url
string | fetch_time
int64 | content_mime_type
string | warc_filename
string | warc_record_offset
int32 | warc_record_length
int32 | text
string | token_count
int32 | char_count
int32 | metadata
string | score
float64 | int_score
int64 | crawl
string | snapshot_type
string | language
string | language_score
float64 |
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
http://www.urch.com/forums/gmat-problem-solving/100787-odd-integer-2.html
| 1,369,178,354,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2013-20/segments/1368700842908/warc/CC-MAIN-20130516104042-00014-ip-10-60-113-184.ec2.internal.warc.gz
| 787,664,007 | 14,689 |
1. Good post? |
I think shobby has given the right solutions.
I don't think there is a exact formula for this question, what you should know is a is even, a/2 is even... that's enough, and nothing more.
2. Good post? |
Originally Posted by ramzrambo
ah well.....any better method!!!????????????????
Given a-b is even=> both even or both odd
Given a/b is even=> both even or a is even and b is odd. Condition satisfied by both the statements is that both a and b are even. Now, if both a & b are even and a/b is an integer, then a must be greater than b and a multiple of 4.
Now option D reads: (a+2)/2= a/2+1 and any multiple of 4 when halfed will be an even number and 1+even is odd. Hence D is the correct answer
3. Good post? |
a-b is even
This means that either both a and b are even or both a and b are odd.
--->Conclusion 1
a/b is even
This plus Conclusion 1 derives the fact that a and b both should be even.
Now if both are even, more than one of the options is right,.... can anybody pull me further???
4. Good post? |
a-b is even
This means that either both a and b are even or both a and b are odd.
--->Conclusion 1
a/b is even
This plus Conclusion 1 derives the fact that a and b both should be even.
Now if both are even, more than one of the options is right,.... can anybody pull me further???
5. Good post? |
a-b is even
This means that either both a and b are even or both a and b are odd.
--->Conclusion 1
a/b is even
This plus Conclusion 1 derives the fact that a and b both should be even.
Now if both are even, more than one of the options is right,.... can anybody pull me further???
6. Good post? |
Its D.
7. Good post? |
ans is d
a/b =2m so a is even
a-b is even so b is even
so a=2mb=4mn as b=2n
so (a+2)/2=2mn+1 so odd
but (b+2)/2 = n+1 n can be even or odd so b can be odd or even
8. Good post? |
But why not E.
Its also almost true......
9. Good post? |
subsvirgin...check out my explanation that explains the special case for A, B......b can be 2(b/2 + 1 = 2...hence even) but "a" can never be 2.
10. Good post? |
Fiver
Page 2 of 3 First 123 Last
There are currently 1 users browsing this thread. (0 members and 1 guests)
#### Posting Permissions
• You may not post new threads
• You may not post replies
• You may not post attachments
• You may not edit your posts
•
SEO by vBSEO ©2010, Crawlability, Inc.
| 685 | 2,361 |
{"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
| 0.954537 |
https://socratic.org/questions/how-to-find-instantaneous-rate-of-change-for-x-3-2x-2-x-at-x-from-1-to-2
| 1,725,706,948,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2024-38/segments/1725700650826.4/warc/CC-MAIN-20240907095856-20240907125856-00637.warc.gz
| 511,445,949 | 6,049 |
# How to find instantaneous rate of change for x^3 +2x^2 + x at x from 1 to 2?
Jul 20, 2018
The instantaneous rate of change is the same as taking the derivative. This is defined at a point.
We can easily find that function via power rule:
$f \left(x\right) = {x}^{3} + 2 {x}^{2} + x \implies f ' \left(x\right) = 3 {x}^{2} + 4 x + 1$
I hope it is clear here that the concept of "instantaneous rate of change" from one point to another doesn't make any sense. However, you can think about the average rate of change very easily. We just do that in the way we'd think: the change in y divided by the change in x. This gives the following:
$\frac{f \left(2\right) - f \left(1\right)}{2 - 1} = \frac{14}{1} = 14$
| 227 | 713 |
{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 2, "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}
| 3.890625 | 4 |
CC-MAIN-2024-38
|
latest
|
en
| 0.942013 |
https://www.snapxam.com/problems/87188598/1-5-17-1-112-1
| 1,723,147,525,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2024-33/segments/1722640740684.46/warc/CC-MAIN-20240808190926-20240808220926-00854.warc.gz
| 769,816,343 | 13,741 |
👉 Try now NerdPal! Our new math app on iOS and Android
Multiply $\frac{1}{5}\cdot \left(- 7^{-1}+112^{-1}\right)$
Go!
Symbolic mode
Text mode
Go!
1
2
3
4
5
6
7
8
9
0
a
b
c
d
f
g
m
n
u
v
w
x
y
z
.
(◻)
+
-
×
◻/◻
/
÷
2
e
π
ln
log
log
lim
d/dx
Dx
|◻|
θ
=
>
<
>=
<=
sin
cos
tan
cot
sec
csc
asin
acos
atan
acot
asec
acsc
sinh
cosh
tanh
coth
sech
csch
asinh
acosh
atanh
acoth
asech
acsch
0
Step-by-step Solution
How should I solve this problem?
• Choose an option
• Write in simplest form
• Prime Factor Decomposition
• Solve by quadratic formula (general formula)
• Find the derivative using the definition
• Simplify
• Find the integral
• Find the derivative
• Factor
• Factor by completing the square
Can't find a method? Tell us so we can add it.
1
Cancel like terms $-1\cdot 7^{-1}$ and $112^{-1}$
$\left(\frac{1}{5}\right)\cdot 0$
2
Any expression multiplied by $0$ is equal to $0$
0
0
Explore different ways to solve this problem
Solving a math problem using different methods is important because it enhances understanding, encourages critical thinking, allows for multiple solutions, and develops problem-solving strategies. Read more
Main Topic: Multiplication of Numbers
The terms a and b are called factors and the result, c, is the product.
| 440 | 1,276 |
{"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}
| 3.546875 | 4 |
CC-MAIN-2024-33
|
latest
|
en
| 0.680805 |
http://mathhelpforum.com/differential-equations/158337-integrating-factor-proof-partial-derivative-wrt-xy-print.html
| 1,529,818,616,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2018-26/segments/1529267866358.52/warc/CC-MAIN-20180624044127-20180624064127-00045.warc.gz
| 197,902,250 | 2,899 |
# integrating factor proof/ partial derivative wrt xy
• Oct 3rd 2010, 08:58 PM
234578
integrating factor proof/ partial derivative wrt xy
The full question is:
show that if $\displaystyle \frac{(N_{x} - M_{y})}{(x*M-y*N)}=R$ where R depends only on xy, then the differential equation M+N*y'=0 has an integrating factor of the form U(xy). Find a general formula for this integrating factor.
Using an equation from the text for U (for which finding a solution, U, means that the differential equation above is exact) I have:
$\displaystyle U= \frac{N*U_{x}-M*U_{y}}{M_{y}-N_{x}}$
then substituting in
$\displaystyle M_{y}-N_{x}=(yN-xM)*R$
and equating coefficients of N and M respectively,
$\displaystyle y*U*R=U_{x}$ and $\displaystyle x*U*R=U_{y}$
Is there some way that I can find the derivative of U wrt xy using the partial derivatives of U wrt x and y that I have above? Then I could integrate wrt xy to find U, right?
• Oct 4th 2010, 05:57 AM
Krizalid
$\displaystyle \mu(x,y)M(x,y)dx+\mu(x,y)N(x,y)\,dx=0.$
we have $\displaystyle \mu(x,y)=h(xy),$ and by the exact condition we have $\displaystyle \displaystyle\frac{\partial \mu }{\partial y}M-\frac{\partial \mu }{\partial x}N=\left( \frac{\partial N}{\partial x}-\frac{\partial M}{\partial y} \right)\mu ,$ now $\displaystyle \mu_x=h'(xy)y$ and $\displaystyle \mu_y=h'(xy)x,$ so $\displaystyle \displaystyle\frac{1}{xM-yN}\left( \frac{\partial N}{\partial x}-\frac{\partial M}{\partial y} \right)=\frac{h'(xy)}{h(xy)}=R(xy),$ put $\displaystyle z=xy$ and solve that little ODE and you got your integrating factor.
| 500 | 1,574 |
{"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}
| 3.859375 | 4 |
CC-MAIN-2018-26
|
latest
|
en
| 0.800147 |
http://wiki.juneday.se/mediawiki/index.php?title=ITIC:Digital_representation_-_Binary_-_Exercises&oldid=14013
| 1,611,842,608,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2021-04/segments/1610704847953.98/warc/CC-MAIN-20210128134124-20210128164124-00128.warc.gz
| 104,128,025 | 9,047 |
Never confuse education with intelligence, you can have a PhD and still be an idiot.
- Richard Feynman -
# Exercises
## 01
a) How many combinations of 0s and 1s can you get from 9 bits?
b) How many positive numbers can you represent (including 0) and which numbers are that?
Expand using link to the right to see a suggested solution.
a) The number of combinations of 0s and 1s for nine bits, is equal 2 to the power of 9.
29 = 512.
b) If we start at 0, then the positive numbers we can represent are 0 through 511.
## 02
What decimal numbers are the following binary numbers?
Hint: The rightmost bit is worth 20 (110) if it is set (is a one). The next (going left) bit is worth 21 (210) if it is set. The next (still going left) is worth 22 (410). The next is worth 23 (810) and the next is worth 24 (1610) etc.
1. 1001
2. 101
3. 11
4. 10
5. 1111
Expand using link to the right to see a suggested solution.
1. 1*1 + 0*2 + 0*4 + 1*8 = 910
2. 1*1 + 0*2 + 1*4 = 5
3. 1*1 + 1*2 = 3
4. 0*1 + 1*2 = 2
5. 1*1 + 1*2 + 1*4 + 1*8 = 15
## 03
Write the following binary numbers in decimal and hexadecimal:
1. 10012
2. 1012
3. 112
4. 102
5. 11112
Expand using link to the right to see a suggested solution.
1. 10012: 910, 916
2. 1012: 510, 516
3. 112: 310, 316
4. 102: 210, 216
5. 11112: 1510, F16
## 04
Write the following hexadecimal values in binary and decimal:
1. B1
2. A
3. F
4. 12
5. 1A
6. 20
Hint: Going from the rightmost and moving left, the first bit is worth 160 (110), the next is worth 161 (1610), the next is worth 162 (25610) etc. Also, remember that A16 is 10, B16 is 11 ... and F16 is 15. So a single C16 is (12*1)10, and C116 is (12*16 + 1 = 193)10.
Hint: You can translate each digit of the hexadecimal number into four bits. B16 is 1110. 1110 is 10112. BB16 is 1011 10112. 3B16 is 0011 10112 (first 3, then B).
Expand using link to the right to see a suggested solution.
Answers (calculation shows hex to decimal):
1. B1: (1011 0001)2, (1 * 1 + 11 * 16 = 177)10
2. A: (1010)2, (10*1)10
3. F: (1111)2, (15*1)10
4. 12: (0001 0010)2, (2*1 + 1*16 = 18)10
5. 1A: (0001 1010)2, (10*1 + 1*16 = 26)10
6. 20: (0010 0000)2, (0*1 + 2*16 = 32)10
## 05
When converting a negative decimal number to binary using two's complement, you first make the leftmost bit into a one (signifying the minus sign), then you represent the number in binary, flip all bits and finally add one. In eight bits, the number -3 becomes:
a) represent the number in binary with the leftmost sign bit set to one:
1000 0011
b) flip the bits of the seven value bits:
1111 1100
1111 1100
+0000 0001
----------
1111 1101
Answer: -3 is in two's complement, eight bits, 1111 1101
Write the following negative decimal numbers in eight bit binary using two's complement:
1. -1
2. -2
3. -3
4. -4
5. -65
6. -127
Expand using link to the right to see a suggested solution.
Suggested solutions:
Write the following negative decimal numbers in eight bit binary using two's complement:
1. -110 is (1111 1111)2
2. -210 is (1111 1110)2
3. -310 is (1111 1101)2
4. -410 is (1111 1100)2
5. -6510 is (1011 1111)2
6. -12710 is (1000 0001)2
To calculate -65, we represent it in binary but with the sign bit set to one:
1100 0001 (1*20 + 1*26)10
Next, we flip the bits of the seven value bits:
1011 1110
1011 1110
+0000 0001
----------
1011 1111
Answer: (-65)10 is (1011 1111)2 in eight bit binary, using two's complement.
To calcualte -127, we get the following steps:
In decimal, 127 is 64 + 32 + 16 + 8 + 4 + 2 + 1, so the bits to set are:
0111 1111
|||| ||| `----2^0, which is 1
|||| |||
|||| || `-----2^1, which is 2
|||| ||
|||| | `------2^2, which is 4
|||| |
|||| `-------2^3, which is 8
||||
||| `---------2^4, which is 16
|||
|| `----------2^5, which is 32
||
| `-----------2^6, which is 64
|
`------------0x2^7, which is 0
Then we flip the bits:
1000 0000
1000 0000
+0000 0001
----------
1000 0001
Answer: -12710 is in eight bit binary, using 2's complement: (1000 0001)2
## 06
Convert the following text to eight bit binary, using the ASCII table.
Hint: Newline is 10, space is 32
Hello World!
In binary.
Expand using link to the right to see a suggested solution.
Suggested solution. Let's write the text, one character per line, with its's ascii value after it:
character ascii number
H 0100 1000
e 0110 0101
l 0110 1100
l 0110 1100
o 0110 1111
SPACE 0010 0000
W 0101 0111
o 0110 1111
r 0111 0010
l 0110 1100
d 0110 0100
! 0010 0001
NEWLINE 0000 1010
I 0100 1001
n 0110 1110
SPACE 0010 0000
b 0110 0010
i 0110 1001
n 0110 1110
a 0110 0001
r 0111 0010
y 0111 1001
. 0010 1110
Hello World!
In binary.
is in eight bit binary (using ASCII):
0100 1000 0110 0101 0110 1100 0110 1100 0110 1111 0010 0000 0101 0111 0110 1111 0111 0010 0110 1100 0110 0100 0010 0001 0000 1010 0100 1001 0110 1110 0010 0000 0110 0010 0110 1001 0110 1110 0110 0001 0111 0010 0111 1001 0010 1110
## 06
Binary numbers use 2 as the base, decimal numbers 10 and hexadecimal numbers 16.
Octal numbers use 8.
Evaluate the following octal numbers to decimal and hexadecimal:
1. 408
2. 508
3. 1018
4. 1728
Expand using link to the right to see a suggested solution.
Evaluate the following octal numbers to decimal and hexadecimal:
1. 408 - 3210 - 2016
2. 508 - 4010 - 2816
3. 1018 - 6510 - 4116
4. 1728 - 12210 - 7A16
Explanations (for a few of the above answers):
408 means, in decimal, (working from right to left) 0*80 + 4*81 = 3210.
We can recognize that (16 * 2 = 32)10, so we get 2016 which is: 0*160 + 2*161.
Answer: 408 is 3210 and 2016.
1728 means 2*80 + 7*81 + 1*82 = 2 + 56 + 64. That is, 12210.
7 * 16 + 10 = 122. So in hex, we get 7A. Remember, 7A means 10*160 + 7*161.
Answer: 1728 is 12210 and 7A16.
## 07
Converting a decimal number to octal can be done by taking the binary representation of the number and dividing it into groups of three.
Example:
3310 = 1000012
Dividing it to groups of three (padding with zeros) 100 001 gives us the decimal numbers 4 and 1. That is also the octal representation of 3310.
Another example:
12710 = 11111112
Dividing into groups of three gives us 001 111 111, so the octal number is 177.
Translate the following decimal numbers to octal:
1. 16
2. 32
3. 64
4. 65
Expand using link to the right to see a suggested solution.
Translate the following decimal numbers to octal:
1. 16
2. 32
3. 64
4. 65
16 is in binary 10000. Grouped in threes: 010 000 which then becomes 208.
32 becomes 100 000, which becomes 408.
64 becomes 001 000 000, which becomes 1008.
65 becomes 001 000 001, which becomes 1018.
| 2,485 | 6,787 |
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 4.34375 | 4 |
CC-MAIN-2021-04
|
latest
|
en
| 0.808907 |
https://www.esaral.com/q/prove-88208
| 1,725,930,096,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2024-38/segments/1725700651164.37/warc/CC-MAIN-20240909233606-20240910023606-00564.warc.gz
| 701,441,787 | 11,264 |
# Prove
Question:
$\int \frac{\sec ^{2} x}{\operatorname{cosec}^{2} x} d x$
Solution:
$\int \frac{\sec ^{2} x}{\operatorname{cosec}^{2} x} d x$
$=\int \frac{1}{\frac{\cos ^{2} x}{1} d x}$
$=\int \frac{\sin ^{2} x}{\cos ^{2} x} d x$
$=\int \tan ^{2} x d x$
$=\int\left(\sec ^{2} x-1\right) d x$
$=\int \sec ^{2} x d x-\int 1 d x$
$=\tan x-x+C$
| 164 | 352 |
{"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 | 4 |
CC-MAIN-2024-38
|
latest
|
en
| 0.253846 |
https://plainmath.net/9978/suppose-vertices-original-figure-example-vertices-clockwise-rotation
| 1,620,726,514,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2021-21/segments/1620243991982.8/warc/CC-MAIN-20210511092245-20210511122245-00135.warc.gz
| 480,604,749 | 9,502 |
Ask question
# Suppose the vertices of the original figure in the example were A(-6,6), B(-2,5), and C(-6,2). What would be the vertices of the image after a 90° clockwise rotation about the origin? A'() B'() C'(___)
Question
Vectors and spaces
asked 2021-02-02
Suppose the vertices of the original figure in the example were A(-6,6), B(-2,5), and C(-6,2). What would be the vertices of the image after a 90° clockwise rotation about the origin?
A'()
B'()
C'(___)
## Answers (1)
2021-02-03
When a point is rotated by 90 clockwise, the values of the x-coordinate and y-coordinate interchange with the x-coordinate changing its sign:
$$\displaystyle{\left({x},{y}\right)}\to{\left({y},-{x}\right)}$$
So, the vertices of the image are: $$\displaystyle{A}{\left(-{6},{6}\right)}\to{A}'{\left({6},{6}\right)}$$
$$\displaystyle{B}{\left(-{2},{5}\right)}\to{B}'{\left({5},{2}\right)}$$
$$\displaystyle{C}{\left(-{6},{2}\right)}\to{C}'{\left({2},{6}\right)}$$
### Relevant Questions
asked 2020-12-25
Case: Dr. Jung’s Diamonds Selection
With Christmas coming, Dr. Jung became interested in buying diamonds for his wife. After perusing the Web, he learned about the “4Cs” of diamonds: cut, color, clarity, and carat. He knew his wife wanted round-cut earrings mounted in white gold settings, so he immediately narrowed his focus to evaluating color, clarity, and carat for that style earring.
After a bit of searching, Dr. Jung located a number of earring sets that he would consider purchasing. But he knew the pricing of diamonds varied considerably. To assist in his decision making, Dr. Jung decided to use regression analysis to develop a model to predict the retail price of different sets of round-cut earrings based on their color, clarity, and carat scores. He assembled the data in the file Diamonds.xls for this purpose. Use this data to answer the following questions for Dr. Jung.
1) Prepare scatter plots showing the relationship between the earring prices (Y) and each of the potential independent variables. What sort of relationship does each plot suggest?
2) Let X1, X2, and X3 represent diamond color, clarity, and carats, respectively. If Dr. Jung wanted to build a linear regression model to estimate earring prices using these variables, which variables would you recommend that he use? Why?
3) Suppose Dr. Jung decides to use clarity (X2) and carats (X3) as independent variables in a regression model to predict earring prices. What is the estimated regression equation? What is the value of the R2 and adjusted-R2 statistics?
4) Use the regression equation identified in the previous question to create estimated prices for each of the earring sets in Dr. Jung’s sample. Which sets of earrings appear to be overpriced and which appear to be bargains? Based on this analysis, which set of earrings would you suggest that Dr. Jung purchase?
5) Dr. Jung now remembers that it sometimes helps to perform a square root transformation on the dependent variable in a regression problem. Modify your spreadsheet to include a new dependent variable that is the square root on the earring prices (use Excel’s SQRT( ) function). If Dr. Jung wanted to build a linear regression model to estimate the square root of earring prices using the same independent variables as before, which variables would you recommend that he use? Why?
1
6) Suppose Dr. Jung decides to use clarity (X2) and carats (X3) as independent variables in a regression model to predict the square root of the earring prices. What is the estimated regression equation? What is the value of the R2 and adjusted-R2 statistics?
7) Use the regression equation identified in the previous question to create estimated prices for each of the earring sets in Dr. Jung’s sample. (Remember, your model estimates the square root of the earring prices. So you must actually square the model’s estimates to convert them to price estimates.) Which sets of earring appears to be overpriced and which appear to be bargains? Based on this analysis, which set of earrings would you suggest that Dr. Jung purchase?
8) Dr. Jung now also remembers that it sometimes helps to include interaction terms in a regression model—where you create a new independent variable as the product of two of the original variables. Modify your spreadsheet to include three new independent variables, X4, X5, and X6, representing interaction terms where: X4 = X1 × X2, X5 = X1 × X3, and X6 = X2 × X3. There are now six potential independent variables. If Dr. Jung wanted to build a linear regression model to estimate the square root of earring prices using the same independent variables as before, which variables would you recommend that he use? Why?
9) Suppose Dr. Jung decides to use color (X1), carats (X3) and the interaction terms X4 (color * clarity) and X5 (color * carats) as independent variables in a regression model to predict the square root of the earring prices. What is the estimated regression equation? What is the value of the R2 and adjusted-R2 statistics?
10) Use the regression equation identified in the previous question to create estimated prices for each of the earring sets in Dr. Jung’s sample. (Remember, your model estimates the square root of the earring prices. So you must square the model’s estimates to convert them to actual price estimates.) Which sets of earrings appear to be overpriced and which appear to be bargains? Based on this analysis, which set of earrings would you suggest that Dr. Jung purchase?
asked 2020-12-01
a) To find:
The images of the following points under under a 90^circ rotation counterclockwise about the origin:
I. $$(2,\ 3)$$
II. $$(-1,\ 2)$$
III, (m,n) interms of m and n
b)To show:
That under a half-turn with the origin as center, the image of a point $$(a,\ b)\ \text{has coordinates}\ (-a,\ -b).$$
c) To find:
The image of $$P (a,\ b)\ text{under the rotation clockwise by} 90^{\circ}$$ about the origin.
asked 2021-03-08
A city planner wanted to place the new town library at site A. The mayor thought that it would be better at site B. What transformations were applied to the building at site A to relocate the building to site B? Did the mayor change the size or orientation of the library?
[Pic]
asked 2021-01-31
Let $$\displaystyle{B}={\left\lbrace{v}{1},{v}{2},\ldots,{v}{m}\right\rbrace}$$ be a basis for Rm. Suppose kvm is a linear combination of v1, v2, ...., vm-1 for some scalar k. What can be said about the possible value(s) of k?
asked 2020-10-27
Draw a graph for the original figure and its dilated image. Check whether the dilation is a similarity transformation or not.
Given:
The given vertices are
original $$\displaystyle\rightarrow{A}{\left({2},{3}\right)},{B}{\left({0},{1}\right)},{C}{\left({3},{0}\right)}$$
image $$\displaystyle\rightarrow{D}{\left({4},{6}\right)},{F}{\left({0},{2}\right)},{G}{\left({6},{0}\right)}$$
asked 2020-10-23
The table below shows the number of people for three different race groups who were shot by police that were either armed or unarmed. These values are very close to the exact numbers. They have been changed slightly for each student to get a unique problem.
Suspect was Armed:
Black - 543
White - 1176
Hispanic - 378
Total - 2097
Suspect was unarmed:
Black - 60
White - 67
Hispanic - 38
Total - 165
Total:
Black - 603
White - 1243
Hispanic - 416
Total - 2262
Give your answer as a decimal to at least three decimal places.
a) What percent are Black?
b) What percent are Unarmed?
c) In order for two variables to be Independent of each other, the P $$(A and B) = P(A) \cdot P(B) P(A and B) = P(A) \cdot P(B).$$
This just means that the percentage of times that both things happen equals the individual percentages multiplied together (Only if they are Independent of each other).
Therefore, if a person's race is independent of whether they were killed being unarmed then the percentage of black people that are killed while being unarmed should equal the percentage of blacks times the percentage of Unarmed. Let's check this. Multiply your answer to part a (percentage of blacks) by your answer to part b (percentage of unarmed).
Remember, the previous answer is only correct if the variables are Independent.
d) Now let's get the real percent that are Black and Unarmed by using the table?
If answer c is "significantly different" than answer d, then that means that there could be a different percentage of unarmed people being shot based on race. We will check this out later in the course.
Let's compare the percentage of unarmed shot for each race.
e) What percent are White and Unarmed?
f) What percent are Hispanic and Unarmed?
If you compare answers d, e and f it shows the highest percentage of unarmed people being shot is most likely white.
Why is that?
This is because there are more white people in the United States than any other race and therefore there are likely to be more white people in the table. Since there are more white people in the table, there most likely would be more white and unarmed people shot by police than any other race. This pulls the percentage of white and unarmed up. In addition, there most likely would be more white and armed shot by police. All the percentages for white people would be higher, because there are more white people. For example, the table contains very few Hispanic people, and the percentage of people in the table that were Hispanic and unarmed is the lowest percentage.
Think of it this way. If you went to a college that was 90% female and 10% male, then females would most likely have the highest percentage of A grades. They would also most likely have the highest percentage of B, C, D and F grades
The correct way to compare is "conditional probability". Conditional probability is getting the probability of something happening, given we are dealing with just the people in a particular group.
g) What percent of blacks shot and killed by police were unarmed?
h) What percent of whites shot and killed by police were unarmed?
i) What percent of Hispanics shot and killed by police were unarmed?
You can see by the answers to part g and h, that the percentage of blacks that were unarmed and killed by police is approximately twice that of whites that were unarmed and killed by police.
j) Why do you believe this is happening?
Do a search on the internet for reasons why blacks are more likely to be killed by police. Read a few articles on the topic. Write your response using the articles as references. Give the websites used in your response. Your answer should be several sentences long with at least one website listed. This part of this problem will be graded after the due date.
asked 2020-11-01
Given f(x)=6x+5, describe how the graph of g compares with the graph of f. g(x)=6(0.2x)+5
Select the correct choice below, and fill in the answer box to complete your choice.
A. The graph of g(x) is translated _ unit(s) to the left compared to the graph of f(x).
B. The graph of g(x) is translated _ unit(s) down compared to graph of f(x).
C. g(x) has a scale factor of _ compared to f(x). Because it scales the vertical direction, the graph is stretched vertically.
D. g(x) has a scale factor of _ compared to f(x). Because it scales the vertical direction, the graph is compressed vertically.
E. g(x) has a scale factor of _ compared to f(x). Because it scales the horizontal direction, the graph is stretched horizontally.
F. g(x) has a scale factor of _ compared to f(x). Because it scales the horizontal direction, the graph is compressed horizontally.
G. The graph of g(x) is translated _ unit(s) to the right compared to the graph of f(x).
H. The graph of g(x) is translated _ unit(s) up compared to graph of f(x).
asked 2020-10-21
The image of the point (2,1) under a translation is (5,-3). Find the coordinates of the image of the point (6,6) under the same translation.
asked 2020-11-12
Finance bonds/dividends/loans exercises, need help or formulas
Some of the exercises, calculating the Ri is clear, but then i got stuck:
A security pays a yearly dividend of 7€ during 5 years, and on the 5th year we could sell it at a price of 75€, market rate is 19%, risk free rate 2%, beta 1,8. What would be its price today? 2.1 And if its dividend growths 1,7% each year along these 5 years-what would be its price?
A security pays a constant dividend of 0,90€ during 5 years and thereafter will be sold at 10 €, market rate 18%, risk free rate 2,5%, beta 1,55, what would be its price today?
At what price have i purchased a security if i already made a 5€ profit, and this security pays dividends as follows: first year 1,50 €, second year 2,25€, third year 3,10€ and on the 3d year i will sell it for 18€. Market rate is 8%, risk free rate 0,90%, beta=2,3.
What is the original maturity (in months) for a ZCB, face value 2500€, required rate of return 16% EAR if we paid 700€ and we bought it 6 month after the issuance, and actually we made an instant profit of 58,97€
You'll need 10 Vespas for your Parcel Delivery Business. Each Vespa has a price of 2850€ fully equipped. Your bank is going to fund this operation with a 5 year loan, 12% nominal rate at the beginning, and after increasing 1% every year. You'll have 5 years to fully amortize this loan. You want tot make monthly installments. At what price should you sell it after 3 1/2 years to lose only 10% of the remaining debt.
asked 2020-11-20
Draw a graph for the original figure and its dilated image. Check whether the dilation is a similarity transformation or not.
Given:
The given vertices are
V(-3,4),W(-5,0),X(1,2),Y(-6,-2),Z(3,1)
...
| 3,396 | 13,582 |
{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 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}
| 3.875 | 4 |
CC-MAIN-2021-21
|
latest
|
en
| 0.917739 |
https://metanumbers.com/3979
| 1,670,333,025,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2022-49/segments/1669446711108.34/warc/CC-MAIN-20221206124909-20221206154909-00127.warc.gz
| 433,893,084 | 7,311 |
# 3979 (number)
3,979 (three thousand nine hundred seventy-nine) is an odd four-digits composite number following 3978 and preceding 3980. In scientific notation, it is written as 3.979 × 103. The sum of its digits is 28. It has a total of 2 prime factors and 4 positive divisors. There are 3,784 positive integers (up to 3979) that are relatively prime to 3979.
## Basic properties
• Is Prime? No
• Number parity Odd
• Number length 4
• Sum of Digits 28
• Digital Root 1
## Name
Short name 3 thousand 979 three thousand nine hundred seventy-nine
## Notation
Scientific notation 3.979 × 103 3.979 × 103
## Prime Factorization of 3979
Prime Factorization 23 × 173
Composite number
Distinct Factors Total Factors Radical ω(n) 2 Total number of distinct prime factors Ω(n) 2 Total number of prime factors rad(n) 3979 Product of the distinct prime numbers λ(n) 1 Returns the parity of Ω(n), such that λ(n) = (-1)Ω(n) μ(n) 1 Returns: 1, if n has an even number of prime factors (and is square free) −1, if n has an odd number of prime factors (and is square free) 0, if n has a squared prime factor Λ(n) 0 Returns log(p) if n is a power pk of any prime p (for any k >= 1), else returns 0
The prime factorization of 3,979 is 23 × 173. Since it has a total of 2 prime factors, 3,979 is a composite number.
## Divisors of 3979
1, 23, 173, 3979
4 divisors
Even divisors 0 4 2 2
Total Divisors Sum of Divisors Aliquot Sum τ(n) 4 Total number of the positive divisors of n σ(n) 4176 Sum of all the positive divisors of n s(n) 197 Sum of the proper positive divisors of n A(n) 1044 Returns the sum of divisors (σ(n)) divided by the total number of divisors (τ(n)) G(n) 63.0793 Returns the nth root of the product of n divisors H(n) 3.8113 Returns the total number of divisors (τ(n)) divided by the sum of the reciprocal of each divisors
The number 3,979 can be divided by 4 positive divisors (out of which 0 are even, and 4 are odd). The sum of these divisors (counting 3,979) is 4,176, the average is 1,044.
## Other Arithmetic Functions (n = 3979)
1 φ(n) n
Euler Totient Carmichael Lambda Prime Pi φ(n) 3784 Total number of positive integers not greater than n that are coprime to n λ(n) 1892 Smallest positive number such that aλ(n) ≡ 1 (mod n) for all a coprime to n π(n) ≈ 552 Total number of primes less than or equal to n r2(n) 0 The number of ways n can be represented as the sum of 2 squares
There are 3,784 positive integers (less than 3,979) that are coprime with 3,979. And there are approximately 552 prime numbers less than or equal to 3,979.
## Divisibility of 3979
m n mod m 2 3 4 5 6 7 8 9 1 1 3 4 1 3 3 1
3,979 is not divisible by any number less than or equal to 9.
## Classification of 3979
• Arithmetic
• Semiprime
• Deficient
• Polite
### By Shape (2D, centered)
• Centered Triangle
• Square Free
### Other numbers
• LucasCarmichael
## Base conversion (3979)
Base System Value
2 Binary 111110001011
3 Ternary 12110101
4 Quaternary 332023
5 Quinary 111404
6 Senary 30231
8 Octal 7613
10 Decimal 3979
12 Duodecimal 2377
20 Vigesimal 9ij
36 Base36 32j
## Basic calculations (n = 3979)
### Multiplication
n×y
n×2 7958 11937 15916 19895
### Division
n÷y
n÷2 1989.5 1326.33 994.75 795.8
### Exponentiation
ny
n2 15832441 62997282739 250666188018481 997400762125535899
### Nth Root
y√n
2√n 63.0793 15.8462 7.94225 5.24753
## 3979 as geometric shapes
### Circle
Diameter 7958 25000.8 4.97391e+07
### Sphere
Volume 2.63882e+11 1.98956e+08 25000.8
### Square
Length = n
Perimeter 15916 1.58324e+07 5627.16
### Cube
Length = n
Surface area 9.49946e+07 6.29973e+10 6891.83
### Equilateral Triangle
Length = n
Perimeter 11937 6.85565e+06 3445.92
### Triangular Pyramid
Length = n
Surface area 2.74226e+07 7.4243e+09 3248.84
## Cryptographic Hash Functions
md5 a58616464d14208b2677a084f5d7456f 4f9e000c9747d9707567bc38cb3ab2d307d48ca0 b003ffe3e5657009f984640507e7857562c7e876ff1144eb696ae1a1a4a908bc 4b43c2b6b9d13b94287a9f877429bdc7418c490ca87853dff8c9666299204b5cd0c2d39cfeb385a2b3f191323d360b11b6b659fda0a6f1dc7e9d8a6ad976971d 74edbc59e6857b794e2ef719998004cb1e9ae407
| 1,458 | 4,133 |
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 3.796875 | 4 |
CC-MAIN-2022-49
|
latest
|
en
| 0.816159 |
https://doquhoracujufuzuv.olivierlile.com/math-fraction-problems-43761ma.html
| 1,627,449,286,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2021-31/segments/1627046153521.1/warc/CC-MAIN-20210728025548-20210728055548-00455.warc.gz
| 234,417,676 | 4,166 |
# Math fraction problems
Division Worksheets Multiplication Chart Trying to memorize the multiplication facts. If you under and overestimated, is the answer in the correct range. Solving fraction problems requires a clear understanding of the process and a mastery of the basic mathematical operations.
Just like in addition, we're not going to change the denominators.
Multiplying Fractions Worksheets with Cross Cancelling These fractions worksheets are great for working on multiplying fractions. These worksheets will generate 10 or 15 mixed number subtraction problems per worksheet. As you learned in Lesson 2we'll multiply the whole number, 2, by the bottom number, 5.
Fact Family Worksheets Long Division Worksheets Introductory long division worksheets, long division worksheets with and without remainders, long division with decimals. You can print them directly from your browser window, but first check how it looks like in the "Print Preview".
Divide the new common denominator,by the old denominator in the first fraction, 6, to get Just like when we added, we'll stack our fractions to keep the numerators lined up. Pay attention to 'shared among' and make sure students don't confuse this phrasing with a subtraction word problem.
Overcoming this early solution bias can be difficult, and it is much better to develop the habit of making a complete pass over the problem before deciding on a path to the solution.
Multiplying and Dividing Multiply fractions by multiplying the two numerators together. Just like before, we're only going to add the numerators. Try underestimating and overestimating, so you know what range the answer is supposed to be in. The fractions worksheets may be selected for four different degrees of difficulty.
In this example, the numerators are 7 and 2.
If you can subtract whole numbersyou can subtract fractions too. These fraction worksheets will generate 10 or 15 problems per worksheet. Negation, Conjunction and Disjunction.
Practice worksheets for comparing fractions. The fraction problems on these sheets require. Bottom line: All adding and subtracting fractions with unlike denominators "problems" start and end with finding an LCD for the denominators, so I advise you to master this skill. Go from Subtracting Fractions With Unlike Denominators to Math Help With Fractions.
The fractions used in these problems have like (common) denominators. If you're seeing this message, it means we're having trouble loading external resources on our website.
If you're behind a web filter, please make sure that the domains *olivierlile.com and *olivierlile.com are unblocked.
Here is a set of practice problems to accompany the Partial Fractions section of the Applications of Integrals chapter of the notes for Paul Dawkins Calculus II course at Lamar University. problem, press the ENTER or RETURN key, or click on the reset problem button to set a new fraction problem.
Click on the check mark at the bottom to keep score! You can choose the number of fraction problems by clicking up and down by the 25 default.
These word problem worksheets place 4th grade math concepts in real world problems that students can relate to. We provide math word problems for addition, subtraction, multiplication, division, time, money, fractions and .
Math fraction problems
Rated 3/5 based on 5 review
Learning Math Problems - Fractions
| 680 | 3,391 |
{"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}
| 4.15625 | 4 |
CC-MAIN-2021-31
|
longest
|
en
| 0.886163 |
https://socratic.org/questions/59c35df411ef6b6ec13d6cfc#478275
| 1,660,611,085,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2022-33/segments/1659882572215.27/warc/CC-MAIN-20220815235954-20220816025954-00501.warc.gz
| 470,241,679 | 7,579 |
Question #d6cfc
Sep 21, 2017
The key word is "experiences". They have gravitational mass in a gravitational field (meaning they have weight), but the sensation of is weightlessness due to their accelerated reference frame.
Explanation:
Apparent weight is a concept that takes into account an accelerated reference frame. For example, if you are in an elevator that accelerates up or down, the amount of force exchange between your feet and the floor changes as the motion of the elevator changes. It is reduced if you are accelerating down and increased if you are accelerating up. For the case of weightlessness, the elevator would accelerate downwards at the rate of g. So there would be no need for a contact force with the ground. One would float in the elevator and feel weightless..
Now regarding a satellite, the capsule in a stable orbit is always accelerating towards the planet at a rate of g (the gravitational field at its location), so one inside the capsule would also need no contact with the "floor" and would feel weightless.
Sep 21, 2017
Because there are no external contact forces pushing or pulling on the astronaut.
Explanation:
There are a few concepts that need to be covered to understand this question. The first is weight. What is it? For an in-depth answer, check out this page, but basically, weight is the force of gravity acting on your mass. That is why you weigh less on the Moon then you do on Earth. Mass is different from weight, since mass isn't dependent on gravity.
Now that you understand what weight is, weightlessness is relatively simple: The absence of the perception of weight. Note that this doesn't mean gravity is non-existent, far from it. To explain weightlessness further, we must first know about contact and non-contact forces.
A contact force is any force which acts upon an object when in contact with that object. For instance, when someone pushes you, a contact force is being acted upon you. Or, when you throw a ball, you apply a contact force to that ball. You are being acted upon by a contact force even when you sit down, as your chair balances the force of gravity with its own upward force, commonly called the normal force.
A non-contact force is any force that acts on an object without being in contact with that object. This includes gravity, which acts on you even when you jump and stop contact with the Earth, and electro-magnetism, which causes magnets to connect with each other without needing to be in contact.
What do these forces have to do with weightlessness? Well, the fact is that weightlessness happens when all contact forces are removed. For example, take skydiving. After you jump out the plane, and before air-resistance starts to properly kick in, you feel weightless. This is because only gravity is working on you, a non-contact force. This is usually called free-fall. Take note that your weight doesn't actually disappear, rather your perception of having weight disappears, since the force of gravity, or weight, can't be felt without any other opposing forces.
How does this relate to astronauts? Well, astronauts orbiting Earth are in a constant state of free-fall towards Earth, and as we've discussed, free-fall equals only gravity acting on an object. Since the pod and astronaut are orbiting at the same speeds, there is no contact force, thus the astronauts experience weightlessness.
http://www.physicsclassroom.com/class/circles/Lesson-4/Weightlessness-in-Orbit
I hope I helped!
| 731 | 3,495 |
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 3.703125 | 4 |
CC-MAIN-2022-33
|
latest
|
en
| 0.964928 |
www.summerhillprimary.com
| 1,561,057,818,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2019-26/segments/1560627999273.24/warc/CC-MAIN-20190620190041-20190620212041-00139.warc.gz
| 857,454,877 | 27,491 |
## Times Tables
As you are probably aware learning their times tables is an imperative part of your child's maths education. Tables are needed for many different areas of maths and form an integral part of their learning.
Below are the new National Curriculum times tables expectations for each year group.
Year 1 - Count in multiples of 2,5 and 10. Recall and use doubles of all numbers to 10 and corresponding halves.
Year 2 - Recall and use multiplication and division facts for the 2, 5 and 10 multiplication tables, including recognising odd and even numbers.
Year 3 - Recall and use multiplication and division facts for the 3, 4 and 8 multiplication tables.
Year 4 - Recall multiplication and division facts for multiplication tables up to 12 × 12.
Year 5 - Revision of all times tables and division facts up to 12 x 12.
Year 6 - Revision of all times tables and division facts up to 12 x 12.
At Summerhill we encourage every child to achieve their class target using many different strategies to help them learn. We have times tables games and songs to sing to help them and we have a great opportunity to show off their new found skills by enabling them to achieve Bronze Silver and Gold status in each of the tables.
Times Tables Testing
Times Tables Testing takes place in the school hall every break time. Each class will be given a specific day that they can go into the hall to be tested. The children are tested by Times Tables Testers ( Children in Year 6 who have achieved Bronze, Silver and Gold in each of the tables up to 12). These children have be trained to listen and ask questions relating to each specific table and make sure that the children answer them in the time specified.
To achieve Bronze status your child must recite the appropriate table quickly and efficiently e.g 1 times 2 is 2, 2 times 2 is 4 etc.
To achieve Silver status your child will have to answer a variety of questions in no particular order on the specific table e.g. What is 7 multiplied by 8?, What are four lots of 7?
To achieve Gold status your child will have to answer a variety of questions about their chosen table that include related division facts e.g. What is 144 divided by 12?
Below are a few video examples of the types of questions the children will be asked and how quick they should answer them.
Please could you encourage your child to learn their year group expectation as it will help them greatly with their basic skills and maths in general.
Timestableathon
Towards the end of the school year we will hold a timestableathon, where children from each class will compete against each other to become the times table champion for their year group. There will also be a reward given to the class with the largest number of children completing their year group target.
| 610 | 2,828 |
{"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-2019-26
|
longest
|
en
| 0.90995 |
https://wiingy.com/learn/math/factors-of-184/
| 1,716,335,809,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2024-22/segments/1715971058522.2/warc/CC-MAIN-20240521214515-20240522004515-00334.warc.gz
| 549,668,894 | 42,830 |
#FutureSTEMLeaders - Wiingy's \$2400 scholarship for School and College Students
Apply Now
Factors
# Factors of 184 | Prime Factorization of 184 | Factor Tree of 184
Written by Prerit Jain
Updated on: 12 Aug 2023
## Factors of 184
Calculate Factors of
The Factors are
https://wiingy.com/learn/math/factors-of-184/
## What are the factors of 184
Have you ever heard of factors, it’s simply the set of numbers that divide a bigger number evenly: say 184 as an example, its factors are 1, 2, 4, 23 46 92, and 184.
But how to find these numbers? Well, you can use two simple methods!
One is division – divide your number by an integer within 0-184 by, Track which integers don’t leave any remains behind which will be all possible divisors!
The second method is Prime Factorization where you will break down a given number into smaller fragments. For instance, in order to get the prime factorization for our choice above [184], we should multiply together prime factors until they add up to our number 2x2x23x2 = 192!
## How to Find Factors of 184
Four methods to find the factors of 184 are:
Factor of 184 using Multiplication Method
Factors of 184 using Division Method
Prime Factorization of 184
Factor tree of 184
## Factors of 184 using the Multiplication Method
Why should we understand factors? They can help us do fun mathematics problem-solving and find secret puzzle codes. Let’s start this with the multiplication method,
Step 1: Let’s list 1 and 184 as two of our factors, every number is always divisible by 1 and themselves, right? Here we found two out of many possible ways
Step 2: Now shall we start dividing by other integers between 1 and 184
Whenever you divide something into groups without leaving remainders then they become one of your factors. Continue this until no numbers can divide evenly without leftovers. They are your factors of 184!
## Factors of 184 Using Division Method
You need a fun way to derive all the factors of a big number, right? Here you go! the division method is an easy and effective way to do that, say 184,
Step 1: Divide 184 by 2; you will get 92 with no remainder; that is 2 can perfectly fit into 184 two times so one of our factors now is 2.
Step 2: Continue dividing until there are no numbers can divide the leftover any further; lets divide 92 by 3, the remainder will be 30 hence 3 fits in five times into 184 making it another factor
Step 3: Keep going until you reach a number where nothing else divides without a remainder. Let us try 7 on 184 the remainder will be 2 meaning 7 is not a factor of 184!
## Prime Factorization of 184
Calculate Prime Factors of
The Prime Factors of 184 =
2 x
2 x
2 x
23
https://wiingy.com/learn/math/factors-of-184/
Prime factorization is a secret code to fragment big numbers into their building parts. Say 184,
Divide 184 by 2 three times every time the answer leaves no remainder; this means the first set of factors are all 2s (or two squared because there are 3)! Continue dividing the answer till you cannot divide it further in this case 23! Now let’s take the last number and multiply it with the first set of factors: The final answer or Prime Factorization is
(2^3) x 23 =184
## Factor tree of 184
https://wiingy.com/learn/math/factors-of-184/
Do you know about factor trees? it is a diagram that helps us find the prime factorization of a number by breaking it down into its factors. come on! Let’s create this fun tree and learn all the prime factors of 184
Step 1: write down your starting number at the top.
Step 2: Divide it by its smallest prime numbers and use those results as branches coming off from the original number.
## Step 3: Repeat this division continuously until you can no longer divide
Let’s use the number 184, shall we?
First of all, start with writing 184 on our paper or whiteboard:
184
Let’s look for its smallest possible prime factors. The first two would be 2×2 = 4. Now draw some lines branching off from the initial “184”:
184 / \ 4 92
Now let’s repeat looking for smaller increments to break apart each section until both sides touch a non-divisible number.
184 /\ 2 92 46 23
## Factor Pairs of 184
Calculate Pair Factors of
1 x 184=184
2 x 92=184
4 x 46=184
8 x 23=184
23 x 8=184
46 x 4=184
92 x 2=184
So Pair Factors of 184 are
(1,184)
(2,92)
(4,46)
(8,23)
(23,8)
(46,4)
(92,2)
https://wiingy.com/learn/math/factors-of-184/
Isn’t it wonderful that big numbers contain a series of smaller numbers fitted in like a puzzle into it? They are called factor pairs. Let’s take our 184, it can be written as four different pairs, (1, 184), (2, 92) there’s (4,46) making (23,8) the final piece. It’s important to note that the pairs 1 x 184 and 184 x 1 are the same pair and are counted once. So if someone wants to solve any problem related to factors quickly – they’ll need two things-to list out their puzzle elements properly from lowest to highest. Carefully explore how these pairs beautifully combine!
## Factors of 184 – Quick Recap
Factors of 184: 1, 2, 4, 8, 23, 46, 92, 184.
Negative Factors of 184: -1, -2, -4, -8, -23, -46, -92, -184.
Prime Factors of 184: 2 × 23
Prime Factorization of 184: 2 × 23
## Fun Facts of Factors of 184
Wanna know exciting facts about 184? 184 isn’t a prime number, which means that when you multiply two or more other numbers together, you get 184. For example, we can divide up 184 into many different factors say 1×184= 14 x 13 = 46 x 4 = 92×2….. By doing “prime factorization” it takes us to 2 special kinds of individual parts that make up our 184- those are 2^3 * 23 times each other. Just know that these values interact with each other and strengthen your skills in adding multiplying and making fractions too!
## Examples of Factor of 184
1. Two numbers have a product of 184. What could those numbers be?
Answer: The two numbers that have a product of 184 are 12 and 15 (12 x 15 = 184).
2. What is the sum of all factors of 184?
Answer: The sum of all factors of 184 is 168 (1 + 2 + 4 + 23 + 46 + 92 = 168).
3. How many distinct prime factors does 184 have?
Answer: 184 has three distinct prime factors (2, 2, and 23).
4. Is the number 184 divisible by 4?
Answer: Yes,184 is divisible by 4 (184 ÷ 4 = 46).
5. If 60 pieces were divided into six equal parts then how many pieces will each part consist of?
Answer: Each part would consist of 10 pieces (60 ÷ 6 = 10).
6. What is the largest prime factor of 184?
Answer: The largest prime factor of 184 is 23.
7. Express 184 as a product of two consecutive integers.
Answer: No it cannot be expressed so (11 x 17 ≠184)
8. Can you find two consecutive even integers whose product equals 184?
Answer: Yes, 12 x 15 =192
9. Explain what an odd factor is in relation to the number 184.
Answer: An odd factor in relation to the number 184 would be any integer that can divide it exactly which has an odd value – so in this case, 23 is an odd factor as it divides 184 exactly and its value is odd (23 ÷ 1 = 23).
10. What are the six distinct factors of the number 184?
Answer: The six distinct factors are 1, 2, 4, 23, 46, and 92.
## Frequently asked questions
### What are all the prime factors of 184?
Answer: The prime factor of 184 is 2 x 2 x23.
### How many factors does 184 have?
Answer: 184 has six distinct factors (1, 2, 4, 23, 46 and 92).
### Is 184 a composite number?
Answer: Yes, 184 is a composite number.
### Does 184 have any odd factors?
Answer: Yes,184 has one odd factor (23).
Written by
Prerit Jain
Share article on
| 2,057 | 7,621 |
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 4.9375 | 5 |
CC-MAIN-2024-22
|
latest
|
en
| 0.886523 |
https://www.cheenta.com/unbiased-pascal-and-mle-isi-mstat-2019-psb-problem-7/
| 1,718,979,805,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2024-26/segments/1718198862125.45/warc/CC-MAIN-20240621125006-20240621155006-00830.warc.gz
| 608,612,062 | 28,373 |
Get inspired by the success stories of our students in IIT JAM MS, ISI MStat, CMI MSc DS. Learn More
# Unbiased, Pascal and MLE | ISI MStat 2019 PSB Problem 7
This is a problem from the ISI MStat Entrance Examination,2019 involving the MLE of the population size and investigating its unbiasedness.
## The Problem:
Suppose an SRSWOR of size n has been drawn from a population labelled $$1,2,3,...,N$$ , where the population size $$N$$ is unknown.
(a)Find the maximum likelihood estimator $$\hat{N}$$ of $$N$$.
(b)Find the probability mass function of $$\hat{N}$$.
(c)Show that $$\frac{n+1}{n}\hat{N} -1$$ is an unbiased estimator of $$N$$.
## Prerequisites:
(a) Simple random sampling (SRSWR/SRSWOR)
(b)Maximum Likelihood estimator and how to find it.
(c)Unbiasedness of an estimator.
(d)Identities involving Binomial coefficients. (For this, you may refer to any standard text on Combinatorics like R.A.Brualdi,Miklos Bona etc.)
## Solution:
(a) Let $$X_1,X_2,..X_n$$ be the sample to be selected. In the SRSWOR scheme,
the selection probability of a sample of size $$n$$ is given by $$P(s)=\frac{1}{{N \choose n}}$$.
As, $$X_1,..,X_n \in \{1,2,...,N \}$$ , we have the maximum among them , that is the $$n$$ th order statistic, $$X_{(n)}$$ is always less than $$N$$.
Now, $${N \choose n}$$ is an increasing function of $$N$$. So, of course, $${X_{(n)} \choose n} \le {N \choose n }$$ , thus on reciprocating, we have $$P(s) \le \frac{1}{ {X_{(n)} \choose n}}$$. Hence the maximum likelihood estimator of $$N$$ i.e. $$\hat{N}$$ is $$X_{(n)}$$.
(b) We need to find the pmf of $$\hat{N}$$.
See that $$P(\hat{N}=m) = \frac{ {m \choose n} - {m-1 \choose n } }{ {N \choose n }}$$ , where $$m=n,n+1,...,N$$.
Can you convince yourself why?
(c) We use a well known identity , the Pascal's Identity to rewrite the distribution of $\hat{N}=X_{(n)}$ a bit more precisely:
We write $$P(\hat{N}=m) = \frac{ {m-1 \choose n-1}}{ {N \choose n} } ; \text{whenever m=n,n+1,...,N }$$
Thus, we have :
\begin{align} E(\hat{N})&=\sum_{m=n}^N m P(\hat{N}=m) =\frac{n}{\binom{N}{n}}\sum_{m=n}^N \frac{m}{n}\binom{m-1}{n-1} =\frac{n}{\binom{N}{n}}\sum_{m=n}^N \binom{m}{n} \end{align}
Also, use the Hockey Stick Identity to see that $$\sum_{m=n}^{N} {m \choose n} = {N+1 \choose n+1}$$
So, we have $$E(\hat{N})=\frac{n}{ {N \choose n}} {N+1 \choose n+1}=\frac{n(N+1)}{n+1}$$.
Thus, we get $$E( \frac{n+1}{n}\hat{N} -1) = N$$
## Useful Exercise:
Look up the many proofs of the Hockey Stick Identity. But make sure you at least learn the proof by a combinatorial argument and an alternative proof involving visualizing the identity via the Pascal's Triangle.
This site uses Akismet to reduce spam. Learn how your comment data is processed.
| 929 | 2,745 |
{"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}
| 4.34375 | 4 |
CC-MAIN-2024-26
|
latest
|
en
| 0.746603 |
https://www.geeksforgeeks.org/count-ways-to-split-array-into-k-non-intersecting-subsets/?ref=lbp
| 1,610,996,383,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2021-04/segments/1610703515235.25/warc/CC-MAIN-20210118185230-20210118215230-00067.warc.gz
| 825,739,643 | 26,510 |
Related Articles
Count ways to split array into K non-intersecting subsets
• Last Updated : 27 Oct, 2020
Given an array, arr[] of size N and an integer K, the task is to split the array into K non-intersecting subsets such that union of all the K subsets is equal to the given array.
Examples:
Input : arr[]= {2, 3}, K=2
Output: 4
Explanations:
Possible ways to partition the array into K(=2) subsets are:{ {{}, {2, 3}}, {{2}, {3}}, {{3}, {2}}, {{2, 3}, {}} }.
Therefore, the required output is 4.
Input: arr[] = {2, 2, 3, 3}, K = 3
Output: 9
Approach: The problem can be solved based on the following observations:
The total number of ways to place an element into any one of the K subsets = K.
Therefore, the total number of ways to place all distinct elements of the given array into K subsets = K × K × …..× K(M times) = KM
Where M = total number of distinct elements in the given array.
Follow the steps below to solve the problem:
Below is the implementation of the above approach:
## C++
`// C++ program to implement` `// the above approach` `#include ` `using` `namespace` `std;` `// Function to get` `// the value of pow(K, M)` `int` `power(``int` `K, ``int` `M)` `{` ` ``// Stores value of pow(K, M)` ` ``int` `res = 1;` ` ``// Calculate value of pow(K, N)` ` ``while` `(M > 0) {` ` ``// If N is odd, update` ` ``// res` ` ``if` `((M & 1) == 1) {` ` ``res = (res * K);` ` ``}` ` ``// Update M to M / 2` ` ``M = M >> 1;` ` ``// Update K` ` ``K = (K * K);` ` ``}` ` ``return` `res;` `}` `// Function to print total ways` `// to split the array that` `// satisfies the given condition` `int` `cntWays(``int` `arr[], ``int` `N,` ` ``int` `K)` `{` ` ``// Stores total ways that` ` ``// satisfies the given` ` ``// condition` ` ``int` `cntways = 0;` ` ``// Stores count of distinct` ` ``// elements in the given arr` ` ``int` `M = 0;` ` ``// Store distinct elements` ` ``// of the given array` ` ``unordered_set<``int``> st;` ` ``// Traverse the given array` ` ``for` `(``int` `i = 0; i < N; i++) {` ` ``// Insert current element` ` ``// into set st.` ` ``st.insert(arr[i]);` ` ``}` ` ``// Update M` ` ``M = st.size();` ` ``// Update cntways` ` ``cntways = power(K, M);` ` ``return` `cntways;` `}` `// Driver Code` `int` `main()` `{` ` ``int` `arr[] = { 2, 3 };` ` ``int` `N = ``sizeof``(arr) / ``sizeof``(arr[0]);` ` ``int` `K = 2;` ` ``cout << cntWays(arr, N, K);` ` ``return` `0;` `}`
## Java
`// Java program to implement ` `// the above approach ` `import` `java.io.*;` `import` `java.util.*; ` `class` `GFG{` ` ` `// Function to get` `// the value of pow(K, M)` `static` `int` `power(``int` `K, ``int` `M)` `{` ` ` ` ``// Stores value of pow(K, M)` ` ``int` `res = ``1``;` ` ` ` ``// Calculate value of pow(K, N)` ` ``while` `(M > ``0``)` ` ``{` ` ` ` ``// If N is odd, update` ` ``// res` ` ``if` `((M & ``1``) == ``1``) ` ` ``{` ` ``res = (res * K);` ` ``}` ` ` ` ``// Update M to M / 2` ` ``M = M >> ``1``;` ` ` ` ``// Update K` ` ``K = (K * K);` ` ``}` ` ``return` `res;` `}` ` ` `// Function to print total ways` `// to split the array that` `// satisfies the given condition` `static` `int` `cntWays(``int` `arr[], ``int` `N,` ` ``int` `K)` `{` ` ` ` ``// Stores total ways that` ` ``// satisfies the given` ` ``// condition` ` ``int` `cntways = ``0``;` ` ` ` ``// Stores count of distinct` ` ``// elements in the given arr` ` ``int` `M = ``0``;` ` ` ` ``// Store distinct elements` ` ``// of the given array` ` ``Set st = ``new` `HashSet(); ` ` ` ` ``// Traverse the given array` ` ``for``(``int` `i = ``0``; i < N; i++) ` ` ``{` ` ` ` ``// Insert current element` ` ``// into set st.` ` ``st.add(arr[i]);` ` ``}` ` ` ` ``// Update M` ` ``M = st.size();` ` ` ` ``// Update cntways` ` ``cntways = power(K, M);` ` ` ` ``return` `cntways;` `}` ` ` `// Driver Code` `public` `static` `void` `main (String[] args) ` `{` ` ``int` `arr[] = { ``2``, ``3` `};` ` ``int` `N = arr.length;` ` ``int` `K = ``2``;` ` ` ` ``System.out.println(cntWays(arr, N, K));` `}` `}` `// This code is contributed by sanjoy_62`
## Python3
`# Python3 program to implement` `# the above approach` `# Function to get` `# the value of pow(K, M)` `def` `power(K, M):` ` ``# Stores value of pow(K, M)` ` ``res ``=` `1` ` ``# Calculate value of pow(K, N)` ` ``while` `(M > ``0``):` ` ``# If N is odd, update` ` ``# res` ` ``if` `((M & ``1``) ``=``=` `1``):` ` ``res ``=` `(res ``*` `K)` ` ``# Update M to M / 2` ` ``M ``=` `M >> ``1` ` ``# Update K` ` ``K ``=` `(K ``*` `K)` ` ` ` ``return` `res` `# Function to print total ways` `# to split the array that` `# satisfies the given condition` `def` `cntWays(arr, N, K):` ` ` ` ``# Stores total ways that` ` ``# satisfies the given` ` ``# condition` ` ``cntways ``=` `0` ` ``# Stores count of distinct` ` ``# elements in the given arr` ` ``M ``=` `0` ` ``# Store distinct elements` ` ``# of the given array` ` ``st ``=` `set``()` ` ``# Traverse the given array` ` ``for` `i ``in` `range``(N):` ` ` ` ``# Insert current element` ` ``# into set st.` ` ``st.add(arr[i])` ` ` ` ``# Update M` ` ``M ``=` `len``(st)` ` ``# Update cntways` ` ``cntways ``=` `power(K, M)` ` ``return` `cntways` `# Driver Code` `if` `__name__ ``=``=` `'__main__'``:` ` ` ` ``arr ``=` `[ ``2``, ``3` `]` ` ``N ``=` `len``(arr)` ` ``K ``=` `2` ` ` ` ``print``(cntWays(arr, N, K))` `# This code is contributed by math_lover`
## C#
`// C# program to implement` `// the above approach ` `using` `System;` `using` `System.Collections.Generic;` `class` `GFG{` ` ` `// Function to get` `// the value of pow(K, M)` `static` `int` `power(``int` `K, ``int` `M)` `{` ` ` ` ``// Stores value of pow(K, M)` ` ``int` `res = 1;` ` ` ` ``// Calculate value of pow(K, N)` ` ``while` `(M > 0)` ` ``{` ` ` ` ``// If N is odd, update` ` ``// res` ` ``if` `((M & 1) == 1) ` ` ``{` ` ``res = (res * K);` ` ``}` ` ` ` ``// Update M to M / 2` ` ``M = M >> 1;` ` ` ` ``// Update K` ` ``K = (K * K);` ` ``}` ` ``return` `res;` `}` ` ` `// Function to print total ways` `// to split the array that` `// satisfies the given condition` `static` `int` `cntWays(``int``[] arr, ``int` `N,` ` ``int` `K)` `{` ` ` ` ``// Stores total ways that` ` ``// satisfies the given` ` ``// condition` ` ``int` `cntways = 0;` ` ` ` ``// Stores count of distinct` ` ``// elements in the given arr` ` ``int` `M = 0;` ` ` ` ``// Store distinct elements` ` ``// of the given array` ` ``HashSet<``int``> st = ``new` `HashSet<``int``>();` ` ` ` ``// Traverse the given array` ` ``for``(``int` `i = 0; i < N; i++) ` ` ``{` ` ` ` ``// Insert current element` ` ``// into set st.` ` ``st.Add(arr[i]);` ` ``}` ` ` ` ``// Update M` ` ``M = st.Count;` ` ` ` ``// Update cntways` ` ``cntways = power(K, M);` ` ` ` ``return` `cntways;` `} ` ` ` `// Driver code` `public` `static` `void` `Main()` `{` ` ``int``[] arr = { 2, 3 };` ` ``int` `N = arr.Length;` ` ``int` `K = 2;` ` ` ` ``Console.WriteLine(cntWays(arr, N, K));` `}` `}` `// This code is contributed by code_hunt`
Output:
```4
```
Time Complexity: O(log N)
Auxiliary Space: O(1)
Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.
My Personal Notes arrow_drop_up
Recommended Articles
Page :
| 3,048 | 8,182 |
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 3.703125 | 4 |
CC-MAIN-2021-04
|
latest
|
en
| 0.630363 |
https://www.coursehero.com/file/5953663/final07sol/
| 1,527,153,168,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2018-22/segments/1526794866107.79/warc/CC-MAIN-20180524073324-20180524093324-00075.warc.gz
| 703,377,964 | 96,265 |
{[ promptMessage ]}
Bookmark it
{[ promptMessage ]}
final07sol
# final07sol - STAT455/855 Fall 2007 Applied Stochastic...
This preview shows pages 1–4. Sign up to view the full content.
STAT455/855 Fall 2007 Applied Stochastic Processes Final Exam, Solutions 1. (15 marks) (a) (11 marks) Let q = P ( X i > m ). This is the same for all X i , and is equal to q = j = m +1 p (1 - p ) j - 1 = (1 - p ) m p j =0 (1 - p ) j = (1 - p ) m p 1 - (1 - p ) = (1 - p ) m . (i) (2 marks) The event { N k = k } occurs if and only if the first k observations are all greater than m . The probability of this is P ( N k = k ) = q k = (1 - p ) mk . (ii) (4 marks) N 1 has a Geometric( q ) distribution and so E [ N 1 ] = 1 q = 1 (1 - p ) m . Conditioning on N 1 , and then on the next observation after N 1 , we have E [ N 2 ] = j =1 E [ N 2 N 1 = j ] P ( N 1 = j ) = j =1 [( j + 1) q + ( j + E [ N 2 ])(1 - q )] P ( N 1 = j ) = j =1 jP ( N 1 = j ) + ( q + (1 - q ) E [ N 2 ]) j =1 P ( N 1 = j ) = E [ N 1 ] + q + (1 - q ) E [ N 2 ] = 1 q + q + (1 - q ) E [ N 2 ] or E [ N 2 ] = 1 + 1 q 2 = 1 + 1 (1 - p ) 2 m .
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
STAT 455/855 -- Final Exam Solutions, 2007 Page 2 of 7 (iii) (5 marks) Conditioning on N k - 1 and the next observation after N k - 1 , we have E [ N k ] = j =1 E [ N k N k - 1 = j ] P ( N k - 1 = j ) = j =1 [( j + 1) q + ( j + E [ N k ])(1 - q )] P ( N k - 1 = j ) = E [ N k - 1 ] + q + (1 - q ) E [ N k ] or E [ N k ] = 1 + 1 q E [ N k - 1 ] . Recursing, we obtain E [ N k ] = 1 + 1 q E [ N k - 1 ] = 1 + 1 q + 1 q 2 E [ N k - 2 ] . . . = 1 + 1 q + 1 q 2 + . . . + 1 q k - 2 + 1 q k - 1 E [ N 1 ] = 1 + 1 q + 1 q 2 + . . . + 1 q k - 2 + 1 q k = 1 - 1 /q k - 1 1 - 1 /q + 1 q k = q k - 1 - 1 q k - 1 - q k - 2 + 1 q k = (1 - p ) m ( k - 1) - 1 (1 - p ) m ( k - 1) - (1 - p ) m ( k - 2) + 1 (1 - p ) mk . (b) (4 marks) Let N i , i = 1 , . . . , 6, denote the number of biased dice that show i , and let X denote the outcome of the fair die. Conditioning on X , we have E [ N ] = 6 i =1 E [ N X = i ] 1 6 = 1 6 6 i =1 E [ N i ] = 1 6 E [ N 1 + N 2 + . . . + N 6 ] = n - 1 6 , since N 1 + N 2 + . . . + N 6 n - 1, the total number of biased dice.
STAT 455/855 -- Final Exam Solutions, 2007 Page 3 of 7 2. (15 marks) (a) (11 marks) The Markov chain is irreducible and aperiodic. We will give the stationary distribution of the chain, which shows that it is positive recurrent.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
This is the end of the preview. Sign up to access the rest of the document.
{[ snackBarMessage ]}
### What students are saying
• As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.
Kiran Temple University Fox School of Business ‘17, Course Hero Intern
• I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.
Dana University of Pennsylvania ‘17, Course Hero Intern
• The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.
Jill Tulane University ‘16, Course Hero Intern
| 1,293 | 3,620 |
{"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}
| 3.703125 | 4 |
CC-MAIN-2018-22
|
latest
|
en
| 0.743739 |
https://www.numbersaplenty.com/9396
| 1,708,556,432,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2024-10/segments/1707947473558.16/warc/CC-MAIN-20240221202132-20240221232132-00895.warc.gz
| 942,129,335 | 3,359 |
Search a number
9396 = 223429
BaseRepresentation
bin10010010110100
3110220000
42102310
5300041
6111300
736252
oct22264
913800
109396
117072
125530
13437a
1435d2
152bb6
hex24b4
9396 has 30 divisors (see below), whose sum is σ = 25410. Its totient is φ = 3024.
The previous prime is 9391. The next prime is 9397. The reversal of 9396 is 6939.
It can be written as a sum of positive squares in only one way, i.e., 8100 + 1296 = 90^2 + 36^2 .
It is a Smith number, since the sum of its digits (27) coincides with the sum of the digits of its prime factors.
It is a Harshad number since it is a multiple of its sum of digits (27).
It is a nude number because it is divisible by every one of its digits.
It is a nialpdrome in base 12.
It is a congruent number.
It is not an unprimeable number, because it can be changed into a prime (9391) by changing a digit.
It is a polite number, since it can be written in 9 ways as a sum of consecutive naturals, for example, 310 + ... + 338.
It is an arithmetic number, because the mean of its divisors is an integer number (847).
29396 is an apocalyptic number.
It is an amenable number.
It is a practical number, because each smaller number is the sum of distinct divisors of 9396, and also a Zumkeller number, because its divisors can be partitioned in two sets with the same sum (12705).
9396 is an abundant number, since it is smaller than the sum of its proper divisors (16014).
It is a pseudoperfect number, because it is the sum of a subset of its proper divisors.
9396 is a wasteful number, since it uses less digits than its factorization.
9396 is an evil number, because the sum of its binary digits is even.
The sum of its prime factors is 45 (or 34 counting only the distinct ones).
The product of its digits is 1458, while the sum is 27.
The square root of 9396 is about 96.9329665284. The cubic root of 9396 is about 21.1015489680.
The spelling of 9396 in words is "nine thousand, three hundred ninety-six".
| 564 | 1,978 |
{"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-2024-10
|
latest
|
en
| 0.927218 |
https://www.jobilize.com/algebra2/course/1-8-the-real-numbers-foundations-by-openstax?qcr=www.quizover.com&page=3
| 1,632,510,801,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2021-39/segments/1631780057564.48/warc/CC-MAIN-20210924171348-20210924201348-00514.warc.gz
| 849,569,677 | 19,932 |
# 1.8 The real numbers (Page 4/13)
Page 4 / 13
Can we simplify $\sqrt{-25}?$ Is there a number whose square is $-25?$
${\left(\text{}\phantom{\rule{0.5em}{0ex}}\right)}^{2}=-25?$
None of the numbers that we have dealt with so far has a square that is $-25.$ Why? Any positive number squared is positive. Any negative number squared is positive. So we say there is no real number equal to $\sqrt{-25}.$
The square root of a negative number is not a real number.
For each number given, identify whether it is a real number or not a real number: $\sqrt{-169}$ $\text{−}\sqrt{64}.$
1. There is no real number whose square is $-169.$ Therefore, $\sqrt{-169}$ is not a real number.
2. Since the negative is in front of the radical, $\text{−}\sqrt{64}$ is $-8,$ Since $-8$ is a real number, $\text{−}\sqrt{64}$ is a real number.
For each number given, identify whether it is a real number or not a real number: $\sqrt{-196}$ $\text{−}\sqrt{81}.$
not a real number real number
For each number given, identify whether it is a real number or not a real number: $\text{−}\sqrt{49}$ $\sqrt{-121}.$
real number not a real number
Given the numbers $-7,\frac{14}{5},8,\sqrt{5},5.9,\text{−}\sqrt{64},$ list the whole numbers integers rational numbers irrational numbers real numbers.
Remember, the whole numbers are 0, 1, 2, 3, … and 8 is the only whole number given.
The integers are the whole numbers, their opposites, and 0. So the whole number 8 is an integer, and $-7$ is the opposite of a whole number so it is an integer, too. Also, notice that 64 is the square of 8 so $\text{−}\sqrt{64}=-8.$ So the integers are $-7,8,\text{−}\sqrt{64}.$
Since all integers are rational, then $-7,8,\text{−}\sqrt{64}$ are rational. Rational numbers also include fractions and decimals that repeat or stop, so $\frac{14}{5}\phantom{\rule{0.2em}{0ex}}\text{and}\phantom{\rule{0.2em}{0ex}}5.9$ are rational. So the list of rational numbers is $-7,\frac{14}{5},8,5.9,-\sqrt{64}.$
Remember that 5 is not a perfect square, so $\sqrt{5}$ is irrational.
All the numbers listed are real numbers.
For the given numbers, list the whole numbers integers rational numbers irrational numbers real numbers: $-3,\text{−}\sqrt{2},0.\stackrel{\text{–}}{3},\frac{9}{5},4,\sqrt{49}.$
$4,\sqrt{49}$ $-3,4,\sqrt{49}$ $-3,0.\stackrel{\text{–}}{3},\frac{9}{5},4,\sqrt{49}$ $\text{−}\sqrt{2}$ $-3,\text{−}\sqrt{2},0.\stackrel{\text{–}}{3},\frac{9}{5},4,\sqrt{49}$
For the given numbers, list the whole numbers integers rational numbers irrational numbers real numbers: $\text{−}\sqrt{25},-\phantom{\rule{0.2em}{0ex}}\frac{3}{8},-1,6,\sqrt{121},2.041975\text{…}$
$6,\sqrt{121}$ $\text{−}\sqrt{25},-1,6,\sqrt{121}$ $\text{−}\sqrt{25},-\phantom{\rule{0.2em}{0ex}}\frac{3}{8},-1,6,\sqrt{121}$ $2.041975\text{…}$ $\text{−}\sqrt{25},-\phantom{\rule{0.2em}{0ex}}\frac{3}{8},-1,6,\sqrt{121},2.041975\text{…}$
## Locate fractions on the number line
The last time we looked at the number line , it only had positive and negative integers on it. We now want to include fraction s and decimals on it.
Doing the Manipulative Mathematics activity “Number Line Part 3” will help you develop a better understanding of the location of fractions on the number line.
Let’s start with fractions and locate $\frac{1}{5},-\phantom{\rule{0.2em}{0ex}}\frac{4}{5},3,\frac{7}{4},-\phantom{\rule{0.2em}{0ex}}\frac{9}{2},-5,\phantom{\rule{0.2em}{0ex}}\text{and}\phantom{\rule{0.2em}{0ex}}\frac{8}{3}$ on the number line.
We’ll start with the whole numbers $3$ and $-5.$ because they are the easiest to plot. See [link] .
The proper fractions listed are $\frac{1}{5}\phantom{\rule{0.2em}{0ex}}\text{and}\phantom{\rule{0.2em}{0ex}}-\phantom{\rule{0.2em}{0ex}}\frac{4}{5}.$ We know the proper fraction $\frac{1}{5}$ has value less than one and so would be located between $\text{0 and 1.}$ The denominator is 5, so we divide the unit from 0 to 1 into 5 equal parts $\frac{1}{5},\frac{2}{5},\frac{3}{5},\frac{4}{5}.$ We plot $\frac{1}{5}.$ See [link] .
Similarly, $-\phantom{\rule{0.2em}{0ex}}\frac{4}{5}$ is between 0 and $-1.$ After dividing the unit into 5 equal parts we plot $-\phantom{\rule{0.2em}{0ex}}\frac{4}{5}.$ See [link] .
Finally, look at the improper fractions $\frac{7}{4},-\phantom{\rule{0.2em}{0ex}}\frac{9}{2},\frac{8}{3}.$ These are fractions in which the numerator is greater than the denominator. Locating these points may be easier if you change each of them to a mixed number. See [link] .
When traveling to Great Britain, Bethany exchanged $602 US dollars into £515 British pounds. How many pounds did she receive for each US dollar? Jakoiya Reply how to reduced echelon form Solomon Reply Jazmine trained for 3 hours on Saturday. She ran 8 miles and then biked 24 miles. Her biking speed is 4 mph faster than her running speed. What is her running speed? Zack Reply d=r×t the equation would be 8/r+24/r+4=3 worked out Sheirtina find the solution to the following functions, check your solutions by substitution. f(x)=x^2-17x+72 Carlos Reply Aziza is solving this equation-2(1+x)=4x+10 Sechabe Reply No. 3^32 -1 has exactly two divisors greater than 75 and less than 85 what is their product? KAJAL Reply x^2+7x-19=0 has Two solutions A and B give your answer to 3 decimal places Adedamola Reply please the answer to the example exercise Patricia Reply 3. When Jenna spent 10 minutes on the elliptical trainer and then did circuit training for20 minutes, her fitness app says she burned 278 calories. When she spent 20 minutes onthe elliptical trainer and 30 minutes circuit training she burned 473 calories. How manycalories does she burn for each minute on the elliptical trainer? How many calories doesshe burn for each minute of circuit training? Edwin Reply .473 Angelita ? Angelita John left his house in Irvine at 8:35 am to drive to a meeting in Los Angeles, 45 miles away. He arrived at the meeting at 9:50. At 3:30 pm, he left the meeting and drove home. He arrived home at 5:18. DaYoungan Reply p-2/3=5/6 how do I solve it with explanation pls Adedamola Reply P=3/2 Vanarith 1/2p2-2/3p=5p/6 James don't understand answer Cindy 4.5 Ruth is y=7/5 a solution of 5y+3=10y-4 Adedamola Reply yes James don't understand answer Cindy Lucinda has a pocketful of dimes and quarters with a value of$6.20. The number of dimes is 18 more than 3 times the number of quarters. How many dimes and how many quarters does Lucinda have?
Find an equation for the line that passes through the point P ( 0 , − 4 ) and has a slope 8/9 .
is that a negative 4 or positive 4?
Felix
y = mx + b
Felix
if negative -4, then -4=8/9(0) + b
Felix
-4=b
Felix
if positive 4, then 4=b
Felix
then plug in y=8/9x - 4 or y=8/9x+4
Felix
Macario is making 12 pounds of nut mixture with macadamia nuts and almonds. macadamia nuts cost $9 per pound and almonds cost$5.25 per pound. how many pounds of macadamia nuts and how many pounds of almonds should macario use for the mixture to cost \$6.50 per pound to make?
| 2,209 | 6,961 |
{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 49, "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-2021-39
|
latest
|
en
| 0.720276 |
https://oeis.org/A301510
| 1,701,263,027,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2023-50/segments/1700679100081.47/warc/CC-MAIN-20231129105306-20231129135306-00592.warc.gz
| 499,152,349 | 4,946 |
The OEIS is supported by the many generous donors to the OEIS Foundation.
Hints (Greetings from The On-Line Encyclopedia of Integer Sequences!)
A301510 Smallest positive number b such that ((b+1)^prime(n) + b^prime(n))/(2*b + 1) is prime, or 0 if no such b exists. 1
1, 1, 1, 1, 1, 1, 1, 1, 4, 1, 3, 16, 1, 11, 6, 37, 1, 9, 120, 9, 1, 2, 67, 16, 1, 26, 103, 12, 60, 1, 239, 4, 40, 2, 44, 174, 33, 1, 3, 260, 114, 1, 161, 70, 1, 3, 2, 3, 50, 45, 472, 228, 183, 66, 37, 7, 122, 235, 68, 102, 294, 8, 13, 1, 40, 62, 143, 1, 61, 7 (list; graph; refs; listen; history; text; internal format)
OFFSET 2,9 COMMENTS Conjecture: a(n) > 0 for every n > 1. Records: 1, 4, 16, 37, 120, 239, 260, 472, 917, 1539, 6633, 7050, 12818, ..., which occur at n = 2, 10, 13, 17, 20, 32, 41, 52, 72, 128, 171, 290, 309, ... - Robert G. Wilson v, Jun 16 2018 LINKS Robert G. Wilson v, Table of n, a(n) for n = 2..345 Richard Fischer, Primzahlen mit der Form [(B+1)^N+B^N]/(2*B+1) Henri Lifchitz & Renaud Lifchitz, Search for: (a^n+b^n)/c FORMULA a(n) = A250201(2*prime(n)) - 1 for n >= 2. - Eric Chen, Jun 06 2018 EXAMPLE a(10) = 4 because (5^29 + 4^29)/9 = 2149818248341 is prime and (2^29 + 1^29)/3, (3^29 + 2^29)/5 and (4^29 + 3^29)/7 are all composite. MATHEMATICA Table[p = Prime[n]; k = 1; While[q = ((b+1)^n+b^n)/(2*b+1); ! PrimeQ[q], k++]; k, {n, 200}] f[n_] := Block[{b = 1, p = Prime@ n}, While[! PrimeQ[((b +1)^p + b^p)/(2b +1)], b++]; b]; Array[f, 70, 2] (* Robert G. Wilson v, Jun 13 2018 *) PROG (PARI) for(n=2, 200, b=0; until(isprime((((b+1)^prime(n)+b^prime(n))/(2*b+1))), b++); print1(b, ", ")) \\ corrected by Eric Chen, Jun 06 2018 CROSSREFS Numbers n such that ((b+1)^n + b^n)/(2*b + 1) is prime for b = 1 to 18: A000978, A057469, A128066, A128335, A128336, A187805, A181141, A187819, A217095, A185239, A213216, A225097, A224984, A221637, A227170, A228573, A227171, A225818. Cf. A058013, A103794, A222119, A247244, A250201. Sequence in context: A162516 A336693 A193793 * A085471 A064221 A229672 Adjacent sequences: A301507 A301508 A301509 * A301511 A301512 A301513 KEYWORD nonn AUTHOR Tim Johannes Ohrtmann, Mar 22 2018 STATUS approved
Lookup | Welcome | Wiki | Register | Music | Plot 2 | Demos | Index | Browse | More | WebCam
Contribute new seq. or comment | Format | Style Sheet | Transforms | Superseeker | Recents
The OEIS Community | Maintained by The OEIS Foundation Inc.
Last modified November 29 07:51 EST 2023. Contains 367429 sequences. (Running on oeis4.)
| 1,057 | 2,472 |
{"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-2023-50
|
latest
|
en
| 0.531429 |
https://chemistry.stackexchange.com/questions/47766/find-the-amount-of-atoms-for-a-given-mass
| 1,653,150,246,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2022-21/segments/1652662539131.21/warc/CC-MAIN-20220521143241-20220521173241-00264.warc.gz
| 207,652,036 | 65,407 |
# Find the amount of atoms for a given mass
The atomic mass of two elements A and B is 40 and 80 respectively. If x g of A contains y atoms, how many atoms are present in 2x g of B?
I have tried to solve by assuming that
• x g of A has y atoms
• so x g of B has 2y atoms, because atomic weight of B is 2 times higher than A
• so 2x g of B contains 4 y atoms.
But the answer key says it was wrong, the correct answer is y. What am I doing wrong?
• >The atomic mass of two elements A and B is 40 and 80 respectively. If x g of A contains y atoms, how many atoms are present in 2x g of B? Since B has a mass double that of A, 2X g og B should have the same amount of atoms as 1 X g og A. The answer is y atoms. May 22, 2021 at 19:50
For the elements A and B, the relation between the number of atoms n, the mass m, and the atomic mass M is given by
$$n_A = \frac{m_A}{M_A}\quad,\quad n_B = \frac{m_B}{M_B}$$
Your assignment states: $$m_B = 2\cdot m_A$$ $$M_B = 2\cdot M_A$$
Solve for $n_B$ ;-)
I seriously suggest to practise these calculations a lot. Ask your tutor for more worksheets or find a workbook (with solutions). Solve it alone, or work in groups. No matter how, you have to practise.
If you are in doubt, where to divide and where to multiply, remember:
| 373 | 1,273 |
{"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}
| 3.828125 | 4 |
CC-MAIN-2022-21
|
longest
|
en
| 0.946038 |
https://gmatclub.com/forum/for-any-integer-p-greater-than-1-p-denotes-the-product-of-all-the-in-260982.html
| 1,548,145,505,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2019-04/segments/1547583831770.96/warc/CC-MAIN-20190122074945-20190122100945-00203.warc.gz
| 501,453,163 | 53,188 |
GMAT Question of the Day - Daily to your Mailbox; hard ones only
It is currently 22 Jan 2019, 00:25
### GMAT Club Daily Prep
#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.
Customized
for You
we will pick new questions that match your level based on your Timer History
Track
every week, we’ll send you an estimated GMAT score based on your performance
Practice
Pays
we will pick new questions that match your level based on your Timer History
## Events & Promotions
###### Events & Promotions in January
PrevNext
SuMoTuWeThFrSa
303112345
6789101112
13141516171819
20212223242526
272829303112
Open Detailed Calendar
• ### The winners of the GMAT game show
January 22, 2019
January 22, 2019
10:00 PM PST
11:00 PM PST
In case you didn’t notice, we recently held the 1st ever GMAT game show and it was awesome! See who won a full GMAT course, and register to the next one.
• ### Key Strategies to Master GMAT SC
January 26, 2019
January 26, 2019
07:00 AM PST
09:00 AM PST
Attend this webinar to learn how to leverage Meaning and Logic to solve the most challenging Sentence Correction Questions.
# For any integer P greater than 1, P! denotes the product of all the in
Author Message
Current Student
Joined: 27 May 2014
Posts: 524
GMAT 1: 730 Q49 V41
For any integer P greater than 1, P! denotes the product of all the in [#permalink]
### Show Tags
07 Mar 2018, 09:44
00:00
Difficulty:
65% (hard)
Question Stats:
57% (01:10) correct 43% (01:35) wrong based on 14 sessions
### HideShow timer Statistics
For any integer P greater than 1, P! denotes the product of all the integers from 1 to P, inclusive. What is the greatest integer m for which 45^m is a factor of 48!?
A) 1
B) 2
C) 5
D) 10
E) 11
--== Message from the GMAT Club Team ==--
THERE IS LIKELY A BETTER DISCUSSION OF THIS EXACT QUESTION.
This discussion does not meet community quality standards. It has been retired.
If you would like to discuss this question please re-post it in the respective forum. Thank you!
To review the GMAT Club's Forums Posting Guidelines, please follow these links: Quantitative | Verbal Please note - we may remove posts that do not follow our posting guidelines. Thank you.
_________________
Retired Moderator
Joined: 25 Feb 2013
Posts: 1220
Location: India
GPA: 3.82
For any integer P greater than 1, P! denotes the product of all the in [#permalink]
### Show Tags
07 Mar 2018, 10:12
1
saswata4s wrote:
For any integer P greater than 1, P! denotes the product of all the integers from 1 to P, inclusive. What is the greatest integer m for which 45^m is a factor of 48!?
A) 1
B) 2
C) 5
D) 10
E) 11
$$45^m=(3^2*5)^m=3^{2m}*5^m$$
so to know the value of $$m$$ we need to know how many powers of $$5$$ are possible in $$48!$$, this can be done as
$$\frac{48}{5}+\frac{48}{5^2}=9+1=10$$. Hence $$m=10$$
Option D
Manager
Joined: 26 Apr 2011
Posts: 60
Location: India
GPA: 3.5
WE: Information Technology (Computer Software)
Re: For any integer P greater than 1, P! denotes the product of all the in [#permalink]
### Show Tags
08 Mar 2018, 03:20
niks18 wrote:
saswata4s wrote:
For any integer P greater than 1, P! denotes the product of all the integers from 1 to P, inclusive. What is the greatest integer m for which 45^m is a factor of 48!?
A) 1
B) 2
C) 5
D) 10
E) 11
$$45^m=(3^2*5)^m=3^{2m}*5^m$$
so to know the value of $$m$$ we need to know how many powers of $$5$$ are possible in $$48!$$, this can be done as
$$\frac{48}{5}+\frac{48}{5^2}=9+1=10$$. Hence $$m=10$$
Option D
Why you did not consider powers of 3?
_________________
Board of Directors
Status: Stepping into my 10 years long dream
Joined: 18 Jul 2015
Posts: 3626
For any integer P greater than 1, P! denotes the product of all the in [#permalink]
### Show Tags
08 Mar 2018, 03:39
1
MrCleantek wrote:
Why you did not consider powers of 3?
Hey MrCleantek ,
3 is not considered because out of 3 and 5, we will always have atleast as many number of 3s as 5s whereas the converse is not true.
Consider an example here: Let's say we have 20! and we need to find out number of 10s. Now every 10 will be composed of one 2 and one 5.
If you find the number of 2's, you will get 18
Number of 5's = 4. Hence, you can see we cannot have more than four 10s because we have only four 4s. Thus finding the number of 5s would do.
Does that make sense?
--== Message from the GMAT Club Team ==--
THERE IS LIKELY A BETTER DISCUSSION OF THIS EXACT QUESTION.
This discussion does not meet community quality standards. It has been retired.
If you would like to discuss this question please re-post it in the respective forum. Thank you!
To review the GMAT Club's Forums Posting Guidelines, please follow these links: Quantitative | Verbal Please note - we may remove posts that do not follow our posting guidelines. Thank you.
_________________
My GMAT Story: From V21 to V40
My MBA Journey: My 10 years long MBA Dream
My Secret Hacks: Best way to use GMATClub | Importance of an Error Log!
Verbal Resources: All SC Resources at one place | All CR Resources at one place
GMAT Club Inbuilt Error Log Functionality - View More.
New Visa Forum - Ask all your Visa Related Questions - here.
New! Best Reply Functionality on GMAT Club!
Find a bug in the new email templates and get rewarded with 2 weeks of GMATClub Tests for free
Check our new About Us Page here.
For any integer P greater than 1, P! denotes the product of all the in &nbs [#permalink] 08 Mar 2018, 03:39
Display posts from previous: Sort by
| 1,642 | 5,639 |
{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 3.921875 | 4 |
CC-MAIN-2019-04
|
latest
|
en
| 0.895851 |
https://www.coursehero.com/file/6691451/M1-practice-questions-w09/
| 1,526,839,629,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2018-22/segments/1526794863662.15/warc/CC-MAIN-20180520170536-20180520190536-00514.warc.gz
| 712,630,730 | 124,163 |
{[ promptMessage ]}
Bookmark it
{[ promptMessage ]}
M1 practice questions w09
# M1 practice questions w09 - MIDTERM 1 PRACTICE WINTER 2009...
This preview shows pages 1–4. Sign up to view the full content.
MIDTERM 1 PRACTICE WINTER 2009 PSTAT 5E Problem 1 Here are 8 data points listed. 96 6 9 1 4 3 11 84 a) Find the mean, median, and mode for the data. b) Explain why the median is different from the mean for this set of data. c) Find the variance and standard deviation for this set of data that fall less of the median.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Problem 2 Circle the T for “true” and the F for “false”. T F If events A and B are independent, then P(AUB) = P(A) + P(B) T F The average of five students’ heights is an example of a continuous variable. T F If two events A and B are mutually exclusive, then they must be independent. T F If two events A and B are independent, then they must be mutually exclusive. T F If a z-score ‘Z’ value is calculated to be 0.25, and we know that the mean is 44 and the x-value ‘X’ is found to be 46, then the standard deviation of x is 8. Problem 3 a) Suppose that X and Y are random variables, and that we have the following information: For this problem, consider only the population case. E[X] = 5 E[Y] = 3 Var(X) = 64 SD(Y) = 12 pXY = 0.5 With this given information, determine the Cov(X,Y). b) Now, if X and Y are both binomial random variables, then, using the same information above, except that the Var(X) and SD(Y) are unknown here, but that a sample size of n=10 is used for X and n=5 is used for Y, determine the Cov(X,Y) for these binomial random variables. (hint: remember the formulas!)
Problem 4 Four fair coins are flipped, one after another. a) Define the sample space for this set of events. b) Find the probability that all 4 of the coins landed on the same value. c) If we know that the first coin landed ‘heads’ for sure, but we do not know which of the next 3 did or did not, find the probability that exactly 3 heads were made.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
This is the end of the preview. Sign up to access the rest of the document.
{[ snackBarMessage ]}
### What students are saying
• As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.
Kiran Temple University Fox School of Business ‘17, Course Hero Intern
• I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.
Dana University of Pennsylvania ‘17, Course Hero Intern
• The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.
Jill Tulane University ‘16, Course Hero Intern
| 800 | 3,216 |
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 4.34375 | 4 |
CC-MAIN-2018-22
|
latest
|
en
| 0.869805 |
http://www.justanswer.com/homework/6593x-explain-0-41-cannot-probability-event.html
| 1,495,502,446,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2017-22/segments/1495463607245.69/warc/CC-MAIN-20170523005639-20170523025639-00264.warc.gz
| 558,739,396 | 19,277 |
• 100% Satisfaction Guarantee
Chris M., M.S.W. Social Work
Category: Homework
Satisfied Customers: 2786
Experience: Master's Degree, strong math and writing skills, experience in one-on-one tutoring (college English)
22285537
Chris M. is online now
# Explain why -0.41 cannot be the probability of some event.
### Resolved Question:
1. Explain why -0.41 cannot be the probability of some event.
2. Explain why 1.21 cannot be the probability of some event.
3. Explain why 120% cannot be the probability of some event?
3. Can the number 0.56 be the probability of an event?
4. If you roll a single die and count the number of dots on top, what is the sample space of all possible outcomes? Are the outcomes equally likely?
5. What is the probability of getting a number less than 5 on a single throw of a die.
6. What is the probability of getting 5 or 6 on a single throw?
7. You roll two fair dice, a green and red one. are the outcomes on the dice independent? Find P(1 on green die and 3 on red die).
8. Consider the following events for a college student selected at random:
A =student is female
B= student is majoring in business
Translate the following phrases into symbols.
a. the probability the student is male or is majoring in business
b. the probability a female student is majoring in business
c. the probability a business major is female
d. the probability the student is female and is not majoring in business
e. The probability the student is female and is majoring in business
Submitted: 5 years ago.
Category: Homework
Expert: Chris M. replied 5 years ago.
Hello, and thanks for the question.
1. Probability can't be expressed as a negative number.
2. Probability can't be expressed as greater than 1.
3. Probability can't be expressed as greater than 100%.
3a. Yes, probability can be expressed as a decimal fraction that is ≥ 0 and ≤ 1.
4. Sample space is {1, 2, 3, 4, 5, 6}. All outcomes are equally likely.
5. 4/6 = 2/3 or 66.67%
6. 2/6 = 1/3 or 33.33%
7. Yes they are independent. P(1 on green die and 3 on red die) = 1/6 *1/6 =1/36.
8.
a. P(A') U P(B)
b. P(B|A) = P(A^B)/P(A)
c. P(A|B) = P(A^B)/P(B)
d. P(A) ^ P(B')
e. P(A) ^ P(B)
Hope this helps!
Customer: replied 5 years ago.
I was also struggling with this question I had.
24. are customers more loyal in the east or west? the columns represent length of customer loyalty (in years) at a primary supermarket. The rows represent regions of the united states.
less than 1 yr 1-2 years 3-4yrs 5-9 yrs 10-14 yrs 15 or more yrs row total
east: 32 54 59 112 77 118 452
midwest: 31 68 68 120 63 173 523
south: 53 92 93 158 106 158 660
west: 41 56 67 78 45 86 373
columntotal:157 270 287 468 291 535 2008
what is the probability that a customer chosen at random?
a). has been loyal 10-14 years?
b). has been loyal 10-14 years, given that he or she is from the east?
c). has been loyal at least 10 years?
d) has been loyal at least 10 years, given that he or she is from the west?
e) is from the west, given that he or she has been loyal less than 1 year?
f) is from the south, given that he or she has been loyal less than 1 year?
g) has been loyal 1 or more years, given that he or she is from the east?
h) has been loyal 1 or more years, given that he or she is from the west?
i) are the events from the east and loyal 15 or more years independent?
Expert: Chris M. replied 5 years ago.
| 987 | 3,389 |
{"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}
| 3.6875 | 4 |
CC-MAIN-2017-22
|
latest
|
en
| 0.926081 |
http://en.wikibooks.org/wiki/Fortran/Fortran_Simple_math
| 1,394,378,563,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2014-10/segments/1393999679512/warc/CC-MAIN-20140305060759-00067-ip-10-183-142-35.ec2.internal.warc.gz
| 62,706,951 | 8,895 |
Fortran/Fortran Simple math
Part of the Fortran WikiBook
Fortran has the following arithmetic operators:
- subtraction
* multiplication
/ division
** exponentiation (associates right-to-left)
Here are some examples of their use:
``` i = 2 + 3 ! sets i equal to 5
i = 2 * 3 ! sets i equal to 6
i = 2 / 3 ! sets i equal to 0, since 2/3 is rounded down to the integer 0, see mixed mode
x = 2 / 3.0 ! sets x approximately equal to 2/3 or 0.666667
i = 2 ** 3 ! sets i equal to 2*2*2 = 8
```
Fortran has a wide range of functions useful in numerical work, such as sin, exp, and log. The argument of a function must have the proper type, and it is enclosed in parentheses:
``` x = sin(3.14159) ! sets x equal to sin(pi), which is zero
```
The intrinsic math functions of Fortran are elemental, meaning that they can take arrays as well as scalars as arguments and return a scalar or an array of the same shape:
``` real :: x(2),pi=3.14159
x = sin((/pi,pi/2/))
```
The above program fragment sets the two elements of array x, x(1) and x(2), equal to sin(pi) and sin(pi/2) respectively.
| 331 | 1,094 |
{"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-2014-10
|
longest
|
en
| 0.879383 |
https://www.build-electronic-circuits.com/how-to-choose-a-power-supply/
| 1,695,946,539,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2023-40/segments/1695233510462.75/warc/CC-MAIN-20230928230810-20230929020810-00596.warc.gz
| 712,996,494 | 28,407 |
# How To Choose a Power Supply
There’s some confusion going on when it comes to how to choose a power supply.
So, I thought I’d clear up some doubts, and give you a simple, beginner-friendly way of choosing a power supply.
Let’s start with the basics. A power supply has a voltage and a current specification:
• The voltage is the output voltage from the power supply.
• The current is the maximum current that the power supply can give.
Confused about voltages and currents? Then check out What You Need to Know About Current, Voltage and Resistance.
## Power Supply Voltage
The first thing to think about when choosing (or building) a power supply is the voltage. What voltage does your device need?
This one is pretty simple.
If your device needs 9V, you need a 9V power supply. If your device needs 5V, you need a 5V supply.
Also, if your device needs a DC voltage (which is the most common), you need a DC output from your power supply. If your device needs AC voltage, you need an AC output.
## Power Supply Current
The second thing to think about is current.
This one is a bit trickier and causes a lot of confusion.
The device you want to power will need a specific amount of current. If you bought the device, it should say how much current it needs somewhere in the technical documentation or by the power connector. If you built the device yourself, you can calculate or measure the amount of current needed.
So, let’s say your device needs 1A (ampere). This means you need to choose a power supply that can give minimum 1A.
## What about Watt?
Sometimes a power supply gives you a number in Watts (W). Watt is actually just the voltage multiplied by the current. So, if you have a 5V power supply that can give 1A of current, you have a 5W power supply.
If your power supply says 100W and 12V, you can figure out the maximum current by a simple calculation:
Current equals the Watt divided by the Voltage. So that’s 100 / 12 = 8.3 A.
## Frequently Asked Questions About How To Choose A Power Supply
Question: I have a 3A power supply, will it break my 500mA circuit?
Answer: No. A 3A power supply will not force 3A into your circuit. A 3A power supply will supply up to 3A of current. It’s your circuit that decides how much of that current it will actually use.
(EDIT: Note that a power supply is not the same as a battery charger. Connecting a power supply with too much current directly to a battery could damage it)
Question: My device says 5V (DC) 1A. What power supply do I need?
Answer: You need a power supply that gives 5V DC and has a maximum current of more than 1A.
Question: My circuit needs 9V, can I use my 12V supply?
Answer: No. If your circuit needs 9V and you connect it to 12V, it’s a good chance it will break.
What questions do you have about how to choose a power supply? Leave your comment question in the comment field below, and I’ll do my best to answer them all.
## More Circuit Building Tutorials
### 38 thoughts on “How To Choose a Power Supply”
1. I bought an item in the USA and was looking to put a UK power supply on it. The current plug details are:
Input:120V, 0.45A (54w)
Output: 12VDC, 1.0A (12w)
When purchasing a UK 12vdc power supply should I be buying it based on the Input or Output wattage?
Reply
• Hey Chris,
You can focus on the output side from the power supply,
If you buy a power supply that gives 12VDC, 1A (12W) it should work perfectly. Note that you can also use a power supply with more than 1A output. You just have to make sure it gives at least 1A.
Best,
Oyvind
Reply
• but what is the power specification of the items. what if the item requires 13v and 1A, and the power supply gives 12v 1A max. i think there will be problem as you have told us earlier. pls i need clarification.
Reply
• Yes, you need to have the same voltage. If the device requires 13V, you need 13V.
For current, if your device requires 1A, your power supply needs to have *at least* 1A or more.
Best,
Oyvind
Reply
2. How can I measure voltage using a multimeter
Reply
• Hey,
First set the multimeter to a voltage setting that is larger than the one you expect to measure. Ex: if you expect to see 9V, use the 20V setting.
Then use the measurement leads to measure the voltage between the two points you want to check.
Touch the positive measurement lead to the most positive point and the negative (COM) to the most negative point.
Reply
• Hello,
Safety wise, should’nt you touch the most negative point first and then the positive ?
Thanks!
JF
Reply
3. Thank you for this helpful post. I appreciate your efforts to clarify issues like this one.
Here is another issue where we could use some insight. I often see ads for phone chargers that purport to be much faster at recharging your phone than the one that originally came with the phone.
Since, as you explained above, the phone draws only the current it needs, how can one charger work faster than another?
Reply
• Hey Tom,
Good question.
Many modern smart phones use USB chargers these days. The USB cable has 4 wires. Two of those is the plus and minus of 5V. The other two are data lines.
The charger will set the voltage of the two data lines to a certain level to tell the phone which current rate it should use.
Best,
Oyvind
Reply
4. If you have a few devices of various voltages, i suggest to use a variable power supply from say DC 0V to 30V or more. You can also build your own variable power supply. I made my own ad it’s nice to do your own projects & is cheaper.
Reply
• Good point!
A simple one can be built from the LM317 chip, for those who’re interested.
Reply
• Hello.
Can you please send me the schematic for that variable power supply using the lm317 thank you
Reply
• You’ll find an example in it’s datasheet.
Reply
• hi wyane that’s awesome. could you please send me your circuit schematic? this is my e-mail :[email protected]
Reply
5. Frequently Asked Questions:
[Question]: I have a 3A power supply, will it break my 500mA circuit?
No. A 3A power supply will not force 3A into your circuit. A 3A power supply will supply *up to* 3A of current. It’s your circuit that decides how much of that current it will actually use.
[Question]: My device says 5V (DC) 1A. What power supply do I need?
You need a power supply that gives 5V DC and has a maximum current of 1A or more.
[Question]: My circuit needs 9V, can I use my 12V supply?
No. If your circuit needs 9V and you connect it to 12V, it’s a good chance it will break.
THIS IS WRONG…. 3A POWER SUPPLY BREAK 500ML CIRCUIT …. AM DAM SURE…. FOR EXAMPLE IF YOUR CELL PHONE IS 5V 500MA POWER SUPPLY??? YOU CAN TRY TO CONNECT WITH 5V 3A POWER SUPPLY.. YOU CAN SEE WHAT WILL HAPPEND.. IF YOU TRY THIS?? PLEASE KEEP A DISTANCE FROM THE EXPERIMENT CELL PHONE.. HEHEHE..
Reply
• Oh yes, connecting a power supply with too much current directly to a battery could damage the battery yes. A battery charger is not the same as a power supply. But most modern phones have the charger part included in the circuitry and make sure that this doesn’t happen. Thanks for the comment!
Reply
6. I am trying to learn whether it is the power supply that pushes the current to the device, and the device regulates it to what it needs, or is it the device that pulls from the power supply the amount of current it needs. Base on my limited understanding of electronics, if the power supply pushes, the device would have to regulate the current to what it needs. But if it is the device that pulls the current it needs then there would be no need for regulation. Am I correct? However the important question is whether it is a push or a pull. Thanks for your insight.
Reply
• Hi,
The device regulates how much current it draws from the power supply.
Best,
Oyvind
Reply
7. Hi. i m doing one project on Peltier cooler, My peltier name is Tec1-12706, (12 valt, 6 ampere), i used my computer SMPS as power supply with 12 valt 18 ampere, when i connect power supply with peltier the wire get hot at that place where i have joined together with peltier and smps wire.
is it happen as more ampere or Wire is thin ?
Reply
• If the wire get hot, it’s too thin for the current yes.
Best,
Oyvind
Reply
8. am looking for a way to make an electric cooling system that runs on electricity. i know peltier will not do the work. so pls i need help. thanks as you help.
Reply
• Hey Abba,
I don’t know much about cooling systems. But maybe someone else has an idea?
Best,
Oyvind
Reply
9. For my application i need 5V, 3A power supply.. so how can i get 3A current at the output? can i put 230-12V, 3A transformer to get 3Amps of current?
Reply
10. Let’s go a little bit slower with the Adapter, like the identification of it’s parts. There is the plug that goes into the wall. I have no name for that but “plug”. Then, there is the output source, usually male, that plugs into whatever device. I have no name for that. Finally, there where the device accepts the input power, usually female, and I have no name for that.
If I want to power a circuit via adapter, I’ve got some interfacing to do. So far, I have only needed YwRobots from Banggood, but that won’t help me forever. BTW, maybe you could do an article on YwRobots.
Reply
11. Let me re-ask my question. I have several power cube adapters spanning back 4 decades that I never threw away. How do I read them. Can I cut off their plugs and make my own DC Circuits powered from an electrical outlet?
Reply
• Hi,
Yes you can. They should have the voltage and maximum current rating written on them, And you can use a multimeter to measure the wires to find plus and minus.
Best,
Oyvind
Reply
12. How can I identify the needed power supply of my device or ciruit created myselfe
Reply
13. Great comments. Since we are discussing power supplies, are constant voltage supplies and constant current supplies both examples of power supplies? Is power supplies just a general name for these types of supplies? Or are there other types of supplies?
Thanks!
Reply
• Hi, yes those two are power supplies. And I can’t think of any other types of power supplies.
Best,
Oyvind
Reply
14. If you have two 5V power supply that require 1A each, what should I be looking for in a power source? 5V 2A?
Reply
• If you have two circuits that need 5V 1A each, yes then you would need a power source of 5V 2A.
Oyvind
Reply
15. If the circuit decides how much current it will use, why do we need resistors to limit current?
Reply
• Resistors are part of the circuit. So you use resistors “in” your circuit to get the current you want.
Reply
16. It´s interesting to know that power supplies need to have the same or more current to be able to provide enough energy to the device you´re trying to turn on. My brother knows more about this than I do, he makes inventions and builds devices himself, and for his birthday I want to give him something he will find useful and I thought about this because I´ve heard he needs it. Thank you for the useful information, I will be looking for a company that sells electric components to buy what he needs.
Reply
17. Dear Friend, article is very nice and clear for beginners, really.
Thanks!
One moment, i think it will be useful to add the info about
Stabilized or not Stabilized DC output.
With the best regards,
Alexey Batin.
Reply
18. the article is very usefull. Thanks for that. I also have some problem. I want 500wh(wotte hours) for my house devices per day. So, I want to charge a battery by using solar pannel for that. How do I select battery. What would be the specifications of the battery which I am going to select? Please give some advice for that.
Thank you.
Reply
19. Hello,
No questions to ask, just wanted to THANK YOU for this article ! it was very useful.
Thank you ^^
Reply
| 2,926 | 11,798 |
{"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-2023-40
|
longest
|
en
| 0.933386 |
http://kwiznet.com/p/takeQuiz.php?ChapterID=10541&CurriculumID=42&Method=Worksheet&NQ=6&Num=1.23
| 1,611,157,996,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2021-04/segments/1610703521139.30/warc/CC-MAIN-20210120151257-20210120181257-00129.warc.gz
| 63,040,794 | 3,391 |
Name: ___________________Date:___________________
Email us to get an instant 20% discount on highly effective K-12 Math & English kwizNET Programs!
### Middle/High School Algebra, Geometry, and Statistics (AGS)1.23 Dependent Equations
Dependent equations are consistent with many solutions. Example: two lines that coincide infinitely many points has many solutions. The equation ax + by + c = 0 is multiplied by a constant k, the resulting equation is kax + kby + kc = 0. The equation ax + by + c = 0 and kax + kby + kc = 0.are known as dependent equations. Example: If 3x + 2y + 3 = 0 and constant 2, write the dependent equation. solution: Multiply the given equation with the given constant term, we get the dependent equation. 2(3x + 2y + 3 = 0) = 6x+ 4y + 6 = 0. Directions: Choose the correct answer. Also write at least ten examples of your own.
Q 1: If 2x + 5y + c = 0 and constant 612x+30y+6c=012x+5y+6c=02x+30y+6c=012x+30y+c=0 Q 2: If 2x + 3y + c = 0 and constant 36x+9y+c = 06x+9y+3c = 02x+9y+3c = 06x+3y+3c = 0 Q 3: If x + 7y + c = 0 and constant 3x+21y+3c=03x+3y+3c=03x+21y+3c=03x+21y+c=0 Q 4: If 3x + 2y - 5 = 0 and constant 39x+6y+15=09x+6y-5=09x+3y-15=09x+6y-15=0 Question 5: This question is available to subscribers only! Question 6: This question is available to subscribers only!
| 474 | 1,306 |
{"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}
| 4.3125 | 4 |
CC-MAIN-2021-04
|
latest
|
en
| 0.868853 |
https://www.vedantu.com/maths/a-cube-b-cube-formula
| 1,721,827,167,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2024-30/segments/1720763518277.99/warc/CC-MAIN-20240724110315-20240724140315-00097.warc.gz
| 880,813,052 | 34,836 |
Courses
Courses for Kids
Free study material
Offline Centres
More
Store
# A Cube Plus B Cube Formula & A Cube Minus B Cube Formula
Reviewed by:
Last updated date: 24th Jul 2024
Total views: 129.3k
Views today: 3.29k
## Introduction
As we all know, maths contains formulas we need to learn for some of our chapters. The A cube B cube formula is one of the most common formulas of maths which is used in almost all chapters for basic solving and simplifying some equations. This comes under basic mathematics and needs to understand well.
In this article, we will understand the concept of a cube b cube formula, its expansion, the proof of this formula, and its types which are a cube plus b cube and a cube minus b cube.
## The A Cube B Cube Formula
The a cube b cube formula is defined as the sum or difference of cube of a and cube of b which are variables. These variables are used to easily put the values of variables and solve the given question.
The a cube b cube formula is in two variations, one is the addition, and the other is the subtraction - a cube plus b cube $\left( {{a}^{3}}+{{b}^{3}} \right)$ and a cube minus b cube $\left( {{a}^{3}}-{{b}^{3}} \right)$.
Now, let's understand how we can apply this formula to the given questions and which questions can be transformed into this formula.
## The $\left( {{a}^{3}}-{{b}^{3}} \right)$ Formula
The a cube minus b cube formula is given as,
$\left( {{a}^{3}}-{{b}^{3}} \right)=\left( a-b \right)\left( {{a}^{2}}+{{b}^{2}}+ab \right)$
## The $\left( {{a}^{3}}-{{b}^{3}} \right)$ Expansion
The $\left( {{a}^{3}}-{{b}^{3}} \right)$ formula can be expanded as,
$\left( {{a}^{3}}-{{b}^{3}} \right)=\left( a-b \right)\left( {{a}^{2}}+{{b}^{2}}+ab \right)$
On multiplying a and b separately,
$\left( {{a}^{3}}-{{b}^{3}} \right)=a\left( {{a}^{2}}+{{b}^{2}}+ab \right)-b\left( {{a}^{2}}+{{b}^{2}}+ab \right)$
$=\left( {{a}^{3}}+a{{b}^{2}}+{{a}^{2}}b \right)-\left( b{{a}^{2}}+{{b}^{3}}+a{{b}^{2}} \right)$
As we can see, simple brackets are opened. The formula of a cube minus b cube is equal to the product of the difference between a and b and the sum of a square, an into b and b square.
## Formula Proof For $\left( {{a}^{3}}-{{b}^{3}} \right)$ Variations a3- b3
The proof of a cube minus b cube is explained in the below steps.
Step-1 To expand the cube minus b cube formula, solve the right-hand side of the formula.
Step-2 Now, first solve the brackets $\left( a-b \right)\left( {{a}^{2}}+{{b}^{2}}+ab \right)$
$=\left( a-b \right)\left( {{a}^{2}}+{{b}^{2}}+ab \right)$
$=a\left( {{a}^{2}}+{{b}^{2}}+ab \right)-b\left( {{a}^{2}}+{{b}^{2}}+ab \right)$ $={{a}^{3}}+a{{b}^{2}}+{{a}^{2}}b-b{{a}^{2}}-{{b}^{3}}-a{{b}^{2}}$
Step-3 On further solving and cancelling the terms, we get the respective formula.
${{a}^{3}}-{{b}^{3}}$
Thus, the left-hand side is equal to the right-hand side of the equation. Hence proved that a cube minus b cube is equal to the product of the difference between a and b and the sum of a square, a into b and b square.
## The $\left( {{a}^{3}}+{{b}^{3}} \right)$ Formula
The cube plus b cube formula is given as,
$\left( {{a}^{3}}+{{b}^{3}} \right)=\left( a+b \right)\left( {{a}^{2}}+{{b}^{2}}-ab \right)$
## The $\left( {{a}^{3}}+{{b}^{3}} \right)$ Expansion
The $\left( {{a}^{3}}+{{b}^{3}} \right)$ formula can be expanded as,
$\left( {{a}^{3}}+{{b}^{3}} \right)=\left( a+b \right)\left( {{a}^{2}}+{{b}^{2}}-ab \right)$
$=a\left( {{a}^{2}}+{{b}^{2}}-ab \right)+b\left( {{a}^{2}}+{{b}^{2}}-ab \right)$
As we can see, simple brackets are opened. The formula of a cube plus b cube is equal to the product of the sum of a and b and the sum of a square and b square minus a into b.
## Formula Proof For $\left( {{a}^{3}}+{{b}^{3}} \right)$ Variations a3+ b3
The proof of a cube plus b cube is explained in the below steps.
Step-1 To expand the cube plus b cube formula, we are considering the right-hand side of the formula
Step-2 Now, first solve the brackets $\left( a+b \right)\left( {{a}^{2}}+{{b}^{2}}-ab \right)$
$\left( {{a}^{3}}+{{b}^{3}} \right)=\left( a+b \right)\left( {{a}^{2}}+{{b}^{2}}-ab \right)$
$=a\left( {{a}^{2}}+{{b}^{2}}-ab \right)+b\left( {{a}^{2}}+{{b}^{2}}-ab \right)$
$={{a}^{3}}+a{{b}^{2}}-{{a}^{2}}b+b{{a}^{2}}+{{b}^{3}}-a{{b}^{2}}$
Step-3 On further solving and cancelling the terms, we get the respective formula.
${{a}^{3}}+{{b}^{3}}$
Thus, the left-hand side is equal to the right-hand side of the equation. Hence proved that a cube plus b cube is equal to the product of the sum of a and b and the sum of a square and b square minus a and b.
## Where to Apply a Cube B Cube Formula
To understand where to apply a cube b cube formula, observe the given question. For example,
For the direct questions, like, 63 - 23, we can simply apply the formula and solve, but for the questions like 8 - 53, we need to figure out if the given number is the cube of some other number. We can see that 8 is the cube of 2. Thus, we can write the given question as 23 - 53 and solve it.
This technique can also be used for a cube plus b cube formula.
This way, we can figure out where to apply the cube minus b cube and a cube plus b cube formula.
## Important Question
1. What will be the factor of 27x3+64?
Ans. Step 1 : write 27x3+64 in the form of $\left( {{a}^{3}}+{{b}^{3}} \right)$.
27 is the cube of 3. It can be written as :
$27{{x}^{3}}+64={{\left( 3x \right)}^{3}}+{{\left( 4 \right)}^{3}}$
On comparing $\left( {{a}^{3}}+{{b}^{3}} \right)$ and ${{\left( 3x \right)}^{3}}+{{\left( 4 \right)}^{3}}$, we get:
a=3x
b=4
According to the formula, $\left( {{a}^{3}}+{{b}^{3}} \right)=\left( a+b \right)\left( {{a}^{2}}+{{b}^{2}}-ab \right)$
${{\left( 3x \right)}^{3}}+{{\left( 4 \right)}^{3}}=\left( 3x+4 \right)\left( {{\left( 3x \right)}^{2}}+{{4}^{2}}+3x\times 4 \right)$
$27{{x}^{3}}+64=\left( 3x+4 \right)\left( 9{{x}^{2}}+16+12x \right)$
## Solved Problem
1. Solve x3 + y3 where values for x are 2 and y is 3.
Ans. We know the formula , $\left( {{x}^{3}}+{{y}^{3}} \right)=\left( x+y \right)\left( {{x}^{2}}+{{y}^{2}}-xy \right)$.
x=2 and y=3 (given)
$\left( {{2}^{3}}+{{3}^{3}} \right)=\left( 2+3 \right)\left( {{2}^{2}}+{{3}^{2}}-2\times 3 \right)$
$=\left( 2+3 \right)\left( 4+9-6 \right)$
$=35$
Therefore, the value for 23+33 is 35.
## Conclusion
The formula of a cube minus b cube is equal to the product of the difference between a and b and the sum of a square into b and b square. The formula of a cube plus b cube is equal to the product of the sum of a and b and the sum of a square and b square minus a into b.
## FAQs on A Cube Plus B Cube Formula & A Cube Minus B Cube Formula
1. Using the a3 - b3 formula, solve 43 - 63.
By using the formula , $\left( {{a}^{3}}-{{b}^{3}} \right)=\left( a-b \right)\left( {{a}^{2}}+{{b}^{2}}+ab \right)$.
a=4 and b=6 (given)
By substituting the values of a and b in the above formula,
${{4}^{3}}-{{6}^{3}}=\left( 4-6 \right)\left( {{4}^{2}}+{{6}^{2}}+4\times 6 \right)$
$=\left( 4-6 \right)\left( 16+36+24 \right)$
$=-152$
Therefore, 43-63= -152.
2. How can you solve a2- b2?
In order to solve we will use the right-hand side of the formula
a2- b2 $=\left( a+b \right)\left( a-b \right)$
$=a\left( a-b \right)+b\left( a-b \right)$
$={{a}^{2}}-ab+ba-{{b}^{2}}$
On cancelling the opposite sign values, we get
${{a}^{2}}-{{b}^{2}}$.
Therefore, the left-hand side is equal to the right-hand side.
3. Find the value of (x+y)2 when x = 4 and y = 3 .
Given : x = 4
Y = 3
We can solve this by using the formula :
${{\left( x+y \right)}^{2}}={{x}^{2}}+{{y}^{2}}+2xy$
${{\left( 4+3 \right)}^{2}}={{4}^{2}}+{{3}^{2}}+2\times 4\times 3$
$=16+9+24$
$=49$
Therefore, (4+3)2 = 49.
| 2,779 | 7,717 |
{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 4.8125 | 5 |
CC-MAIN-2024-30
|
latest
|
en
| 0.798758 |
http://mathhelpforum.com/algebra/77978-finding-ratio-geometric-sequence.html
| 1,481,306,422,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2016-50/segments/1480698542714.38/warc/CC-MAIN-20161202170902-00215-ip-10-31-129-80.ec2.internal.warc.gz
| 171,266,580 | 10,045 |
# Thread: Finding the ratio of the geometric sequence?
1. ## Finding the ratio of the geometric sequence?
i, -1, i
So the ratio if sqrt(-1)/-1? There doesn't seem to be a common ratio...
2. Originally Posted by puzzledwithpolynomials
i, -1, i
So the ratio if sqrt(-1)/-1? There doesn't seem to be a common ratio...
$\frac{t_2}{t_1} = \frac{i}{-1} = -i$.
$\frac{t_3}{t_2} = \frac{-1}{i} = \frac{-1}{i}\cdot\frac{-i}{-i} = \frac{-(-i)}{-i^2} = \frac{i}{-(-1)} = \frac{i}{1} = i$.
You're right, there's not a common ratio.
3. Originally Posted by puzzledwithpolynomials
i, -1, i
So the ratio if sqrt(-1)/-1? There doesn't seem to be a common ratio...
As written, I must agree with you. It is no geometric.
If it were $i\;,\, - 1\;,\, \color{red}{- i}$ then $i$ works.
| 274 | 771 |
{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 4, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 3.796875 | 4 |
CC-MAIN-2016-50
|
longest
|
en
| 0.868767 |
https://web.njit.edu/~kappraff/chinese.html
| 1,561,641,845,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2019-26/segments/1560628001138.93/warc/CC-MAIN-20190627115818-20190627141818-00014.warc.gz
| 633,324,762 | 2,034 |
Home | AcademicPersonal | Research | Site Map | Search Web | Email | Links
Back to Class Notes (M326)
Chinese Remainder Theorem:
Ask someone to pick a number from 1 to 1000. they must tell you the remainder that they get when they divide their number by first 7, then 11, then 13. You then have enough information to tell what their original number was. Here is how to do it.
Let's say the remainders after dividing by 7, 11, and 13 respectively were a, b, and c. Compute the expression:
715a + 364b + 924c
This number has the desired properties but it is generally larger than 1000. To determine your friend's number, keep subracting 1001 from this result until you get a number less than 1000.
Example:
Let's say your friend's number were 200.
Then a=4, b=2, and c=5
Therefore: 715(4) + 364(2) + 924(5) = 8208
If you subtract 8 groups of 1001, then you get,
8208 - 8008 = 200
Your problem is to figure out how this trick works. The answer lies in the following tables of modular arithmetic mod 7, mod 11, and mod 13.
143 = 11x13 Multiples of 143 Value Mod 7 91 = 7x13 Multiples of 91 Value Mod 11 77 = 7x11 Multiples of 77 Value Mod 13 143 286 429 572 715 858 1001 3 6 2 5 1 4 0 91 182 273 364 455 546 637 728 819 910 1001 3 6 9 1 4 7 10 2 5 8 0 77 154 231 308 385 462 539 616 693 770 847 924 1001 12 11 10 9 8 7 6 5 4 3 2 1 0
Home | AcademicPersonal | Research | Site Map | Search Web | Email | Links
| 516 | 1,454 |
{"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-2019-26
|
latest
|
en
| 0.702832 |
https://www.teacherspayteachers.com/Product/Geometry-Enrichment-Projects-Choice-Board-5th-Grade-2692122
| 1,531,748,358,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2018-30/segments/1531676589270.3/warc/CC-MAIN-20180716115452-20180716135452-00368.warc.gz
| 1,015,609,577 | 19,747 |
# Geometry Enrichment Projects Choice Board – 5th Grade
Subject
Resource Type
Common Core Standards
Product Rating
4.0
8 Ratings
File Type
PDF (Acrobat) Document File
4 MB|14 pages
Share
Get this as part of a bundle:
1. Math Enrichment Choice Board Bundle – All Fifth Grade Standards– These five enrichment menu projects are an amazing differentiation tool that not only empowers students through choice but also meets their individual needs. This resource includes all five of my 5th grade math enrichment choice bo
Product Description
Geometry Choice Board – Fifth Grade– This enrichment menu project is an amazing differentiation tool that not only empowers students through choice but also meets their individual needs.
You will find that the geometry enrichment board contains three leveled activities for each standard: appetizer, entrée, and dessert. No two activities are alike!
Students can either choose to complete activities from the project menu or design their own projects with the approval of their teachers by completing the project proposal.
This resource includes…
• Sample Lesson Plans
• Project Proposal
• Project Rubric
• Presentation Rubric
This choice board enrichment project is especially wonderful for both gifted and reluctant learners because it gives the students a greater sense of ownership, the ability to work at their own pace, and the freedom to choose or design activities based on their own interests and readiness.
Standards:
5.G.A.1 - Use a pair of perpendicular number lines, called axes, to define a coordinate system, and I understand that the first number indicates how far to travel from the origin in the direction of one axis and the second number indicates how far to travel in the direction of the second axis.
5.G.A.2 - Represent real world and mathematical problems by graphing points in the first quadrant of the coordinate plane, and interpret coordinate values of points in the context of the situation.
5.G.B.3 - Understand that attributes belonging to a category of two-dimensional figures also belong to all subcategories of that category.
5.G.B.4 - Classify two-dimensional figures in a hierarchy based on properties.
* * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * *
CHECK OUT THESE OTHER CHOICE BOARD MENUS (click on each to view):
Operations and Algebraic Thinking Choice Board – Fifth Grade
Numbers and Operations Choice Board – Fifth Grade
Measurement & Data Choice Board – Fifth Grade
Fraction Choice Board – Fifth Grade
Math Enrichment Choice Board Bundle – All Fifth Grade Standards
* * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * *
Please click the "Follow Me" link located under my picture and store name on the right-hand column of this page to keep updated on all my new offerings!
Pinterest
Simon Says School Blog
Total Pages
14 pages
Rubric only
Teaching Duration
N/A
Report this Resource
\$1.00
| 623 | 2,917 |
{"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-2018-30
|
latest
|
en
| 0.905955 |
https://www.hackmath.net/en/word-math-problems/units?page_num=110
| 1,607,116,849,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2020-50/segments/1606141743438.76/warc/CC-MAIN-20201204193220-20201204223220-00108.warc.gz
| 686,684,959 | 10,454 |
# Units - math word problems
#### Number of problems found: 3477
• The half life
The half-life of a radioactive isotope is the time it takes for a quantity of the isotope to be reduced to half its initial mass. Starting with 145 grams of a radioactive isotope, how much will be left after 3 half-lives?
• Viewing angle
The observer sees a straight fence 60 m long at a viewing angle of 30°. It is 102 m away from one end of the enclosure. How far is the observer from the other end of the enclosure?
• KLMN trapezoid
The KLMN trapezoid has bases KL 40cm and MN 16cm. On the KL base is point P. The segment NP divides the trapezoid into units with the same area. What is the distance of point P from point K?
• Pyramid 4sides
Calculate the volume and the surface of a regular quadrangular pyramid when the edge of the base is 4 cm long and the height of the pyramid is 7 cm.
• Force meter
We put the statuette on the force meter. The force meter showed a value 25 N. Then we placed the statuette on the force meter completely immersed in the water. The force meter showed a value 17 N. What is the volume of the statuette?
• Rectangular field
A rectangular field has a diagonal of length 169m. If the length and width are in the ratio 12:5. Find the dimensions of the field, the perimeter of the field and the area of the field.
• Brick weight
The brick weighs 2 kg and a half bricks. How much does one brick weigh?
• Bath
In the bath is 30 liters of hot water. Then added 36 liters of cold water at temperature of 19 °C decreased temperature of water to 41 °C. What was the initial temperature of the hot water?
• Mixing Celsius and Fahrenheit
Add up three temperatures: 5°F +6°F +0°C
• A car
A car weighing 1.05 tonnes driving at the maximum allowed speed in the village (50 km/h) hit a solid concrete bulkhead. Calculate height it would have to fall on the concrete surface to make the impact intensity the same as in the first case!
• Family
Family has 4 children. Ondra is 3 years older than Matthew and Karlos 5 years older than the youngest Jane. We know that they are together 30 years and 3 years ago they were together 19 years. Determine how old the children are.
• Driver
The driver of the car at a speed of 100 km/h faced the obstacle and began to brake with a slowing of 5 m/s². What is the path to stopping the car when the driver has registered the obstacle with a delay of 0.7 s?
• The pond
We can see the pond at an angle 65°37'. Its end points are 155 m and 177 m away from the observer. What is the width of the pond?
• Inflow - outflow
The tank will fill by inflow for 143 minutes and empties by outflow for 150 minutes. How long take to fill tank if it is also opened the inflow and outflow?
• Eight
Eight small Christmas balls with a radius of 1 cm have the same volume as one large Christmas ball. What has a bigger surface: eight small balls, or one big ball?
• Trio ratio
Hans, Alena and Thomas have a total of 740 USD. Hans and Alena split in the ratio 5: 6 and Alena and Thomas in the ratio 4: 5. How much will everyone get?
• Sand
The maximum weight of the car is 5000 kg. 10 m3 of sand must be transferred. How many times does a car have to go? (density of sand is 1500 kg/m3)
• Aluminum wire
Aluminum wire of 3 mm diameter has a total weight of 1909 kg and a density of 2700 kg/m3. How long is the wire bundle?
• Thales
Thales is 1 m from the hole. The eyes are 150 cm above the ground and look into the hole with a diameter of 120 cm as shown. Calculate the depth of the hole.
• Three siblings
Three siblings have birthday in one day-today. Together they have 35 years today. The youngest is three years younger than middle and the oldest is 5 years older than middle. How old is each?
Do you have an interesting mathematical word problem that you can't solve it? Submit a math problem, and we can try to solve it.
We will send a solution to your e-mail address. Solved examples are also published here. Please enter the e-mail correctly and check whether you don't have a full mailbox.
Please do not submit problems from current active competitions such as Mathematical Olympiad, correspondence seminars etc...
| 1,046 | 4,147 |
{"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-2020-50
|
latest
|
en
| 0.889861 |
https://wiki.seg.org/wiki/Dipping_refractor
| 1,721,077,789,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2024-30/segments/1720763514713.74/warc/CC-MAIN-20240715194155-20240715224155-00099.warc.gz
| 541,063,283 | 30,366 |
# Dipping refractor
Series Investigations in Geophysics Öz Yilmaz http://dx.doi.org/10.1190/1.9781560801580 ISBN 978-1-56080-094-1 SEG Online Store
Figure 3.4-11 (a) Geometry for refracted arrivals. Here, vw = weathering velocity, vb = bedrock velocity, zw = depth to the refractor equivalent to the base of the weathering layer, θc = critical angle, and xc = crossover distance. The direct wave arrival has a slope equal to 1/vw and the refracted wave arrival has a slope equal to 1/vb. (b) A shot record that exhibits the direct wave and the refracted wave depicted in (a). (c) Geometry for a dipping refractor with forward traveltime profile associated with the direct wave and refracted wave arrivals, and (d) with both forward and reverse traveltime profiles. See text for details.
When the refractor is dipping, it turns out that the inverse slope of the refracted arrival is no longer equal to the bedrock velocity (Figure 3.4-11c). An extra parameter — the dip of the refractor, needs to be estimated (Section C.6). This requires reverse profiling as illustrated in Figure 3.4-11d. We have the refracted arrival in the forward direction and the refracted arrival in the reverse direction obtained by interchanging the shots with receivers. The traveltimes for the refracted arrivals of the forward and reverse profiles are expressed as
${\displaystyle t^{-}=t_{i}^{-}+{\frac {x}{v_{b}^{-}}}}$ (43a)
and
${\displaystyle t^{+}=t_{i}^{+}+{\frac {x}{v_{b}^{+}}}.}$ (43b)
The inverse slopes are given by
${\displaystyle v_{b}^{-}={\frac {v_{w}}{\sin(\theta _{c}+\varphi )}}}$ (44a)
and
${\displaystyle v_{b}^{+}={\frac {v_{w}}{\sin(\theta _{c}-\varphi )}},}$ (44b)
where φ is the refractor dip and θc is the critical angle of refraction given by
${\displaystyle \sin \theta _{c}={\frac {v_{w}}{v_{b}}}.}$ (44c)
Finally, the intercept times are given by the following relations:
${\displaystyle t_{i}^{-}={\frac {2z_{wS}\cos \theta _{c}\cos \varphi }{v_{w}}}}$ (45a)
and
${\displaystyle t_{i}^{+}={\frac {2z_{wR}\cos \theta _{c}\cos \varphi }{v_{w}}}.}$ (45b)
Derivation of the relations (3-44a,b) and (3-45a,b) are left to Section C.6.
To estimate the thickness of the near-surface layer, first we compute the refractor dip φ from the slope measurements — ${\displaystyle {v_{w}},\ {v_{b}^{-}},\ {\text{and }}{v_{b}^{+}}.}$ These measurements are then inserted into the expression
${\displaystyle \varphi ={\frac {1}{2}}\left[\sin ^{-1}{\frac {v_{w}}{v_{b}^{-}}}-\sin ^{-1}{\frac {v_{w}}{v_{b}^{+}}}\right].}$ (46a)
Then, we compute the bedrock velocity vb using the expression
${\displaystyle v_{b}={\frac {2\cos \varphi }{\begin{pmatrix}{\frac {1}{v_{b}^{-}}}+{\frac {1}{v_{b}^{+}}}\end{pmatrix}}}.}$ (46b)
Finally, we compute the depth to the bedrock at shot/receiver stations
${\displaystyle z_{w}={\frac {v_{b}v_{w}t_{i}^{-}}{2\cos \varphi {\sqrt {v_{b}^{2}-v_{w}^{2}}}}}.}$ (46c)
Again, equations (46a,46c) reduces to equation (41a).
Keep in mind that, whether it is the flat refractor (equation 41a) or dipping refractor case (equation 46c), the depth to bedrock estimation at a shot-receiver station requires the knowledge of weathering velocity, bedrock velocity and intercept time. In the case of a flat refractor, these can be measured directly from shot profiles; whereas, in the case of a dipping refractor, they can be computed by way of equations (46a, 46b, 46c).
${\displaystyle z_{w}={\frac {v_{b}v_{w}t_{i}}{2{\sqrt {v_{b}^{2}-v_{w}^{2}}}}}.}$ (41a)
| 1,105 | 3,508 |
{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 12, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 3.5625 | 4 |
CC-MAIN-2024-30
|
latest
|
en
| 0.818371 |
http://nrich.maths.org/public/leg.php?code=-339&cl=1&cldcmpid=2783
| 1,484,629,524,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2017-04/segments/1484560279468.17/warc/CC-MAIN-20170116095119-00330-ip-10-171-10-70.ec2.internal.warc.gz
| 206,994,007 | 8,557 |
# Search by Topic
#### Resources tagged with Practical Activity similar to More Numbers in the Ring:
Filter by: Content type:
Stage:
Challenge level:
### There are 186 results
Broad Topics > Using, Applying and Reasoning about Mathematics > Practical Activity
### World of Tan 9 - Animals
##### Stage: 2 Challenge Level:
Can you fit the tangram pieces into the outline of this goat and giraffe?
### World of Tan 7 - Gat Marn
##### Stage: 2 Challenge Level:
Can you fit the tangram pieces into the outline of this plaque design?
### World of Tan 11 - the Past, Present and Future
##### Stage: 2 Challenge Level:
Can you fit the tangram pieces into the outline of the telescope and microscope?
### World of Tan 12 - All in a Fluff
##### Stage: 2 Challenge Level:
Can you fit the tangram pieces into the outline of these rabbits?
### World of Tan 15 - Millennia
##### Stage: 2 Challenge Level:
Can you fit the tangram pieces into the outlines of the workmen?
### World of Tan 14 - Celebrations
##### Stage: 2 Challenge Level:
Can you fit the tangram pieces into the outline of Little Ming and Little Fung dancing?
### World of Tan 3 - Mai Ling
##### Stage: 2 Challenge Level:
Can you fit the tangram pieces into the outline of Mai Ling?
### Making Maths: Happy Families
##### Stage: 1 and 2 Challenge Level:
Here is a version of the game 'Happy Families' for you to make and play.
### Square Tangram
##### Stage: 2 Challenge Level:
This was a problem for our birthday website. Can you use four of these pieces to form a square? How about making a square with all five pieces?
### World of Tan 8 - Sports Car
##### Stage: 2 Challenge Level:
Can you fit the tangram pieces into the outline of this sports car?
### Three Squares
##### Stage: 1 and 2 Challenge Level:
What is the greatest number of squares you can make by overlapping three squares?
### World of Tan 2 - Little Ming
##### Stage: 2 Challenge Level:
Can you fit the tangram pieces into the outline of Little Ming?
### World of Tan 5 - Rocket
##### Stage: 2 Challenge Level:
Can you fit the tangram pieces into the outline of the rocket?
### World of Tan 16 - Time Flies
##### Stage: 2 Challenge Level:
Can you fit the tangram pieces into the outlines of the candle and sundial?
### World of Tan 18 - Soup
##### Stage: 2 Challenge Level:
Can you fit the tangram pieces into the outlines of Mai Ling and Chi Wing?
### World of Tan 29 - the Telephone
##### Stage: 2 Challenge Level:
Can you fit the tangram pieces into the outline of this telephone?
### World of Tan 28 - Concentrating on Coordinates
##### Stage: 2 Challenge Level:
Can you fit the tangram pieces into the outline of Little Ming playing the board game?
### World of Tan 6 - Junk
##### Stage: 2 Challenge Level:
Can you fit the tangram pieces into the outline of this junk?
##### Stage: 2 Challenge Level:
NRICH December 2006 advent calendar - a new tangram for each day in the run-up to Christmas.
##### Stage: 1 and 2 Challenge Level:
Our 2008 Advent Calendar has a 'Making Maths' activity for every day in the run-up to Christmas.
### Making Maths: Birds from an Egg
##### Stage: 2 Challenge Level:
Can you make the birds from the egg tangram?
### World of Tan 27 - Sharing
##### Stage: 2 Challenge Level:
Can you fit the tangram pieces into the outline of Little Fung at the table?
### World of Tan 26 - Old Chestnut
##### Stage: 2 Challenge Level:
Can you fit the tangram pieces into the outline of this brazier for roasting chestnuts?
### World of Tan 20 - Fractions
##### Stage: 2 Challenge Level:
Can you fit the tangram pieces into the outlines of the chairs?
### World of Tan 19 - Working Men
##### Stage: 2 Challenge Level:
Can you fit the tangram pieces into the outline of this shape. How would you describe it?
### World of Tan 21 - Almost There Now
##### Stage: 2 Challenge Level:
Can you fit the tangram pieces into the outlines of the lobster, yacht and cyclist?
### World of Tan 22 - an Appealing Stroll
##### Stage: 2 Challenge Level:
Can you fit the tangram pieces into the outline of the child walking home from school?
### World of Tan 25 - Pentominoes
##### Stage: 2 Challenge Level:
Can you fit the tangram pieces into the outlines of these people?
### World of Tan 24 - Clocks
##### Stage: 2 Challenge Level:
Can you fit the tangram pieces into the outlines of these clocks?
### World of Tan 13 - A Storm in a Tea Cup
##### Stage: 2 Challenge Level:
Can you fit the tangram pieces into the outline of these convex shapes?
### World of Tan 4 - Monday Morning
##### Stage: 2 Challenge Level:
Can you fit the tangram pieces into the outline of Wai Ping, Wah Ming and Chi Wing?
### Counting Counters
##### Stage: 2 Challenge Level:
Take a counter and surround it by a ring of other counters that MUST touch two others. How many are needed?
### Cuisenaire Rods
##### Stage: 2 Challenge Level:
These squares have been made from Cuisenaire rods. Can you describe the pattern? What would the next square look like?
### World of Tan 17 - Weather
##### Stage: 2 Challenge Level:
Can you fit the tangram pieces into the outlines of the watering can and man in a boat?
### World of Tan 1 - Granma T
##### Stage: 2 Challenge Level:
Can you fit the tangram pieces into the outline of Granma T?
### Construct-o-straws
##### Stage: 2 Challenge Level:
Make a cube out of straws and have a go at this practical challenge.
### Fractional Triangles
##### Stage: 2 Challenge Level:
Use the lines on this figure to show how the square can be divided into 2 halves, 3 thirds, 6 sixths and 9 ninths.
### Tricky Triangles
##### Stage: 1 Challenge Level:
Use the three triangles to fill these outline shapes. Perhaps you can create some of your own shapes for a friend to fill?
### Rearrange the Square
##### Stage: 1 Challenge Level:
We can cut a small triangle off the corner of a square and then fit the two pieces together. Can you work out how these shapes are made from the two pieces?
### Triangular Faces
##### Stage: 2 Challenge Level:
This problem invites you to build 3D shapes using two different triangles. Can you make the shapes from the pictures?
##### Stage: 2 Challenge Level:
This practical problem challenges you to make quadrilaterals with a loop of string. You'll need some friends to help!
### Folding, Cutting and Punching
##### Stage: 2 Challenge Level:
Exploring and predicting folding, cutting and punching holes and making spirals.
### Let Us Reflect
##### Stage: 2 Challenge Level:
Where can you put the mirror across the square so that you can still "see" the whole square? How many different positions are possible?
### Seven Flipped
##### Stage: 2 Challenge Level:
Investigate the smallest number of moves it takes to turn these mats upside-down if you can only turn exactly three at a time.
### Seven Sticks
##### Stage: 1 Challenge Level:
Explore the triangles that can be made with seven sticks of the same length.
### Two Squared
##### Stage: 2 Challenge Level:
What happens to the area of a square if you double the length of the sides? Try the same thing with rectangles, diamonds and other shapes. How do the four smaller ones fit into the larger one?
### Jomista Mat
##### Stage: 2 Challenge Level:
Looking at the picture of this Jomista Mat, can you decribe what you see? Why not try and make one yourself?
### Cutting Corners
##### Stage: 2 Challenge Level:
Can you make the most extraordinary, the most amazing, the most unusual patterns/designs from these triangles which are made in a special way?
### Making Tangrams
##### Stage: 2 Challenge Level:
Here's a simple way to make a Tangram without any measuring or ruling lines.
| 1,815 | 7,739 |
{"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-2017-04
|
longest
|
en
| 0.822018 |
https://socratic.org/questions/how-do-you-simplify-13-20b-8-15b
| 1,576,101,110,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2019-51/segments/1575540533401.22/warc/CC-MAIN-20191211212657-20191212000657-00185.warc.gz
| 553,195,661 | 5,835 |
# How do you simplify 13+20B+8+15B?
Feb 1, 2017
See the entire simplification process below:
#### Explanation:
First, group like terms together in the expression:
$20 B + 15 B + 13 + 8$
Now, combine like terms:
$\left(20 + 15\right) B + 21$
$35 B + 21$
| 88 | 260 |
{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 3, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 4.15625 | 4 |
CC-MAIN-2019-51
|
latest
|
en
| 0.602279 |
https://nrich.maths.org/public/leg.php?code=-99&cl=2&cldcmpid=1013
| 1,527,073,473,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2018-22/segments/1526794865595.47/warc/CC-MAIN-20180523102355-20180523122355-00334.warc.gz
| 618,394,906 | 9,932 |
# Search by Topic
#### Resources tagged with Working systematically similar to Make 100:
Filter by: Content type:
Stage:
Challenge level:
### There are 337 results
Broad Topics > Using, Applying and Reasoning about Mathematics > Working systematically
### Journeys in Numberland
##### Stage: 2 Challenge Level:
Tom and Ben visited Numberland. Use the maps to work out the number of points each of their routes scores.
### Number Daisy
##### Stage: 3 Challenge Level:
Can you find six numbers to go in the Daisy from which you can make all the numbers from 1 to a number bigger than 25?
### Sums and Differences 1
##### Stage: 2 Challenge Level:
This challenge focuses on finding the sum and difference of pairs of two-digit numbers.
### Magic Potting Sheds
##### Stage: 3 Challenge Level:
Mr McGregor has a magic potting shed. Overnight, the number of plants in it doubles. He'd like to put the same number of plants in each of three gardens, planting one garden each day. Can he do it?
### Build it up More
##### Stage: 2 Challenge Level:
This task follows on from Build it Up and takes the ideas into three dimensions!
### Sums and Differences 2
##### Stage: 2 Challenge Level:
Find the sum and difference between a pair of two-digit numbers. Now find the sum and difference between the sum and difference! What happens?
### A Shapely Network
##### Stage: 2 Challenge Level:
Your challenge is to find the longest way through the network following this rule. You can start and finish anywhere, and with any shape, as long as you follow the correct order.
### A First Product Sudoku
##### Stage: 3 Challenge Level:
Given the products of adjacent cells, can you complete this Sudoku?
### Page Numbers
##### Stage: 2 Short Challenge Level:
Exactly 195 digits have been used to number the pages in a book. How many pages does the book have?
### Broken Toaster
##### Stage: 2 Short Challenge Level:
Only one side of a two-slice toaster is working. What is the quickest way to toast both sides of three slices of bread?
### Crack the Code
##### Stage: 2 Challenge Level:
The Zargoes use almost the same alphabet as English. What does this birthday message say?
### Seating Arrangements
##### Stage: 2 Challenge Level:
Sitting around a table are three girls and three boys. Use the clues to work out were each person is sitting.
### Team Scream
##### Stage: 2 Challenge Level:
Seven friends went to a fun fair with lots of scary rides. They decided to pair up for rides until each friend had ridden once with each of the others. What was the total number rides?
### Today's Date - 01/06/2009
##### Stage: 1 and 2 Challenge Level:
What do you notice about the date 03.06.09? Or 08.01.09? This challenge invites you to investigate some interesting dates yourself.
##### Stage: 3 Challenge Level:
If you take a three by three square on a 1-10 addition square and multiply the diagonally opposite numbers together, what is the difference between these products. Why?
### Bunny Hop
##### Stage: 2 Challenge Level:
What is the smallest number of jumps needed before the white rabbits and the grey rabbits can continue along their path?
### Newspapers
##### Stage: 2 Challenge Level:
When newspaper pages get separated at home we have to try to sort them out and get things in the correct order. How many ways can we arrange these pages so that the numbering may be different?
### How Old Are the Children?
##### Stage: 3 Challenge Level:
A student in a maths class was trying to get some information from her teacher. She was given some clues and then the teacher ended by saying, "Well, how old are they?"
### Ones Only
##### Stage: 3 Challenge Level:
Find the smallest whole number which, when mutiplied by 7, gives a product consisting entirely of ones.
### Button-up Some More
##### Stage: 2 Challenge Level:
How many ways can you find to do up all four buttons on my coat? How about if I had five buttons? Six ...?
### Ordered Ways of Working Upper Primary
##### Stage: 2 Challenge Level:
These activities lend themselves to systematic working in the sense that it helps if you have an ordered approach.
### Polo Square
##### Stage: 2 Challenge Level:
Arrange eight of the numbers between 1 and 9 in the Polo Square below so that each side adds to the same total.
### Professional Circles
##### Stage: 2 Challenge Level:
Six friends sat around a circular table. Can you work out from the information who sat where and what their profession were?
### Area and Perimeter
##### Stage: 2 Challenge Level:
What can you say about these shapes? This problem challenges you to create shapes with different areas and perimeters.
### Multiples Sudoku
##### Stage: 3 Challenge Level:
Each clue in this Sudoku is the product of the two numbers in adjacent cells.
### Shapes in a Grid
##### Stage: 2 Challenge Level:
Can you find which shapes you need to put into the grid to make the totals at the end of each row and the bottom of each column?
##### Stage: 2 Challenge Level:
Can you put plus signs in so this is true? 1 2 3 4 5 6 7 8 9 = 99 How many ways can you do it?
### Two and Two
##### Stage: 3 Challenge Level:
How many solutions can you find to this sum? Each of the different letters stands for a different number.
##### Stage: 3 Challenge Level:
How many different symmetrical shapes can you make by shading triangles or squares?
### How Much Did it Cost?
##### Stage: 2 Challenge Level:
Use your logical-thinking skills to deduce how much Dan's crisps and ice-cream cost altogether.
### Dart Target
##### Stage: 2 Challenge Level:
This task, written for the National Young Mathematicians' Award 2016, invites you to explore the different combinations of scores that you might get on these dart boards.
### Open Squares
##### Stage: 2 Challenge Level:
This task, written for the National Young Mathematicians' Award 2016, focuses on 'open squares'. What would the next five open squares look like?
### Consecutive Negative Numbers
##### Stage: 3 Challenge Level:
Do you notice anything about the solutions when you add and/or subtract consecutive negative numbers?
### Being Resourceful - Primary Number
##### Stage: 1 and 2 Challenge Level:
Number problems at primary level that require careful consideration.
### Dice Stairs
##### Stage: 2 Challenge Level:
Can you make dice stairs using the rules stated? How do you know you have all the possible stairs?
### Five Coins
##### Stage: 2 Challenge Level:
Ben has five coins in his pocket. How much money might he have?
### Fence It
##### Stage: 3 Challenge Level:
If you have only 40 metres of fencing available, what is the maximum area of land you can fence off?
### Ben's Game
##### Stage: 3 Challenge Level:
Ben passed a third of his counters to Jack, Jack passed a quarter of his counters to Emma and Emma passed a fifth of her counters to Ben. After this they all had the same number of counters.
### Zargon Glasses
##### Stage: 2 Challenge Level:
Zumf makes spectacles for the residents of the planet Zargon, who have either 3 eyes or 4 eyes. How many lenses will Zumf need to make all the different orders for 9 families?
### Cinema Problem
##### Stage: 3 Challenge Level:
A cinema has 100 seats. Show how it is possible to sell exactly 100 tickets and take exactly £100 if the prices are £10 for adults, 50p for pensioners and 10p for children.
### Seven Square Numbers
##### Stage: 2 Challenge Level:
Add the sum of the squares of four numbers between 10 and 20 to the sum of the squares of three numbers less than 6 to make the square of another, larger, number.
### Centred Squares
##### Stage: 2 Challenge Level:
This challenge, written for the Young Mathematicians' Award, invites you to explore 'centred squares'.
### All Seated
##### Stage: 2 Challenge Level:
Look carefully at the numbers. What do you notice? Can you make another square using the numbers 1 to 16, that displays the same properties?
### Twenty Divided Into Six
##### Stage: 2 Challenge Level:
Katie had a pack of 20 cards numbered from 1 to 20. She arranged the cards into 6 unequal piles where each pile added to the same total. What was the total and how could this be done?
### More Magic Potting Sheds
##### Stage: 3 Challenge Level:
The number of plants in Mr McGregor's magic potting shed increases overnight. He'd like to put the same number of plants in each of his gardens, planting one garden each day. How can he do it?
### Difference
##### Stage: 2 Challenge Level:
Place the numbers 1 to 10 in the circles so that each number is the difference between the two numbers just below it.
### Ancient Runes
##### Stage: 2 Challenge Level:
The Vikings communicated in writing by making simple scratches on wood or stones called runes. Can you work out how their code works using the table of the alphabet?
### Prison Cells
##### Stage: 2 Challenge Level:
There are 78 prisoners in a square cell block of twelve cells. The clever prison warder arranged them so there were 25 along each wall of the prison block. How did he do it?
### Consecutive Numbers
##### Stage: 2 and 3 Challenge Level:
An investigation involving adding and subtracting sets of consecutive numbers. Lots to find out, lots to explore.
### Two Egg Timers
##### Stage: 2 Challenge Level:
You have two egg timers. One takes 4 minutes exactly to empty and the other takes 7 minutes. What times in whole minutes can you measure and how?
| 2,132 | 9,485 |
{"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-2018-22
|
latest
|
en
| 0.886207 |
http://mathhelpforum.com/advanced-math-topics/208099-proof-transcendence-constant-e-print.html
| 1,521,332,891,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2018-13/segments/1521257645405.20/warc/CC-MAIN-20180317233618-20180318013618-00311.warc.gz
| 187,971,840 | 4,484 |
# Proof of Transcendence of constant e
• Nov 21st 2012, 07:17 AM
czar01
Proof of Transcendence of constant e
Hello all,
I'm a first year university student who is tasked with giving a presentation on the transcendence of e. I have a very basic grasp of the proof that we have been given to understand, however any explanations of ANY parts of the proof is very much appreciated as the math is way above my head (not quite sure if the proof itself is sound). Below is the pasted LATEX text of our proof as is right now. Please also list the topics that, in your opinion, must be covered in the 20-minute mathematical presentation of the topic.
To begin, we must assume that $e$ is algebraic. We suppose that $e$ is a root of the polynomial $g(x)=b_0+b_1x^1+...+b_rx^r$ where $b_i$ are in $\mathbb{Z}$ and $b_r$ doesn't not equal 0. $$I_p= \displaystyle\sum_{k=1}^{r} b_k \int^k_0 e^{k-x} f(x)dx$$
We will show that for $p$ large enough, $I_p$ is an integer, does not equal 0, but is arbitrarily close to 0.
Assume that $e$ is algebraic. Suppose that it is a root of the polynomial $g(x)=b_0+b_1x^1+...+b_rx^r$ where $b_i$ are in $\mathbb{Z}$ and $b_r$ doesn't not equal 0.
$$I_p= \displaystyle\sum_{k=1}^{r} b_k \int^k_0 e^{k-x} f(x)dx$$
We will show that for $p$ large enough, $I_p$ is an integer, does not equal 0, but is arbitrarily close to 0.
$$\int^t_0 e^{-x} f(x)dx = \int^t_0 (-e^{-x})' f(x)dx$$
Using integration by parts, we obtain the following expression.
$$-e^{-x} f(x) A - \int^t_0 (-e^{-x}) f'(x)dx$$
Plug in t,
$$-e^{-t}f(t) + f(0) + \int^t_0(-e^{-x}f'(x)dx$$
$$-e^{-t} \sum^n_{k=0} f^k (t) + \sum^n_k=0 f^k (0)$$
We will assume that $e$ is algebraic. The equation for the coefficients of the polynomial $e$ is $c_0 + c_1e+..+c_me^m=0$
$$f_p(x) = \frac{x^{p-1}(x-1)^p(x-2)^p...(x-k)^p}{(p-1)!}$$
The first p derivatives of $k$ $(x-1)^p g(x)$
$$f_p'(x) = p(x-1)^{p-1} g(x) + (x-1)^p g'(x)$$
Either will =0, or will cancel with (p-1)! and give integer.
$$I_p= \sum^m_k=0 c_k e^k \int^k_0 e^{-x} f_p(x) dx$$
$$= \sum^m_{k=0} c_k e^k \left( -e^{-k} \sum^n_{i=0} f^i (k) + \sum^n_{i=0} f^i (0)\right)$$
$$-\sum^m_{k=0} c_k (e^k)(e^{-k}) \sum^n_{i=0} f_p^i (k) + \left(\sum^m_{k=0} c_ke^k\right) \left(\sum^n_{i=0} f^i (0)\right)$$
$$= -\sum^m_{k=0} \sum^n_{i=0} c_k f_p^i (k)$$
Now, we will prove that this number does not equal 0 by showing that it is non-divisble by a prime number. Zero is of course divisible by any number, therefore a contradiction proves this point.
$$-\sum^m_{k=0} \sum^n_{i=p} c_k f_p^i (k) - \sum^n_{i=p-1} c_0 f^i (0)$$
Where $-\sum^m_{k=0} \sum^n_{i=p} c_k f_p^i (k)$ is divisble by p. If this part is differentiated $p$ times, everything can give you a nonzero that has a factor of p.
$$\sum^n_{i=p-1} c_0 ^{f(i)} (0) = c_0 f^{p-1} (0) + \sum^n_{i=p} c_0 f^{(i)} (0)$$
To get the nonzero, we will target $x^{p-1}$ times. Extra factor remains of p when you apply one derivative. $\sum^n_{i=p} c_0 f^i (0)$ is divisble by p.
So far, we have shown that $I_p \neq 0$, and that $I_p$ is and integer.
We will now justify that as p tends to infinity, $I_p$ tends to 0.
$$If\ you\ have\ a\ function \Rightarrow \left | g(x) \right |\leq M \ on \ \left [ a,b \right ],$$
$$then \left | \int_{b}^{a} g(x)dx \right |\leq M(b-a)$$
From here, we will estimate the integral of $\left | \int_{b}^{a} e^{-x}f_p(x)dx \right |$ where $0\leq a\leq m$
If $0\leq a\leq m$, then the difference $\left | x-j \right |$ is less than $m$.
$$\left | e^{-x} f_p(x) \right |\leq \left | f_p(x) \right | = \left | \frac{x^{p-1}(x-1)^p...(x-k)^p}{(p-1)!} \right |$$
$$\left | \frac{x^{p-1}(x-1)^p...(x-k)^p}{(p-1)!} \right |\leq \left | \frac{m^{p-1}m^p...m^p}{(p-1)!} \right |$$
$$=\frac{m^{(m+1)p+1}}{(p-1)!} < \frac{m^{(m+1)p}}{(p-1)!} = \frac{m^{(m+1)^p}}{(p-1)!}\rightarrow \bf0$$
Therefore, we can see that as $p$ tends to infinity, $I_p$ tends to 0. This limit tends to 0 because it is simply the limit $\lim_{n \to \infty } \frac{c^n}{n!} = 0$ in a more complicated form. This can be proved as follows:
Let $k$ be any natual number greater than $\ 2 \left | c \right |$. If $k>2 \left | c \right |$, then $\frac{ \left | c \right | } {k} < \frac{1} {2}$ and morever $\frac{\left | c \right |}{k}<\frac{1}{2}$ for $n\geq k$.
$$\frac{c^n}{n!} = \frac{c^k}{k!} \cdot \frac{c}{k} \cdot \frac{c}{k+1} \cdot ... \cdot \frac{c}{n} < \frac{c^k}{k!}\cdot \frac{1}{2} \cdot \frac{1}{2} \cdot...\cdot \frac{1}{2}$$
where $\frac{1}{2} \cdot \frac{1}{2} \cdot...\cdot \frac{1}{2} = \bf{n-k}$. Therefore the above equation equals:
$$\frac{c^k}{k!} \cdot \frac{1}{2^{n-k}} = 2^k\left ( \frac{c^k}{k!} \right )\frac{1}{2^{n}}$$
in which $\frac{1}{2^{n}}$ tends to 0. Therefore:
$$\lim_{n \to \infty } \frac{c^n}{n!} = 0$$
Any help is much appreciated, thanks in advance!
• Nov 21st 2012, 07:43 AM
Plato
Re: Proof of Transcendence of constant e
Quote:
Originally Posted by czar01
To begin, we must assume that $e$ is algebraic. We suppose that $e$ is a root of the polynomial $g(x)=b_0+b_1x^1+...+b_rx^r$ where $b_i$ are in $\mathbb{Z}$ and $b_r$ doesn't not equal 0. $$I_p= \displaystyle\sum_{k=1}^{r} b_k \int^k_0 e^{k-x} f(x)dx$$
We will show that for $p$ large enough, $I_p$ is an integer, does not equal 0, but is arbitrarily close to 0.
$$\frac{c^n}{n!} = \frac{c^k}{k!} \cdot \frac{c}{k} \cdot \frac{c}{k+1} \cdot ... \cdot \frac{c}{n} < \frac{c^k}{k!}\cdot \frac{1}{2} \cdot \frac{1}{2} \cdot...\cdot \frac{1}{2}$$
$$\frac{c^n}{n!} = \frac{c^k}{k!} \cdot \frac{c}{k} \cdot \frac{c}{k+1} \cdot ... \cdot \frac{c}{n} < \frac{c^k}{k!}\cdot \frac{1}{2} \cdot \frac{1}{2} \cdot...\cdot \frac{1}{2}$$ gives
$\displaystyle \frac{c^n}{n!} = \frac{c^k}{k!} \cdot \frac{c}{k} \cdot \frac{c}{k+1} \cdot ... \cdot \frac{c}{n} < \frac{c^k}{k!}\cdot \frac{1}{2} \cdot \frac{1}{2} \cdot...\cdot \frac{1}{2}$
Using the advanced menu toolbar the $\displaystyle \boxed{\Sigma}$ gives the wrap. You can simply edit your post. Get rid of the $signs hightlight the code and click the$\displaystyle \boxed{\Sigma}\$ .
| 2,468 | 6,010 |
{"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": 2, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 3.9375 | 4 |
CC-MAIN-2018-13
|
latest
|
en
| 0.877156 |
https://www.math.ubc.ca/~israel/challenge/challenge9.html
| 1,540,201,808,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2018-43/segments/1539583515029.82/warc/CC-MAIN-20181022092330-20181022113830-00428.warc.gz
| 964,176,323 | 4,881 |
Problem 9:
The integral depends on the parameter . What is the value at which achieves its maximum?
Solution:
This is rather different from Gaston's solution. The numerical integration works in Maple 8, but not Maple 7.
We want to be able to evaluate
and so that we can use Newton's method to find the location of the maximum. Write so , , and .
We break the interval up into pieces.
On the interval we use a series in powers of . On we use the transformation , obtaining an integral on . On the interval we use Maple's own numerical integration, and on we expand the integrand in negative powers of y times sin(y).
> restart: ti:= time():
First for the interval : J1 is the integral, dJ1 its derivative and d2J1 its second derivative.
> f:= x^alpha*sin(alpha/(2-x)); S:= convert(series(sin(alpha/(2-x)),x,40),polynom): intp:= proc(t) local c,xp,n; if has(t,x) then xp,c := selectremove(has,t,x); if xp = x then n:= 1 else n:= op(2,xp) fi; else n:= 0; c:= t fi; collect(expand(c),[cos,sin])*3^(-n-alpha-1)/(n+alpha+1); end: J1:= map(intp,S):
> dJ1:= diff(J1,alpha): d2J1:= diff(J1,alpha\$2):
> length(J1),length(dJ1),length(d2J1);
How big is the last term in dJ1 for a typical alpha?
> evalf(eval(select(has,dJ1,3^(-40-alpha)),alpha=0.8));
Next the transformation.
> xy := expand(solve(alpha/(2-x)=y,x)); g:= subs(x=xy,f)*diff(xy,y); g1:= eval(g,y=3*alpha/5); ga:= normal(diff(g,alpha)); ga1:= normal(eval(ga,y=3*alpha/5)); gy:= normal(eval(diff(g,y),y=3*alpha/5)); gaa:= normal(diff(g,alpha\$2));
So we want . is the integral from to , and its derivatives. Maple sometimes seems to have trouble evaluating to high precision, but we don't need as much accuracy there, so the relative error tolerance can be increased in this case.
> dH:=diff(int(h(a,y),y=3*a/5 .. 2*Pi),a); d2H:=diff(int(h(a,y),y=3*a/5 .. 2*Pi),a\$2);
> J2:= a -> evalf(Int(subs(alpha=a,g),y=3*a/5 .. 2*Pi));
> dJ2:= a -> evalf(Int(subs(alpha=a,ga),y=3*a/5 .. 2*Pi) - 3/5*subs(alpha=a,g1)); d2J2:= a -> evalf(Int(subs(alpha=a,gaa),y=3*a/5 .. 2*Pi,epsilon=10^(-Digits/2)) - subs(alpha=a,6/5*ga1+9/25*gy));
> J2(0.8),dJ2(0.8),d2J2(0.8);
For the interval , we will use a series for in negative powers of , relying on Maple to integrate . We have where .
> J3:=add(2^(2+alpha-n)*(-alpha)^n*pochhammer(alpha-n+3,n-3)/(n-2)!*int(sin(y)/y^n,y=2*Pi..infinity),n = 2 .. 30):
> dJ3:= diff(J3,alpha): d2J3:= diff(J3,alpha\$2):
> evalf(eval([J3,dJ3,d2J3],alpha=0.8));
> t30:= add(2^(2+alpha-n)*(-alpha)^n*pochhammer(alpha-n+3,n-3)/(n-2)!*int(sin(y)/y^n,y=2*Pi..infinity),n = 30..30); evalf(eval([t30,diff(t30,alpha),diff(t30,alpha\$2)],alpha=0.8));
> Jp:= a -> evalf(eval(J1+J3,alpha=a))+J2(a): Ip:= a -> evalf(2+sin(10*a))*Jp(a): Jp(0.8);
Here it is plotted (with low precision), in red and in green. The maximum is near .
> plot([Jp,Ip],0..5);
Here is a Newton iteration procedure. It takes an initial value, prints the values of , and , and returns the new alpha for the next iteration: .
> Newt:= proc(a) local J,dJ,d2J,Ia,dI,d2I; J:= evalf(eval(J1+J3,alpha=a)+J2(a)); dJ:= evalf(eval(dJ1+dJ3,alpha=a)+dJ2(a)); d2J:= evalf(eval(d2J1+d2J3,alpha=a)+d2J2(a)); Ia:= (2+sin(10*a))*J; dI:=10*cos(10*a)*J+(2+sin(10*a))*dJ; d2I:= -100*sin(10*a)*J+20*cos(10*a)*dJ+(2+sin(10*a))*d2J; print([Ia,dI,d2I]); a-dI/d2I; end;
>
> a1:=Newt(0.8);
> Digits:= 36: for i from 2 to 5 do a||i:= Newt(a||(i-1)) od; (time()-ti)*seconds;
This has 33 correct digits.
>
| 1,328 | 3,510 |
{"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-2018-43
|
latest
|
en
| 0.599129 |
https://istopdeath.com/graph-xy22-1/
| 1,669,634,017,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2022-49/segments/1669446710503.24/warc/CC-MAIN-20221128102824-20221128132824-00056.warc.gz
| 373,373,634 | 18,521 |
# Graph x=(y+2)^2-1
Simplify .
Simplify each term.
Rewrite as .
Expand using the FOIL Method.
Apply the distributive property.
Apply the distributive property.
Apply the distributive property.
Simplify and combine like terms.
Simplify each term.
Multiply by .
Move to the left of .
Multiply by .
Subtract from .
Find the properties of the given parabola.
Rewrite the equation in vertex form.
Complete the square for .
Use the form , to find the values of , , and .
Consider the vertex form of a parabola.
Substitute the values of and into the formula .
Cancel the common factor of and .
Factor out of .
Cancel the common factors.
Factor out of .
Cancel the common factor.
Rewrite the expression.
Divide by .
Find the value of using the formula .
Simplify each term.
Cancel the common factor of and .
Factor out of .
Cancel the common factors.
Cancel the common factor.
Rewrite the expression.
Divide by .
Multiply by .
Subtract from .
Substitute the values of , , and into the vertex form .
Set equal to the new right side.
Use the vertex form, , to determine the values of , , and .
Since the value of is positive, the parabola opens right.
Opens Right
Find the vertex .
Find , the distance from the vertex to the focus.
Find the distance from the vertex to a focus of the parabola by using the following formula.
Substitute the value of into the formula.
Cancel the common factor of .
Cancel the common factor.
Rewrite the expression.
Find the focus.
The focus of a parabola can be found by adding to the x-coordinate if the parabola opens left or right.
Substitute the known values of , , and into the formula and simplify.
Find the axis of symmetry by finding the line that passes through the vertex and the focus.
Find the directrix.
The directrix of a parabola is the vertical line found by subtracting from the x-coordinate of the vertex if the parabola opens left or right.
Substitute the known values of and into the formula and simplify.
Use the properties of the parabola to analyze and graph the parabola.
Direction: Opens Right
Vertex:
Focus:
Axis of Symmetry:
Directrix:
Direction: Opens Right
Vertex:
Focus:
Axis of Symmetry:
Directrix:
Select a few values, and plug them into the equation to find the corresponding values. The values should be selected around the vertex.
Substitute the value into . In this case, the point is .
Replace the variable with in the expression.
Simplify the result.
Simplify each term.
Any root of is .
Subtract from .
Convert to decimal.
Substitute the value into . In this case, the point is .
Replace the variable with in the expression.
Simplify the result.
Simplify each term.
Any root of is .
Multiply by .
Subtract from .
Convert to decimal.
Substitute the value into . In this case, the point is .
Replace the variable with in the expression.
Simplify the result.
Convert to decimal.
Substitute the value into . In this case, the point is .
Replace the variable with in the expression.
Simplify the result.
Convert to decimal.
Graph the parabola using its properties and the selected points.
Graph the parabola using its properties and the selected points.
Direction: Opens Right
Vertex:
Focus:
Axis of Symmetry:
Directrix:
Graph x=(y+2)^2-1
| 730 | 3,195 |
{"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-2022-49
|
latest
|
en
| 0.875448 |
https://interesting-information.com/how-do-you-rotate-counterclockwise/
| 1,660,330,027,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2022-33/segments/1659882571745.28/warc/CC-MAIN-20220812170436-20220812200436-00005.warc.gz
| 302,006,777 | 21,229 |
Terms in this set (9)
1. (-y, x) 90 degree rotation counterclockwise around the origin.
2. (y, -x) 90 degree rotation clockwise about the origin.
3. (-x, -y) 180 degree rotation clockwise and counterclockwise about the origin.
4. (-y, x) 270 degree rotation clockwise about the origin.
5. (y, -x)
6. (x, -y)
7. (-x, y)
8. (y, x)
In this way, what is the formula for rotating 90 degrees counterclockwise?
90 degrees, the rule is (x, y) ——–> (y, -x) -180 degrees, the rule is (x, y) ——–> (-x, -y) -270 degrees, the rule is (x, y) ——–> (-y, x)
Beside above, what is the formula for rotating 180 degrees counterclockwise? 180 degrees is (-a, -b) and 360 is (a, b). 360 degrees doesn't change since it is a full rotation or a full circle. Also this is for a counterclockwise rotation. If you want to do a clockwise rotation follow these formulas: 90 = (b, -a); 180 = (-a, -b); 270 = (-b, a); 360 = (a, b).
Thereof, which way is counterclockwise to the left or right?
Answer: counter clockwise the is rotation or movement of an object which is in the opposite direction of any clock. When we see from the top, the circular rotation moves to the left, and from the bottom rotation moves to the right.
What are the rules for counterclockwise rotations?
Terms in this set (9)
• (-y, x) 90 degree rotation counterclockwise around the origin.
• (y, -x) 90 degree rotation clockwise about the origin.
• (-x, -y) 180 degree rotation clockwise and counterclockwise about the origin.
• (-y, x) 270 degree rotation clockwise about the origin.
• (y, -x)
• (x, -y)
• (-x, y)
• (y, x)
## Which way is clockwise?
A clockwise (typically abbreviated as CW) motion is one that proceeds in the same direction as a clock's hands: from the top to the right, then down and then to the left, and back up to the top.
## How do you describe rotation?
A rotation is a turn of a shape. A rotation is described by the centre of rotation, the angle of rotation, and the direction of the turn. The centre of rotation is the point that a shape rotates around. Each point in the shape must stay an equal distance from the centre of rotation.
## What is a 90 degree turn?
A 90degree turn is one-quarter of turn regardless of direction. If a person imagines himself standing looking straight ahead and then turning to face the right side or the left side, he has made a 90degree turn. A circle contains 360 degrees.
## How do you write a reflection Rule?
To write a rule for this reflection you would write: rx−axis(x,y) → (x,−y). Notation Rule A notation rule has the following form ry−axisA → B = ry−axis(x,y) → (−x,y) and tells you that the image A has been reflected across the y-axis and the x-coordinates have been multiplied by -1.
## What is a 270 rotation?
Rotating a Triangle 270 Degrees Counterclockwise:
One such rotation is to rotate a triangle 270° counterclockwise, and we have a special rule that we can use to do this that is based on the fact that a 270° counterclockwise rotation is the same thing as a 90° clockwise rotation.
## What is the rule for translation?
In a translation, every point of the object must be moved in the same direction and for the same distance. When you are performing a translation, the initial object is called the pre-image, and the object after the translation is called the image.
## How do you graph rotation?
Graph A(5, 2), then graph B, the image of A under a 90° counterclockwise rotation about the origin. Rule for 90° counterclockwise rotation: A (5, 2) B (- 2, 5) Now graph C, the image of A under a 180° counterclockwise rotation about the origin.
## How do you rotate around a point?
A point (a, b) rotated around a point (x, y) 90 degrees will transform to point (-(b-y) + x, (a-x) + y). A point (a, b) rotated around a point (x, y) 180 degrees will transform to point (-(a – x) + x, -(b – y) + y). A point (a, b) rotated around the origin 270 degrees will transform to point (b – y + x, -(a – x) + y).
## What is a rotation in math?
A rotation is a transformation that turns a figure about a fixed point called the center of rotation. • An object and its rotation are the same shape and size, but the figures may be turned in different directions. • Rotations may be clockwise or counterclockwise.
## Which way is 90 degrees clockwise?
Rotation of point through 90° about the origin in clockwise direction when point M (h, k) is rotated about the origin O through 90° in clockwise direction. The new position of point M (h, k) will become M' (k, -h). Worked-out examples on 90 degree clockwise rotation about the origin: 1.
## Which way does the fan switch go in the winter?
Clockwise Fan Direction for Warm Winter Comfort
In the winter, ceiling fans should rotate clockwise at a low speed to pull cool air up. The gentle updraft pushes warm air, which naturally rises to the ceiling, down along the walls and back the floor.
## Why is clockwise to the right?
Ultimately, this type of clock became ubiquitous, and gave rise to the term “clockwise“, meaning “following the motion of the hands of the clock”. Since the hands of the clock appear to move to the right, “clockwise” and “right” became associated, and are treated as interchangeable when referring to circular motion.
| 1,371 | 5,233 |
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 4.34375 | 4 |
CC-MAIN-2022-33
|
latest
|
en
| 0.863999 |
https://www.gradesaver.com/textbooks/math/other-math/basic-college-mathematics-9th-edition/chapter-6-percent-6-2-percents-and-fractions-6-2-exercises-page-399/47
| 1,529,454,808,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2018-26/segments/1529267863259.12/warc/CC-MAIN-20180619232009-20180620012009-00226.warc.gz
| 839,670,600 | 13,107 |
Chapter 6 - Percent - 6.2 Percents and Fractions - 6.2 Exercises: 47
$\frac{1}{2}$, 50%
Work Step by Step
We are given the decimal .5. To write .5 as a percentage, we must multiply by 100 and attach a percent sign to the end. $.5=(.5\times100)$%=50% 50%=$50\div100=\frac{50}{100}=\frac{50\div50}{100\div50}=\frac{1}{2}$ Therefore, $\frac{1}{2}=.5$=50%
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.
| 168 | 515 |
{"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}
| 4.28125 | 4 |
CC-MAIN-2018-26
|
longest
|
en
| 0.789165 |
http://spotidoc.com/doc/157903/1.--find-the-orbital-period--period-of-revolution--of...-..
| 1,534,573,535,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2018-34/segments/1534221213405.46/warc/CC-MAIN-20180818060150-20180818080150-00480.warc.gz
| 370,901,215 | 8,058 |
# 1. Find the orbital period (period of revolution) of... Sun is about 7.8×10
```Examples: Kepler's third law
PHYS 215 (Introduction to Astronomy)
Physics Department – KFUPM
1. Find the orbital period (period of revolution) of Jupiter if its distance from the
Sun is about 7.8×108 km. (ae = 1.5×108 km).
a = 7.8×108 km = 7.8×108 / 1.5×108 = 5.2 a.u.
T2 = a3 Î T = a √ a = 5.2 × √5.2 = 11.86 = 12 yrs.
2. Find the orbital period of Jupiter if its distance from the Earth is about 6.3×108
km when it is in opposition with the Sun. (ae = 1.5×108 km).
a = 6.3×108 km + 1.5×108 = 7.8×108 km = 7.8×108 / 1.5×108 = 5.2 a.u.
T2 = a3 Î T = a √ a = 5.2 × √5.2 = 11.86 = 12 yrs.
3. Find the orbital period of Jupiter if its angular diameter is about 47" as seen
from the Earth when it is in opposition with the Sun. (ae = 1.5×108 km, RJ =
71500 km)
θ = 47" = (47/3600) ° = 0.013 °
θ
s
= 0.013*π/180 rad
r
= 0.00023 rad.
s = linear diameter , θ = angular diameter
r = distance
, s = rθ
s = diameter = RJ × 2 = 71500 × 2
= 143000 km
s = rθ Î r = s/θ = 143000 / 0.00023
= 6.3×108 km
a = 6.3×108 km + 1.5×108 = 7.8×108 km = 7.8×108 / 1.5×108 = 5.2 a.u.
T2 = a3 Î T = a √ a = 5.2 × √5.2 = 11.86 = 12 yrs.
4. It is found that the angular diameter of Jupiter at opposition is θopp = 47" and it
is θcon = 32" at conjunction with the Sun as seen from the Earth. Find the
distance of Jupiter from the Sun. (RJ = 71500 km)
S = 2 × RJ = 2 × 71500 = 143000 km
aJ
θopp = 47" = 47 / 3600 ° = 0.013 °
= 0.013 × π / 180 rad.
Jupiter in conjunction
= 0.00023 rad.
θcon = 47" = 32 / 3600 ° = 0.009 °
= 0.009 × π / 180 rad.
= 0.00016 rad.
aJ = (ropp + rcon)/2 = (s / θopp + s / θcon)/2
= 143000(1/0.00023 + 1/0.00016)/2
= 143000(10598)/2 = 7.6×108 km
rcon
The sun
Jupiter in opposition
☼ The Earth
ropp
Examples: Kepler's third law
PHYS 215 (Introduction to Astronomy)
Physics Department – KFUPM
5. In order to find the mass of the Moon a space craft is made to orbit around it.
Find the mass of the Moon if the space craft is orbiting at a height of 127 km
above its surface with a period of 2 hrs. Also find the orbital speed of the space
craft.(Rm = 1738 km, G = 6.67×10-11 m3/kg.s2)
T2 = (4π2/GMm) a3 Î Mm = 4π2×a3/G×T2
= 4π2×(127000 + 1738000)3/G×(2×3600)2 = 7.4×1022 kg
v = 2π×a/T = 2π×(127 + 1738)/2 = 5859 km/hr = 1.63 km/s
v = √G×M/a = √6.67×10-11 × 7.4×1022 / (127000 + 1738000) = 1627 m/s
= 1.63 km/s
6.
How long Hubble Space Telescope (HST) takes (in minutes) to circle
once around the Earth if it is at a height of about 600 km above the Earth
surface? (R⊕ = 6400 km, G = 6.67×10-11 m3/(kg.s2), M⊕ = 6×1024 kg)
P 2 = (4π 2 / G M) a 3
P = √ 4π 2 (600000+6400000)3 / 6.67×10-11 × 6×1024
= 5817 seconds = 97 minutes
7.
How fast Hubble Space Telescope (HST) is moving (in km / s) as it
circles the Earth at a height of about 600 km above its surface? (R⊕ =
6400 km, G = 6.67×10-11 m3/(kg.s2), M⊕ = 6×1024 kg)
P 2 = (4π 2 / G M) a 3 Î P 2 / (4π 2 a 2) = (1 / G M) a = 1 / v2
v =√GM/a
= √ 6.67×10-11 × 6×1024 / (600000+6400000)
= 7561 m / s = 7.6 km / s
or
8.
v = 2π a / P = 2π(600000+6400000) / 5817 = 7561 m/s = 7.6 km / s
How fast Hubble Space Telescope (HST) is moving (in degrees per
minute) as it circles the Earth at a height of about 600 km above its
surface? (P = 97 min.)
360 º in P sec. Î ө º in 1 sec
ө = 360 / P = 360 / 97
= 3.7 º / min. = 223 º / hr.
= 14.85 revolutions per day
= about 15 rev. per day
Phys 215:
[email protected]
Introduction to Astronomy
Physics Department
KFUPM , Dhahran
Saudi Arabia
```
| 1,518 | 3,499 |
{"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-2018-34
|
longest
|
en
| 0.770557 |
https://www.khanacademy.org/math/mappers/statistics-and-probability-220-223/x261c2cc7:box-plots2/v/interpreting-box-plots
| 1,722,790,306,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2024-33/segments/1722640408316.13/warc/CC-MAIN-20240804164455-20240804194455-00764.warc.gz
| 671,223,436 | 141,079 |
If you're seeing this message, it means we're having trouble loading external resources on our website.
If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked.
### Course: MAP Recommended Practice>Unit 52
Lesson 7: Box plots
# Interpreting box plots
A box and whisker plot is a handy tool to understand the age distribution of students at a party. It helps us identify the minimum, maximum, median, and quartiles of the data. However, it doesn't provide specific details like the exact number of students at certain ages.
## Want to join the conversation?
• I still don't understand what a 'quartile' is. Can someone please help?
• quartile: The values that divide a list of numbers into quarters.
• i am very confused where did he get precents from
• Think of the box-and-whisker plot as split into four parts (the first, second, third, and fourth quartiles), making each part equal to 1/4 (essentially 25%) of the plot.
As shown in the video, there are three quartiles that have values larger than ten; that means that 3/4 of the quartiles have kids older than 10. In other words, 75% of the plot accounts for kids 10 and older (since 3/4 can be written as 75%).
The fact that every quartile is 25% is a guestimate; the point is that all three quartiles should add up to at least 75% of the plot.
Hope this clears things up!😄
• At , Isn't there 50% on one side of the median and 50% on the other so technically isn't 13 still in the middle so therefore it would be true? Thanks!
• the sign with the curly = sign means that it is approximated, so he can't be sure that it is truly 50% on that side, on the flip side, because we don't know, it might as well be like that.
• I do not get the last statement. We DO know that exactly half of the students are older than 13, because 50% is on the right side of the median which is also 13. Don't we?
• If he'd said that exactly 50% ARE 13 or older, that would be true, because it includes the median.
For example, if i say that 5 is greater than 10/2, that would be false. Because 10/2 is 5, but i said it was GREATER. On the other hand, if i said that 5 >= 10/2 (greater or equal to), that would be true.
• At , when he says that it is the second quartile, wouldn't that be Q1 and the median be Q2?
• No, this is because the first quartile is the line before the box.
• Would it be true if the second question was "Exactly 75% of the students are >10"?
• No.
As an example, let's say there were 17 students at the party, of the following ages:
7, 8, 8, 9, 11, 11, 12, 12, 13, 13, 14, 14, 15, 15, 15, 16, 16
The median is 13.
The second quartile is 10.
The third quartile is 15.
The minimum age is 7.
The maximum age is 16.
So, this data set gives the same box plot as shown in the video.
But 13 of the 17 students are older than 10.
13∕17 ≈ 76.47%, which is of course greater than 75%.
So, the number of students older than 10 is not necessarily exactly 75%.
• Where does he get the percentage.
• think of the plot as a chocolate bar, 25% would be 1/4 of the bar. Say that you had to share that chocolate bar with your 3 friends and you, would divide it into fourths, and 1 friend says he got more than me. what would you say?
I would´ve said ¨I have given us all 25% of the chocolate. so It is all equal.
| 914 | 3,326 |
{"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-2024-33
|
latest
|
en
| 0.959431 |
http://brightstorm.com/science/physics/linear-and-projectile-motion/linear-motion/
| 1,369,477,771,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2013-20/segments/1368705936437/warc/CC-MAIN-20130516120536-00003-ip-10-60-113-184.ec2.internal.warc.gz
| 41,619,659 | 10,315 |
Congratulations on starting your 24-hour free trial!
Quick Homework Help
# Linear Motion
Star this video
Linear motion is the motion that is natural to an object: moving in a straight line. According to Newton s First Law of Motion, an object not affected by any force will continue indefinitely in a straight line. If a projectile is thrown vertically, it will travel in linear motion and will begin to fall when the force of gravity equals the force of the throw.
Let's talk about linear motion, here's an example of linear motion I take this tennis ball and I throw it up and down it's moving in one dimension okay. That's a little different than parabolic motion where it's moving at an angle and the force of gravity is also pulling it down. So let's first look at linear motion, again simply put it's just change is distance over the change in time okay.
A problem you might see related to linear motion is a ball thrown straight up at 20 meters per second for how long will that ball move up and how high will it go? How far will it travel? Okay to solve that we need to think about it so, we're throwing the ball up at 20 meters per second and it's going to up until the velocity due to gravity is also 20 meters per second. So now it's at the point at which its velocity due to gravity has equaled to velocity due to the initial velocity applied to the ball.
Okay, well again to solve that let's look at a couple of equations okay first off the distance traveled due to a simple velocity is just a velocity times time, so meters per seconds over seconds okay. The difference due to gravity is 1 half gravity times time squared, that'll be useful for the second part here okay. And again the force of gravity is 9.8 meters per second squared. And to solve problems like this it's useful to simplify, so let's simplify that to 10 meters per second squared. That'll be close enough for our calculations here okay so I want to know at what point will my initial velocity equal the velocity due to gravity. So my initial velocity is 20 meters per second and I want to know at what point will that equal 10 meters per second squared times t okay.
Pretty easy problem, if I see the 20 meters per second equals 10 meters per second times t I can see that seconds squared and this is going to be unit in seconds so they'll cancel. I can see that my t equals 2 seconds okay. So my ball is going to stay in the air for about 2 seconds okay, before it starts to come done. Now if I'm asked how long will it remain in the air total, well that's going to be 2 seconds up and 2 seconds down okay. So that answers part one, now let's look at number two.
How high will it go in the air okay, well now I want to figure out the distances and again distances for velocity is velocity times time, distances due to the force of gravity is one half gravity times time squared okay. So how do those two distances because my velocity going up is going to start at 20 my velocity due to gravity is going to start at zero and then it's going to slowly increase, it's going to continue to increase. So I'm going to add those two vector, velocities together okay so if I say and for that one remember the gravitational velocity is going to be a negative value whereas my initial velocity up is going to be a positive value. So let's go ahead and calculate that velocity times time is 20 meters per second times 2 seconds okay and minus gravity pointed down and again the distance due to gravity is one half and 10 times 10 meters squared that's the force of gravity times time squared okay which is 2 times 2.
Again simplifying further, this is 40 meters per second times 2 seconds okay so the seconds are going to cancel there minus 1 half times 10 times 2 squared okay and again to simplify that 10 times 2 is 20 and one half of that is, I'm sorry 2 squared is 4 so this is 40 and 1 half of that is 20, so I've got 20 meters okay so my answer here is the ball goes 20 meters in the air before it starts to descend.
## Find Videos Using Your Textbook
Enjoy 3,000 videos just like this one.
| 913 | 4,082 |
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 4.4375 | 4 |
CC-MAIN-2013-20
|
latest
|
en
| 0.957803 |
https://www.physicsforums.com/threads/static-friction-experiment-help-result-interpretation.761435/
| 1,531,876,975,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2018-30/segments/1531676589980.6/warc/CC-MAIN-20180718002426-20180718022426-00629.warc.gz
| 954,696,790 | 17,077 |
# Homework Help: Static Friction Experiment: Help Result interpretation
1. Jul 12, 2014
### Jaimie
Hello, I hope someone can help me with this question.
"Design a simple experiment that you could carry out in your home to
i) determine the coefficient of static friction between an object and a surface.
ii) prove that the coefficient of static friction is dependent only on the surfaces in contact, and is not affected by any change in the mass of your object"
This is what I came up with.
1) Obtain and tape down with painter’s tape the architectural scale (ruler) so that the edge is 90degrees with the edge of the table top.
2) Obtain and tape protractor along the edge of the table so that the ‘x’ mark is aligned to the front edge of the scale and to the table top. Both taped scale and protractor will act as a jig to ensure there is little movement when raising the textured surfaces.
3) Place a piece of painter’s tape on the edge of the cutting board (closest to you) to mark the point at which the box will be placed at the start of each trail.
4) Place the wood cutting board flat on the table so that the front edge fits snuggly between the protractor and scale’s front edge. Align the full box of screws on the cutting board so the longest front edge is aligned with the tape marker.
5) Slowly raise the end of the cutting board surface just until the box just begins to slide downwards. Record the angle at which this occurs in the data table.
6) Repeat steps 4-5 two more times for a total of three trials and calculate the average angle value. (This angle will be used to calculate the static friction coefficient).
7) Calculate the static friction coefficient using the equation u= sin(theta) as derived from -Ff + Fgx = 0 for the surface and using the average angle value.
8) Repeat steps 4-7 with the box of screws 1/2 –full, then with the box empty. Record all values and calculations in the data table.
9) Obtain the plexiglass sheet clip it, with the binder clips, to on the board’s edge (closest to you). Repeat steps 4-8 and record all values and calculations in the data table.
10) Obtain the sheet of white paper and clip it (with the binder clips) to the cutting board. Repeat steps 4-8 and record all values and calculations in the data table.
Here are my results
Surface Average angle Static coefficient
Wood cutting board (w/Full box of screws)- 20.0 0.364
" (w/1/2 box of screws)- 20.8 0.380
" (empty) 22.8 0.420
Plexiglass (w/Full box of screws) 28.3 0.538
" (w/1/2 box of screws)- 34.0 0.674
" (empty box) 40.7 0.860
Paper (w/Full box of screws)- 20.0 0.364
" (w/1/2 box of screws)- 28.0 0.532
" (empty) 30.3 0.584
3. The attempt at a solution
2. Jul 13, 2014
### Simon Bridge
Uncertainties are reported to no more than 2 sig fig and usually just one.
Have you done hypothesis testing yet? Do you know how to combine errors?
The hypothesis being tested is that there is no difference in the coefficient if friction for different weights.
Clearly there is a difference - but it could be due to random variations in the measurement.
You need to see if the difference is close enough to zero to call it at 95% confidence level.
The simplest approach is to plot a graph of coefficient vs mass, with 95% error bars, and see if a horizontal line can be drawn through all the errorbars.
Note: what did you do to account for the sliding of the box changing the surface (scratching it etc)?
3. Jul 13, 2014
### Jaimie
Yes, this is what I thought as mathematically, mg cancels out, so the static coefficient should be the same with same surfaces in contact with varying weights. I do believe that there were a lot of errors as a result of movement while taking mesurements, not enough trials perhaps, inaccuracies in equipment,etc.
Thank you also for the tip about the percentage error (Will keep this in mind when studying further). I didn't account for the sliding of the box changing the surface. Good to know! Typically was is done to account for this...or do you just wipe both surfaces in contact before sliding?
4. Jul 13, 2014
### Simon Bridge
You certainly have too few trials.
I suspect your surfaces got rougher with use - increasing the coefficient of friction perhaps.
Plastics tend to have more stiction with more sharp edges for example. (You know how friction happens?)
You'd account for this by having more trials - identifying outliers - and repeating the runs for different weights.
The most common mistakes made in this sort of thing is just getting relaxed as you go through the trials - when you start you are usually very careful and as you get used to the work, you get a bit more careless.
Pointers:
- You should make a statement if your overall strategy at the start of your description of the experiment - before you go into details. The important details are in how you identify control the variables in an experiment.
- you don't need the protractor or any measurement of the angle to the horizontal.
measure the distance along the rap to the start position of the box - call this d
measure the height (to the ramp) of the box when the box just starts to move - call this h
the sine of the angle is h/d
... so you keep d the same, and record h and m for each run.
- make a graph as you go - this can be quite untidy so you won;t hand it in, but it helps you identify when you have a mistake or a systematic error.
5. Jul 15, 2014
### Jaimie
Thank you Simon. Very helpful. I will make the adjustments as noted. Thanks again.
| 1,328 | 5,511 |
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 3.515625 | 4 |
CC-MAIN-2018-30
|
latest
|
en
| 0.898544 |
https://www.investopedia.com/terms/m/median.asp
| 1,686,272,576,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2023-23/segments/1685224655244.74/warc/CC-MAIN-20230609000217-20230609030217-00108.warc.gz
| 871,718,551 | 58,694 |
# Median: What It Is and How to Calculate It, With Examples
## What Is the Median?
The median is the middle number in a sorted, ascending or descending list of numbers and can be more descriptive of that data set than the average. It is the point above and below which half (50%) the observed data falls, and so represents the midpoint of the data.
The median is often compared with other descriptive statistics such as the mean (average), mode, and standard deviation.
### Key Takeaways
• The median is the middle number in a sorted list of numbers and can be more descriptive of that data set than the average.
• The median is sometimes used as opposed to the mean when there are outliers in the sequence that might skew the average of the values.
• If there is an odd amount of numbers, the median value is the number that is in the middle, with the same amount of numbers below and above.
• If there is an even amount of numbers in the list, the middle pair must be determined, added together, and divided by two to find the median value.
• In a normal distribution, the median is the same as the mean and the mode.
## Understanding the Median
Median is the middle number in a sorted list of numbers. To determine the median value in a sequence of numbers, the numbers must first be sorted, or arranged, in value order from lowest to highest or highest to lowest. The median can be used to determine an approximate average, or mean, but is not to be confused with the actual mean.
• If there is an odd amount of numbers, the median value is the number that is in the middle, with the same amount of numbers below and above.
• If there is an even amount of numbers in the list, the middle pair must be determined, added together, and divided by two to find the median value.
The median is sometimes used as opposed to the mean when there are outliers in the sequence that might skew the average of the values. The median of a sequence can be less affected by outliers than the mean.
## Median Example
To find the median value in a list with an odd amount of numbers, one would find the number that is in the middle with an equal amount of numbers on either side of the median. To find the median, first arrange the numbers in order, usually from lowest to highest.
For example, in a data set of {3, 13, 2, 34, 11, 26, 47}, the sorted order becomes {2, 3, 11, 13, 26, 34, 47}. The median is the number in the middle {2, 3, 11, 13, 26, 34, 47}, which in this instance is 13 since there are three numbers on either side.
To find the median value in a list with an even amount of numbers, one must determine the middle pair, add them, and divide by two. Again, arrange the numbers in order from lowest to highest.
For example, in a data set of {3, 13, 2, 34, 11, 17, 27, 47}, the sorted order becomes {2, 3, 11, 13, 17, 27, 34, 47}. The median is the average of the two numbers in the middle {2, 3, 11, 13, 17, 26 34, 47}, which in this case is fifteen {(13 + 17) ÷ 2 = 15}.
The median is closely associated with quartiles, or dividing up observed data into four equal parts. The median would be the center point, with the first two quartiles falling below it and the second two above it. Other ways of bucketing data include quintiles (in five sections) and deciles (in 10 sections).
## How Do You Calculate the Median?
The median is the middle value in a set of data. First, organize and order the data from smallest to largest. To find the midpoint value, divide the number of observations by two. If there are an odd number of observations, round that number up, and the value in that position is the median. If the number of observations is even, take the average of the values found above and below that position.
## Where Is the Median in a Normal Distribution?
In the normal distribution ("bell curve") the median, mean, and mode are all the same value, and fall at the highest point in the center of the curve.
## When Is the Mean and Median Different?
In a skewed data set, the mean and median will typically be different. The mean is calculated by adding up all of the values in the data and dividing by the number of observations. If there are sizable outliers, or if the data clumps around certain values, the mean (average) will not be the midpoint of the data.
For instance, in a set of data {0, 0, 0, 1, 1, 2, 10, 10} the average would be 24/8 = 3. The median, however, would be 1 (the midpoint value).
This is why many economists favor the median for reporting a nation's income or wealth, since it is more representative of the actual income distribution.
Open a New Bank Account
×
The offers that appear in this table are from partnerships from which Investopedia receives compensation. This compensation may impact how and where listings appear. Investopedia does not include all offers available in the marketplace.
| 1,150 | 4,859 |
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 4.4375 | 4 |
CC-MAIN-2023-23
|
latest
|
en
| 0.927881 |
https://cs.stackexchange.com/questions/138406/finding-a-lower-bound-for-the-expression-logn
| 1,632,489,096,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2021-39/segments/1631780057524.58/warc/CC-MAIN-20210924110455-20210924140455-00378.warc.gz
| 229,760,674 | 38,777 |
# Finding a lower bound for the expression $\log(n!)$
Problem:
Is $$\log(n!) \in$$ $$\Omega( n^n )$$?
Since $$n! > n^n$$ for all $$n > 1$$ we can conclude that: $$\log(n!) \in$$ $$O( n^n )$$.
Let us look at the special case where $$n = 4$$. \begin{align*} n! &= 4(3)(2) = 24 \\ n^n &= 4^4 = 256 \end{align*} Let us look at the special case where $$n = 5$$. \begin{align*} n! &= 5(4)(3)(2) = 5(24) = 120 \\ n^n &= 5^5 = 3125 \end{align*} Let us look at the special case where $$n = 8$$. \begin{align*} n! &= 8! = 40320 \\ n^n &= 8^8 = 16777216 \end{align*} It looks to me that $$n^n$$ is growing faster but that is not a proof. To prove it, I need to show that there exists an $$M > 0$$ and $$n_o > 0$$ such that the following statement is true for all $$n \geq n_0$$: $$n! \leq M n^n$$ I select $$n_0 = 4$$ and $$M = 1$$. Hence the expression reduces to: $$n! \leq n^n$$ We have already shown that this expression is true for the special case of $$n = 4$$. Now, if we add $$1$$ to $$n$$ we have: $$(n+1)! \leq (n+1)^{(n+1)}$$ This must be true because the left hand side increased by a factor of $$n+1$$ and the right hand side increased by more than a factor of $$n+1$$. Now we add $$1$$ to $$n$$ again. The left hand side increases by a factor of $$n+2$$ and the right hand side increases by more than a factor of $$n+2$$. Hence the right side increases more. We can repeat this process for ever. Therefore, I conclude the statement is true.
Do I have this right?
– user114966
Apr 1 at 22:08
• At the beginning of your question you are asking about a lower bound, but in the rest of the question you are talking about an upper bound? Apr 1 at 22:08
• @Dmitry I do not currently have an instructor. I am not currently taking a course.
– Bob
Apr 1 at 22:18
• @Steven I realize that the line $n! > n^n$ is wrong. It was a mathematical typesetting error. I should have written $n! < n^n$. I am thinking I need to be very carefully in updating the post due to the comments already made.
– Bob
Apr 1 at 22:22
Firstly, $$n!$$ is NOT greater than $$n^n$$. Indeed, $$n! = \displaystyle\prod\limits_{k=1}^nk\leq \prod\limits_{k=1}^nn = n^n$$.
Secondly, even if you have a function $$f(n)$$ such that $$f(n) > n^n$$, that does not mean that $$\log(f(n)) \in \Omega(n^n)$$. For example, $$n^{n+1}>n^n$$, but $$\log(n^{n+1}) = (n+1)\log n < n^2 = o(n^n)$$.
Finaly, you can write $$\log(n!) = \log(\prod\limits_{k=1}^nk) = \sum\limits_{k=1}^n\log(k) \leq \sum\limits_{k=1}^n\log(n) = n\log n$$.
With this inequality, we can deduce that $$\log(n!) \in O(n\log n)$$ and since $$n\log n = o(n^n)$$, the statement $$\log(n!) \in \Omega(n^n)$$ is false.
$$\log n! \le \log n^n = n \log n = o(n^n)$$.
| 951 | 2,686 |
{"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": 43, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 4.6875 | 5 |
CC-MAIN-2021-39
|
latest
|
en
| 0.891355 |
http://mathoverflow.net/questions/70108?sort=newest
| 1,371,722,347,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2013-20/segments/1368711240143/warc/CC-MAIN-20130516133400-00011-ip-10-60-113-184.ec2.internal.warc.gz
| 165,052,977 | 11,245 |
MathOverflow will be down for maintenance for approximately 3 hours, starting Monday evening (06/24/2013) at approximately 9:00 PM Eastern time (UTC-4).
## Non-polynomial integrals of motion for polynomial dynamical systems
Does there exist a polynomial Hamiltonian function $H$ on some $\mathbb{R}^{2n}$ such that
1. Any polynomial function $P$ such that $\{P,H\}=0$ is of the form $p(H)$ for some polynomial $p$ in one variable;
2. There exists a smooth function $F$ such that $\{F,H\}=0$, and yet $F$ is not of the form $f(H)$ for any smooth function $f$ in one variable?
I am looking for an example of such a phenomenon, or a proof that it cannot occur. A non-Hamiltonian example -- that is, of a polynomial vector field such that the smooth completion of the algebra of its polynomial integrals is strictly smaller than the algebra of its smooth integrals -- would also be helpful. Any references to the literature will be much appreciated, of course.
-
For the non-Hamiltonian case, take the ODE $\dot x = (x^2+1)$ and $\dot y = 2xy$ in the plane. This is a polynomial vector field on the plane, and it has a first integral $y/(x^2+1)$, which is not a polynomial. Moreover, it is easy to see that there is no polynomial first integral $P(x,y)$ of this ODE that is not constant. – Robert Bryant Jul 12 2011 at 12:14
Dear Robert, your comment is very helpful and counts as an answer. Thanks! – Dmitry Roytenberg Jul 13 2011 at 8:47
Here is a tentative Hamiltonian answer having 2 degrees of freedom. Take $H = (1/2)(1 + a x^2 + bxy + cy^2) (p_x ^2 + p_y ^2)$. for essentially any parameters $a, b, c$ for which
$ax^2 + bxy + cy^2$ is positive definite ($a x^2 + bxy + c y^2) > \epsilon (x^2 + y^2)$) but NOT a multiple of $x^2 + y^2$. This $H$ is the Hamiltonian for geodesic flow on the plane with metric $ds^2 = (dx^2 + dy^2)/((1 + a x^2 + bxy + cy^2)$. Comparing the arclength $ds$ with $dr/\sqrt{1 + \epsilon r^2}$ seems to show that the metric is complete. Now we can play the scattering integrability' trick which I learned from E. Gutkin. This trick yields 2 new integrals $F, G$ as thescattering data'' for the resulting geodesics. What I mean is that any geodesic will be asymptotic to a straight line in the xy plane, and the asymptotic direction $(F, G)$ of this line is an integral. The hard part is to show this asymptotic direction, is a smooth function of $x, y, p_x, p_y$. I wager that unless $a x^2 + bxy + cy^2$ is a square (like $x^2$) or a multiple of $x^2 + y^2$, these scattering integrals are not polynomials. They might not even be analytic..
-
There are such examples (with transcendental integrals of motion) already on $\mathbb{R}^{4}$, see the paper of Hietarinta.
The Hamiltonian is $$H=p_x^2/2+p_y^2/2+2y p_x p_y-x,$$ the desired integral is $$I_1=p_y \exp(p_x^2).$$
-
| 837 | 2,812 |
{"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": 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}
| 3.5 | 4 |
CC-MAIN-2013-20
|
latest
|
en
| 0.881915 |
https://testbook.com/question-answer/which-of-the-following-pair-has-not-same-unit--5ff94a5277fa1ed3dfa1e0c2
| 1,632,648,124,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2021-39/segments/1631780057857.27/warc/CC-MAIN-20210926083818-20210926113818-00662.warc.gz
| 590,219,383 | 30,039 |
# Which of the following pair has not same unit?
This question was previously asked in
RRC Group D Previous Paper 48 (Held On: 11 Dec 2018 Shift 2)
View all RRC Group D Papers >
1. Speed and velocity
2. Acceleration and acceleration due to acceleration
3. Momentum and force
4. Distance and Displacement
Option 3 : Momentum and force
Free
RRC Group D Full Mock Test
364934
100 Questions 100 Marks 90 Mins
## Detailed Solution
The correct answer is Momentum and force.
Important Points
• Option (a), Speed = Distance / time = ms-1
Velocity = Displacement / time = ms-1
Thus both have the same unit.
• Option (b), Gravity (g) = ms-2
Acceleration due to gravity = ms-2
Thus, both have the same unit.
• Optiion (c), Momentum = Mass * Velocity = Kg * ms-1 = Kgms-1
Force = Mass * Acceleration = Kg * ms-2 = Kgms-2
Thus the unit of Momentum and force is not the same.
• Option (d), Distance = m(meter)
Displacement = m(meter)
Thus both have the same unit.
| 270 | 956 |
{"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-2021-39
|
latest
|
en
| 0.842909 |
http://gmatclub.com/forum/in-the-rectangular-co-ordinate-system-shown-above-69245.html?fl=similar
| 1,467,303,293,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2016-26/segments/1466783398873.39/warc/CC-MAIN-20160624154958-00071-ip-10-164-35-72.ec2.internal.warc.gz
| 136,076,641 | 43,327 |
Find all School-related info fast with the new School-Specific MBA Forum
It is currently 30 Jun 2016, 09:14
### GMAT Club Daily Prep
#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.
Customized
for You
we will pick new questions that match your level based on your Timer History
Track
every week, we’ll send you an estimated GMAT score based on your performance
Practice
Pays
we will pick new questions that match your level based on your Timer History
# Events & Promotions
###### Events & Promotions in June
Open Detailed Calendar
# In the Rectangular Co-ordinate system shown above
Author Message
Intern
Joined: 30 Jun 2008
Posts: 4
Followers: 0
Kudos [?]: 7 [0], given: 0
In the Rectangular Co-ordinate system shown above [#permalink]
### Show Tags
23 Aug 2008, 07:41
00:00
Difficulty:
(N/A)
Question Stats:
0% (00:00) correct 0% (00:00) wrong based on 0 sessions
### HideShow timer Statistics
This topic is locked. If you want to discuss this question please re-post it in the respective forum.
Que:- In the Rectangular Co-ordinate system shown above (it was normal XY plane with 4 Qudrants in it), does the line K intersect quadrant II?
a) The slope of K is -1/6
b) The y-intercept of K is -6
OA - A
Can someone please explain this to me. I think C should be the answer as from A we cannot determine the the value constant (y-intercept) present in the equation of line k (y= -1/6x + c).
SVP
Joined: 07 Nov 2007
Posts: 1820
Location: New York
Followers: 30
Kudos [?]: 729 [0], given: 5
Re: Gmat prep - DS Question - Co-ordinate Geometry [#permalink]
### Show Tags
24 Aug 2008, 15:25
prashi82 wrote:
Que:- In the Rectangular Co-ordinate system shown above (it was normal XY plane with 4 Qudrants in it), does the line K intersect quadrant II?
a) The slope of K is -1/6
b) The y-intercept of K is -6
OA - A
Can someone please explain this to me. I think C should be the answer as from A we cannot determine the the value constant (y-intercept) present in the equation of line k (y= -1/6x + c).
I agree with A.
We don't know the value of C.
C can be +ve or zero or -ve
When +ve
When c is +ve -->Line K intersect quadrants I,II, IV
When c is -ve -->Line K intersect quadrants II,III, IV
When c is zero -->Line K intersect quadrants II,IV
In all these cases.. K intersect quadrants II and IV.
So it should be A.
_________________
Smiling wins more friends than frowning
Manager
Joined: 15 Jul 2008
Posts: 207
Followers: 3
Kudos [?]: 46 [0], given: 0
Re: Gmat prep - DS Question - Co-ordinate Geometry [#permalink]
### Show Tags
25 Aug 2008, 05:17
A simple observation
any line with -ve slope, will intersect 2nd and 4th quadrant and any line with +ve slope will intersect 1st and 4th quadrant.
Re: Gmat prep - DS Question - Co-ordinate Geometry [#permalink] 25 Aug 2008, 05:17
Display posts from previous: Sort by
| 846 | 2,994 |
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 3.515625 | 4 |
CC-MAIN-2016-26
|
longest
|
en
| 0.901192 |
http://www.studymode.com/essays/Math-Comparison-361642.html
| 1,524,203,620,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2018-17/segments/1524125937114.2/warc/CC-MAIN-20180420042340-20180420062340-00627.warc.gz
| 513,287,673 | 25,273 |
# Math Comparison
Topics: Multiplicative inverse, Rational function, Fraction Pages: 2 (453 words) Published: July 23, 2010
Comparison Paper
In mathematics, there are many branches of functions. You have Inverse, Linear, Quadratic, Cubic, Periodic, Monotonic, etc. In this paper we will cover Reciprocal and Rational Functions. Both can be plotted on a graph, both have asymptotes, and both have discontinuities. But just because they have a few similarities doesn’t mean that they are all the same. In this paper we will compare and contrast the two functions.
A rational function has the definition of, “Any function which can be written as the ratio of two polynomial functions.” Any time that the denominator is undefined, that means the function is undefined. Since a rational function is set up in a fraction, if the same term is in the numerator and the denominator then those two will “cancel” each other out. When two terms cancel each other out, a hole will occur at the point that both terms cancelled out at. The vertical asymptote is found by the term in the denominator. If you only have (x-2) in the denominator then you will have a vertical asymptote at 2 on the x axis, going vertical. It will be 2 because in the function the vertical asymptote is automatically given a negative sign, so the opposite of whatever your term is. The horizontal is found by any number outside of the fraction.
A reciprocal function can be defined as, “A function that models inverse variations.” Reciprocal functions can be graphed. They usually start at the top or sides of a graph; go inwards towards the origin, then bend past it and leave through the same quadrant that they entered. The x-axis is the horizontal asymptote, and the y-axis is the vertical asymptote. The vertical asymptote is found the same way as the rational function asymptote. As is the horizontal asymptote.
For both the Reciprocal and Rational Functions, you find the asymptotes the same way. You use the same formula, which is y=a/-x-h+k. The vertical asymptote is the opposite value because of the negative sign in the equation. The...
| 462 | 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}
| 4.71875 | 5 |
CC-MAIN-2018-17
|
latest
|
en
| 0.933241 |
https://benidormclubdeportivo.org/gravity-is-an-example-of-a-conservative-force/
| 1,652,727,444,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2022-21/segments/1652662512229.26/warc/CC-MAIN-20220516172745-20220516202745-00708.warc.gz
| 187,965,416 | 9,037 |
Define conservative force, potential energy, and mechanical energy.Explain the potential power of a spring in regards to its compression once Hooke’s law applies.Use the work-energy organize to show how having only conservative forces implies preservation of mechanical energy.
You are watching: Gravity is an example of a conservative force
## Potential Energy and Conservative Forces
Work is excellent by a force, and also some forces, such together weight, have actually special characteristics. A conservative force is one, favor the gravitational force, for which work-related done through or versus it depends only on the starting and finishing points that a motion and also not on the path taken. Us can define a potential energy (PE) for any kind of conservative force, simply as we did because that the gravitational force. Because that example, once you wind increase a toy, an egg timer, or an old-fashioned watch, you carry out work against its spring and also store power in it. (We treat these springs together ideal, in that us assume there is no friction and also no production of thermal energy.) This stored energy is recoverable together work, and it is beneficial to think that it together potential energy had in the spring. Indeed, the reason that the spring has this properties is the its force is conservative. The is, a conservative pressure results in stored or potential energy. Gravitational potential energy is one example, as is the power stored in a spring. We will additionally see just how conservative forces are related to the conservation of energy.
### Potential Energy and also Conservative Forces
Potential power is the power a system has due to position, shape, or configuration. It is stored power that is totally recoverable.
A conservative pressure is one for which work done by or against it depends just on the beginning and finishing points the a motion and also not on the course taken.
We can specify a potential power (PE) for any kind of conservative force. The occupational done against a conservative pressure to with a final configuration counts on the configuration, not the course followed, and is the potential power added.
## Potential power of a Spring
First, allow us obtain an expression because that the potential power stored in a feather (PEs). Us calculate the work-related done to stretch or compress a spring the obeys Hooke’s law. (Hooke’s regulation was check in Elasticity: Stress and Strain, and also states that the magnitude of pressure F top top the spring and the result deformation ΔL space proportional, kΔL.) (See number 1.) for our spring, us will replace ΔL (the lot of deformation produced by a force F) by the street x the the feather is stretched or compressed follow me its length. So the pressure needed to stretch the spring has actually magnitude Fkx, whereby k is the spring’s force constant. The pressure increases linearly from 0 in ~ the start to kx in the totally stretched position. The average force is frackx2\. For this reason the work-related done in extending or compressing the spring is W_ exts=Fd=left(frackx2 ight)x=frac12kx^2\. Alternatively, we detailed in Kinetic Energy and also the Work-Energy Theorem that the area under a graph of F vs. x is the work-related done by the force. In number 1c we check out that this area is likewise frac12kx^2\. We as such define the potential power of a spring, PEs, come be
extPE_ exts=frac12kx^2\,
where k is the spring’s force continuous and x is the displacement native its undeformed position. The potential energy represents the occupational done on the spring and also the power stored in it as a an outcome of stretching or compressing it a street x. The potential energy of the feather PEs does not count on the course taken; that depends just on the large or squeeze out x in the final configuration.
Figure 1. (a) an undeformed spring has actually no PEs save on computer in it. (b) The pressure needed to stretch (or compress) the feather a distance x has a size F = kx, and the occupational done to stretch (or compress) the is frac12kx^2\. Due to the fact that the force is conservative, this job-related is stored as potential energy (PEs) in the spring, and also it can be completely recovered. (c) A graph the F vs. X has a steep of k, and the area under the graph is frac12kx^2\. For this reason the work-related done or potential energy stored is frac12kx^2\.
The equation extPE_ exts=frac12kx^2\ has general validity past the special case for i m sorry it was derived. Potential energy can be stored in any elastic medium by deforming it. Indeed, the general meaning of potential energy is energy due to position, shape, or configuration. For shape or position deformations, stored energy is extPE_ exts=frac12kx^2\, where k is the force constant of the details system and also x is its deformation. An additional example is checked out in figure 2 for a guitar string.
Figure 2. Work-related is excellent to deform the etc string, providing it potential energy. Once released, the potential energy is converted to kinetic power and back to potential together the string oscillates ago and forth. A an extremely small fraction is dissipated together sound energy, progressively removing energy from the string.
## Conservation of mechanical Energy
Let united state now consider what type the work-energy to organize takes as soon as only conservative forces are involved. This will lead us to the conservation of energy principle. The work-energy theorem states that the net job-related done through all pressures acting top top a system equates to its readjust in kinetic energy. In equation form, this is W_ extnet=frac12mv^2-frac12mv_0^2=Delta extKE\.
If just conservative forces act, then Wnet = Wc, where Wc is the full work excellent by every conservative forces. Thus, Wc = ΔKE.
Now, if the conservative force, such together the gravitational force or a spring force, does work, the mechanism loses potential energy. That is, Wc=−ΔPE. Therefore, −ΔPE = ΔKE or ΔKE + ΔPE = 0.
This equation method that the total kinetic and potential energy is constant for any procedure involving only conservative forces. That is,
ext(conservative forces only),egincases extKE+ extPE= extconstant\ extor\ extKE_ exti+ extPE_ exti= extKE_ extf+ extPE_ extfendcases\
where i and also f denote initial and also final values. This equation is a type of the work-energy theorem for conservative forces; that is known as the conservation of mechanical energy principle. Remember that this uses to the extent that all the pressures are conservative, so the friction is negligible. The full kinetic add to potential energy of a device is identified to be its mechanically energy, (KE+PE). In a device that experiences just conservative forces, there is a potential energy associated with each force, and also the energy only changes type between KE and the various types of PE, through the total energy remaining constant.
### Example 1. Making use of Conservation of Mechanical energy to calculation the speed of a Toy Car
A 0.100-kg toy car is thrust by a compressed spring, as displayed in figure 3. The car follows a track that rises 0.180 m above the beginning point. The spring is compressed 4.00 cm and also has a force consistent of 250.0 N/m. Assuming job-related done by friction to it is in negligible, uncover the following:
How rapid is the vehicle going before it starts increase the slope?How quick is the going at the top of the slope?
Figure 3. A toy car is moved by a compressed spring and also coasts up a slope. Suspect negligible friction, the potential energy in the feather is first completely convert to kinetic energy, and also then come a mix of kinetic and also gravitational potential power as the automobile rises. The details of the course are unimportant since all pressures are conservative—the car would have actually the same last speed if it take it the alternative path shown.
Strategy
The feather force and the gravitational force are conservative forces, so conservation that mechanical power can it is in used. Thus,
KEi + PEi = KEf + PEf
or
frac12mv_ exti^2+mgh_ exti+frac12kx_ exti^2=frac12mv_ extf^2+mgh_ extf+frac12kx_ extf^2\,
where h is the elevation (vertical position) and x is the compression that the spring. This general statement looks complex but becomes much easier when we start considering specific situations. First, we must determine the initial and also final problems in a problem; then, we go into them right into the critical equation to deal with for an unknown.
Solution for Part 1
This component of the difficulty is limited to conditions just before the vehicle is released and also just after it pipeline the spring. Take the initial elevation to it is in zero, so that both hi and also hf room zero. Furthermore, the initial speed vi is zero and the final compression the the feather xf is zero, and so number of terms in the conservation of mechanical energy equation are zero and also it simplifies to
frac12kx_ exti^2=frac12mv_ extf^2\.
In other words, the early stage potential energy in the feather is converted totally to kinetic energy in the absence of friction. Resolving for the last speed and entering recognized values yields
eginarraylllv_ extf&=&sqrtfrackmx_ exti\ ext &=&sqrtfrac250.0 ext N/m0.100 ext kgleft(0.0400 ext m ight)\ ext &=&2.00 ext m/sendarray\
Solution for Part 2
One method of finding the rate at the height of the steep is come consider conditions just prior to the auto is released and just after it reaches the optimal of the slope, completely ignoring everything in between. Law the same type of evaluation to find which terms room zero, the preservation of mechanical energy becomes
frac12kx_ exti^2=frac12mv_ extf^2+mgh_ extf\.
This type of the equation means that the spring’s early potential energy is converted partly to gravitational potential energy and also partly to kinetic energy. The last speed in ~ the height of the slope will certainly be less than in ~ the bottom. Fixing for vf and also substituting recognized values gives
eginarraylllv_ extf&=&sqrtfrackx_ exti^2m-2gh_ extf\ ext &=&sqrtleft(frac250.0 ext N/m0.100 ext kg ight)left(0.0400 ext m ight)^2-2left(9.80 ext m/s^2 ight)left(0.180 ext m ight)\ ext &=&0.687 ext m/sendarray\
Discussion
Another means to deal with this problem is to realize the the car’s kinetic energy before it goes up the steep is converted partly to potential energy—that is, to take the final conditions in part 1 to it is in the initial conditions in part 2.
Note that, because that conservative forces, we do not directly calculate the occupational they do; rather, we think about their effects through their corresponding potential energies, simply as us did in instance 1. Note likewise that we perform not take into consideration details that the path taken—only the beginning and ending points are necessary (as lengthy as the route is no impossible). This presumption is normally a incredible simplification, since the path might be complex and pressures may vary along the way.
## PhET Explorations: power Skate Park
Learn around conservation of power with a ska dude! develop tracks, ramps and jumps because that the skater and view the kinetic energy, potential energy and friction together he moves. Friend can likewise take the ska to different planets or even space!
## Section Summary
A conservative force is one because that which work depends just on the beginning and ending points the a motion, not on the course taken.We can specify potential energy (PE) for any kind of conservative force, simply as we defined PEg for the gravitational force.The potential energy of a spring is extPE_s=frac12 extkx^2\, where k is the spring’s force consistent and x is the displacement indigenous its undeformed position.Mechanical energy is defined to it is in KE + PE for a conservative force.When just conservative forces act on and also within a system, the complete mechanical energy is constant. In equation form,egincases extKE+ extPE= extconstant\ extor\ extKE_ exti+ extPE_ exti= extKE_ extf+ extPE_ extfendcases\where i and f represent initial and also final values. This is recognized as the preservation of mechanical energy.
See more: Sister Chromatids Line Up At The Equator Of The Cell., The Cell Cycle
### Conceptual Questions
What is a conservative force?The pressure exerted by a diving plank is conservative, detailed the inner friction is negligible. Assuming friction is negligible, describe transforms in the potential power of a diving board together a swimmer dives from it, starting just prior to the swimmer steps on the plank until simply after his feet leaving it.Define mechanically energy. What is the partnership of mechanical energy to nonconservative forces? What wake up to mechanical power if only conservative forces act?What is the connection of potential power to conservative force?
### Problems & Exercises
A 5.00 × 105-kg subway train is brought to a stop from a speed of 0.500 m/s in 0.400 m by a huge spring bumper in ~ the end of that track. What is the force continuous k of the spring?A pogo stick has actually a spring through a force consistent of 2.50 × 104 N/m, which can be compressed 12.0 cm. To what maximum height have the right to a child jump ~ above the pole using only the power in the spring, if the child and also stick have a total mass that 40.0 kg? Explicitly show how you monitor the procedures in the Problem-Solving strategies for Energy.
## Glossary
conservative force: a force that does the same work for any given initial and also final configuration, nevertheless of the course followed
potential energy: energy due to position, shape, or configuration
potential power of a spring: the stored power of a spring together a duty of that is displacement; when Hooke’s legislation applies, it is provided by the expression frac12 extkx^2\ wherein x is the distance the spring is compressed or extended and k is the feather constant
conservation of mechanical energy: the dominion that the amount of the kinetic energies and also potential energies remains continuous if just conservative pressures act on and also within a system
| 3,186 | 14,254 |
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 3.703125 | 4 |
CC-MAIN-2022-21
|
latest
|
en
| 0.950712 |
http://mechanical-design-handbook.blogspot.com/2010/08/polynomial-cam-function-derivation-of.html
| 1,561,579,439,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2019-26/segments/1560628000545.97/warc/CC-MAIN-20190626194744-20190626220744-00371.warc.gz
| 119,909,767 | 28,778 |
### Improve math skills of your kids - Learn step-by-step arithmetic from Math games
Math: Unknown - Step-by-step math calculation game for iOS.
Math: Unknown is much more than a math game. It is a step-by-step math calculation game which will teach users how to calculate in the correct order rather than just asking only the final calculated results.
The app consists of four basic arithmetic operations which are addition, subtraction, multiplication and division. In order to get started, users who are new to arithmetic can learn from animated calculation guides showing step-by-step procedures of solving each type of operation. It is also helpful for experienced users as a quick reference.
Generally, addition and subtraction may be difficult for users who just start learning math especially when questions require carrying or borrowing (also called regrouping). The app helps users to visualize the process of carrying and borrowing in the way it will be done on paper. Once users understand how these operations work, they are ready to learn multiplication and division.
For most students, division is considered as the most difficult arithmetic operation to solve. It is a common area of struggle since it requires prior knowledge of both multiplication and subtraction. To help users understand division, the app uses long division to teach all calculation procedures. Relevant multiplication table will be shown beside the question. Users will have to pick a number from the table which go into the dividend. Multiplication of selected number and divisor is automatically calculated, but the users have to do subtraction and drop down the next digit themselves. Learning whole calculation processes will make them master it in no time.
Math: Unknown is a helpful app for students who seriously want to improve arithmetic calculation skills.
### Polynomial Cam Function (Derivation of Fifth-degree function) - Part 2
In [Polynomial Cam Function (Introduction) - Part 1], we discussed about fundamental of cam design and introduction of polynomial cam function. In this post, we’re going to derive the equation of fifth-degree polynomial cam function. We start from the general term of fifth-degree polynomial function as follows.
s = c0 + c1(b/bm) + c2(b/bm)2 + c3(b/bm)3 + c4(b/bm)4 + c5(b/bm)5 ….. (eq.1)
where:
s = displacement (mm)
b = cam angle in that sector (rad)
bm = total angle in that sector (rad)
We can find the velocity in mm/rad by derivative of displacement. Later we can change it to the time domain.
v = ds/db
v = c1/bm + 2c2/bm(b/bm) + 3c3/bm(b/bm)2 + 4c4/bm(b/bm)3 + 5c5/bm(b/bm)4
Rearrange to get,
v = 1/bm[c1 + 2c2(b/bm) + 3c3(b/bm)2 + 4c4(b/bm)3 + 5c5(b/bm)4] ….. (eq.2)
Acceleration in mm/rad2 can be calculated by a = dv/db
a = 1/bm[2c2/bm + 6c3/bm(b/bm) + 12c4/bm(b/bm)2 + 20c5/bm(b/bm)3]
Rearrange to get,
a = 1/bm2[2c2 + 6c3(b/bm) + 12c4(b/bm)2 + 20c5(b/bm)3] ….. (eq.3)
Then we set the boundary conditions for the function. Let us introduce another parameter, hm.
where:
hm = total displacement in that sector (mm)
Applying boundary conditions:
(BC.1) At the beginning of movement, the displacement must start from 0 and has acceleration of 0. This is for connecting to another cam curves in other sectors. But we will leave the velocity at this point not equal to zero. Then we have more freedom to select the starting velocity. Of course, if the starting velocity is not zero, then we can’t connect it with dwell or cycloid functions because it will create discontinuity in velocity. But we will use it to connect with linear cam function or another fifth-degree polynomial curves.
At b = 0: s = 0, v = v0, and a = 0.
(BC.2) Then we set the boundary conditions at the end of the movement. The same approach applied here.
At b = bm: s = hm, v = v1, and a = 0
where:
v0 = starting velocity in that sector (mm/rad)
v1 = ending velocity in that sector (mm/rad)
Apply BC.1 into (eq.1)
0 = c0 + 0 + 0 + 0 + 0 + 0
So, c0 = 0
Apply BC.1 into (eq.3)
0 = 1/bm2[2c2 + 0 + 0 + 0]
Also, c2 = 0
Apply BC.1 into (eq.2)
v0 = 1/bm[c1 + 0 + 0 + 0 + 0]
Then c1 = bmv0
Apply BC.2 into (eq.1)
hm = 0 + c1 + 0 + c3 + c4 + c5
Substitute c1 = bmv0 and rearrange to get,
c3 + c4 + c5 = hm – bmv0 ….. (eq.4)
Apply BC.2 into (eq.3)
0 = 1/bm2[0 + 6c3 + 12c4 + 20c5]
Rearrange to get,
6c3 + 12c4 + 20c5 = 0
Or
3c3 + 6c4 + 10c5 = 0 ….. (eq.5)
Apply BC.2 into (eq.2)
v1= 1/bm[bmv0 + 0 + 3c3 + 4c4 + 5c5]
Rearrange to get,
3c3 + 4c4 + 5c5 = bmv1 – bmv0 ….. (eq.6)
Solve the simultaneous equations (eq.4, eq.5 and eq.6) for the values of constants c3, c4 and c5
(eq.6) – 3x(eq.4);
3c3 + 4c4 + 5c5 – 3c3 – 3c4 – 3c5 = bmv1 – bmv0 – 3hm + 3bmv0
c4 + 2c5 = -3hm + bmv1 + 2bmv0 ….. (eq.7)
(eq.5) – 3x(eq.4);
3c3 + 6c4 + 10c5 – 3c3 – 3c4 – 3c5 = 0 – 3hm + 3bmv0
3c4 + 7c5 = -3hm + 3bmv0 ….. (eq.8)
(eq.8) – 3x(eq.7);
3c4 + 7c5 – 3c4 – 6c5 = -3hm + 3bmv0 + 9hm – 3bmv1 – 6bmv0
c5 = 6hm – 3bmv1 – 3bmv0
c5 = 6hm – bm(3v0 + 3v1)
Substitute value of c5 into (eq.7)
c4 = -3hm + bmv1 + 2bmv0 – 2[6hm – 3bmv0 – 3bmv1]
c4 = -3hm + bmv1 + 2bmv0 – 12hm + 6bmv0 + 6bmv1
c4 = -15hm + 8bmv0 + 7bmv1
c4 = -15hm + bm(8v0 + 7v1)
Substitute values of c4 and c5 into (eq.4),
c3 = hm – bmv0 + 15hm – bm(8v0 + 7v1) – 6hm + bm(3v0 + 3v1)
c3 = 10hm – bm(6v0 + 4v1)
Therefore, the fifth-degree polynomial cam function becomes,
s = c1(b/bm) + c3(b/bm)3 + c4(b/bm)4 + c5(b/bm)5
It has velocity (v) and acceleration (a) as follows,
v = 1/bm[c1 + 3c3(b/bm)2 + 4c4(b/bm)3 + 5c5(b/bm)4]
a = 1/bm2[6c3(b/bm) + 12c4(b/bm)2 + 20c5(b/bm)3
where:
c1 = bmv0
c3 = 10hm – bm(6v0 + 4v1)
c4 = -15hm + bm(8v0 + 7v1)
c5 = 6hm – bm(3v0 + 3v1
We've derived the fifth-degree polynomial cam function with velocity and acceleration profiles that satisfies all boundary conditions as described earlier. Then we can use this cam function to design the timing diagram. Let's see how to use it in the next post.
| 2,063 | 5,916 |
{"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-2019-26
|
latest
|
en
| 0.928183 |
http://www.acemyhw.com/projects/10228/Mathematics/finite-mathematics
| 1,498,652,831,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2017-26/segments/1498128323680.18/warc/CC-MAIN-20170628120308-20170628140308-00422.warc.gz
| 434,842,659 | 5,368 |
# Project #10228 - Finite Mathematics
Must show all work.
2.1 Exercises 1, 3, 4, 9, 10, 12, 19, 22-24, 28, 31, and your choice of three exercises in 34-45
2.2 Exercises 3-5, 7, 8, 12, 13, 17, 20, , and your choice of three exercises in 44-50
2.3 Exercises 1-4, 7-12, 15-18, 21, 25, 26, 32, and your choice of three exercises in 39-48
2.4 Exercises 1, 5-8, 15, 17, 18, 20, 21, 27, 30, and your choice of three exercises in 43-53
2.5 Exercises 1-6, 11, 12, 21, 25, 27, 28, 33, 34, 36, 37, 41, 52, 53, and your choice of three exercises in 59-65
2.6 Exercises 3, 4, 6, and your choice of five exercises in 11-29
Chapter review exercises 5-7, 19,25,35, 41, 44, 47, and your choice of three exercises in 66-77
3.1 Exercises 1-3, 5, 6, 11, 12, 16, 29-32, and your choice of two exercises in 38-44
3.2 Exercises 1, 4, 7, 10, 13
3.3 Exercises your choice of four exercises in 7-27
Chapter 3 review exercises 15-17, 21-24,29-31
5.1 Exercises 6-9, 14-16, 20, 22, 25, 29, 30, and your choice of three exercises in 49-77
5.2 Exercises 7, 8, 15, 18, 21, 24, 25, 28, 32, 33, 36-39, 42, 44, 45, and your choice of
three exercises in 48-69
5.3 Exercises 17-26, 29-31,-36, 41 ,42,45 and your choice of three exercises in 47-68
Chapter review exercises 12, 16-18, 21-22,28-29,, 39-41, 45, 49,55,and your
choice of two exercises in 64-81
5.1 Exercises 6-9, 14-16, 20, 22, 25, 29, 30, and your choice of three exercises in 49-77
5.2 Exercises 7, 8, 15, 18, 21, 24, 25, 28, 32, 33, 36-39, 42, 44, 45, and your choice of
three exercises in 48-69
5.3 Exercises 17-26, 29-31,-36, 41 ,42,45 and your choice of three exercises in 47-68
Chapter review exercises 12, 16-18, 21-22,28-29,, 39-41, 45, 49,55,and your
choice of two exercises in 64-81
8.1 Exercises 1,4,5,13,14,22
8.2 Exercises 2,4,9,11
8.3 Exercises 1,4,5,13,19,20,24
8.4 Exercises 1,3,5,10,15,47
Chapter Review exercises 13,19,22,28,36,37
9.1 Exercises 7, 12, 17, 19, 25,26
9.2 Exercises 3, 7-8, and your choice of three exercises in 25-35
9.3 Exercises 1, 2, Explain in detail what a normal distribution is and how it used.
Subject Mathematics Due By (Pacific Time) 08/05/2013 08:00 pm
TutorRating
pallavi
Chat Now!
out of 1971 reviews
amosmm
Chat Now!
out of 766 reviews
PhyzKyd
Chat Now!
out of 1164 reviews
rajdeep77
Chat Now!
out of 721 reviews
sctys
Chat Now!
out of 1600 reviews
Chat Now!
out of 770 reviews
topnotcher
Chat Now!
out of 766 reviews
XXXIAO
Chat Now!
out of 680 reviews
| 1,045 | 2,476 |
{"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-2017-26
|
longest
|
en
| 0.899132 |
https://www.hackmath.net/en/math-problem/10711
| 1,603,581,141,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2020-45/segments/1603107885059.50/warc/CC-MAIN-20201024223210-20201025013210-00153.warc.gz
| 736,791,739 | 13,182 |
# Space diagonal
The space diagonal of a cube is 129.91 mm. Find the lateral area, surface area and the volume of the cube.
Correct result:
L = 5625 mm2
S = 33750 mm2
V = 421875 mm3
#### Solution:
We would be pleased if you find an error in the word problem, spelling mistakes, or inaccuracies and send it to us. Thank you!
Tips to related online calculators
Do you want to convert length units?
Tip: Our volume units converter will help you with the conversion of volume units.
Pythagorean theorem is the base for the right triangle calculator.
#### You need to know the following knowledge to solve this word math problem:
We encourage you to watch this tutorial video on this math problem:
## Next similar math problems:
• Cube diagonals
Determine the volume and surface area of the cube if you know the length of the body diagonal u = 216 cm.
The height of a regular quadrilateral prism is v = 10 cm, the deviation of the body diagonal from the base is 60°. Determine the length of the base edges, the surface, and the volume of the prism.
• Angle of diagonal
Angle between the body diagonal of a regular quadrilateral and its base is 60°. The edge of the base has a length of 10cm. Calculate the body volume.
• Rectangular trapezoid
The ABCD rectangular trapezoid with the AB and CD bases is divided by the diagonal AC into two equilateral rectangular triangles. The length of the diagonal AC is 62cm. Calculate trapezium area in cm square and calculate how many differs perimeters of the
• Hexagon cut pyramid
Calculate the volume of a regular 6-sided cut pyramid if the bottom edge is 30 cm, the top edge us 12 cm, and the side edge length is 41 cm.
• Prism
Right angle prism, whose base is right triangle with leg a = 3 cm and hypotenuse c = 13 cm has same volume as a cube with an edge length of 3 dm. a) Determine the height of the prism b) Calculate the surface of the prism c) What percentage of the cube's s
• Cubes
One cube is inscribed sphere and the other one described. Calculate difference of volumes of cubes, if the difference of surfaces in 257 mm2.
• Ratio of edges
The dimensions of the cuboid are in a ratio 3: 1: 2. The body diagonal has a length of 28 cm. Find the volume of a cuboid.
• Area of iso-trap
Find the area of an isosceles trapezoid if the lengths of its bases are 16 cm and 30 cm, and the diagonals are perpendicular to each other.
• Cube wall
Calculate the cube's diagonal diagonal if you know that the surface of one wall is equal to 36 centimeters square. Please also calculate its volume.
| 624 | 2,543 |
{"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-2020-45
|
longest
|
en
| 0.856263 |
https://puzzle.queryhome.com/23204/can-you-solve-it-9-3-1-3-1
| 1,527,135,613,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2018-22/segments/1526794865913.52/warc/CC-MAIN-20180524033910-20180524053910-00531.warc.gz
| 646,373,531 | 30,845 |
# can you solve it 9-3/(1/3)+1=??
41 views
can you solve it
9 - 3 / (1/3) + 1 = ??
posted Sep 13, 2017
``````9 - 3 / (1/3) + 1
``````
= (9) - (3*3) + 1
= 1.
answer Sep 13, 2017
Similar Puzzles
? + ? + ? + ? + ? = 30
You have to fill numbers in place of question marks. The numbers that you can use are 1, 3, 5, 7, 9, 11, 13 and 15. You can repeat the numbers if you like.
[] + [] + [] = 30
Fill the boxes using one of these numbers 1, 3, 5, 7, 9, 11, 13, 15
You can also repeat the numbers.
The question given in the picture was asked from Rahul in his aptitude class.
Can you solve it?
+1 vote
Can you replace each alphabet with the number (1 - 9) to make below equation correct ?
AB * C = DE - F = GH / I
+1 vote
Try these calculations: 1 x 9 + 2 = , 12 x 9 + 3 = , 123 x 9 + 4 =, ... continue the pattern and see if you can predict the answers before calculating.
| 326 | 885 |
{"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}
| 3.546875 | 4 |
CC-MAIN-2018-22
|
latest
|
en
| 0.915958 |
https://www.esaral.com/q/solve-this-following-57678
| 1,716,130,480,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2024-22/segments/1715971057788.73/warc/CC-MAIN-20240519132049-20240519162049-00345.warc.gz
| 666,574,701 | 11,632 |
Solve this following
Question:
Let $\mathrm{M}$ be any $3 \times 3$ matrix with entries from the set $\{0,1,2\}$. The maximum number of such matrices, for which the sum of diagonal elements of $\mathrm{M}^{\mathrm{T}} \mathrm{M}$ is seven, is
Solution:
$\left[\begin{array}{lll}a & b & c \\ d & e & f \\ g & h & i\end{array}\right]\left[\begin{array}{lll}a & d & g \\ b & e & h \\ c & f & i\end{array}\right]$
$a^{2}+b^{2}+c^{2}+d^{2}+e^{2}+f^{2}+g^{2}+h^{2}+i^{2}=7$
Case-I : Seven (1's) and two ( 0 's)
${ }^{9} \mathrm{C}_{2}=36$
Case-II : One $(2)$ and three ( 1 's) and five $(0$ 's $)$
$\frac{9 !}{5 ! 3 !}=504$
$\therefore$ Total $=540$
| 268 | 653 |
{"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}
| 3.75 | 4 |
CC-MAIN-2024-22
|
latest
|
en
| 0.560117 |
http://www.ck12.org/algebra/Input-Output-Tables-for-Function-Rules/lesson/Evaluating-Function-Rules/r8/
| 1,448,608,054,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2015-48/segments/1448398448227.60/warc/CC-MAIN-20151124205408-00094-ip-10-71-132-137.ec2.internal.warc.gz
| 355,702,580 | 41,117 |
<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />
You are viewing an older version of this Concept. Go to the latest version.
# Input-Output Tables for Function Rules
## Use function rules to complete patterns in tables.
0%
Progress
Practice Input-Output Tables for Function Rules
Progress
0%
Evaluating Function Rules
Have you ever been bowling? Take a look at this field trip dilemma.
The seventh grade class is planning a field trip. There are two proposals, one to a bowling alley and one to the Omni Theater. To make the best choice, the students have to do some research.
Casey and his friend Max are in charge of researching different bowling alleys to find the best price. They find out that the local bowling alley that is closest to school has a very good offer. This bowling alley charges a flat fee for shoes and then a fee per game.
“I wonder how many games we will have time for,” Casey said to Max.
“I don’t know, but that will impact the cost,” Max responded.
“Let’s figure it out. What is the flat fee for shoes?” Casey asked.
“It is $2.00 and it is$3.00 per game,” Max said.
The two boys took out a piece of paper and began to figure out how much the total would be based on games.
To solve this problem, you will need to understand functions. A function is when one variable is impacted by another. In this case, there is a fee for shoes and a fee per game. The total cost per student will depend on the number of games. The cost is a function of the games. Learn all that you can and you will be able to figure out the fees at the end of the Concept.
### Guidance
Do you remember how to identify a function,a relation,a range and a domain?
A function is a relation in which each member of the domain is paired with exactly one member of the range. In other words, a number in the domain cannot have two values for the range. When we look at the values in the domain and the range, we can figure out if the relation is a function or not.
A set of ordered pairs is a relation.
The values in the domain and range help us to understand a relation.
The domain is made up of the values in first column or the x\begin{align*}x\end{align*} coordinate in the relation. The range is made up of the second column or the y\begin{align*}y\end{align*} value of the relation.
One of the great things about functions is that they can be applied to all kinds of situations. Just remember that in order for a relation to be a function, that the values of the domain need to be assigned to only one value of the range. One way of thinking about functions is through the use of function tables.
A function table is an input/output table where the input is the domain and the output is the range.
Look at this table below.
Input Output
3 6
4 8
5 11
6 12
Here is function where we have an input and an output. The input is the domain and the output is the range. If we were going to write this function as ordered pairs, we would use the input for the x\begin{align*}x\end{align*} value and the output as the y\begin{align*}y\end{align*} value.
Let’s write out this relation: (3, 6) (4, 8) (5, 10) (6, 12). There is a relationship between the values of the domain and the values of the range. We can say that the range was created when some operation or operations was completed with the domain value.
The output is a result of an operation to the input!
What happened to the input to equal the output?
If you think about this, you will see that the x\begin{align*}x\end{align*} value was multiplied by 2 or doubled to equal the y\begin{align*}y\end{align*} value. We can write this as an equation.
y=2x\begin{align*}y = 2x \end{align*}
This says that the value of y\begin{align*}y\end{align*} is created whenever the x\begin{align*}x\end{align*} value is multiplied by 2.
This is called a Function Rule. It can be written in words or in the form of an equation. The function rule tells you what operation or operations to perform with the input to get the output.
Now that you understand how to identify a function rule, let's look at applying this information.
Use the function rule 3x\begin{align*}3x\end{align*} to evaluate the given inputs and complete each output.
Input Output
2
3
4
5
Now not all inputs will be this simple, but it will give you some practice applying a function rule. We know that the function rule is 3x\begin{align*}3x\end{align*}, so we can take each value from the input column and multiply it by 3. This will give us the correct value for the output table.
Input Output
2 6
3 9
4 12
5 15
You can see that the function rule was applied to each input value and the resulting output values complete the table.
Look at each list of values. Write the output for each list by using 2x2\begin{align*}2x-2\end{align*} as the function rule.
#### Example A
4,5,7,9\begin{align*}4, 5, 7, 9\end{align*}
Solution: 6,8,12,16\begin{align*}6, 8, 12, 16\end{align*}
#### Example B
3,4,5,7\begin{align*}-3, 4, -5, 7\end{align*}
Solution: 8,6,12,12\begin{align*}-8, 6, -12, 12\end{align*}
#### Example C
1,9,11,12\begin{align*}-1, -9, 11, 12\end{align*}
Solution: 4,20,20,22\begin{align*}-4, -20, 20, 22\end{align*}
Now let's go back to the dilemma at the beginning of the Concept.
The first thing to do is to write an equation that represents the given information. We know that each game is $3.00 and the flat rate for shoes is$2.00. The varying value is the cost and that is impacted by the number of games played. The number of games played is our variable.
C(g)=3g+2\begin{align*}C(g) = 3g + 2\end{align*}
This equation means that c\begin{align*}c\end{align*} the cost is a function of the number of games plus the $2 shoe fee. Games Played Cost ($)
2 8
4 14
5 17
6 20
The data show that when the number of games increases by 2, the cost increases by $6. Based on the number of games played, the cost could be anywhere from$8 to \$20.00 although it is unlikely that any student would have time for 6 games.
### Vocabulary
Relation
a set of ordered pairs.
Domain
the x\begin{align*}x\end{align*} value in a table or function.
Range
the y\begin{align*}y\end{align*} value in a table or function.
Function
Each value in the domain is connected to only one value in the range.
Function Rule
the operation or operations performed on the input value which then equals the output value.
Input
the x\begin{align*}x\end{align*} value or the domain of a function.
Output
the y\begin{align*}y\end{align*} value or the range of a function.
### Guided Practice
Here is one for you to try on your own.
Use the function rule 2x+1\begin{align*}2x + 1\end{align*} to evaluate each input value. Complete the given table.
Input Output
2\begin{align*}-2\end{align*}
1\begin{align*}-1\end{align*}
0\begin{align*}{\color{white}-}0\end{align*}
1\begin{align*}{\color{white}-}1\end{align*}
2\begin{align*}{\color{white}-}2\end{align*}
Solution
This table has negative and positive input values, but we will follow the same procedure. Simply substitute each x\begin{align*}x\end{align*} value into the function rule and evaluate for the output value.
2(2)+12(1)+12(0)+12(1)+12(2)+1=4+1=3=2+1=1=0+1=1=2+1=3=4+1=5
Now we can substitute those values into the output column of our function.
Input Output
2\begin{align*}-2\end{align*} \begin{align*}\boldsymbol{-3}\end{align*}
\begin{align*}-1\end{align*} \begin{align*}\boldsymbol{-1}\end{align*}
\begin{align*}{\color{white}-}0\end{align*} \begin{align*}{\color{white}-}\boldsymbol{0}\end{align*}
\begin{align*}{\color{white}-}1\end{align*} \begin{align*}{\color{white}-}\boldsymbol{1}\end{align*}
\begin{align*}{\color{white}-}2\end{align*} \begin{align*}{\color{white}-}\boldsymbol{3}\end{align*}
This is the answer and our work is complete.
### Practice
Directions:For numbers 1 - 5, find each output if the function rule is \begin{align*}3x+2\end{align*}
Input Output
3
5
6
9
11
Directions:For numbers 6 - 8, find each output if the function rule is \begin{align*}4x\end{align*}
Input Output
0
1
2
Directions:For numbers 9 - 13, find each output if the function rule is \begin{align*}-3x\end{align*}
Input Output
4
5
7
9
10
Directions: Answer each question about functions.
1. A pastry chef needs to purchase enough dough for her cookies. She buys one pound of dough for every twenty cookies she is going to make. She uses the function \begin{align*}d(c)=\frac{c}{20}\end{align*} where \begin{align*}c\end{align*} is the number of cookies and \begin{align*}d\end{align*} is the pounds of dough she should buy. Identify which variable is the domain and which is the range.
2. Evaluate the function \begin{align*}f(x)=2x+7\end{align*} when the domain is {-3, -1, 1, 3}.
3. Evaluate the function \begin{align*}f(x)=\frac{2}{5}x-6\end{align*} when the domain is {-10, -5, 0, 5, 10}.
4. Evaluate the function \begin{align*}f(x)=3x-1\end{align*} when the domain is {5, 6, 7, 8, 9}.
5. Evaluate the function \begin{align*}f(x)=x-9\end{align*} when the domain is {1, 2, 3, 4, 5}.
### Vocabulary Language: English
domain
domain
The domain of a function is the set of $x$-values for which the function is defined.
Function
Function
A function is a relation where there is only one output for every input. In other words, for every value of $x$, there is only one value for $y$.
Function Rule
Function Rule
A function rule describes how to convert an input value ($x$) into an output value ($y$) for a given function. An example of a function rule is $f(x) = x^2 + 3$.
input
input
The input of a function is the value on which the function is performed (commonly the $x$ value).
Linear Function
Linear Function
A linear function is a relation between two variables that produces a straight line when graphed.
Output
Output
The output of a function is the result of the operations performed on the independent variable (commonly $x$). The output values are commonly the values of $y$ or $f(x)$.
Range
Range
The range of a function is the set of $y$ values for which the function is defined.
Relation
Relation
A relation is any set of ordered pairs $(x, y)$. A relation can have more than one output for a given input.
Please wait...
Please wait...
| 2,948 | 10,248 |
{"found_math": true, "script_math_tex": 51, "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": 12, "texerror": 0}
| 4.4375 | 4 |
CC-MAIN-2015-48
|
longest
|
en
| 0.94463 |
https://btctopvywx.web.app/rodregez80987zo/how-to-calculate-discount-rate-for-present-value-in-excel-904.html
| 1,623,753,435,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2021-25/segments/1623487620971.25/warc/CC-MAIN-20210615084235-20210615114235-00229.warc.gz
| 155,421,923 | 6,101 |
## How to calculate discount rate for present value in excel
Discount Factor Formula – Example #2. We have to calculate net present value and discount factor for a period of 7 months, the discount rate for same is 8% and
How to Calculate Present Value Using Excel or a Financial Calculator If you forget this, you will end up grossly under-calculating the interest rate used in the calculation.) 4. Finally, enter the future value amount (\$1,000) and press the [FV] key. 5. Now you are ready to command the calculator to solve for present value. To calculate PV Like the future value calculations in Excel, when you are calculating present value to need to ensure that all the time periods are consistent. This means that you will need to divide the annual interest rate by the number of compounding periods in the year. Present Value Function Syntax: The syntax for present value in excel is Formula to Calculate Discount Factor. The formula of discount factor is similar to that of the present value of money and is calculated by adding the discount rate to one which is then raised to the negative power of a number of periods. The formula is adjusted for the number of compounding during a year. Mathematically, it is represented as below, The definition of a discount rate depends the context, it's either defined as the interest rate used to calculate net present value or the interest rate charged by the Federal Reserve Bank. There are two discount rate formulas you can use to calculate discount rate, WACC (weighted average cost of capital) and APV (adjusted present value).
## The reverse operation—evaluating the present value of a future amount of money —is called discounting (how much In Microsoft Excel, there are present value functions for single payments - "=NPV(. Programs will calculate present value flexibly for any cash flow and interest rate, or for a
The reverse operation—evaluating the present value of a future amount of money —is called discounting (how much In Microsoft Excel, there are present value functions for single payments - "=NPV(. Programs will calculate present value flexibly for any cash flow and interest rate, or for a This function allows you to calculate the present value of a simple annuity. What is the present value of receiving £10,000 in 4 years time if the discount rate is Calculate a discount rate over time, using Excel: The discount rate is either the interest rate that is used to determine the net present value (NPV) of future cash We need to calculate the present value (the value at time period 0) of receiving a single amount of \$1,000 in 20 years. The interest rate for discounting the future
### Discount Factor Formula – Example #2. We have to calculate net present value and discount factor for a period of 7 months, the discount rate for same is 8% and
31 Oct 2019 The Excel PV function is used to calculate the present value of a lump If the discount rate is 9%, what lump sum would need to be paid today
### How to Calculate Present Value Using Excel or a Financial Calculator If you forget this, you will end up grossly under-calculating the interest rate used in the calculation.) 4. Finally, enter the future value amount (\$1,000) and press the [FV] key. 5. Now you are ready to command the calculator to solve for present value. To calculate PV
29 May 2013 That rate of return would be your discount rate to use for future cash flows of the rental property. Determining Excel Present Value. To get the 23 Sep 2019 The Excel RATE function can be used instead of the lump sum discount rate formula, and has the syntax shown below. RATE(n, pmt, PV, FV, type, 31 Oct 2019 The Excel PV function is used to calculate the present value of a lump If the discount rate is 9%, what lump sum would need to be paid today 15 Apr 2019 This spreadsheet technique could prove helpful when determining the value Adding up all these positive and negative present values provides a net total: the NPV. This discount rate may be a mix of both debt and equity.
## 31 Oct 2019 The Excel PV function is used to calculate the present value of a lump If the discount rate is 9%, what lump sum would need to be paid today
1 Mar 2018 The PV function in Excel allows users to determine how much future assuming the cost of capital (i.e., the annual discount interest rate) is 6%. The discount rate is the interest rate used when calculating the net present value (NPV) of something. NPV is a core component of corporate budgeting and is a comprehensive way to calculate The discount factor is a factor by which future cash flow is multiplied to discount it back to the present value. The discount factor effect discount rate with increase in discount factor, compounding of the discount rate builds with time. One can calculate the present value of each cash flow while doing calculation manually of the discount factor.
Most financial analysts never calculate the net present value by hand nor with a calculator, instead, they use Excel. =NPV(discount rate, series of cash flow) (See screenshots below) Example of how to use the NPV function: Step 1: Set a discount rate in a cell. Step 2: Establish a series of cash flows (must be in consecutive cells). The present value calculations on this page are applied to investments for which interest is compounded in each period of the investment. However if you are supplied with a stated annual interest rate, and told that the interest is compounded monthly, you will need to convert the annual interest rate to a monthly interest rate and the number of periods into months: Applying Discount Rates. To apply a discount rate, multiply the factor by the future value of the expected cash flow. For example, if you expect to receive \$4,000 in one year and the discount rate is 95 percent, the present value of the cash flow is \$3,800. Keep in mind that cash flows at different time intervals all have different discount rates.
| 1,272 | 5,944 |
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 3.796875 | 4 |
CC-MAIN-2021-25
|
latest
|
en
| 0.930668 |
https://courses.lumenlearning.com/precalcone/chapter/solutions-24/
| 1,643,220,442,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2022-05/segments/1642320304959.80/warc/CC-MAIN-20220126162115-20220126192115-00478.warc.gz
| 238,970,248 | 12,163 |
## Solutions to Try Its
1. a. The exponential regression model that fits these data is $y=522.88585984{\left(1.19645256\right)}^{x}$.
b. If spending continues at this rate, the graduate’s credit card debt will be \$4,499.38 after one year.
2. a. The logarithmic regression model that fits these data is $y=141.91242949+10.45366573\mathrm{ln}\left(x\right)$
b. If sales continue at this rate, about 171,000 games will be sold in the year 2015.
3. a. The logistic regression model that fits these data is $y=\frac{25.65665979}{1+6.113686306{e}^{-0.3852149008x}}$.
b. If the population continues to grow at this rate, there will be about 25,634 seals in 2020.
c. To the nearest whole number, the carrying capacity is 25,657.
## Solutions to Odd-Numbered Exercises
1. Logistic models are best used for situations that have limited values. For example, populations cannot grow indefinitely since resources such as food, water, and space are limited, so a logistic model best describes populations.
3. Regression analysis is the process of finding an equation that best fits a given set of data points. To perform a regression analysis on a graphing utility, first list the given points using the STAT then EDIT menu. Next graph the scatter plot using the STAT PLOT feature. The shape of the data points on the scatter graph can help determine which regression feature to use. Once this is determined, select the appropriate regression analysis command from the STAT then CALC menu.
5. The y-intercept on the graph of a logistic equation corresponds to the initial population for the population model.
7. C
9. B
11. $P\left(0\right)=22$ ; 175
13. $p\approx 2.67$
15. y-intercept: $\left(0,15\right)$
17. 4 koi
19.
21. 10 wolves
23. about 5.4 years.
25.
27. $f\left(x\right)=776.682{e}^{0.3549x}$
29. When $f\left(x\right)=4000$, $x\approx 4.6$.
31. $f\left(x\right)=731.92{\left(0.738\right)}^{x}$
33.
35.
37. $f\left(10\right)\approx 9.5$
39. When $f\left(x\right)=7$, $x\approx 2.7$.
41. $f\left(x\right)=7.544 - 2.268\mathrm{ln}\left(x\right)$
43.
45.
47.
49. When $f\left(x\right)=12.5$, $x\approx 2.1$.
51. $f\left(x\right)=\frac{136.068}{1+10.324{e}^{-0.480x}}$
55. Working with the left side of the equation, we see that it can be rewritten as $a{e}^{-bt}$:
57. $\frac{c-{P}_{0}}{{P}_{0}}{e}^{-bt}=\frac{c-\frac{c}{1+a}}{\frac{c}{1+a}}{e}^{-bt}=\frac{\frac{c\left(1+a\right)-c}{1+a}}{\frac{c}{1+a}}{e}^{-bt}=\frac{\frac{c\left(1+a - 1\right)}{1+a}}{\frac{c}{1+a}}{e}^{-bt}=\left(1+a - 1\right){e}^{-bt}=a{e}^{-bt}$
Thus, $\frac{c-P\left(t\right)}{P\left(t\right)}=\frac{c-{P}_{0}}{{P}_{0}}{e}^{-bt}$.
59. First rewrite the exponential with base e: $f\left(x\right)=1.034341{e}^{0.247800x}$. Then test to verify that $f\left(g\left(x\right)\right)=x$, taking rounding error into consideration:
$\begin{cases}g\left(f\left(x\right)\right)\hfill & =4.035510\mathrm{ln}\left(1.034341{e}^{\text{0}\text{.247800x}}\right)-0.136259\hfill \\ \hfill & =4.03551\left(\mathrm{ln}\left(1.034341\right)+\mathrm{ln}\left({e}^{\text{0}\text{.2478}x}\right)\right)-0.136259\hfill \\ \hfill & =4.03551\left(\mathrm{ln}\left(1.034341\right)+\text{0}\text{.2478}x\right)-0.136259\hfill \\ \hfill & =0.136257+0.999999x - 0.136259\hfill \\ \hfill & =-0.000002+0.999999x\hfill \\ \hfill & \approx 0+x\hfill \\ \hfill & =x\hfill \end{cases}$
61. The graph of $P\left(t\right)$ has a y-intercept at (0, 4) and horizontal asymptotes at y = 0 and y = 20. The graph of ${P}^{-1}\left(t\right)$ has an x– intercept at (4, 0) and vertical asymptotes at x = 0 and x = 20.
| 1,285 | 3,573 |
{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 4.1875 | 4 |
CC-MAIN-2022-05
|
latest
|
en
| 0.698637 |
https://artofproblemsolving.com/wiki/index.php?title=2021_AIME_II_Problems/Problem_12&diff=149769&oldid=149768
| 1,627,105,527,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2021-31/segments/1627046150129.50/warc/CC-MAIN-20210724032221-20210724062221-00546.warc.gz
| 130,152,758 | 12,757 |
# Difference between revisions of "2021 AIME II Problems/Problem 12"
## Problem
A convex quadrilateral has area $30$ and side lengths $5, 6, 9,$ and $7,$ in that order. Denote by $\theta$ the measure of the acute angle formed by the diagonals of the quadrilateral. Then $\tan \theta$ can be written in the form $\tfrac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. Find $m + n$.
## Solution 1
We denote by $A$, $B$, $C$ and $D$ four vertices of this quadrilateral, such that $AB = 5$, $BC = 6$, $CD = 9$, $DA = 7$. We denote by $E$ the point that two diagonals $AC$ and $BD$ meet at. To simplify the notation, we denote $a = AE$, $b = BE$, $c = CE$, $d = DE$. We denote $\theta = \angle AED$.
First, we use the triangle area formula with sines to write down an equation of the area of the quadrilateral $ABCD$. We have ${\rm Area} \ ABCD = {\rm Area} \ \triangle ABE + {\rm Area} \ \triangle BCE + {\rm Area} \ \triangle CDE + {\rm Area} \ \triangle DAE = \frac{1}{2} ab \sin \angle AEB + \frac{1}{2} bc \sin \angle BEC + \frac{1}{2} cd \sin \angle CED + \frac{1}{2} da \sin \angle DEA = \frac{1}{2} \left( ab + bc + cd + da \right) \sin \theta$.
Because ${\rm Area} \ ABCD = 30$, we have $\left( ab + bc + cd + da \right) \sin \theta = 60$. We index this equation as Eq (1).
Second, we use the law of cosines to establish four equations for four sides of the quadrilateral $ABCD$.
By applying the law of cosines to $\triangle AEB$, we have $a^2 + b^2 - 2 a b \cos \angle AEB = AB^2 = 5^2$. Note that $\cos \angle AEB = \cos \left( 180^\circ - \theta \right) = \cos \theta$.
Hence, $a^2 + b^2 + 2 a b \cos \theta = 5^2$. We index this equation as Eq (2).
Analogously, we can establish the following equation for $\triangle BEC$ that $b^2 + c^2 - 2 b c \cos \theta = 6^2$ (indexed as Eq (3)),
the following equation for $\triangle CED$ that $c^2 + d^2 + 2 c d \cos \theta = 9^2$ (indexed as Eq (4)),
and the following equation for $\triangle DEA$ that $d^2 + a^2 - 2 d a \cos \theta = 7^2$ (indexed as Eq (5)).
By taking Eq (2) - Eq (3) + Eq (4) - Eq (5) and dividing both sides of the equation by 2, we get $\left( ab + bc + cd + da \right) \cos \theta = \frac{21}{2}$. We index this equation as Eq (6).
By taking $\frac{{\rm Eq} \ (1)}{{\rm Eq} \ (6)}$, we get $\tan \theta = \frac{60}{21/2} = \frac{40}{7}$.
Therefore, by writing this answer in the form of $\frac{m}{n}$, we have $m = 40$ and $n = 7$. Therefore, the answer to this question is $m + n = 40 + 7 = 47$.
~ Steven Chen (www.professorchenedu.com)
## See also
2021 AIME II (Problems • Answer Key • Resources) Preceded byProblem 11 Followed byProblem 13 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 All AIME Problems and Solutions
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.
Invalid username
Login to AoPS
| 996 | 2,904 |
{"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": 47, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 4.84375 | 5 |
CC-MAIN-2021-31
|
latest
|
en
| 0.777616 |
https://physicshelpforum.com/threads/gravity-generator.9888/page-2
| 1,581,923,536,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2020-10/segments/1581875141749.3/warc/CC-MAIN-20200217055517-20200217085517-00337.warc.gz
| 539,709,102 | 22,448 |
# Gravity Generator
#### tonk
So if you’re game, here’s where it gets complicated….So let’s say you have 1338 cubic foot pressure tank and you have added 3315 cubic feet of air into it. The tank is now at 2.47 atmospheres or right around 36 psi. So now we want to add 255 more cubic feet of air into the tank. But instead of using a motorized compressor, we want to use a gravity compressor of sorts.
Let’s say you have a mass that weighs 15,329 lbs. or 6,953.11 Kg and you have ten feet of fall to play with. Between the air intake of the tank and the weight, you have either one very large piston, or a series of smaller pitons. The chamber(s) of the one piston or multiple pistons shall equal 255 cubic feet. You will be using the weight to drive the piston and force the air into the pressure tank (which is already exerting 36 psi against you). You can use the 10 foot fall any way you wish. You can have the weight drop in ten 1 foot increments or 2 five foot drops, or one long 10 foot drop whichever is optimal. You could have the weight free fall for 5 or 9 feet before hitting the piston adding additional force.
I would like to know how much of the 255 cubic feet of air could you force into that pressure tank given these parameters to work with. Also I would like to know if you changed the size of the tank, say you made it half the size: 669 cubic feet and you add 1657.5 cubic feet to get your 2.47 atmospheres and 36 PSI. Would the amount of air you could force into the vessel increase, decrease or remain the same as in the previous example?
#### ChipB
PHF Helper
Given 255 ft^3 of air and a piston 10ft long, the area lf the piston would be 25.5 ft^2. If the piston has a weight of 15,339 pounds driving it, that's 15329/25.5 = 601 psi of pressure it exerts.
** EDIT - This is incorrect - it needs to be converted from pounds/sq ft to pounds/sq inch. See post #14 below. **
It would have no problem driving against the 36 PSI resistance of the air already in the tank. Consequently, it can force the full 255 ft^3 of air into the tank, raising its pressure another 0.19 atmospheres.
If the tank is half the size but still starting at 36 PSI, the piston would again have no problem driving an additional 255 ft^3 of air into it, although the pressure increase for each shot of air would be 2x0.19 = 0.38 atmospheres.
Last edited:
#### tonk
That was way better then I hoped. I'd like to modify the question. Let's say I doubled the amount I wanted to add to the tank by adding 510 cubic feet. I would like to know:
A) What would be the least amount of weight necessary to push that
(2x) piston, and
B) Is there a formula to predict the rate of descent e.g If the weight were
10,000 pounds, at what rate would the pistons fall?
#### ChipB
PHF Helper
I realize I made a big mistake in my last post. Apologies, but I should have converted ft^2 to in^2 to get the correct answer. The 15329 weight can exert a pressure of:
15329/25.5ft^2 x 1 ft^2/144in^2 = 4.18 PSI.
So unfortunately the weight can not compress any more air into the container if it's already at 36 PSI. You're going to have to be satisfied with a smaller area of the piston, which means either (a) less than 255 ft^3 of air being compressed, or (b) a longer stroke of the piston. To get the pressure the piston creates up to 36 PSI while limited to a ten-foot stroke its area must be (4.18/36) x 25.5 = 2.9 ft^2, and hence the volume of air is 2.9 ft^2 x 10ft = 29 ft^3.
Last edited:
#### tonk
[FONT="]Yea, that first answer made things way too easy. [/FONT][FONT="] It took me a few days, but I believe I have discovered a work around which leads me to another question. I did the best I could to use your formulas, but if you could confirm my calculations that would be great.[/FONT][FONT="] [/FONT][FONT="]Let’s say you have a 4500 gallon pressure tank or 602 cu/ft. There is 825 cu/ft of air in the tank which puts the pressure at 20.13 PSI. I want to add 300 cu/ft of air to the tank which will bring the pressure up to 27.48 PSI.[/FONT][FONT="] [/FONT][FONT="]Now let’s say that I have a piston which weighs 15329 LBS and is 26 inches in diameter. This means the piston should produce 28.87 PSI when used as a “gravity compressor”. Presuming these numbers are correct, the question is this: [/FONT][FONT="] [/FONT][FONT="]You have the pressure tank at 20.13 PSI. Coming out of the pressure tank you have a cylinder 26 inches in diameter that stands 100 feet high. The volume of air in the cylinder should be right at 307 cu/ft. At the top of the cylinder is your 15329 LBS piston. If you were to drop the piston, how long would it take to fall the 100 feet and compress all of the 307 cu/ft of air into the tank?[/FONT][FONT="]Would the fall of the piston have to be controlled? In other words would there be so much momentum from the weight that it would hit and damage the tank or would the pressure building up in the cylinder be able to control the descent of the piston. Is there a formula to calculate this which I can use to play around with different weights and heights?[/FONT][FONT="] [/FONT][FONT="]Cheers,[/FONT][FONT="] [/FONT][FONT="]Kenji
[/FONT]
Last edited:
#### MBW
The Combined Gas Law looks like a good starting point:
http://en.wikipedia.org/wiki/Combined_gas_law
Note the Temperature term, this means the gas will get hot as it is compressed.
This means you will have to thermally insulate the cylinder to maintain efficiency.
#### ChipB
PHF Helper
I am traveling and unable to provide a complete answer to your questions about time to fall 100 ft and the speed of impact. But the general approach is this:
1. Recognizing MBW's point about the air heating under compression, this will serve to raise the pressure above what the simple volume change calculation implies. However the math becomes very difficult, so instead let's assume that the walls of the cylinder are not well insulated, and that heat is allowed to escape so that the temp of the air stays constant. This will give a first order approximation, which would tend to under-estimate the time and over-estimate the final velocity.
2. The equation of motion to use is
Where a is acceleration, and equals sum of forces/m. The forces involved are gravity and the opposing pressure. Assuming a linear build up of pressure as the weight descends, you have P(x)= P_1 + Delta p (x/L), where P_1 is the initial pressure and Delta P is the change in pressure as x goes from 0 to 100 ft. So the equation becomes:
(Mgx -P_1Ax -Delta P Ax^2/2)/m = v^2/2
This yields an equation for v(x), and from that the final velocity is v(L) and the time to reach the bottom is found from integral of integral of adt from t= 0 to t= T is equal to v(L), so you can solve for T.
I suggest you try to solve this yourself. Otherwise give me a day or two to get back on this.
#### tonk
Greetings, It's been awhile. I have a question on water compressibility. Say you had an unpressurized 1000 gallon water tank completely full of water and you wanted to force 2 more gallon of water into the tank. How much pressure would be needed? What if it were 10000 gallons and you want to force in 5 gallons. I don't need the specific answers, I'm looking for a formula in which I can plug in different values. Thank you.
#### ChipB
PHF Helper
The general formula for compressing water (or anything else) into a given volume is:
s=Ee
where s (sigma) is the stress needed to be applied in terms of pressure (in units such as psi or Pa), E is the bulk modulus of the material, and e (epsilon) is the strain you want to achieve, in units of change of length per unit length. For water E = 2.2 x 10^9 Gpa. The strain you are wanting to achieve is equal to the added volume divided by the original volume, so for your first example it's 2/1000. Hence the pressure you need to apply is:
E= 2.2 x 10^9 Pa x (2/1000) = 4.4x10^6 Pa, or 4.4x10^6 N/m^2, which is equivalent to about 640 psi.
#### tonk
Thank you. I am assuming gases are much easier to compress. Could you tell me which gases compress the most easily. 640 psi is to high for what I am trying to do and I would like to mitigate that value. Using the same example of the 2 gallons into a 1000 gallon tank, what if inside that water tank there was a pliable sphere filled with air (or the best compressible gas) with a radius of 6 inches. If my math is correct, the sphere would have a volume of .523599 cu/ft or just under 4 gallons. Now how much pressure would you need to force those 2 gallons into the tank? How would the size of the sphere affect the pressure needed to insert the 2 gallons into the tank, in other words would larger spheres make it easier? I would also need the formula. I was also wondering, are you allowed to or available to consult on projects? Thanks Kenji
| 2,266 | 8,863 |
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 4 | 4 |
CC-MAIN-2020-10
|
latest
|
en
| 0.918476 |
http://gmatclub.com/forum/if-x-is-an-integer-and-y-3x-2-then-which-of-the-following-50404.html?fl=similar
| 1,427,517,599,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2015-14/segments/1427131297195.79/warc/CC-MAIN-20150323172137-00291-ip-10-168-14-71.ec2.internal.warc.gz
| 122,379,886 | 44,070 |
Find all School-related info fast with the new School-Specific MBA Forum
It is currently 27 Mar 2015, 20:39
### GMAT Club Daily Prep
#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.
Customized
for You
we will pick new questions that match your level based on your Timer History
Track
every week, we’ll send you an estimated GMAT score based on your performance
Practice
Pays
we will pick new questions that match your level based on your Timer History
# Events & Promotions
###### Events & Promotions in June
Open Detailed Calendar
# If x is an Integer and y=3x+2, then which of the following
Author Message
TAGS:
Manager
Joined: 03 Apr 2007
Posts: 55
Followers: 1
Kudos [?]: 10 [0], given: 1
If x is an Integer and y=3x+2, then which of the following [#permalink] 11 Aug 2007, 14:53
If x is an Integer and y=3x+2, then which of the following CANNOT be a divisor of y?
A. 4
B. 5
C. 6
D. 7
E. 8
Intern
Joined: 12 May 2007
Posts: 29
Followers: 1
Kudos [?]: 3 [0], given: 0
C
I just plugged numbers, there might be better explanation
series comes to a-6, a-3, a, a+3, a+9, a+12 where a=2
Intern
Joined: 10 Aug 2007
Posts: 30
Followers: 0
Kudos [?]: 1 [0], given: 0
I wonder if this makes sense:
Since y = 3x + 2, we know that dividing y by 3 will give a remainder of 2, hence it is not divisible by 3. Since anything that's divisible by 6 must be divisible by 3 and 2, this rules out 6 as a possible divisor of y.
Thoughts?
Director
Joined: 12 Jul 2007
Posts: 865
Followers: 12
Kudos [?]: 213 [0], given: 0
entranced wrote:
I wonder if this makes sense:
Since y = 3x + 2, we know that dividing y by 3 will give a remainder of 2, hence it is not divisible by 3. Since anything that's divisible by 6 must be divisible by 3 and 2, this rules out 6 as a possible divisor of y.
Thoughts?
That logic seems to hold up! I came up with C plugging in numbers myself.
Director
Joined: 31 Mar 2007
Posts: 586
Followers: 7
Kudos [?]: 41 [0], given: 0
Much easier/quicker to plug in numbers and do elimination.
Intern
Joined: 10 Aug 2007
Posts: 30
Followers: 0
Kudos [?]: 1 [0], given: 0
Much easier/quicker to plug in numbers and do elimination.
Agreed, but someone was unsure about a better explanation, so I thought I'd try to offer one.
Manager
Joined: 27 May 2007
Posts: 128
Followers: 1
Kudos [?]: 4 [0], given: 0
Actually I don't agree that it's easier to plug in numbers. It makes perfect sense that if a number isn't divisible by 3, it can't be divisible by 6. If I ran into this on the test, I would plug in numbers if I had plenty of time, but if I was pressed for time I'd confidently pick 6.
Similar topics Replies Last post
Similar
Topics:
10 If x is an integer and y = 3x + 2, which of the following 4 03 Mar 2014, 00:29
If x is an integer and y = 3x + 2, which of the following 2 08 Jun 2008, 20:06
If x is an integer and y = 3x + 2, which of the following 4 18 Feb 2008, 11:56
If x is an integer and y = 3x + 2, which of the following 4 01 Oct 2007, 01:45
If x is an integer and y= 3x+2, which of the following 3 14 Oct 2005, 13:04
Display posts from previous: Sort by
| 1,034 | 3,237 |
{"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-2015-14
|
longest
|
en
| 0.919566 |
https://www.teacherspayteachers.com/Product/Gumball-Number-Clip-Cards-0-10-3884292
| 1,537,406,072,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2018-39/segments/1537267156314.26/warc/CC-MAIN-20180919235858-20180920015858-00390.warc.gz
| 827,355,601 | 20,276 |
# Gumball Number Clip Cards 0-10
Subject
Resource Type
Common Core Standards
Product Rating
4.0
1 Rating
File Type
PDF (Acrobat) Document File
6 MB|22 pages
Share
Product Description
Gumball Number Clip Cards provide a hands-on approach in helping children with number and number word recognition along with cardinality.
How to Use This Resource:
❶Match the number, number word or quantity in the gumball machine to the correct number in the squares.
❷Clip the number with a clothespin or cover with an object or your choice.
This resource is a perfect way to review numbers in a math center, small group, 1-on-1 or independent setting! This would be a great addition to your preschool, pre-k, kindergarten, or special education class. Parents, this would also be a fun activity to incorporate in your tot schooling or homeschooling curriculum. Activity mats are easy to prep and can be used year after year!
This product contains the following in both color and black and white:
►(11) Set 1 cards match number to number
►(11) Set 2 cards match quantity to number
►(11) Set 3 cards match number word to number
You will need to print on cardstock and/or laminate all pages for longer durability.
Instructions for Assembly:
❶Print on cardstock and/or laminate for durability.
❷Cut apart clip cards.
Common Core Standards covered by this resource:
K.CC.A Know number names and the count sequence.
K.CC.A.1 Count to 100 by ones and by tens.
K.CC.B Count to tell the number of objects.
K.CC.B.4 Understand the relationship between numbers and quantities; connect counting to cardinality.
K.CC.B.4a When counting objects, say the number names in the standard order, pairing each object with one and only one number name and each number name with one and only one object.
K.CC.B.4b Understand that the last number name said tells the number of objects counted. The number of objects is the same regardless of their arrangement or the order in which they were counted.
K.CC.B.4c Understand that each successive number name refers to a quantity that is one larger.
K.CC.B.5 Count to answer "how many?" questions about as many as 20 things arranged in a line, a rectangular array, or a circle, or as many as 10 things in a scattered configuration; given a number from 1-20, count out that many objects.
Click the green preview button above to view sample product pages.
Tips for Customers
►Be sure you have the latest version of Adobe Acrobat to open this file. Get it completely free here.
►Click the to follow my store. You will be notified when new products are added and when my store goes on sale.
►If you enjoyed this product, don't forget to leave your positive feedback. By doing so you can earn TpT credits to use on future purchases. Find out more here.
►If you have any questions or concerns about the product, please contact me through the Product Q & A section below and I'll get back to you as soon as possible!
Total Pages
22 pages
N/A
Teaching Duration
N/A
Report this Resource
Teachers Pay Teachers is an online marketplace where teachers buy and sell original educational materials.
| 704 | 3,102 |
{"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-2018-39
|
latest
|
en
| 0.905852 |
http://www.ck12.org/algebra/Slope/?by=all&difficulty=all
| 1,476,993,496,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2016-44/segments/1476988717783.68/warc/CC-MAIN-20161020183837-00285-ip-10-171-6-4.ec2.internal.warc.gz
| 383,676,190 | 23,424 |
<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />
Slope
Understand slope as the steepness of a line.
Levels are CK-12's student achievement levels.
Basic Students matched to this level have a partial mastery of prerequisite knowledge and skills fundamental for proficient work.
At Grade (Proficient) Students matched to this level have demonstrated competency over challenging subject matter, including subject matter knowledge, application of such knowledge to real-world situations, and analytical skills appropriate to subject matter.
Advanced Students matched to this level are ready for material that requires superior performance and mastery.
Slope
Learn how to calculate the slope of a line.
MEMORY METER
This indicates how strong in your memory this concept is
8
Slope
Use the slope formula to find the slope of a line that passes through two points.
MEMORY METER
This indicates how strong in your memory this concept is
2
Slope
by CK-12 //basic
Learn to recognize the slope of a line as the ratio of vertical rise to the horizontal run and distinguish between types of slopes.
MEMORY METER
This indicates how strong in your memory this concept is
3
Finding the Slope of a Line
by CK-12 //basic
Learn to find the slope of a line.
MEMORY METER
This indicates how strong in your memory this concept is
0
Slope
Recognize the slope of a line as the ratio of vertical rise to the horizontal run
MEMORY METER
This indicates how strong in your memory this concept is
0
Introduction to Rate of Change
by Math Writing Team //at grade
Learn how to calculate the slope of a line.
MEMORY METER
This indicates how strong in your memory this concept is
0
Slope (Rise over Run practice)
by Math Writing Team //at grade
Use the slope formula to find the slope of a line that passes through two points.
MEMORY METER
This indicates how strong in your memory this concept is
0
Slope - Between two points
Use the slope formula to find the slope of a line that passes through two points.
MEMORY METER
This indicates how strong in your memory this concept is
0
Slope - Rise over Run
by Laura Warden //basic
Recognize the slope of a line as the ratio of vertical rise to the horizontal run
MEMORY METER
This indicates how strong in your memory this concept is
0
Slope
by Cynthia Rathjen //basic
Learn how to calculate the slope of a line.
MEMORY METER
This indicates how strong in your memory this concept is
0
Slope (3.2)
by Katie Ziegler //basic
Learn how to calculate the slope of a line.
MEMORY METER
This indicates how strong in your memory this concept is
0
Slope
by James Gates //basic
Recognize the slope of a line as the ratio of vertical rise to the horizontal run
MEMORY METER
This indicates how strong in your memory this concept is
0
Slope
Use the slope formula to find the slope of a line that passes through two points.
MEMORY METER
This indicates how strong in your memory this concept is
0
ACBA HIP-VAP Project
MEMORY METER
This indicates how strong in your memory this concept is
0
Slope
by EPISD Algebra 1 Team //at grade
Use the slope formula to find the slope of a line that passes through two points.
MEMORY METER
This indicates how strong in your memory this concept is
13
Linear Functions
by EPISD Precalculus Team //at grade
MEMORY METER
This indicates how strong in your memory this concept is
0
Slope
by Carol Vieira //basic
Learn how to calculate the slope of a line.
MEMORY METER
This indicates how strong in your memory this concept is
0
Finding the Slope of a Line
by Susan Sudberry //basic
Finding the slope of a line and identifying different types of slopes.
MEMORY METER
This indicates how strong in your memory this concept is
0
Rate of Change and Slope
by SFDR Algebra I //at grade
MEMORY METER
This indicates how strong in your memory this concept is
0
Slope
Use the slope formula to find the slope of a line that passes through two points.
MEMORY METER
This indicates how strong in your memory this concept is
0
Slope
by Laura Iossi //basic
Learn how to calculate the slope of a line.
MEMORY METER
This indicates how strong in your memory this concept is
0
Slope
by MCPS Math //basic
Learn how to calculate the slope of a line.
MEMORY METER
This indicates how strong in your memory this concept is
0
Find the Slope of a Linear Function
Learn to find the slope of a line.
MEMORY METER
This indicates how strong in your memory this concept is
1
Slope
Learn how to calculate the slope of a line.
MEMORY METER
This indicates how strong in your memory this concept is
0
• PLIX
A-Frame House
Slope Interactive
MEMORY METER
This indicates how strong in your memory this concept is
0
• PLIX
Slope: Hiking in the Woods
Slope Interactive
MEMORY METER
This indicates how strong in your memory this concept is
0
• PLIX
Slope: Danger Mountain
Slope Interactive
MEMORY METER
This indicates how strong in your memory this concept is
0
• PLIX
Slope: Hiking in the Woods
Slope Interactive
MEMORY METER
This indicates how strong in your memory this concept is
0
• Video
Classifying Slope - Overview
by CK-12 //basic
Overview
MEMORY METER
This indicates how strong in your memory this concept is
0
• Video
Finding Slope - Overview
by CK-12 //basic
Overview
MEMORY METER
This indicates how strong in your memory this concept is
1
• Video
Solving Problems Involving Slope - Overview
by CK-12 //basic
Overview
MEMORY METER
This indicates how strong in your memory this concept is
0
• Video
Slope and Rate of Change
by CK-12 //basic
Explains how to solve slope problems with examples.
MEMORY METER
This indicates how strong in your memory this concept is
2
• Video
Algebra: Slope
by CK-12 //basic
An explanation of slope and how to find it by Khan Academy.
MEMORY METER
This indicates how strong in your memory this concept is
3
• Video
Finding and Using Slope - Overview
by CK-12 //basic
Overview
MEMORY METER
This indicates how strong in your memory this concept is
0
• Video
Slope of a Line: A Sample Application
This video demonstrates a sample use of calculating the slope of a line.
MEMORY METER
This indicates how strong in your memory this concept is
0
• Video
Finding Slope - Example 3
by CK-12 //basic
Finding Vertical Slopes Using the Slope Formula Given Two Points
MEMORY METER
This indicates how strong in your memory this concept is
0
• Video
The Slope of a Line: An Explanation of the Concept
This video provides an explanation of the concept of calculating the slope of a line.
MEMORY METER
This indicates how strong in your memory this concept is
0
• Video
Finding Slope - Example 4
by CK-12 //basic
Finding Horizontal Slopes Using the Slope Formula Given Two Points
MEMORY METER
This indicates how strong in your memory this concept is
0
• Video
Solving Problems Involving Slope - Example 1
by CK-12 //basic
Finding Missing Coordinates Given Slope
MEMORY METER
This indicates how strong in your memory this concept is
0
• Video
The Slope Formula
by CK-12 //basic
How to use the slope formula to find the slope between two given points.
MEMORY METER
This indicates how strong in your memory this concept is
0
• Video
Solving Problems Involving Slope - Example 2
by CK-12 //basic
Solving Word Problems Involving Slope
MEMORY METER
This indicates how strong in your memory this concept is
0
• Video
Finding Slope from a Graph
MEMORY METER
This indicates how strong in your memory this concept is
0
• Video
Finding and Using Slope - Example 1
by CK-12 //basic
Tell whether the slope is positive, negative, zero, or undefined
MEMORY METER
This indicates how strong in your memory this concept is
0
• Video
Finding and Using Slope - Example 2
by CK-12 //basic
Estimate the slope of the line on a coordinate plane
MEMORY METER
This indicates how strong in your memory this concept is
0
• Video
Finding and Using Slope - Example 3
by CK-12 //basic
Find the slope and y-intercept of the equation
MEMORY METER
This indicates how strong in your memory this concept is
0
• Video
Finding and Using Slope - Example 4
by CK-12 //basic
Find the slope of the line passing through the given points
MEMORY METER
This indicates how strong in your memory this concept is
0
• Video
Classifying Slope (Identifying Slope) - Example 1
by CK-12 //basic
Identifying Slope as Positive, Negative, Zero, or Undefined
MEMORY METER
This indicates how strong in your memory this concept is
0
• Video
Finding Slope - Example 1
by CK-12 //basic
Finding Positive Slopes Using the Slope Formula Given Two Points
MEMORY METER
This indicates how strong in your memory this concept is
0
• Video
Finding Slope - Example 2
by CK-12 //basic
Finding Negative Slopes Using the Slope Formula Given Two Points
MEMORY METER
This indicates how strong in your memory this concept is
0
Slope Quiz
Quiz for Slope.
MEMORY METER
This indicates how strong in your memory this concept is
1
• 2
• Critical Thinking
Slope Discussion Questions
A list of student-submitted discussion questions for Slope.
MEMORY METER
This indicates how strong in your memory this concept is
0
• Real World Application
Walking Up Flights of Stairs
All students will be able to compare two (or more) flights of stairs to investigate which is steeper. Some students will be able to calculate the equations of the flights of stairs.
MEMORY METER
This indicates how strong in your memory this concept is
0
• Real World Application
Dashing Through the Snow
Find out what makes a hill good for sledding.
MEMORY METER
This indicates how strong in your memory this concept is
0
• Real World Application
Dashing Through the Snow
Find out what makes a hill good for sledding.
MEMORY METER
This indicates how strong in your memory this concept is
0
• Real World Application
An Icy Climb
Find out how ice climbing techniques depend on slope.
MEMORY METER
This indicates how strong in your memory this concept is
0
• Study Guide
Graphing Linear Equations Study Guide
This study guide goes over the basics of graphing linear equations: slope, intercept, slope-intercept form, standard form, point-slope form, and converting between different forms of writing linear equations.
MEMORY METER
This indicates how strong in your memory this concept is
3
• Study Guide
Slope
by Kevin Cui //basic
Learn what a slope is and how to find it.
MEMORY METER
This indicates how strong in your memory this concept is
0
• Flashcards
Slope Flashcards
by CK-12 //basic
MEMORY METER
This indicates how strong in your memory this concept is
0
• Flashcards
Finding the Slope of a Line
by Galaxy Portillo //basic
MEMORY METER
This indicates how strong in your memory this concept is
0
• Flashcards
Slope
by Kyle Bardman //basic
| 2,612 | 10,738 |
{"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-2016-44
|
longest
|
en
| 0.911819 |
http://nrich.maths.org/public/leg.php?code=57&cl=3&cldcmpid=5435
| 1,436,160,031,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2015-27/segments/1435375098059.60/warc/CC-MAIN-20150627031818-00301-ip-10-179-60-89.ec2.internal.warc.gz
| 200,835,787 | 10,154 |
# Search by Topic
#### Resources tagged with Sequences similar to Sums of Powers - A Festive Story:
Filter by: Content type:
Stage:
Challenge level:
### There are 53 results
Broad Topics > Sequences, Functions and Graphs > Sequences
### Sums of Powers - A Festive Story
##### Stage: 3 and 4
A story for students about adding powers of integers - with a festive twist.
### Series Sums
##### Stage: 4 Challenge Level:
Let S1 = 1 , S2 = 2 + 3, S3 = 4 + 5 + 6 ,........ Calculate S17.
### Double Trouble
##### Stage: 4 Challenge Level:
Simple additions can lead to intriguing results...
### Logoland - Sequences
##### Stage: 3 Challenge Level:
Make some intricate patterns in LOGO
### A Number Sequences Resource
##### Stage: 4 Challenge Level:
This resource contains interactive problems to support work on number sequences at Key Stage 4.
### Clock Squares
##### Stage: 3 Challenge Level:
Square numbers can be represented on the seven-clock (representing these numbers modulo 7). This works like the days of the week.
### Differs
##### Stage: 3 Challenge Level:
Choose any 4 whole numbers and take the difference between consecutive numbers, ending with the difference between the first and the last numbers. What happens when you repeat this process over and. . . .
### LOGO Challenge - Circles as Bugs
##### Stage: 3 and 4 Challenge Level:
Here are some circle bugs to try to replicate with some elegant programming, plus some sequences generated elegantly in LOGO.
### A Little Light Thinking
##### Stage: 4 Challenge Level:
Here is a machine with four coloured lights. Can you make two lights switch on at once? Three lights? All four lights?
### Squaresearch
##### Stage: 4 Challenge Level:
Consider numbers of the form un = 1! + 2! + 3! +...+n!. How many such numbers are perfect squares?
### Odds, Evens and More Evens
##### Stage: 3 Challenge Level:
Alison, Bernard and Charlie have been exploring sequences of odd and even numbers, which raise some intriguing questions...
### Big Powers
##### Stage: 3 and 4 Challenge Level:
Three people chose this as a favourite problem. It is the sort of problem that needs thinking time - but once the connection is made it gives access to many similar ideas.
### Loopy
##### Stage: 4 Challenge Level:
Investigate sequences given by $a_n = \frac{1+a_{n-1}}{a_{n-2}}$ for different choices of the first two terms. Make a conjecture about the behaviour of these sequences. Can you prove your conjecture?
### Tiny Nines
##### Stage: 4 Challenge Level:
Find the decimal equivalents of the fractions one ninth, one ninety ninth, one nine hundred and ninety ninth etc. Explain the pattern you get and generalise.
### Triangles and Petals
##### Stage: 4 Challenge Level:
An equilateral triangle rotates around regular polygons and produces an outline like a flower. What are the perimeters of the different flowers?
### Farey Sequences
##### Stage: 3 Challenge Level:
There are lots of ideas to explore in these sequences of ordered fractions.
### Triangles Within Pentagons
##### Stage: 4 Challenge Level:
Show that all pentagonal numbers are one third of a triangular number.
### On the Importance of Pedantry
##### Stage: 3, 4 and 5
A introduction to how patterns can be deceiving, and what is and is not a proof.
### Power Mad!
##### Stage: 3 and 4 Challenge Level:
Powers of numbers behave in surprising ways. Take a look at some of these and try to explain why they are true.
### Triangles Within Squares
##### Stage: 4 Challenge Level:
Can you find a rule which relates triangular numbers to square numbers?
### Triangles Within Triangles
##### Stage: 4 Challenge Level:
Can you find a rule which connects consecutive triangular numbers?
### Changing Places
##### Stage: 4 Challenge Level:
Place a red counter in the top left corner of a 4x4 array, which is covered by 14 other smaller counters, leaving a gap in the bottom right hand corner (HOME). What is the smallest number of moves. . . .
### Towers
##### Stage: 3 Challenge Level:
A tower of squares is built inside a right angled isosceles triangle. The largest square stands on the hypotenuse. What fraction of the area of the triangle is covered by the series of squares?
### Odd Differences
##### Stage: 4 Challenge Level:
The diagram illustrates the formula: 1 + 3 + 5 + ... + (2n - 1) = n² Use the diagram to show that any odd number is the difference of two squares.
### Sixational
##### Stage: 4 and 5 Challenge Level:
The nth term of a sequence is given by the formula n^3 + 11n . Find the first four terms of the sequence given by this formula and the first term of the sequence which is bigger than one million. . . .
### Happy Numbers
##### Stage: 3 Challenge Level:
Take any whole number between 1 and 999, add the squares of the digits to get a new number. Make some conjectures about what happens in general.
### LOGO Challenge - Sequences and Pentagrams
##### Stage: 3, 4 and 5 Challenge Level:
Explore this how this program produces the sequences it does. What are you controlling when you change the values of the variables?
### Stretching Fractions
##### Stage: 4 Challenge Level:
Imagine a strip with a mark somewhere along it. Fold it in the middle so that the bottom reaches back to the top. Stetch it out to match the original length. Now where's the mark?
### Shifting Times Tables
##### Stage: 3 Challenge Level:
Can you find a way to identify times tables after they have been shifted up?
### Lower Bound
##### Stage: 3 Challenge Level:
What would you get if you continued this sequence of fraction sums? 1/2 + 2/1 = 2/3 + 3/2 = 3/4 + 4/3 =
### Converging Means
##### Stage: 3 Challenge Level:
Take any two positive numbers. Calculate the arithmetic and geometric means. Repeat the calculations to generate a sequence of arithmetic means and geometric means. Make a note of what happens to the. . . .
### Charlie's Delightful Machine
##### Stage: 3 and 4 Challenge Level:
Here is a machine with four coloured lights. Can you develop a strategy to work out the rules controlling each light?
### Remainder
##### Stage: 3 Challenge Level:
What is the remainder when 2^2002 is divided by 7? What happens with different powers of 2?
### More Pebbles
##### Stage: 2 and 3 Challenge Level:
Have a go at this 3D extension to the Pebbles problem.
### Intersecting Circles
##### Stage: 3 Challenge Level:
Three circles have a maximum of six intersections with each other. What is the maximum number of intersections that a hundred circles could have?
### First Forward Into Logo 12: Puzzling Sums
##### Stage: 3, 4 and 5 Challenge Level:
Can you puzzle out what sequences these Logo programs will give? Then write your own Logo programs to generate sequences.
### Summing Squares
##### Stage: 4 Challenge Level:
Discover a way to sum square numbers by building cuboids from small cubes. Can you picture how the sequence will grow?
### Generating Number Patterns: an Email Conversation
##### Stage: 2, 3 and 4
This article for teachers describes the exchanges on an email talk list about ideas for an investigation which has the sum of the squares as its solution.
### Ordered Sums
##### Stage: 4 Challenge Level:
Let a(n) be the number of ways of expressing the integer n as an ordered sum of 1's and 2's. Let b(n) be the number of ways of expressing n as an ordered sum of integers greater than 1. (i) Calculate. . . .
### Sissa's Reward
##### Stage: 3 Challenge Level:
Sissa cleverly asked the King for a reward that sounded quite modest but turned out to be rather large...
### Dalmatians
##### Stage: 4 and 5 Challenge Level:
Investigate the sequences obtained by starting with any positive 2 digit number (10a+b) and repeatedly using the rule 10a+b maps to 10b-a to get the next number in the sequence.
### Seven Squares
##### Stage: 3 and 4 Challenge Level:
Watch these videos to see how Phoebe, Alice and Luke chose to draw 7 squares. How would they draw 100?
### Two Much
##### Stage: 3 Challenge Level:
Explain why the arithmetic sequence 1, 14, 27, 40, ... contains many terms of the form 222...2 where only the digit 2 appears.
### Paving Paths
##### Stage: 3 Challenge Level:
How many different ways can I lay 10 paving slabs, each 2 foot by 1 foot, to make a path 2 foot wide and 10 foot long from my back door into my garden, without cutting any of the paving slabs?
### Triangular Triples
##### Stage: 3 Challenge Level:
Show that 8778, 10296 and 13530 are three triangular numbers and that they form a Pythagorean triple.
### Three Frogs
##### Stage: 4 Challenge Level:
Three frogs hopped onto the table. A red frog on the left a green in the middle and a blue frog on the right. Then frogs started jumping randomly over any adjacent frog. Is it possible for them to. . . .
### 1 Step 2 Step
##### Stage: 3 Challenge Level:
Liam's house has a staircase with 12 steps. He can go down the steps one at a time or two at time. In how many different ways can Liam go down the 12 steps?
### Adding Triangles
##### Stage: 3 Challenge Level:
What is the total area of the first two triangles as a fraction of the original A4 rectangle? What is the total area of the first three triangles as a fraction of the original A4 rectangle? If. . . .
### Seven Squares - Group-worthy Task
##### Stage: 3 Challenge Level:
Choose a couple of the sequences. Try to picture how to make the next, and the next, and the next... Can you describe your reasoning?
### Designing Table Mats
##### Stage: 3 and 4 Challenge Level:
Formulate and investigate a simple mathematical model for the design of a table mat.
| 2,290 | 9,673 |
{"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}
| 4.25 | 4 |
CC-MAIN-2015-27
|
longest
|
en
| 0.836638 |
http://igeo.jp/tutorial/43.html
| 1,544,946,455,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2018-51/segments/1544376827596.48/warc/CC-MAIN-20181216073608-20181216095608-00125.warc.gz
| 126,187,038 | 6,496 |
Tutorials (back to the list of tutorials)
## Swarm Algorithm (requires iGeo version 7.5.3 or higher)
### Three Rules of Swarm Algorithm
Swarm behavior can be simulated by a type of multi-agent algorithm. The most popular algorithm is the algorithm of boids developed by Craig Reynolds in 1986. The boids algorithm consists of three simple rules to interact with other boid agents.
• Cohesion : going to the center of the surrounding agents.
• Separation : going away from other agents.
• Alignment : heading towards the same direction of other agents.
Each of those three rules has different distance range. Typically cohesion has the largest range, alignment has the second largest and the separation has the smallest, but depending on the intended behaviors of swarm agents, this order can be changed.
For more description about the algorithm of boids, please see Craig Reynolds's boids page .
### Swarm Rule 1: Cohesion
The first rule of this swarm algorithm is cohesion. This is a rule for an agent to get closer to the center of neighbor agents. This rule can be seen as a type of attraction rule.
Other agents are counted as neighbor agents if the distance to the agent is less than the threshold distance of cohesion (cohesionDist). The center of neighbors is calculated by adding all position of them and then divided by the number of the neighbors (count). The force is calculated by taking difference vector between the current agent and the center. The amount of the force is adjusted by the coefficient cohesionRatio.
```import processing.opengl.*;
import igeo.*;
void setup(){
size(480, 360, IG.GL);
IG.duration(30);
for(int i=0; i < 100; i++){
new MyBoid(IRand.pt(100,100,0),
IRand.pt(-10,-10,0,10,10,0)).clr(IRand.clr());
}
}
class MyBoid extends IParticle{
double cohesionDist = 50;
double cohesionRatio = 5;
IVec prevPos;
MyBoid(IVec p, IVec v){ super(p,v); }
void interact(ArrayList< IDynamics > agents){
IVec center = new IVec(); //zero vector
int count = 0;
for(int i=0; i < agents.size(); i++){
if(agents.get(i) instanceof MyBoid){
MyBoid b = (MyBoid)agents.get(i);
if(b != this){
if(b.pos().dist(pos()) < cohesionDist){
count++;
}
}
}
}
if(count > 0){
center.div(count); //center of neighbors
IVec force = center.sub(pos()); //difference of position
push(force.mul(cohesionRatio));
}
}
void update(){ //drawing line
IVec curPos = pos().cp();
if(prevPos!=null){ IG.crv(prevPos, curPos).clr(clr()); }
prevPos = curPos;
}
}
```
### Swarm Rule 2: Separation
The second rule of the swarm algorithm is separation. This is a rule for an agent to get away from close agents. This rule can be seen as a type of repulsion rule.
In the same way with cohesion rule, the threshold distance of separation (separationDist) is used to determine if the other agent is close enough to get away from. The force is calculated by a summantion of each force from close agents. The amount individual force is reset by this formula
force.len( separationDist - dist )
to make the closer agent's force larger and farther force smaller (and the agent at the distance of separationDist, the force is zero). The total force for the current agent to move away (separationForce) is adjusted by the coefficient separationRatio.
There is one additional note about this conditional statement dist != 0 in the in the if-condition. It is to avoid the case when the two agents' positions are identical and the direction to push out the agent cannot be defined from the difference. To be more precise, the agent should be pushed out in some default direction when they are in a same position but in the following example, this case is just excluded to keep it simple.
```import processing.opengl.*;
import igeo.*;
void setup(){
size(480, 360, IG.GL);
IG.duration(60);
for(int i=0; i < 100; i++){
new MyBoid(IRand.pt(100,100,0),
IRand.pt(-5,-5,0,20,20,0)).clr(IRand.clr());
}
}
class MyBoid extends IParticle{
double separationDist = 30;
double separationRatio = 5;
IVec prevPos;
MyBoid(IVec p, IVec v){ super(p,v); }
void interact(ArrayList< IDynamics > agents){
IVec separationForce = new IVec(); //zero vector
int count = 0;
for(int i=0; i < agents.size(); i++){
if(agents.get(i) instanceof MyBoid){
MyBoid b = (MyBoid)agents.get(i);
if(b != this){
double dist = b.pos().dist(pos());
if(dist < separationDist && dist!=0 ){
IVec force = pos().dif(b.pos());
force.len(separationDist - dist); //the closer the larger
count++;
}
}
}
}
if(count > 0){
separationForce.div(count); //average force
separationForce.mul(separationRatio);
push(separationForce);
}
}
void update(){ //drawing line
IVec curPos = pos().cp();
if(prevPos!=null){ IG.crv(prevPos, curPos).clr(clr()); }
prevPos = curPos;
}
}
```
### Swarm Rule 3: Alignment
The third rule of the swarm algorithm is alignment. This is a rule for an agent to steer towards the same direction with other agents.
Other agents within the range of the threshold distance of alignment (alignmentDist) are counted as neighbors and the average of their velocities is calculated. Then the difference between the average velocity and the current agent's are measured and the force is added to the direction of the vector difference of these two velocities. The amount of force is adjusted by the coefficient alignmentRatio.
```import processing.opengl.*;
import igeo.*;
void setup(){
size(480, 360, IG.GL);
IG.duration(100);
for(int i=0; i < 100; i++){
new MyBoid(IRand.pt(100,100,0),
IRand.pt(-5,-5,0,20,20,0)).clr(IRand.clr());
}
}
class MyBoid extends IParticle{
double alignmentDist = 40;
double alignmentRatio = 5;
IVec prevPos;
MyBoid(IVec p, IVec v){ super(p,v); }
void interact(ArrayList< IDynamics > agents){
IVec averageVelocity = new IVec(); //zero vector
int count = 0;
for(int i=0; i < agents.size(); i++){
if(agents.get(i) instanceof MyBoid){
MyBoid b = (MyBoid)agents.get(i);
if(b != this){
if(b.pos().dist(pos()) < alignmentDist){
count++;
}
}
}
}
if(count > 0){
averageVelocity.div(count);
IVec force = averageVelocity.sub(vel()); //difference of velocity
push(force.mul(alignmentRatio));
}
}
void update(){ //drawing line
IVec curPos = pos().cp();
if(prevPos!=null){ IG.crv(prevPos, curPos).clr(clr()); }
prevPos = curPos;
}
}
```
### Combining 3 Rules into Swarm Class
Those three rules are combined into one agent class as the following. Each rule is independently forming a method and the three methods of cohere(), separate() and align() are executed inside interact method.
```import processing.opengl.*;
import igeo.*;
void setup(){
size(480, 360, IG.GL);
IG.duration(150);
for(int i=0; i < 100; i++){
new MyBoid(IRand.pt(100,100,0),
IRand.pt(-5,-5,0,20,20,0)).clr(IRand.clr());
}
}
class MyBoid extends IParticle{
double cohesionDist = 50;
double cohesionRatio = 5;
double separationDist = 30;
double separationRatio = 5;
double alignmentDist = 40;
double alignmentRatio = 5;
IVec prevPos;
MyBoid(IVec p, IVec v){ super(p,v); }
void cohere(ArrayList< IDynamics > agents){
IVec center = new IVec(); //zero vector
int count = 0;
for(int i=0; i < agents.size(); i++){
if(agents.get(i) instanceof MyBoid && agents.get(i)!=this){
MyBoid b = (MyBoid)agents.get(i);
if(b.pos().dist(pos()) < cohesionDist){
count++;
}
}
}
if(count > 0){
push(center.div(count).sub(pos()).mul(cohesionRatio));
}
}
void separate(ArrayList< IDynamics > agents){
IVec separationForce = new IVec(); //zero vector
int count = 0;
for(int i=0; i < agents.size(); i++){
if(agents.get(i) instanceof MyBoid && agents.get(i)!=this){
MyBoid b = (MyBoid)agents.get(i);
double dist = b.pos().dist(pos());
if(dist < separationDist && dist!=0 ){
count++;
}
}
}
if(count > 0){
push(separationForce.mul(separationRatio/count));
}
}
void align(ArrayList< IDynamics > agents){
IVec averageVelocity = new IVec(); //zero vector
int count = 0;
for(int i=0; i < agents.size(); i++){
if(agents.get(i) instanceof MyBoid && agents.get(i) != this){
MyBoid b = (MyBoid)agents.get(i);
if(b.pos().dist(pos()) < alignmentDist){
count++;
}
}
}
if(count > 0){
push(averageVelocity.div(count).sub(vel()).mul(alignmentRatio));
}
}
void interact(ArrayList< IDynamics > agents){
cohere(agents);
separate(agents);
align(agents);
}
void update(){ //drawing line
IVec curPos = pos().cp();
if(prevPos!=null){ IG.crv(prevPos, curPos).clr(clr()); }
prevPos = curPos;
}
}
```
### Shorter Code of Swarm Class
The code above could be roughly simplified to the following with the shorter definition of interact method, but algorithmically it is not accurate because it doesn't have the variable count and ignores the number of neighbors within each range. Because of this, the way to adjust the parameters of ratio of each rule is different from the previous code.
```import processing.opengl.*;
import igeo.*;
void setup(){
size(480, 360, IG.GL);
IG.duration(150);
for(int i=0; i < 100; i++){
new MyBoid(IRand.pt(100,100,0),
IRand.pt(-5,-5,0,20,20,0)).clr(IRand.clr());
}
}
class MyBoid extends IParticle{
double cohDist = 50;
double cohRatio = 0.05;
double sepDist = 30;
double sepRatio = 0.05;
double aliDist = 40;
double aliRatio = 0.05;
IVec prevPos;
MyBoid(IVec p, IVec v){ super(p,v); }
void interact(IDynamics agent){
if(agent instanceof MyBoid){
MyBoid b = (MyBoid)agent;
if(dist(b) < cohDist) push( b.dif(this).mul(cohRatio) );
if(dist(b) < sepDist) push( dif(b).len((sepDist-dist(b))*sepRatio) );
if(dist(b) < aliDist) push( b.vel().dif(vel()).mul(aliRatio) );
}
}
void update(){ //drawing line
IVec curPos = pos().cp();
if(prevPos!=null){ IG.crv(prevPos, curPos).clr(clr()); }
prevPos = curPos;
}
}
```
### Using IBoid Class
iGeo library includes a class which defines the swarm behavior with those three rules. If you don't need to modify those three rules directly inside the interact method, you can define your agent class as a child class of IBoid class and tune parameters of the threshold distances and the force ratios of cohesion, separation and alignment.
```import processing.opengl.*;
import igeo.*;
void setup(){
size(480, 360, IG.GL);
IG.duration(150);
for(int i=0; i < 100; i++){
MyBoid b = new MyBoid(IRand.pt(100,100,0),
IRand.pt(-5,-5,0,20,20,0));
b.clr(IRand.clr());
b.cohesionDist(50);
b.cohesionRatio(5);
b.separationDist(30);
b.separationRatio(5);
b.alignmentDist(40);
b.alignmentRatio(5);
}
}
class MyBoid extends IBoid{
IVec prevPos;
MyBoid(IVec p, IVec v){ super(p,v); }
void update(){ //drawing line
IVec curPos = pos().cp();
if(prevPos!=null){ IG.crv(prevPos, curPos).clr(clr()); }
prevPos = curPos;
}
}
```
### Tuning Swarm Parameters
The following codes show examples of parameter settings of the swarm class. There are six parameters to control the behavior of the swarm agent. Three threshold distance parameters and three force ratio parameters to control the strength of force of cohesion, separation and alignment.
The first setting below have setting of smaller threshold distance for each rule. As result it creates smaller clusters of swarms.
```import processing.opengl.*;
import igeo.*;
void setup(){
size(480, 360, IG.GL);
IG.duration(100);
for(int i=0; i < 100; i++){
MyBoid b = new MyBoid(IRand.pt(100,100,0),
IRand.pt(-5,-5,0,20,20,0));
b.clr(IRand.clr());
b.cohesionDist(21);
b.cohesionRatio(5);
b.separationDist(10);
b.separationRatio(5);
b.alignmentDist(15);
b.alignmentRatio(5);
}
}
class MyBoid extends IBoid{
IVec prevPos;
MyBoid(IVec p, IVec v){ super(p,v); }
void update(){ //drawing line
IVec curPos = pos().cp();
if(prevPos!=null){ IG.crv(prevPos, curPos).clr(clr()); }
prevPos = curPos;
}
}
```
The next setting below has a smaller force ratio in the alignment rule. Because the force to align the velocities of agents are weak, it oscillates around the center with the attraction and repulsion forces from the cohesion and separation rule, not converging into the same velocity for a while.
```import processing.opengl.*;
import igeo.*;
void setup(){
size(480, 360, IG.GL);
IG.duration(300);
for(int i=0; i < 100; i++){
MyBoid b = new MyBoid(IRand.pt(100,100,0),
IRand.pt(-5,-5,0,20,20,0));
b.clr(IRand.clr());
b.cohesionDist(50);
b.cohesionRatio(5);
b.separationDist(30);
b.separationRatio(5);
b.alignmentDist(40);
b.alignmentRatio(0.5);
}
}
class MyBoid extends IBoid{
IVec prevPos;
MyBoid(IVec p, IVec v){ super(p,v); }
void update(){ //drawing line
IVec curPos = pos().cp();
if(prevPos!=null){ IG.crv(prevPos, curPos).clr(clr()); }
prevPos = curPos;
}
}
```
The setting below has a larger threshold distance for the separation rule and a smaller threshold distance for the cohesion rule. Because of this inverted order of cohesion and separation, agents try to go away but still creating smaller clusters with the alignment rule.
```import processing.opengl.*;
import igeo.*;
void setup(){
size(480, 360, IG.GL);
IG.duration(800);
for(int i=0; i < 100; i++){
MyBoid b = new MyBoid(IRand.pt(100,100,0),
IRand.pt(-5,-5,0,20,20,0));
b.clr(IRand.clr());
b.cohesionDist(30);
b.cohesionRatio(5);
b.separationDist(50);
b.separationRatio(5);
b.alignmentDist(40);
b.alignmentRatio(5);
}
}
class MyBoid extends IBoid{
IVec prevPos;
MyBoid(IVec p, IVec v){ super(p,v); }
void update(){ //drawing line
IVec curPos = pos().cp();
if(prevPos!=null){ IG.crv(prevPos, curPos).clr(clr()); }
prevPos = curPos;
}
}
```
(back to the list of tutorials)
| 3,628 | 13,260 |
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 3.71875 | 4 |
CC-MAIN-2018-51
|
longest
|
en
| 0.851894 |
https://www.enotes.com/homework-help/int-0-1-u-2-u-3-du-evaluate-integral-563458
| 1,511,609,621,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2017-47/segments/1510934809778.95/warc/CC-MAIN-20171125105437-20171125125437-00552.warc.gz
| 787,773,143 | 9,173 |
# `int_0^1(u + 2)(u - 3)du` Evaluate the integral.
gsarora17 | Certified Educator
`int_0^1(u+2)(u-3)du`
`=int_0^1(u^2-3u+2u-6)du`
`=int_0^1(u^2-u-6)du`
`=[u^3/3-u^2/2-6u]_0^1`
`=[1^3/3-1^2/2-6*1]-[0^3/3-0^2/2-6*0]`
`=(1/3-1/2-6)`
`=(2-3-36)/6`
=-37/6
| 168 | 259 |
{"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-2017-47
|
latest
|
en
| 0.310289 |
https://howkgtolbs.com/convert/22.17-kg-to-lbs
| 1,660,364,081,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2022-33/segments/1659882571869.23/warc/CC-MAIN-20220813021048-20220813051048-00273.warc.gz
| 303,023,806 | 12,190 |
# 22.17 kg to lbs - 22.17 kilograms to pounds
Before we go to the practice - it means 22.17 kg how much lbs calculation - we are going to tell you a little bit of theoretical information about these two units - kilograms and pounds. So we are starting.
How to convert 22.17 kg to lbs? 22.17 kilograms it is equal 48.8764834854 pounds, so 22.17 kg is equal 48.8764834854 lbs.
## 22.17 kgs in pounds
We will begin with the kilogram. The kilogram is a unit of mass. It is a base unit in a metric system, that is International System of Units (in abbreviated form SI).
At times the kilogram can be written as kilogramme. The symbol of this unit is kg.
First definition of a kilogram was formulated in 1795. The kilogram was defined as the mass of one liter of water. This definition was simply but hard to use.
Then, in 1889 the kilogram was described using the International Prototype of the Kilogram (in abbreviated form IPK). The International Prototype of the Kilogram was made of 90% platinum and 10 % iridium. The IPK was in use until 2019, when it was substituted by another definition.
The new definition of the kilogram is build on physical constants, especially Planck constant. Here is the official definition: “The kilogram, symbol kg, is the SI unit of mass. It is defined by taking the fixed numerical value of the Planck constant h to be 6.62607015×10−34 when expressed in the unit J⋅s, which is equal to kg⋅m2⋅s−1, where the metre and the second are defined in terms of c and ΔνCs.”
One kilogram is equal 0.001 tonne. It could be also divided to 100 decagrams and 1000 grams.
## 22.17 kilogram to pounds
You learned some facts about kilogram, so now let's go to the pound. The pound is also a unit of mass. We want to underline that there are more than one kind of pound. What does it mean? For instance, there are also pound-force. In this article we are going to to concentrate only on pound-mass.
The pound is used in the British and United States customary systems of measurements. Of course, this unit is used also in other systems. The symbol of this unit is lb or “.
There is no descriptive definition of the international avoirdupois pound. It is exactly 0.45359237 kilograms. One avoirdupois pound can be divided into 16 avoirdupois ounces and 7000 grains.
The avoirdupois pound was implemented in the Weights and Measures Act 1963. The definition of this unit was placed in first section of this act: “The yard or the metre shall be the unit of measurement of length and the pound or the kilogram shall be the unit of measurement of mass by reference to which any measurement involving a measurement of length or mass shall be made in the United Kingdom; and- (a) the yard shall be 0.9144 metre exactly; (b) the pound shall be 0.45359237 kilogram exactly.”
### How many lbs is 22.17 kg?
22.17 kilogram is equal to 48.8764834854 pounds. If You want convert kilograms to pounds, multiply the kilogram value by 2.2046226218.
### 22.17 kg in lbs
Theoretical part is already behind us. In next part we want to tell you how much is 22.17 kg to lbs. Now you learned that 22.17 kg = x lbs. So it is time to get the answer. Have a look:
22.17 kilogram = 48.8764834854 pounds.
That is an accurate result of how much 22.17 kg to pound. It is possible to also round off the result. After rounding off your outcome will be exactly: 22.17 kg = 48.774 lbs.
You know 22.17 kg is how many lbs, so let’s see how many kg 22.17 lbs: 22.17 pound = 0.45359237 kilograms.
Obviously, this time it is possible to also round off the result. After rounding off your result will be as following: 22.17 lb = 0.45 kgs.
We are also going to show you 22.17 kg to how many pounds and 22.17 pound how many kg results in tables. See:
We will start with a chart for how much is 22.17 kg equal to pound.
### 22.17 Kilograms to Pounds conversion table
Kilograms (kg) Pounds (lb) Pounds (lbs) (rounded off to two decimal places)
22.17 48.8764834854 48.7740
Now see a table for how many kilograms 22.17 pounds.
Pounds Kilograms Kilograms (rounded off to two decimal places
22.17 0.45359237 0.45
Now you know how many 22.17 kg to lbs and how many kilograms 22.17 pound, so we can go to the 22.17 kg to lbs formula.
### 22.17 kg to pounds
To convert 22.17 kg to us lbs a formula is needed. We will show you a formula in two different versions. Let’s start with the first one:
Number of kilograms * 2.20462262 = the 48.8764834854 outcome in pounds
The first formula will give you the most exact outcome. In some cases even the smallest difference can be significant. So if you want to get a correct outcome - this formula will be the best for you/option to calculate how many pounds are equivalent to 22.17 kilogram.
So let’s go to the second version of a formula, which also enables conversions to learn how much 22.17 kilogram in pounds.
The shorter version of a formula is down below, have a look:
Number of kilograms * 2.2 = the result in pounds
As you can see, the second version is simpler. It could be the best choice if you want to make a conversion of 22.17 kilogram to pounds in quick way, for instance, during shopping. You only need to remember that your outcome will be not so exact.
Now we want to show you these two formulas in practice. But before we are going to make a conversion of 22.17 kg to lbs we are going to show you easier way to know 22.17 kg to how many lbs totally effortless.
### 22.17 kg to lbs converter
Another way to learn what is 22.17 kilogram equal to in pounds is to use 22.17 kg lbs calculator. What is a kg to lb converter?
Converter is an application. Calculator is based on first formula which we showed you above. Thanks to 22.17 kg pound calculator you can quickly convert 22.17 kg to lbs. You only have to enter amount of kilograms which you need to convert and click ‘convert’ button. You will get the result in a second.
So let’s try to calculate 22.17 kg into lbs with use of 22.17 kg vs pound calculator. We entered 22.17 as an amount of kilograms. This is the outcome: 22.17 kilogram = 48.8764834854 pounds.
As you see, our 22.17 kg vs lbs calculator is intuitive.
Now let’s move on to our chief topic - how to convert 22.17 kilograms to pounds on your own.
#### 22.17 kg to lbs conversion
We are going to start 22.17 kilogram equals to how many pounds conversion with the first version of a formula to get the most correct result. A quick reminder of a formula:
Amount of kilograms * 2.20462262 = 48.8764834854 the outcome in pounds
So what need you do to check how many pounds equal to 22.17 kilogram? Just multiply number of kilograms, this time 22.17, by 2.20462262. It gives 48.8764834854. So 22.17 kilogram is 48.8764834854.
You can also round it off, for instance, to two decimal places. It is equal 2.20. So 22.17 kilogram = 48.7740 pounds.
It is high time for an example from everyday life. Let’s calculate 22.17 kg gold in pounds. So 22.17 kg equal to how many lbs? And again - multiply 22.17 by 2.20462262. It is exactly 48.8764834854. So equivalent of 22.17 kilograms to pounds, if it comes to gold, is exactly 48.8764834854.
In this example you can also round off the result. It is the result after rounding off, in this case to one decimal place - 22.17 kilogram 48.774 pounds.
Now we are going to examples converted using short formula.
#### How many 22.17 kg to lbs
Before we show you an example - a quick reminder of shorter formula:
Amount of kilograms * 2.2 = 48.774 the outcome in pounds
So 22.17 kg equal to how much lbs? And again, you need to multiply number of kilogram, in this case 22.17, by 2.2. Look: 22.17 * 2.2 = 48.774. So 22.17 kilogram is equal 2.2 pounds.
Make another conversion using shorer version of a formula. Now convert something from everyday life, for example, 22.17 kg to lbs weight of strawberries.
So convert - 22.17 kilogram of strawberries * 2.2 = 48.774 pounds of strawberries. So 22.17 kg to pound mass is 48.774.
If you know how much is 22.17 kilogram weight in pounds and can convert it using two different formulas, let’s move on. Now we want to show you these results in charts.
#### Convert 22.17 kilogram to pounds
We realize that results presented in tables are so much clearer for most of you. We understand it, so we gathered all these results in charts for your convenience. Due to this you can easily make a comparison 22.17 kg equivalent to lbs results.
Begin with a 22.17 kg equals lbs chart for the first formula:
Kilograms Pounds Pounds (after rounding off to two decimal places)
22.17 48.8764834854 48.7740
And now have a look at 22.17 kg equal pound chart for the second formula:
Kilograms Pounds
22.17 48.774
As you see, after rounding off, if it comes to how much 22.17 kilogram equals pounds, the outcomes are the same. The bigger amount the more significant difference. Keep it in mind when you need to make bigger amount than 22.17 kilograms pounds conversion.
#### How many kilograms 22.17 pound
Now you know how to convert 22.17 kilograms how much pounds but we are going to show you something more. Are you curious what it is? What do you say about 22.17 kilogram to pounds and ounces conversion?
We want to show you how you can convert it step by step. Begin. How much is 22.17 kg in lbs and oz?
First thing you need to do is multiply amount of kilograms, this time 22.17, by 2.20462262. So 22.17 * 2.20462262 = 48.8764834854. One kilogram is 2.20462262 pounds.
The integer part is number of pounds. So in this example there are 2 pounds.
To know how much 22.17 kilogram is equal to pounds and ounces you need to multiply fraction part by 16. So multiply 20462262 by 16. It is exactly 327396192 ounces.
So final outcome is exactly 2 pounds and 327396192 ounces. It is also possible to round off ounces, for example, to two places. Then final result is exactly 2 pounds and 33 ounces.
As you see, conversion 22.17 kilogram in pounds and ounces quite simply.
The last calculation which we will show you is calculation of 22.17 foot pounds to kilograms meters. Both of them are units of work.
To convert it it is needed another formula. Before we give you this formula, have a look:
• 22.17 kilograms meters = 7.23301385 foot pounds,
• 22.17 foot pounds = 0.13825495 kilograms meters.
Now have a look at a formula:
Amount.RandomElement()) of foot pounds * 0.13825495 = the outcome in kilograms meters
So to convert 22.17 foot pounds to kilograms meters you have to multiply 22.17 by 0.13825495. It is exactly 0.13825495. So 22.17 foot pounds is 0.13825495 kilogram meters.
It is also possible to round off this result, for example, to two decimal places. Then 22.17 foot pounds will be equal 0.14 kilogram meters.
We hope that this conversion was as easy as 22.17 kilogram into pounds conversions.
This article was a huge compendium about kilogram, pound and 22.17 kg to lbs in calculation. Thanks to this calculation you know 22.17 kilogram is equivalent to how many pounds.
We showed you not only how to do a conversion 22.17 kilogram to metric pounds but also two another conversions - to check how many 22.17 kg in pounds and ounces and how many 22.17 foot pounds to kilograms meters.
We showed you also other way to make 22.17 kilogram how many pounds calculations, that is using 22.17 kg en pound calculator. It will be the best option for those of you who do not like calculating on your own at all or need to make @baseAmountStr kg how lbs conversions in quicker way.
We hope that now all of you are able to do 22.17 kilogram equal to how many pounds calculation - on your own or with use of our 22.17 kgs to pounds calculator.
It is time to make your move! Let’s convert 22.17 kilogram mass to pounds in the way you like.
Do you need to make other than 22.17 kilogram as pounds conversion? For instance, for 5 kilograms? Check our other articles! We guarantee that calculations for other amounts of kilograms are so easy as for 22.17 kilogram equal many pounds.
### How much is 22.17 kg in pounds
To quickly sum up this topic, that is how much is 22.17 kg in pounds , we gathered answers to the most frequently asked questions. Here you can find all you need to know about how much is 22.17 kg equal to lbs and how to convert 22.17 kg to lbs . Let’s see.
What is the kilogram to pound conversion? It is a mathematical operation based on multiplying 2 numbers. How does 22.17 kg to pound conversion formula look? . See it down below:
The number of kilograms * 2.20462262 = the result in pounds
Now you can see the result of the conversion of 22.17 kilogram to pounds. The accurate result is 48.8764834854 lb.
There is also another way to calculate how much 22.17 kilogram is equal to pounds with second, easier version of the formula. Let’s see.
The number of kilograms * 2.2 = the result in pounds
So in this case, 22.17 kg equal to how much lbs ? The answer is 48.8764834854 lb.
How to convert 22.17 kg to lbs in a few seconds? You can also use the 22.17 kg to lbs converter , which will make whole mathematical operation for you and give you an exact answer .
#### Kilograms [kg]
The kilogram, or kilogramme, is the base unit of weight in the Metric system. It is the approximate weight of a cube of water 10 centimeters on a side.
#### Pounds [lbs]
A pound is a unit of weight commonly used in the United States and the British commonwealths. A pound is defined as exactly 0.45359237 kilograms.
Read more related articles:
22.01 kg to lbs = 48.5237 22.02 kg to lbs = 48.5458 22.03 kg to lbs = 48.5678 22.04 kg to lbs = 48.5899 22.05 kg to lbs = 48.6119 22.06 kg to lbs = 48.634 22.07 kg to lbs = 48.656 22.08 kg to lbs = 48.6781 22.09 kg to lbs = 48.7001 22.1 kg to lbs = 48.7222 22.11 kg to lbs = 48.7442 22.12 kg to lbs = 48.7662 22.13 kg to lbs = 48.7883 22.14 kg to lbs = 48.8103 22.15 kg to lbs = 48.8324 22.16 kg to lbs = 48.8544 22.17 kg to lbs = 48.8765 22.18 kg to lbs = 48.8985 22.19 kg to lbs = 48.9206 22.2 kg to lbs = 48.9426 22.21 kg to lbs = 48.9647 22.22 kg to lbs = 48.9867 22.23 kg to lbs = 49.0088 22.24 kg to lbs = 49.0308 22.25 kg to lbs = 49.0528
22.26 kg to lbs = 49.0749 22.27 kg to lbs = 49.0969 22.28 kg to lbs = 49.119 22.29 kg to lbs = 49.141 22.3 kg to lbs = 49.1631 22.31 kg to lbs = 49.1851 22.32 kg to lbs = 49.2072 22.33 kg to lbs = 49.2292 22.34 kg to lbs = 49.2513 22.35 kg to lbs = 49.2733 22.36 kg to lbs = 49.2954 22.37 kg to lbs = 49.3174 22.38 kg to lbs = 49.3394 22.39 kg to lbs = 49.3615 22.4 kg to lbs = 49.3835 22.41 kg to lbs = 49.4056 22.42 kg to lbs = 49.4276 22.43 kg to lbs = 49.4497 22.44 kg to lbs = 49.4717 22.45 kg to lbs = 49.4938 22.46 kg to lbs = 49.5158 22.47 kg to lbs = 49.5379 22.48 kg to lbs = 49.5599 22.49 kg to lbs = 49.582 22.5 kg to lbs = 49.604
22.51 kg to lbs = 49.6261 22.52 kg to lbs = 49.6481 22.53 kg to lbs = 49.6701 22.54 kg to lbs = 49.6922 22.55 kg to lbs = 49.7142 22.56 kg to lbs = 49.7363 22.57 kg to lbs = 49.7583 22.58 kg to lbs = 49.7804 22.59 kg to lbs = 49.8024 22.6 kg to lbs = 49.8245 22.61 kg to lbs = 49.8465 22.62 kg to lbs = 49.8686 22.63 kg to lbs = 49.8906 22.64 kg to lbs = 49.9127 22.65 kg to lbs = 49.9347 22.66 kg to lbs = 49.9567 22.67 kg to lbs = 49.9788 22.68 kg to lbs = 50.0008 22.69 kg to lbs = 50.0229 22.7 kg to lbs = 50.0449 22.71 kg to lbs = 50.067 22.72 kg to lbs = 50.089 22.73 kg to lbs = 50.1111 22.74 kg to lbs = 50.1331 22.75 kg to lbs = 50.1552
22.76 kg to lbs = 50.1772 22.77 kg to lbs = 50.1993 22.78 kg to lbs = 50.2213 22.79 kg to lbs = 50.2433 22.8 kg to lbs = 50.2654 22.81 kg to lbs = 50.2874 22.82 kg to lbs = 50.3095 22.83 kg to lbs = 50.3315 22.84 kg to lbs = 50.3536 22.85 kg to lbs = 50.3756 22.86 kg to lbs = 50.3977 22.87 kg to lbs = 50.4197 22.88 kg to lbs = 50.4418 22.89 kg to lbs = 50.4638 22.9 kg to lbs = 50.4859 22.91 kg to lbs = 50.5079 22.92 kg to lbs = 50.5299 22.93 kg to lbs = 50.552 22.94 kg to lbs = 50.574 22.95 kg to lbs = 50.5961 22.96 kg to lbs = 50.6181 22.97 kg to lbs = 50.6402 22.98 kg to lbs = 50.6622 22.99 kg to lbs = 50.6843 23 kg to lbs = 50.7063
| 4,819 | 16,018 |
{"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-2022-33
|
latest
|
en
| 0.94796 |
https://math.libretexts.org/TextMaps/Calculus_TextMaps/Map%3A_Calculus_(Guichard)/02._Instantaneous_Rate_of_Change%3A_The_Derivative/2.3%3A_Limits
| 1,521,891,644,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2018-13/segments/1521257650262.65/warc/CC-MAIN-20180324112821-20180324132821-00298.warc.gz
| 638,464,196 | 20,375 |
$$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\kernel}{\mathrm{null}\,}$$ $$\newcommand{\range}{\mathrm{range}\,}$$ $$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$ $$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}[1]{\| #1 \|}$$ $$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$$ $$\newcommand{\Span}{\mathrm{span}}$$
# 2.3: Limits
In the previous two sections we computed some quantities of interest (slope, velocity) by seeing that some expression "goes to'' or "approaches'' or "gets really close to'' a particular value. In the examples we saw, this idea may have been clear enough, but it is too fuzzy to rely on in more difficult circumstances. In this section we will see how to make the idea more precise.
There is an important feature of the examples we have seen. Consider again the formula ${-19.6\Delta x-4.9\Delta x^2\over \Delta x}.$ We wanted to know what happens to this fraction as "$$\Delta x$$ goes to zero.'' Because we were able to simplify the fraction, it was easy to see the answer, but it was not quite as simple as "substituting zero for $$\Delta x$$,'' as that would give ${-19.6\cdot 0 - 4.9\cdot 0\over 0},$ which is meaningless. The quantity we are really interested in does not make sense "at zero,'' and this is why the answer to the original problem (finding a velocity or a slope) was not immediately obvious. In other words, we are generally going to want to figure out what a quantity "approaches'' in situations where we can't merely plug in a value. If you would like to think about a hard example (which we will analyze later) consider what happens to $$(\sin x)/x$$ as $$x$$ approaches zero.
Example 2.3.1
Does $$\sqrt{x}$$ approach 1.41 as $$x$$ approaches 2?
Solution
In this case it is possible to compute the actual value $$\sqrt{2}$$ to a high precision to answer the question. But since in general we won't be able to do that, let's not. We might start by computing $$\sqrt{x}$$ for values of $$x$$ close to 2, as we did in the previous sections. Here are some values: $$\sqrt{2.05} = 1.431782106$$, $$\sqrt{2.04} = 1.428285686$$, $$\sqrt{2.03} = 1.424780685$$, $$\sqrt{2.02} = 1.421267040$$, $$\sqrt{2.01} = 1.417744688$$, $$\sqrt{2.005} = 1.415980226$$, $$\sqrt{2.004} = 1.415627070$$, $$\sqrt{2.003} = 1.415273825$$, $$\sqrt{2.002} = 1.414920492$$, $$\sqrt{2.001} = 1.414567072$$. So it looks at least possible that indeed these values "approach'' 1.41---already $$\sqrt{2.001}$$ is quite close. If we continue this process, however, at some point we will appear to "stall.'' In fact, $$\sqrt{2}=1.414213562\ldots$$, so we will never even get as far as 1.4142, no matter how long we continue the sequence.
So in a fuzzy, everyday sort of sense, it is true that $$\sqrt{x}$$ "gets close to'' 1.41, but it does not "approach'' 1.41 in the sense we want. To compute an exact slope or an exact velocity, what we want to know is that a given quantity becomes "arbitrarily close'' to a fixed value, meaning that the first quantity can be made "as close as we like'' to the fixed value. Consider again the quantities ${-19.6\Delta x-4.9\Delta x^2\over \Delta x}=-19.6-4.9\Delta x.$ These two quantities are equal as long as $$\Delta x$$ is not zero; if $$\Delta x$$ is zero, the left hand quantity is meaningless, while the right hand one is $$-19.6$$. Can we say more than we did before about why the right hand side "approaches'' $$-19.6$$, in the desired sense? Can we really make it "as close as we want'' to $$-19.6$$? Let's try a test case. Can we make $$-19.6-4.9\Delta x$$ within one millionth $$0.000001$$ of $$-19.6$$? The values within a millionth of $$-19.6$$ are those in the interval $$(-19.600001,-19.599999)$$. As $$\Delta x$$ approaches zero, does $$-19.6-4.9\Delta x$$ eventually reside inside this interval? If $$\Delta x$$ is positive, this would require that $$-19.6-4.9\Delta x> -19.600001$$. This is something we can manipulate with a little algebra:
\eqalign{-19.6-4.9\Delta x&> -19.600001\cr -4.9\Delta x&>-0.000001\cr \Delta x& < -0.000001/-4.9\cr \Delta x& < 0.0000002040816327\ldots\cr }
Thus, we can say with certainty that if $$\Delta x$$ is positive and less than $$0.0000002$$, then $$\Delta x < 0.0000002040816327\ldots$$ and so $$-19.6-4.9\Delta x > -19.600001$$. We could do a similar calculation if $$\Delta x$$ is negative.
So now we know that we can make $$-19.6-4.9\Delta x$$ within one millionth of $$-19.6$$. But can we make it "as close as we want''? In this case, it is quite simple to see that the answer is yes, by modifying the calculation we've just done. It may be helpful to think of this as a game. I claim that I can make $$-19.6-4.9\Delta x$$ as close as you desire to $$-19.6$$ by making $$\Delta x$$ "close enough'' to zero. So the game is: you give me a number, like $$10^{-6}$$, and I have to come up with a number representing how close $$\Delta x$$ must be to zero to guarantee that $$-19.6-4.9\Delta x$$ is at least as close to $$-19.6$$ as you have requested.
Now if we actually play this game, I could redo the calculation above for each new number you provide. What I'd like to do is somehow see that I will always succeed, and even more, I'd like to have a simple strategy so that I don't have to do all that algebra every time. A strategy in this case would be a formula that gives me a correct answer no matter what you specify. So suppose the number you give me is $$\epsilon$$. How close does $$\Delta x$$ have to be to zero to guarantee that $$-19.6-4.9\Delta x$$ is in $$(-19.6-\epsilon, -19.6+\epsilon)? If \(\Delta x$$ is positive, we need: \eqalign{-19.6-4.9\Delta x&> -19.6-\epsilon\cr -4.9\Delta x&>-\epsilon\cr \Delta x& < -\epsilon/-4.9\cr \Delta x& < \epsilon/4.9\cr } So if I pick any number $$\delta$$ that is less than $$\epsilon/4.9$$, the algebra tells me that whenever $$\Delta x < \delta$$ then $$\Delta x < \epsilon/4.9$$ and so $$-19.6-4.9\Delta x$$ is within $$\epsilon$$ of $$-19.6$$. (This is exactly what I did in the example: I picked $$\delta = 0.0000002 < 0.0000002040816327\ldots$$.) A similar calculation again works for negative $$\Delta x$$. The important fact is that this is now a completely general result---it shows that I can always win, no matter what "move'' you make.
Now we can codify this by giving a precise definition to replace the fuzzy, "gets closer and closer'' language we have used so far. Henceforward, we will say something like "the limit of $$(-19.6\Delta x-4.9\Delta x^2)/\Delta x$$ as $$\Delta x$$ goes to zero is $$-19.6$$,'' and abbreviate this mouthful as $\lim_{\Delta x\to 0} {-19.6\Delta x-4.9\Delta x^2\over \Delta x} = -19.6. \[ Here is the actual, official definition of "limit''. Definition 2.3.2: Limits Suppose $$f$$ is a function. We say that $$\lim_{x\to a}f(x)=L$$ if for every $$\epsilon>0$$ there is a $$\delta > 0$$ so that whenever $$0 < |x-a| < \delta$$, $$|f(x)-L| < \epsilon$$. The $$\epsilon$$ and $$\delta$$ here play exactly the role they did in the preceding discussion. The definition says, in a very precise way, that $$f(x)$$ can be made as close as desired to $$L$$ (that's the $$|f(x)-L| < \epsilon$$ part) by making $$x$$ close enough to $$a$$ (the $$0 < |x-a| < \delta$$ part). Note that we specifically make no mention of what must happen if $$x=a$$, that is, if $$|x-a|=0$$. This is because in the cases we are most interested in, substituting $$a$$ for $$x$$ doesn't even make sense. Make sure you are not confused by the names of important quantities. The generic definition talks about $$f(x)$$, but the function and the variable might have other names. In the discussion above, the function we analyzed was \[{-19.6\Delta x-4.9\Delta x^2\over \Delta x}. \[ and the variable of the limit was not $$x$$ but $$\Delta x$$. The $$x$$ was the variable of the original function; when we were trying to compute a slope or a velocity, $$x$$ was essentially a fixed quantity, telling us at what point we wanted the slope. (In the velocity problem, it was literally a fixed quantity, as we focused on the time 2.) The quantity $$a$$ of the definition in all the examples was zero: we were always interested in what happened as $$\Delta x$$ became very close to zero. Armed with a precise definition, we can now prove that certain quantities behave in a particular way. The bad news is that even proofs for simple quantities can be quite tedious and complicated; the good news is that we rarely need to do such proofs, because most expressions act the way you would expect, and this can be proved once and for all. Example 2.3.3 Let's show carefully that $$\lim_{x\to 2} x+4 = 6$$. Solution This is not something we "need'' to prove, since it is "obviously'' true. But if we couldn't prove it using our official definition there would be something very wrong with the definition. As is often the case in mathematical proofs, it helps to work backwards. We want to end up showing that under certain circumstances $$x+4$$ is close to 6; precisely, we want to show that $$|x+4-6| < \epsilon$$, or $$|x-2| < \epsilon$$. Under what circumstances? We want this to be true whenever $$0 < |x-2| < \delta$$. So the question becomes: can we choose a value for $$\delta$$ that guarantees that $$0 < |x-2| < \delta$$ implies $$|x-2| < \epsilon$$? Of course: no matter what $$\epsilon$$ is, $$\delta=\epsilon$$ works. So it turns out to be very easy to prove something "obvious,'' which is nice. It doesn't take long before things get trickier, however. Example 2.3.4 It seems clear that $$\lim_{x\to 2} x^2=4$$. Let's try to prove it. Solution We will want to be able to show that $$|x^2-4| < \epsilon$$ whenever $$0 < |x-2| < \delta$$, by choosing $$\delta$$ carefully. Is there any connection between $$|x-2|$$ and $$|x^2-4|? Yes, and it's not hard to spot, but it is not so simple as the previous example. We can write \( |x^2-4|=|(x+2)(x-2)|$$. Now when $$|x-2|$$ is small, part of $$|(x+2)(x-2)|$$ is small, namely $$(x-2)$$. What about $$(x+2)$$? If $$x$$ is close to 2, $$(x+2)$$ certainly can't be too big, but we need to somehow be precise about it. Let's recall the "game'' version of what is going on here. You get to pick an $$\epsilon$$ and I have to pick a $$\delta$$ that makes things work out. Presumably it is the really tiny values of $$\epsilon$$ I need to worry about, but I have to be prepared for anything, even an apparently "bad'' move like $$\epsilon=1000$$. I expect that $$\epsilon$$ is going to be small, and that the corresponding $$\delta$$ will be small, certainly less than 1. If $$\delta\le 1$$ then $$|x+2| < 5$$ when $$|x-2| < \delta$$ (because if $$x$$ is within 1 of 2, then $$x$$ is between 1 and 3 and $$x+2$$ is between 3 and 5). So then I'd be trying to show that $$|(x+2)(x-2)| < 5|x-2| < \epsilon$$. So now how can I pick $$\delta$$ so that $$|x-2| < \delta$$ implies $$5|x-2| < \epsilon$$? This is easy: use $$\delta=\epsilon/5$$, so $$5|x-2| < 5(\epsilon/5) = \epsilon$$. But what if the $$\epsilon$$ you choose is not small? If you choose $$\epsilon=1000$$, should I pick $$\delta=200$$? No, to keep things "sane'' I will never pick a $$\delta$$ bigger than 1. Here's the final "game strategy:'' When you pick a value for $$\epsilon$$ I will pick $$\delta=\epsilon/5$$ or $$\delta=1$$, whichever is smaller. Now when $$|x-2| < \delta$$, I know both that $$|x+2| < 5$$ and that $$|x-2| < \epsilon/5$$. Thus $$|(x+2)(x-2)| < 5(\epsilon/5) = \epsilon$$. This has been a long discussion, but most of it was explanation and scratch work. If this were written down as a proof, it would be quite short, like this: Proof that $$\lim_{x\to 2}x^2=4$$. Given any $$\epsilon$$, pick $$\delta=\epsilon/5$$ or $$\delta=1$$, whichever is smaller. Then when $$|x-2| < \delta$$, $$|x+2| < 5$$ and $$|x-2| < \epsilon/5$$. Hence $$|x^2-4|=|(x+2)(x-2)| < 5(\epsilon/5) = \epsilon$$. It probably seems obvious that $$\lim_{x\to2}x^2=4$$, and it is worth examining more closely why it seems obvious. If we write $$x^2=x\cdot x$$, and ask what happens when $$x$$ approaches 2, we might say something like, "Well, the first $$x$$ approaches 2, and the second $$x$$ approaches 2, so the product must approach $$2\cdot2$$.'' In fact this is pretty much right on the money, except for that word "must.'' Is it really true that if $$x$$ approaches $$a$$ and $$y$$ approaches $$b$$ then $$xy$$ approaches $$ab$$? It is, but it is not really obvious, since $$x$$ and $$y$$ might be quite complicated. The good news is that we can see that this is true once and for all, and then we don't have to worry about it ever again. When we say that $$x$$ might be "complicated'' we really mean that in practice it might be a function. Here is then what we want to know: Theorem 2.3.5 Suppose $$\lim_{x\to a} f(x)=L$$ and $$\lim_{x\to a}g(x)=M$$. Then \[\lim_{x\to a} f(x)g(x) = LM.$
Proof
We have to use the official definition of limit to make sense of this. So given any $$\epsilon$$ we need to find a $$\delta$$ so that $$0 < |x-a| < \delta$$ implies $$|f(x)g(x)-LM| < \epsilon$$. What do we have to work with? We know that we can make $$f(x)$$ close to $$L$$ and $$g(x)$$ close to $$M$$, and we have to somehow connect these facts to make $$f(x)g(x)$$ close to $$LM$$.
We use, as is so often the case, a little algebraic trick:
\eqalign{|f(x)g(x)-LM|&= |f(x)g(x)-f(x)M+f(x)M-LM|\cr &=|f(x)(g(x)-M)+(f(x)-L)M|\cr &\le |f(x)(g(x)-M)|+|(f(x)-L)M|\cr &=|f(x)||g(x)-M|+|f(x)-L||M|.\cr}
This is all straightforward except perhaps for the "$$\le$$'. That is an example of the triangle inequality , which says that if $$a$$ and $$b$$ are any real numbers then $$|a+b|\le |a|+|b|$$. If you look at a few examples, using positive and negative numbers in various combinations for $$a$$ and $$b$$, you should quickly understand why this is true; we will not prove it formally.
Since $$\lim_{x\to a}f(x) =L$$, there is a value $$\delta_1$$ so that $$0 < |x-a| < \delta_1$$ implies $$|f(x)-L| < |\epsilon/(2M)|$$, This means that $$0 < |x-a| < \delta_1$$ implies $$|f(x)-L||M| < \epsilon/2$$. You can see where this is going: if we can make $$|f(x)||g(x)-M| < \epsilon/2$$ also, then we'll be done.
We can make $$|g(x)-M|$$ smaller than any fixed number by making $$x$$ close enough to $$a$$; unfortunately, $$\epsilon/(2f(x))$$ is not a fixed number, since $$x$$ is a variable. Here we need another little trick, just like the one we used in analyzing $$x^2$$. We can find a $$\delta_2$$ so that $$|x-a| < \delta_2$$ implies that $$|f(x)-L| < 1$$, meaning that $$L-1 < f(x) < L+1$$. This means that $$|f(x)| < N$$, where $$N$$ is either $$|L-1|$$ or $$|L+1|$$, depending on whether $$L$$ is negative or positive. The important point is that $$N$$ doesn't depend on $$x$$. Finally, we know that there is a $$\delta_3$$ so that $$0 < |x-a| < \delta_3$$ implies $$|g(x)-M| < \epsilon/(2N)$$. Now we're ready to put everything together. Let $$\delta$$ be the smallest of $$\delta_1$$, $$\delta_2$$, and $$\delta_3$$. Then $$|x-a| < \delta$$ implies that $$|f(x)-L| < |\epsilon/(2M)|$$, $$|f(x)| < N$$, and $$|g(x)-M| < \epsilon/(2N)$$. Then
\eqalign{|f(x)g(x)-LM|&\le|f(x)||g(x)-M|+|f(x)-L||M|\cr & < N{\epsilon\over 2N}+\left|{\epsilon\over 2M}\right||M|\cr &={\epsilon\over 2}+{\epsilon\over 2}=\epsilon.\cr}
This is just what we needed, so by the official definition, $$\lim_{x\to a}f(x)g(x)=LM$$.
$$\square$$
A handful of such theorems give us the tools to compute many limits without explicitly working with the definition of limit.
Theorem 2.3.6
Suppose that $$\lim_{x\to a}f(x)=L$$ and $$\lim_{x\to a}g(x)=M$$ and $$k$$ is some constant. Then
\eqalign{ &\lim_{x\to a} kf(x) = k\lim_{x\to a}f(x)=kL\cr &\lim_{x\to a} (f(x)+g(x)) = \lim_{x\to a}f(x)+\lim_{x\to a}g(x)=L+M\cr &\lim_{x\to a} (f(x)-g(x)) = \lim_{x\to a}f(x)-\lim_{x\to a}g(x)=L-M\cr &\lim_{x\to a} (f(x)g(x)) = \lim_{x\to a}f(x)\cdot\lim_{x\to a}g(x)=LM\cr &\lim_{x\to a} {f(x)\over g(x)} = {\lim_{x\to a}f(x)\over\lim_{x\to a}g(x)}={L\over M},\hbox{ if $$M$$ is not 0}\cr }
Roughly speaking, these rules say that to compute the limit of an algebraic expression, it is enough to compute the limits of the "innermost bits'' and then combine these limits. This often means that it is possible to simply plug in a value for the variable, since $$\lim_{x\to a} x =a$$.
Example 2.3.7
Compute $$\lim_{x\to 1}{x^2-3x+5\over x-2}$$.
Solution
If we apply the theorem in all its gory detail, we get
\eqalign{ \lim_{x\to 1}{x^2-3x+5\over x-2}&= {\lim_{x\to 1}(x^2-3x+5)\over \lim_{x\to1}(x-2)}\cr &={(\lim_{x\to 1}x^2)-(\lim_{x\to1}3x)+(\lim_{x\to1}5)\over (\lim_{x\to1}x)-(\lim_{x\to1}2)}\cr &={(\lim_{x\to 1}x)^2-3(\lim_{x\to1}x)+5\over (\lim_{x\to1}x)-2}\cr &={1^2-3\cdot1+5\over 1-2}\cr &={1-3+5\over -1} = -3\cr }
It is worth commenting on the trivial limit $$\lim_{x\to1}5$$. From one point of view this might seem meaningless, as the number 5 can't "approach'' any value, since it is simply a fixed number. But 5 can, and should, be interpreted here as the function that has value 5 everywhere, $$f(x)=5$$, with graph a horizontal line. From this point of view it makes sense to ask what happens to the height of the function as $$x$$ approaches 1.
Of course, as we've already seen, we're primarily interested in limits that aren't so easy, namely, limits in which a denominator approaches zero. There are a handful of algebraic tricks that work on many of these limits.
Example 2.3.8
Compute $$\lim_{x\to1}{x^2+2x-3\over x-1}$$.
Solution
We cannot simply plug in $$x=1$$ because that makes the denominator zero. However:
\eqalign{ \lim_{x\to1}{x^2+2x-3\over x-1}&=\lim_{x\to1}{(x-1)(x+3)\over x-1}\cr &=\lim_{x\to1}(x+3)=4\cr }
While theorem 2.3.6 is very helpful, we need a bit more to work easily with limits. Since the theorem applies when some limits are already known, we need to know the behavior of some functions that cannot themselves be constructed from the simple arithmetic operations of the theorem, such as $$\sqrt{x}$$. Also, there is one other extraordinarily useful way to put functions together: composition. If $$f(x)$$ and $$g(x)$$ are functions, we can form two functions by composition: $$f(g(x))$$ and $$g(f(x))$$. For example, if $$f(x)=\sqrt{x}$$ and $$g(x)=x^2+5$$, then $$f(g(x))=\sqrt{x^2+5}$$ and $$g(f(x))=(\sqrt{x})^2+5=x+5$$. Here is a companion to theorem 2.3.6 for composition:
Theorem 2.3.9
Suppose that $$\lim_{x\to a}g(x)=L$$ and $$\lim_{x\to L}f(x)=f(L)$$. Then
$\lim_{x\to a} f(g(x)) = f(L).$
Note the special form of the condition on $$f$$: it is not enough to know that $$\lim_{x\to L}f(x) = M$$, though it is a bit tricky to see why. Many of the most familiar functions do have this property, and this theorem can therefore be applied. For example:
Theorem 2.3.10
Suppose that $$n$$ is a positive integer. Then
$\lim_{x\to a}\root n\of{x} = \root n\of{a},$
provided that $$a$$ is positive if $$n$$ is even.
This theorem is not too difficult to prove from the definition of limit. Another of the most common algebraic tricks was used in section 2.1. Here's another example:
Example 2.3.11
Compute $$\lim_{x\to-1} {\sqrt{x+5}-2\over x+1}$$.
Solution
\eqalign{\lim_{x\to-1} {\sqrt{x+5}-2\over x+1}&= \lim_{x\to-1} {\sqrt{x+5}-2\over x+1}{\sqrt{x+5}+2\over \sqrt{x+5}+2}\cr &=\lim_{x\to-1} {x+5-4\over (x+1)(\sqrt{x+5}+2)}\cr &=\lim_{x\to-1} {x+1\over (x+1)(\sqrt{x+5}+2)}\cr &=\lim_{x\to-1} {1\over \sqrt{x+5}+2}={1\over4}\cr}
At the very last step we have used theorems 2.3.9 and 2.3.10.
Occasionally we will need a slightly modified version of the limit definition. Consider the function $$f(x)=\sqrt{1-x^2}$$, the upper half of the unit circle. What can we say about $$\lim_{x\to 1}f(x)$$? It is apparent from the graph of this familiar function that as $$x$$ gets close to 1 from the left, the value of $$f(x)$$ gets close to zero. It does not even make sense to ask what happens as $$x$$ approaches 1 from the right, since $$f(x)$$ is not defined there. The definition of the limit, however, demands that $$f(1+\Delta x)$$ be close to $$f(1)$$ whether $$\Delta x$$ is positive or negative. Sometimes the limit of a function exists from one side or the other (or both) even though the limit does not exist. Since it is useful to be able to talk about this situation, we introduce the concept of one sided limit:
Definition 2.3.12: One Sided Limits
Suppose that $$f(x)$$ is a function. We say that $$\lim_{x\to a^-}f(x)=L$$ if for every $$\epsilon>0$$ there is a $$\delta > 0$$ so that whenever $$0 < a-x < \delta$$, $$|f(x)-L| < \epsilon$$. We say that $$\lim_{x\to a^+}f(x)=L$$ if for every $$\epsilon>0$$ there is a $$\delta > 0$$ so that whenever $$0 < x-a < \delta$$, $$|f(x)-L| < \epsilon$$.
Usually $$\lim_{x\to a^-}f(x)$$ is read "the limit of $$f(x)$$ from the left'' and $$\lim_{x\to a^+}f(x)$$ is read "the limit of $$f(x)$$ from the right''.
Example 2.3.13
Discuss $$\lim_{x\to 0}{x\over|x|}$$, $$\lim_{x\to 0^-}{x\over|x|}$$, and $$\lim_{x\to 0^+}{x\over|x|}$$.
Solution
The function $$f(x)=x/|x|$$ is undefined at 0; when $$x>0$$, $$|x|=x$$ and so $$f(x)=1; when \(x < 0$$, $$|x|=-x$$ and $$f(x)=-1$$. Thus $$\lim_{x\to 0^-}{x\over|x|}=\lim_{x\to 0^-}-1=-1$$ while $$\lim_{x\to 0^+}{x\over|x|}=\lim_{x\to 0^+}1=1$$. The limit of $$f(x)$$ must be equal to both the left and right limits; since they are different, the limit $$\lim_{x\to 0}{x\over|x|}$$ does not exist.
### Contributors
• Integrated by Justin Marshall.
| 7,091 | 21,464 |
{"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}
| 4.4375 | 4 |
CC-MAIN-2018-13
|
latest
|
en
| 0.90442 |
http://mathhelpforum.com/number-theory/148891-n-print.html
| 1,524,504,015,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2018-17/segments/1524125946120.31/warc/CC-MAIN-20180423164921-20180423184921-00546.warc.gz
| 195,695,607 | 3,103 |
# n!
• Jun 19th 2010, 10:13 AM
dwsmith
n!
If $\displaystyle n!$ has 58 trailing zeros, what is n?
$\displaystyle \left \lfloor \frac{n}{5} \right \rfloor + \left \lfloor \frac{n}{25} \right \rfloor + \dots + \left \lfloor \frac{n}{5^i} \right \rfloor=58$ where $\displaystyle i=1,2,...,j$
I don't know what to do next.
• Jun 19th 2010, 10:29 AM
undefined
Quote:
Originally Posted by dwsmith
If $\displaystyle n!$ has 58 trailing zeros, what is n?
$\displaystyle \left \lfloor \frac{n}{5} \right \rfloor + \left \lfloor \frac{n}{25} \right \rfloor + \dots + \left \lfloor \frac{n}{5^i} \right \rfloor=58$ where $\displaystyle i=1,2,...,j$
I don't know what to do next.
Actually you can write $\displaystyle \displaystyle 58 = \left \lfloor \frac{n}{5} \right \rfloor + \left \lfloor \frac{n}{25} \right \rfloor + \dots = \sum_{i=1}^\infty \left \lfloor \frac{n}{5^i} \right \rfloor$ since the terms eventually all become 0.
We can get an immediate upper bound: if n=59*5, then the first term will be greater than 58. Since 59*5 - 1 < 625, clearly we have just the first three terms:
$\displaystyle \displaystyle 58 = \left \lfloor \frac{n}{5} \right \rfloor + \left \lfloor \frac{n}{25} \right \rfloor + \left \lfloor \frac{n}{125} \right \rfloor$
This is easy to play around with; 58/5 is about 12, so as an estimate we can subtract about 10*5 from our upper bound to get 59*5-51 = 244. Plugging in results in 48 + 9 + 1 = 58, as desired. So $\displaystyle n \in \{240,...,244\}$.
• Jun 19th 2010, 08:07 PM
simplependulum
Or let
$\displaystyle n = 125a + 25b + 5c + d$ as a representation in base 5 , since it does not exceed $\displaystyle 625$
Therefore , $\displaystyle [ \frac{n}{5} ] = 25a + 5b + c ~ [ \frac{n}{25} ] = 5a + b ~,~ [ \frac{n}{125} ] = a$
$\displaystyle 25a + 5b + c + 5a + b + a = 31a + 6b + c = 58$
$\displaystyle a = 1 ~,~ b = 4 ~,~ c =3$ and $\displaystyle d$ ranges from $\displaystyle 0$ to $\displaystyle 4$ .
$\displaystyle 125(1) + 25(4) + 5(3) + 0 \leq n \leq 125(1) + 25(4) + 5(3) + 4$
$\displaystyle 240 \leq n \leq 244$ .
| 781 | 2,070 |
{"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}
| 4.34375 | 4 |
CC-MAIN-2018-17
|
latest
|
en
| 0.658144 |
https://hoven-in.appspot.com/Home/Aptitude/Pipes-and-Cisterns/maths-aptitude-quiz-questions-and-mock-test-on-pipes-and-cisterns-006.html
| 1,660,275,356,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2022-33/segments/1659882571538.36/warc/CC-MAIN-20220812014923-20220812044923-00334.warc.gz
| 303,173,439 | 6,750 |
# Pipes and Cisterns Quiz Set 006
### Question 1
A tank is filled in \$1{2/33}\$ minutes by three taps running together. Times taken by the three taps to independently fill the tank are in an AP[Arithmetic Progression]. If the first tap is a leakage tap and the second tap takes 1 minute to fill the tank, then, the common difference of the AP can be?
A
6.
B
7.
C
5.
D
8.
Soln.
Ans: a
Let the times taken by the three taps be 1 - d, 1 and 1 + d. The time taken by the first tap will be negative because it is a leakage tap. Then \${35/33}\$ minutes work of all the taps should add to 1. So we have, \${35/33}\$ × \$(1/{1 - d} + 1/1 + 1/{1 + d})\$ = 1, which is same as \$2/{1 - d^2} + 1\$ = \${33/35}\$. Solving we get d = ±6.
### Question 2
A tank is (2/5)th filled with water. When 44 liters of water are added, it becomes (8/9)th filled. What is the capacity of the tank?
A
90 liters.
B
100 liters.
C
110 liters.
D
120 liters.
Soln.
Ans: a
Let x be the capacity in liters. \${2x}/5 + 44 = {8x}/9\$. Solving, x = 90 liters.
### Question 3
One tap can fill a tank 2 times faster than the other. If they together fill it in 9 minutes, how much time does the slower alone take to fill the tank?
A
27 mins.
B
3 mins.
C
4 mins.
D
5 mins.
Soln.
Ans: a
Let the one minute work of the taps be 1/x and 2/x. We have \$1/x + 2/x = 1/9\$, which gives x = 3 × 9 = 27 mins.
### Question 4
A tank is filled in 11 minutes by three taps running together. Times taken by the three taps independently are in an AP[Arithmetic Progression], whose first term is a and common difference d. Then, a and d satisfy the relation?
A
a3 - 33a2 - ad2 + 11d2 = 0.
B
a3 - 22a2 + ad2 + 11d2 = 0.
C
a3 - 11a2 - ad2 + 11d2 = 0.
D
a3 - 55a2 + ad2 + 11d2 = 0.
Soln.
Ans: a
Let the times taken by the three taps be a - d, a and a + d. Then 11 minutes work of all the taps should add to 1. So we have, \$11 × 1/{a - d} + 11 × 1/a + 11 × 1/{a + d}\$ = 1, which is same as a3 - 33a2 - ad2 + 11d2 = 0.
### Question 5
Tap X can fill the tank in 11 mins. Tap Y can empty it in 5 mins. In how many minutes will the tank be emptied if both the taps are opened together when the tank is \$8/10\$th full of water?
A
\$7{1/3}\$ mins.
B
\$12{1/2}\$ mins.
C
\$6{1/3}\$ mins.
D
\$6{1/5}\$ mins.
Soln.
Ans: a
1 filled tank can be emptied in \${11 × 5}/{11 - 5}\$ mins. So 8/10 can be emptied in \${11 × 5}/{11 - 5}\$ × \$8/10\$ = \${22/3}\$, which is same as: \$7{1/3}\$ mins.
This Blog Post/Article "Pipes and Cisterns Quiz Set 006" by Parveen (Hoven) is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License.
Updated on 2020-02-07. Published on: 2016-05-05
Posted by Parveen(Hoven),
Aptitude Trainer
| 971 | 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}
| 4.4375 | 4 |
CC-MAIN-2022-33
|
latest
|
en
| 0.889655 |
https://brainly.com/question/178032
| 1,484,726,475,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2017-04/segments/1484560280242.65/warc/CC-MAIN-20170116095120-00317-ip-10-171-10-70.ec2.internal.warc.gz
| 794,722,777 | 8,924 |
2014-11-07T00:55:08-05:00
### This Is a Certified Answer
Certified answers contain reliable, trustworthy information vouched for by a hand-picked team of experts. Brainly has millions of high quality answers, all of them carefully moderated by our most trusted community members, but certified answers are the finest of the finest.
13+4x=1-x
4x=-12-x (minus 13 from each side)
5x=-12 (add x to each side)
You divide -12 by 5x
-12÷5x=2⇒2.4
x=2.4
2014-11-07T01:09:01-05:00
Simplifying 13 + 4x = 1 + -1x
Solving 13 + 4x = 1 + -1x
Solving for variable 'x'.
Move all terms containing x to the left, all other terms to the right.
Add 'x' to each side of the equation.13 + 4x + x = 1 + -1x + x
Combine like terms: 4x + x = 5x 13 + 5x = 1 + -1x + x
Combine like terms: -1x + x = 0 13 + 5x = 1 + 0 13 + 5x = 1
Add '-13' to each side of the equation. 13 + -13 + 5x = 1 + -13
Combine like terms: 13 + -13 = 0 0 + 5x = 1 + -13 5x = 1 + -13
Combine like terms: 1 + -13 = -12 5x = -12 Divide each side by '5'. x = -2.4 Simplifying x = -2.4
| 401 | 1,029 |
{"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}
| 4.15625 | 4 |
CC-MAIN-2017-04
|
latest
|
en
| 0.868623 |
http://brainden.com/forum/index.php?/topic/411-take-me-there/
| 1,438,170,740,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2015-32/segments/1438042986423.95/warc/CC-MAIN-20150728002306-00210-ip-10-236-191-2.ec2.internal.warc.gz
| 34,482,276 | 16,951 |
Followers 0
# Take me there
## 18 posts in this topic
Posted · Report post
A 79 kgs man was having a problem crossing the bridge as he read the sign board content on the bridge. "This bridge capacity is up to 80 kgs only, if you exceed this limit the bridge will fall".
He has two pieces of water melon weighing 1 kg each and he need to bring home the fruit as he promised with his kids.
Can you suggest an idea how the man can cross the bridge?
0
##### Share on other sites
Posted · Report post
take 1 drop it on the otha side then go bak n gt the otha
0
##### Share on other sites
Posted · Report post
Nope!
0
##### Share on other sites
Posted · Report post
Take one in hand throw one in air and continue the same like throwing balls trick
0
##### Share on other sites
Posted · Report post
you've got the trick.
0
##### Share on other sites
Posted · Report post
Take one in hand throw one in air and continue the same like throwing balls trick
I do not believe this would work. If the absolute capacity of the bridge is 80 kg, the man weighs exactly 79 kg and each piece of watermelon weighs exactly 1 kg, the bridge would collapse from the force of the man throwing and catching the watermelon. It is actually very unlikely that the man could cross the bridge at all, even with just one piece of watermelon, as the force of each step would exceed the 80 kg limit. If the limit takes this extra force into consideration, then the man should be able to cross with both pieces by stepping very gently, and thus generating less down force.
0
##### Share on other sites
Posted · Report post
yes you're right. but it doesn't matter on the question. i didn't emphasize this to take into consideration. this was based on a simple math.
0
##### Share on other sites
Posted · Report post
Interesting take on the answer, but riddles should only have one possible answer. There are many different ways I could think of getting across with both watermelons.
One way of accomplishing this is stating in the riddle things like the man can not use any other objects not mentioned in the riddle, and that he can only cross the bridge once without it falling.
0
##### Share on other sites
Posted · Report post
but riddles should only have one possible answer.
Who says?
0
##### Share on other sites
Posted · Report post
I agree with irninjabob about the part where the riddle should specify what objects can be used. Our man could use a helicopter to fly home. Or he could get a really really really really really really really really really really really really really really really really bouncy trampoline. Or, take the long way and find another bridge. Or use his cell phone to call up his old pal Joe who is 78Kg to carry them across for him, and when Joe is finished crossing he goes himself.
0
##### Share on other sites
Posted · Report post
but riddles should only have one possible answer.
Who says?
Good point. There can be riddles that have different answers. What I meant was when you have a riddle with only one answer, it should only be possible to get that answer. If you are going to say "no" to an answer that technically would work simply because it wasn't the one you had in mind, then you intended there to be only one answer. To accomplish this, you add the extra information specifying what you can and can't do. Other than that, I thought this was a neat riddle. With that extra information this riddle would have been a lot of fun to work out.
0
##### Share on other sites
Posted · Report post
yes you're right. but it doesn't matter on the question. i didn't emphasize this to take into consideration. this was based on a simple math.
Sure it matters. Riddari is correct that juggling will not allow the man to cross the bridge and that's what matters. Here is a good discussion on the riddle.
0
##### Share on other sites
Posted · Report post
Doing the math and explaining that downward forces are needed to produce upward acceleration, etc. is absolutely correct; even tho the numbers change depending on how, and how high, he throws them. One might be tempted to suggest he could throw them very gently upward ... this approach fails, but more math is needed to prove it.
A general argument states: If the total weight of the juggler and the watermelons is NOT being supported by the bridge [the only supporting structure present] then some or all of them will fall, and juggler will fail to get them across. Thus, the bridge DOES support the entire weight, and juggling is seen not to be a solution.
... unless ... the bridge is so short he can toss them into the air before getting on the bridge -- and catch them after he has finished crossing it.
0
##### Share on other sites
Posted · Report post
Doing the math and explaining that downward forces are needed to produce upward acceleration, etc. is absolutely correct; even tho the numbers change depending on how, and how high, he throws them. One might be tempted to suggest he could throw them very gently upward ... this approach fails, but more math is needed to prove it.
Throwing them "very gently upward" will not work - it does not change the amount of force needed - just the amount of time the watermelons spend in the air. Think of it this way: if he just stands there with a watermelon in his hand, he exerts 80kg down force (79 for himself, 1 for the watermelon). To move the watermelon upward - at any speed - requires additional force be exerted downward, exceeding the 80kg limit and collapsing the bridge.
(In reality, kgs are not a unit of force. The bridge is capable of supporting 80kg x 9.8N/kg = 784 Newtons of force. The man exerts 774.2N and the watermelon exerts 9.8N. We can just stick with kgs for simplicity)
... unless ... the bridge is so short he can toss them into the air before getting on the bridge -- and catch them after he has finished crossing it.
I like that solution. I had thought about horizontal juggling (yes, it exists - I do it myself). Basically, think of regular juggling of three items, but reduce the height. Keep reducing it until the majority of the force used in throwing the ball is along the horizontal plane rather than vertical. At this point you are juggling very quickly, but with very little upward force. Unless the bridge has really bad horizontal stability, this would allow juggling to get the melons across. But to do this with watermelons would be extremely difficult. And he must still walk very slowly, so as not to exert more than 9.8N while lifting his body weight or placing his foot. That would be unlikely.
I wonder if he could bowl the watermelons across and then walk? I guess that depends on the length of the bridge.
I say that he should walk downriver to a bridge that has been properly constructed, get is watermelons across, then call the newspaper to report on the shoddy construction practices, lack of inspections and obvious safety concerns raised by his recent ordeal. Someone should be doing jail-time over this!
0
##### Share on other sites
Posted · Report post
The man should exercise vigorously for about an hour, Boxers seen to find skipping does the trick, once he has lost 2 KG, he will be able to casually walk across proudly holding on to his watermelons (is it just me or does that sound like a euphemism, perhaps he hurt them skipping?)
Or if he was a Catholic, he could just give up religion, he would then be able to get across easily (less Mass)
0
##### Share on other sites
Posted · Report post
I think that "IrNinjaBob" is quite intelligent, and quite correct. In many riddles, there are more than one answer (some sarcastic nd unlikely, but nonetheless able to solve the puzzle), and there are some which there is only an absolute one answer. This is because of the information given. If the information given is too broad, like "What can you put in a bucket to make it lighter?" can have multiple answers. If the person posting a question wants merely one answer, than the person should put more information, else he or she will get many responses.
For the puzzle, I think that throwing the watermelons would increase the weight, as "Riddari" had said, and the bridge would crack (as well as the watermelons would most likely break, but that is out of question and unneeded).
My answer is that the man carry one water melon with him, weighing a perfect 80kg, put down the water melon on the other side, than walk back to the other watermelond, weighing 79kg, and then carry that watermelon back to the other side, weighing a perfect 80kg. This answer is plausible, reasonable, and simple, is it not?
0
##### Share on other sites
Posted · Report post
... the man carry one water melon with him, weighing a perfect 80kg, put down the water melon on the other side, than walk back to the other watermelond, weighing 79kg, and then carry that watermelon back to the other side, weighing a perfect 80kg. This answer is plausible, reasonable, and simple, is it not?
Plausible, reasonable and simple. Compelling, even.
Oh wait. The problem states:
"This bridge capacity is up to 80 kgs only, if you exceed this limit the bridge will fall".
"up to 80 kg only" [an open interval] does not include 80. Therefore "a perfect 80kg" "exceed this limit," and so "the bridge will fall."
0
##### Share on other sites
Posted · Report post
"This bridge capacity is up to 80 kgs only, if you exceed this limit the bridge will fall".
"up to 80 kg only" [an open interval] does not include 80. Therefore "a perfect 80kg" "exceed this limit," and so "the bridge will fall."
You are correct, my mistake.
Seeing this as not an option, pursuing other ideas:
Throwing it - This would work, theoretically, if it does not matter if the watermelon is damaged.
Rolling it - This would have identical effects, only the water melon would not be broken beyond holding, but it may have a few dents.
the bridge is so short he can toss them into the air before getting on the bridge -- and catch them after he has finished crossing it.
Yes, this couldwork as well, as the length of the bridge is unspecified.
Well done, bonanova.
0
## Create an account
Register a new account
| 2,323 | 10,216 |
{"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-2015-32
|
latest
|
en
| 0.950611 |
https://www.tutorialspoint.com/p-factorize-the-expression-a-2-49b-2-p
| 1,701,248,741,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2023-50/segments/1700679100057.69/warc/CC-MAIN-20231129073519-20231129103519-00160.warc.gz
| 1,173,267,780 | 20,383 |
# Factorize the expression: $a^2-49b^2$.
Given: Expression: $a^2-49b^2$.
To do: To factorize $a^2-49b^2$.
Solution:
$a^2-49b^2$
$=( a)^2-( 7b)^2$
$=( a-7b)( a+7b)$
Tutorialspoint
Simply Easy Learning
Updated on: 10-Oct-2022
34 Views
##### Kickstart Your Career
Get certified by completing the course
| 119 | 311 |
{"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-2023-50
|
latest
|
en
| 0.388302 |
https://origin.geeksforgeeks.org/check-number-power-k-using-base-changing-method/
| 1,685,505,588,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2023-23/segments/1685224646257.46/warc/CC-MAIN-20230531022541-20230531052541-00372.warc.gz
| 455,151,762 | 47,189 |
Get the best out of our app
GFG App
Open App
Browser
Continue
# Check if a number is power of k using base changing method
This program checks whether a number n can be expressed as power of k and if yes, then to what power should k be raised to make it n. Following example will clarify :
Examples:
```Input : n = 16, k = 2
Output : yes : 4
Explanation : Answer is yes because 16 can
be expressed as power of 2.
Input : n = 27, k = 3
Output : yes : 3
Explanation : Answer is yes as 27 can be
expressed as power of 3.
Input : n = 20, k = 5
Output : No
Explanation : Answer is No as 20 cannot
be expressed as power of 5. ```
We have discussed two methods in below post
:Check if a number is a power of another number
In this post, a new Base Changing method is discussed.
In Base Changing Method, we simply change the base of number n to k and check if the first digit of Changed number is 1 and remaining all are zero.
Example for this : Let’s take n = 16 and k = 2.
Change 16 to base 2. i.e. (10000)2. Since first digit is 1 and remaining are zero. Hence 16 can be expressed as power of 2. Count the length of (10000)2 and subtract 1 from it, that’ll be the number to which 2 must be raised to make 16. In this case 5 – 1 = 4.
Another example : Let’s take n = 20 and k = 3.
20 in base 3 is (202)3. Since there are two non-zero digit, hence 20 cannot be expressed as power of 3.
## C++
`// CPP program to check if a number can be` `// raised to k` `#include ` `#include ` `using` `namespace` `std;` `bool` `isPowerOfK(unsigned ``int` `n, unsigned ``int` `k)` `{` ` ``// loop to change base n to base = k` ` ``bool` `oneSeen = ``false``;` ` ``while` `(n > 0) {` ` ``// Find current digit in base k` ` ``int` `digit = n % k;` ` ``// If digit is neither 0 nor 1 ` ` ``if` `(digit > 1)` ` ``return` `false``;` ` ``// Make sure that only one 1` ` ``// is present. ` ` ``if` `(digit == 1)` ` ``{` ` ``if` `(oneSeen)` ` ``return` `false``;` ` ``oneSeen = ``true``;` ` ``} ` ` ``n /= k;` ` ``}` ` ` ` ``return` `true``; ` `}` `// Driver code` `int` `main()` `{` ` ``int` `n = 64, k = 4;` ` ``if` `(isPowerOfK(n ,k))` ` ``cout << ``"Yes"``;` ` ``else` ` ``cout << ``"No"``;` `}`
## Java
`// Java program to check if a number can be` `// raised to k` `class` `GFG` `{` ` ``static` `boolean` `isPowerOfK(``int` `n,``int` `k)` ` ``{` ` ``// loop to change base n to base = k` ` ``boolean` `oneSeen = ``false``;` ` ``while` `(n > ``0``) ` ` ``{` ` ` ` ``// Find current digit in base k` ` ``int` `digit = n % k;` ` ` ` ``// If digit is neither 0 nor 1 ` ` ``if` `(digit > ``1``)` ` ``return` `false``;` ` ` ` ``// Make sure that only one 1` ` ``// is present. ` ` ``if` `(digit == ``1``)` ` ``{` ` ``if` `(oneSeen)` ` ``return` `false``;` ` ``oneSeen = ``true``;` ` ``} ` ` ` ` ``n /= k;` ` ``}` ` ` ` ``return` `true``; ` ` ``}` ` ` ` ``// Driver code` ` ``public` `static` `void` `main (String[] args) ` ` ``{` ` ``int` `n = ``64``, k = ``4``;` ` ` ` ``if` `(isPowerOfK(n ,k))` ` ``System.out.print(``"Yes"``);` ` ``else` ` ``System.out.print(``"No"``);` ` ``}` `}` `// This code is contributed by Anant Agarwal.`
## Python3
`# Python program to` `# check if a number can be` `# raised to k` `def` `isPowerOfK(n, k):` ` ``# loop to change base` ` ``# n to base = k` ` ``oneSeen ``=` `False` ` ``while` `(n > ``0``):` ` ` ` ``# Find current digit in base k` ` ``digit ``=` `n ``%` `k` ` ` ` ``# If digit is neither 0 nor 1 ` ` ``if` `(digit > ``1``):` ` ``return` `False` ` ` ` ``# Make sure that only one 1` ` ``# is present. ` ` ``if` `(digit ``=``=` `1``):` ` ` ` ``if` `(oneSeen):` ` ``return` `False` ` ``oneSeen ``=` `True` ` ` ` ``n ``/``/``=` `k` ` ` ` ``return` `True` ` ` `# Driver code` `n ``=` `64` `k ``=` `4` ` ` `if` `(isPowerOfK(n , k)):` ` ``print``(``"Yes"``)` `else``:` ` ``print``(``"No"``)` `# This code is contributed` `# by Anant Agarwal.`
## C#
`// C# program to check if a number can be` `// raised to k` `using` `System;` `class` `GFG {` ` ` ` ``static` `bool` `isPowerOfK(``int` `n, ``int` `k)` ` ``{` ` ` ` ``// loop to change base n to base = k` ` ``bool` `oneSeen = ``false``;` ` ``while` `(n > 0) ` ` ``{` ` ` ` ``// Find current digit in base k` ` ``int` `digit = n % k;` ` ` ` ``// If digit is neither 0 nor 1 ` ` ``if` `(digit > 1)` ` ``return` `false``;` ` ` ` ``// Make sure that only one 1` ` ``// is present. ` ` ``if` `(digit == 1)` ` ``{` ` ``if` `(oneSeen)` ` ``return` `false``;` ` ` ` ``oneSeen = ``true``;` ` ``} ` ` ` ` ``n /= k;` ` ``}` ` ` ` ``return` `true``; ` ` ``}` ` ` ` ``// Driver code` ` ``public` `static` `void` `Main () ` ` ``{` ` ``int` `n = 64, k = 4;` ` ` ` ``if` `(isPowerOfK(n ,k))` ` ``Console.WriteLine(``"Yes"``);` ` ``else` ` ``Console.WriteLine(``"No"``);` ` ``}` `}` `// This code is contributed by vt_m.`
## PHP
` 0) ` ` ``{` ` ``// Find current ` ` ``// digit in base k` ` ``\$digit` `= ``\$n` `% ``\$k``;` ` ``// If digit is ` ` ``// neither 0 nor 1 ` ` ``if` `(``\$digit` `> 1)` ` ``return` `false;` ` ``// Make sure that` ` ``// only one 1` ` ``// is present. ` ` ``if` `(``\$digit` `== 1)` ` ``{` ` ``if` `(``\$oneSeen``)` ` ``return` `false;` ` ``\$oneSeen` `= true;` ` ``} ` ` ``\$n` `= (int)``\$n` `/ ``\$k``;` ` ``}` ` ` ` ``return` `true; ` `}` `// Driver code` `\$n` `= 64;` `\$k` `= 4;` `if` `(isPowerOfK(``\$n``, ``\$k``))` ` ``echo` `"Yes"``;` `else` ` ``echo` `"No"``;` `// This code is contributed ` `// by ajit` `?>`
## Javascript
``
Output:
`Yes`
Time Complexity: O(logK n)
Space Complexity: O(1)
Optimized Approach:
This approach avoids the need to convert n to base k and check whether it can be represented using only the digits 0 and 1. It also avoids the need to track whether a 1 has already been seen. This results in a simpler and more efficient algorithm.
Here’s a step-by-step explanation of the code:
1. Define the isPrime function which takes an integer n as input and returns true if n is prime, and false otherwise.
2. Define the isSumOfPrimes function with parameter n.
3. Loop over all numbers from 2 to n/2 (inclusive) as potential prime numbers, and check whether each one is a prime and whether the difference between n and that number is also a prime. The loop continues until the first pair of primes is found.
4. If a pair of primes is found, return true. Otherwise, return false.
5. In the main function, set n to the desired value.
6. Call the isSumOfPrimes function with n.
7. If the function returns true, print “Yes” to the console, indicating that n can be expressed as the sum of two prime numbers. Otherwise, print “No”.
## C++
`#include ` `using` `namespace` `std;` `bool` `isPowerOfK(``int` `n, ``int` `k) {` ` ``// Check for base cases` ` ``if` `(n == 0 || k == 0 || k == 1) {` ` ``return` `false``;` ` ``}` ` ``// Check if n is a power of k` ` ``while` `(n % k == 0) {` ` ``n /= k;` ` ``}` ` ``return` `n == 1;` `}` `int` `main() {` ` ``int` `n = 64, k = 4;` ` ``if` `(isPowerOfK(n, k)) {` ` ``cout << ``"Yes"``;` ` ``} ``else` `{` ` ``cout << ``"No"``;` ` ``}` ` ``return` `0;` `}`
## Java
`class` `GFG {` ` ``static` `boolean` `isPowerOfK(``int` `n, ``int` `k) {` ` ``// Check for base cases` ` ``if` `(n == ``0` `|| k == ``0` `|| k == ``1``) {` ` ``return` `false``;` ` ``}` ` ``// Check if n is a power of k` ` ``while` `(n % k == ``0``) {` ` ``n /= k;` ` ``}` ` ``return` `n == ``1``;` ` ``}` ` ``public` `static` `void` `main(String[] args) {` ` ``int` `n = ``64``, k = ``4``;` ` ``if` `(isPowerOfK(n, k)) {` ` ``System.out.print(``"Yes"``);` ` ``} ``else` `{` ` ``System.out.print(``"No"``);` ` ``}` ` ``}` `}`
## Python3
`def` `isPowerOfK(n, k):` ` ``# Check for base cases` ` ``if` `n ``=``=` `0` `or` `k ``=``=` `0` `or` `k ``=``=` `1``:` ` ``return` `False` ` ``# Check if n is a power of k` ` ``while` `n ``%` `k ``=``=` `0``:` ` ``n ``/``/``=` `k` ` ``return` `n ``=``=` `1` `n ``=` `64` `k ``=` `4` `if` `isPowerOfK(n, k):` ` ``print``(``"Yes"``)` `else``:` ` ``print``(``"No"``)`
## C#
`using` `System;` `class` `GFG {` ` ``static` `bool` `IsPowerOfK(``int` `n, ``int` `k)` ` ``{` ` ` ` ``// Check for base cases` ` ``if` `(n == 0 || k == 0 || k == 1) {` ` ``return` `false``;` ` ``}` ` ``// Check if n is a power of k` ` ``while` `(n % k == 0) {` ` ``n /= k;` ` ``}` ` ``return` `n == 1;` ` ``}` ` ``static` `void` `Main(``string``[] args) {` ` ``int` `n = 64, k = 4;` ` ``if` `(IsPowerOfK(n, k)) {` ` ``Console.Write(``"Yes"``);` ` ``} ``else` `{` ` ``Console.Write(``"No"``);` ` ``}` ` ``}` `}`
## Javascript
`function` `isPowerOfK(n, k) {` ` ``// Check for base cases` ` ``if` `(n === 0 || k === 0 || k === 1) {` ` ``return` `false``;` ` ``}` ` ``// Check if n is a power of k` ` ``while` `(n % k === 0) {` ` ``n = Math.floor(n / k);` ` ``}` ` ``return` `n === 1;` `}` `let n = 64;` `let k = 4;` `if` `(isPowerOfK(n, k)) {` ` ``console.log(``"Yes"``);` `} ``else` `{` ` ``console.log(``"No"``);` `}`
OUTPUT:
`YES`
Time Complexity: O(logK n)
Space Complexity: O(1)
This article is contributed by Shubham Rana. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to [email protected]. See your article appearing on the GeeksforGeeks main page and help other Geeks.
Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.
My Personal Notes arrow_drop_up
Similar Reads
Related Tutorials
| 3,996 | 10,986 |
{"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-2023-23
|
latest
|
en
| 0.821967 |
http://slideshowes.com/doc/213763/context-free-languages---university-of-virginia
| 1,493,084,467,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2017-17/segments/1492917120001.0/warc/CC-MAIN-20170423031200-00559-ip-10-145-167-34.ec2.internal.warc.gz
| 355,496,620 | 13,571 |
Lecture 11:
Parsimonious
Parsing
cs302: Theory of Computation
University of Virginia
Computer Science
David Evans
http://www.cs.virginia.edu/evans
• Fix proof from last class
• Interpretive Dance!
• Parsimonious Parsing (Parsimoniously)
PS3 Comments Available Today
PS3 will be returned Tuesday
Lecture 11: Parsimonious Parsing
2
Closure Properties of CFLs
If A and B are context free languages then:
AR is a context-free language TRUE
A* is a context-free language TRUE
A is a context-free language (complement)?
A B is a context-free language TRUE
A B is a context-free language ?
Lecture 11: Parsimonious Parsing
3
Complementing Non-CFLs
Lww = {ww | w Σ* } is not a CFL.
Is its complement?
What is
the actual
language?
Yes. This CFG recognizes is:
S 0S0 | 1S1 | 0X1 | 1X0
X 0X0 | 1X1 | 0X1 | 1X0 | 0 | 1 | ε
Bogus Proof!
S 0X1 01X01 0101 Lww
Lecture 11: Parsimonious Parsing
4
CFG for Lww (L )
ww
All odd length strings are in Lww
S SOdd | SEven
SEven XY | YX
X ZXZ | 0
Y ZYZ | 1
Z0|1
SOdd 0R | 1R | 0 | 1
R 0SOdd | 1SOdd
How can we prove this is correct?
Lecture 11: Parsimonious Parsing
5
Sodd generates all
odd-length strings
SOdd 0R | 1R | 0 | 1
R 0SOdd | 1SOdd
Proof by induction on the length of the string.
Basis. SOdd generates all odd-length strings of
length 1. There are two possible strings: 0 and 1.
They are produces from the 3rd and 4th rules.
Induction. Assume SOdd generates all odd-length
strings of length n for n = 2k+1, k 0. Show it can
generate all odd-length string of length n+2.
Lecture 11: Parsimonious Parsing
6
SOdd generates all
odd-length strings
SOdd 0R | 1R | 0 | 1
R 0SOdd | 1SOdd
Induction. Assume SOdd generates all odd-length strings
of length n for n = 2k+1, k 0. Show it can generate all
odd-length string of length n+2.
All n+2 length strings are of the form abt where t is an nlength string and a {0, 1}, b {0, 1}. There is some
derivation from SOdd * t (by the induction hypothesis). We
can generate all four possibilities for a and b:
00t: SOdd 0R 00SOdd * 00t
01t: SOdd 0R 01SOdd * 01t
10t: SOdd 1R 10SOdd * 10t
11t: SOdd 1R 11SOdd * 01t
Lecture 11: Parsimonious Parsing
7
CFG for Lww (L )
ww
S SOdd | SEven
SEven XY | YX
X ZXZ | 0
Y ZYZ | 1
Z0|1
SOdd 0R | 1R | 0 | 1
R 0SOdd | 1SOdd
?
Proof-by-leaving-as-“Challenge
Problem” (note: you cannot use this
Lecture 11: Parsimonious Parsing
8
Even Strings
Show S generates the set
of all even-length strings
that are not in Lww.
Even
SEven XY | YX
X ZXZ | 0
Y ZYZ | 1
Z0|1
Proof by induction on the length of the string.
Basis. SEven generates all even-length strings of
length 0 that are not in Lww. The only length 0
string is ε. ε is in Lww since ε = εε, so ε should not be
generated by SEven. Since SEven does not contain any right
sides that go to ε, this is correct.
Lecture 11: Parsimonious Parsing
9
Closure Properties of CFLs
If A and B are context free languages then:
AR is a context-free language TRUE
A* is a context-free language TRUE
A is not necessarily a context-free
language (complement)
A B is a context-free language TRUE
A B is a context-free language ?
Lecture 11: Parsimonious Parsing
10
Left for you to solve
(possibly on Exam 1)
Where is English?
0n1n
0n
Described by DFA, NFA,
RegExp, RegGram
w
Regular Languages
Context-Free Languages
Lecture 11: Parsimonious Parsing
11
0n1n2n
A
ww
English Regular Languages
The cat likes fish.
The cat the dog chased likes fish.
The cat the dog the rat bit chased likes fish.
…
This is a pumping lemma proof!
Lecture 11: Parsimonious Parsing
12
= DFA
= CFG
Chomsky’s
(Syntactic
Structures,
1957)
Lecture 11: Parsimonious Parsing
13
• Most linguists argue that most
natural languages are not contextfree
• But, it is hard to really answer this
question:
e.g.,
“The cat the dog the rat bit chased likes fish.” English?
Lecture 11: Parsimonious Parsing
14
Where is Java?
0n1n
0n
Described by DFA, NFA,
RegExp, RegGram
w
Regular Languages
Context-Free Languages
Lecture 11: Parsimonious Parsing
15
0n1n2n
A
ww
Interpretive Dance
Lecture 11: Parsimonious Parsing
16
Where is Java?
0n1n
0n
Described by DFA, NFA,
RegExp, RegGram
w
Regular Languages
Context-Free Languages
Lecture 11: Parsimonious Parsing
17
0n1n2n
A
ww
What is the Java Language?
public class Test {
In the Java
public static void main(String [] a) {
Language
println("Hello World!");
}
Test.java:3: cannot resolve symbol
}
symbol : method println (java.lang.String)
// C:\users\luser\Test.java
public class Test {
Not in the Java
public static void main(String [] a) {
Language
println ("Hello Universe!");
}
Test.java:1: illegal unicode escape
}}
// C:\users\luser\Test.java
Lecture 11: Parsimonious Parsing
18
// C:\users\luser\Test.java
public class Test {
public static void main(String [] a) {
println ("Hello Universe!");
> javac Test.java
}
Test.java:1: illegal unicode escape
}}
// C:\users\luser\Test.java
Scanning error
^
Test.java:6: 'class' or 'interface' expected
}}
^
Parsing errors
Test.java:7: 'class' or 'interface' expected
Static semantic errors
Lecture 11: Parsimonious Parsing
^
Test.java:4: cannot resolve symbol
symbol : method println (java.lang.String)
location: class Test
println ("Hello World");
^
4 errors
19
Defining the Java Language
{ w | w can be generated by the CFG
for Java in the Java Language
Specification }
{ w | a correct Java compiler can build
a parse tree for w }
Lecture 11: Parsimonious Parsing
20
Parsing
Lecture 11: Parsimonious Parsing
+
M
M
M
T
T
3
2
3
*
T
1
+ 2 * 1
21
Parsing
Programming
languages
are (should be)
designed to make
parsing easy,
efficient, and
unambiguous.
S
Derivation
S S+M|M
MM*T | T
T (S) | number
S
Unambiguous
S S + S | S * S | (S) | number
S
S
3
S
+
S
S
2
3
*
S
S
1
3
+ 2 * 1
3
Lecture 11: Parsimonious Parsing
*
S
22
+
S
S
1
2
+ 2 * 1
Ambiguity
How can one determine if a CFG is ambiguous?
Super-duper-challenge problem: create a program that
solve the “is this CFG ambiguous” problem:
Input: CFG
Output: “Yes” (ambiguous)/“No” (unambiguous)
Warning: Undecidable Problem Alert!
(Not only can you not do this, it is impossible
for any program to do this.)
(We will cover undecidable
problems after Spring Break)
Lecture 11: Parsimonious Parsing
23
Parsing
Lecture 11: Parsimonious Parsing
+
M
M
M
T
T
3
2
3
*
T
1
+ 2 * 1
24
Parsing
Programming
languages
are (should be)
designed to make
parsing easy,
efficient, and
unambiguous.
S
Derivation
S S+M|M
MM*T | T
T (S) | number
S
“Easy” and “Efficient”
• “Easy” - we can automate the
process of building a parser from a
description of a grammar
• “Efficient” – the resulting parser can
build a parse tree quickly (linear time
in the length of the input)
Lecture 11: Parsimonious Parsing
25
Recursive Descent
Parsing
S S+M|M
MM*T | T
T (S) | number
Parse() { S(); }
S() {
• Easy to produce
try { S(); expect(“+”); M(); }
and understand
catch { backup(); }
• Can be done for
try { M(); } catch {backup(); }
any CFG
error(); }
M() {
try { M(); expect(“*”); T(); } catch … Problems:
• Inefficient (might
try { T(); } catch { backup(); }
not even finish)
error (); }
• “Nondeterministic”
T() {
try { expect(“(“); S(); expect(“)”); } catch …;
try { number(); } catch …; }
Lecture 11: Parsimonious Parsing
26
• A CFG is an LL(k) grammar if it can
be parser deterministically with
S S+M|M
MM*T | T
T (S) | number
1
S S+M
SM
LL(1) grammar
Lecture 11: Parsimonious Parsing
27
+
S S+M
2
Parse() { S(); }
S() {
if (lookahead(1, “+”)) { S(); eat(“+”); M(); }
else { M();}
M() {
if (lookahead(1, “*”)) { M(); eat(“*”); T(); }
else { T(); } }
T() {
if (lookahead(0, “(“)) { eat(“(“); S(); eat(“)”); }
else { number();}
S S+M|M
MM*T | T
T (S) | number
Lecture 11: Parsimonious Parsing
28
JavaCC
https://javacc.dev.java.net/
• Input: Grammar specification
• Output: A Java program that is a
recursive descent parser for the
specified grammar
Doesn’t work for all CFGs: only for LL(k) grammars
Lecture 11: Parsimonious Parsing
29
Language Classes
Scheme
0n1n
0n1n2n
ww
0n
Described by DFA, NFA,
RegExp, RegGram
w
Regular Languages
Java
Python
Lecture 11: Parsimonious Parsing
30
Next Week
• Monday (2): Office Hours
(Qi Mi in 226D)
• Monday (5:30): TA help session
• Tuesday’s class (Pieter Hooimeyer): starting
to get outside the yellow circle: using
grammars to solve security problems
• Wednesday (9:30am): Office Hours (Qi Mi
in 226D)
• Wednesday (6pm): TAs’ Exam Review
• Thursday: exam in class
Lecture 11: Parsimonious Parsing
31
| 2,961 | 8,477 |
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 3.78125 | 4 |
CC-MAIN-2017-17
|
longest
|
en
| 0.79575 |
http://mathhelpforum.com/calculus/158135-limit-infinity-equaling-infinity.html
| 1,571,499,524,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2019-43/segments/1570986696339.42/warc/CC-MAIN-20191019141654-20191019165154-00311.warc.gz
| 133,283,588 | 11,827 |
# Thread: Limit at infinity equaling infinity?
1. ## Limit at infinity equaling infinity?
Hi guys,
I'm doing limit exercises and I got this question:
Evaluate the following limit:
lim
x → −∞ √(x2 + 5 x + 1) - x
After working it out by multiplying by conjugate I get 5/0 which is undefined, but I know the answer is infinity. So where did I go wrong? This is the last step which I deduced 5/0 from:
(5 + (0))/-sqrt(1 + (0) + (0)) + 1
Any clarification is appreciated
2. Hello, DannyMath!
$\displaystyle \displaystyle \text{Evaluate: }\;\lim_{x\to-\infty}\sqrt{x^2+5x+1} - x$
$\displaystyle \text{. . . but I know the answer is infinity.}$ .?
Multiply numerator and denominator by the conjugate:
. . $\displaystyle \displaystyle \frac{\sqrt{x^2+5x+1} - x}{1}\cdot\frac{\sqrt{x^2+5x+1} + x}{\sqrt{x^2+5x+1} + x}$
. . $\displaystyle \displaystyle =\;\frac{(x^2+5x +1) - x^2}{\sqrt{x^2+5x+1} + x} \;=\; \frac{5x+1}{\sqrt{x^2+5x+1} + x}$
$\displaystyle \displaystyle \text{Divide numerator and denominator by }x\!:\;\; \displaystyle \frac{\dfrac{5x+1}{x}}{\dfrac{\sqrt{x^2+5x+1} + x}{x}}\;\;\bf[1]$
$\displaystyle \displaystyle \text{The numerator is: }\;\frac{5x}{x} + \frac{1}{x} \;=\;5 + \frac{1}{x}$
$\displaystyle \displaystyle \text{The denominator is: }\;\frac{\sqrt{x^2+5x+1}}{x} + \frac{x}{x}$
. . $\displaystyle \displaystyle \;=\;\frac{\sqrt{x^2+5x+1}}{\sqrt{x^2}} + 1\;=\;\sqrt{\frac{x^2+5x+1}{x^2}} + 1$
. . $\displaystyle \displaystyle =\;\sqrt{\frac{x^2}{x^2} + \frac{5x}{x^2} + \frac{1}{x^2}} + 1 \;=\; \sqrt{1 + \frac{5}{x} + \frac{1}{x^2}} + 1$
$\displaystyle \displaystyle \text{Then }{\bf[1]}\text{ becomes: }\; \frac{5 + \frac{1}{x}}{\sqrt{1+\frac{5}{x} + \frac{1}{x^2}} + 1}$
$\displaystyle \displaystyle \text{Therefore: }\;\lim_{x\to-\infty}\frac{5+\frac{1}{x}}{\sqrt{1 + \frac{5}{x} + \frac{1}{x^2}} + 1} \;=\;\frac{5 + 0}{\sqrt{1+0+0} + 1} \;=\;\frac{5}{2}$
3. Hey, thanks for the response! This was actually a 2 part exercise I was given for homework and I only posted the part (b). Part (a) was the exact same question but x goes to infinity rather than negative infinity. For that one I got 5/2 and the system told me I was right. But for the one I posted I got 5/0 but my graphing program showed it going to no particular horizontal asymptote so I had a sneaking suspicion that it was infinity. So that's what I submitted and it said I was correct :S ??
Also, in the denominator, since you're turning x into sqrt(x^2), aren't you losing a negative sign if you don't write it -sqrt(x^2)? Since x would be negative as x >> infinity?
4. Originally Posted by Soroban
Hello, DannyMath!
Multiply numerator and denominator by the conjugate:
. . $\displaystyle \displaystyle \frac{\sqrt{x^2+5x+1} - x}{1}\cdot\frac{\sqrt{x^2+5x+1} + x}{\sqrt{x^2+5x+1} + x}$
. . $\displaystyle \displaystyle =\;\frac{(x^2+5x +1) - x^2}{\sqrt{x^2+5x+1} + x} \;=\; \frac{5x+1}{\sqrt{x^2+5x+1} + x}$
$\displaystyle \displaystyle \text{Divide numerator and denominator by }x\!:\;\; \displaystyle \frac{\dfrac{5x+1}{x}}{\dfrac{\sqrt{x^2+5x+1} + x}{x}}\;\;\bf[1]$
$\displaystyle \displaystyle \text{The numerator is: }\;\frac{5x}{x} + \frac{1}{x} \;=\;5 + \frac{1}{x}$
$\displaystyle \displaystyle \text{The denominator is: }\;\frac{\sqrt{x^2+5x+1}}{x} + \frac{x}{x}$
. . $\displaystyle \displaystyle \;=\;\frac{\sqrt{x^2+5x+1}}{\sqrt{x^2}} + 1\;=\;\sqrt{\frac{x^2+5x+1}{x^2}} + 1$
You can't do the above since $\displaystyle x<0$...it can't go into the square root. What has to be done is:
$\displaystyle \frac{\sqrt{x^2+5x+1}}{x} + \frac{x}{x}=-\frac{\sqrt{x^2+5x+1}}{-x} + \frac{x}{x}=-\sqrt{\frac{x^2}{x^2} + \frac{5x}{x^2} + \frac{1}{x^2}} + 1 \;=-\; \sqrt{1 + \frac{5}{x} + \frac{1}{x^2}} + 1$...etc.
At the end the limit indeed is $\displaystyle \infty$
Tonio
. . $\displaystyle \displaystyle =\;\sqrt{\frac{x^2}{x^2} + \frac{5x}{x^2} + \frac{1}{x^2}} + 1 \;=\; \sqrt{1 + \frac{5}{x} + \frac{1}{x^2}} + 1$
$\displaystyle \displaystyle \text{Then }{\bf[1]}\text{ becomes: }\; \frac{5 + \frac{1}{x}}{\sqrt{1+\frac{5}{x} + \frac{1}{x^2}} + 1}$
$\displaystyle \displaystyle \text{Therefore: }\;\lim_{x\to-\infty}\frac{5+\frac{1}{x}}{\sqrt{1 + \frac{5}{x} + \frac{1}{x^2}} + 1} \;=\;\frac{5 + 0}{\sqrt{1+0+0} + 1} \;=\;\frac{5}{2}$
.
5. Thank you Tonio, I did indeed get that but doesn't that end up being (5 + (0))/-sqrt(1 + (0) + (0)) + 1 which turns into 5/0? I've just started limits so I don't know if that is the correct way to finish the problem (obviously it's not since it gives wrong answer :P)
6. Originally Posted by DannyMath
Thank you Tonio, I did indeed get that but doesn't that end up being (5 + (0))/-sqrt(1 + (0) + (0)) + 1 which turns into 5/0? I've just started limits so I don't know if that is the correct way to finish the problem (obviously it's not since it gives wrong answer :P)
Well, as you know you can't divide by zero...
What you actually get is an expression of the form $\displaystyle \frac{c}{g(x)}$ , with $\displaystyle c$ a constant and $\displaystyle g(x)$ a
function which converges to zero. It's not hard to see, even by means of the $\displaystyle \epsilon,\,\delta$ definition, that the limit of
this expression has as limit $\displaystyle \pm\infty$ , depending on the constant's and the function's signs.
Tonio
7. Oh ok I will work on understanding that. We skipped the precise delta/epsilon description in our books and were just told to divide by greatest x degree in denominator. I just tried to go as far as I could with the definition. Maybe one day it will be as easy to see for me as it is for you guys! HAHA! Thanks!
| 2,049 | 5,611 |
{"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}
| 4.25 | 4 |
CC-MAIN-2019-43
|
latest
|
en
| 0.573875 |
https://www.smore.com/wj6v
| 1,540,277,101,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2018-43/segments/1539583516071.83/warc/CC-MAIN-20181023044407-20181023065907-00350.warc.gz
| 1,066,044,271 | 11,793 |
# Matrix
### part of math
The orgins of mathematical matrices lie with the study of systems of simultaneous linear equations. The term "matrix" for such arrangements was introduced in 1850 by James Joseph Sylvester.Olga Taussky Todd (1906-1995), who began by using matrices to analyze vibrations on airplanes during World War II and became the torchbearer for matrix theory.
## Uses of Matrix
-solving systems of simultaneous linear equations -describing the quantum mechanics of atomic structure -designing computer game graphics -analyzing relationships -plotting complicated dance steps
## Example of Matrix
x + y + z =6 2y + 5z = -4 2x + 5y - z = 27
We can call the matrices "A", "X" and "B" and the equation becomes: AX=B Where: *A is the 3x3 matrix of x , y and z coefficients *X is x, y and z, *B is 6, -4 and 27 Then the solution is this: X=BA^-1
X=5 Y=3 Z=-2
| 248 | 907 |
{"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-2018-43
|
latest
|
en
| 0.827414 |
https://www.numerade.com/questions/in-section-92-youll-see-the-identity-sin-2-xfrac12-frac12-cos-2-x-use-this-identity-to-graph-the-fun/
| 1,620,569,991,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2021-21/segments/1620243988986.98/warc/CC-MAIN-20210509122756-20210509152756-00408.warc.gz
| 934,101,417 | 25,724 |
Our Discord hit 10K members! 🎉 Meet students and ask top educators your questions.Join Here!
# In Section 9.2 you'll see the identity $\sin ^{2} x=\frac{1}{2}-\frac{1}{2} \cos 2 x$. Use this identity to graph the function $y=\sin ^{2} x$ for one period.
### Discussion
You must be signed in to discuss.
##### Catherine R.
Missouri State University
##### Kristen K.
University of Michigan - Ann Arbor
##### Michael J.
Idaho State University
Lectures
Join Bootcamp
### Video Transcript
so we need to draw the graph off. Why is he called to sine squared off X? And the question says we that sine squared off X can also be written as 1/2 minus one over to co sign off two weeks. So basically, we need to draw the graph. Of course, I'm two X, with an aptitude of 1/2 and a vertical shift off 1/2. So let's ah, start by setting up our access. So we have our basic X and y axes. Secondly, since there's a vertical shift off 1/2 therefore, to start things off with, just draw a horizontal line at why is equal to about half next. We know that the amplitude is 1/2 so we draw a line 1/2 minutes above this, which would be at one and another line at 1/2 minutes below it. So these are the values or while between which are graph would be plotted. So we have said next because it says coastline two X there for the period would be to pie divided by two, which is equal to pi. So here we have pipe or four. Next, I over to next. We have three pi over four and then the F pie. So those are extract use that we would need. Now, this is a negative coasting to X graph. So it's gonna start here at the origin and then go through this point Piper for 1/2 and then go on to if I were to one and then three pi over 4 1/2 and then finally he Hi is you. So that would be the graph that wouldn't draw for this question.
Other Schools
##### Catherine R.
Missouri State University
##### Kristen K.
University of Michigan - Ann Arbor
##### Michael J.
Idaho State University
| 504 | 1,969 |
{"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 | 4 |
CC-MAIN-2021-21
|
latest
|
en
| 0.94819 |
https://www.shaalaa.com/textbook-solutions/c/selina-solutions-icse-concise-mathematics-class-10-2019-2020-chapter-13-section-mid-point-formula_880
| 1,566,570,126,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2019-35/segments/1566027318421.65/warc/CC-MAIN-20190823130046-20190823152046-00364.warc.gz
| 979,517,143 | 12,319 |
Share
Books Shortlist
# Selina solutions for Class 10 Mathematics chapter 13 - Section and Mid-Point Formula
## ICSE Concise Mathematics for Class 10 (2019-2020)
#### Chapters
Chapter 2: Banking (Recurring Deposit Account)
Chapter 3: Shares and Dividends
Chapter 4: Linear Inequations (in one variable)
Chapter 6: Solving (simple) Problmes (Based on Quadratic Equations)
Chapter 7: Ratio and Proportion (Including Properties and Uses)
Chapter 8: Remainder And Factor Theorems
Chapter 9: Matrices
Chapter 10: Arithmetic Progression
Chapter 11: Geometric Progression
Chapter 12: Reflection (In x-axis, y-axis, x=a, y=a and the origin ; Invariant Points)
Chapter 13: Section and Mid-Point Formula
Chapter 14: Equation of a Line
Chapter 15: Similarity (With Applications to Maps and Models)
Chapter 16: Loci (Locus and its Constructions)
Chapter 17: Circles
Chapter 18: Tangents and Intersecting Chords
Chapter 19: Constructions (Circles)
Chapter 20: Cylinder, Cone and Sphere (Surface Area and Volume)
Chapter 21: Trigonometrical Identities (Including Trigonometrical Ratios of Complementary Angles and Use of Four Figure Trigonometrical Tables)
Chapter 22: Heights and Distances
Chapter 23: Graphical Representation (Histograms and Ogives)
Chapter 24: Measures of Central Tendency (Mean, Median, Quartiles and Mode)
Chapter 25: Probability
## Selina solutions for Class 10 Mathematics chapter 13 - Section and Mid-Point Formula
Selina solutions for Class 10 Maths chapter 13 (Section and Mid-Point Formula) include all questions with solution and detail explanation. This will clear students doubts about any question and improve application skills while preparing for board exams. The detailed, step-by-step solutions will help you understand the concepts better and clear your confusions, if any. Shaalaa.com has the CISCE ICSE Concise Mathematics for Class 10 (2019-2020) solutions in a manner that help students grasp basic concepts better and faster.
Further, we at Shaalaa.com are providing such solutions so that students can prepare for written exams. Selina textbook solutions can be a core help for self-study and acts as a perfect self-help guidance for students.
Concepts covered in Class 10 Mathematics chapter 13 Section and Mid-Point Formula are Mid-point Formula, Section Formula, Distance Formula, Co-ordinates Expressed as (x,y).
Using Selina Class 10 solutions Section and Mid-Point Formula exercise by students are an easy way to prepare for the exams, as they involve solutions arranged chapter-wise also page wise. The questions involved in Selina Solutions are important questions that can be asked in the final exam. Maximum students of CISCE Class 10 prefer Selina Textbook Solutions to score more in exam.
Get the free view of chapter 13 Section and Mid-Point Formula Class 10 extra questions for Maths and can use Shaalaa.com to keep it handy for your exam preparation
S
| 676 | 2,929 |
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 3.5625 | 4 |
CC-MAIN-2019-35
|
latest
|
en
| 0.842286 |
http://math.stackexchange.com/questions/123056/converting-from-one-to-another-numeral-system
| 1,469,535,921,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2016-30/segments/1469257824853.47/warc/CC-MAIN-20160723071024-00228-ip-10-185-27-174.ec2.internal.warc.gz
| 158,708,482 | 17,108 |
# Converting from one to another numeral system
In a book for C# I read that there is direct way to convert from binary to hexadecimal:
1. convert to nibbles with leading zeros
2. replace with the hexadecimal representation of the nibble
The rule is divide the binary number into groups containing the number of the power to which we want to convert. 16 = 2 to power 4, so 4 digits.
I tried from base 4 to base 16: 13 33 33 = 7 f f. 4 to the power 2 is 16.
So when the "power" relation is missing (relation is 3/2), is there a way to convert from quaternary (133130) to octal(3734)?
-
conversion in binary comes to mind : 11 111 011 100 – Raymond Manzoni Mar 21 '12 at 21:37
I was wandering if this is possible, without binary conversion. – Bakudan Mar 21 '12 at 21:42
Ross' method is fine for powers $p_i$ of the same base here 2 : convert $133_4\to 37_8$ and $130_4\to 34_8$ say using a lookup table with $2^{LCM(p_1,p_2)}$ entries (64 here) if the LCM of the powers is not too high. – Raymond Manzoni Mar 21 '12 at 22:41
You can group the digits. As an octal digit accounts for 3 binary bits and a quaternary digit accounts for 2, the LCM of these is 6. So you can take pairs octal digits and convert them to three quaternary digits (like $33_8=123_4$) or vice versa. In a sense, you are going through binary, but in a hidden way. This still depends upon both bases of interest being powers of the same number (here, 2)
-
| 428 | 1,432 |
{"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}
| 3.53125 | 4 |
CC-MAIN-2016-30
|
latest
|
en
| 0.865472 |
https://www.jobilize.com/online/course/1-1-x-and-y-intercepts-prolegomena-by-openstax?qcr=www.quizover.com
| 1,569,184,912,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2019-39/segments/1568514575674.3/warc/CC-MAIN-20190922201055-20190922223055-00231.warc.gz
| 899,049,883 | 19,708 |
# 1.1 X and y-intercepts
Page 1 / 1
Finding x&y intercepts
A rational function is a function of the form $R(x)=\frac{p(x)}{q(x)}$ , where p and q are polynomial functions and $q\neq 0$ .
The domain is all real numbers except for numbers that make the denominator = 0.
x-intercepts are the points at which the graph crosses the x-axis. They are also known as roots, zeros, or solutions.
To find x-intercepts, let y (or f(x)) = 0 and solve for x. In rational functions, this means that you are multiplying by 0 so to find the x-intercept, just set the numerator (the top of the fraction) equal to 0 and solve for x.
Remember: x-intercepts are points that look like (x,0)
For $y=\frac{x-1}{x-2}$ find the x-intercept
The x-intercept is (1,0) since $x-1=0$ , $x=1$
The y-intercept is the point where the graph crosses the y-axis. If the graph is a function, there is only one y-intercept (and it only has ONE name)
To find the y-intercept (this is easier than the x-intercept), let x = 0. Plug in 0 for x in the equation and simplify.
Remember: y-intercepts are points that look like (0,y)
For $y=\frac{x+1}{x-2}$ find the y-intercept
The y-intercept is (0, $\frac{-1}{2}$ ) since $\frac{0+1}{0-2}=\frac{-1}{2}$
Find the x- and y-intercepts of the following:
$y=\frac{1}{x+2}$
x-intercept: None since $1\neq 0$
y-intercept: (0, $\frac{1}{2}$ ) since $\frac{1}{0+2}=\frac{1}{2}$
$y=\frac{1-3x}{1-x}$
x-intercept: ( $\frac{1}{3}$ ,0) since $1-3x=0$ , $-3x=-1$ , $x=\frac{1}{3}$
y-intercept: (0,1) since $\frac{1-3\times 0}{1-0}=1$
$y=\frac{x^{2}}{x^{2}+9}$
x-intercept: (0,0) since $x^{2}=0$ , $x=0$
y-intercept: (0,0) since $\frac{0^{2}}{0^{2}+9}=\frac{0}{9}=0$ or because the x-intercept is (0,0)
$y=\frac{\sqrt{x+1}}{(x-2)^{2}}$
x-intercept: (-1,0) since $\sqrt{x+1}=0$ , $x+1=0$ , $x=-1$
y-intercept: (0, $\frac{1}{4}$ ) since $\frac{\sqrt{0+1}}{(0-2)^{2}}=\frac{\sqrt{1}}{-2^{2}}=\frac{1}{4}$
$y=\frac{3x}{x^{2}-x-2}$
x-intercept: (0,0) since $3x=0$ , $x=0$
y-intercept: (0,0) since the x-intercept is (0,0)
$y=\frac{1}{x-3}+1$
x-intercept: (2,0) since $\frac{1}{x-3}+1=0$ , $\frac{1}{x-3}=-1$ , $-x+3=1$ , $-x=-2$ , $x=2$
y-intercept: (0, $\frac{2}{3}$ ) since $\frac{1}{0-3}+1=\frac{-1}{3}+1=\frac{2}{3}$
$y=\frac{x^{2}-4}{\sqrt{x+1}}$
x-intercepts: (-2,0), (2,0) since $x^{2}-4=0$ , $x^{2}=4$ , $x=-2$ , $x=2$
y-intercept: (0, -4) since $\frac{0^{2}-4}{\sqrt{0+1}}=\frac{-4}{\sqrt{1}}=-4$
$y=4+\frac{5}{x^{2}+2}$
x-intercept: None since $4+\frac{5}{x^{2}+2}=0$ , $\frac{5}{x^{2}+2}=-4$ , $-4x^{2}-8=5$ , $-4x^{2}=13$ , $x^{2}=\frac{13}{4}$ , a number squared will never be a negative number, so there is no x-intercept
y-intercept: (0, $\frac{13}{2}$ ) since $4+\frac{5}{0^{2}+2}=4+\frac{5}{2}=\frac{13}{2}$
$y=\frac{\sqrt{5x-2}}{x-3}$
x-intercept: ( $\frac{2}{5}$ , 0) since $\sqrt{5x-2}=0$ , $5x-2=0$ , $5x=2$ , $x=\frac{2}{5}$
y-intercept: None since $y=\frac{\sqrt{5\times 0-2}}{0-3}$ takes the square root of a negative number.
$y=\frac{x^{3}-8}{x^{2}+1}$
x-intercept: (2,0) since $x^{3}-8=0$ , $(x-2)(x^{2}+2x+4)=0$ , $x=2$
y-intercept: (0,-8) since $\frac{0^{3}-8}{0^{2}+1}=\frac{-8}{1}=-8$
Is there any normative that regulates the use of silver nanoparticles?
what king of growth are you checking .?
Renato
What fields keep nano created devices from performing or assimulating ? Magnetic fields ? Are do they assimilate ?
why we need to study biomolecules, molecular biology in nanotechnology?
?
Kyle
yes I'm doing my masters in nanotechnology, we are being studying all these domains as well..
why?
what school?
Kyle
biomolecules are e building blocks of every organics and inorganic materials.
Joe
anyone know any internet site where one can find nanotechnology papers?
research.net
kanaga
sciencedirect big data base
Ernesto
Introduction about quantum dots in nanotechnology
what does nano mean?
nano basically means 10^(-9). nanometer is a unit to measure length.
Bharti
do you think it's worthwhile in the long term to study the effects and possibilities of nanotechnology on viral treatment?
absolutely yes
Daniel
how to know photocatalytic properties of tio2 nanoparticles...what to do now
it is a goid question and i want to know the answer as well
Maciej
Abigail
for teaching engĺish at school how nano technology help us
Anassong
Do somebody tell me a best nano engineering book for beginners?
there is no specific books for beginners but there is book called principle of nanotechnology
NANO
what is fullerene does it is used to make bukky balls
are you nano engineer ?
s.
fullerene is a bucky ball aka Carbon 60 molecule. It was name by the architect Fuller. He design the geodesic dome. it resembles a soccer ball.
Tarell
what is the actual application of fullerenes nowadays?
Damian
That is a great question Damian. best way to answer that question is to Google it. there are hundreds of applications for buck minister fullerenes, from medical to aerospace. you can also find plenty of research papers that will give you great detail on the potential applications of fullerenes.
Tarell
what is the Synthesis, properties,and applications of carbon nano chemistry
Mostly, they use nano carbon for electronics and for materials to be strengthened.
Virgil
is Bucky paper clear?
CYNTHIA
carbon nanotubes has various application in fuel cells membrane, current research on cancer drug,and in electronics MEMS and NEMS etc
NANO
so some one know about replacing silicon atom with phosphorous in semiconductors device?
Yeah, it is a pain to say the least. You basically have to heat the substarte up to around 1000 degrees celcius then pass phosphene gas over top of it, which is explosive and toxic by the way, under very low pressure.
Harper
Do you know which machine is used to that process?
s.
how to fabricate graphene ink ?
for screen printed electrodes ?
SUYASH
What is lattice structure?
of graphene you mean?
Ebrahim
or in general
Ebrahim
in general
s.
Graphene has a hexagonal structure
tahir
On having this app for quite a bit time, Haven't realised there's a chat room in it.
Cied
what is biological synthesis of nanoparticles
Got questions? Join the online conversation and get instant answers!
| 2,041 | 6,169 |
{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 65, "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-2019-39
|
longest
|
en
| 0.816875 |
http://mathhelpforum.com/trigonometry/186462-proving-questions-leading-hence-solve-x.html
| 1,529,392,161,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2018-26/segments/1529267861980.33/warc/CC-MAIN-20180619060647-20180619080647-00069.warc.gz
| 198,339,389 | 10,268 |
1. ## Proving Questions leading to 'Hence' solve for x
I have a question to ask, would anyone of you explain to me how to solve this?
1. Prove that (1+cosθ+sin2θ)/(1-cos2θ+sin2θ)=cotθ
I'm able to prove it but,,
Then the following part asks me solve the equation for 0≤θ≤2π
1+sin2θ=3cos2θ
2. ## Re: Proving Questions leading to 'Hence' solve for x
Is the first part supposed to be $\displaystyle (1 + cos(2\theta) +sin(2\theta))$ for the numerator?
If not, can you show me your proof
For the second part, wait till someone more experienced comes along; however, while you're waiting, here is my attempt at it...
Let $\displaystyle t = tan(\theta)$
$\displaystyle 1 + sin(2\theta) = 3cos(2\theta)$
$\displaystyle \equiv~$ $\displaystyle ~\frac{1+t^2}{1+t^2} + \frac{2t}{1+t^2} - \frac{3(1-t^2)}{1+t^2} = 0$
Solve for t...
$\displaystyle tan(\theta) = \frac{1}{2}$
$\displaystyle or$
$\displaystyle tan(\theta) = -1$
For $\displaystyle tan(\theta) = -1:$
$\displaystyle \theta = arctan(-1) = -\frac{\pi}{4}$ therefore $\displaystyle V=\{\frac{3\pi}{4}, \frac{7\pi}{4}\}$ are some solutions. Not 100% on how to solve for 1/2.
3. ## Re: Proving Questions leading to 'Hence' solve for x
Probably you made a mistake in the numerator because I think the identity has to be $\displaystyle \frac{1+\cos(2x)+\sin(2x)}{1-\cos(2x)+\sin(2x)}=\cot(x)$ (like terrorsquid already noticed).
You can prove it this way (I use $\displaystyle x$ in stead of $\displaystyle \theta$):
$\displaystyle \frac{1+\cos(2x)+\sin(2x)}{1-\cos(2x)+\sin(2x)}=\frac{1+(2\cos^2(x)-1)+2\sin(x)\cos(x)}{1-(1-2\sin^2(x))+2\sin(x)\cos(x)}$
$\displaystyle =\frac{2\cos^2(x)+2\sin(x)\cos(x)}{2\sin^2(x)+2\si n(x) \cos(x)}$
$\displaystyle =\frac{2\cos(x)[\cos(x)+\sin(x)]}{2\sin(x)[\sin(x)+\cos(x)]}=\frac{\cos(x)}{\sin(x)}=\cot(x)$
The method Terrorsquid used is very useful.
For the other solution:
$\displaystyle \tan(\theta)=\frac{1}{2}$ that means:
$\displaystyle \theta=\arctan\left(\frac{1}{2}\right)+k\pi$
| 721 | 1,991 |
{"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}
| 4.1875 | 4 |
CC-MAIN-2018-26
|
latest
|
en
| 0.710142 |
https://www.proprofs.com/quiz-school/story.php?title=4-english
| 1,726,586,009,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2024-38/segments/1725700651800.83/warc/CC-MAIN-20240917140525-20240917170525-00499.warc.gz
| 864,423,941 | 100,438 |
# Class 4 Th Mathematics
Approved & Edited by ProProfs Editorial Team
The editorial team at ProProfs Quizzes consists of a select group of subject experts, trivia writers, and quiz masters who have authored over 10,000 quizzes taken by more than 100 million users. This team includes our in-house seasoned quiz moderators and subject matter experts. Our editorial experts, spread across the world, are rigorously trained using our comprehensive guidelines to ensure that you receive the highest quality quizzes.
| By Rathoddj11
R
Rathoddj11
Community Contributor
Quizzes Created: 1 | Total Attempts: 911
Questions: 10 | Attempts: 913
Settings
श्री डि जे राठोड धामणगांव रेल्वे
• 1.
### 543 x 75=
• A.
505040
• B.
35780
• C.
40725
• D.
34567
C. 40725
Explanation
The correct answer is 40725 because when you multiply 543 by 75, you get the product of 40725.
Rate this question:
• 2.
• A.
487
• B.
587
• C.
657
• D.
345
A. 487
• 3.
• A.
1235
• B.
1798
• C.
1636
• D.
2023
C. 1636
• 4.
### Aman came 8 meters to sew the uniform, which is how many centimeters of cloth he brought ?
• A.
400 centimeter
• B.
550 centimeter
• C.
800 centimeter
• D.
850 centimeter
C. 800 centimeter
Explanation
Aman came 8 meters to sew the uniform. Since there are 100 centimeters in one meter, we can convert meters to centimeters by multiplying by 100. Therefore, 8 meters is equal to 800 centimeters.
Rate this question:
• 5.
### Measure the perimeter of triangle ?
• A.
11 cm
• B.
17 cm
• C.
18 cm
• D.
12 cm
D. 12 cm
Explanation
The correct answer is 12 cm because the perimeter of a triangle is the sum of the lengths of its three sides. Since the given options are all different lengths, we can determine that the perimeter of the triangle is 12 cm.
Rate this question:
• 6.
### How many total Rupees ?
• A.
₹ 100
• B.
₹ 250
• C.
₹ 150
• D.
₹ 200
C. ₹ 150
Explanation
The correct answer is ₹ 150 because it is the only option that matches the question's prompt of "How many total Rupees?" The other options are irrelevant as they represent different amounts of money.
Rate this question:
• 7.
### Make the number
• A.
2033
• B.
2034
• C.
2133
• D.
2134
B. 2034
Explanation
The given sequence consists of four numbers: 2033, 2034, 2133, and 2134. The pattern in the sequence is that the first two digits remain the same in each number, while the last digit increases by one for the first two numbers and then resets to the original value for the last two numbers. Therefore, the next number in the sequence would follow the same pattern, resulting in 2034.
Rate this question:
• 8.
### Write the proper fraction ?
• A.
1/3, 1/4
• B.
1/3, 1/2
• C.
1/2, 1/3
• D.
1/2, 1/3
A. 1/3, 1/4
Explanation
The correct answer is 1/3, 1/4. This is because a proper fraction is a fraction where the numerator (the top number) is smaller than the denominator (the bottom number). In the given options, both 1/3 and 1/4 fit this criteria, making them proper fractions.
Rate this question:
• 9.
### Write the proper sign blank 👉 5775 --- 5757
• A.
<
• B.
=
• C.
>
• D.
None of the above
C. >
Explanation
The answer is ">" because the number 5775 is greater than the number 5757.
Rate this question:
• 10.
### How many Farmers takes Vegetables in their crops ?
• A.
50
• B.
30
• C.
40
• D.
60
C. 40
Quiz Review Timeline +
Our quizzes are rigorously reviewed, monitored and continuously updated by our expert board to maintain accuracy, relevance, and timeliness.
• Current Version
• Mar 21, 2023
Quiz Edited by
ProProfs Editorial Team
• Apr 11, 2020
Quiz Created by
Rathoddj11
Related Topics
×
Wait!
Here's an interesting quiz for you.
| 1,074 | 3,721 |
{"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}
| 4.53125 | 5 |
CC-MAIN-2024-38
|
latest
|
en
| 0.840116 |
https://help.matheass.eu/en/E204-Special_lines_in_a_triangle.html
| 1,723,642,961,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2024-33/segments/1722641113960.89/warc/CC-MAIN-20240814123926-20240814153926-00103.warc.gz
| 228,501,823 | 2,775 |
## Special lines in a triangle
If the coordinates of the three vertices of a triangle are entered, the program calculates the equations of the perpendicular bisectors[1], of the side bisectors[2], of the angle bisectors[3] and of the altitudes[4]. In addition, the centers and radii of the circumcircle[5], of the incircle[6] and of the three excircles[7].
A list of check boxes can be used to select which objects are to be calculated and drawn.
Perpendicular bisec.
Medians
Angle bisectors
Altitudes
Incircle
Circumcircle
Excircles
### Example 1: incircle and excircles of a triangle
```Given:
¯¯¯¯¯¯
Vertices: A(1|0) B(5|1) C(3|6)
Results:
¯¯¯¯¯¯¯
Sides: a : 5·x + 2·y = 27
b : 3·x - y = 3
c : x - 4·y = 1
Incircle: Mi(3,119|1,962) r i = 1,390
Excircles: Ma(7,626|6,136) ra = 4,346
Mb(-4,356|5,784) rb = 6,910
Mc(3,248|-2,427) rc = 2,900```
The center of the incircle (green) lies on the bisector of the three interior angles. The centers of the excircles (red) are each on the bisector of an inner angle and on the bisector of the outside angle of the other two triangle angles. These construction lines are also drawn.
### Example 2: Altitudes in an obtuse-angled triangle
```Given:
¯¯¯¯¯¯
Vertices: A(7|3) B(16|10) C(8|9)
Results:
¯¯¯¯¯¯¯
Sides: a : -x + 8·y = 64
b : 6·x - y = 39
c : 7·x - 9·y = 22
Altitudes: ha : 8·x + y = 59
hb : x + 6·y = 76
hc : 9·x + 7·y = 135
Perp. feet: Ha(6,277|8,785) Hb(8,378|11,27)
Hc(10,53|5,746
Orthocenter: H(11,05|8,26)```
The intersection of the altitudes of an obtuse-angled triangle lies outside the triangle. The construction lines are also drawn. In order to make them more visible, the grid lines have been hidden.
| 604 | 1,723 |
{"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}
| 4.15625 | 4 |
CC-MAIN-2024-33
|
latest
|
en
| 0.781847 |
http://deepmarketing.io/2018/02/are-the-digits-of-pi-truly-random/
| 1,548,005,136,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2019-04/segments/1547583728901.52/warc/CC-MAIN-20190120163942-20190120185942-00186.warc.gz
| 60,629,681 | 14,552 |
# Are the Digits of Pi Truly Random?
Are the Digits of Pi Truly Random?
This article covers far more than the title suggests. It is written in simple English and accessible to quantitative professionals from a variety of backgrounds. Deep mathematical and data science research (including a result about the randomness of Pi, which is just a particular case) are presented here, without using arcane terminology or complicated equations.
The topic discussed here, under a unified framework, is at the intersection of mathematics, probability theory, chaotic systems, stochastic processes, data and computer science. Many exotic objects are investigated, such as an unusual version of the logistic map, nested square roots, and representation of a number in a fractional or irrational base system.
The article is also useful to anyone interested in learning these topics, whether they have any interest in the randomness or Pi or not, because of the numerous potential applications. I hope the style is refreshing, and I believe that you will find plenty of material rarely if ever discussed in textbooks or in the classroom. The requirements to understand this material are minimal, as I went to great lengths (over a period of years) to make it accessible to a large audience.
The randomness of the digits of Pi is one of the most fascinating, unsolved mathematical problems of all times, having been investigated by many million of people over several hundred years. The scope of this article encompasses this particular problem as part of a far more general framework. More questions are asked than answered, making this document a stepping stone for future research.
1. General Framework
We are dealing with sequences x(n) of positive real numbers defined by a recurrence relationship x(n+1) = g(n) for some function g, and initiated with a seed x(1) = x. We are interested only in functions g that produce chaos and preserve the domain. For instance, if x(n) is anywhere in [0, 1] we want x(n+1) to also be anywhere in [0.1].
We also study another sequence derived from x(n), and defined by a(n) = h(x(n)) for some function h. The function h is typically very simple and produces only small integer values such as (say) 0, 1, … , 9 regardless of x(n). The sequence a(n) is called the digits of x represented in the chaotic system defined by the function f.
Using appropriate functions for g, the sequence x(n) behaves as a realization of a stochastic (random) process, each term x(n) acting as a random deviate of some random variable X(n). To be of any value here, the function g must be associated with random variables that are identically distributed, though not necessarily independent. Instead of using the notation X(1), X(2) and so on, we simply write X as all these random variables have the same distribution. The distribution of X is called the equilibrium distribution (or fixed-point distribution) of the system, and it is the solution of a simple stochastic integral equation. Read this article for some illustrations, especially section 3. Likewise, the a(n) sequence has a statistical distribution associated with it, called the digits distribution, and easily derived from the equilibrium distribution.
Examples of functions g producing these patterns are investigated in details in this article. Such systems are usually referred to as chaotic systems. It is a special case of (ergodic) stochastic process or time series. Note that in practice, for a given chaos-inducing function g, not all seeds — the initial value x(1) — produce a sequence x(n) with the desired properties. Seeds failing to do so are called bad seeds and result in equilibrium and digits distributions that are degenerate. Even though all the systems discussed here have infinitely many bad seeds, these seeds are so rare in practice (compared with the good seeds) that the probability to stumble upon one by chance for a given chaos-inducing g, is zero.
At this point, the initiated reader probably knows what I am going to talk about, even though much of it is original research published for the first time. In the grand scheme of things, the randomness of Pi (and I will discuss it later in this article) appears as a very peculiar case, certainly not the most noteworthy result to prove, though still very fascinating.
Questions, Properties and Notations about Chaotic Sequences Investigated Here
For the remaining of this article, we are going to focus on the following, applied to a few different types of chaos-inducing functions g:
The digits a(n) are denoted as [a(1), a(2), a(3)… ] or {a(n)} and uniquely represents the number x. We want to build sequences {a(n)} based on the functions h and g, such that if x and x’ are two distinct numbers, then the respective representations {a(n)} and {a'(n)} are different.
There is a function f, usually straightforward (but not always) such that for a given chaos-inducing g and a given h, we have x = f(a(1), a(2), a(3)…).
For a given g and h, can the algorithm producing {a(n)} generate any arbitrary {a(n)} or only some specific types of {a(n)} with some constraints on the digits? Sometimes yes (example: decimal representations), sometimes no (example: continuous fractions).
Are the sequences {x(n)} and {a(n)} auto-correlated? Sometimes yes, sometimes no. What are the implications? For the representation of a number in any integer base b such as b = 10, {x(n)} is auto-correlated while {a(n)} is not. If the base b is not an integer, {x(n)} is much less auto-correlated, but {a(n)} becomes auto-correlated. Also if the base b is not an integer, the digits distribution is no longer uniform.
Can we compute the equilibrium and digits distributions? Yes using an iterative algorithm, but in all the examples discussed in the next section, exact solutions in closed form are known and featured, and I will show you how they were obtained.
Potential Applications
The sequences {x(n)} and {a(n)} can be use for continuous / discrete number generation and in cryptographic applications. See also this article with applications to modeling population demographics, and physics. They can also be used for machine precision benchmarking, as numerical errors accumulate so fast that after 50 iterations, regardless of the sequence, all digits are wrong unless you use high precision computing.
2. Examples of Chaotic Sequences Representing Numbers
All the g functions presented here are chaos-inducing. This is necessary if one wants, with a given g, to represent all numbers, not just a small fraction of them. For each case, we present, whenever possible, the following features:
The function f discussed earlier, to reconstruct the original number x (the seed)
The (continuous) equilibrium distribution and its support domain
The (discrete) digits distribution
The functions g and h
Other properties such as auto-correlations or exact formula for x(n) when available
Three types of chaotic processes are discussed
Tradition representation of a number in base b, with b > 1 (integer or not) and denoted as Base(b)
Logistic map of exponent p, denoted as LM(p)
Nested powers of exponent q, denoted as NP(q)
In all cases, the seed x(1) = x is a positive real number. The equilibrium distribution (if it exists) is always solution of the stochastic integral equation
P(X < y) = P(g(X) < y).
A way to solve this apparently complex equation is described in details, and in very simple words, in this article, with an illustration for the logistic map LM(1/2) discussed below. It involves two steps:
Data Science Step: Gather data to compute the empirical percentile distribution for x(n), and perform various regressions (polynomial, logarithmic and so on) until you discover one with an incredibly fantastic fit with data — like a perfect fit.
Mathematical Step: Plug that tentative distribution into the stochastic integral equation, and see if it solves the equation.
Also, by definition, as discussed earlier, x(n+1) = g(x(n)) and a(n) = h(x(n)). This can be used to design a trivial algorithm to compute all the digits in any system powered a chaos-inducing function g. The seed x can be reconstructed using the formula x = f({a(n)}. Here the notation INT represents the integer part, sometimes called floor function. In the formula displayed as images, the floor function is represented by left and right brackets: this is the standard convention.
The recursions
Numbers in Base 2, 10, 3/2 or Pi
This system (the traditional decimal system in base b) is characterized by:
with b > 1. The seed x must be in [0,1] so that all x(n)’s are also in [0.1]. If x = Pi, use x-3 instead.
We distinguishes two cases:
If b is an integer: The equilibrium distribution is uniform on [0,1] and the digits distribution is uniform on {0, 1, … , b} for all but very rare bad seeds x such as rational numbers. This fact has been “known” for millennia (at least for b = 10) and it is taken for granted; it can be proved rigorously by solving the stochastic integral equation P(X < y) = P(g(X) < y) attached to this system. However, in general, such systems do not produce a uniform distribution for the digits: this is the only exception in the examples discussed in this article. The auto-correlation between x(n+1) and x(n) is equal to 1 / b, while the auto-correlation between a(n+1) and a(n) is equal to 0. So the h function actually decorrelates the sequence {x(n)}, to use the vocabulary of time series theory.
If b is not an integer: The equilibrium distribution is NOT uniform on [0,1]. The auto-correlation between x(n+1) and x(n) is not 0, but much closer to 0 than in the above case. This time, the auto-correlation between a(n+1) and a(n) is NOT equal to 0. The digits take values in {0, 1, … , INT(b)}. I did not test if the digits distribution was uniform on that domain, but my gut feelings tells me that this is not the case.
Computation details for a customizable base b, are available in this spreadsheet. Rather than computing auto-correlations based on a single seed (that is, one instance of the process) using large values of n, I computed them using a large number of seeds (all random numbers) computing the auto-correlations for a small, fixed n but across many seeds. Due to the nature of the process (it is ergodic) both computations yield the same conclusions. The reason for using many seeds and a small n is because large n (above 20) lead to total loss of numerical accuracy.
Nested Square Roots
This is a particular case of a far more general model described in section 3 in the following article. It is characterized byThis system is related to continued fractions, except that a(n) can only be equal to 0, 1, or 2, as opposed to taking any arbitrary positive integer value. The seed x must be in [1, 2] so that all x(n)’s are also in [1, 2]. If x = Pi, use x-1 instead.
The equilibrium distribution is given by the following formula:
The digits distribution is given by the following formula:
Again, these results are easily derived from the stochastic integral equation. Note that the sequence {x(n)} exhibits strong auto-correlations in this case, though auto-correlations for {a(n)} are close to (but different from) 0.
Logistic Map LM(p)
The logistic map has many applications, for instance in social sciences. Here we are interested in the logistic map of exponent p, with p = 1 corresponding to the standard, well known version. We also discuss the case p = 1/2, as these are two cases (possibly the only two cases) that provide a chaotic-inducing g covering all real numbers (one that meets the criteria outlined in section 1) with known equilibrium distribution that can be expressed with a simple closed form formula. The function h associated with this system was chosen arbitrarily, but this is also the simplest function h that one could come up with. I don’t know yet if there is a simple formula for the function f.
This system is characterized by
The seed x must be in [0,1] so that all x(n)’s are also in [0.1]. If x = Pi, use x-3 or x/4 instead. A remarkable fact, for the case p = 1, is that the auto-correlations for {x(n)} are all equal to 0, making it, in some way, more random than the decimal system or other systems discussed in this article. For the case p = 1/2 the auto-correlation between x(n+1) and x(n) is equal to -1/2, see proof here (check out section 3 after clicking on the link.) By contrast, it is equal to 1/2 for the decimal system in base 2.
The equilibrium distributions, defined for y in [0, 1], are equal to:
As usual, this applies to good seeds only (the vast majority of seeds). A proof of the formula for p = 1/2 can be found here. The integral for the case p = 1 can be explicitly computed, see here.
For the case p =1, the digits distribution is uniform on {0, 1}. The probability for a digit to be 0 is equal to P( X < 1/2) = 0.5, based on this computation. This is not true for the logistic map with p = 1/2.
3. About the Randomness of the Digits of Pi
The discussion here is not just about Pi, but about any number. The digits {a(n)} of a number x, in a specific system defined by g and h, are said to be random, if they have the proper digits distribution, as well as the proper auto-correlation structure associated with that system. In short, any finite set of digits of x must have the proper joint distribution that all but a tiny subset of numbers (of Lebesgue measure zero) have in that system. The target digits distribution can be computed from the equilibrium distribution, as discussed in the previous section. Numbers satisfying this criteria are called good seeds earlier this article.
For instance, in the decimal system of base b, if b is an integer, the distribution in question is uniform, auto-correlation between successive digits are zero, and the digits are independently distributed with that same uniform distribution. If the base b is not an integer (say b = 3/2), the theoretical auto-correlation between a(n+1) and a(n) is clearly not zero, the digits distribution is clearly not uniform, and both can be computed exactly using the same mathematical arguments as in this article (see section 3.)
In the literature, the randomness of the digits is defined as normalcy. For the traditional decimal system, the definitions (random digits, good seed, normal number) are similar though good seed is stronger in the sense that it takes the absence of auto-correlation into account. Anyway, for Pi or SQRT(2), even proving that the digit 5 appears infinitely many times — a very weak result indeed — would be a phenomenal discovery.
The Digits of Pi are Random in the Logistic Map System
We show here why Pi is a good seed in that system, with its digits behaving exactly like those of pretty much all numbers in that system. The result is based on the fact that for the standard logistic map (corresponding to p = 1 in the previous section), an exact formula is available for {x(n)} and thus for {a(n)} as well, see here. It is given by
It implies that bad seeds (all bad seeds?) are numbers x that make w(x) a rational number. Pi/4 is not one of them, so Pi/4 must be a good seed.
While this result about the randomness of the digits of Pi may seem spectacular at first glance, there are still big hurdles to overcome to formally prove it. In particular:
Are the digits a(n) properly defined in the logistic map system? No function x = f({a(n)}) has been found yet, unlike in the decimal or nested square root systems. Can two different numbers have the same digits?
Are the only bad seeds those that are very easy to spot in the exact formula for x(n)? If difficult to prove, we may be back to square one.
Probably the biggest hurdle is to prove that w(Pi/4) is irrational.
Note that for the logistic map system, the digits distribution (for good seeds, that is for pretty much all numbers) is uniform on {0, 1} like for the decimal system of base 2. This was proved in the previous section. Also, the a(n)’s are independent since the x(n)’s are. Thus the digits produced by the logistic map system must be sound, solving hurdle #1. Finally, since x must be in [0, 1], here we use Pi/4, rather than Pi.
Paths to Proving Randomness in the Decimal System
The problem with the decimal system (in any base) is the opposite of the logistic map system. The function x = f({a(n)}) is known, but there is no explicit, closed-form formula for a(n). This makes things more complicated.
However, we could try one of the following approaches:
Find an approximation for x(n) or maybe an infinite series involving terms similar to w(x), making it easier to spot the bad seeds — all of them. Today, the only bad seeds known are rational numbers and artificially manufactured numbers such as x = 0.101101110…
Study the problem in a base b that is not an integer. Try a sequence of bases that converge to (say) b =2 or b = 10.
Use a transformation of {x(n)} that more easily leads to a solution.
The process {x(n)} is chaotic. In number theory, working with the cumulative process, in this case y(n) = x(n) + x(n-1), which is much smoother, sometimes makes things easier.
Find a mapping between the logistic map and decimal system, for x(n). However, such mappings may be very complicated if they exist at all.
These are all very challenging problems.
Finally, the algorithm used here to compute the digits of Pi or any other number, in any system, is not practical. On must machines, x(n) becomes totally wrong in as little as 30 iterations due to round-off errors accumulating exponentially fast. This is discussed here. Nevertheless, it would be interesting to see how far you can go, with high precision computing, until the digits being computed are no longer correct. Better, use a pre-computed table of digits. For Pi, more than a trillion digits are known.
In this article, the focus was never on a single number like Pi, but on all numbers. The processes investigated here being all ergodic, it is easier to identify general statistical properties by focusing on a large collection of numbers, using small values of n and digits correctly computed, rather than on tons of digits computed for an handful of numbers, with pretty much all digits incorrectly computed due to round-off errors.
DSC Resources
Services: Hire a Data Scientist | Search DSC | Classifieds | Find a Job
Contributors: Post a Blog | Ask a Question
| 4,174 | 18,372 |
{"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-2019-04
|
latest
|
en
| 0.945532 |
http://gmatclub.com/forum/in-a-battle-70-of-the-combatants-lost-an-eye-80-an-ear-61531.html
| 1,430,908,907,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2015-18/segments/1430458524551.85/warc/CC-MAIN-20150501053524-00046-ip-10-235-10-82.ec2.internal.warc.gz
| 96,706,876 | 45,741 |
Find all School-related info fast with the new School-Specific MBA Forum
It is currently 06 May 2015, 02:41
### GMAT Club Daily Prep
#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.
Customized
for You
we will pick new questions that match your level based on your Timer History
Track
every week, we’ll send you an estimated GMAT score based on your performance
Practice
Pays
we will pick new questions that match your level based on your Timer History
# Events & Promotions
###### Events & Promotions in June
Open Detailed Calendar
# In a battle 70% of the combatants lost an eye, 80% an ear,
Author Message
TAGS:
Manager
Joined: 01 Nov 2007
Posts: 101
Followers: 1
Kudos [?]: 25 [0], given: 0
In a battle 70% of the combatants lost an eye, 80% an ear, [#permalink] 19 Mar 2008, 22:59
00:00
Difficulty:
(N/A)
Question Stats:
0% (00:00) correct 0% (00:00) wrong based on 0 sessions
In a battle 70% of the combatants lost an eye, 80% an ear, 75% an arm, 85% a leg, x% lost all four limbs. The minimum value of x is?
Manager
Joined: 02 Mar 2008
Posts: 211
Concentration: Finance, Strategy
Followers: 1
Kudos [?]: 40 [0], given: 1
Re: PS (sets) [#permalink] 19 Mar 2008, 23:53
85%
leg is subset of 4 limbs ( i suppose ^^). So those lost 4 limbs sure lost leg
CEO
Joined: 17 Nov 2007
Posts: 3578
Concentration: Entrepreneurship, Other
Schools: Chicago (Booth) - Class of 2011
GMAT 1: 750 Q50 V40
Followers: 407
Kudos [?]: 2149 [0], given: 359
Re: PS (sets) [#permalink] 20 Mar 2008, 00:15
Expert's post
az780 wrote:
In a battle 70% of the combatants lost an eye, 80% an ear, 75% an arm, 85% a leg, x% lost all four limbs. The minimum value of x is?
terrible
it is a strange question...
So, 70% an eye, 80% an ear, 75% an arm, 85% a leg.
1. an eye and and an ear: p=[50%,70%]
2. (an eye and and an ear) and an arm: p=[25%,70%]
3. ((an eye and and an ear) and an arm) and a leg: p=[10%,70%]
Therefore, min: 10%, max: 70%
_________________
HOT! GMAT TOOLKIT 2 (iOS) / GMAT TOOLKIT (Android) - The OFFICIAL GMAT CLUB PREP APP, a must-have app especially if you aim at 700+ | PrepGame
SVP
Joined: 04 May 2006
Posts: 1936
Schools: CBS, Kellogg
Followers: 19
Kudos [?]: 441 [0], given: 1
Re: PS (sets) [#permalink] 20 Mar 2008, 01:15
walker wrote:
az780 wrote:
In a battle 70% of the combatants lost an eye, 80% an ear, 75% an arm, 85% a leg, x% lost all four limbs. The minimum value of x is?
terrible
it is a strange question...
So, 70% an eye, 80% an ear, 75% an arm, 85% a leg.
1. an eye and and an ear: p=[50%,70%]
2. (an eye and and an ear) and an arm: p=[25%,70%]
3. ((an eye and and an ear) and an arm) and a leg: p=[10%,70%]
Therefore, min: 10%, max: 70%
This is really funny problem. I can not stand when I read it.
Anyway, Walker! I try to but can not follow your logic! Sets always make me confused! Please give me some detail. Your approach strange also
_________________
CEO
Joined: 17 Nov 2007
Posts: 3578
Concentration: Entrepreneurship, Other
Schools: Chicago (Booth) - Class of 2011
GMAT 1: 750 Q50 V40
Followers: 407
Kudos [?]: 2149 [0], given: 359
Re: PS (sets) [#permalink] 20 Mar 2008, 01:39
Expert's post
sondenso wrote:
This is really funny problem. I can not stand when I read it.
Anyway, Walker! I try to but can not follow your logic! Sets always make me confused! Please give me some detail. Your approach strange also
1. Let we have two attributes: A and B.
2. Let a% of people have A and b% of people have B.
3. we will try to find max and min ab% (the percentage of people that have two attributes)
I'll try to draw:
the maximum configuration:
|aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa------------------| 100%
|bbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbb--------------| 100%
max=min(a,b)
the minimum configuration:
|aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa------------------| 100%
|---------------bbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbb| 100%
min=b-(100-a)=a+b-100. if min<0 then min have to be 0
for example, a=75%, b=87%
max=min(75%,87%)=75%
min=87%-(100%-75%)=62%
_________________
HOT! GMAT TOOLKIT 2 (iOS) / GMAT TOOLKIT (Android) - The OFFICIAL GMAT CLUB PREP APP, a must-have app especially if you aim at 700+ | PrepGame
SVP
Joined: 04 May 2006
Posts: 1936
Schools: CBS, Kellogg
Followers: 19
Kudos [?]: 441 [0], given: 1
Re: PS (sets) [#permalink] 20 Mar 2008, 18:34
walker wrote:
sondenso wrote:
This is really funny problem. I can not stand when I read it.
Anyway, Walker! I try to but can not follow your logic! Sets always make me confused! Please give me some detail. Your approach strange also
1. Let we have two attributes: A and B.
2. Let a% of people have A and b% of people have B.
3. we will try to find max and min ab% (the percentage of people that have two attributes)
I'll try to draw:
the maximum configuration:
|aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa------------------| 100%
|bbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbb--------------| 100%
max=min(a,b)
the minimum configuration:
|aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa------------------| 100%
|---------------bbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbb| 100%
min=b-(100-a)=a+b-100. if min<0 then min have to be 0
for example, a=75%, b=87%
max=min(75%,87%)=75%
min=87%-(100%-75%)=62%
Thank you Walker, you are great!
_________________
Re: PS (sets) [#permalink] 20 Mar 2008, 18:34
Similar topics Replies Last post
Similar
Topics:
GMAT 1: Lost the battle, but I have to win the war 3 02 Jul 2013, 03:17
1 During 2001, a stock lost 80 percent of its value 2 03 Mar 2012, 20:45
%age=(Income 80 – Income70)/Income70 4 18 May 2008, 12:12
I lost my battle against GMAT - 580 ( Q 42, V 28) 7 28 Dec 2006, 08:20
Amount Earned/Day $96$84 $80$70 \$48 1 13 Apr 2006, 15:12
Display posts from previous: Sort by
| 1,891 | 5,845 |
{"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}
| 3.6875 | 4 |
CC-MAIN-2015-18
|
longest
|
en
| 0.852654 |
https://smartconversion.com/ConversionExample/meter-per-square-second-to-foot-per-(minute%E2%80%A2second)/418/399
| 1,695,950,526,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2023-40/segments/1695233510462.75/warc/CC-MAIN-20230928230810-20230929020810-00092.warc.gz
| 566,104,507 | 5,532 |
# Convert meter / square second to foot / (minute • second)
Learn how to convert 1 meter / square second to foot / (minute • second) step by step.
## Calculation Breakdown
Set up the equation
$$1.0\left(\dfrac{meter}{square \text{ } second}\right)={\color{rgb(20,165,174)} x}\left(\dfrac{foot}{minute \times second}\right)$$
Define the base values of the selected units in relation to the SI unit $$\left(\dfrac{meter}{square \text{ } second}\right)$$
$$\text{Left side: 1.0 } \left(\dfrac{meter}{square \text{ } second}\right) = {\color{rgb(89,182,91)} 1.0\left(\dfrac{meter}{square \text{ } second}\right)} = {\color{rgb(89,182,91)} 1.0\left(\dfrac{m}{s^{2}}\right)}$$
$$\text{Right side: 1.0 } \left(\dfrac{foot}{minute \times second}\right) = {\color{rgb(125,164,120)} \dfrac{0.3048}{60.0}\left(\dfrac{meter}{square \text{ } second}\right)} = {\color{rgb(125,164,120)} \dfrac{0.3048}{60.0}\left(\dfrac{m}{s^{2}}\right)}$$
Insert known values into the conversion equation to determine $${\color{rgb(20,165,174)} x}$$
$$1.0\left(\dfrac{meter}{square \text{ } second}\right)={\color{rgb(20,165,174)} x}\left(\dfrac{foot}{minute \times second}\right)$$
$$\text{Insert known values } =>$$
$$1.0 \times {\color{rgb(89,182,91)} 1.0} \times {\color{rgb(89,182,91)} \left(\dfrac{meter}{square \text{ } second}\right)} = {\color{rgb(20,165,174)} x} \times {\color{rgb(125,164,120)} {\color{rgb(125,164,120)} \dfrac{0.3048}{60.0}}} \times {\color{rgb(125,164,120)} \left(\dfrac{meter}{square \text{ } second}\right)}$$
$$\text{Or}$$
$$1.0 \cdot {\color{rgb(89,182,91)} 1.0} \cdot {\color{rgb(89,182,91)} \left(\dfrac{m}{s^{2}}\right)} = {\color{rgb(20,165,174)} x} \cdot {\color{rgb(125,164,120)} \dfrac{0.3048}{60.0}} \cdot {\color{rgb(125,164,120)} \left(\dfrac{m}{s^{2}}\right)}$$
$$\text{Cancel SI units}$$
$$1.0 \times {\color{rgb(89,182,91)} 1.0} \cdot {\color{rgb(89,182,91)} \cancel{\left(\dfrac{m}{s^{2}}\right)}} = {\color{rgb(20,165,174)} x} \times {\color{rgb(125,164,120)} \dfrac{0.3048}{60.0}} \times {\color{rgb(125,164,120)} \cancel{\left(\dfrac{m}{s^{2}}\right)}}$$
$$\text{Conversion Equation}$$
$$1.0 = {\color{rgb(20,165,174)} x} \times \dfrac{0.3048}{60.0}$$
$$\text{Simplify}$$
$$1.0 = {\color{rgb(20,165,174)} x} \times \dfrac{0.3048}{60.0}$$
Switch sides
$${\color{rgb(20,165,174)} x} \times \dfrac{0.3048}{60.0} = 1.0$$
Isolate $${\color{rgb(20,165,174)} x}$$
Multiply both sides by $$\left(\dfrac{60.0}{0.3048}\right)$$
$${\color{rgb(20,165,174)} x} \times \dfrac{0.3048}{60.0} \times \dfrac{60.0}{0.3048} = 1.0 \times \dfrac{60.0}{0.3048}$$
$$\text{Cancel}$$
$${\color{rgb(20,165,174)} x} \times \dfrac{{\color{rgb(255,204,153)} \cancel{0.3048}} \times {\color{rgb(99,194,222)} \cancel{60.0}}}{{\color{rgb(99,194,222)} \cancel{60.0}} \times {\color{rgb(255,204,153)} \cancel{0.3048}}} = 1.0 \times \dfrac{60.0}{0.3048}$$
$$\text{Simplify}$$
$${\color{rgb(20,165,174)} x} = \dfrac{60.0}{0.3048}$$
Solve $${\color{rgb(20,165,174)} x}$$
$${\color{rgb(20,165,174)} x}\approx196.8503937\approx1.9685 \times 10^{2}$$
$$\text{Conversion Equation}$$
$$1.0\left(\dfrac{meter}{square \text{ } second}\right)\approx{\color{rgb(20,165,174)} 1.9685 \times 10^{2}}\left(\dfrac{foot}{minute \times second}\right)$$
| 1,354 | 3,222 |
{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 4.28125 | 4 |
CC-MAIN-2023-40
|
latest
|
en
| 0.314271 |
http://www.physicsforums.com/showthread.php?t=209332
| 1,409,329,485,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2014-35/segments/1408500832662.33/warc/CC-MAIN-20140820021352-00309-ip-10-180-136-8.ec2.internal.warc.gz
| 543,416,075 | 10,004 |
# Work done against gravity, what am i missing
by lignocaine
Tags: gravity, missing, work
P: 3 Hi everyone. Im studying basic physics for my anaesthesia primary exams and came across the subject of work. I am however finding some difficulty grasping the concept of work done against gravity. The example given in my literature for work done against gravity is the following: lifting an apple of 102grams 1m vertically will require 1joule of energy. here it is explained that because the apple exerts a downward force of 1N (0.102kg x 9.8m/s) that an equally opposite force of 1N will need to be exerted upwards to move the object against gravity. But, if 1N is exerted in an upward direction, the apple will remain still in your hand because the net forces acting on the apple are zero. So in order to move the apple against gravity, a force of greater than 1N will have to be exerted by your hand. An example would be using a force of 1.5N over a distance of 1m which would effectively yield a force of 0.5N acting upwards and if this example is used, the work done to move the apple would be 0.5N x 1m = 05J If a higher upward acting force was used, then the resulting work done to move the object 1m would be even higher So, the work done to move a fixed mass object 1m against gravity can vary depending on the net force acting on the object? Is my logic flawed and if yes, where?
Sci Advisor HW Helper PF Gold P: 12,016 Do not mix together acceleration and velocity!! Balance of forces (i.e, net force equalling zero!) means zero acceleration. But you can perfectly well have zero acceleration and a non-zero, constant velocity. Besides, when we think of work done BY a force, we are interested in finding that particular force's contribution to the total work done. Therefore, it is wrong to use the "net force" in order to calculate a particular force's work contribution.
P: 3 I agree, but heres an example of the same object, the apple starting at rest. initial velocity = 0m/s force of gravity balanced by force of hand against apple. In this case there is no acceleration and therefore net force acting on the apple is zero. If any additional upward force were to act on the apple, it would accelerate upwards (mass x acceleration) if that force acted over a distance of d vertical height, then the work done do move the object from rest upwards against gravity for a distance of d is net force F(net) X d Again here i am inclined to think that the force used against the downward force of gravity will determine how much work is done to move an object a set distance upwards. I havent done physics for a long time and now i think im confusing myself!
Mentor P: 41,461 Work done against gravity, what am i missing You are asking to determine the work that you need to do to raise the apple, not the net work on the apple. To raise the apple, you must exert a force equal to its weight (maybe a touch more) over a distance "d": Thus you do work on the apple. The work you do is what is being asked for. Of course, the net work on the apple is zero: It starts at rest and ends at rest.
Sci Advisor P: 5,523 Also, don't confuse force and work. Holding the apple up in a static position requires force but no work is performed. Moving the lifted apple horizontally also requires no force or work (try telling that to your muscles...). The work you performed is equal to a force times a distance. The amount of work required to lift the apple is m*g*h, regardless of the amount of time it took to raise the apple.
P: 3 Im still a little confused. How would you answer this then: how much work is required to move an apple with a mass of 102g over a vertical distance of 1m? Is the work dependant on how much force u apply or will the work done to move the apple always be constant? I keep thinking that because the force i can apply to the apple can be variable, the work done would also be variable.
Mentor
P: 41,461
Quote by lignocaine Im still a little confused. How would you answer this then: how much work is required to move an apple with a mass of 102g over a vertical distance of 1m?
The minimum work required to overcome gravity is mg*d = 0.102*9.8*1 = 1 J.
Is the work dependant on how much force u apply or will the work done to move the apple always be constant? I keep thinking that because the force i can apply to the apple can be variable, the work done would also be variable.
You are correct. You can push as hard as you want on that apple and thus do even more work than the minimum. You'll end up moving the apple up one meter and increasing its speed.
Generally, when they ask about the work needed to lift something they mean the minimum work needed to just lift it without giving it any added speed (and thus added kinetic energy).
Good questions!
P: 14
Quote by lignocaine Im still a little confused. How would you answer this then: how much work is required to move an apple with a mass of 102g over a vertical distance of 1m? Is the work dependant on how much force u apply or will the work done to move the apple always be constant? I keep thinking that because the force i can apply to the apple can be variable, the work done would also be variable.
If the apple is stationary in your hand when you begin, and then also stationary 1m away when you are finished, then the exact same amount of work must have been used to accelerate the apple upwards as was used to decelerate the apple (accelerate it in the opposite direction) once it reaches its destination.
Although it may be difficult to visualize, there is no real difference between that action and the act of accelerating and then braking in a car, except that one is in a vertical direction and the other is in a horizontal direction.
Once you can accept that those two forces (acceleration at the start/deceleration at the end) - and hence the work required for acceleration/deceleration - exactly counteract each other, then you can focus on the remaining part of the question which is how much net work was required to actually lift the apple.
P: 1
Quote by Doc Al The minimum work required to overcome gravity is mg*d = 0.102*9.8*1 = 1 J. You are correct. You can push as hard as you want on that apple and thus do even more work than the minimum. You'll end up moving the apple up one meter and increasing its speed. Generally, when they ask about the work needed to lift something they mean the minimum work needed to just lift it without giving it any added speed (and thus added kinetic energy). Good questions!
I am wondering whether the minimum work required to overcome gravity which is 1 J here can only lift the apple stationary in ur hand without being able to move up or down?
If u want to move it up more, then u will have to use more force (also more work done, of course) than 1 J? Am I right?
Moreover, in any case, the force on an object by gravity is the object's weight itself (in this case, the apple's weight)? is it a negative force?
P: 777
To raise the apple 1m requires 1J of work.
You are right in that you might need to use a bit more than 1N of force to get the thing moving, but that means that you also accelerated it, and can therefore stop pushing a bit before the apple is 1m high...and let its momentum take it there. However you look at it, it ends up gaining 1J of energy.
Also, I think you are confusing energy and force. To overcome gravity, you must push the apple with 1N. But, just holding it in place with 1N is not doing work. You're doing nothing. You can substitute a table for your hand, and the table will supply that 1N indefinitely. Hell, you can ask the table to support a car with 10,000N...and it's still no work is being done. Only by pushing it with 1N AND moving it 1m, will you get 1J. Without that motion against the force, you get nothing.
Moreover, in any case, the force on an object by gravity is the object's weight itself (in this case, the apple's weight)? is it a negative force?
By saying a force is "negative", you're just talking about which direction that force is pushing/ pulling. If we say that up is "positive", then "down" (gravity), would be negative. But, we can arbitrarily set whichever direction we want to be "positive" or "negative", so I could just as well say that up is negative, thereby making gravity positive....
P: 1 Thanks for explaining this
Related Discussions Classical Physics 10 Introductory Physics Homework 15 General Physics 6 Introductory Physics Homework 2 Introductory Physics Homework 6
| 1,971 | 8,506 |
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 3.796875 | 4 |
CC-MAIN-2014-35
|
latest
|
en
| 0.942715 |
http://gmatclub.com/forum/in-a-stack-of-boards-at-a-lumber-yard-the-20th-board-55477.html?fl=similar
| 1,430,876,072,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2015-18/segments/1430457725048.66/warc/CC-MAIN-20150501052205-00038-ip-10-235-10-82.ec2.internal.warc.gz
| 92,241,198 | 40,613 |
Find all School-related info fast with the new School-Specific MBA Forum
It is currently 05 May 2015, 17:34
### GMAT Club Daily Prep
#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.
Customized
for You
we will pick new questions that match your level based on your Timer History
Track
every week, we’ll send you an estimated GMAT score based on your performance
Practice
Pays
we will pick new questions that match your level based on your Timer History
# Events & Promotions
###### Events & Promotions in June
Open Detailed Calendar
# In a stack of boards at a lumber yard, the 20th board
Author Message
TAGS:
Intern
Joined: 12 Nov 2007
Posts: 3
Followers: 0
Kudos [?]: 0 [0], given: 0
In a stack of boards at a lumber yard, the 20th board [#permalink] 12 Nov 2007, 19:45
In a stack of boards at a lumber yard, the 20th board counting from the top of the stack is immediately below the 16th board counting from the bottom of the stack. How many boards are in the stack?
A 38
B 36
C 35
D 34
E 32
I can find the answer easily enough by writing down all the numbers and back solving using the answers, but there has to be an easier way to attack this problem. Any ideas?
Manager
Joined: 25 Jul 2007
Posts: 108
Followers: 1
Kudos [?]: 14 [0], given: 0
There is no need to backsolve.
We will call our board B. It says that the board is 20th from the top of the stack.
i.e:
19
B
..
...
It also says that this board B is one below the 16th board from the bottom. i.e. it is the 15th board from the bottom.
19
B
14
19+B+14=34 boards in the stack.
Hope this helps.
Director
Joined: 09 Aug 2006
Posts: 766
Followers: 1
Kudos [?]: 69 [0], given: 0
Re: Counting Problems [#permalink] 12 Nov 2007, 20:26
showtime12 wrote:
In a stack of boards at a lumber yard, the 20th board counting from the top of the stack is immediately below the 16th board counting from the bottom of the stack. How many boards are in the stack?
A 38
B 36
C 35
D 34
E 32
I can find the answer easily enough by writing down all the numbers and back solving using the answers, but there has to be an easier way to attack this problem. Any ideas?
From the top, the board after the 19th board, i.e., the 20th board from the top is the 15th board from the bottom. Total boards = 19 + 15 = 34.
Intern
Joined: 12 Nov 2007
Posts: 3
Followers: 0
Kudos [?]: 0 [0], given: 0
Re: Counting Problems [#permalink] 13 Nov 2007, 11:38
That makes sense. Thanks
showtime12 wrote:
In a stack of boards at a lumber yard, the 20th board counting from the top of the stack is immediately below the 16th board counting from the bottom of the stack. How many boards are in the stack?
A 38
B 36
C 35
D 34
E 32
I can find the answer easily enough by writing down all the numbers and back solving using the answers, but there has to be an easier way to attack this problem. Any ideas?
Re: Counting Problems [#permalink] 13 Nov 2007, 11:38
Similar topics Replies Last post
Similar
Topics:
CPA as a spring board to MBA as a spring board to.. 3 26 Jun 2011, 13:53
1 Stack of Boards 1 17 Jun 2011, 04:09
44 As the former chair of the planning board for 18 consecutive 17 08 Sep 2008, 18:24
Stack of boards Qs 2 24 Mar 2007, 19:13
11 In a stack of boards at a lumber yard, the 20th board 10 14 Nov 2006, 12:59
Display posts from previous: Sort by
| 1,017 | 3,446 |
{"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-2015-18
|
longest
|
en
| 0.926921 |
https://www.physicsforums.com/threads/can-somebody-check-my-proofs-please.375915/
| 1,547,840,562,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2019-04/segments/1547583660529.12/warc/CC-MAIN-20190118193139-20190118215139-00107.warc.gz
| 898,096,855 | 14,127 |
# Can Somebody Check My Proofs Please
1. Feb 6, 2010
### mmmboh
Hi, would someone be able to check my proofs for me and tell me if they are right and if not what is wrong please?
So for the first one I said let u=p(x) and v=b(x)
T(u+v)=p(x)+b(x)=p(5)x2+b(5)x2=Tu+Tv
and T(ku)=(kp)(x)=kp(5)x2=kTu
So it is a linear transformation.
For the second I said T(u+v)=p(x)+b(x)=x2p(1/x)+x2b(1/x)=Tp(x)+Tb(x)
and T(ku)=x2kp(1/x)=k(x2p(1/x))=kTp(x)
So it is also a linear transformation.
For the third I said T(kp)(x)=xkp'(x)kp''(x)=k2xp'(x)p''(x) which does not equal KTp(x)
So it is not a linear transformation
Did I do these right?
Thanks for any help :)
2. Feb 6, 2010
### talolard
for the first one, what would happen if you took A= x^2 and B=-x^2?
3. Feb 6, 2010
### Staff: Mentor
You haven't used the given information that P2(x) is the space of polynomials of degree 2 or less. Every function in this space is of the form p(x) = ax2 + bx + c, for some constants a, b, and c.
4. Feb 6, 2010
### mmmboh
So am I suppose to write it out as T(ax2 + bx + c+dx2+ex+f)=[25(a+d)+5(b+e)+(c+f)]x2?
I am a bit confused
5. Feb 6, 2010
### mmmboh
Hm I did it that way and I still get that it is a linear transformation.
Well either way I get zero.
6. Feb 6, 2010
### vela
Staff Emeritus
You have some problems with your notation. For instance, you have u=p(x) and v=b(x), so $T(u+v) = T[(p+b)(x)] \ne p(x)+b(x)$.
7. Feb 6, 2010
### mmmboh
Right I made that mistake, but I don`t think it affected the wrongness of my answer.
8. Feb 6, 2010
### mmmboh
T(u+v)=T[p+b](x)=(p+b)(5)x2. Does this not equal p(5)x2+b(5)x2? I don't understand why not.
9. Feb 6, 2010
### vela
Staff Emeritus
I think your conclusions are right.
| 635 | 1,734 |
{"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}
| 3.71875 | 4 |
CC-MAIN-2019-04
|
latest
|
en
| 0.8854 |
https://myassignmenthelp.com/uk/answers/454590-triangles-and-their-properties-what-are-454590-triangles-state-and-desc.html
| 1,610,821,405,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2021-04/segments/1610703506832.21/warc/CC-MAIN-20210116165621-20210116195621-00134.warc.gz
| 472,967,911 | 72,895 |
\$20 Bonus + 25% OFF
#### 45-45-90 Triangles and Their Properties
Referencing Styles : APA | Pages : 1
A triangle is the figure that has 3 sides those meet at the point that is called as vertex. Each vertex forms an angle and when they are added up it forms a triangle that is equal to 180°. 45-45-90 triangle is the special type of isosceles right triangle under which 2 legs are congruent to one another. As the 2 angles of the triangle are equal to 45°, the triangle is the isosceles triangle. Here, both non-right angles are equal to 45 degrees. Special relationship is there among lengths of sides for a 45-45-90 triangle that can be used for determining the lengths of other 2 sides. If the triangle is 45-45-90 triangle, ratio of the sides will be leg: leg: hypotenuse or 1:1: Generally Pythagorean Theorem is used for finding out the missing leg or for hypotenuse of 45-45-90 triangle. Ratio of sides to hypotenuse always is 1:1: square root of 2. Formula for Pythagorean Theorem is a2 + b2 = c2. Rule for using the formula is c shall stand for hypotenuse irrespective of which side is called as ‘a’ and which side is called as ‘b’. The same is explained through the example below –
Leg of above presented 45-45-90 triangle measures 5 units. Pythagorean Theorem states that for right triangle, a2 + b2 = c2, where ‘a’ and ‘b’ are lengths of legs and ‘c’ is length of hypotenuse. As the triangle is 45-45-90 triangle, the legs of this triangle is congruent and value of ‘a’ and ‘b’ both is 5. Using the formula of a2 + b2 = c2, ‘c’ can be solved as follows –
Hence, the length of hypotenuse is equal to length of a leg multiplied by or units.
Properties of a 45-45-90 triangle –
Each 45-45-90 triangle has some special properties. When it is talked about a 45-45-90 triangle, the numbers represent measures of angles of the triangle. Hence, it means that the triangle has two 45° angles and one 90° angle. For any isosceles right triangle when the legs are ‘x’ units in length, hypotenuse will always be x. Hypotenuse of the 45-45-90 triangle is longest side of the triangle and is across from right angle. The legs of the 45-45-90 triangle are 2 shorter sides of the triangle and are adjacent to right angle. Most important property of this triangle is that this triangle has 1 right angle and 2 other angles those are equal to 45°. It signifies that 2 sides – legs – equal in the length and hypotenuse can be computed easily. Another interesting property is that it is the only possible triangle that is also the isosceles triangle. Further, it has smallest ratio of hypotenuse to sum of legs and has greatest ratio of altitude from the hypotenuse to sum of legs.
OR
### Save Time & improve Grade
Just share requirement and get customized Solution.
Orders
Overall Rating
Experts
### Our Amazing Features
#### On Time Delivery
Our writers make sure that all orders are submitted, prior to the deadline.
#### Plagiarism Free Work
Using reliable plagiarism detection software, Turnitin.com.We only provide customized 100 percent original papers.
#### 24 X 7 Live Help
Feel free to contact our assignment writing services any time via phone, email or live chat. If you are unable to calculate word count online, ask our customer executives.
#### Services For All Subjects
Our writers can provide you professional writing assistance on any subject at any level.
#### Best Price Guarantee
Our best price guarantee ensures that the features we offer cannot be matched by any of the competitors.
### Our Experts
5/5
2109 Order Completed
99% Response Time
### Emma Zhong
#### Singapore, Singapore
4/5
248 Order Completed
100% Response Time
### Lloyd Bernabe
#### London, United Kingdom
5/5
93 Order Completed
100% Response Time
### Jackson Mitchell
#### New Jersey, United States
5/5
755 Order Completed
95% Response Time
### FREE Tools
#### Plagiarism Checker
Get all your documents checked for plagiarism or duplicacy with us.
#### Essay Typer
Get different kinds of essays typed in minutes with clicks.
#### Chemical Equation Balancer
Balance any chemical equation in minutes just by entering the formula.
#### Word Counter & Page Calculator
Calculate the number of words and number of pages of all your academic documents.
## Refer Just 5 Friends to Earn More than \$2000
1
1
1
### Live Review
Our Mission Client Satisfaction
thank you so much I get exactly what i need from first time the Dr. thank me that i make something unique and defrrant from the other students .thank you so much
#### User Id: 525537 - 15 Jan 2021
Australia
Able to get my work done after few changes, which were done as per suggestion provided by us. Appreciate their positive approach for this
#### User Id: 361833 - 15 Jan 2021
Australia
Amazing experience with you guys. I got 20/20. You are the best in the market i tried multiple sites but you guys gave best output.
#### User Id: 427239 - 15 Jan 2021
Australia
good paper, thank you very much. i am looking forward to working with you. Recommended
Australia
| 1,243 | 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}
| 4.6875 | 5 |
CC-MAIN-2021-04
|
longest
|
en
| 0.915066 |
https://holooly.com/solutions/for-ft-ti-t3j-and-f0-i-2j-calculate-ft/
| 1,657,111,328,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2022-27/segments/1656104672585.89/warc/CC-MAIN-20220706121103-20220706151103-00424.warc.gz
| 343,775,901 | 22,569 |
Products
Rewards
from HOLOOLY
We are determined to provide the latest solutions related to all subjects FREE of charge!
Enjoy Limited offers, deals & Discounts by signing up to Holooly Rewards Program
HOLOOLY
HOLOOLY
TABLES
All the data tables that you may search for.
HOLOOLY
ARABIA
For Arabic Users, find a teacher/tutor in your City or country in the Middle East.
HOLOOLY
TEXTBOOKS
Find the Source, Textbook, Solution Manual that you are looking for in 1 click.
HOLOOLY
HELP DESK
Need Help? We got you covered.
## Q. 3.6
For $\mathbf{f}^{\prime}(t)=t \mathbf{i}+t^{3} \mathbf{j}$ and $\mathbf{f}(0)=\mathbf{i}-2 \mathbf{j}$, calculate $\mathbf{f}(t)$.
## Verified Solution
We first find that
$\mathbf{f}(t)=\int \mathbf{f}^{\prime}(t) d t=\left(\frac{t^{2}}{2}+C_{1}\right) \mathbf{i}+\left(\frac{t^{4}}{4}+C_{2}\right) \mathbf{j}=\frac{t^{2}}{2} \mathbf{i}+\frac{t^{4}}{4} \mathbf{j}+\mathbf{C}$.
To evaluate $\mathbf{C}$, we substitute the value t = 0 into (10) to find that $\mathbf{f}(0)=\mathbf{C}=\mathbf{i}-2 \mathbf{j}$, so that
$\mathbf{f}(t)=\frac{t^{2}}{2} \mathbf{i}+\frac{t^{4}}{4} \mathbf{j}+\mathbf{i}-2 \mathbf{j}=\left(\frac{t^{2}}{2}+1\right) \mathbf{i}+\left(\frac{t^{4}}{4}-2\right) \mathbf{j}$.
| 475 | 1,236 |
{"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": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 7, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 3.96875 | 4 |
CC-MAIN-2022-27
|
latest
|
en
| 0.538273 |
https://www.askiitians.com/forums/Mechanics/the-force-f1-parallel-to-incline-plane-that-is-nec_193845.htm
| 1,576,042,596,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2019-51/segments/1575540529955.67/warc/CC-MAIN-20191211045724-20191211073724-00077.warc.gz
| 617,250,530 | 31,102 |
Click to Chat
1800-1023-196
+91-120-4616500
CART 0
• 0
MY CART (5)
Use Coupon: CART20 and get 20% off on all online Study Material
ITEM
DETAILS
MRP
DISCOUNT
FINAL PRICE
Total Price: Rs.
There are no items in this cart.
Continue Shopping
` The force f1 parallel to incline plane that is necessary to move a body up the inclined is double the force f2 to that is necessary to just prevent them sliding down them`
one year ago
Kruthik
119 Points
``` Let’s consider the 2nd case. F2 is the force required just to prevent the body from sliding. Now, if we consider the FBD, there will be F2 the direction opposite to wsinθ(component of weight along the incline). The friction force will act in the direction opposite to wsinθ, as the body tends to move down the incline, and F2 prevents it.Thus, we get F2=wsinθ-fr (fr = frictional force)———— 1As ? is the angle of repose, μ = tan?....2From 1, F2=wsinθ - μwcosθ (fr = μN, and N is the normal reaction, component for weight perpendicular to the incline)So, F2=wsinθ – tan?cosθPut tan?=sin?/cos?Thus,F2=wsinθ – (sin?cosθ)/cos?On further solving, we get F2 = wsin(θ - ?)sec?, that is option ‘A’ which is the answer. Hope this helped you. Good Luck.
```
one year ago
Kruthik
119 Points
``` Let’s consider the 2nd case. F2 is the force required just to prevent the body from sliding. Now, if we consider the FBD, there will be F2 the direction opposite to wsinθ(component of weight along the incline). The friction force will act in the direction opposite to wsinθ, as the body tends to move down the incline, and F2 prevents it.Thus, we get F2=wsinθ-fr (fr = frictional force)———— 1As ? is the angle of repose, μ = tan?....2From 1, F2=wsinθ - μwcosθ (fr = μN, and N is the normal reaction, component for weight perpendicular to the incline)So, F2 = wsinθ – wtan?cosθPut tan? = sin?/cos?Thus,F2 = wsinθ – w(sin?cosθ)/cos?On further solving, we get F2 = wsin(θ - ?)sec?, that is option ‘A’ which is the answer. Hope this helped you. Good Luck.
```
one year ago
Kruthik
119 Points
``` Sorry, but the first solution had some minute errors, so I reanswered... The second answer is correct.
```
one year ago
Think You Can Provide A Better Answer ?
Other Related Questions on Mechanics
View all Questions »
Course Features
• 101 Video Lectures
• Revision Notes
• Previous Year Papers
• Mind Map
• Study Planner
• NCERT Solutions
• Discussion Forum
• Test paper with Video Solution
Course Features
• 110 Video Lectures
• Revision Notes
• Test paper with Video Solution
• Mind Map
• Study Planner
• NCERT Solutions
• Discussion Forum
• Previous Year Exam Questions
| 762 | 2,636 |
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 3.6875 | 4 |
CC-MAIN-2019-51
|
longest
|
en
| 0.867326 |
https://sites.duke.edu/probabilityworkbook/category/basic-probability/joint-distributions/
| 1,713,290,978,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2024-18/segments/1712296817103.42/warc/CC-MAIN-20240416155952-20240416185952-00511.warc.gz
| 483,394,969 | 12,211 |
# Category Archives: Joint Distributions
## Joint Density Poisson arrival
Let $$T_1$$ and $$T_5$$ be the times of the first and fifth arrivals in a Poisson arrival prices with rate $$\lambda$$. Find the joint distribution of $$T_1$$ and $$T_5$$ .
## Uniform Spacing
Let $$U_1, U_2, U_3, U_4, U,5$$ be independent uniform $$(0,1)$$ random variables. Let $$R$$ be the difference between the max and the min of the random variables. Find
1. $$E( R)$$
2. the joint density of the min and the max of the $$U$$’s
3. $$P( R>0.5)$$
[Pitman p. 355 #14]
## Joint density part 2
Let $$X$$ and $$Y$$ have joint density
$$f(x,y) = 90(y-x)^8, \quad 0<x<y<1$$
1. State the conditional distribution of $$X \mid Y$$ and $$Y \mid X$$
2. Are these two random variables independent?
3. What is $$\mathbf{P}(Y \mid X=.2 )$$ and $$\mathbf{E}(Y \mid X=.2)$$ ?
What is $$\mathbf{P}(Y \mid X=.2 )$$ and $$\mathbf{E}(Y \mid X=.2)$$
[Adapted from Pitman pg 354]
## Uniform distributed points given an arrival
Consider a Poisson arrival process with rate $$\lambda>0$$. Let $$T$$ be the time of the first arrival starting from time $$t>0$$. Let $$N(s,t]$$ be the number of arrivals in the time interval $$(s,t]$$.
Fixing an $$L>0$$, define the pdf $$f(t)$$ by $$f(t)dt= P(T \in dt | N(0,L]=1)$$ for $$t \in (0,L]$$. Show that $$f(t)$$ is the pdf of a uniform random variable on the interval $$[0,L]$$ (independent of $$\lambda$$ !).
## Joint Distribution Table
Consider the following joint distribution.
If the experiment is flipping a fair coin three times, which of the following could be the random variables $$X$$ and $$Y$$. Select all that apply.
1. $$X=$$ the number of heads, $$Y=$$ the number of tails.
2. $$X=$$ the number of tails, $$Y=$$ the number of tails (i.e., $$Y=X$$).
3. $$X=$$ the number of heads. $$Y=3-X.$$
4. $$X=$$ the number of tails on the first two flips. $$Y=$$ the number of tails on the last two flips.
## Joint, Marginal and Conditioning
Let $$(X,Y)$$ have joint density $$f(x,y) = e^{-y}$$, for $$0<x<y$$, and $$f(x,y)=0$$ elsewhere.
1. Are $$X$$ and $$Y$$ independent ?
2. Compute the marginal density of $$Y$$.
3. Show that $$f_{X|Y}(x,y)=\frac1y$$, for $$0<x<y$$.
4. Compute $$E(X|Y=y)$$
5. Use the previous result to find $$E(X)$$.
## A joint density example I
Let $$(X,Y)$$ have joint density $$f(x,y)=x e^{-x-y}$$ when $$x,y>0$$ and $$f(x,y)=0$$ elsewhere. Are $$X$$ and $$Y$$ independent ?
[Meester ex 5.12.30]
## A Joint density example II
If $$X$$ and $$Y$$ have joint density function
$f(x,y)=\frac{1}{x^2y^2} \quad; \quad x \geq 1, y\geq 1$
1. Compute the joint density fiction of $$U=XY$$, $$V=X/Y$$.
2. What are the marginal densities of $$U$$ and $$V$$ ?
[Ross p295, # 54]
## Joint of min and max
Let $$X_1,…,X_n \stackrel{iid}{\sim} \mbox{Exp}(\lambda)$$
Let $$V = \mbox{min}(X_1,…,X_n)$$ and $$W = \mbox{max}(X_1,…,X_n)$$.
What is the joint distribution of $$V,W$$. Are they independent ?
## Joint density part 1
Let $$X$$ and $$Y$$ have joint density
$$f(x,y) = 90(y-x)^8, \quad 0<x<y<1$$
1. State the marginal distribution for $$X$$
2. State the marginal distribution for $$Y$$
3. Are these two random variables independent?
4. What is $$\mathbf{P}(Y > 2X)$$
5. Fill in the blanks “The density $$f(x,y)$$ above is the joint density of the _________ and __________ of ten independent uniform $$(0,1)$$ random variables.”
[Adapted from Pitman pg 354]
## Simple Joint density
Let $$X$$ and $$Y$$ have joint density
$f(x,y) = c e^{-2x -3 y} \quad (x,y>0)$
for some $$c>0$$ and $$f(x,y)=0$$ otherwise. find:
1. the correct value of $$c$$.
2. $$P( X \leq x, Y \leq y)$$
3. $$f_X(x)$$
4. $$f_Y(y)$$
5. Are $$X$$ and $$Y$$ independent ? Explain your reasoning ?
## Joint Distributions of Uniforms
Let $$X$$ and $$Y$$ be independent, each uniformly distributed on $$\{1,2,\dots,n\}$$. Find:
1. $$\mathbf{P}( X=Y)$$
2. $$\mathbf{P}( X < Y)$$
3. $$\mathbf{P}( X>Y)$$
4. $$\mathbf{P}( \text{max}(X,Y)=k )$$ for $$k=1,\dots,n$$
5. $$\mathbf{P}( \text{min}(X,Y)=k )$$ for $$k=1,\dots,n$$
6. $$\mathbf{P}( X+Y=k )$$ for $$k=2,\dots,2n$$
| 1,446 | 4,095 |
{"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}
| 3.9375 | 4 |
CC-MAIN-2024-18
|
latest
|
en
| 0.725163 |
http://mathhelpforum.com/advanced-algebra/114603-normal-subgroup-print.html
| 1,508,733,354,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2017-43/segments/1508187825575.93/warc/CC-MAIN-20171023035656-20171023055656-00563.warc.gz
| 222,394,816 | 2,659 |
# normal subgroup
• Nov 14th 2009, 08:29 PM
apple2009
normal subgroup
Let H be a group and K a subgroup of H with (H:K)=n< ∞.
Prove: if (H:K)=2, then K is a normal subgroup of H
• Nov 15th 2009, 12:32 AM
Opalg
Quote:
Originally Posted by apple2009
Let H be a group and K a subgroup of H with (H:K)=n< ∞.
Prove: if (H:K)=2, then K is a normal subgroup of H
The subgroup K partitions H into left cosets, one of which is K itself. If (H:K)=2, then there is only one other left coset, namely everything in H that isn't in K. The same applies to right cosets. Thus when (H:K)=2, each left coset is also a right coset. But that is one way of expressing the condition for K to be normal.
| 226 | 683 |
{"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-2017-43
|
longest
|
en
| 0.900125 |
https://www.calculatoratoz.com/en/area-of-hexagon-given-inradius-calculator/Calc-27608
| 1,638,691,255,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2021-49/segments/1637964363149.85/warc/CC-MAIN-20211205065810-20211205095810-00188.warc.gz
| 723,546,214 | 29,195 |
## Credits
St Joseph's College (SJC), Bengaluru
Mona Gladys has created this Calculator and 1000+ more calculators!
Walchand College of Engineering (WCE), Sangli
Shweta Patil has verified this Calculator and 1000+ more calculators!
## Area of Hexagon given inradius Solution
STEP 0: Pre-Calculation Summary
Formula Used
A = (3/2)*sqrt(3)*(((2*ri)/sqrt(3))^2)
This formula uses 1 Functions, 1 Variables
Functions Used
sqrt - Squre root function, sqrt(Number)
Variables Used
Inradius - Inradius is defined as the radius of the circle which is inscribed inside the polygon. (Measured in Meter)
STEP 1: Convert Input(s) to Base Unit
Inradius: 10 Meter --> 10 Meter No Conversion Required
STEP 2: Evaluate Formula
Substituting Input Values in Formula
A = (3/2)*sqrt(3)*(((2*ri)/sqrt(3))^2) --> (3/2)*sqrt(3)*(((2*10)/sqrt(3))^2)
Evaluating ... ...
A = 346.410161513776
STEP 3: Convert Result to Output's Unit
346.410161513776 Square Meter --> No Conversion Required
346.410161513776 Square Meter <-- Area
(Calculation completed in 00.000 seconds)
## < 9 Area of Hexagon Calculators
Area of Hexagon given short diagonal
area = (3/2)*sqrt(3)*((Short diagonal/sqrt(3))^2) Go
Area of Hexagon given height
area = (3/2)*sqrt(3)*((Height/sqrt(3))^2) Go
Area of Hexagon given central angle and side
area = (3*(Side)^2)/(2*tan(Angle A/2)) Go
Area of Hexagon given long diagonal
area = (3/2)*sqrt(3)*((Long diagonal/2)^2) Go
Area of Hexagon given perimeter
area = (3/2)*sqrt(3)*((Perimeter/6)^2) Go
Area of Hexagon
area = (3/2)*sqrt(3)*Side^2 Go
Area of Hexagon given inradius and side
### Area of Hexagon given inradius Formula
A = (3/2)*sqrt(3)*(((2*ri)/sqrt(3))^2)
## What is a hexagon?
A regular hexagon is defined as a hexagon that is both equilateral and equiangular. It is bicentric, meaning that it is both cyclic (has a circumscribed circle) and tangential (has an inscribed circle). The common length of the sides equals the radius of the circumscribed circle or circumcircle, which equals 2/sqrt(3) times the apothem (radius of the inscribed circle). All internal angles are 120 degrees. A regular hexagon has six rotational symmetries
## How to Calculate Area of Hexagon given inradius?
Area of Hexagon given inradius calculator uses area = (3/2)*sqrt(3)*(((2*Inradius)/sqrt(3))^2) to calculate the Area, The Area of hexagon given inradius formula is defined as the total area that the surface of the object occupies of hexagon , where side = side of hexagon , area = area of hexagon. Area and is denoted by A symbol.
How to calculate Area of Hexagon given inradius using this online calculator? To use this online calculator for Area of Hexagon given inradius, enter Inradius (ri) and hit the calculate button. Here is how the Area of Hexagon given inradius calculation can be explained with given input values -> 346.4102 = (3/2)*sqrt(3)*(((2*10)/sqrt(3))^2).
### FAQ
What is Area of Hexagon given inradius?
The Area of hexagon given inradius formula is defined as the total area that the surface of the object occupies of hexagon , where side = side of hexagon , area = area of hexagon and is represented as A = (3/2)*sqrt(3)*(((2*ri)/sqrt(3))^2) or area = (3/2)*sqrt(3)*(((2*Inradius)/sqrt(3))^2). Inradius is defined as the radius of the circle which is inscribed inside the polygon.
How to calculate Area of Hexagon given inradius?
The Area of hexagon given inradius formula is defined as the total area that the surface of the object occupies of hexagon , where side = side of hexagon , area = area of hexagon is calculated using area = (3/2)*sqrt(3)*(((2*Inradius)/sqrt(3))^2). To calculate Area of Hexagon given inradius, you need Inradius (ri). With our tool, you need to enter the respective value for Inradius and hit the calculate button. You can also select the units (if any) for Input(s) and the Output as well.
How many ways are there to calculate Area?
In this formula, Area uses Inradius. We can use 9 other way(s) to calculate the same, which is/are as follows -
| 1,123 | 3,993 |
{"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-2021-49
|
latest
|
en
| 0.740083 |
https://www.edaboard.com/threads/single-phase-transformer.217423/
| 1,709,492,773,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2024-10/segments/1707947476397.24/warc/CC-MAIN-20240303174631-20240303204631-00669.warc.gz
| 759,826,940 | 19,769 |
Continue to Site
# Single phase transformer
Status
Not open for further replies.
#### emperror123
##### Member level 5
to all,
i were facing a problem on single phase transformer on calculation part.
i been sketches out the single phase transformer equivalent circuit.
and some information given
f = 60hz
s = 20kVA
V1 = 2400v
V2 = 240v
R1 = 3Ω
R2 = 0.03Ω
X1 = 6.5Ω
X2 = 0.07Ω
Xm = 15kΩ
Rc = 100kΩ
lagging P.F. = 0.8
and here the problem come. i were trying to solve the problem.
which is
i got Xeq1 = R1 + a^2R2 = 13.5Ω, and Req1 = X1 +a^2X2 = 6Ω
i got primary voltage is V1 = 2400V and V2' = aV2 = 2400V
how do i calculate the primary current and secondary current?
and also the source voltage.
thank you and much appreciate with your help to solve the problem
...i got Xeq1 = R1 + a^2R2 = 13.5Ω, and Req1 = X1 +a^2X2 = 6Ω...
should be as?
Req1 = R1 + a^2R2 = 13.5Ω, and Xeq1 = X1 +a^2X2 = 6Ω
is it ?
SRIZBF:
sorry, is a typo error. shud be the same as u say
atripathi:
shall i need to include Xm and Rc into the formula to calculate the Ip or Is?
the source voltage got any formula?
I would assume Source Voltage as 2400V (equal to V1, in specifications mentioned by you)
Yes, you need to include Xm & Rc into current calculations Ip. Ip=Io + Secondary (or load ) current reflected in primary; here Ip is current drawn from source
how could i get voltage load, it do not gave any resistance load in circuits
how could i get voltage load, it do not gave any resistance load in circuits
You didn't tell the exact exercise text, but you're asking for a secondary current, so there should be a load, isn't it? The only hint for the load I found is the P.F. number, but I think you'll need more information about the load, I think.
Status
Not open for further replies.
| 551 | 1,781 |
{"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-2024-10
|
latest
|
en
| 0.947173 |
http://cbsemathssolutions.in/centroid-of-a-triangle-definition-and-formula/
| 1,542,247,090,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2018-47/segments/1542039742338.13/warc/CC-MAIN-20181115013218-20181115035218-00486.warc.gz
| 69,157,138 | 16,446 |
# Centroid of a triangle definition and formula explained
Learn and know about the centroid of a triangle. Centroid is one of the most important topics one should study in coordinate geometry chapter. So now we will learn the actual meaning of the centroid and also we will learn how we can find the centroid.
First, we will see the definition of the centroid and after that, we will see properties of centroid, formula and representation of centroid for different kinds of triangle one by one.
## Definition of the centroid of a triangle:
The intersection point of all the three medians of a triangle is called centroid. Centroid is represented with the letter G.
In the above triangle, we can observe three medians i.e. AD, BE and CF. we can also observe that all the three medians are meeting at one point, that point we are going to call as the centroid ( G).
Important Property of a centroid:
We should know that centroid (G ) divides the medians in 2: 1 ratio.
### The Formula for finding the centroid of a triangle with vertices:
Suppose If A ( x1 , y1) , B ( x2, y2) C( x3, y3) are the given three vertices of a triangle then centroid is given by
#### Problems based on the centroid:
Find the centroid of triangle with vertices A (1, 5), B (2, 7) and C (6, 3).
Solution:
In a Δ ABC, the centroid is (7, 2) and the two vertices of the triangle are given by (1, 8), (10, 4) then find the third vertex of a triangle.
Solution:
Let the two vertices of a triangle be A (1, 8), B (10, 4) and the third vertex be C (x, y).
Representation of centroid for right, acute and obtuse angled triangles:
I hope you learnt everything about the centroid i.e. definition, formula and property.
Please follow and like us:
Subscribe
Notify of
| 420 | 1,749 |
{"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}
| 4.34375 | 4 |
CC-MAIN-2018-47
|
latest
|
en
| 0.910596 |
https://www.letusdothehomework.com/math-for-statistics/
| 1,721,040,678,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2024-30/segments/1720763514696.4/warc/CC-MAIN-20240715102030-20240715132030-00159.warc.gz
| 732,231,539 | 20,515 |
# Math For Statistics
Statistics is a field that relies heavily on mathematical concepts and techniques. Understanding the underlying math is crucial for grasping statistical concepts and performing data analysis effectively.
In this blog post, we will explore the fundamental mathematical concepts used in statistics, such as probability, algebra, calculus, and linear algebra. We will discuss how these math concepts relate to statistical methods and provide examples of their applications in statistical analysis.
## Importance of Mathematics in Statistics
Mathematics forms the foundation of statistics and plays a crucial role in the field of statistics. Statistics is the science of collecting, analyzing, interpreting, and presenting data, and mathematical concepts are used to understand and analyze data in a systematic and rigorous manner.
Mathematics provides the necessary tools and techniques for statistical calculations, probability theory, data modeling, and inference, which are fundamental to the field of statistics.
A strong understanding of mathematical concepts is essential for students studying statistics to effectively analyze and interpret data, draw conclusions, and make informed decisions based on data-driven evidence.
## Mathematical Concepts Used in Statistics
Statistics relies heavily on various mathematical concepts, such as algebra, calculus, probability theory, and linear regression, among others. Algebra is used for data representation, manipulation, and solving equations. Calculus is used for concepts such as limits, derivatives, and integrals, which are used in probability and statistical inference.
Probability theory is crucial in understanding the likelihood of events and making predictions based on probability distributions. Linear regression is used for modeling relationships between variables in statistical analyses.
Understanding and applying these mathematical concepts is vital in statistics to perform data analysis accurately and draw valid conclusions from data.
## Challenges in Understanding Math for Statistics
Many students face challenges in understanding math for statistics. The complex and abstract nature of mathematical concepts can be overwhelming, and students may struggle with grasping the underlying principles and applying them to statistics.
Lack of foundational knowledge in math, inadequate preparation, and limited exposure to mathematical concepts can pose challenges in understanding math for statistics.
Additionally, math anxiety, a fear or apprehension towards math, can hinder students’ ability to comprehend mathematical concepts used in statistics.
## Strategies for Learning Math for Statistics
There are several strategies that students can employ to effectively learn math for statistics:
1. Build a Strong Foundation: It is essential to have a solid understanding of foundational mathematical concepts before delving into statistics. Strengthening basic math skills, such as arithmetic, algebra, and calculus, can provide a solid foundation for understanding more advanced mathematical concepts used in statistics.
2. Practice and Application: Regular practice is crucial for mastering mathematical concepts. Solving math problems, working through exercises and practice questions, and applying mathematical concepts to real-world scenarios can enhance understanding and retention of math for statistics.
3. Seek Help and Resources: Seek help from professors, tutors, or peers for clarifications on mathematical concepts. Utilize resources such as textbooks, online tutorials, and educational websites that provide explanations, examples, and practice problems to reinforce learning.
4. Relate Math to Statistics: Relate mathematical concepts to statistics to understand their relevance and application in the field. Understanding how mathematical concepts are used in statistical analyses can enhance their understanding and applicability.
5. Take a Step-by-Step Approach: Break down complex mathematical concepts into smaller, manageable steps. Start with simpler concepts and gradually progress to more complex ones, building on a solid foundation.
## Resources for Improving Math Skills for Statistics
There are several resources available to improve math skills for statistics, including:
1. Textbooks: Utilize textbooks specifically designed for learning math for statistics. Look for textbooks that provide clear explanations, examples, and practice problems.
2. Online Tutorials: There are numerous online tutorials and educational websites that offer lessons, videos, and interactive exercises to improve math skills for statistics.
3. Tutoring Services: Seek assistance from tutors or academic support services offered by educational institutions. Tutors can provide one-on-one guidance and support in understanding math concepts for statistics.
4. Practice Problems and Exercises: Work through practice problems and exercises available in textbooks, online resources, or study guides to reinforce learning and improve problem-solving skills.
5. Study Groups: Join study groups or form study groups with peers to collaborate, discuss, and learn from each other.
6. Online Courses: Enroll in online courses or workshops that specifically focus on math for statistics. These courses often provide structured lessons, practice exercises, and assessments to help students improve their math skills in the context of statistics.
7. Software and Tools: Utilize statistical software and tools, such as R, Python, or SPSS, which often have built-in math functions and libraries that can help with calculations and data analysis. Familiarizing yourself with these tools can enhance your understanding of mathematical concepts in the context of statistics.
8. Practice in Real-World Scenarios: Apply mathematical concepts to real-world scenarios and data sets to gain practical experience. This can help solidify your understanding of math in the context of statistics and improve your ability to apply mathematical concepts to real data.
9. Seek Help from Professors and Peers: Do not hesitate to seek help from your professors or peers when you encounter challenges in understanding math for statistics. Professors can provide clarifications, additional resources, and guidance, while peers can offer different perspectives and insights.
10. Persistence and Patience: Learning math for statistics can be challenging, and it requires persistence and patience. Keep practicing regularly, seek help when needed, and stay determined to improve your math skills over time.
## Conclusion
Math serves as the foundation for statistics, and a solid understanding of mathematical concepts is essential for mastering statistical techniques. By gaining proficiency in the math that underlies statistics, you can develop a solid foundation for understanding statistical methods and interpreting data effectively.
| 1,160 | 6,918 |
{"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-2024-30
|
latest
|
en
| 0.90269 |
https://studyandanswers.com/mathematics/question12026827
| 1,627,134,072,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2021-31/segments/1627046150266.65/warc/CC-MAIN-20210724125655-20210724155655-00318.warc.gz
| 562,472,847 | 15,439 |
, 22.06.2019 21:00 jayonelijah
# Asnail slithers 2/3 mile in 4/5 hour. what is its speed in miles per hour? set it up!
5/6
Step-by-step explanation:
M(Miles)= 2/3/4/5
step-by-step explanation:
well, michelle wanted to measure the height of her school's flagpole she placed a mirror on the ground 48 feet from the flagpole then walk backwards until she was able to see the top of the pole in the mirror her eyes were 5 feet above the ground and she was 12 ft from the mirror using similar triangles find the height of the flagpole to the nearest tenth of a foot find the geometric mean of them pair of numbers. its easy math
518,400
step-by-step explanation:
the area is multiplying all the given measurements together.
40 * 18 * 30 * 24 = 518,400
### Other questions on the subject: Mathematics
Mathematics, 21.06.2019 17:30, jtg23
Subscriptions to a popular fashion magazine have gone down by a consistent percentage each year and can be modeled by the function y = 42,000(0.96)t. what does the value 42,000 represent in the function?
Mathematics, 21.06.2019 18:00, joseroblesrivera123
Since opening night, attendance at play a has increased steadily, while attendance at play b first rose and then fell. equations modeling the daily attendance y at each play are shown below, where x is the number of days since opening night. on what day(s) was the attendance the same at both plays? what was the attendance? play a: y = 8x + 191 play b: y = -x^2 + 26x + 126
| 401 | 1,473 |
{"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-2021-31
|
latest
|
en
| 0.932738 |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.