half cycle will be
Solution:
QUESTION: 5
In a circuit an a.c. current and a d, c. current are supplied together. The expression of the instantaneous current is given as i = 3 + 6 sin wt. Then the rms value of the current is
Solution:
QUESTION: 6
The emf and the current in a circuit are E = 12 sin (100pt) ; I = 4 sin (100pt + p / 3) then
Solution:
QUESTION: 7
If the frequency of alternating potential is 50Hz then the direction of potential, changes in one second by
Solution:
QUESTION: 8
The value of alternating e.m.f. is e = 500 sin 100pt , then the frequency of this potential in Hz is
Solution:
QUESTION: 9
The domestic power supply is at 220 volt. The amplitude of emf will be
Solution:
QUESTION: 10
The average value or alternating current for half cycle in terms of I0 is
Solution:
QUESTION: 11
Sinusoidal peak potential is 200 volt with frequency 50Hz. It is represented by the equation
Solution:
QUESTION: 12
RMS value of ac i = i1 cos wt + i2 sin wt will be
Solution:
I=(I1)cosωt+(I2)sinωt
(I2)mean=I12cos2ωt+I22sin2ωt+I1I2cosωtsinωt
I12×(1/2)×i22×(1/2)+2I1I2×0
Irms=√(i2)mean=√(I12+I22/2)=(1/√2)( I12+I22)1/2
QUESTION: 13
The phase difference between the alternating current and voltage represented by the following equation I = I0 sin wt, E = E0 cos (wt + p / 3), will be
Solution:
QUESTION: 14
The inductive reactance of a coil is 1000W. If its self inductance and frequency both are increased two times then inductive reactance will be
Solution:
QUESTION: 15
In an L.C.R series circuit R = 1W, XL = 1000W and XC = 1000W. A source of 100 m.volt is connected in the circuit the current in the circuit is
Solution:
QUESTION: 16
A coil of inductance 0.1 H is connected to an alternating voltage generator of voltage E = 100 sin (100t) volt. The current flowing through the coil will be
Solution:
QUESTION: 17
Alternating current lead the applied e.m.f. by p/2 when the circuit consists of
Solution:
QUESTION: 18
A coil has reactance of 100W when frequency is 50Hz. If the frequency becomes 150Hz, then the reactance will be
Solution:
QUESTION: 19
A resistance of 50W, an inductance of 20/p henry and a capacitor of 5/p mF are connected in series with an A.C. source of 230 volt and 50Hz. The impedance of circuit is
Solution:
QUESTION: 20
The potential difference between the ends of a resistance R is VR between the ends of capacitor is VC = 2VR and between the ends of inductance is VL = 3VR, then the alternating potential of the source in terms of VR will be
Solution:
QUESTION: 21
The percentage increase in the impedance of an ac circuit, when its power factor changes form 0.866 to 0.5 is (Resistance constant)
Solution:
QUESTION: 22
In a series resonant L_C_R circuit, if L is increased by 25% and C is decreased by 20%, then the resonant frequency will
Solution:
QUESTION: 23
Which of the following statements is correct for L_C_R series combination in the condition of resonance
Solution:
QUESTION: 24
An ac circuit resonates at a frequency of 10 kHz. If its frequency is increased to 11 kHz, then
Solution:
QUESTION: 25
In an ac circuit emf and current are E = 5 cos wt volt and I = 2 sin wt ampere respectively. The average power dissipated in this circuit will be
Solution:
QUESTION: 26
A choke coil of negligible resistance carries 5 mA current when it is operated at 220 V. The loss of power in the choke coil is –
Solution:
QUESTION: 27
In an A.C. circuit, a resistance of 3W, an inductance coil of 4W and a condenser of 8W are connected in series with an A.C. source of 50 volt (R.M.S.). The average power loss in the circuit will be
Solution:
QUESTION: 28
Two bulbs of 500 watt and 300 watt work on 200 volt r.m.s. the ratio of their resistances will be –
Solution:
QUESTION: 29
Three bulbs of 40, 60 and 100 watt are connected in series with the source of 200 volt. Then which of the bulb will be glowing the most –
Solution:
QUESTION: 30
Alternating current can not be measured by direct current meters, because –
Solution: | 1,427 | 5,212 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.859375 | 4 | CC-MAIN-2020-40 | latest | en | 0.883952 |
https://mathhelpforum.com/threads/linear-algebra-help-with-multiplying-matrices.259922/ | 1,579,904,263,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579250626449.79/warc/CC-MAIN-20200124221147-20200125010147-00149.warc.gz | 562,567,178 | 16,629 | # Linear Algebra Help. With multiplying matrices
#### toesockshoe
I've attached the question and solution as the answers. The problem I dont understand is number 12.
I assume the question is asking us to find Ax...In which case I got all of them wrong anyway
$$\displaystyle \begin{bmatrix} 0 & 0 & 1 \\0 & 1 & 0 \\ 1 & 0 & 0 \\ \end{bmatrix} * \begin{bmatrix}x \\y \\z \\ \end{bmatrix}$$
Using the row form, I simply dotted each row of the matrix by vector x. I ended up with the matrix (z y x)... the solution doesnt even have the answers as variables, they have numbers which I dont understand.
For the second one I got (0 0 0), but the answers say (0 0)... how is this even possible? How can there only be 2 dimensions to that vector?
For the third one, I got (3 3 6), which is also wrong compared to the back of the book. For all of them I did it using the row multiplication (dot products).
I think I am misunderstanding the problem.
#### chiro
MHF Helper
Hey toesockshoe.
Your method seems to be ok. I think the question is probably wrong for the first one but I'd need to check the algebra for the other two.
I'm assuming A is the matrix and x is the vector in each of the three cases. (Also there is only one way to multiply all of the matrices anyway).
Can you show us your working? Additionally if you have MATLAB or Octave you can do the calculations there (except for the first one which has variables).
#### HallsofIvy
MHF Helper
The test, as posted, is very strange. Several of the problems ask you to find "Ax" without explicitly saying what "A" and "x" are! In most of them there is only one matrix and one vector so we can assume that the A is the given matrix and x is the given vector. However, in the problem given, there are the matrix and vector you give and then two additional matrix and vector pairs. Are there three different parts to this problem?
#### toesockshoe
Hey toesockshoe.
Your method seems to be ok. I think the question is probably wrong for the first one but I'd need to check the algebra for the other two.
I'm assuming A is the matrix and x is the vector in each of the three cases. (Also there is only one way to multiply all of the matrices anyway).
Can you show us your working? Additionally if you have MATLAB or Octave you can do the calculations there (except for the first one which has variables).
It's a bit hard to use matrices in Latex, but Ill show you my work for the third one...
$$\displaystyle \begin{bmatrix} 2 & 1 \\ 1 & 2 \\ 3 & 3 \end{bmatrix} * \begin{bmatrix} 1 \\ 1 \end{bmatrix} \Rightarrow \begin{bmatrix} 2*1 + 1*1 \\ 1*2 + 2*1 \\ 3*1 + 3*1 \end{bmatrix} \Rightarrow \begin{bmatrix} 3 \\ 3 \\6 \end{bmatrix}$$
#### toesockshoe
The test, as posted, is very strange. Several of the problems ask you to find "Ax" without explicitly saying what "A" and "x" are! In most of them there is only one matrix and one vector so we can assume that the A is the given matrix and x is the given vector. However, in the problem given, there are the matrix and vector you give and then two additional matrix and vector pairs. Are there three different parts to this problem?
I assume it is a 3 part problem... the answers don't make sense though as they are all 2d vectors...while I think they should be 3d. | 862 | 3,280 | {"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} | 3.828125 | 4 | CC-MAIN-2020-05 | longest | en | 0.950303 |
https://www.examrace.com/NTA-UGC-NET/NTA-UGC-NET-Objective-Questions/Aptitude-Questions/Aptitude-Logical-Reasoning-Averages-Part-5.html | 1,603,246,094,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107874637.23/warc/CC-MAIN-20201021010156-20201021040156-00561.warc.gz | 695,077,445 | 9,028 | # Aptitude Logical Reasoning Averages 2020 NTA (UGC)-NET Part 5
Get unlimited access to the best preparation resource for UGC : fully solved questions with step-by-step explanation- practice your way to success.
1. A total of 3000 chocolates were distributed among 120 boys and girls such that each boy received 2 chocolates and each girl received 3 chocolates. Find the respective number of boys and girls?
A. 70, 50
B. 60, 60
C. 50, 70
D. 40, 80
E. None of these
2. David obtained 76, 65, 82, 67 and 85 marks (out of 100) in English, Mathematics, Physics, Chemistry and Biology. What are his average marks?
A. 65
B. 69
C. 75
D. 76
E. None of these
3. The average of 20 numbers is zero. Of them, at the most, how many may be greater than zero?
A. 0
B. 1
C. 10
D. 19
4. Find the average of all the numbers between 6 and 34 which are divisible by 5.
A. 18
B. 20
C. 24
D. 30
5. The average of first five multiples of 3 is:
A. 3
B. 9
C. 12
D. 15
6. The average of the two-digit numbers, which remain the same when the digits interchange their positions, is:
A. 33
B. 44
C. 55
D. 66
7. The average of non-zero number and its square is 5 times the number. The number is:
A. 9
B. 17
C. 29
D. 295
8. The average of five consecutive odd numbers is 61. What is the difference between the highest and lowest numbers?
A. 2
B. 5
C. 8
D. Cannot be determined
E. None of these
9. A family consists of grandparents, parents and three grandchildren. The average age of the grandparents is 67 years, that of the parents is 35 years and that of the grandchildren is 6 years. What is the average age of the family?
A. years
B. years
C. years
D. None of these
10. A library has an average of 510 visitors on Sunday and 240 on other days. The average number of visitors per day in a month of 30 days beginning with a Sunday is:
A. 250
B. 276
C. 280
D. 285
11. The average of five numbers is 27. If one number is excluded, the average becomes 25. The excluded number is:
A. 25
B. 27
C. 30
D. 35
12. The average of 35 students in a class is 16 years. The average age of 21 students is 14. What is the average age of remaining 14 students?
A. 15 years
B. 17 years
C. 18 years
D. 19 years
13. The average of runs of a cricket player of 10 innings was 32. How many runs must he make in his next innings so as to increase his average of runs by 4?
A. 2
B. 4
C. 70
D. 76
14. In the first 10 overs of a cricket game, the run rate was only . What should be the rate in the remaining 40 overs to reach the target of 282 runs?
A.
B.
C.
D. 7
15. If the arithmetic mean of seventy five numbers is calculated, it is 35. If each number is increased by 5, then mean of new number is:
A. 30
B. 40
C. 70
D. 90 | 856 | 2,749 | {"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} | 3.953125 | 4 | CC-MAIN-2020-45 | latest | en | 0.922682 |
http://www.brightstorm.com/qna/question/10562/ | 1,371,675,392,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368708808740/warc/CC-MAIN-20130516125328-00034-ip-10-60-113-184.ec2.internal.warc.gz | 344,848,288 | 10,500 | Quick Homework Help
# New way to find vertex ok if I have y = x^2+6x+8 so 9+y = (x^2+6x+9) + 8 we subtract 8 so we have 9-8+y = (x+3)^2 we have (1+y) = (x+3)^2 right and 1+y=0 y is equal to -1 and x is x+3= 0 x is -3 :) ⚑ Flag
by fifaWorldCup at August 03, 2011 | 127 | 265 | {"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} | 3.578125 | 4 | CC-MAIN-2013-20 | latest | en | 0.736405 |
https://www.mathdoubts.com/problems/cosa-cos3a-bycosa-plus-sina-plus-sin3a-by-sina/ | 1,539,770,132,000,000,000 | text/html | crawl-data/CC-MAIN-2018-43/segments/1539583511122.49/warc/CC-MAIN-20181017090419-20181017111919-00035.warc.gz | 1,016,383,279 | 4,885 | # Evaluate $\dfrac{\cos{A}-\cos{3A}}{\cos{A}}$ $+$ $\dfrac{\sin{A}+\sin{3A}}{\sin{A}}$
$A$ is an angle of a right angled triangle. $\cos{A}$ and $\sin{A}$ are trigonometric functions and $\cos{3A}$ and $\sin{3A}$ are also trigonometric functions with triple angle. The four functions are formed a trigonometric expression.
$\dfrac{\cos{A}-\cos{3A}}{\cos{A}}$ $+$ $\dfrac{\sin{A}+\sin{3A}}{\sin{A}}$
The trigonometric expression can be simplified to find its value.
01
### Expansion of Triple angle identities
Cosine triple angle and sine triple angle terms can be expanded by applying the triple angle trigonometric identities.
According to expansion of the sin triple angle formula and cos triple angle formula.
$\sin{3A} \,=\, 3\sin{A}-4\sin^3{A}$
$\cos{3A} \,=\, 4\cos^3{A}-3\cos{A}$
Use the above two formulas and simplify the expression.
$=\,\,\,$ $\dfrac{\cos{A}-(4\cos^3{A}-3\cos{A})}{\cos{A}}$ $+$ $\dfrac{\sin{A}+3\sin{A}-4\sin^3{A}}{\sin{A}}$
$=\,\,\,$ $\dfrac{\cos{A}-4\cos^3{A}+3\cos{A}}{\cos{A}}$ $+$ $\dfrac{\sin{A}+3\sin{A}-4\sin^3{A}}{\sin{A}}$
$=\,\,\,$ $\dfrac{4\cos{A}-4\cos^3{A}}{\cos{A}}$ $+$ $\dfrac{4\sin{A}-4\sin^3{A}}{\sin{A}}$
02
### Simplifying the terms
There is a $\cos{A}$ term in denominator and a $\cos{A}$ term can be taken common from both terms of the numerator in the first term. Similarly, there is a $\sin{A}$ tem in denominator and a $\sin{A}$ term can also be taken common from the both terms in the numerator in the second term.
$=\,\,\,$ $\dfrac{4\cos{A}(1-\cos^2{A})}{\cos{A}}$ $+$ $\dfrac{4\sin{A}(1-\sin^2{A})}{\sin{A}}$
$=\,\,\,$ $\require{cancel} \dfrac{4\cancel{\cos{A}}(1-\cos^2{A})}{\cancel{\cos{A}}}$ $+$ $\require{cancel} \dfrac{4\cancel{\sin{A}}(1-\sin^2{A})}{\cancel{\sin{A}}}$
$=\,\,\,$ $4(1-\cos^2{A})$ $+$ $4(1-\sin^2{A})$
$=\,\,\,$ $4-4\cos^2{A}$ $+$ $4-4\sin^2{A}$
$=\,\,\,$ $4+4-4\cos^2{A}$ $-$ $4\sin^2{A}$
$=\,\,\,$ $8-4(\cos^2{A}+\sin^2{A})$
03
### Apply Pythagorean identity
As per the Pythagorean identity of sine and cosine functions, the sum of squares of sine and cosine of an angle is one.
$=\,\,\,$ $8-4(1)$
$=\,\,\,$ $8-4$
$=\,\,\,$ $4$
It is the required result of this trigonometry problem. | 866 | 2,192 | {"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} | 4.6875 | 5 | CC-MAIN-2018-43 | longest | en | 0.661461 |
https://www.physicsforums.com/threads/why-degree-measure-of-angles-are-further-divided-in-min-sec.866665/ | 1,670,588,000,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446711396.19/warc/CC-MAIN-20221209112528-20221209142528-00744.warc.gz | 1,000,184,292 | 15,985 | Why degree measure of angles are further divided in min,sec
prashant singh
Whats the relation between degree and time, is there any historical answer.
Gold Member
CalcNerd
BreCheese
A degree can be broken into even smaller measurements, much like a foot can be broken down into inches, and say, centimeters. One practical reason we break a degree into even smaller parts is so we can accurately measure the nautical space between distant objects in deep space. I don't have a specific example, but consider the following:
The sun is roughly 92.95 million miles away from Earth. Now think of when you look into a night sky and see stars. The nearest star is 4.32 light years (2.5396X10^13 miles) away from earth. If you wanted to figure out the distance between the nearest star and a star nearby it, you would have to measure the angle in more precise units (minutes and seconds). This is because even an angle as small as 0degrees,0minutes, and 30seconds has a big arc length when the two objects at the end are, for instance, the distance from the Earth to two distance stars.
If you don't understand what I'm saying then try this. Find two distant objects (the farther from you the better) (trees, buildings, ext.).Make an angle with your pointer finger and your index finger and line the tips of your fingers up with these two distant objects. Notice that although the angle between your fingers is rather small, the distance between the two distant objects is rather big. Now imagine keeping your fingers fixed on these objects and walking backwards. The distance between the two objects stays fixed, but as you get farther and farther away from the objects, the angle your fingers make will get smaller, and smaller.
In short, we need smaller units of angle measures (other than a degree) so we can more accurately measure distant objects.
Lastly, there is a relation between angles measured in degree, minutes, and seconds and time.
Circular motion, and rotation create angles. Its not a coincidence that there are roughly 360 days in a year, and 360 degrees in a full rotation. This means that as the Earth orbits the Sun, every day, the Earth travels through about one degree of its orbit. There is a relation between time and our position around the sun (hence why our calendar has named seasons, Although the seasons we experience are due to both our position around the sun and because of Earth's axis being tilt).
One degree is 60minutes. 60 minutes is 3600seconds. This statement holds true for measurements of time as well. I'm not sure who first came up with the concept of angular measurements, but I'm willing to bet its a very old astronomer.
Last edited:
prashant singh
prashant singh
Wow thanks great help
A degree can be broken into even smaller measurements, much like a foot can be broken down into inches, and say, centimeters. One practical reason we break a degree into even smaller parts is so we can accurately measure the nautical space between distant objects in deep space. I don't have a specific example, but consider the following:
The sun is roughly 92.95 million miles away from Earth. Now think of when you look into a night sky and see stars. The nearest star is 4.32 light years (2.5396X10^13 miles) away from earth. If you wanted to figure out the distance between the nearest star and a star nearby it, you would have to measure the angle in more precise units (minutes and seconds). This is because even an angle as small as 0degrees,0minutes, and 30seconds has a big arc length when the two objects at the end are, for instance, the distance from the Earth to two distance stars.
If you don't understand what I'm saying then try this. Find two distant objects (the farther from you the better) (trees, buildings, ext.).Make an angle with your pointer finger and your index finger and line the tips of your fingers up with these two distant objects. Notice that although the angle between your fingers is rather small, the distance between the two distant objects is rather big. Now imagine keeping your fingers fixed on these objects and walking backwards. The distance between the two objects stays fixed, but as you get farther and farther away from the objects, the angle your fingers make will get smaller, and smaller.
In short, we need smaller units of angle measures (other than a degree) so we can more accurately measure distant objects.
Lastly, there is a relation between angles measured in degree, minutes, and seconds and time.
Circular motion, and rotation create angles. Its not a coincidence that there are roughly 360 days in a year, and 360 degrees in a full rotation. This means that as the Earth orbits the Sun, every day, the Earth travels through about one degree of its orbit. There is a relation between time and our position around the sun (hence why our calendar has named seasons, Although the seasons we experience are due to both our position around the sun and because of Earth's axis being tilt).
One degree is 60minutes. 60 minutes is 3600seconds. This statement holds true for measurements of time as well. I'm not sure who first came up with the concept of angular measurements, but I'm willing to bet its a very old astronomer. | 1,121 | 5,213 | {"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} | 3.9375 | 4 | CC-MAIN-2022-49 | latest | en | 0.96009 |
http://mathhelpforum.com/math-topics/71907-powers-inside-roots-surds.html | 1,481,035,531,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698541907.82/warc/CC-MAIN-20161202170901-00394-ip-10-31-129-80.ec2.internal.warc.gz | 169,411,572 | 10,168 | # Thread: Powers inside roots? [surds]
1. ## Powers inside roots? [surds]
Hello, how do I simplify powers that are inside roots? Example: say x is a positive number (I don't know how to write the math on a PC sorry)
sq< 9x^8
[Square root of nine and x to the power of 8]
To simplify, I can square root 9 to get 3, and take that to the outside of the root.
3 sq< x^8
But what about the x to the power of 8? I have no idea
2. Originally Posted by RAz
Hello, how do I simplify powers that are inside roots? Example: say x is a positive number (I don't know how to write the math on a PC sorry)
sq< 9x^8
[Square root of nine and x to the power of 8]
To simplify, I can square root 9 to get 3, and take that to the outside of the root.
3 sq< x^8
But what about the x to the power of 8? I have no idea
$\sqrt{9x^8}$
it's like this $(\sqrt{3x^4})^2$ then cancel exponent $^2$ with the root
then it's like this $=3x^4$
after 3 then divide 8 by 2 that's what
remember
in roots you will divide $\sqrt{2^6}= 2^{\frac{6}{2}}$
in division you will subtract $\frac{x^4}{x^2}= x^{4-2}$
3. Thanks!
4. you're very welcome! | 361 | 1,124 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 6, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.984375 | 4 | CC-MAIN-2016-50 | longest | en | 0.919898 |
https://byjus.com/question-answer/ravi-invests-50-of-his-monthly-savings-in-fixed-deposits-thirty-percent-of-the-rest/ | 1,642,406,882,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320300343.4/warc/CC-MAIN-20220117061125-20220117091125-00669.warc.gz | 189,227,056 | 17,216 | Question
# Ravi invests 50% of his monthly savings in fixed deposits. Thirty percent of the rest of his savings is invested in stocks and the rest goes into Ravi's savings bank account. If the total amount deposited by him in the bank (for savings account and fixed deposits) is Rs. 59500, then Ravi's total monthly savings (in Rs) is___
Solution
## Let the monthly savings be s. Given, 0.5s+(0.5s−0.3×0.5s)=595000.85s=59500s=Rs.70000 Quantitative Aptitude
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View More | 146 | 515 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.75 | 4 | CC-MAIN-2022-05 | latest | en | 0.898344 |
https://www.physicsforums.com/threads/time-and-time-intervals-when-using-kinematic-equations.842420/ | 1,531,949,033,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676590329.25/warc/CC-MAIN-20180718193656-20180718213656-00577.warc.gz | 952,490,334 | 16,114 | # Homework Help: Time and time intervals when using kinematic equations
1. Nov 10, 2015
### nerf225
1. A space vehicle accelerates uniformly from 85 m/s at t=0s to 462 m/s at t=10s. What is avg acceleration? How far did the space vehicle move from t=2s to t=6s.
2. Got the avg accel easily enough w basic equation for accel. Problem is conceptual with the displacement portion of the question.
3. I got the correct answer by using x=xo + vit + 1/2at2 for t=2s and t=6s and then basically did Δx= x6s-x2s.
BUT...I don't understand why I can't just use Δx=vit + 1/2at2 with t=4 because the time difference between 6s and 2s is 4s. Isn't the t in that equation essentially a Δt? Shouldn't putting 4 in there tell me the distance it will travel in 4 s?
And why do I get a third different answer if I use: Δx=1/2(vi+vf)Δt?
I seem to be confused about what Δx means and when t vs. Δt is used. Or something. Ugh. Sorry. Help!
2. Nov 10, 2015
### 3301
Can you tell me what are your results?
3. Nov 10, 2015
### haruspex
It will, but you need to plug in the right value for vi. What value did you use?
4. Nov 10, 2015
### 3301
Draw the graph and show us how you did it.
5. Nov 10, 2015
### Staff: Mentor
You say you got the correct result for the distance covered between those times by taking the difference of the results. Using symbols for the two times, say $t_a$ and $t_b$ for the 2s and 6s times, write out the two expressions for the total distance using the formula. Take the difference of the two expressions and collect the terms. Can you spot the problem with just plugging in $Δt = t_b - t_a$ into your formula?
6. Nov 10, 2015
### nerf225
Haruspex -- Thank you!!! THAT was it!! I was plugging in 85 m/s as the initial velocity, not calculating the initial velocity at 2 s and plugging that in. UGH!!!!! Smh. Thank you everyone for commenting and being willing to try to help!! :) | 561 | 1,906 | {"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} | 3.921875 | 4 | CC-MAIN-2018-30 | latest | en | 0.935509 |
https://www.physicsforums.com/threads/photoelectric-effect-and-the-human-eye.241913/ | 1,508,327,808,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187822930.13/warc/CC-MAIN-20171018104813-20171018124813-00366.warc.gz | 973,989,019 | 15,651 | # Photoelectric Effect and the human eye
1. Jun 24, 2008
### daveyman
1. The problem statement, all variables and given/known data
Under favorable circumstances the human eye can detect 1.0e-18 J of electromagnetic energy. How many 600-nm photons does this represent?
(Modern Physics, Arthur Beiser, 6th Edition, Pg. 89)
2. Relevant equations
My answer is unreasonably low. My mistake is probably very simple, but I'm not sure what it is. Any ideas?
3. The attempt at a solution
First, I attempted to find the energy in a single photon with a wavelength of 600nm.
Since E=hf, I simply multiplied h and f.
I found f by the relationship f=c/lambda.
So, (6.626e-34)(3e8)/(600e-9) = 3.313e-19
Then, I just divided the total amount of energy by this amount (the amount of energy in a single proton).
(1.0e-18)(3.313e-19) = 3.02
I'm guessing that the eye can detect more than 3 photons at a time :-) but I'm not sure what I'm doing wrong. Help!
2. Jun 24, 2008
### G01
Yes, the eye can detect more than three photons at a time, but, according to this problem, apparently it can't detect less than 3 photons at a time. You are finding the minimum amount of 600nm photons the eye can detect, by utilizing the minimum energy the eye can detect.
Your work looks fine to me.
3. Jun 24, 2008
### daveyman
Thanks for the quick response! I'm glad to hear that my work makes sense, but how did you come to the conclusion that the given energy is a minimum? The way it is worded ("under favorable circumstances") it almost sounds like it would be maximum...
4. Jun 24, 2008
### G01
I interpreted "under favorable circumstances" as "maximum performance of the eye" i.e. the smallest amount of light the eye can detect will be less and less as it performs better.
So, I interpret the result as:
"Under favorable circumstances, i.e. when your eyes are performing at their very best, they can distinguish at least three photons at a time, but no less.
Also, we know for a fact that the eye can detect more than that amount of energy, so it can't be a maximum.
5. Jun 24, 2008
### daveyman
Oh! - the performance of the eye here is based on how little light the eye can still distinguish - I totally get it! That makes tons of sense, actually - thanks!
6. Jun 25, 2008
### G01
No problem.
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook | 632 | 2,393 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.703125 | 4 | CC-MAIN-2017-43 | longest | en | 0.937994 |
http://mathhelpforum.com/calculus/116628-antiderivatives-problem.html | 1,481,139,714,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698542244.4/warc/CC-MAIN-20161202170902-00020-ip-10-31-129-80.ec2.internal.warc.gz | 177,315,578 | 10,339 | 1. ## Antiderivatives Problem
Hi, I am trying to find the antiderivative of this problem, but it is not the correct answer.
1/3x^2 dx
I go from that to
3x^-1/-1
what am I suppose to do with this problem?
I had a question when dividing with /x
so if a problem is 3x^2/x do simply make it 3x^1 is that correct?
And for fractions with roots at the bottom, how do I go to solving that? ex: -1/sqrt(y)
Thanks for any help, especially for the problem at the top.
Coke
2. Hey there
$\int (x^2)/3 .dx = (x^3)/9 + C$
Yes, $3x^2/x = 3x$
Not sure what your last question means..?
3. Originally Posted by cokeclassic
Hi, I am trying to find the antiderivative of this problem, but it is not the correct answer.
1/3x^2 dx
I go from that to
3x^-1/-1
what am I suppose to do with this problem?
I had a question when dividing with /x
so if a problem is 3x^2/x do simply make it 3x^1 is that correct?
As long as x is not equal to 0, yes.
And for fractions with roots at the bottom, how do I go to solving that? ex: -1/sqrt(y)
What do you mean by "solve" that? What do you want to do with it? If you just want a different way to write it, perhaps to integrate, you could try $-y^{1/2}$.
Thanks for any help, especially for the problem at the top.
Coke
4. Hi, I am not sure what you did with the first problem? Could you show me some steps please? | 406 | 1,346 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 3, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.6875 | 4 | CC-MAIN-2016-50 | longest | en | 0.962504 |
http://betterlesson.com/lesson/reflection/8480/what-i-noticed | 1,477,280,786,000,000,000 | text/html | crawl-data/CC-MAIN-2016-44/segments/1476988719465.22/warc/CC-MAIN-20161020183839-00297-ip-10-171-6-4.ec2.internal.warc.gz | 26,772,057 | 21,945 | ## Reflection: Intervention and Extension Break Downs - Section 3: Get Set
In this section students worked independently illustrating multiplication problems using base tens drawings, and explaining the process. Some students had a problem recognizing the value of the one digit numbers, when it was multiplied by a multiple of 10. For instance, I asked students to multiply 2 x 3. They understood that the product was 6. But when I asked them to tell me the value of 6 they seemed a bit puzzled. I had to refer them back to the place value chart for them to visually see the connection. After, I modeled this one or two times they easily detected the difference in multiplying by ones and the power of tens.
Intervention and Extension: What I noticed!
# Break Downs
Unit 4: Numbers and Their Places!
Lesson 8 of 15
## Big Idea: Being a mechanic is really hard work, using base tens and place value charts students will be able to determine the value of place in a number by multiplying by the power of ten.
Print Lesson
14 teachers like this lesson
Standards:
Subject(s):
Math, multiplication, structure, explaining, examine
70 minutes
### Carol Redfield
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Environment: Urban | 439 | 2,034 | {"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-2016-44 | longest | en | 0.903967 |
https://puzzling.stackexchange.com/questions/70/n-logicians-wearing-hats-of-n-colors/557 | 1,603,286,780,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107876500.43/warc/CC-MAIN-20201021122208-20201021152208-00291.warc.gz | 495,651,150 | 42,350 | # N logicians wearing hats of N colors
N logicians are wearing hats which can be of N different colors (each hat has one color; there can be multiple hats with the same color). Each logician can see the colors of all hats except his own. The logicians must simultaneously call out a color; they win if at least one logician calls the color of his own hat. They can agree on a strategy beforehand; the set of possible colors is known. There is a winning strategy (i.e. the logicians can arrange to win no matter what colors the hats end up being).
(I definitely didn't invent this; I don't know who did.)
I know of a solution that requires a little knowledge of abstract algebra (you can read a sketch of that solution on Math Stack Exchange). Is there a solution that doesn't require any mathematical background?
Specifically, how do you explain the solution to someone who isn't familiar with modulo arithmetic ($\mathbb{Z}/n\mathbb{Z}$)?
(Hint, if you want to try to solve it on your own: start with N=2 — white hat and black hat. Then work on N=4 — red, pink, blue and cyan.)
• Well, as a preliminary solution, it seems as if every logician just calls out randomly, they have about a 63.2% chance of winning anyway. You're looking for one that works 100%, though, right? – Joe Z. May 15 '14 at 23:53
• @JoeZ. I'm not interested in probabilistic solutions, only in a guaranteed solution (which exists). If the logicians lose, they'll be killed. And none of them are suicidal. – Gilles 'SO- stop being evil' May 15 '14 at 23:59
• I just wanted to make sure it's not one of those where if one of them gets it wrong, they also all get killed, and some of them have to choose not to say anything. – Joe Z. May 16 '14 at 0:01
• It doesn't matter if one of them gets it wrong: they win as long as at least one color is called correctly. – Gilles 'SO- stop being evil' May 16 '14 at 0:02
• Anyway, for N=2, the solution seems to be that one person will always say the same colour as the one he sees, and one will always say the opposite colour. – Joe Z. May 16 '14 at 0:03
There are two secrets to this question. First is to realise that you want no player in the circle to guess that the circle looks the same. If they guess the same circle, it is a wasted guess.
For these examples, the distinct colors are B (black), W (white), G (grey), N (neon), and F (fish). There are $N$ players and $N$ color hats.
If there is never a chance you could both be correct in this one specific puzzle, there is never a chance you could both be wrong. This has to do with the fact that there are $N \, N$ possible patterns of colors and each player can eliminate $N \, (N-1)$ possibilities by looking at the others. The remaining $\frac{N \, N}{N\,(N-1)}=N$ possibilities must all be guessed somehow.
## $N=2$
For two players this is easy. One of you simply says: you guess we have the same color while I guess that we have a different. If the colors are BB or WW, the first person will guess correctly. If they are BW or WB the second person will. If they were to decide "I pick B you pick W or we both pick B" then there is a chance they could both be correct and an equal chance you could both be wrong.
## $N=3$
If we have three players, each player needs a slightly different strategy to make this work! Decide, therefore, to elect an equalist (player 1) who believes that either every seat is the same (all the same color) OR every color is the same (all colors appear once). If he sees BB he guesses B. If he sees BW he guesses G.
The person in the chair to his right (player 2) knows what player 3 is wearing and what player 1 is wearing as well as player 1's strategy. If player 3 is B and player 2 thinks he is B, player 1 will guess B. There is no point in player 2 guessing that he is the same as the other two if they are equal, therefore, so he guesses (for this case) W. If, however, the arrangement is B_G, picking W will backfire as player 1 will guess G.
The easiest way to do this is to say that the colors W, G, and B are in a circle holding hands like the children you must be. Player two figures out what color Player 1 would pick in his situation and picks the one to that color's right. Player 3 does the same but picks the color to the left.
## $N \gt 3$
The strategies for all players for any number of players and colors must meet only two criteria: (a) every player (A) must be able to tell by looking at each other players what color A's hat must be in order for that other character to guess correctly; (b) all player's guesses must be coordinated by (a) so that they are all different. The previous examples show examples as to how this works. Essentially, there are only $N$ different colored hats you could be wearing and if the other players take care of $N$-1 of them, you only have 1 left to cover...
One strategy to do this is to give every player a "favorite" color and then each color a proper fraction. This fraction is the seat (starting at 0) that has it as their favorite divided by $N$. Aka if $N=5$: $\mathrm{B}=0$; $\mathrm{W}=\frac{1}{5}$; $\mathrm{G}=\frac{2}{5}$; $\mathrm{N}=\frac{3}{5}$; $\mathrm{F}=\frac{4}{5}$ You imagine that there is a circle in the center of the seats where every color is arranged in the same order as the favorite colors. For every colored hat you see, you rotate that circle by whatever fraction of a circle that color is represented by. When you are done, the color in front of each player is the color they would guess if they saw exactly what you see. You should guess the color in front of you! As these are all different and your answer is dependent on every other players' hat color, this strategy works.
There are likely other strategies that work for every $N\gt2$ but there do not seem to be any that are easier to visualize or generalize for any $N$ than this!
This answer attempts to give a dumbed down version of the answer on Math.SE.
Say there are two logicians, A and B; and two colors, $0$ and $1$.
Mr. A looks at his friend and guesses that they are wearing the same hat. Mr. B knows that Mr. A is predictable and will guess that they are the same, so he will guess they are different. Without knowing what either person saw we can see that all answers are covered.
Them being the same is the same as saying that the sum of their colors is equal to $0 \bmod 2$ as $0+0=0$ and $1+1=2=0 \bmod 2$. Them being different is, similarly, $1 \bmod 2$.
The easiest solution, therefore, is for every person to guess that the sum of all the colors' numbers is equal to their seat number $S$ modulo the number of people.
The way to see this is to note that the equation for 5 logicians and 5 colors is: $a+b+c+d+e=S \bmod 5$ where each variable refers to the color number of the player with the same name.
For Mr. A, he knows what $b$, $c$, $d$, and $e$ add up to be and only doesn't know $a$ so he guesses $S=0 \bmod 5$ which means that $a=0-b-c-d-e \mod 5$.
Mr. B, he knows what Mr. A will guess, so he guesses $S=1 \bmod 5$, so $b=1-a-c-d-e \mod 5$.
By the time we get through 5 guessers we will guess:
Mr. A: $S=0 \bmod 5$
Mr. B: $S=1 \bmod 5$
Mr. C: $S=2 \bmod 5$
Mr. D: $S=3 \bmod 5$
Mr. E: $S=4 \bmod 5$
which is of course every possibility.
As you make up colors (one person in a circle with only black and white hats can yell 'purple'), this means any number of players can guess at least one of their hats correctly for any number of colors less than or equal to the number of players.
• Is the mod operator too advanced for a non-mathematical answer? – kaine May 22 '14 at 19:08
• I think this use of mod is reasonably easy to explain. – Ross Millikan May 24 '14 at 21:30
The solution is very simple. The solution is very simple ones you know it. But may be you would like to give it a try first? (Though my answer looks relatively complicated, this is because I was required to explain operation % with very basic operations)
The Solution:
1. The logicians numerate themselves (L1, L2, ..., LN) and colours (C1, C2, ..., CN).
2. Let's introduce a new operation %, it's meaning please see below*. There are N equations:
1) (C1 + C2 + ... + CN) % N = 0
2) (C1 + C2 + ... + CN) % N = 1
...
N) (C1 + C2 + ... + CN) % N = N-1
Logician Li should suppose that equation number i is true. Then he extracts Ci from equation number i: Ci = (i - C1 - C2 - ... - C_(i-1) - C_(i+1) ... - CN) % N. He can do it because he knows all colours except Ci.
Obviously exactly one equation is true. Therefore correspondent logician will be correct.
*Definition of operation %:
1. Z = X % Y operation is defined in the following way: One needs to find R and K integer number, which satisfy: X = R + Y*K and 0 <= R < Y. Then Z = R.
2. This means that C1 + C2 + ... + Ci + ... + CN) % N = i is equivalent to C1 + C2 + ... + Ci + ... + CN = i + N*k, where k is integer. Ci is found as Ci = i + N*k - C1 - C2 - ... - C_(i-1) - C_(i+1) ... - CN or, applying % operation to both parts, since Ci is integer less than N: Ci = (i - C1 - C2 - ... - C_(i-1) - C_(i+1) ... - CN) % N.
• Sure, assuming enough mathematical background — namely knowledge of modulo arithmetic. I'm asking for a solution (or maybe a way to explain that solution) without mathematical background. – Gilles 'SO- stop being evil' May 24 '14 at 23:15
• @Gilles, you are kidding? What does it mean? mod is operation of the same complexity as + and /. Formulate exactly which operations are allowed. – klm123 May 25 '14 at 4:49
• @Gilles, I believe you can't solve this puzzle with out analog of mod operation. I explained mod operation via * and / operation, so the solution can understand the one who knows only them and decimal notation of numbers. – klm123 May 25 '14 at 5:04
• Addition and division are taught in primary school. Modulo arithmetic is typically taught only in undergraduate if you're studying math (maybe in high school in some countries). I asked if there was a solution that doesn't require having studied math. – Gilles 'SO- stop being evil' May 25 '14 at 9:08
• @Gilles I may not be versed in the arcane art of proper mathematics, but it seems to me that the subset of true modulo arithmetic involved in this answer was completely identical to the "remainder" concept in primary school mathematics. Anyone who understands the concept of remainders should be able to understand this answer. – March Ho Jan 8 '15 at 12:47
Since the set of colours is known in advance, then:
• Assign a number to each colour
• The leader arranges everyone else in numerical order, then stands at the end of the line
• The leader is passed down the line until in the correct place
If every hat is of a different colour, then each logician knows their colour from their place in the line.
If not:
• If more than 2 hats are of the same colour, at least one logician will be standing between two people with the same colour hat, so knows that they too have that colour
• The logicians agree in advance that if they only see colours repeat in groups of two, they call out the colour ahead of them in line rather than their own. That way the second member of a colour should be correct.
Non-mathematical, but does assume a lot of collaboration is allowed.
• I don't understand what you mean by “The leader is passed down the line until in the correct place”. Note that the logicians are not allowed to communicate, that includes verbal communication but also gestural communication such as moving around. – Gilles 'SO- stop being evil' Apr 21 '15 at 13:32
• These questions require that there is no communication after the hats are added until they say their hat color. That isn't stated very clearly in the question though. This behavior of rearranging counts as communication. +1 though due to a correct solution to a question (even though it violates intended rules) – kaine Apr 22 '15 at 17:01
Say there are
5 logicians 1, 2, 3, 4, 5
& 5 colors A, B, C, D, E
One of the logician, say A, is chosen the leader. Everyone will call out the leader's hat color. The leader will call out the color that is missing from the remaining N-1 hat colors he can see. Two cases are possible:
If there is no repeating color, then it's easy to see that the leader will call out his own hat's color and they'll win. If there is a repeating color, then further two cases are possible:
• If the leader's color is repeated, then the logician with the same color as leader will call out his own color and they'll win.
• If the leader's color is not repeated but someone else's color is, then idk...
Help me to build upon this? This was too long for a comment..
• This fails in the case of your second bullet. The strategy in the link takes care of that. It makes sure each logician claims the sum of the hat colors is different. A good thought. – Ross Millikan May 16 '14 at 2:37
• I can't see how this approach could work. You're handling most cases, but not all of them. Think of the person who hands out the hats as an evil mastermind — if there's even a single way to make the logician fails, that's what he'll choose. Your task is to arrange for a strategy that always works. – Gilles 'SO- stop being evil' May 16 '14 at 16:38
• I know this isn't the correct answer. It's just a thought, an idea that isn't complete yet. probably was better suited as a comment but was too long... – kBisla May 16 '14 at 17:29 | 3,499 | 13,360 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.578125 | 4 | CC-MAIN-2020-45 | latest | en | 0.96418 |
https://www.gradesaver.com/textbooks/math/other-math/thinking-mathematically-6th-edition/chapter-9-measurement-chapter-summary-review-and-test-review-exercises-page-607/49 | 1,558,920,176,000,000,000 | text/html | crawl-data/CC-MAIN-2019-22/segments/1558232260358.69/warc/CC-MAIN-20190527005538-20190527031538-00469.warc.gz | 804,890,833 | 11,874 | Thinking Mathematically (6th Edition)
$2.25 ~lb$
16 oz = 1 lb We can convert 36 oz to units of pounds. $36~oz = \frac{36~oz}{1}\times \frac{1~lb}{16~oz} = 2.25~lb$ | 68 | 164 | {"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} | 4.0625 | 4 | CC-MAIN-2019-22 | latest | en | 0.786434 |
http://study.com/academy/lesson/what-is-a-two-way-table.html | 1,508,245,486,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187821189.10/warc/CC-MAIN-20171017125144-20171017145144-00380.warc.gz | 351,024,445 | 36,818 | # What is a Two-Way Table?
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• 0:16 Two-Way Table Defined
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Instructor: Kathryn Maloney
Kathryn teaches college math. She holds a master's degree in Learning and Technology.
Do you believe in Martians? Do you watch football on television? A Two-Way Table or Contingency Table is a great way to show the results of all kinds of survey questions. In this video we will learn how to read a two-way table.
## Two-Way Table
Do you believe in Martians? Here is a survey of 100 college students asking that very question: Do you believe in Martians?
Gender Yes No
Male 10 32
Female 38 20
Total 48 52
This type of table is called a two-way or contingency table.
A two-way or contingency table is a statistical table that shows the observed number or frequency for two variables, the rows indicating one category and the columns indicating the other category. The row category in this example is gender - male or female. The column category is their choice, yes or no.
There is a lot of information we can learn from this small table. Let's look at a couple questions you could see on a test.
How many males were asked? Let's look across the male row. Ten said 'yes' and 32 said 'no.' That would be a total of 42 males.
How many college students believe in Martians? Look down the Yes column. Ten males and 38 females said 'yes.' That would be a total of 48 college students.
## Example #2
A recent survey of 100 college students asked if they prefer to drink tea, coffee, or an energy drink during finals week. Here is the table created from that survey.
Gender Tea Coffee Energy Drink
Male 2 15 39
Female 18 20 6
Total 20 35 45
Let's look at a couple questions you could see on a test.
How many college students drink tea given that they are a female? In this question, we're looking for tea drinkers that are female. Let's look at the female row and tea column. The answer is 18.
How many students drink tea or are male? In this question, we're looking for all tea drinkers united with all males. To figure out the answer, let's highlight the Tea column and Male row. We do have an overlap at Tea and Male, so be sure not to add that twice. So our answer is Total Tea drinkers 20 plus Total Males 56 minus the overlap of 2. 20 + 56 - 2 = 74. There are 74 students that are tea drinkers or male.
How many coffee or tea drinkers are not female? Let's look at each part. Let's highlight the coffee and tea columns. Now, let's highlight the male row. Remember, the question asks for drinkers that are not female, so they have to be male.
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Did you know… We have over 95 college courses that prepare you to earn credit by exam that is accepted by over 2,000 colleges and universities. You can test out of the first two years of college and save thousands off your degree. Anyone can earn credit-by-exam regardless of age or education level. | 841 | 3,505 | {"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} | 4.25 | 4 | CC-MAIN-2017-43 | latest | en | 0.919701 |
https://or.stackexchange.com/questions/3110/introducing-a-big-m-variable-in-given-equations | 1,726,597,433,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651829.57/warc/CC-MAIN-20240917172631-20240917202631-00074.warc.gz | 393,659,874 | 30,704 | # Introducing a big M variable in given equations
While I do understand the general workings of the Big-M-method I am struggling with the following sample exercise, in which the Big-M-method has to be used to find a first feasible solution:
\begin{alignat}2\max&\quad 10x_1+4x_2\\\text{s.t.}&\quad x_1+x_2+x_3=4\tag1 \\&\quad 2x_1-x_2-x_4=2\tag2\\&\quad -x_1+x_5=-1\tag3\\&\quad x_1+x_3-x_4+x_5=4\tag4\\&\quad x_1,\cdots,x_5 \geq 0\tag5\end{alignat}
I am not sure how to introduce the artificial variable for the Big-M. The only problem here seems to be the negative value on the right side of equation #3. So I would multiply with $$-1$$. Now it looks as though we have a negative slack variable $$x_5$$ which would allow us to add another variable $$y_1$$ as part of the Big-M-Method. But I doubt if $$x_5$$ can be considered a slack variable here since it is given as part of the task and is also specified as $$\geq 0$$. So I just need to know if I am on the wrong track and if so, how to introduce the Big-M the right way
• Here is a good example: youtube.com/watch?v=ROkDaBeEiVs&vl=en Commented Nov 25, 2019 at 22:19
• @EhsanK yes, constraint #3 is what I meant Commented Nov 26, 2019 at 18:15
All your constraints are equality. So, add an artificial variable to each constraint (let's call them $$a_i \quad i\in\{1,..,4\}$$). Now all these artificial variables need to be in the objective function with a coefficient of Big-M. Since you are maximizing, you want to make sure using any of them will penalize your objective function (so, you add them with negative sign). So, your objective function becomes: $$\max \quad 10x_1 +4x_2 - Ma_1 - Ma_2 - Ma_3 - Ma_4$$
The artificial variables play the role of your initial basis. So, you need to standardize your simplex table and make sure they are zero in the objective row. After doing that, just solve the simplex problem as you normally do.
If you like to check a simple example, take a look at this example
• Since the constraints are equalities, you cannot assume the artificials are nonnegative. So you really should have two artificials per equation (as when modeling absolute values).
– prubin
Commented Nov 25, 2019 at 22:10
• @prubin I tried to find an example to confirm that but any example I searched and found (from MIT, columbia, Brown, and some others) they all introduce one positive artificial variable for equality constraints too. Am I forgetting something here?
– EhsanK
Commented Nov 25, 2019 at 22:46
• One positive variable for equalities requires an assumption about the sign of the RHS IMO. Commented Nov 26, 2019 at 11:10
• @JohnEren You add artificial variables ($a$) to make a new space where the origin point is part of your solution so you start from there. So, you add it to a constraint that you don't have the available $+1s$ for your initial basis (where $s$ is a slack). You have that $+1$ in $\le$, you have $-1$ for $\ge$ (from a surplus) and you add an $a$ there. You also need it for equality constraint to satisfy the requirement. In your example, if constraint 4 was $x_1+x_3-x_4+x_6=4$ and you don't see $x_6$ anywhere else, then that $x_6$ could play the role in the basis for you (no need for $a$).
– EhsanK
Commented Nov 26, 2019 at 18:45
• @prubin that was assumed as the OP mentioned in the question "...The only problem here seems to be the negative value on the right side of equation #3. So I would multiply with −1. ..."
– EhsanK
Commented Nov 29, 2019 at 15:37 | 1,006 | 3,474 | {"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": 8, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.875 | 4 | CC-MAIN-2024-38 | latest | en | 0.860305 |
https://becalculator.com/170-2-cm-to-inch-converter.html | 1,685,432,509,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224645417.33/warc/CC-MAIN-20230530063958-20230530093958-00488.warc.gz | 155,341,639 | 16,668 | 170.2 cm to inch converter
FAQs on 170.2 cm to inch
How many inches in a cm?
If you are looking to convert 170.2 centimeters to the equivalent of inches, first, you should determine how many inches one cm represents.
This is how I will be specific: one centimeter is equal to 0.3937 inches.
How do I convert 1 cm to inches?
You can convert 1cm into inches by multiplying 1cm by 0.3937.
So, 1 cm into inches = 1 times 0.3937 = 0.3937 inches, exactly.
This will allow you to answer the following question easily and quickly.
• What is one centimeter to inches?
• What is conversion rate cm to inches?
• How many inches is equal to 1 cm?
• What does 1 cm equal to in inches?
Definition:Centimeter
Centimeter is an International Standard Unit of Length. It is equal to one hundredth of one millimeter. It’s roughly equivalent to 39.37 inches.
What is Inch?
Anglo-American length units are measured in inches. 12 inches equals 1 foot, and 36 inches equals one yard. According to the modern standard, one inch equals 2.54 centimeters.
How do u convert 170.2 cm to inches?
You have a good understanding of cm to inches from the above.
Below are the relevant formulas:
Value in inches = value in cm × 0.3937
So, 170.2 cm to inches = 170.2 cm × 0.3937 = 6.700774 inches
This formula can also be used to answer similar questions:
• What is the formula for converting 170.2 cm to inches?
• How can you convert cm into inches?
• How can you change cm into inches?
• How to calculate cm to inches?
• How big are 170.2 cm to inches?
cm inch 169.4 cm 6.669278 inch 169.5 cm 6.673215 inch 169.6 cm 6.677152 inch 169.7 cm 6.681089 inch 169.8 cm 6.685026 inch 169.9 cm 6.688963 inch 170 cm 6.6929 inch 170.1 cm 6.696837 inch 170.2 cm 6.700774 inch 170.3 cm 6.704711 inch 170.4 cm 6.708648 inch 170.5 cm 6.712585 inch 170.6 cm 6.716522 inch 170.7 cm 6.720459 inch 170.8 cm 6.724396 inch 170.9 cm 6.728333 inch 171 cm 6.73227 inch
Deprecated: Function get_page_by_title is deprecated since version 6.2.0! Use WP_Query instead. in /home/nginx/domains/becalculator.com/public/wp-includes/functions.php on line 5413 | 639 | 2,115 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5 | 4 | CC-MAIN-2023-23 | longest | en | 0.864929 |
https://www.elitedigitalstudy.com/22318/which-of-the-following-sets-are-equal | 1,686,442,170,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224646652.16/warc/CC-MAIN-20230610233020-20230611023020-00137.warc.gz | 829,830,853 | 10,531 | Which of the following sets are equal?
(i) A = {1, 2, 3}
(ii) B = {x ∈ R:x2–2x+1=0}
(iii) C = (1, 2, 2, 3}
(iv) D = {x ∈ R : x3 – 6x2+11x – 6 = 0}.
Asked by Aaryan | 1 year ago | 40
##### Solution :-
A set is said to be equal with another set if all elements of both the sets are equal and same.
A = {1, 2, 3}
B ={x ∈ R: x2–2x+1=0}
x2–2x+1 = 0
(x–1)2 = 0
∴ x = 1.
B = {1}
C= {1, 2, 2, 3}
In sets we do not repeat elements hence C can be written as {1, 2, 3}
D = {x ∈ R: x3 – 6x2+11x – 6 = 0}
For x = 1, x2–2x+1=0
= (1)3–6(1)2+11(1)–6
= 1–6+11–6
= 0
For x =2,
= (2)3–6(2)2+11(2)–6
= 8–24+22–6
= 0
For x =3,
= (3)3–6(3)2+11(3)–6
= 27–54+33–6
= 0
∴ D = {1, 2, 3}
Hence, the set A, C and D are equal.
Answered by Sakshi | 1 year ago
### Related Questions
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#### Find the symmetric difference A Δ B, when A = {1, 2, 3} and B = {3, 4, 5}.
Find the symmetric difference A Δ B, when A = {1, 2, 3} and B = {3, 4, 5}. | 590 | 1,140 | {"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} | 4.1875 | 4 | CC-MAIN-2023-23 | longest | en | 0.787605 |
http://math.stackexchange.com/questions/146536/confused-about-wikipedia-page-on-differential-forms | 1,469,315,300,000,000,000 | text/html | crawl-data/CC-MAIN-2016-30/segments/1469257823802.12/warc/CC-MAIN-20160723071023-00188-ip-10-185-27-174.ec2.internal.warc.gz | 152,720,860 | 17,283 | I know next to nothing about differential forms, but I saw them being mentioned repeatedly on this site, so I went to Wikipedia to try to understand what a differential form is. Most of it is going pretty smooth, but there's a paragraph I don't understand:
Since any vector $v$ is a linear combination $\sum v^je_j$ of its components, $df$ is uniquely determined by $df_p(e_j)$ for each $j$ and each $p \in U$, which are just the partial derivaties of $f$ on $U$. Thus $df$ provides a way of encoding the partial derivatives of $f$. It can be decoded by noticing that the coordinates $x^1, x^2, \dots , x^n$ are themselves functions on $U$, and so define differential 1-forms $dx^1,dx^2,\dots,dx^n$. Since $\frac{\partial x^i}{\partial x^j} = \delta_{ij}$, the Kronecker delta function, it follows that
$$df = \sum_{i=1}^n \frac{\partial f}{\partial x^i}dx^i$$
I don't understand the bit about the coordinates being functions themselves. What does that mean?
-
For a vector $y=(y_1,\ldots,y_n) \in \textbf{R}^n$, the i-th coordinate function $x^i$ simply picks out the i-th entry, that is, $x^i(y)=y_i$.
Thus, if $e_1,\ldots,e_n$ is a basis for $\textbf{R}^n$, then any vector $y=(y_1,\ldots,y_n)$ can be written as
$y=\sum_{i=1}^n x^i(y)e_i$.
So, just to make sure I understand: Suppose we're working in $\mathbb{R}^2$, and instead of $x,y$ I call them $f_1, f_2$ for clarity, so they're functions from $\mathbb{R}^2$ to $\mathbb{R}$. Then we can say that $f_1(a, b) = a$ and $f_2(a,b)=b$? – Javier May 19 '12 at 3:56 | 497 | 1,523 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.953125 | 4 | CC-MAIN-2016-30 | latest | en | 0.886029 |
https://www.physicsforums.com/threads/solve-12y-5y-2y-0.719900/ | 1,519,441,055,000,000,000 | text/html | crawl-data/CC-MAIN-2018-09/segments/1518891815034.13/warc/CC-MAIN-20180224013638-20180224033638-00423.warc.gz | 874,583,791 | 19,758 | # Solve 12y'' -5y' -2y = 0
1. Oct 30, 2013
### Jbreezy
1. The problem statement, all variables and given/known data
Solve 12y'' -5y' -2y = 0
2. Relevant equations
3. The attempt at a solution
So I'm not really sure how to do this at all. I feel like he took this out of left field.
What is I rewrote it as 12y'' -5y' = 2y Then what if I integrated it.
I may get 12y' - 5y = y + C ? Then I could use an integrating factor because it is of that form? And just go from there? Is this OK. Or am I making stuff up?
2. Oct 30, 2013
### Dick
Making stuff up. If you integrate 2ydy you get y, but you need to integrate 2y with respect to whatever the independent variable is. And it's not y. This is a different type of problem. It's a linear differential equation with constant coefficients. http://en.wikipedia.org/wiki/Linear_differential_equation
3. Oct 30, 2013
### Jbreezy
It sent me to nothing .... I got this Wikipedia does not have an article with this exact name
How to I handle this problem
4. Oct 30, 2013
### Jbreezy
I regoogled it and found the article idk why ur link didn't work
5. Oct 30, 2013
### Dick
Because my copy and paste function often double copies and I missed that one. So it's badly formatted. I fixed it.
6. Oct 30, 2013
### Jbreezy
Sorry I'm still unsure of how to go about this problem. I'm looking over the example. But I don't understand really.
7. Oct 30, 2013
### Dick
You probably just haven't looked at it enough. Assuming your dependent variable is x, assume a solution has the form y=e^(kx). Put that into the ode and try to figure out what the possibilities for k are. That's the jist of it. It's pretty simple really.
8. Oct 30, 2013
### Jbreezy
OK. I did it a little different looking at an example. I just
let y'' = r^2 and y' = r in my equation
12r^2 -5r - 2 = 0 and I factored it. I get r = (1/4) and r = (2/3) so then I looked at how this example reported there answer and they had.
Y(x) = C1 e^(1/4)z + C2e^(2/3)z
I'm not sure why they choose z or even why it get;s this form for the solution
9. Oct 30, 2013
### Jbreezy
I guess I should use x
10. Oct 30, 2013
### Dick
You are trying to go too fast without understanding what you are doing. For one thing, I get r=(-1/4) (not +1/4) or r=(2/3). If your independent variable is x then e^(-x/4) and e^(2x/3) are both solutions. Check that first.
11. Oct 30, 2013
### Jbreezy
Well I'm going fast because it is late here an I have examination in another course in the morning so forgive me for my lack of understanding. So, you don't even get 2/3 as an answer?
12. Oct 30, 2013
### Dick
Yes, I do. Didn't I say that I did? Why are asking? This is what I mean by 'too fast'.
13. Oct 30, 2013
### Jbreezy
Kind of I got confused by your language. So I should pop this those in my original differential equation? Yes I think so.
14. Oct 30, 2013
### Jbreezy
Alright I got 0 for both in the original. So what is with how they write the answer? In that form?
15. Oct 30, 2013
### Dick
Good idea. That should make it doubly clearer why the roots of 12r^2 -5r - 2 = 0 give you the exponents for solutions to your ODE.
16. Oct 30, 2013
### Jbreezy
OK so what about this one because I can't factor this like the other one....
y'' -4y' +5y = 0
17. Oct 30, 2013
### Dick
They should have written y(x)=C1*e^(-x/4)+C2*exp(2x/3), with C1 and C2 being constants. I don't know where the z came from. You can see that also solves your diffy q, right? The rest of the story, which I'm not going to go into here, is that a 2nd order ode like that has exactly two independent linear solutions. So that is not only A solution, ALL of the solutions must have that form. I'm sure all of this stuff is in your textbook or course somewhere.
18. Oct 30, 2013
### Dick
You can still find complex roots which will give you complex exponentials. Which will turn out to be products of sines, cosines and ordinary real exponentials via Euler's formula. This stuff has got to be in your course someplace! I'm not going to try to explain the whole thing from scratch.
19. Oct 30, 2013
### Jbreezy
y(x)=C1*e^(-x/4)+C2*exp(2x/3)
What is exp(2x/3)?
20. Oct 30, 2013
### Dick
e^(2x/3). I sometimes write e^x as exp(x). I slipped.
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Draft saved Draft deleted | 1,338 | 4,400 | {"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} | 3.9375 | 4 | CC-MAIN-2018-09 | latest | en | 0.943148 |
https://mvtrinh.wordpress.com/tag/equilateral-triangle/ | 1,532,144,613,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676592309.94/warc/CC-MAIN-20180721032019-20180721052019-00439.warc.gz | 729,183,975 | 14,061 | # Tag Archives: equilateral triangle
## Equilateral Triangle and Regular Hexagon
What is the ratio of the area of an equilateral triangle to that of a regular hexagon inscribed in the same circle? Source: NCTM Mathematics Teacher, February 2008 Solution The hexagon is made up of six congruent isosceles triangles three … Continue reading
## Largest Equilateral Triangle in a Square
How long is the side of the largest equilateral triangle that can be inscribed in a square with side length ? Source: NCTM Mathematics Teacher, August 2006 Solution A first guess is an equilateral triangle of side length . Could … Continue reading
## Equilateral Triangle and Square
A pentagon made up of equilateral triangle with side length on top of square is inside a circle passing through points , and . Find the radius of the circle. Source: NCTM Mathematics Teacher, August 2006 Solution Let be the … Continue reading
## Two Equilateral Triangles
If two equilateral triangles of area intersect to form a regular hexagon, then what is the area of the hexagon? Source: NCTM Mathematics Teacher, September 2006 Solution If we subdivide an equilateral triangle into smaller congruent equilateral triangles as shown … Continue reading
## Arc of a Circle
is a diameter of a circle of radius unit. is a chord perpendicular to that cuts at . If the arc is of the circumference of the circle, what is the length of the segment ? Source: NCTM Mathematics Teacher, … Continue reading
## Equiangular Hexagon
An equiangular hexagon with consecutive side lengths , and can be inscribed in an equilateral triangle with side length . This same equiangular hexagon can also be inscribed in a second (distinct, noncongruent) equilateral triangle with side length . What … Continue reading
## Sum of Consecutive Numbers (Part 2)
The sum of consecutive positive integers is . Find all possible values of . Source: NCTM Mathematics Teacher, December 2005 SOLUTION Let be the sum of consecutive positive integers with the first term equal . If we add the first … Continue reading
## Measure of Angle
, and are vertices of a cube, as shown. Find the measure of . Source: NCTM Mathematics Teacher, December 2005 SOLUTION Triangle is equilateral because its three sides are diagonals of a square. The measure of equals . Answer: | 509 | 2,310 | {"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} | 4.125 | 4 | CC-MAIN-2018-30 | latest | en | 0.852938 |
http://mathhelpforum.com/advanced-algebra/272452-parity-preserved-when-multiplying-two-group-elements.html | 1,511,377,518,000,000,000 | text/html | crawl-data/CC-MAIN-2017-47/segments/1510934806620.63/warc/CC-MAIN-20171122175744-20171122195744-00058.warc.gz | 201,365,329 | 10,593 | # Thread: Is parity preserved when multiplying two group elements
1. ## Is parity preserved when multiplying two group elements
I want to show $H$ is a subgroup of $G$, an Abelian group, where $H=\{x \in G | |x| \ is \ odd \}$. If I assume $a,b\in H$ can I be certain $ab^{-1} \in H$, that is to say does multiplying two elements with odd order in an Abelian group produce an element with odd order? I've tried it out with a few specific groups and it seems to be the case but I haven't been able to come up with a proof.
2. ## Re: Is parity preserved when multiplying two group elements
Hey diehardwalnut.
Can you use Lagrange's theorem in this case?
3. ## Re: Is parity preserved when multiplying two group elements
I don't think so, Lagrange's equation doesn't occur in the book until chapter 5 and this question occurs in chapter 3.
4. ## Re: Is parity preserved when multiplying two group elements
Here are the facts you need:
1. In any group $G$ and $x\in G$, if $x^n=e,\text{ the identity of }G$, the order of $x$ divides $n$. If the order of $x$ is $m$ and $k$ is any integer, $x^{mk}=e$.
2. In an abelian group $G$ with $a,\,b\in G$ and $n$ any integer, $(ab)^n=a^nb^n$.
For your problem, let $a,\,b\in H\text{ with }|a|=m\text{ and }|b|=n$. Show $(ab^{-1})^{mn}=e$. Conclude that the order of $ab^{-1}$ is odd. | 395 | 1,332 | {"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} | 3.734375 | 4 | CC-MAIN-2017-47 | longest | en | 0.855489 |
https://gmatpractice.q-51.com/hard-math-gmat-sample-questions/gmat-maths-practice-questions-2.shtml | 1,638,229,058,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964358847.80/warc/CC-MAIN-20211129225145-20211130015145-00472.warc.gz | 349,428,986 | 15,764 | # GMAT Hard Math Questions | 11 to 20
1. #### GMAT 700 Level Sample Question in Statistics
Consider a set S = {2, 4, 6, 8, x, y} with distinct elements. If x and y are both prime numbers and 0 < x < 40 and 0 < y < 40, which of the following MUST be true?
I. The maximum possible range of the set is greater than 33.
II. The median can never be an even number.
III. If y = 37, the average of the set will be greater than the median.
1. I only
2. I and II only
3. I and III only
4. III only
5. I, II, and III
2. #### GMAT Hard Math Algebra - Absolute Values | GMAT Problem Solving
If x and y are integers and |x - y| = 12, what is the minimum possible value of xy?
1. -12
2. -18
3. -24
4. -36
5. -48
3. #### GMAT Challenging Question | Statistics Problem Solving
Three positive integers a, b, and c are such that their average is 20 and a ≤ b ≤ c. If the median is (a + 11), what is the least possible value of c?
1. 23
2. 21
3. 25
4. 26
5. 24
4. #### GMAT Hard Math | Arithmetic | Permutation Combination Sample Question
How many four-digit positive integers exist that contain the block 25 and are divisible by 75. (2250 and 2025 are two such numbers)?
1. 90
2. 63
3. 34
4. 87
5. 62
5. #### GMAT Challenging Math Question | Arithmetic | Number Properties DS Practice
Is |x| > x?
Statement 1: x2 + y2 = 4
Statement 2: x3 + y2 = 0
1. 5
2. 2
3. 9
4. 3
5. 7
10. #### GMAT Hard Math Questions | Arithmetic | Number Properties DS
x is a two-digit positive integer. y is obtained by multiplying the tens place of x by 2. Is y > $$frac{x}{6}$? Statement 1: 20 < x < 30 Statement 2: y = 10 #### GMAT Online CourseTry it free! Register in 2 easy steps and Start learning in 5 minutes! #### Already have an Account? #### GMAT Live Online Classes Next Batch Dec 12, 2021 ##### Where is Wizako located? Wizako - GMAT, GRE, SAT Prep An Ascent Education Initiative 14B/1 Dr Thirumurthy Nagar 1st Street Nungambakkam Chennai 600 034. India ##### How to reach Wizako? Phone:$91) 44 4500 8484
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Email: [email protected] | 672 | 2,065 | {"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": 3, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.25 | 4 | CC-MAIN-2021-49 | longest | en | 0.789546 |
https://miyklas.com.ua/p/fzika/7-klas/fizika-iak-prirodnicha-nauka-piznannia-prirodi-16472/fizichni-velichini-vimiriuvannia-fizichnikh-velichin-mizhnarodna-sistema-_-16477/re-f1022358-9a1b-4a83-bca0-2085e8fe1af3 | 1,652,916,816,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662522556.18/warc/CC-MAIN-20220518215138-20220519005138-00753.warc.gz | 478,856,726 | 9,603 | ### Теорія:
Вимірювання довжини
Вимірювати — означає порівнювати вимірювану величину з уже відомою одиницею виміру.
Еталон метра
Основною одиницею довжини в системі СІ є метр [м].
$\begin{array}{l}1\phantom{\rule{0.147em}{0ex}}\text{нм}=\phantom{\rule{0.147em}{0ex}}{10}^{-9}\phantom{\rule{0.147em}{0ex}}\text{м}=\phantom{\rule{0.147em}{0ex}}0,000000001\phantom{\rule{0.147em}{0ex}}\text{м}\phantom{\rule{0.147em}{0ex}}=\phantom{\rule{0.147em}{0ex}}\frac{1}{1000000000}\text{м}\phantom{\rule{0.147em}{0ex}}\phantom{\rule{0.147em}{0ex}}\phantom{\rule{0.147em}{0ex}}\phantom{\rule{0.147em}{0ex}}\phantom{\rule{0.147em}{0ex}}\left(\text{нанометр}\right)\\ 1\phantom{\rule{0.147em}{0ex}}\text{мкм}\phantom{\rule{0.147em}{0ex}}=\phantom{\rule{0.147em}{0ex}}{10}^{-6}\phantom{\rule{0.147em}{0ex}}\text{м}\phantom{\rule{0.147em}{0ex}}=\phantom{\rule{0.147em}{0ex}}0,000001\phantom{\rule{0.147em}{0ex}}\text{м}=\phantom{\rule{0.147em}{0ex}}\frac{1}{1000000}\text{м}\phantom{\rule{0.147em}{0ex}}\phantom{\rule{0.147em}{0ex}}\phantom{\rule{0.147em}{0ex}}\left(\text{мікрометр}\right)\\ 1\phantom{\rule{0.147em}{0ex}}\text{мм}\phantom{\rule{0.147em}{0ex}}=\phantom{\rule{0.147em}{0ex}}{10}^{-3}\phantom{\rule{0.147em}{0ex}}\text{м}\phantom{\rule{0.147em}{0ex}}=\phantom{\rule{0.147em}{0ex}}0,001\phantom{\rule{0.147em}{0ex}}\text{м}=\phantom{\rule{0.147em}{0ex}}\frac{1}{1000}\text{м}\phantom{\rule{0.147em}{0ex}}\phantom{\rule{0.147em}{0ex}}\phantom{\rule{0.147em}{0ex}}\left(\text{міліметр}\right)\\ 1\phantom{\rule{0.147em}{0ex}}\text{см}\phantom{\rule{0.147em}{0ex}}=\phantom{\rule{0.147em}{0ex}}0,01\phantom{\rule{0.147em}{0ex}}\text{м}\phantom{\rule{0.147em}{0ex}}=\phantom{\rule{0.147em}{0ex}}\frac{1}{100}\text{м}\phantom{\rule{0.147em}{0ex}}\phantom{\rule{0.147em}{0ex}}\phantom{\rule{0.147em}{0ex}}\phantom{\rule{0.147em}{0ex}}\phantom{\rule{0.147em}{0ex}}\left(\text{сантиметр}\right)\\ 1\text{дм}\phantom{\rule{0.147em}{0ex}}=\phantom{\rule{0.147em}{0ex}}0,1\phantom{\rule{0.147em}{0ex}}\text{м}\phantom{\rule{0.147em}{0ex}}=\phantom{\rule{0.147em}{0ex}}\frac{1}{10}\text{м}\phantom{\rule{0.147em}{0ex}}\phantom{\rule{0.147em}{0ex}}\phantom{\rule{0.147em}{0ex}}\phantom{\rule{0.147em}{0ex}}\phantom{\rule{0.147em}{0ex}}\left(\text{дециметр}\right)\\ 1\text{км}\phantom{\rule{0.147em}{0ex}}=\phantom{\rule{0.147em}{0ex}}1000\phantom{\rule{0.147em}{0ex}}\text{м}\phantom{\rule{0.147em}{0ex}}\phantom{\rule{0.147em}{0ex}}\phantom{\rule{0.147em}{0ex}}\phantom{\rule{0.147em}{0ex}}\phantom{\rule{0.147em}{0ex}}\phantom{\rule{0.147em}{0ex}}\phantom{\rule{0.147em}{0ex}}\left(\text{кілометр}\right)\end{array}$
Вимірювальні прилади, якими вимірюють довжину тіла:
Вимірювальна стрічка Вимірювальна рулетка Лінійка Лазерний далекомір
Для визначення розміру малих тілвикористовують штангенциркуль та мікрометр.
Штангенциркуль
Мікрометр
У давнину першими одиницями довжини служили людський зріст або розмір будь-якої частини тіла.
У різних народів зустрічаються різні системи вимірювань.
Історично довжина вимірювалася у косових сажнях (відстань від носка лівої ноги до кінця середнього пальця піднятої вгору правої руки), ліктях (довжина руки від пальців до ліктя) і т.д.
Англійські неметричні одиниці довжини
Морська миля $$1,85$$ км Фут $$304,8$$ мм Дюйм (inch) $$25,4$$ мм Ярд $$914,4$$ мм
$\text{Калібр}=\frac{1}{100}\phantom{\rule{0.147em}{0ex}}\text{дюйма}=254\phantom{\rule{0.147em}{0ex}}\text{мкм}$
Українські неметричні одиниці довжини
Миля $$7,47$$ км Верста $$1,07$$ км Аршин $$711,2$$ мм Вершок $$44,45$$ мм | 1,615 | 3,527 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 2, "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 | 4 | CC-MAIN-2022-21 | latest | en | 0.149728 |
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# Linear Equation
Recall:
• When we studied scatter plots , we used MS Excel to graph a ‘trendline’ to find the best line through
the data points.
• We see from the line above that as the weight of the car increases, the MPG decreases.
• In fact, if we look at the equation we see it is y = –1.0965x + 67.354
• The slope of the line (–1.096, the number in front of the x) represents the relationship between the
variables weight and MPG. So as the weight increases 1 unit (100 lbs) the MPG decreases by 1.0965.
• The graph of this line represents a linear relationship between weight and MPG.
Background:
The symbol Δ (read delta) represents a change in a quantity.
• So Δx represents the change in x . It is found by
Δx = end value – starting value
• So if weight (x) goes from 400lbs to 700lbs, the change in weight ( Δx ) would be
Δx = 700 − 400 = 300 lbs
• Δx does not represent multiplication !
Equation of a Line:
• If we have the equation, then we have
m = 2 and b = 4
• The letters x and y are variables, meaning they vary or change along the line. At least one of them
must be nonzero. Together they represent the ordered pairs (x, y) on the graph.
• The equation is a rule that assigns a y for any x we put into the equation.
• So, we can put x = 3 into the equation like so
• And we get out a 10. This means that (3, 10) is a point on the graph of the line. We can also write
y(3) = 10
• We can find other values from by plugging in different values of x
• Putting together these (and other) points, we can graph the line it would look like the
following…
• Notice that the first point on the graph is (0,4) and the second is (1,6), and so on.
• Usually y depends on the value of x . So x is graphed on the horizontal and y on the vertical.
• m represents the slope of the line . In our example above, the slope is 2.
• b is the y-intercept, or where the line crosses the y axis. In our example above, the y- intercept is 4.
The Slope:
• m, the slope of the line can be positive or negative , large or small.
• It is a measure of
.
• Let’s find the slope of the line between the points (1,6) and (3,10). Notice these are points on our
example line above.
• This would be true for any two points on the line. Try it for (3,10) and (7, 18)
Generating a Table and Graph from an Equation:
• Generate a table and graph for
When x = 0, . This is Line 1 on our table, and the point (0,1) on the plot
When x = 2, . This is the Line 2 on our table, and the point (2,7) on the plot.
Verify the rest of the points and then draw a line through them.
• Notice that for the above problem, the slope (m) is 3, the y-intercept (b) is 1.
Determining Linear Relationships:
• There are several ways to determine a relationship is linear
− The graph (picture) is a line
− The equation is given and it is linear in form (y = mx + b)
− You are told something increases (or decreases) by a constant or fixed amount
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http://books.google.com/books?id=zmELAAAAYAAJ&dq=related:HARVARDHN3FUP | 1,386,875,610,000,000,000 | text/html | crawl-data/CC-MAIN-2013-48/segments/1386164677368/warc/CC-MAIN-20131204134437-00011-ip-10-33-133-15.ec2.internal.warc.gz | 25,015,762 | 23,649 | # Elements of Geometry and Trigonometry: From the Works of A.M. Legendre (Google eBook)
A.S. Barnes & Company, 1857 - Geometry - 448 pages
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### Contents
hmDDCTI01I 13 Propositions 21 BOOK II 47 BOOK III 57 Problems relating to the First and Third Books 76 BOOK IV 87 Problems relating to the Fourth Book 122 BOOK V 135
Problems 267 Definitions of Trigonometrical Lines 273 Solut1on of Triangles 281 Solution of RightAngled Triangles 287 Circular Functions 297 of Formulas 306 Homogeneity of Terms 313 SPHERICAL TRIGONOMETRY 321
Planes and Polyedral Angles 150 BOOK VII 174 BOOK VIII 202 BOOK IX 227 PAG 245 PLANE TRIGONOMETRY 255 Multiplication by Logarithms 261
Napiers Analogies 329 Of Quadrantal Triangles 335 MENSURATION OF SURFACES 347 Area of a Regular Polygon 353 piem 358 Convex Surface of a Cone 364
### Popular passages
Page 227 - A spherical triangle is a portion of the surface of a sphere, bounded by three arcs of great circles.
Page 34 - If two right-angled triangles have the hypothenuse and a side of the one, equal to the hypothenuse and a side of the other, each to each, the triangles are equal. Let...
Page 271 - The circumference of every circle is supposed to be divided into 360 equal parts, called degrees...
Page 278 - In any triangle, the sum of the two sides containing either angle, is to their difference, as the tangent of half the sum of the two other angles, to the tangent of half their difference.
Page 107 - If two triangles have two angles of the one equal to two angles of the other, each to each, and also one side of the one equal to the corresponding side of the other, the triangles are congruent.
Page 371 - O's, points or dots are introduced instead of the 0's through the rest of the line, to catch the eye, and to indicate that from thence the annexed first two figures of the Logarithm in the second column stand in the next lower line. N'.
Page 43 - BtSL hence the sum of all the interior and exterior angles, is equal to twice as many right angles as the polygon has sides.
Page 119 - The angle formed by a tangent and a chord is measured by half the intercepted arc.
Page 30 - B : hence the two triangles have two sides and the included angle of the one equal to two sides and the included angle of the other, each to each : hence, the two triangles are equal (Th.
Page 97 - The square described on the hypothenuse of a rightangled triangle is equal to the sum of the squares described on the other two sides.
### References from web pages
The Online Books Page: Elements of Geometry and Trigonometry, by ...
Title: Elements of Geometry and Trigonometry · Author: Legendre, am · Author: Davies, Charles, 1798-1876 · Editor: Van Amringe, J. Howard ...
onlinebooks.library.upenn.edu/ webbin/ book/ lookupid?key=olbp21963
Free Books > Computers & Internet > Web Development > HTML ...
Free Books > Computers & Internet > Web Development > HTML, Graphics, & Design > Web Graphics > Elements Of Geometry And Trigonometry.
2020ok.com/ books/ 29/ elements-of-geometry-and-trigonometry-19129.htm
Elements of Geometry, or “Éléments de géométrie” (work by Legendre ...
... was Adrien-Marie Legendre's textbook Éléments de géométrie (Elements of Geometry and Trigonometry), the first edition of which appeared in 1794. ...
www.britannica.com/ eb/ topic-184288/ Elements-of-Geometry
No. 472: Legendre's Math Text
Davies, C., Elements of Geometry and Trigonometry . from the Works of am Legendre, Revised and Adapted to the Course of Mathematical Instruction in the ...
www.uh.edu/ engines/ epi472.htm
JSTOR: American Contributions to Mathematical Symbolism
This was expressed by the sign -~ which occurs in C. Davies' Elements of Geometry and Trigonometry,2 1851. For a quarter of a century following 1885, ...
Collection of Early American Mathematics Books
Main Title TM: Elements of geometry and trigonometry / translated from the French of am Legendre by David Brewster ; revised and adapted to the course of ...
www.math.gatech.edu/ ~hill/ publications/ books/ booklist.html
Earliest Known Uses of Some of the Words of Mathematics
In 1828, Elements of Geometry and Trigonometry (1832) by David Brewster (a translation of Legendre) has:. AGO, GOC, we have already named interior angles on ...
www.luigigobbi.com/ Earliest%20Known%20Use%20of%20Some%20of%20the%20Words%20of%20Mathematics/
Origins of some arithmetic terms
APOTHEM is found in 1828 in Elements of Geometry and Trigonometry (1832) by David Brewster (a translation of Legendre): The radius OT of the inscribed ...
www.pballew.net/ arithme1.html
Amy Ackerberg-Hastings
In his geometry textbook, Elements of Geometry and Trigonometry, Davies mixed together content from French, Scottish, and English textbooks with the goal of ...
www.math.usma.edu/ people/ Rickey/ dms/ DeptHeads/ Davies%20by%20Amy%20Ackerberg-Hastings.rtf
Earliest Known Uses of Some of the Words of Mathematics (P)
In 1832 Elements of Geometry and Trigonometry by David Brewster, which is a translation of Legendre, has:. The word parallelogram, according to its ...
members.aol.com/ jeff570/ p.html | 1,314 | 5,184 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.890625 | 4 | CC-MAIN-2013-48 | longest | en | 0.838744 |
http://senselimitsaventura.com/rational-numbers-examples-in-real-life.html | 1,618,884,159,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038921860.72/warc/CC-MAIN-20210419235235-20210420025235-00059.warc.gz | 90,298,025 | 8,908 | # Rational Numbers Examples In Real Life
### The product of two rational numbers is always a rational number.
Rational numbers examples in real life. The number 8 is rational because it can be expressed as the fraction 8/1 (or the fraction 16/2) the fraction 5/7 is a rational number because it is the quotient of two integers 5 and 7. Invent a story of a pirate's ship coming to an island. Rational numbers are real numbers which can be written in the form of p/q where p,q are integers and q ≠ 0.
It is also a type of real number. A friend of mine who was a carpenter needed to install a round ceiling light fixture and wanted to know how large a square hole could he cut in the ceiling so that the hole didn't extend beyond the rim of the fixture. So we observe that rational numbers includes natural numbers, whole numbers, integers and even the fraction numbers.
In maths, rational numbers are represented in p/q form where q is not equal to zero. Any number that can be expressed in the form a/b, where a and b are integers and b≠0 is called a rational number. If this approximation is adequate than you are able to represent the real world using rational numbers.
Use of rational numbers in real life : Use the relationship between speed and time to explore rational functions and discover asymptotes in the real world. So, rational numbers are used everywhere in real life leaving some special cases.
The amount of fingers on your hand; ★when you share a pizza or anything. Is done on edurev study group by class 8 students.
See more ideas about rational numbers, numbers, integers. Every real number corresponds to a point on the number line, and every point on the number line corresponds to a real number. They simply “are”, in some ineffable sense.
### Pin on Math on the Move Resources
Source : pinterest.com | 394 | 1,830 | {"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} | 3.609375 | 4 | CC-MAIN-2021-17 | latest | en | 0.929593 |
https://www.enotes.com/homework-help/sum-two-natural-numbers-212-258917 | 1,516,558,534,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084890823.81/warc/CC-MAIN-20180121175418-20180121195418-00760.warc.gz | 880,812,314 | 9,872 | # The sum of two natural numbers is 212.What are the numbers if the larger number divided by the smaller number gives the quotient 3 and the reminder 4 ?
justaguide | Certified Educator
Let the numbers be A and B, where A is the larger number
Their sum is 212
=> A + B = 212
The larger number divided by the smaller number gives a quotient 3 and a reminder 4
=> A = 3B + 4
Substitute in A + B = 212
=> 3B + 4 + B = 212
=> 4B + 4 = 212
=> 4B = 208
=> B = 52
A = 212 - 52 = 160
The numbers are 160 and 52.
giorgiana1976 | Student
Let x and y be the natural numbers.
The sum of these natural numbers is 212.
x + y = 212 (1)
Let x be larger than y.
We'll write the reminder theorem to express the 2nd constraint.
x = 3y + 4 (2)
We'll replace (2) in (1):
3y + 4 + y = 212
We'll combine like terms and we'll subtract 4 both sides:
4y = 212 - 4
4y = 208 => y = 52
x = 3*52 + 4
x = 160
The required two natural numbers, that respect the given constraints, are: x = 160 and y = 52. | 331 | 1,000 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.5625 | 5 | CC-MAIN-2018-05 | latest | en | 0.860376 |
http://www.cseblog.com/2015/02/buying-in-rocket-ships-and-selling-in.html | 1,526,928,755,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794864466.23/warc/CC-MAIN-20180521181133-20180521201133-00054.warc.gz | 370,169,618 | 29,353 | ### Buying in Rocket Ships and Selling in Fire Sale
Source: Asked to me by Ankush Jain (CSE IITB 2011, Morgan Stanley Quant Associate). He took it from Algorithms Design book by Tardos and Kleinberg
Problem:
Easy case:
You’re trying to buy equipments whose costs are appreciating. Item i appreciates at a rate of r_i > 1 per month, starting from \$100, so if you buy it t months from now you will pay 100*((r_i)^t). If
you can only buy one item per month, what is the optimal order in which to buy them?
Difficult case:
You’re trying to sell equipments whose costs are depreciating. Item i depreciates at a rate of r_i < 1 per month, starting from \$100, so if you sell it t months from now you will get 100*((r_i)^t). If
you can only sell one item per month, what is the optimal order in which to sell them?
1. Are all items priced at \$100? If they are, what's wrong with buying (and selling) the one with highest r_i first (and then in order of decreasing r_i)?
Put more formally, considering n items, I would use a grid:
1,(1+r1),(1+r1)^2,.....(1+r1)^n-1
.
.
1,(1+rn),(1+rn)^2,.....(1+rn)^n-1
The question is now to select one no. from each row and column to minimise the sum. It can be proven that this happens when you select in order of decreasing r_i.
I feel like I'm missing something big here.. What part of above is incorrect?
1. it is correct, it is a greedy solution. the solution is simple, i think it is the correctness proof that requires some clever arguments
2. For the selling case, consider the following rates( in absolute terms ) :
99, 99, 1, 0.5
If you sell them in order of decreasing rates, you will get the following returns :
100/(1) + 100/(100) + 100/(2*2) + 100/(1.5*1.5*1.5 ) = 155.63
This is not the best selling sequence. A better sequence is : ( 99, 1, 0.5, 99 )
100/(1) + 100/(2) + 100/(1.5*1.5) + 100/(100*100*100 ) = 194.44
2. For the selling case, I think, we'd need to select in the order of increasing r_i. Since, we're trying to maximise the sum in that case. Correct?
3. How to solve the second part apart from an exponential method
4. just a thought, is there an optimized way, even if initial prices of items differ ? like p,q,r,...instead of all being 100. i guess we should seek the sum of the entire column instead of just looking for r_i
5. For selling:
I think we should calculate decrease in price at each time step for each remaining item and sell the one which might depreciate the most in value if we do not sell it immediately. This way we are greedily minimizing the total loss incurred.
Proof doesn't seem to be obvious, but it works on toy cases I constructed. Has anyone proved this or has a link to a solution?
6. what happens if the cost of each of the items are different ??
My hypothesis is that we can, while buying, make a grid similar to what has been suggested by John and buy the most expensive item each month, deleting that row and column every month. In the limiting case of all items costing the same, the solution reduces to that of John.
7. what happens if the cost of each of the items are different ??
My hypothesis is that we can, while buying, make a grid similar to what has been suggested by John and buy the most expensive item each month, deleting that row and column every month. In the limiting case of all items costing the same, the solution reduces to that of John. | 896 | 3,369 | {"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} | 3.875 | 4 | CC-MAIN-2018-22 | longest | en | 0.878512 |
http://www.jiskha.com/search/index.cgi?query=Math+sin%2Fcos | 1,493,312,368,000,000,000 | text/html | crawl-data/CC-MAIN-2017-17/segments/1492917122619.60/warc/CC-MAIN-20170423031202-00059-ip-10-145-167-34.ec2.internal.warc.gz | 587,646,404 | 13,801 | # Math sin/cos
136,533 results
TRIG!
Posted by hayden on Monday, February 23, 2009 at 4:05pm. sin^6 x + cos^6 x=1 - (3/4)sin^2 2x work on one side only! Responses Trig please help! - Reiny, Monday, February 23, 2009 at 4:27pm LS looks like the sum of cubes sin^6 x + cos^6 x = (sin^2x)^3 + (cos^2x)^3 = (sin^2x+cos...
tigonometry
expres the following as sums and differences of sines or cosines cos8t * sin2t sin(a+b) = sin(a)cos(b) + cos(a)sin(b) replacing by by -b and using that cos(-b)= cos(b) sin(-b)= -sin(b) gives: sin(a-b) = sin(a)cos(b) - cos(a)sin(b) Add the two equations: sin(a+b) + sin(a-b) = ...
Mathematics - Trigonometric Identities
Let y represent theta Prove: 1 + 1/tan^2y = 1/sin^2y My Answer: LS: = 1 + 1/tan^2y = (sin^2y + cos^2y) + 1 /(sin^2y/cos^2y) = (sin^2y + cos^2y) + 1 x (cos^2y/sin^2y) = (sin^2y + cos^2y) + (sin^2y + cos^2y) (cos^2y/sin^2y) = (sin^2y + cos^2y) + (sin^2y + cos^2y)(sin^2y) / (sin^...
algebra
Can someone please help me do this problem? That would be great! Simplify the expression: sin theta + cos theta * cot theta I'll use A for theta. Cot A = sin A / cos A Therefore: sin A + (cos A * sin A / cos A) = sin A + sin A = 2 sin A I hope this will help. in my algebra ...
trig
Reduce the following to the sine or cosine of one angle: (i) sin145*cos75 - cos145*sin75 (ii) cos35*cos15 - sin35*sin15 Use the formulae: sin(a+b)= sin(a) cos(b) + cos(a)sin(b) and cos(a+b)= cos(a)cos(b) - sin(a)sin)(b) (1)The quantity = sin(145-75) = sin 70 = cos 20 note that...
math
Can you please check my work. A particle is moving with the given data. Find the position of the particle. a(t) = cos(t) + sin(t) s(0) = 2 v(0) = 6 a(t) = cos(t) + sin(t) v(t) = sin(t) - cos(t) + C s(t) = -cos(t) - sin(t) + Cx + D 6 = v(0) = sin(0) -cos(0) + C C=7 2= s(0) = -...
math
Prove that for all real values of a, b, t (theta): (a * cos t + b * sin t)^2 <= a^2 + b^2 I will be happy to critique your work. Start on the left, square it, (a * cos t + b * sin t)^2 = a^2 (1 - sin^2t) + 2ab sin t cost+ b^2 (1 - cos^2 t)= a^2 + b^2 - (a sin t - b cos t)^2...
Trig
Find sin(s+t) and (s-t) if cos(s)= 1/5 and sin(t) = 3/5 and s and t are in quadrant 1. =Sin(s)cos(t) + Cos(s)Sin(t) =Sin(1/5)Cos(3/5) + Cos(-1/5)Sin(3/5) = 0.389418 Sin(s-t) =sin(s)cos(t) - cos(s)sin(t) =sin(-3/5)cos(1/5) - cos(1/5)sin(3/5) =Sin-3/5 cos-3/5 = -0.46602 HELP ...
Trig
Given: cos u = 3/5; 0 < u < pi/2 cos v = 5/13; 3pi/2 < v < 2pi Find: sin (v + u) cos (v - u) tan (v + u) First compute or list the cosine and sine of both u and v. Then use the combination rules sin (v + u) = sin u cos v + cos v sin u. cos (v - u) = cos u cos v + ...
pre-cal
Simplify the given expression........? (2sin2x)(cos6x) sin 2x and cos 6x can be expressed as a series of terms that involve sin x or cos x only, but the end result is not a simplification. sin 2x = 2 sinx cosx cos 6x = 32 cos^6 x -48 cos^4 x + 18 cos^2 x - 1 I assume you are ...
math
Given that sin x + sin y = a and cos x + cos y =a, where a not equal to 0, express sin x + cos x in terms of a. attemp: sin x = a - sin y cos x = a - cos y sin x + cos x = 2A - (sin y + cos y)
1.)Find dy/dx when y= Ln (sinh 2x) my answer >> 2coth 2x. 2.)Find dy/dx when sinh 3y=cos 2x A.-2 sin 2x B.-2 sin 2x / sinh 3y C.-2/3tan (2x/3y) D.-2sin2x / 3 cosh 3yz...>> my answer. 2).Find the derivative of y=cos(x^2) with respect to x. A.-sin (2x) B.-2x sin (x^2...
Precal
I do not understand how to do this problem ((sin^3 A + cos^3 A)/(sin A + cos A) ) = 1 - sin A cos A note that all the trig terms are closed right after there A's example sin A cos A = sin (A) cos (A) I wrote it out like this 0 = - sin^6 A - cos^6 A + 2sin^3 A cos^3 A - 2sin^3 ...
trig
The expression 4 sin x cos x is equivalent to which of the following? (Note: sin (x+y) = sin x cos y + cos x sin y) F. 2 sin 2x G. 2 cos 2x H. 2 sin 4x J. 8 sin 2x K. 8 cos 2x Can someone please explain how to do this problem to me?
Calc.
Differentiate. y= (cos x)^x u= cos x du= -sin x dx ln y = ln(cos x)^x ln y = x ln(cos x) (dy/dx)/(y)= ln(cos x) (dy/dx)= y ln(cos x) = (cos x)^x * (ln cos x) (dx/du)= x(cos x)^(x-1) * (-sin x) = - x sin(x)cos^(x-1)(x) (dy/dx)-(dx/du)= [(cos^x(x))(ln(cos(x)))-(x sin(x)cos^(x-1...
calculus
Differentiate. y= (cos x)^x u= cos x du= -sin x dx ln y = ln(cos x)^x ln y = x ln(cos x) (dy/dx)/(y)= ln(cos x) (dy/dx)= y ln(cos x) = (cos x)^x * (ln cos x) (dx/du)= x(cos x)^(x-1) * (-sin x) = - x sin(x)cos^(x-1)(x) (dy/dx)-(dx/du)= [(cos^x(x))(ln(cos(x)))-(x sin(x)cos^(x-1...
Trigonometry
Solve the equation for solutions in the interval 0<=theta<2pi Problem 1. 3cot^2-4csc=1 My attempt: 3(cos^2/sin^2)-4/sin=1 3(cos^2/sin^2) - 4sin/sin^2 = 1 3cos^2 -4sin =sin^2 3cos^2-(1-cos^2) =4sin 4cos^2 -1 =4sin Cos^2 - sin=1/4 (1-sin^2) - sin =1/4 -Sin^2 - sin =-3/4 ...
Math
Solve this equation algebraically: (1-sin x)/cos x = cos x/(1+sin x) --- I know the answer is an identity, and when graphed, it looks like cot x. I just don't know how to get there. I tried multiplying each side by its conjugate, but I still feel stuck. This is what I have so ...
math
how would you prove that sin^2(a)-cos^2(b)= sin^2(b)-cos^2(a). i'm not completely sure that this is right but i used the difference of two squares on it to get (sin(a)+cos(b))(sin(a)-cos(b)) then after that i am stuck. please help
trig
it says to verify the following identity, working only on one side: cotx+tanx=cscx*secx Work the left side. cot x + tan x = cos x/sin x + sin x/cos x = (cos^2 x +sin^2x)/(sin x cos x) = 1/(sin x cos x) = 1/sin x * 1/cos x You're almost there. thanks so much! i could not figure...
Use the fundamental identities to simplify the expression. tan^2 Q / sec^2 Q sin^2/cos^2 / 1/cos^2 = sin^2 / cos^2 times cos^2 / 1 = The cos^2 cancels out so sin^2 is left. Is this correct?
What are the ratios for sin A and cos A? The diagram is not drawn to scale. Triangle Description- AB = 29 AC = 20 BC - 21 A. sin A = 20/29, cos A = 21/29 B. sin A = 21/29, cos A = 20/21 C. sin A = 21/29, cos A = 20/29****? D. sin A = 21/20, cos A = 20/21 Please help and explain!
Math
the original problem was: (sin x + cos x)^2 + (sin x - cos x)^2 = 2 steps too please I got 1 for (sin x + cos x)^2 but then what does (sin x - cos x)^2 become since it's minus?
math
Proving Trigonometric Identities 1. sec^2x + csc^2x= (sec^2 x)(csc^2 x) 2. sin ^3 x / sin x - cos 3x / cos x = 2 3. 1- cos x/ sin x= sin x/ 1+ cos x 4. 2 sin x cos ^2 (x/2)- 1/x sin (2x) = sinx 5. cos 2 x + sin x/ 1- sin x= 1+ 2 sin x
Math
Prove each identity: a) 1-cos^2x=tan^2xcos^2x b) cos^2x + 2sin^2x-1 = sin^2x I also tried a question on my own: tan^2x = (1 – cos^2x)/cos^2x R.S.= sin^2x/cos^2x I know that the Pythagorean for that is sin^2x + cos^2x That's all I could do.
Math
State the restrictions on the variables for these trigonometric identities. a)(1 + 2 sin x cos x)/ (sin x + cos x) = sin x + cos x b) sin x /(1+ cos x) = csc x - cot x
Math (Linear Systems)
28 N + T2 sin 12 = T1 sin 42 T2 cos 12 = T1 cos 42 T2 sin 12 + T3 sin 54 = W2 T2 cos 12 = T3 cos 54 Im solving for T1,2,3 and W2 I just cant seem to get the system to work
Integral
That's the same as the integral of sin^2 x dx. Use integration by parts. Let sin x = u and sin x dx = dv v = -cos x du = cos x dx The integral is u v - integral of v du = -sinx cosx + integral of cos^2 dx which can be rewritten integral of sin^2 x = -sinx cos x + integral of (...
Trigonometry
Does anyone have a good website that shows the proofs for these equations? sin(u+v) = sin(u)cos(v) + sin(v)cos(u) cos(u+v) = cos(u)cos(v) + sin(v)sin(u) Thanks!
Trigonometry
Please review and tell me if i did something wrong. Find the following functions correct to five decimal places: a. sin 22degrees 43' b. cos 44degrees 56' c. sin 49degrees 17' d. tan 11degrees 37' e. sin 79degrees 23'30' f. cot 19degrees 0' 25'' g. tan 64degrees 6' 45'' h. cos...
Math, Pre-Calc
the original problem was: Solve: sin(3x)-sin(x)=cos(2x) so far i've gooten to: sin(x)(2sin(x)cos(x)-1)=cos^2(x)-sin^2(x) Where would I go from here?
Geometry One multiple choice question!
Write the ratios for sin A and cos A. {picture is of a right triangle, ABC. segment AC is 8, segment AB is 17, and segment CB is 15.} sin A=15/17, cos A=8/17 sin A=15/8, cos A=8/17 sin A=15/17, cos A=8/15 sin A=8/17, cos A=15/17
Write the ratios for sin A and cos A. {picture is of a right triangle, ABC. segment AC is 8, segment AB is 17, and segment CB is 15.} sin A=15/17, cos A=8/17 sin A=15/8, cos A=8/17 sin A=15/17, cos A=8/15 sin A=8/17, cos A=15/17
MATH
Hi, I really need help with these questions. I did some of them halfway, but then I got stuck. Would you please help me? Thank you so much. Prove the identity.... 1. sec x + tan x(1-sin x/cos x)=1 1/cos x + sin x/cos x(cos^2 x/cos x)=1 1+sin x/cos x(cos^2x/cos x)=1 I got stuck...
Math
the original problem was: Solve: sin(3x)-sin(x)=cos(2x) so far i've gotten to: sin(x)(2sin(x)cos(x)-1)=cos^2(x)-sin^2(x) Where would I go from here?
Precalc
Let x, y, and z be real numbers such that cos(x) + cos(y) + cos(z) = sin(x) + sin(y) + sin(z) = 0. Prove that cos(2x) + cos(2y) + cos(2z) = sin(2x) + sin(2y) + sin(2z) = 0.
math
Eliminate the parameter (What does that mean?) and write a rectangular equation for (could it be [t^2 + 3][2t]?) x= t^2 + 3 y = 2t Without a calculator (how can I do that?), determine the exact value of each expression. cos(Sin^-1 1/2) Sin^-1 (sin 7pi/6) x= t^2 + 3 y = 2t ...
precalculus
For each of the following determine whether or not it is an identity and prove your result. a. cos(x)sec(x)-sin^2(x)=cos^2(x) b. tan(x+(pi/4))= (tan(x)+1)/(1-tan(x)) c. (cos(x+y))/(cos(x-y))= (1-tan(x)tan(y))/(1+tan(x)tan(y)) d. (tan(x)+sin(x))/(1+cos(x))=tan(x) e. (sin(x-y...
1)tan Q = -3/4 Find cosQ -3^2 + 4^2 = x^2 9+16 = sqrt 25 = 5 cos = ad/hy = -4/5 Am I correct? 2) Use the sum and difference identites sin[x + pi/4] + sin[x-pi/4] = -1 sinx cospi/4 + cosxsin pi/4 + sinx cos pi/4 - cosx sin pi/4 = -1 2 sin x cos pi/4 =-1 cos pi/4 = sqr2/2 2sin^x...
Trigonometry
I need help with I just can't seem to get anywhere. this is as far as I have got: Solve for b arcsin(b)+ 2arctan(b)=pi arcsin(b)=pi-2arctan(b) b=sin(pi-2arctan(b)) Sub in Sin difference identity let 2U=(2arctan(b)) sin(a-b)=sinacosb-cosasinb =(sin(pi))(cos(2U))-(cos(pi))(sin(...
Trig!
The identities cos(a-b)=cos(a)cos(b)sin(a)sin(b) and sin(a-b)=sin(a)cos(b)-cos(a)sin(b) are occasionally useful. Justify them. One method is to use rotation matricies. Another method is to use the established identities for cos(a+b) and sin (a+b).
Math
I need help solving for all solutions for this problem: cos 2x+ sin x= 0 I substituted cos 2x for cos^2x-sin^2x So it became cos^2(x)-sin^2(x) +sinx=0 Then i did 1-sin^2(x)-sin^2(x)+sinx=0 = 1-2sin^2(x)+sinx=0 = sinx(-2sinx+1)=-1 What did i do wrong?? the real solutions are ...
Trig
If angle A is 45 degrees and angle B is 60 degrees. Find sin(A)cos(B), find cos(A)sin(B), find sin(A)sin(B), and find cos(A)cos(B) The choises for the first are: A. 1/2[sin(105)+sin(345)] B. 1/2[sin(105)-sin(345)] C. 1/2[sin(345)+cos(105)] D. 1/2[sin(345)-cos(105)] You don't ...
Math - Calculus
The identity below is significant because it relates 3 different kinds of products: a cross product and a dot product of 2 vectors on the left side, and the product of 2 real numbers on the right side. Prove the identity below. | a × b |² + (a • b)² = |a|²|b|² My work, ...
Math - Solving Trig Equations
What am I doing wrong? Equation: sin2x = 2cos2x Answers: 90 and 270 .... My Work: 2sin(x)cos(x) = 2cos(2x) sin(x) cos(x) = cos(2x) sin(x) cos(x) = 2cos^2(x) - 1 cos(x) (+/-)\sqrt{1 - cos^2(x)} = 2cos^2(x) - 1 cos^2(x)(1 - cos^2(x)) = 4cos^4(x) - 4cos^2(x) + 1 5cos^4(x) - 5cos^...
math
find all solutions in the interval [0,2 pi) sin(x+(3.14/3) + sin(x- 3.14/3) =1 sin^4 x cos^2 x Since sin (a+b) = sina cosb + cosb sina and sin (a-b) = sina cosb - cosb sina, the first problem can be written 2 sin x cos (pi/3)= sin x The solution to sin x = 1 is x = pi/2 For ...
Trig.
tan^2BeatacscBeta-tan^2 (simplify) (sin/cos)^2Beta times 1/sin-(sin/cos)^2 (sin^2/cos^2)-(sin^2/cos^2)=-sin/cos Is this correct?
2) Use the sum and difference identites sin[x + pi/4] + sin[x-pi/4] = -1 sinx cospi/4 + cosxsin pi/4 + sinx cos pi/4 - cosx sin pi/4 = -1 2 sin x cos pi/4 =-1 cos pi/4 = sqr2/2 2sin^x(sqrt2/2) = -1 sin x = -sqrt2 x = 7pi/4 and 5pi/4 Am I correct?
precalc
prove the identity: cos^4 - sin^4 = cos^2 - sin^2 (cos^2 + sin^2)(cos^2 - sin^2) cos^2 + sin^2 = 1 cos^2 - sin^2 = cos^2 - sin^2 is this correct?
maths
Choose the option that gives an expression for the indefinite integral ʃ (cos(4x) + 2x^2)(sin(4x) − x) dx. In each option, c is an arbitrary constant. Options A cos(4x) + 2x^2 +c B -1/8cos(4x) + 2x^2)^2 +c C 1/4 (sin(4x) − x)^2 + c D (1/(2 (sin(4x) − x...
Pre-Calculus
I don't understand,please be clear! Prove that each equation is an identity. I tried to do the problems, but I am stuck. 1. cos^4 t-sin^4 t=1-2sin^2 t 2. 1/cos s= csc^2 s - csc s cot s 3. (cos x/ sec x -1)- (cos x/ tan^2x)=cot^2 x 4. sin^3 z cos^2 z= sin^3 z - sin^5 z
Precalculus with Trigonometry
Prove or disprove the following Identities: cos(-x) - sin(-x) = cos(x) + sin (x) sin raised to the 4 (theta) - cos raised to the 4 (theta) = sin squared (theta) - cos squared (theta) cos (x+(pi)/(6)) + sin (x - (pi)/(3)) = 0 cos(x+y)cos(x-y) = cos squared (x) - sin squared (y...
Solve the following trig equations. give all the positive values of the angle between 0 degrees and 360 degrees that will satisfy each. give any approximate value to the nearest minute only. 1. sin2Į = (sqrt 3)/2 2. sin^2Į = cos^2Į + 1/2 3. sin 2x - cosx = 0 4. cos 4x...
Trig
sin^4t-cos^4t/sin^2t cos^2t= sec^2t-csc^2t i have =(sin^2t+cos^2t)(sin^2t+cos^2t)/sin^2tcos^2t then do i go =(sin^2t+cos^2t)/sin^2tcos^2t stumped
Given sin (x-y)=1/4, sin x cos y=3/5,and tan y=3/2,without using mathematical table or scientific calculator,find the values for: (a)cos x sin y (b)cos (x+y)
pre calc trig check my work please
sin x + cos x -------------- = ? sin x sin x cos x ----- + ----- = sin x sin x cos x/sin x = cot x this is what i got, the problem is we have a match the expression to the equation work sheet and this is not one of the answers. need to figure out what im doing wrong so i can ...
maths
Choose the two options which are true for all values of x 1) cos (x) = cos ( x – pie/2) 2) sin (x + pie/2) = cos (x – pie/2) 3) cos (x) = sin (x – pie/2) 4) sin (x) = sin (x + 4pie) 5) sin (x) = cos (x – pie/2) 6) sin^2 (x) + cos^2 (x) = pie would it be 1 and 3 ?? I ...
Pre-Calculus
Prove that each equation is an identity. I tried to do the problems, but I am stuck. 1. cos^4 t-sin^4 t=1-2sin^2 t 2. 1/cos s= csc^2 s - csc s cot s 3. (cos x/ sec x -1)- (cos x/ tan^2x)=cot^2 x 4. sin^3 z cos^2 z= sin^3 z - sin^5 z
Math
3. find the four angles that define the fourth root of z1=1+ sqrt3*i z = 2 * (1/2 + i * sqrt(3)/2) z = 2 * (cos(pi/3 + 2pi * k) + i * sin(pi/3 + 2pi * k)) z = 2 * (cos((pi/3) * (1 + 6k)) + i * sin((pi/3) * (1 + 6k))) z^(1/4) = 2^(1/4) * (cos((pi/12) * (1 + 6k)) + i * sin((pi/...
Math-Trigonometry
Show that if x, y, and z are consecutive terms of an arithmetic sequence, and tan y is defined, then (sin x + sin y + sin z) / (cos x + cos y + cos z) = tan y. So I tried letting x = y-k (since x,y,z are consecutive terms of an arithmetic sequence), then z= y+k So we get (sin(...
calculus
Find the points on the curve y= (cos x)/(2 + sin x) at which the tangent is horizontal. I am not sure, but would I find the derivative first: y'= [(2 + sin x)(-sin x) - (cos x)(cos x)]/(2 + sin x)^2 But then I don't know what to do or if that is even correct??? Would I ...
math
Express the following in simplest form: (complex fraction) Sin(x) - Cos(x) Cos(x) Sin(x) _______________ 1 - 1 Cos(x) Sin(x)
precalculus
I don't understand this problem: (Tanө + cos ө)/ (sec ө + cot ө) so I start off like this: ={(sinө / cos ө)+cosө}{cos ө + (sinө/cosө)} =[(sin ө +cos^2ө) (cos^2ө +sin ө)]/ cos ө but what ...
math
A trigonmetric polynomial of order n is t(x) = c0 + c1 * cos x + c2 * cos 2x + ... + cn * cos nx + d1 * sin x + d2 * sin 2x + ... + dn * sin nx The output vector space of such a function has the vector basis: { 1, cos x, cos 2x, ..., cos nx, sin x, sin 2x, ..., sin nx } Use ...
verifying trigonometric identities
How do I do these problems? Verify the identity. a= alpha, b=beta, t= theta 1. (1 + sin a) (1 - sin a)= cos^2a 2. cos^2b - sin^2b = 2cos^2b - 1 3. sin^2a - sin^4a = cos^2a - cos^4a 4. (csc^2 t / cot t) = csc t sec t 5. (cot^2 t / csc t) = csc t = sin t
these must be written as a single trig expression, in the form sin ax or cos bx. a)2 sin 4x cos4x b)2 cos^2 3x-1 c)1-2 sin^2 4x I need to learn this!! if you can show me the steps and solve it so I can learn I'd be grateful!!! 1) apply the formula for sin 2z 2)3) cos^2z + sin^...
Mathematics - Trigonometric Identities
Prove: sin^2x - sin^4x = cos^2x - cos^4x What I have, LS = (sinx - sin^2x) (sinx + sin^2x) = (sinx - 1 -cos^2x) (sinx + 1 - cos^2x) = sin^2x + sinx - sinx - cos^2xsinx - cos^2xsinx - 1 - 1 + cos^4x = sin^2x - 2cos^2xsinx - 2 + cos^4x Where did I go wrong? Can anyone please ...
trigonometry
find without tables or calculator. (1) sin^2 (22 1/2)- cos^2 (22 1/2) (2) sin 60 cos 30 + sin 30 cos 60 (3) cos(90-y) = sin 56degree 47^i
Math (trigonometric identities)
I was given 21 questions for homework and I can't get the last few no matter how hard and how many times I try. 17. Sinx-1/sinx+1 = -cos^2x/(sinx+1)^2 18. Sin^4x + 2sin^2xcos^2x + cos^4x = 1 19. 4/cos^2x - 5 = 4tan^2x - 1 20. Cosx - sinx - cos^3x/Cosx = sin^2 - tanx 21. Sin^2x...
Trigonometry
I need to prove that the following is true. Thanks. csc^2(A/2)=2secA/secA-1 Right Side=(2/cosA)/(1/cosA - 1) = (2/cosA)/[(1-cosA)/cosA] =2/cosA x (cosA)/(1-cosA) =2/(1-cosA) now recall cos 2X = cos^2 X - sin^2 X and we could say cos A = cos^2 A/2 - sin^2 A/2 and of course sin^...
Mathematics - Trigonometric Identities - Reiny
Mathematics - Trigonometric Identities - Reiny, Friday, November 9, 2007 at 10:30pm (sinx - 1 -cos^2x) (sinx + 1 - cos^2x) should have been (sinx - 1 + cos^2x) (sinx + 1 - cos^2x) and then the next line should be sin^2x + sinx - cos^2xsinx - sinx - 1 + cos^2 x + cos^2xsinx - ...
Math
Show using integration by parts that: e^3x sin(2x)dx = 4/26 e^3x (3/2 sin(2x) - cos(2x)) +c Bit stuck on this. Using rule f udv = uv - f vdu u = e^3x dv + sin(2x)dx f dv = v du/dx = 3e^3x v = -1/2 cos(2x) so uv - f vdu: = (e^3x)(-1/2 cos(2x)) - (-1/2 cos(2x))(3e^3x) Don't know...
math
Determine exact value of cos(cos^-1(19 pi)). is this the cos (a+b)= cos a cos b- sina sin b? or is it something different. When plugging it in the calculator, do we enter it with cos and then the (cos^-1(19 pi)).
MATH...THE GRAPHS OF SINE, COSINE AND TANGENT
anyone explain how this cos 2A cos A – sin 2A sin A can become to this cos (2A + A) = cos 3A
calculus
Find complete length of curve r=a sin^3(theta/3). I have gone thus- (theta written as t) r^2= a^2 sin^6 t/3 and (dr/dt)^2=a^2 sin^4(t/3)cos^2(t/3) s=Int Sqrt[a^2 sin^6 t/3+a^2 sin^4(t/3)cos^2(t/3)]dt =a Int Sqrt[sin^4(t/3){(sin^2(t/3)+cos^2(t/3)}]dt=a Int Sqrt[sin^4(t/3)dt =a ...
Math
Eliminate the parameter and write the corresponding rectangular equation whose graph represents the curve. x = sec Q y = cos Q x^2 + y^2 = 1/cos^2 + sin^2/cos^2 = x^2(1 +sin^2) = x^2(2-cos^2) x^2(2-1/x^2) = 2x^2 - 1 x^2 - y^2 = 1 My teacher said to use secant*cosine = 1. ...
math
How would you establish this identity: (1+sec(beta))/(sec(beta))=(sin^2(beta))/(1-cos(beta)) on the right, sin^2 = 1-cos^2, that factor to 1-cos * `1+cos, then the denominator makes the entire right side 1+cosB which is 1+1/sec which is 1/sec (sec+1) qed using sec(beta) = 1/...
Trig Help!
Question: Trying to find cos π/12, if cos π/6 = square root 3 over 2, how to find cos π/12 using DOUBLE angle formula? This is what I got so far.. cos 2(π/6) = cos (π/6 + π/6) = (cos π/6)(cos π/6) - (sin π/6)(sin π/6) = cos^2...
Limit Calculas
Evaluate lim->4 sin(2y)/tan(5y) Here is what I have so far. I am not sure the next steps. Can someone help me? 1. sin(2y)/(sin(5y)*cos(5y)) 2. (sin(2y)*cos(5y))/sin(5y)
calculus
Evaluate lim->4 sin(2y)/tan(5y) Here is what I have so far. I am not sure the next steps. Can someone help me? 1. sin(2y)/(sin(5y)*cos(5y)) 2. (sin(2y)*cos(5y))/sin(5y)
Pre-Calc
How do I solve this? My work has led me to a dead end. tan(45-x) + cot(45-x) =4 my work: (tan45 - tanx)/(1+ tan45tanx) + (cot45 - cotx)/(1 + cot45cotx) = 4 (1-tanx)/(1+tanx) + (1-cotx)/(1+cotx) = 4 Then I found a common denominator, giving me this: (2-2cotxtanx)/(1+cotx+tanx+...
math
Use the exact values of the sin, cos and tan of pi/3 and pi/6, and the symmetry of the graphs of sin, cos and tan, to find the exact values of si -pi/6, cos 5/3pi and tan 4pi/3. I have found the answers to the first three using the special tables sin(ƒÎ/6) = cos(ƒÎ/3) = 1/...
math
Use the exact values of the sin, cos and tan of pi/3 and pi/6, and the symmetry of the graphs of sin, cos and tan, to find the exact values of sin -pi/6, cos 5/3pi and tan 4pi/3. I have found the answers to the first three using the special tables sin pi/6 = cos pi/3 = 1/2 cos...
Math
verify the following identity used in calculus: cos(x+h)-cos(x)/h=cos(x)[cos(h)-1/h]-sin(x)[sin(h)/h]
Trig
Solve in terms of sine and cosine: sec(x) csc(x)- sec(x) sin(x) so far I have: 1/cos(x) 1/sin(x) - 1/cos(x) sin(x) I am not sure where to go to from there. The book says the answer is cot(x) or cos(x)/sin(x) Thank you in advance.
Trigonometry desperate help, clueless girl here
2. solve cos 2x-3sin x cos 2x=0 for the principal values to two decimal places. 3. solve tan^2 + tan x-1= 0 for the principal values to two decimal places. 4. Prove that tan^2(x) -1 + cos^2(x) = tan^2(x) sin^2 (x). 5.Prove that tan(x) sin(x) + cos(x)= sec(x) 6.Prove that tan(x...
Trigonometry
1.Solve tan^2x + tan x – 1 = 0 for the principal value(s) to two decimal places. 6.Prove that tan y cos^2 y + sin^2y/sin y = cos y + sin y 10.Prove that 1+tanθ/1-tanθ = sec^2θ+2tanθ/1-tan^2θ 17.Prove that sin^2w-cos^2w/tan w sin w + cos w tan w = ...
Math, derivatives
Let g(x) = sin (cos x^3) Find g ' (x): The choices are a) -3x^2sinx^3cos(cos x^3) b) -3x^2sinx^3sin(cos x^3) c) -3x^2cosx^3sin(cos x^3) d) 3x^2sin^2(cos x^3) I'm not exactly sure where I should start. Should I begin with d/dx of sin? Or do the inside derivative first...and do ...
Calculus
Integrate 1/sinx dx using the identity sinx=2(sin(x/2)cos(x/2)). I rewrote the integral to 1/2 ∫ 1/(sin(x/2)cos(x/2))dx, but I don't know how to continue. Thanks for the help. Calculus - Steve, Tuesday, January 12, 2016 at 12:45am 1/2 ∫ 1/(sin(x/2)cos(x/2))dx let u...
Math - Solving for Trig Equations
Solve the following equation for 0 less than and/or equal to "x" less than and/or equal to 360 -- cos^2x - 1 = sin^2x -- Attempt: cos^2x - 1 - sin^2x = 0 cos^2x - 1 - (1 - cos^2x) = 0 cos^2x - 1 - 1 + cos^2x = 0 2cos^2x - 2 = 0 (2cos^2x/2)= (-2/2) cos^2x = -1 cosx = square ...
Cos-Derivative
y= (cos^3 x) (cos 3x) I got -3 sin(3x) cos^3x - 3 sin(x) cos (3x) cos^2 (x) using the product rule Is this right? Thanks.
Pre Calculus
Use one of the identities cos(t + 2ðk) = cos t or sin(t + 2ðk) = sin t to evaluate each expression. (Enter your answers in exact form.) (a) sin(17ð/4) (b) sin(−17ð/4) (c) cos(17ð) (d) cos(45ð/4) (e) tan(−3ð/4) (f) cos(7ð/4) (g) sec(ð/6+2ð) (h) csc(2ð &#...
Pre Calculus
Use one of the identities cos(t + 2ðk) = cos t or sin(t + 2ðk) = sin t to evaluate each expression. (Enter your answers in exact form.) (a) sin(17ð/4) (b) sin(−17ð/4) (c) cos(17ð) (d) cos(45ð/4) (e) tan(−3ð/4) (f) cos(7ð/4) (g) sec(ð/6+2ð) (h) csc(2ð &#...
Math
Which of the following are inverse functions? 1. Arcsin x and sin x 2. cos^-1 x and cos x 3. csc x and sin x 4. e^x and ln x 5. x^2 and +/- sqrt x 6. x^3 and cubic root of x 7. cot x and tan x 8. sin x and cos x 9. log x/3 and 3^x I believe the answers are 2, 4, 6, and 9, but ...
math (repost)
Which of the following are inverse functions? 1. Arcsin x and sin x 2. cos^-1 x and cos x 3. csc x and sin x 4. e^x and ln x 5. x^2 and +/- sqrt x 6. x^3 and cubic root of x 7. cot x and tan x 8. sin x and cos x 9. log x/3 and 3^x I believe the answers are 2, 4, 6, and 9, but ...
Trig
Simplify sin x cos^2x-sinx Here's my book's explanation which I don't totally follow sin x cos^2x-sinx=sinx(cos^2x-1) =-sinx(1-cos^2x) =-sinx(sin^2x) (Where does sine come from and what happend to cosine?) =-sin^3x
Math (Trig)
sorry, another I can't figure out Show that (1-cot^2x)/(tan^2x-1)=cot^2x I started by factoring both as difference of squares. Would I be better served by writing in terms of sine and cosine? Such as: [1-(cos^2x/sin^2x)]/[(sin^2x/cos^2x)-1]=(cos^2x/sin^2x)
Trigonometry
For 0<x<pi/2, sin x and cos x are both less than 1 and greater than 0 (easy to see). We are also given that sin^2x+cos^2x=1. Use this to show that sin^7x+cos^7x<1 for 0<x<pi/2. Unsure on how to proceed?
Math
Evaluate *Note - We have to find the exact value of these. That I know to do. For example sin5π/12 will be broken into sin (π/6) + (π/4) So... sin 5π/12 sin (π/6) + (π/4) sin π/6 cos π/4 + cos π/6 sin π/4 I get all those steps...
Maths
Please solve this trigonometric identity proof problem. I have completed 20; this is the hardest one. Many thanks (sin^3(x)-cos^3(x))/sin(x)-cos(x)=1+sin(x)cos(x)
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Post a New Question | 10,181 | 25,620 | {"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} | 4.03125 | 4 | CC-MAIN-2017-17 | latest | en | 0.690721 |
http://mathtutorkk1969.blogspot.com/2015/10/a-random-sample-of-size-49-is-taken.html | 1,527,376,127,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794867949.10/warc/CC-MAIN-20180526225551-20180527005551-00061.warc.gz | 183,285,499 | 12,607 | ## Friday, October 23, 2015
A random sample of size 49 is taken from a population with mean 31and standard deviation of 12.
What are the expected value and the standard deviation of the sample mean ?
Describe the probability distribution of .
What is the probability that the sample mean is greater than 32?
What is the probability that the sample mean falls between 27 and 29?
What is the probability that the sample mean will be within ±3 of the population mean?
The distribution is the normal distribution which is symmetrical about the mean. Symmetric means it looks the same on both sides of the mean.
P(sample mean > 32)
We need to find a Z-score, which shows how many standard deviations away from the mean a value is.
Z = (sample mean - population mean)/(standard deviation/square root(n))
Z = (32 - 31)/(12/square root (49))
Z = 1/(12/7)
Z = 0.58
Now find Z(0.58) on the standard normal distribution chart. Since the chart shows probabilities less than, we need to take 1- Z(0.58)
Here is the chat I used for all these. http://www.regentsprep.org/regents/math/algtrig/ats7/zchart.htm
Z(0.58) = .7190
1- .7190 = .2010
The probability it falls between 27 and 29 ... P(27 < X < 29)
We need 2 Z-scores now, one for 27 and one for 29
Z = (27 -31)/(12/sqrt(49))
Z = -4/(12/7)
Z = -2.33
and
Z = (29 - 31)/(12/sqrt(49))
Z = -2/(12/7)
Z = -1.17
Now we get Z(-1.17) and Z(-2.33) and subtract them
.1210 - .0099 = .1111
Probability sample mean is within +/- 3standard deviations we simply find Z(3) and Z(-3) and subtract
So we get .9987 - .0013 = .9974 | 466 | 1,554 | {"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-2018-22 | latest | en | 0.8831 |
https://fr.scribd.com/presentation/414353901/Lesson-2-GenMath-1stqrt-pptx | 1,568,871,495,000,000,000 | text/html | crawl-data/CC-MAIN-2019-39/segments/1568514573439.64/warc/CC-MAIN-20190919040032-20190919062032-00510.warc.gz | 497,763,748 | 72,671 | Vous êtes sur la page 1sur 21
# LESSON 2:
EVALUATING
FUNCTIONS
Evaluating a function means
replacing the variable in the
function, in this case x, with a
value from the function’s domain
and computing for the result. To
denote that we are evaluating f at
a for some a in the domain of f,
we write f(a).
Example 1. Evaluate the following
functions at x=1.5:
1.) f(x)=2x+1
2.) q(x)=x²-2x+2
3.) g(x)= 𝑥 + 1
2𝑥+1
4.) r(x)=
𝑥−1
1.) f(x)=4
2.)q(x)=1.25
3.) g(x)= 2.5
4.) r(x)= 8
EXAMPLE 2:
Find g(-4) for g(x)= 𝑥 + 1 , and r(1)
2𝑥+1
for r(x)= .
𝑥−1
Solution: This is not possible
because -4 is not in the
domain of g(x) and 1 is not
in the domain of r(x).
EXAMPLE 3:
- Evaluate the following
functions.
a.) f(3x-1) using f(x)=2x+1
b.)q(2x+3) using q(x)=x²-2x+2
a.) 6x-1
b.)4x²+8x+5
SOLVED
EXAMPLES:
1.) Evaluate the following functions at x=3.
a.)f(x)=x-3
b.)g(x)=x²-3x+5
3
c.)h(x)= 𝑥³ + 𝑥 + 3
𝑥²+1
d.)p(x) =
𝑥−4
## e.) f(x)=│x-5│ where │x-5│ means the
absolute value of x-5.
a.)f(3)=0
b.)g(3)=5
3
c.)h(3)= 33
d.) p(3)=-10
e.)f(3)=2
2.)For what values of x
can we not evaluate the
𝑥+3
function f(x)= ?
𝑥²−4
Solution:
The domain of the function
is given by {x: xԐR, x≠±2}.
Since 2 and -2 are not in
the domain, we cannot
evaluate the function at
x=2,-2.
3.) Evaluate f(a+b) where
f(x) = 4x²-3x.
f(a+b)=4a²-3a+8ab-3b+4b².
SUPPLEMENTARY
EXERCISES:
1.) Evaluate the following functions at x=-4.
a.) f(x)=x³-64
b.)g(x)=│x³-3x²+3x-1│
c.)r(x)= 5 − 𝑥
𝑥+3
d.)q(x)=
𝑥²+7𝑥+12
9 − 𝑥², 𝑖𝑓 𝑥 < 2
2.) Given f(x) = 𝑥 + 7, 𝑖𝑓 2 ≤ 𝑥 < 10 ,
│𝑥 − 4│, 𝑖𝑓 𝑥 ≥ 10
give the values of the following:
a.) f(2)
b.)f(12.5)
c.)f(-3)
d.)f(5)
e.)f(1.5)
3.)Given f(x)=x²-4x+4. Solve
for
a.) f(3)
b.) f(x+3)
c.) Is f(x+3) the same as
f(x)+f(3) ?
4.) A computer shop charges P20 per
hour (or a fraction of an hour) for the
first 2 hours and an additional P10 per
hour for each succeeding hour. Find
how much you would pay if you used
one of their computers for:
a.) 40 minutes
b.) 3 hours
c.) 150 minutes
5.) Under certain circumstances, a rumor | 943 | 2,011 | {"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} | 4.625 | 5 | CC-MAIN-2019-39 | latest | en | 0.500134 |
https://id.scribd.com/document/363361043/PEMBAHASAN-SOAL-UTS-1 | 1,597,503,127,000,000,000 | text/html | crawl-data/CC-MAIN-2020-34/segments/1596439740848.39/warc/CC-MAIN-20200815124541-20200815154541-00133.warc.gz | 349,768,380 | 59,078 | Anda di halaman 1dari 3
# A.
Rata-rata hitung
.
x =
3031
= 50
= 60,62
b. Rata-rata ukur
.log
Log u=
87,74331
= 50
= 1,7548662
U = log 1,7548662
= 0,244244009
## c. rata rata harmonik
H=
50
= 0,95301
= 52,4654
MEDIAN:
2
Me = b + p (
)
50
211
= 61,5 + 11 ( )
8
= 61,5 + 11 (14)
= 61,5 + 154
= 215,5
MODUS :
1
Mo = b +p ( )
1+2
0
= 61,5 +11 ( )
0+3
= 61,5 + 33
= 94,5
KUARTIL :
Nilai Q1
1
4.
Q = b + p (
)
1
4.5011
= 61,5 + 11 ( 8
)
12,511
=61,5 + 11 ( 8 )
16,5
= 61,5 + 8
= 61,5 + 2,0625
= 64,125
2
4.
Q = b + p (
)
2
4.5011
= 61,5 + 11 ( )
8
2511
=61,5 + 11 ( 8 )
14
= 61,5 + 8
= 61,5 + 1,75
= 63,25
3
4.
Q = b + p ( )
3
4.5011
= 61,5 + 11 ( 8
)
37,511
=61,5 + 11 ( 8 )
26,5
= 61,5 +
8
= 61,5 + 3,3125
= 64,8125
Simpangan Baku :
Deviasi standarv:
= ( )/N
= (492 60,62)/50
= (431,38)2 /50
=(186088,7044)/50
= 3721,774088
= 61,00634465 | 513 | 871 | {"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} | 3.8125 | 4 | CC-MAIN-2020-34 | latest | en | 0.117883 |
https://math.stackexchange.com/questions/2441410/transformation-of-fx-y-z | 1,596,997,385,000,000,000 | text/html | crawl-data/CC-MAIN-2020-34/segments/1596439738562.5/warc/CC-MAIN-20200809162458-20200809192458-00001.warc.gz | 386,591,125 | 32,999 | Transformation of F(x,y,z)
I have to prove that by rotation of axes the expression $F(x,y,z) = ax^2 + by^2 + cz^2 +2fyz + 2gzx + 2hxy$ transforms to $λ_1x^2 + λ_2y^2 + λ_3z^2 =0$ where $λ_1, λ_2,λ_3$ are roots of cubic
$\begin{vmatrix} a-λ&h&g\\ h&b-λ&f\\ g&f&c-λ\\ \end{vmatrix}=0$
Solution in book says that $ax^2 + by^2 + cz^2 +2fyz + 2gzx + 2hxy - λ(x^2 +y^2+z^2)$ should reduce to $λ_1x^2 + λ_2y^2 + λ_3z^2-λ(x^2 +y^2+z^2)$ . Hence, book says, both these expressions will be product of linear factors for same value of λ. I don't understand why this is necessary.
Further, it is said that if $(a-λ)x^2 + (b-λ)y^2 + (c-λ)z^2 +2fyz + 2gzx + 2hxy$ is a product of two linear factors then $\begin{vmatrix} a-λ&h&g\\ h&b-λ&f\\ g&f&c-λ\\ \end{vmatrix}=0$
I also don't get how this condition is arrived at. Any help in clearing these two doubts will be helpful.
• Do you mean maybe three degree equation? – Michael Rozenberg Sep 23 '17 at 8:19
• No, condition is derived for general equation of second degree in x,y, and z – LoneCuriousWolf Sep 23 '17 at 8:21
• @MichaelRozenberg are you able to understand the question? – LoneCuriousWolf Sep 23 '17 at 10:29
• Yes of course! It should be $(a_1x+b_1y+c_1z+d_1)(a_2x+b_2y+c_2z+d_2)=0$, but I don't see something interesting for this, otherwise, just to write $f(x,y,z)=(a_1x+b_1y+c_1z+d_1)(a_2x+b_2y+c_2z+d_2)$ and to solve the system. – Michael Rozenberg Sep 23 '17 at 10:33
• @MichaelRozenberg I have edited and added some background to the question. Let me know if you could find anything interesting now. – LoneCuriousWolf Sep 23 '17 at 11:25 | 616 | 1,599 | {"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} | 3.890625 | 4 | CC-MAIN-2020-34 | latest | en | 0.830293 |
https://www.everipedia.com/Pythagoras_tree_%28fractal%29/ | 1,498,455,323,000,000,000 | text/html | crawl-data/CC-MAIN-2017-26/segments/1498128320679.64/warc/CC-MAIN-20170626050425-20170626070425-00396.warc.gz | 852,852,683 | 57,973 | # Pythagoras tree (fractal)
27
The Pythagoras tree is a planefractal constructed from squares. Invented by the Dutchmathematics teacher Albert E. Bosman in 1942,[2] it is named after the ancient Greek mathematician Pythagoras because each triple of touching squares encloses a right triangle, in a configuration traditionally used to depict the Pythagorean theorem. If the largest square has a size of L × L, the entire Pythagoras tree fits snugly inside a box of size 6L × 4L.[3][5] The finer details of the tree resemble the Lévy C curve.
## Construction
The construction of the Pythagoras tree begins with a square. Upon this square are constructed two squares, each scaled down by a linear factor of ½√2, such that the corners of the squares coincide pairwise. The same procedure is then applied recursively to the two smaller squares, ad infinitum. The illustration below shows the first few iterations in the construction process.[3][5]
Order 0 Order 1 Order 2 Order 3
## Area
Iteration n in the construction adds 2n squares of size , for a total area of 1. Thus the area of the tree might seem to grow without bound in the limit as n → ∞. However, some of the squares overlap starting at the order 5 iteration, and the tree actually has a finite area because it fits inside a 6×4 box.[3]
It can be shown easily that the area A of the Pythagoras tree must be in the range 5 < A < 18, which can be narrowed down further with extra effort. Little seems to be known about the actual value of A.
## Varying the angle
An interesting set of variations can be constructed by maintaining an isosceles triangle but changing the base angle (90 degrees for the standard Pythagoras tree). In particular, when the base half-angle is set to (30°) = arcsin(0.5), it is easily seen that the size of the squares remains constant. The first overlap occurs at the fourth iteration. The general pattern produced is the rhombitrihexagonal tiling, an array of hexagons bordered by the constructing squares.
Order 4 Order 10
In the limit where the half-angle is 90 degrees, there is obviously no overlap, and the total area is twice the area of the base square. It would be interesting to know if there's an algorithmic relationship between the value of the base half-angle and the iteration at which the squares first overlap each other.
## History
Pythagoras tree was first constructed by Albert E. Bosman (1891–1961), Dutch mathematics teacher, in 1942.[3]
This is a discussion about the above article. Concerns about the topic, its accuracy, inclusion of information etc. should be discussed here. Off-topic discussion not pertaining to the topic of the page will be removed.
• Angel Ordaz • 4 months, 1 week ago
So beautiful. Endless Branch mutations.
• Nikka4 months, 1 week ago | 665 | 2,785 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.921875 | 4 | CC-MAIN-2017-26 | latest | en | 0.908925 |
https://www.adventuresinmachinelearning.com/mastering-convolution-essential-mathematics-in-digital-signal-processing/ | 1,725,779,302,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700650960.90/warc/CC-MAIN-20240908052321-20240908082321-00202.warc.gz | 610,892,828 | 16,922 | # Mastering Convolution: Essential Mathematics in Digital Signal Processing
## Introduction to Convolution in Digital Signal Processing
Digital Signal Processing (DSP) is an essential aspect of modern electronics, particularly in the realm of communication systems and digital audio processing. One of the most important mathematical techniques used in DSP is convolution.
Convolution allows for the manipulation and processing of signals, which are at the very core of DSP. In this article, we will delve into what convolution is, how it works, and its importance in DSP.
## Explanation of Convolution
Convolution is a mathematical operation that takes two functions and combines them into a third function, providing a description of how one function modifies the other. In the realm of signal processing, convolution is used to calculate the output of a linear time-invariant system given its input.
For instance, consider a system with a transfer function H(s) and an input signal x(t). By performing the convolution operation, we can obtain the output signal y(t), which is described as:
y(t) = H(s) * x(t)
where * denotes the convolve operation.
The operation is linear and time-invariant, meaning that its output depends only on its input and is not affected by the time at which it is applied.
## Importance of Convolution in Digital Signal Processing
Convolution is essential in DSP, as it allows for the manipulation and processing of signals, which are at the very core of DSP. Digital electronics is all about processing signals, and convolution is one of the main ways in which that processing happens.
Convolution is used in various aspects of signal processing, including filtering, feature extraction, and signal analysis. The importance of convolution is further highlighted by the extensive use of the numpy library, which provides an implementation of convolution in Python.
By taking advantage of the numpy library, DSP engineers can perform convolution operations in a fast and efficient manner.
## Understanding the convolve() Function
### Overview of the convolve() Function
The numpy library provides Python programmers and digital signal processing engineers with a convolve() function that can be used to perform convolution. This function takes two one-dimensional arrays and returns their convolution.
The convolve() function is a mathematical technique used to combine two sequences and can be used for filtering, convolution, correlation, and more.
### Linear Convolution
Convolution can be split into two types, linear and circular. Linear convolution is used when working with finite-length signals and can be implemented using the convolve() function.
### Syntax of the convolve() Function
The general syntax for the convolve() function is:
``numpy.convolve(sequence1, sequence2, mode='full')``
Here, sequence1 and sequence2 are the two one-dimensional arrays that are to be convolved, and mode is an optional parameter that can be used to specify the output mode.
### Different Modes of Convolution
The convolve() function supports three different modes – full mode, same mode, and valid mode.
Full mode returns the full discrete linear convolution of the inputs, which is an array of length M+N-1, where M and N are the lengths of sequence1 and sequence2, respectively.
Same mode returns an output array of the same shape as sequence1, but with edges trimmed off, such that it has the same length as sequence1. Valid mode only returns the part of the convolution sequence that overlaps with the original sequence, resulting in an array of length max(M, N) – min(M, N) + 1.
## Conclusion
Convolution is a fundamental mathematical technique used in digital signal processing that allows for the processing and manipulation of signals. In this article, we provided a brief overview of what convolution is, how it works, its importance in digital signal processing, and how the convolve() function can be used to perform convolution operations in Python using the numpy library.
With the knowledge gained, you can now take on more complex tasks in signal processing using convolution.
## Use Cases for the convolve() Function
The convolve() function provided by the numpy library can be used in various signal processing applications. In this expansion, we will explore some use cases for the convolve() function.
### Creating Input Arrays
Before performing convolution using the convolve() function, it is necessary first to create the inputs – one-dimensional arrays. These arrays are created using the np.array() function from the numpy library.
For instance, if we want to create an array of ones with length 5, we can run:
``````import numpy as np
input_array = np.array([1, 1, 1, 1, 1])``````
Similarly, we can create another array with values [0, 1, 2, 3, 4]:
``input_kernal = np.array([0, 1, 2, 3, 4])``
Once the inputs are created, we can then use the convolve() function to perform convolution.
### Default Convolution Output
The default mode of the convolve() function returns the convolution product of the two input arrays. For instance, let us compute the convolution of the two arrays, input_array and input_kernal:
``convolution = np.convolve(input_array, input_kernal)``
The output of the above code will be the convolution product of both arrays, which is [0, 1, 3, 6, 9, 8, 4].
The length of the convolution product is equal to the length of the input arrays minus 1. In this case, both arrays have a length of 5, and the output has a length of 5+5-1 = 9.
### Same Mode Output
The same mode of the convolve() function returns an output array of the same shape as the first input array, but with the edges trimmed off. This output array is the convolution of the two input arrays where the output is centered with respect to the middle of the input array.
Additionally, the output has the same number of points as the first input array. To use the same mode of the convolve() function, we include the mode=’same’ parameter in the function call.
For example, using the input_array and input_kernal created earlier, we have:
``same_mode_convolution = np.convolve(input_array, input_kernal, mode='same')``
The output will be the convolution of both arrays with an output array of the same length as the first input array. The output of this convolution is [3, 6, 9, 8, 4].
### Valid Mode Output
The valid mode of the convolve() function returns only the part of the convolution sequence that overlaps with the original sequence. The output array has the length equal to abs(len(x) – len(h)) + 1, where x and h are the two input arrays.
This mode is useful when we don’t want the output to include the edges of the input array. To use the valid mode of the convolve() function, we include the mode=’valid’ parameter in the function call.
For example, using the same input_array and input_kernal, we have:
``valid_mode_convolution = np.convolve(input_array, input_kernal, mode='valid')``
The output of this convolution is [6, 9, 8], which is of length abs(len(input_array) – len(input_kernal)) + 1 = 3.
## Conclusion
In conclusion, the convolve() function is an essential tool in digital signal processing. With the numpy library, we can generate one-dimensional arrays for use as inputs, and then use the convolve() function to perform convolution operations.
We have explored some of the use cases for the convolve() function, including the default convolution output, same mode output, and valid mode output. By applying these use cases, engineers can manipulate the signals in the digital domain, making signal processing in DSP more efficient and accurate.
In summary, convolution is a powerful mathematical technique that has a crucial role in digital signal processing. The numpy library’s convolve() function provides a simple and efficient means to apply the convolution operation to one-dimensional arrays in Python.
We explored the different modes of the convolve() function, including the default mode, same mode, and valid mode, each of which has specific use cases. As digital signal processing continues to evolve, the role of convolution and the applications of the convolve() function will become increasingly important.
As engineers and programmers in this field, understanding and employing the convolve() function is essential to our success in signal processing. | 1,755 | 8,389 | {"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} | 3.90625 | 4 | CC-MAIN-2024-38 | latest | en | 0.920887 |
http://www.algebra.com/cgi-bin/show-question-source.mpl?solution=72720 | 1,369,206,028,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368701445114/warc/CC-MAIN-20130516105045-00000-ip-10-60-113-184.ec2.internal.warc.gz | 316,608,177 | 772 | ```Question 99911
16(20+X)/2=224
(320+16X)/2=224 CROSS MULTIPLY
320+16X=448
16X=448-320
16X=128
X=128/16
X=8 IS THE LENGTH OF THE OTHER SIDE.
PROOF
16(20+8)/2=224
(16*28)/2=224
448/2=224
224=224``` | 98 | 197 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.53125 | 4 | CC-MAIN-2013-20 | latest | en | 0.370469 |
https://golylyparize.regardbouddhiste.com/excercise-14-problems-47568pe.html | 1,606,288,650,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141181482.18/warc/CC-MAIN-20201125071137-20201125101137-00336.warc.gz | 324,944,317 | 3,709 | # Excercise 14 problems
For each exercise, a link to a possible solution is provided. Each solution includes a discussion of how a programmer might approach the problem and interesting points raised by the problem or its solution, as well as complete source code of the solution.
Trigometric Ratios We can use Pythagoras' Theorem see above to calculate the length of an unknown side in a right-angled triangle when we are given information about the lengths of 2 other sides.
However, if we are given information about an angle other than the 90 degrees and one side and need to calculate the length of another side then we use the trig ratios. This first tutorial takes you through naming the sides of a triangle and an introduction to the trig ratios.
## Best Types of Exercise for Asthma - Health
This tutorial is suitable for students in Year 10 or We can use trig ratios to calulate the length of a side on a triangle if we know the length of one side and the size of one angle.
You will need a scientific calculator for this tutorial This tutorial is suitable for students in Year 11 or Calculating angles using the trigometric ratios. OK, so probably saw this coming. If you know the length of the two sides of a right-angle triangle then you can caculate the size of the other angles inside the triangle.
We're really getting to know stuff about right-angle triangles.Physical Geography Laboratory Manual for McKnight’s Physical Geography: A Landscape Appreciation, Eleventh Edition offers a comprehensive set of lab exercises to accompany any physical geography class.
The manual is organized to meet your needs, providing the flexibility to pick and choose a. Student Resources For more information on how to order these items, Chapter Exercise Exercise Exercise Exercise Problem A. Working Papers Plus for Select Exercises and Problems Chapters ISBN: Shop ProForm online.
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What is the molarity of an HCl solution. Math []. 1. Write a function to calculate if a number is prime. Return 1 if it is prime and 0 if it is not a prime.
2. Write a function to determine the number of prime numbers below n. C programming Exercises, Practice, Solution: C is a general-purpose, imperative computer programming language, supporting structured programming, lexical variable scope and recursion, while a static type system prevents many unintended operations.
14 Leg Exercises for Men - Elite Men's Guide | 554 | 2,755 | {"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-2020-50 | latest | en | 0.896775 |
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# If the sum of the interior angles of a regular polygon...
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If the sum of the interior angles of a regular polygon... [#permalink] 05 Oct 2010, 04:40
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If the sum of the interior angles of a regular polygon measures up to 1440 degrees, how many sides does the polygon have?
1. 10 sides
2. 8 sides
3. 12 sides
4. 9 sides
5. None of these
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Re: If the sum of the interior angles of a regular polygon... [#permalink] 05 Oct 2010, 05:16
10 sides,
Formula to find out number of sides of a polygon is: (n-2)*180
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Re: If the sum of the interior angles of a regular polygon... [#permalink] 05 Oct 2010, 05:18
Expert's post
pzazz12 wrote:
If the sum of the interior angles of a regular polygon measures up to 1440 degrees, how many sides does the polygon have?
1. 10 sides
2. 8 sides
3. 12 sides
4. 9 sides
5. None of these
Sum of Interior Angles of a polygon is $$180(n-2)$$ where $$n$$ is the number of sides.
So $$180(n-2)=1440$$ --> $$n=10$$.
For more on this check Polygons chapter of Math Book: math-polygons-87336.html
Hope it helps.
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Re: If the sum of the interior angles of a regular polygon... [#permalink] 05 Oct 2010, 05:19
Hi
The formula for interior angle is (n-2)*180
(10-2)*180 = 1440.
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Re: If the sum of the interior angles of a regular polygon... [#permalink] 05 Oct 2010, 05:56
If you don't know the formula you can get to the answer this way. A little more involved but probably still doable in the crunch:
Start with the second to highest number in the answers (i.e. 10) so you can go higher or lower depending on what you find.
Draw or visualize a decagon, and run the five diagonals that go through the center, so that you end up with 10 identical 'wedges' or 'pizza slices' which will be isosceles triangles.
All 10 triangles will have the center, where all diagonals cross, in common. You know the sum of all those angles is 360, so since you have 10 of them, that angle is 360/10 = 36 for each triangle. Now you can see that for each slice, the sum of the other triangles must be 180 - 36 = 144. Therefore, since you have 10 triangles, the total sum of the angles that make the decagon is 144 *10 = 1440. Voila.
If this had been too low, you know the polygon would have to have more sides, if it had been too big, you would need less. so you could work that way.
Still, formula is MUCH simpler so this tells me I need to brush up!!
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Re: If the sum of the interior angles of a regular polygon... [#permalink] 05 Oct 2010, 09:58
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Re: If the sum of the interior angles of a regular polygon... [#permalink] 05 Oct 2010, 11:15
Sum of interior angles of a regular polygon = 180 * (n-2), where n is the number of sides.
180 * (n-2) = 1440.
Solve for n.
Re: If the sum of the interior angles of a regular polygon... [#permalink] 05 Oct 2010, 11:15
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# Tutor profile: Matt M.
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## Questions
### Subject:Physics (Newtonian Mechanics)
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A 2 kilogram block slides from rest down a 2 meter tall ramp, at an angle of 25 degrees above the ground. The coefficient of kinetic friction $$\mu_k$$ between the block and the ramp is 0.3. What is the block's speed at the bottom of the ramp?
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Here we consider the energy of the earth-block system. The initial gravitational potential energy is $(U_i=mgh_i=(2 kg)(9.8 m/s^2)(2 m)=39.2 J$) and the final gravitational potential energy is $(U_f=mgh_f=(2 kg)(9.8 m/s^2)(0 m)=0 J.$)Since the block starts at rest, the initial kinetic energy is 0 J, and the final kinetic energy will be $(K_f=\frac{1}{2}mv_f^2$)where $$v_f$$ is the final speed we want to calculate. Additionally, energy is not conserved in this system, but some energy is removed by the external work done by the friction of the ramp. This work is given by $$W_f=\overrightarrow{F_f}\cdot \overrightarrow{d}$$, where $$\overrightarrow{d}$$ is the block's displacement and$$\overrightarrow{F_f}$$ is the force of friction. The force is in the opposite direction of the displacement, so the work is negative (this makes sense, as we have already said energy is being removed from the system), and the two vectors are parallel, so the dot product is simply the product of their magnitudes. The magnitude of the force of friction is $(F_f = \mu_k F_N= \mu_k m g cos(\theta) = (0.3)(2 kg)(9.8 m/s^2)(cos(25 ^{\circ}) = 5.33 N$)where $$F_N$$ is the normal force, and the magnitude of the displacement is $( d=\frac{h}{sin(\theta)} = \frac{2m}{sin(25 ^{\circ})} = 4.73 m$) making the work done by friction $( W_f = -F_f \cdot d = -(5.33 N)(4.73 m) = -25.21 J.$) Now we put this all together: $( U_i + K_i +W_f = U_f + K_f$)$( 39.2 J + 0 J - 25.21 J = 0 J + \frac{1}{2}mv_f^2$)$( \frac{1}{2}(2 kg)v_f^2 = 13.99 J$)$(v_f=3.74 m/s$)
### Subject:Physics (Electricity and Magnetism)
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Question:
Using Gauss's Law, find the electric field produced by a sphere of radius $$R$$, with a charge $$Q$$ uniformly distributed through the volume, as a function of the distance $$r$$ from the center of the sphere.
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Matt M.
Gauss's Law tells us that the electric flux through an arbitrary closed surface (a so-called "Gaussian" surface) is proportional to the charge contained within the surface: $(\Phi_{e}=\frac{Q_{enc}}{\varepsilon_0}$) where $$\varepsilon_0 \approx 8.85 \times 10^{-12} F \cdot m^{-1}$$ is a fundamental constant called the permittivity of free space. Since we have a spherical system, we choose a spherical Gaussian surface centered at the center of the charged sphere - this will simplify things greatly. The definition of electric flux through a surface $$S$$ is $( \Phi_e = \int_{S} \overrightarrow{E} \cdot d\overrightarrow{A}$), but for a spherical surface and a spherical system, the electric field $$\overrightarrow{E}$$ and the area element $$d\overrightarrow{A}$$ are parallel, pointing directly away from the sphere's center, and the field is constant at a given radius, so we do not need to worry about the integral and instead simply multiply by the Gaussian surface's area: $( \Phi_e = E*A_{S} = E*(4\pi r^2) = \frac{Q_{enc}}{\varepsilon_0}$) So $( E=\frac{Q_{enc}}{4\pi r^2\varepsilon_0}$) and now all we need to find is $$Q_{enc}$$, and there are two possible cases: Case 1: $$r>R$$ Here $$Q_{enc}=Q$$ (since all the charge is inside the Gaussian surface) and $( E=\frac{Q}{4\pi r^2\varepsilon_0}$) Note that this simply looks the same as if all the charge was concentrated as a point charge at $$r=0$$. Case 2: $$r \leq R$$ Here not all of the charge $$Q$$ is inside the Gaussian surface. Instead, because the charge is uniformly distributed, we can use a ratio of the volume of the Gaussian sphere to the physical sphere: $( \frac{Q_{enc}}{Q} = \frac{V_{enc}}{V}=\frac{\frac{4}{3}\pi r^3}{\frac{4}{3}\pi R^3} = \frac{r^3}{R^3} \Longrightarrow Q_{enc}=Q\frac{r^3}{R^3}$) Substitute this into the equation for $$E$$ derived above: $( E=\frac{Q}{4\pi r^2\varepsilon_0} \frac{r^3}{R^3} = \frac{Q}{4\pi R^3\varepsilon_0}r$) Now we have the expression for the electric field both inside and outside of the sphere.
### Subject:Calculus
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Question:
Find the indefinite integral $$\int x^2 sin(x) dx$$ using integration by parts.
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In integration by parts, we recognize an integrand that is the product of an easy-to-differentiate function $$u$$ and an easy-to-integrate function $$v'$$, and use the relationship $$\int uv' = uv-\int u'v$$, which is derived from the chain rule. Here we observe that $$u=x^2$$ is easy to differentiate, and $$v'=sin(x)$$ is easy to integrate, and we get $$u'=2x$$ and $$v=-cos(x)$$, respectively. (For now we neglect the constant of integration, which only needs to be added at the end.) Then by substituting into the above relationship, we obtain $( \int x^2 sin(x) dx = -x^2cos(x) -\int 2x (-cos(x))dx = -x^2cos(x) + 2\int xcos(x)dx$). Here the second integral $$\int xcos(x)dx$$ can again be solved using integration by parts, using $$u=x$$ and $$v'=cos(x)$$, corresponding to $$u'=1$$ and $$v=sin(x)$$. This gives $(\int xcos(x)dx = xsin(x) -\int sin(x)dx = xsin(x) +cos(x) \space \space (+C)$), where the second integration is simple. Substituting this back into our above solution gives the final answer: $( \int x^2 sin(x) dx = -x^2cos(x) +2xsin(x) +2cos(x)+C$)
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https://purecalculators.com/root-calculator | 1,716,321,837,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058512.80/warc/CC-MAIN-20240521183800-20240521213800-00678.warc.gz | 429,226,000 | 29,368 | Mathematical Calculators
## Mathematical Root Calculator (square Root Calculator)
This free calculator calculates the second, third and higher exponents and roots. Formula is also available.
Result
√? = ?
Result
∛? = ?
## General root calculator
Result
√? = ?
◦About root calculator ◦How does square root calculator work? ◦How to use root calculator? ◦What is a root? ◦Learn about root of a number ◦3 reasons to study mathematics
## How does square root calculator work?
A square root calculator is a handy tool that helps you calculate square roots quickly and easily. It works by entering the input, like 3.14, and then pressing the “calculate” button. The calculator will then display the square root of this number as well as other related information, like the decimal point and percent value. This is a great tool to have on hand when you need to calculate a square root quickly and accurately.
## How to use root calculator?
Our root calculator is really simple to use. You just have to fill your desired number and the root you want to calculate, and you will get correct results.
You can use our premade square root calculator, cube calculator, or general root calculator.
Square root calculator uses second root.
Cube root calculator uses third root.
And general root calculator you can choose to which root you want to calculate the number.
## What is a root?
Root of an number is an simple mathematical concept. The root of an number Y is an number, which is multiplied by itself as many times as given number, equals to Y.
For example, cube root of an number 27 is 3.
3 x 3 x 3 = 27.
Root of a number
## Learn about root of a number
Root numbers are one of the fascinating things you find in mathematics.
Root of a number at Wolfram
## 3 reasons to study mathematics
Mathematics are very usefull in everyday life. Sometimes you might forget how much we use mathmetics in different places.
1. Healthy exercise for your brain
Mathmetics are very healthy for the brains. Studying mathmetical principles and calculating mathmetical equations developes your brain muscles. Mathematics also develop abstract thinking and teach new ways of solving problems.
2. Understanding the world
Big part of our world uses mathmetical principles and concepts. Therefore understanding mathmetics helps you to understand logic behind different kind of structures, such as architectural buildings and computers.
3. Universal beauty
Mathmetics are called "universal language", and not for light reasons! Different societies throughout history have independently founded mathemetics and applied it to their everyday lives. Also big part of mathmetics comes from studying nature, and expressing it's beauty in mathmetical patterns.
Article author
Angelica Miller
Angelica is a psychology student and a content writer. She loves nature and wathing documentaries and educational YouTube videos.
###### Mathematical Root Calculator (square Root Calculator) English
Published: Mon Aug 09 2021
In category Mathematical calculators
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Convert a fraction to a mixed number calculator using our simple tool | 2,564 | 13,440 | {"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} | 3.953125 | 4 | CC-MAIN-2024-22 | latest | en | 0.901906 |
https://discuss.codechef.com/t/beautgar-editorial/20989 | 1,685,401,637,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224644913.39/warc/CC-MAIN-20230529205037-20230529235037-00264.warc.gz | 259,493,524 | 9,071 | # BEAUTGAR - EDITORIAL
Setter: Praveen Dhinwa
Tester: Teja Vardhan Reddy
Editorialist: Taranpreet Singh
Simple
Observations.
### PROBLEM:
Given a garland represented by string S consisting of two colored flowers Red and Blue, determine, whether we can attain a beautiful garland, defined as a garland having no two consecutive flowers of the same color, by applying the following operation at most once.
Make two cuts on garland to obtain two parts, reverse one and join it back.
### SUPER QUICK EXPLANATION
• Only even length garlands with the same number of red and blue flowers have the possibility of being a part of beautiful garland, except garland of length 1.
• Such garland is beautiful, if the number of positions having current character same as next character, is either zero or two.
### EXPLANATION
Let us see a beautiful garland first.
Let us consider a garland of length 1. We know, it has only one flower, so it can never have two consecutive flowers of the same color, hence, is a beautiful garland. We answer yes in this case.
Now, we have garland length \geq 2.
Now, we need to end up with a beautiful garland, which, we can notice, has the same number of red and blue colored flowers. Hence, the garland, which does not have the same number of flowers can never end up as a beautiful garland, even after performing operations.
This way, all garlands we are left with are the garlands of even length (simple to prove) having the same number of flowers of both colors.
First of all, check if the garland is already beautiful and report if it is. Now, we are left with a non-beautiful garland and we have to make a cut, trying to make it beautiful.
Let us understand the consequence of performing the operation. See, we make the cut, reverse a string, and attach it back. Except for border flowers of both pieces of garland, the rest of the flowers continue to have same adjacent flowers as before.
Consider an example RRGGRRGG. We make first cut between first and second flower while the second cut between seventh and eighth flower. The resultant garland becomes RGRRGGRG. We see, that only pairs of positions around both cuts are affected by the operation.
Hence, if we call a pair of adjacent same-colored flowers as mismatch positions, then, we can prove, that we can remove exactly two mismatches using one swap.
See, since only the two pairs of positions enclosing either of the cut are affected by a cut, we cannot remove more than 2 mismatch positions using one operation.
This way, the solution is just to count the number of mismatch positions in garland, and we can make the garland beautiful, if there are either zero or exactly two mismatch positions, after handling the case of garland consisting of a single flower.
Exercise
Prove that the number of mismatch positions in an even length garland having the same number of both colored flowers, cannot have an odd number of mismatch positions.
### Time Complexity
Time complexity is O(|S|) since we iterate over the string.
### AUTHORâS AND TESTERâS SOLUTIONS:
Feel free to Share your approach, If it differs. Suggestions are always welcomed.
6 Likes
``````Let us consider a garland of length 1. We know, it has only one flower, so it can never have two consecutive flowers of the same color, hence, is a beautiful garland. We answer yes in this case.
``````
Length of garland is at least 2 in the question :p. Good that you told about the case though, because many times I have seen people get confused in such cases (one case in point is case of N=1 in Beautiful Array)
A very good exercise would be, to prove that, if we can do the operation infinite times, we can always make the garland beautiful if it has equal R and G's
(Hint: Try obtaining string of form RRR...GGGG... and proceed henceforth.)
1 Like
My try at the exercise:
Letâs call contiguous same-coloured flowers a block. So, e.g., blocks will look like R, RRR, GG etc. My proof assumes that the maximum length of any block of flowers in the garland is 2. So the garland may look like RRGRGGRG etc.
Let the number of blocks of red flowers of length 1 be R_1 and of length 2 be R_2, and similarly for green flowers, let the numbers of blocks be G_1 and G_2 respectively.
Since the garland is made of alternating red and green blocks, the number of red blocks is equal to the number of green blocks, i.e.,
R_1 + R_2 = G_1 + G_2
Also, since the number of red and green flowers is equal, we have
R_1 + 2R_2 = G_1 + 2G_2
Solving the above equations, we get R_2 = G_2, i.e., the number of red and green blocks of length 2 is the same. So, the total number of blocks of length 2 is 2R_2 (or 2G_2), which is also the total number of mismatch positions. Hence the total number of mismatch positions is even.
1 Like
I missed the âequal coloursâ criterion because I was thinking about the general problem with unlimited different colours. So when I modified it to just red and green for the competition it was a little clumsy.
The unlimited-colours solution requires that
(a) there are no more than two pairs
(b) if two pairs, they are different colours
© if only one, there are two consecutive flowers somewhere neither of which is the pair colour.
© is a little tricky, but can be tested (by the pigeonhole principle) by checking that the pair colour isnât dominant; that is, accounts for not more than half of the flowers. So a garland like GRBRYRR canât be made beautiful - every cut has R on one side, so the R pair will always survive. And of course it has 4 R out of 7 flowers.
My solution to this extended problem
2 Likes
Wonât the exercise is proved by the following statement :
If we start with a mismatch and then move on taking alternating flowers , we are bound to make one more mismatch so that the count of R and G is same .
For example , let us begin with
RR , Now
RRG , //R is one more than G
RRGRG, //R is one more than G
RRGRGG //We will have to make another mismatch to equalise R and G!
Similar string will be obtained in case we get a mismatch in between or with the pair s[0] and s[n-1].
1 Like
1. find count (g) of consecutive Green ones, increment g by 1 if first and last are green
2. find count Âź of consecutive Red ones, increment r by 1 if first and last are red
3. `if (r == 1 and g == 1) or (r == 0 and g == 0): print('yes') else: print('no')`
4. you have to make just 2 cuts, one for RR and another for GG, so there should be exactly 1 GG & 1 RR or none.
2 Likes
Analyzing a bit more we can conclude that since we are only allowed to make one pair of cut, so the no. of consecutive flower pairs must be exactly 2 and that too the pairs must be of different colors(this justifies why the number of red and green flowers must be equal), else even after exchanging there will be at least one pair which would still be of same color.
https://www.codechef.com/viewsolution/21673749
i have applied the concept that if length of garland is 1 then garland is beautiful else i hae counted number of R and G and if they are equal and their sum is even and then i have counted the mismatched pair and printed the result accordingly.
Quite an overkill for this problem.
1 Like
I thought the same thing
But then after some time I saw question again and got that it has only two colorsâŠ
1 Like
@l_returns yes, I knew there were only two colours, but you can see in my competition code the fossils of the extended-problem thinking above.
You forgot to mention, that number of R and G should be equal. Rest fine
@taran_1407 âR and G equalâ follows automatically from the above assessment.
1 Like
Fortunately the case for only one flower is excluded from the problem, since an argument could be made for it being either beautiful or not beautiful. It depends on the criteria behind the âno pair of adjacent flowers with identical coloursâ definition. A simple reading allows it to be beautiful, but if the criterion is derived from needing variety, or considering a traverse along the garland and wanting a switch each time, a lone flower will not be beautiful.
I am well into over-thinking territory here.
1 Like
/* package codechef; // donât place package name! */
import java.util.;
import java.lang.
;
import java.io.*;
/* Name of the class has to be âMainâ only if the class is public. */
class Codechef
{
public static void main (String[] args) throws java.lang.Exception
{
Scanner sc=new Scanner(System.in);
int t=sc.nextInt();
while(tâ>0)
{
String s=sc.next();
if(s.length()==1)
System.out.println(âyesâ);
else if(s.length()%2==1)
System.out.println(ânoâ);
else
{ int xr=0,xg=0;
for(int i=0;i<s.length()-1;i++)
{ if(xr==2 || xg==2)
break;
if(s.charAt(i)==s.charAt(i+1) && s.charAt(i)==âGâ)
xg++;
else if(s.charAt(i)==s.charAt(i+1) && s.charAt(i)==âRâ)
xr++;
}
if((xg==0 && xr==0) || (xg==1 && xr==1))
System.out.println(âyesâ);
else
System.out.println(ânoâ);
}
}
}
}
kindly help me by telling the problem in my code
Man , I feel so dumb after coding a simulation based solution | 2,269 | 9,072 | {"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} | 4.1875 | 4 | CC-MAIN-2023-23 | latest | en | 0.938898 |
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