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https://wizardofodds.com/article/rothstein-betting-system/ | 1,718,819,657,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861828.24/warc/CC-MAIN-20240619154358-20240619184358-00118.warc.gz | 550,505,170 | 11,154 | # Rothstein Betting System
## Introduction
The Rothstein is a betting system closely related to the Martingale. The difference is with the Rothstein, the player increases his bets more aggressively, ensuring one unit of profit for every bet made, as long as he doesn't go broke. This is unlike the Martingale, where the player has a net win of one unit for every bet won. The tradeoff is the player will go broke faster if he doesn't see a win.
## Rules
The only difference between the Rothstein and the Martingale is the Rothstein adds one more unit to each bet after a loss, compared to the Martingale. In other words, instead of doubling the previous bet after a loss, you double it and then add one more unit. Following are the details.
1. As with most betting systems, it is best explained and applied on games with even money bets.
2. The player must choose a unit size and a bankroll. The first ten bankroll sizes in units are: 1,4,11,26,57,120,247,502,1013,2036. Note how as they get large they are slightly less than powers of 2. To be specific, they can all be expressed as 2n+1-(n+2), where n is the maximum number of consecutive losses.
3. Player starts by making a one-unit bet.
4. If last bet made wins, quit and walk away with a net win of one unit for every bet made.
5. If last bet made loses, then bet two times the previous bet plus one more unit next time.
6. Repeat until you either win a bet or go bust trying.
Note that the bet sizes (in units) will be powers of 2 minus one: 1, 3, 7, 15, 31, 63, etc..
## Five Step Analysis
The following table shows my analysis based on the pass bet in craps where the player can make up to five bets. The worst case scenario, with five losses, is a loss of 57 units.
### Five Step Analysis — Craps Pass Bet
Loses Net Win Probability Expected Win
0 1 0.492929 0.492929
1 2 0.249950 0.499900
2 3 0.126742 0.380227
3 4 0.064267 0.257069
4 5 0.032588 0.162940
5 -57 0.033523 -1.910810
Total 1.000000 -0.117744
The following is my summary of the table above.
• Expected net win = -0.117744 units
• Expected total bet = 8.326175 units
• Ratio of expected win to expected total bet = -1.41%
The following table shows my analysis based on the don't pass bet in craps where the player can make up to five bets. The worst case scenario, with five losses, is a loss of 57 units. For purposes of my analysis, I don't count pushes (12 on the come out roll) as a bet resolved.
### Five Step Analysis — Craps Don't Pass Bet
Loses Net Win Probability Expected Win
0 1 0.493182 0.493182
1 2 0.249954 0.499907
2 3 0.126681 0.380043
3 4 0.064204 0.256817
4 5 0.032540 0.162699
5 -57 0.033440 -1.906057
Total 1.000000 -0.113409
The following is my summary of the table above.
• Expected net win = -0.113409 units
• Expected total bet = 8.316626 units
• Ratio of expected win to expected total bet = -1.36%
The following table shows my analysis based on any even money bet in single-zero roulette where the player can make up to five bets. The worst case scenario, with five losses, is a loss of 57 units.
### Five Step Analysis — Single-Zero Roulette Even Money Bets
Loses Net Win Probability Expected Win
0 1 0.486486 0.486486
1 2 0.249817 0.499635
2 3 0.128285 0.384854
3 4 0.065876 0.263504
4 5 0.033828 0.169141
5 -57 0.035707 -2.035327
Total 1.000000 -0.231708
The following is my summary of the table above.
• Expected net win = -0.231708 units
• Expected total bet = 8.573191 units
• Ratio of expected win to expected total bet = -2.70%
The following table shows my analysis based on any even money bet in double-zero roulette where the player can make up to five bets. The worst case scenario, with five losses, is a loss of 57 units.
### Five Step Analysis — Double-Zero Roulette Even Money Bets
Loses Net Win Probability Expected Win
0 1 0.473684 0.473684
1 2 0.249307 0.498615
2 3 0.131214 0.393643
3 4 0.069060 0.276241
4 5 0.036347 0.181737
5 -57 0.040386 -2.302008
Total 1.000000 -0.478087
The following is my summary of the table above.
• Expected net win = -0.478087 units
• Expected total bet = 9.083655 units
• Ratio of expected win to expected total bet = -5.26%
The following table shows my analysis based on the Player bet in baccarat where the player can make up to five bets. The worst case scenario, with five losses, is a loss of 57 units. I do not count tie outcomes as a bet resolved.
### Five Step Analysis — Player Bet in Baccarat
Loses Net Win Probability Expected Win
0 1 0.493175 0.493175
1 2 0.249953 0.499907
2 3 0.126683 0.380048
3 4 0.064206 0.256824
4 5 0.032541 0.162706
5 -57 0.033442 -1.906182
Total 1.000000 -0.113523
The following is my summary of the table above.
• Expected net win = -0.113523 units
• Expected total bet = 8.316877 units
• Ratio of expected win to expected total bet = -1.36%
Please note that 1.36% is the house edge on the Player bet if ties are not counted as a bet resolved.
## Ten Step Analysis
The following table shows my analysis based on the pass bet in craps where the player can make up to ten bets. The worst case scenario, with ten losses, is a loss of 2036 units.
### Ten Step Analysis — Craps Pass Bet
Loses Net Win Probability Expected Win
0 1 0.492929 0.492929
1 2 0.249950 0.499900
2 3 0.126742 0.380227
3 4 0.064267 0.257069
4 5 0.032588 0.162940
5 6 0.016524 0.099147
6 7 0.008379 0.058653
7 8 0.004249 0.033990
8 9 0.002154 0.019390
9 10 0.001092 0.010924
10 -2036 0.001124 -2.288037
Total 1.000000 -0.272866
The following is my summary of the table above.
• Expected net win = -0.272866 units
• Expected total bet = 19.295522 units
• Ratio of expected win to expected total bet = -1.41%
The following table shows my analysis based on the don't pass bet in craps where the player can make up to ten bets. The worst case scenario, with ten losses, is a loss of 2036 units. For purposes of my analysis, I don't count pushes (12 on the come out roll) as a bet resolved.
### Ten Step Analysis — Craps Don't Pass Bet
Loses Net Win Probability Expected Win
0 1 0.493182 0.493182
1 2 0.249954 0.499907
2 3 0.126681 0.380043
3 4 0.064204 0.256817
4 5 0.032540 0.162699
5 6 0.016492 0.098951
6 7 0.008358 0.058508
7 8 0.004236 0.033889
8 9 0.002147 0.019323
9 10 0.001088 0.010881
10 -2036 0.001118 -2.276668
Total 1.000000 -0.262467
The following is my summary of the table above.
• Expected net win = -0.262467 units
• Expected total bet = 19.247601
• Ratio of expected win to expected total bet = -1.36%
The following table shows my analysis based on any even money bet in single-zero roulette where the player can make up to ten bets. The worst case scenario, with ten losses, is a loss of 2036 units.
### Ten Step Analysis — Single-Zero Roulette Even Money Bets
Loses Net Win Probability Expected Win
0 1 0.486486 0.486486
1 2 0.249817 0.499635
2 3 0.128285 0.384854
3 4 0.065876 0.263504
4 5 0.033828 0.169141
5 6 0.017371 0.104227
6 7 0.008920 0.062442
7 8 0.004581 0.036646
8 9 0.002352 0.021170
9 10 0.001208 0.012079
10 -2036 0.001275 -2.595951
Total 1.000000 -0.555767
The following is my summary of the table above.
• Expected net win = -0.555767 units
• Expected total bet = 20.563375 units
• Ratio of expected win to expected total bet = -2.70%
The following table shows my analysis based on any even money bet in double-zero roulette where the player can make up to ten bets. The worst case scenario, with ten losses, is a loss of 2036 units.
### Ten Step Analysis — Double-Zero Roulette Even Money Bets
Loses Net Win Probability Expected Win
0 1 0.473684 0.473684
1 2 0.249307 0.498615
2 3 0.131214 0.393643
3 4 0.069060 0.276241
4 5 0.036347 0.181737
5 6 0.019130 0.114782
6 7 0.010069 0.070480
7 8 0.005299 0.042394
8 9 0.002789 0.025102
9 10 0.001468 0.014679
10 -2036 0.001631 -3.320793
Total 1.000000 -1.229435
The following is my summary of the table above.
• Expected net win = -1.229435 units
• Expected total bet = 23.359270 units
• Ratio of expected win to expected total bet = -5.26%
The following table shows my analysis based on the Player bet in baccarat where the player can make up to ten bets. The worst case scenario, with ten losses, is a loss of 2036 units. I do not count a tie as a bet resolved.
### Ten Step Analysis — Player Bet in Baccarat
Loses Net Win Probability Expected Win
0 1 0.493175 0.493175
1 2 0.249953 0.499907
2 3 0.126683 0.380048
3 4 0.064206 0.256824
4 5 0.032541 0.162706
5 -57 0.033442 -1.906182
Total 1.000000 -0.113523
The following is my summary of the table above.
• Expected net win = -0.262740 units
• Expected total bet = 19.248861 units
• Ratio of expected win to expected total bet = -1.36%
Please note the house edge on the Player bet is 1.36% if ties are not counted as a bet resolved.
## Conclusion
My conclusion is the same as my conclusion for every betting system. The ratio of the expected win to the expected amount bet always equals how much the player can expected to lose on each individual bet. In other words, the product of -1 and the house edge. Measure this way, as how much the player can expect to lose compared to total amount bet, all betting systems are equally worthless.
That said, if you must use a betting system, please use a free one, like the Rothstein, as opposed to paying some quack/con artist for one.
## Video
Please enjoy my video on the Rothstein betting system. | 3,035 | 9,364 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.0625 | 4 | CC-MAIN-2024-26 | latest | en | 0.924209 |
https://en.wikibooks.org/wiki/Template:High_School_Mathematics_Extensions/Matrices | 1,524,818,983,000,000,000 | text/html | crawl-data/CC-MAIN-2018-17/segments/1524127095762.40/warc/CC-MAIN-20180427075937-20180427095937-00041.warc.gz | 602,701,080 | 27,572 | # Template:High School Mathematics Extensions/Matrices
## Introduction
"Matrix" may be more popularly known as a giant computer simulation, but in mathematics it is a totally different thing. To be more precise, a matrix (plural matrices) is a rectangular array of numbers. For example, below is a typical way to write a matrix, with numbers arranged in rows and columns and with round brackets around the numbers:
${\displaystyle {\begin{pmatrix}1&5&10&20\\1&-3&-5&9\\3&-1&-1&-1\\3&2&4&-5\end{pmatrix}}}$
The above matrix has 4 rows and 4 columns, so we call it a 4 × 4 (4 by 4) matrix. Also, we can have matrices of many different shapes. The shape of a matrix is the name for the dimensions of the matrix (m by n, where m is the number of rows and n the number of columns). Here are some more examples of matrices:
This is an example of a 3 × 3 matrix:
${\displaystyle {\begin{pmatrix}1&2&3\\4&5&6\\7&8&9\\\end{pmatrix}}}$
This is an example of a 5 × 4 matrix:
${\displaystyle {\begin{pmatrix}a&b&c&d\\h&g&f&e\\i&j&k&l\\p&o&n&m\\q&r&s&t\\\end{pmatrix}}}$
This is an example of a 1 × 6 matrix:
${\displaystyle {\begin{pmatrix}1&2&3&4&5&6\\\end{pmatrix}}}$
The theory of matrices is intimately connected with that of (linear) simultaneous equations. The ancient Chinese had established a systematic way to solve simultaneous equations. The theory of simultaneous equations was furthered in the east by the Japanese mathematician, Seki and a little later by Leibniz, Newton's greatest rival. Later, Gauss (1777 - 1855), one of the three giants of modern mathematics, popularised the use of Gaussian elimination, which is a simple step by step algorithm for solving any number of linear simultaneous equations. By then the use of matrices to represent simultaneous equation neatly on paper (as discussed above) had become quite common1.
Consider the simultaneous equations:
${\displaystyle x+y=10}$
${\displaystyle x-y=4}$
it has the solution x = 7 and y = 3, and the usual way to solve it is to add the two equations together to eliminate the y. Matrix theory offers us another way to solve the above simultaneous equations via matrix multiplication (covered below). We will study the widely accepted way to multiply two matrices together. In theory with matrix multiplication we can solve any number of simultaneous equations, but we shall mainly restrict our attention to 2 × 2 matrices. But even with that restriction, we have opened up doors to topics simultaneous equations could never offer us. Two such examples are
1. using matrices to solve linear recurrence relations which can be used to model population growth, and
2. encrypting messages with matrices.
We shall commence our study by learning some of the more fundamental concepts of matrices. Once we have a firm grasp of the basics, we shall move on to study the real meat of this chapter, matrix multiplication.
### Elements
An element of a matrix is a particular number inside the matrix, and it is uniquely located with a pair of numbers. E.g. let the following matrix be denoted by A, or symbolically:
${\displaystyle A={\begin{pmatrix}1&2&3\\4&5&6\\7&8&9\\\end{pmatrix}}}$
the (2,2)th entry of A is 5; the (1,1)th entry of A is 1, the (3,3) entry of A is 9 and the (3,2)th entry of A is 8. The (i , j)th entry of A is usually denoted ai,j and the (i , j)th entry of a matrix B is usually denoted by bi,j and so on.
### Summary
• A matrix is an array of numbers
• A m×n matrix has m rows and n columns
• The shape of a matrix is determined by its number of rows and columns
• The (i,j)th element of a matrix is located in ith row and jth column
### Matrix addition & Multiplication by a scalar
Matrices can be added together. But only the matrices of the same shape can be added. This is very natural. E.g.
${\displaystyle A={\begin{pmatrix}1&2&3\\4&5&6\\7&8&9\\\end{pmatrix}}}$
${\displaystyle B={\begin{pmatrix}2&9&8\\0&-1&8\\4&6&7\\\end{pmatrix}}}$
then
${\displaystyle A+B={\begin{pmatrix}1&2&3\\4&5&6\\7&8&9\\\end{pmatrix}}+{\begin{pmatrix}2&9&8\\0&-1&8\\4&6&7\\\end{pmatrix}}={\begin{pmatrix}1+2&2+9&3+8\\4+0&5+(-1)&6+8\\7+4&8+6&9+7\\\end{pmatrix}}={\begin{pmatrix}3&11&11\\4&4&14\\11&14&16\\\end{pmatrix}}}$
Similarly matrices can be multiplied by a number. We call the number a scalar to distinguish it from a matrix. The reader need not worry about the definition here, just remember that a scalar is simply a number.
${\displaystyle 5A=A+A+A+A+A=5{\begin{pmatrix}1&2&3\\4&5&6\\7&8&9\\\end{pmatrix}}={\begin{pmatrix}5&10&15\\20&25&30\\35&40&45\\\end{pmatrix}}}$
in this case the scalar value is 5. In general, when we do s × A , where s is a scalar and A a matrix, we multiply each entry of A by s.
## Matrix Multiplication
The widely accepted way to multiply two matrices together is definitely non-intuitive. As mentioned above, multiplication can help with solving simultaneous equations. We will now give a brief outline of how this can be done. Firstly, any system of linear simultaneous equations can be written as a matrix of coefficients multiplied by a matrix of unknowns equaling a matrix of results. This description may sound a little complicated, but in symbolic form it is quite clear. The previous statement simply says that if A, x and b are matrices, then Ax = b, can be used to represent some system of simultaneous equations. The beautiful thing about matrix multiplications is that some matrices can have multiplicative inverses, that is we can multiply both sides of the equation by A-1 to get x = A-1b, which effectively solves the simultaneous equations.
The reader will surely come to understand matrix multiplication better as this chapter progresses. For now we should consider the simplest case of matrix multiplication, multiplying vectors. We will see a few examples and then we will explain process of multiplication
${\displaystyle {\begin{matrix}A_{2\times 1}={\begin{pmatrix}2\\9\\\end{pmatrix}}&,&B_{1\times 2}={\begin{pmatrix}3&5\end{pmatrix}}\end{matrix}}}$
then
${\displaystyle B_{1\times 2}\times A_{2\times 1}={\begin{pmatrix}3&5\end{pmatrix}}\times {\begin{pmatrix}2\\9\\\end{pmatrix}}={\begin{pmatrix}(3\times 2)+(5\times 9)\end{pmatrix}}={\begin{pmatrix}51\end{pmatrix}}}$
Similarly if:
${\displaystyle {\begin{matrix}A_{3\times 1}={\begin{pmatrix}1\\2\\3\end{pmatrix}}&,&B_{1\times 3}={\begin{pmatrix}4&5&6\end{pmatrix}}\end{matrix}}}$
then
${\displaystyle B_{1\times 3}\times A_{3\times 1}={\begin{pmatrix}4&5&6\end{pmatrix}}\times {\begin{pmatrix}1\\2\\3\\\end{pmatrix}}={\begin{pmatrix}(4\times 1)+(5\times 2)+(6\times 3)\end{pmatrix}}={\begin{pmatrix}32\end{pmatrix}}}$
A matrix with just one row is called a row vector, similarly a matrix with just one column is called a column vector. When we multiply a row vector A, with a column vector B, we multiply the element in the first column of A by the element in the first row of B and add to that the product of the second column of A and second row of B and so on. More generally we multiply a1,i by bi,1 (where i ranges from 1 to n, the number of rows/columns) and sum up all of the products. Symbolically:
${\displaystyle A_{1\times n}\times B_{n\times 1}=(\sum _{i=1}^{n}a_{1,i}\times b_{i,1})}$ (for information on the ${\displaystyle \sum }$ sign, see Summation_Sign)
where n is the number of rows/columns.
In words: the product of a column vector and a row vector is the sum of the product of item 1,i from the row vector and i,1 from the column vector where i is from 1 to the width/height of these vectors.
Note: The product of matrices is also a matrix. The product of a row vector and column vector is a 1 by 1 matrix, not a scalar.
### Exercises
Multiply:
${\displaystyle {\begin{pmatrix}1&2\end{pmatrix}}{\begin{pmatrix}1\\2\end{pmatrix}}}$
${\displaystyle {\begin{pmatrix}1\\2\end{pmatrix}}{\begin{pmatrix}1&2\end{pmatrix}}}$
${\displaystyle {\begin{pmatrix}{\frac {1}{8}}&9\end{pmatrix}}{\begin{pmatrix}16\\2\end{pmatrix}}}$
${\displaystyle {\begin{pmatrix}a&b\end{pmatrix}}{\begin{pmatrix}d\\e\end{pmatrix}}}$
${\displaystyle {\begin{pmatrix}6+6b&3-b\end{pmatrix}}{\begin{pmatrix}0\\0\end{pmatrix}}}$
${\displaystyle {\begin{pmatrix}0&abc\end{pmatrix}}{\begin{pmatrix}a\\0\end{pmatrix}}}$
### Multiplication of non-vector matrices
Suppose ${\displaystyle A_{m\times n}B_{n\times p}=C_{m\times p}}$ where A, B and C are matrices. We multiply the ith row of A with the jth column of B as if they are vector-matrices. The resulting number is the (i,j)th element of C. Symbolically:
${\displaystyle c_{i,j}=\sum _{k=1}^{n}a_{i,k}\times b_{k,j}}$
Example 1
Evaluate AB = C and BA'= D, where
${\displaystyle A={\begin{pmatrix}3&2\\5&6\\\end{pmatrix}}}$
and
${\displaystyle B={\begin{pmatrix}2&6\\8&7\\\end{pmatrix}}}$
Solution
${\displaystyle c_{1,1}={\begin{pmatrix}3&2\end{pmatrix}}{\begin{pmatrix}2\\8\end{pmatrix}}=(3\times 2+2\times 8)=22}$
${\displaystyle c_{1,2}={\begin{pmatrix}3&2\end{pmatrix}}{\begin{pmatrix}6\\7\end{pmatrix}}=(3\times 6+2\times 7)=32}$
${\displaystyle c_{2,1}={\begin{pmatrix}5&6\end{pmatrix}}{\begin{pmatrix}2\\8\end{pmatrix}}=(5\times 2+6\times 8)=58}$
${\displaystyle c_{2,2}={\begin{pmatrix}5&6\end{pmatrix}}{\begin{pmatrix}6\\7\end{pmatrix}}=(5\times 6+6\times 7)=72}$
i.e.
${\displaystyle C={\begin{pmatrix}22&32\\58&72\end{pmatrix}}}$
${\displaystyle d_{1,1}={\begin{pmatrix}2&6\end{pmatrix}}{\begin{pmatrix}3\\5\end{pmatrix}}=(2\times 3+6\times 5)=36}$
${\displaystyle d_{1,2}={\begin{pmatrix}2&6\end{pmatrix}}{\begin{pmatrix}2\\6\end{pmatrix}}=(2\times 2+6\times 6)=40}$
${\displaystyle d_{2,1}={\begin{pmatrix}8&7\end{pmatrix}}{\begin{pmatrix}3\\5\end{pmatrix}}=(8\times 3+7\times 5)=59}$
${\displaystyle d_{2,2}={\begin{pmatrix}8&7\end{pmatrix}}{\begin{pmatrix}2\\6\end{pmatrix}}=(8\times 2+7\times 6)=58}$
i.e.
${\displaystyle D={\begin{pmatrix}36&40\\59&58\end{pmatrix}}}$
Example 2 Evaluate AB and BA where
${\displaystyle A={\begin{pmatrix}5&17\\2&7\end{pmatrix}}}$
${\displaystyle B={\begin{pmatrix}7&-17\\-2&5\end{pmatrix}}}$
Solution
${\displaystyle {\begin{pmatrix}5&17\\2&7\end{pmatrix}}{\begin{pmatrix}7&-17\\-2&5\end{pmatrix}}={\begin{pmatrix}1&0\\0&1\end{pmatrix}}}$
${\displaystyle {\begin{pmatrix}7&-17\\-2&5\end{pmatrix}}{\begin{pmatrix}5&17\\2&7\end{pmatrix}}={\begin{pmatrix}1&0\\0&1\end{pmatrix}}}$
Example 3 Evaluate AB and BA where
${\displaystyle A={\begin{pmatrix}2&6\\0&5\end{pmatrix}}}$
${\displaystyle B={\begin{pmatrix}5&-6\\0&2\end{pmatrix}}}$
Solution
${\displaystyle {\begin{pmatrix}2&6\\0&5\end{pmatrix}}{\begin{pmatrix}5&-6\\0&2\end{pmatrix}}={\begin{pmatrix}10&0\\0&10\end{pmatrix}}}$
${\displaystyle {\begin{pmatrix}5&-6\\0&2\end{pmatrix}}{\begin{pmatrix}2&6\\0&5\end{pmatrix}}={\begin{pmatrix}10&0\\0&10\end{pmatrix}}}$
Example 4 Evaluate the following multiplication:
${\displaystyle {\begin{pmatrix}a\\b\end{pmatrix}}{\begin{pmatrix}c&d\\\end{pmatrix}}}$
Solution
Note that:
${\displaystyle {\begin{pmatrix}a\\b\end{pmatrix}}}$
is a 2 by 1 matrix and
${\displaystyle {\begin{pmatrix}c&d\\\end{pmatrix}}}$
is a 1 by 2 matrix. So the multiplication makes sense and the product should be a 2 by 2 matrix.
${\displaystyle {\begin{pmatrix}a\\b\end{pmatrix}}{\begin{pmatrix}c&d\\\end{pmatrix}}={\begin{pmatrix}ac&ad\\bc&bd\\\end{pmatrix}}}$
Example 5 Evaluate the following multiplication:
${\displaystyle {\begin{pmatrix}1\\2\end{pmatrix}}{\begin{pmatrix}3&4\\\end{pmatrix}}}$
Solution
${\displaystyle {\begin{pmatrix}1\\2\end{pmatrix}}{\begin{pmatrix}3&4\\\end{pmatrix}}={\begin{pmatrix}1\times 3&1\times 4\\2\times 3&2\times 4\\\end{pmatrix}}={\begin{pmatrix}3&4\\6&8\\\end{pmatrix}}}$
Example 6 Evaluate the following multiplication:
${\displaystyle {\begin{pmatrix}a&0\\0&b\end{pmatrix}}{\begin{pmatrix}c&0\\0&d\end{pmatrix}}}$
Solution ${\displaystyle {\begin{pmatrix}a&0\\0&b\end{pmatrix}}{\begin{pmatrix}c&0\\0&d\end{pmatrix}}={\begin{pmatrix}ac&0\\0&bd\end{pmatrix}}}$
Example 7 Evaluate the following multiplication:
${\displaystyle {\begin{pmatrix}a&b\\c&d\end{pmatrix}}{\begin{pmatrix}x\\y\end{pmatrix}}}$
Solution ${\displaystyle {\begin{pmatrix}a&b\\c&d\end{pmatrix}}{\begin{pmatrix}x\\y\end{pmatrix}}={\begin{pmatrix}ax+by\\cx+dy\end{pmatrix}}}$
Note Multiplication of matrices is generally not commutative, i.e. generally ABBA.
#### Diagonal matrices
A diagonal matrix is a matrix with zero entries everywhere except possibly down the diagonal. Multiplying diagonal matrices is really convenient, as you need only to multiply the diagonal entries together.
Examples
The following are all diagonal matrices ${\displaystyle {\begin{pmatrix}a&0\\0&b\end{pmatrix}}{\begin{pmatrix}c&0\\0&d\end{pmatrix}}{\begin{pmatrix}1&0\\0&2\end{pmatrix}}{\begin{pmatrix}0&0\\0&0\end{pmatrix}}{\begin{pmatrix}a&0&0\\0&c&0\\0&0&0\end{pmatrix}}}$
Example 1 ${\displaystyle {\begin{pmatrix}a&0\\0&b\end{pmatrix}}{\begin{pmatrix}e&0\\0&f\end{pmatrix}}{\begin{pmatrix}h&0\\0&i\\\end{pmatrix}}={\begin{pmatrix}aeh&0\\0&bfi\\\end{pmatrix}}}$
Example 2 ${\displaystyle {\begin{pmatrix}a&0\\0&b\end{pmatrix}}{\begin{pmatrix}a&0\\0&b\end{pmatrix}}{\begin{pmatrix}a&0\\0&b\end{pmatrix}}={\begin{pmatrix}a^{3}&0\\0&b^{3}\end{pmatrix}}}$
The above examples show that if D is a diagonal matrix then Dk is very easy to compute, all we need to do is to take the diagonal entries to the kth power. This will be an extremely useful fact later on, when we learn how to compute the nth Fibonacci number using matrices.
### Exercises
1. State the dimensions of C
a) C = An×pBp×m
b) ${\displaystyle C={\begin{pmatrix}10^{10}&20\\5000&0\end{pmatrix}}{\begin{pmatrix}1&2&3&4\\2&5&6&6\end{pmatrix}}}$
2. Evaluate. Please note that in matrix multiplication (AB)C = A(BC) i.e. the order in which you do the multiplications does not matter (proved later).
a)
${\displaystyle {\begin{pmatrix}1&1\\0&1\\\end{pmatrix}}{\begin{pmatrix}1&1\\0&1\\\end{pmatrix}}{\begin{pmatrix}1&1\\0&1\\\end{pmatrix}}{\begin{pmatrix}1\\1\\\end{pmatrix}}}$
b)
${\displaystyle {\begin{pmatrix}3&1\\2&8\\\end{pmatrix}}{\begin{pmatrix}1&1\\0&2\\\end{pmatrix}}{\begin{pmatrix}1&1\\0&1\\\end{pmatrix}}{\begin{pmatrix}1\\1\\\end{pmatrix}}}$
3. Performing the following multiplications:
${\displaystyle C={\begin{pmatrix}1&2\\4&5\end{pmatrix}}{\begin{pmatrix}1&0\\0&1\\\end{pmatrix}}}$
${\displaystyle D={\begin{pmatrix}1&0\\0&1\end{pmatrix}}{\begin{pmatrix}1&2\\4&5\\\end{pmatrix}}}$
What do you notice?
## The Identity & multiplication laws
The exercise above showed us that the matrix:
${\displaystyle {\begin{pmatrix}1&0\\0&1\end{pmatrix}}}$
is a very special. It is called the 2 by 2 identity matrix. An identity matrix is a square matrix, whose diagonal entries are 1's and all other entries are zero. The identity matrix, I, has the following very special properties
1. ${\displaystyle A\times I=A}$
2. ${\displaystyle I\times A=A}$
for all matrices A. We don't usually specify the shape of the identity because it's obvious from the context, and in this chapter we will only deal with the 2 by 2 identity matrix. In the real number system, the number 1 satisfies: r × 1 = r = 1 × r, so it's clear that the identity matrix is analogous to "1".
Associativity, distributivity and (non)-commutativity
Matrix multiplication is a great deal different to the multiplication we know from multiplying real numbers. So it is comforting to know that many of the laws the real numbers satisfy also carries over to the matrix world. But with one big exception, in general ABBA.
Let A, B, and C be matrices. Associativity means
(AB)C = A(BC)
i.e. the order in which you multiply the matrices is unimportant, because the final result you get is the same regardless of the order which you do the multiplications.
On the other hand, distributivity means
A(B + C) = AB + AC
and
(A + B)C = AC + BC
Note: The commutative property of the real numbers (i.e. ab = ba), does not carry over to the matrix world.
Convince yourself
For all 2 by 2 matrices A, B and C. And I the identity matrix.
1. Convince yourself that in the 2 by 2 case:
A(B + C) = AB + AC
and
(A + B)C = AC + BC
2. Convince yourself that in the 2 by 2 case:
A(BC) = (AB)C
3. Convince yourself that:
${\displaystyle AB\neq BA}$
in general. When does AB = BA? Name at least one case.
Note that all of the above are true for all matrices (of any dimension/shape).
## Determinant and Inverses
We shall consider the simultaneous equations:
ax + by = α (1)
cx + dy = β (2)
where a, b, c, d, α and β are constants. We want to determine the necessary conditions for (1) and (2) to have a unique solution for x and y. We proceed:
Let (1') = (1) × c
Let (2') = (2) × a
i.e.
acx + bcy = cα (1')
acx + ady = aβ (2')
Now
let (3) = (2') - (1')
(ad - bc)y = aβ - cα (3)
Now y can be uniquely determined if and only if (ad - bc) ≠ 0. So the necessary condition for (1) and (2) to have a unique solution depends on all four of the coefficients of x and y. We call this number (ad - bc) the determinant, because it tells us whether there is a unique solution to two simultaneous equations of 2 variables. In summary
if (ad - bc) = 0 then there is no unique solution
if (ad - bc) ≠ 0 then there is a unique solution.
Note: Unique, we can not emphasise this word enough. If the determinant is zero, it doesn't necessarily mean that there is no solution to the simultaneous equations! Consider:
x + y = 2
7x + 7y = 14
the above set of equations has determinant zero, but there is obviously a solution, namely x = y = 1. In fact there are infiinitely many solutions! On the other hand consider also:
x + y = 1
x + y = 2
this set of equations has determinant zero, and there is no solution at all. So if determinant is zero then there is either no solution or infinitely many solutions.
Determinant of a matrix
We define the determinant of a 2 × 2 matrix
${\displaystyle A={\begin{pmatrix}a&b\\c&d\end{pmatrix}}}$
to be
${\displaystyle \det(A)=ad-bc\!}$
### Inverses
It is perhaps, at this stage, not very clear what the use is of the det(A). But it's intimately connected with the idea of an inverse. Consider in the real number system a number b, it has (multiplicative) inverse 1/b, i.e. b(1/b) = (1/b)b = 1. We know that 1/b does not exist when b = 0.
In the world of matrices, a matrix A may or may not have an inverse depending on the value of the determinant det(A)! How is this so? Let's suppose A (known) does have an inverse B (i.e. AB = I = BA). So we aim to find B. Let's suppose further that
${\displaystyle A={\begin{pmatrix}a&b\\c&d\end{pmatrix}}}$
and
${\displaystyle B={\begin{pmatrix}w&x\\y&z\end{pmatrix}}}$
we need to solve four simultaneous equations to get the values of w, x, y and z in terms of a, b, c, d and det(A).
aw + by = 1
cw + dy = 0
ax + bz = 0
cx + dz = 1
the reader can try to solve the above by him/herself. The required answer is
${\displaystyle B={\frac {1}{\det(A)}}{\begin{pmatrix}d&-b\\-c&a\end{pmatrix}}}$
In here we assumed that A has an inverse, but this doesn't make sense if det(A) = 0, as we can not divide by zero. So A-1 (the inverse of A) exists if and only if det(A) ≠ 0.
Summary
If AB = BA = I, then we say B is the inverse of A. We denote the inverse of A by A-1. The inverse of a 2 × 2 matrix
${\displaystyle A={\begin{pmatrix}a&b\\c&d\end{pmatrix}}}$
is
${\displaystyle A^{-1}={\frac {1}{\det(A)}}{\begin{pmatrix}d&-b\\-c&a\end{pmatrix}}}$
provided the determinant of A is not zero.
### Solving simultaneous equations
Suppose we are to solve:
ax + by = α
cx + dy = β
We let
${\displaystyle A={\begin{pmatrix}a&b\\c&d\end{pmatrix}}}$
${\displaystyle w={\begin{pmatrix}x\\y\end{pmatrix}}}$
${\displaystyle \gamma ={\begin{pmatrix}\alpha \\\beta \end{pmatrix}}}$
we can translate it into matrix form
${\displaystyle {\begin{pmatrix}a&b\\c&d\end{pmatrix}}{\begin{pmatrix}x\\y\end{pmatrix}}={\begin{pmatrix}\alpha \\\beta \end{pmatrix}}}$
i.e
${\displaystyle Aw=\gamma }$
If A's determinant is not zero, then we can pre-multiply both sides by A-1 (the inverse of A)
${\displaystyle {\begin{matrix}A^{-1}Aw&=&A^{-1}\gamma \\Iw&=&A^{-1}\gamma \\w&=&A^{-1}\gamma \\\end{matrix}}}$
i.e.
${\displaystyle {\begin{pmatrix}x\\y\end{pmatrix}}={\frac {1}{ad-bc}}{\begin{pmatrix}d&-b\\-c&a\end{pmatrix}}{\begin{pmatrix}\alpha \\\beta \end{pmatrix}}}$
which implies that x and y are unique.
Examples
Find the inverse of A, if it exists
a) ${\displaystyle A={\begin{pmatrix}1&5\\2&3\end{pmatrix}}}$
b) ${\displaystyle A={\begin{pmatrix}10&2\\2&7\end{pmatrix}}}$
c) ${\displaystyle A={\begin{pmatrix}a&b\\3a&3b\end{pmatrix}}}$
d) ${\displaystyle A={\begin{pmatrix}3&5\\5&3\end{pmatrix}}}$
Solutions
a) ${\displaystyle A^{-1}={\frac {1}{-7}}{\begin{pmatrix}3&-5\\-2&1\end{pmatrix}}}$
b) ${\displaystyle A^{-1}={\frac {1}{66}}{\begin{pmatrix}7&-2\\-2&10\end{pmatrix}}}$
c) No solution, as det(A) = 3ab - 3ab = 0
d) ${\displaystyle A^{-1}={\frac {1}{-16}}{\begin{pmatrix}3&-5\\-5&3\end{pmatrix}}}$
Exercises
1. Find the determinant of
${\displaystyle A={\begin{pmatrix}{\frac {2}{5}}&{\frac {2}{3}}\\\\{\frac {3}{2}}&{\frac {5}{2}}\end{pmatrix}}}$. Using the determinant of A, decide whether there's a unique solution to the following simultaneous equations
${\displaystyle {\begin{matrix}{\frac {2}{5}}x+{\frac {2}{3}}y=0\\{\frac {3}{2}}x+{\frac {5}{2}}y=0\end{matrix}}}$
2. Suppose
C = AB
show that
det(C) = det(A)det(B)
for the 2 × 2 case. Note: it's true for all cases.
3. Show that if you swap the rows of A to get A' , then det(A) = -det(A' )
4. Using the result of 2
a) Prove that if:
${\displaystyle A=P^{-1}BP}$
then det(A) = det(B)
b) Prove that if:
Ak = 0
for some positive integer k, then det(A) = 0.
5. a) Compute A5, i.e. multiply A by itself 5 times, where
${\displaystyle A={\begin{pmatrix}-1&6\\-1&4\\\end{pmatrix}}}$
b) Find the inverse of P where
${\displaystyle P={\begin{pmatrix}1&-2\\-1&3\\\end{pmatrix}}}$
c) Verify that
${\displaystyle A=P^{-1}{\begin{pmatrix}1&0\\0&2\\\end{pmatrix}}P}$
d) Compute A5 by using part (b) and (c).
e) Compute A100 | 7,378 | 22,037 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 94, "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} | 4.59375 | 5 | CC-MAIN-2018-17 | longest | en | 0.890077 |
https://dzintarsshop-spb.ru/science/ap-gp-hp-formulas.php | 1,624,194,473,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623487662882.61/warc/CC-MAIN-20210620114611-20210620144611-00518.warc.gz | 206,823,683 | 8,669 | # AP GP HP FORMULAS PDF
Notifications Mark All Read. A series of number is termed to be in arithmetic progression when the difference between two consecutive numbers remain the same. It also means that the next number can be obtained by adding or subtracting the constant number to the previous in the sequence. Therefore, this constant number is known as the common difference d.
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Notifications Mark All Read. A series of number is termed to be in arithmetic progression when the difference between two consecutive numbers remain the same. It also means that the next number can be obtained by adding or subtracting the constant number to the previous in the sequence.
Therefore, this constant number is known as the common difference d. For example, 3, 6, 9, 12 is an AP as the difference between two consecutive terms is three which is fixed.
If the ratio of any two successive terms is invariably similar, then the sequence is termed a geometric progression. Naturally, it can be defined as a series in which the next number in the series can be obtained by multiplying a constant to the previous number in the series.
Therefore, this fixed term is known as the common ratio. Quantities are termed as GP if they tend to increase and decrease by a common factor.
For example, 2, 4, 8, 16 can be a GP because the ratio between two consecutive numbers is same, i. Harmonic progression is the series when the reciprocal of the terms are in AP. Toggle navigation. P and H. P A series of number is termed to be in arithmetic progression when the difference between two consecutive numbers remain the same.
Formulas of Arithmetic Progressions A. Definition of Geometric Progression G. P If the ratio of any two successive terms is invariably similar, then the sequence is termed a geometric progression. Formulas of Geometric Progression G. Definition of Harmonic Progression H. P Harmonic progression is the series when the reciprocal of the terms are in AP. Formulas of Harmonic Progression H. Log in to Reply. Hozaifa Bin Imtiyaz awesome. PrepInsta Instagram. Link to.
APICECTOMIA TECNICA PDF | 531 | 2,381 | {"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-2021-25 | latest | en | 0.943488 |
https://chemistry.stackexchange.com/questions/44368/enthalpy-change-with-temperature | 1,558,855,916,000,000,000 | text/html | crawl-data/CC-MAIN-2019-22/segments/1558232258862.99/warc/CC-MAIN-20190526065059-20190526091059-00069.warc.gz | 431,257,056 | 31,624 | Enthalpy change with temperature
Recently I stumbled across a book that describes enthalpy change with temperature, specifically for a product of combustion process (e.g. CO as a result of solid carbon combustion). The book says that enthalpy of formation at $298$ K is $-110.5~\rm\frac{kJ}{mol}$, meaning that this is the amount of energy that gets released during combustion. But in a very next paragraph there are two opposite calculation for enthalpy at 2000K: one says it is $-53~\rm\frac{kJ}{mol}$ (less energy get released during combustion), and the other it is $+167~\rm\frac{kJ}{mol}$ (the energy must be brought into system)?
Can you please clarify which number is correct and does energy goes out or into the system?
• Well, perhaps you could summarize the 'two opposite calculations' to give us a hint about them and why they might be different? – Jon Custer Jan 28 '16 at 14:05
• What it says is that the enthalpy at 2000 K (value of enthalpy in the book: -53 kJ/mol or +167 kJ/mol?) is the enthalpy of formation at 298 K (value = -110 kJ/mol) + change in enthalpy from 298 K to the 2000K (value = 57 kJ/mol). My guess is that authors ment this is enthalpy for a case where CO2 had to be first produced through the combustion at referent temperature 298 K (resulting in negative enthalpy of -110 kJ/mol) and then heated to 2000 K by adding 57 kJ/mol extra energy from enviroment. Resulting -167 kJ/mol would be just an error and -53 kJ is the right value. Does it sound ok? – Marin Jan 28 '16 at 15:13 | 414 | 1,518 | {"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} | 3.21875 | 3 | CC-MAIN-2019-22 | longest | en | 0.921708 |
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The effect of concentration on potato discs
Extracts from this document...
Introduction
John Saunders - 10R 23/9/01 What is the effect of concentration on the size of potato discs? Research The main process that will take place in this experiment is osmosis. To explain osmosis more, I've conducted some research to help me decide what will happen in the experiment, and create a prediction, and also perhaps help me set up a good method to get the best results from my experiment. What osmosis is: Osmosis is the movement of water from an area of high concentration to an area of low concentration through a semi-permeable membrane. The process of osmosis: Salt Water High concentration of salt Semi-permeable High concentration of water Low concentration of water membrane Low concentration of salt This generally shows the osmosis process. The semi-permeable membrane isn't a solid wall, so molecules can get through it, but only small molecules are deceptive enough to do it. In this case, only the water can get through it, and the salt solution can't. Osmosis is basically where the cell tries to form an equilibrium between itself and the solvent or solute around it. In the case of this diagram above, the side with the lower concentration of salt solution (higher concentration with water), passes on water to the side with the higher salt solution (lower concentration with water), to dilute it. This is to try and form an equilibrium between the two sides of the membrane. ...read more.
Middle
I will do this by using exactly the same weight of potato discs, and using exactly the same amount of concentrated solutions. In this case, I'm using 7 pieces of potato, each weighing 0.6g, and 7 x 30ml of concentrated solution. If I test fairly, this should mean that I wouldn't need to check on the statistics I found on the discs as closely. I will use 7 different concentrations, ranging from 0.05, up to 0.25, so I have a good range of statistics I can look at. From this, I may be able to see if I can find a set pattern between the relationship of concentration and the size of the discs. Depending on the time we have for the experiment, I will probably spend longer setting the experiment up, as this experiment needs time to develop. This would mean that I wouldn't need to wait another 3 or 4 days finding results if I repeat the experiment, for if I did something wrong the first time in setting the experiment up. I will record the results in a table, so that they are easier to compare and contrast, and so that I can comment on my findings a lot easier. Because there are seven different results to compare, I don't want to waste valuable time looking through paragraphs for data, which could be so easily found in a table. ...read more.
Conclusion
This could have affected how much it took or gave within the salt solution concentration, so we shouldn't have left some skin on one or two of them if there was any on. I did measure all the measurements exactly, so the only other factors I can think of are how much fluid was on the discs when we took them out of the concentrations, or, how much fluid was on the weighing scales if any excess water from earlier weight measurements was on them. This is especially true because were talking about tenths of a gram with this experiment. One particular result that I thought that went wrong was the 0.15 and the 0.175 results. This is because the trend is that the weights gradually decrease as the concentrations get higher. These trends break the sequence, as they gain weight dramatically from the 0.125 weight, which in comparison, weighed just over half of the 0.175 weight, even though it is a higher concentration. I personally don't think the conclusion earlier on in the analysis of the experiment is enough to support or undermine my experiment, for the simple reason that the experiment has so many flaws as far as the wild results are concerned. Although there do seem to be some trends within the weights, these still don't follow any exact trends, and we can't say they would follow any strict pattern if range of the concentrations was greater. 7 1 ...read more.
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## 1836. Babel Fish
Time limit: 0.5 second
Memory limit: 64 MB
The Babel fish is a rare and very useful creature, which saved Arthur Dent in his travels around the Galaxy more than once. If you insert it into your ear, you'll be able to understand representatives of any race in the Galaxy as if they spoke your own language.
Upon his return to the Earth, Arthur took the fish out of his ear and decided to keep it in case his friend Ford Prefect arrived unexpectedly and dragged him off on a new travel. Arthur thought it would be best to keep the fish in the tank the intelligent dolphins had given him. Then he would be able to have the tank with him even if he went far away from home in his car.
The tank was a rectangular parallelepiped with a square base and open top. To control the amount of water in the tank, Arthur mounted a sensor for registering the water level at each of its four vertical edges.
Since the state of British country roads was far from ideal, the car tilted often and the water surface was not always parallel to the bottom of the tank, which produced unequal readings of the sensors. Arthur had to invent some method of calculating the amount of water in the tank.
### Input
The first line contains the number of tests t (1 ≤ t ≤ 104). In each of the following t lines you are given five integers: the length of the base square of the tank and the water levels at the vertical edges. All the integers are nonnegative and do not exceed 106. The edges are described in the order shown in the figure.
### Output
For each test output the volume of water in the tank if it is determined uniquely, “error” if the data are inconsistent, and “ambiguous” if the answer cannot be found uniquely. All the numbers should be given with absolute or relative error of at most 10−6.
### Sample
inputoutput
```2
10 2 3 4 3
10 2 3 3 3
```
```300.000
error
```
Problem Author: Denis Dublennykh
Problem Source: Ural Championship 2011 | 508 | 2,173 | {"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} | 2.921875 | 3 | CC-MAIN-2023-50 | latest | en | 0.959707 |
https://www.studyebooks.com/2021/09/problems-in-elementary-physics-1978-pdf.html | 1,723,444,344,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641036271.72/warc/CC-MAIN-20240812061749-20240812091749-00176.warc.gz | 754,729,624 | 47,200 | # Problems In Elementary Physics - (1978) PDF ebook by B. Bukhovtsev
## Problems In Elementary Physics
This collection of 816 problems is based on the textbook "Elementary Physics" edited by Academician G. S. Landsberg. For this reason, the content and nature of the problems and their arrangement mainly conform with this textbook.
There is no section devoted to "Atomic Physics", however, since the exercises in Landsberg's book illustrate the relevant material in sufficient detail. Some problems on this subject have been included in other chapters.
The problems, most of which are unique, require a fundamental knowledge of the basic laws of physics, and the ability to apply them in the most diverse conditions. A number of problems in the book have been revised from those used at the annual contests organized by the Physics faculty of the Moscow University. The solutions to all the difficult problems are given in great detail. Solutions are also given for some of the simpler ones. The book is recommended for self-education of senior pupils of general and special secondary and technical schools. Many problems will be useful for first and second-year students of higher schools.
#### Contents:
Contents
Chapter 1. Mechanics (Problems: 7 | Answers and Solutions: 160)
1-1. Kinematics of Uniform Rectilinear Motion
1-2. Kinematics of Non-Uniform and Uniformly Variable Rectilinear Motion
1-3. Dynamics of Rectilinear Motion
1-4. The Law of Conservation of Momentum
1-5. Statics
1-6. Work and Energy
1-7. Kinematics of Curvilinear Motion
1-8. Dynamics of Curvilinear Motion
1-9. The Law of Gravitation
1-10. Hydro and Aerostatics
1-11. Hydro and Aerodynamics
Chapter 2. Heat. Molecular Physics (Problems: 68| Answers and Solutions: 268)
2-1. Thermal Expansion of Solids and Liquids
2-2. The Law of Conservation of Energy. Thermal Conductivity
2-3. Properties of Gases
2-4. Properties of Liquids
2-5. Mutual Conversion of Liquids. and Solids
2-6. Elasticity and Strength
2-7. Properties of Vapours
Chapter 3. Electricity and Magnetism (Problems: 87| Answers and Solutions: 294)
3-1. Electrostatics
3-2. Direct Current
3-3. Electric Current in Gases and a Vacuum
3-4. Magnetic Field of a Current. The action of a Magnetic Field on a Current and Moving Charges
3-5. Electromagnetic Induction. Alternating Current
3-6. Electrical Machines
Chapter 4. Oscillations and Waves (Problems: 127| Answers and Solutions: 365)
4-1. Mechanical Oscillations
4-2. Electrical Oscillations
4-3. Waves
Chapter 5. Geometrical Optics (Problems: 135| Answers and Solutions: 380)
5-1. Photometry
5-2. Fundamental Laws of Optics
5-3. Lenses and Spherical Mirrors
5-4. Optical Systems and Devices
Chapter 6. Physical Optics (Problems: 151| Answers and Solutions: 421)
6-1. Interference of Light
6-2. Diffraction of Light
6-3. Dispersion of Light and Colours of Bodies
mirtitles
book details :
• Author::B. Bukhovtsev -V. Krivchenkov, G. Myakishev, and V. Shalnov
• Publication date: 1971 | 779 | 2,979 | {"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} | 2.609375 | 3 | CC-MAIN-2024-33 | latest | en | 0.886039 |
http://mathoverflow.net/revisions/117210/list | 1,368,968,307,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368697503739/warc/CC-MAIN-20130516094503-00079-ip-10-60-113-184.ec2.internal.warc.gz | 165,314,387 | 5,489 | 5 deleted 6 characters in body
Suppose you need to gift-wrap a cylinder (e.g., a can of tennis balls, or a large candle) of height $h$ and radius $r$. Here wrap is the natural sense of covering the surface area of the cylinder completely, without cutting the square, creasing however needed. What is the smallest square that suffices for a given $h$ and $r$? For example, a rectangle of dimensions $(h+2r) \times (2 \pi r)$ suffices for how one might wrap a can of tennis balls or a stout candlelike this:
In this $h=3$ and $r=1$ case, the rectangle has dimenions $5 \times 6.28$, and so a square of side $2 \pi$ suffices. But is that optimal?
Merry Christmas!
4 added 9 characters in body; edited title; deleted 3 characters in body; edited title
Suppose you need to gift-wrap a cylinder (e.g., a can of tennis balls, or a large candle) of height $h$ and radius $r$. Here wrap is the natural sense of covering the surface area of the cylinder completely, without cutting the square, creasing however needed. What is the smallest square that suffices for a given $h$ and $r$? For example, a rectangle of dimensions $(h+2r) \times (2 \pi r)$ suffices for how one might wrap a can of tennis balls or a stout candle like this:
In this $h=3$ and $r=1$ case, the rectangle has dimenions $5 \times 6.28$, and so a square of side $2 \pi$ suffices. But is that optimal?
Merry Christmas!
3 Rollback to Revision 1
Suppose you need to gift-wrap a cylinder (e.g., a can of tennis balls, or a large candle) of height $h$ and radius $r$. Here wrap is the natural sense of covering the surface area of the cylinder completely, without cutting the square, creasing however needed. What is the smallest rectangle square that suffices for a given $h$ and $r$? For example, a rectangle of dimensions $(h+2r) \times (2 \pi r)$ suffices for how one might wrap a can of tennis balls or a stout candlelike this:
In this $h=3$ and $r=1$ case, the rectangle has dimenions $5 \times 6.28$, and so a square of side $2 \pi$ suffices. But is that optimal?
Merry Christmas!
2 added 9 characters in body
1 | 564 | 2,086 | {"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-2013-20 | latest | en | 0.859484 |
https://jeesunkim.com/maths/techniques/differentiation_techniques/ | 1,721,522,866,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763517544.48/warc/CC-MAIN-20240720235600-20240721025600-00125.warc.gz | 275,827,363 | 33,563 | # Differentiation Techniques
### 1. Intuitive Limits
The limit of a function as $x \to 0$, this function can be replaced so that calculating the limit is much simpler. The following replacements can be considered.
• $\sin x \implies x$,
• $\sin kx \implies kx$, $\; \sin^n x \implies x^n$
• $\tan x \implies x$,
• $\tan kx \implies kx$, $\; \tan^n x \implies x^n$
• $\sin^{-1} x \implies x$,
• $\sin^{-1} kx \implies kx$, $\; (\sin^{-1} kx)^n \implies (kx)^n$
• $\tan^{-1} x \implies x$,
• $\tan^{-1} kx \implies kx$, $\; (\tan^{-1} kx)^n \implies (kx)^n$
• $x^3 + x \implies x$.
• $x^5 - x^3 + x \implies x$, so in this case addition/subtraction does not matter.
When $x$ is large enough, the following comparison is also useful. \begin{aligned} \vert \sin x \vert < \ln x < x^n < 2^x < x! < x^x \end{aligned}
### 2. Limit of Exponential Functions
There are a few ways to find the limits of the form of $0^0$, $\infty^0$, or $1^\infty$.
• Given $y = f(x)^{g(x)}$, taking a logarithm of both sides of the function may be helpful. In other words, the limit of $y$ can be found after calculating the limit of $\ln y = g(x) \ln f(x)$.
• The property $a^b = e^{b \ln a}$ may be used.
\begin{aligned} \begin{cases} \; \displaystyle \lim_{x \to a} f(x)^{g(x)} = \displaystyle \lim_{x \to a} e^{g(x) \ln {f(x)}} \\\\ \; \displaystyle \lim_{x \to a} f(x)^{\frac{1}{g(x)}} = \displaystyle \lim_{x \to a} e^{\frac{\ln {f(x)}}{g(x)}} \end{cases} \end{aligned}
• Especially, in the case of $1^\infty$, the property $a^b = e^{b (a-1)}$ may be used as well. Assuming that $f(a) = 1$ and $g(a) = 0$,
\begin{aligned} \lim_{x \to a} \cfrac{\ln {f(x)}}{g(x)} = \lim_{x \to a} \cfrac{\cfrac{f'(x)}{f(x)}}{g'(x)} &= \lim_{x \to a} \cfrac{f'(x)}{g'(x)} = \lim_{x \to a} \cfrac{\cfrac{f(x) - f(a)}{x - a}}{\cfrac{g(x) - g(a)}{x - a}} = \lim_{x \to a} \cfrac{f(x) - 1}{g(x)} \\\\ \implies \lim_{x \to a} f(x)^{\frac{1}{g(x)}} &= \lim_{x \to a} e^{\frac{\ln {f(x)}}{g(x)}} = \lim_{x \to a} e^{\frac{f(x) - 1}{g(x)}} \end{aligned}
• The following results are also useful.
\begin{aligned} \begin{cases} \; \displaystyle \lim_{x \to 0} (kx)^{hx} = 1 \\\\ \; \displaystyle \lim_{x \to \infty} (kx)^{\frac{h}{x}} = 1 \end{cases}, \qquad \begin{cases} \; \displaystyle \lim_{x \to 0} \left(\cfrac{k}{x}\right)^{hx} = 1 \\\\ \; \displaystyle \lim_{x \to \infty} \left(\cfrac{k}{x}\right)^{\frac{h}{x}} = 1 \end{cases} \end{aligned}
### 3. Inverse Trigonometric Functions
Derivatives of inverse trigonometric functions are as follows.
• $(\sin^{-1} x)' = \cfrac{1}{\sqrt{1 - x^2}} \quad \left( -\cfrac{\pi}{2} \leq \sin^{-1} x \leq \cfrac{\pi}{2} \right) \;$,
• $(\cos^{-1} x)' = \cfrac{-1}{\sqrt{1 - x^2}} \quad \left( 0 \leq \cos^{-1} x \leq \pi \right) \;$,
• $(\tan^{-1} x)' = \cfrac{1}{1 x^2} \quad \left( -\cfrac{\pi}{2} < \sin^{-1} x < \cfrac{\pi}{2} \right) \;$,
• $(\cot^{-1} x)' = -\cfrac{1}{1 + x^2} \quad \left( 0 < \cot^{-1} x < \pi \right) \;$,
• $(\sec^{-1} x)' = \cfrac{1}{\vert x \vert \sqrt{x^2 - 1}} \quad \left( 0 \leq \sec^{-1} x \leq \pi, \; \sec^{-1} x \ne \cfrac{\pi}{2} \right) \;$,
• $(\csc^{-1} x)' = -\cfrac{1}{\vert x \vert \sqrt{x^2 - 1}} \quad \left( -\cfrac{\pi}{2} \leq \csc^{-1} x \leq \cfrac{\pi}{2}, \; \csc^{-1} x \ne 0 \right) \;$.
In addition, the following are important identities of inverse trigonometric functions.
• $\sin^{-1}x + \cos^{-1}x = \cfrac{\pi}{2} \;$,
• $\sin^{-1}(-x) = -\sin^{-1}x \;$,
• $\cos^{-1}(-x) = \pi - \cos^{-1}x \;$,
• $\csc^{-1} x = \sin^{-1}\left( \cfrac{1}{x} \right) \;$.
### 4. Derivative of an Absolute Value Function
Usually, although this can be done by range, the following formula is noteworthy. \begin{aligned} (\vert f(x) \vert)' = \cfrac{f(x) f(x)'}{\vert f(x) \vert} \end{aligned}
### 5. Derivative & Errors
By definition, the derivative of $f(x)$ is as follows. \begin{aligned} \lim_{\Delta x \to 0} \cfrac{\Delta y}{\Delta x} = f'(x) \end{aligned}
Since the average rate of change is not the same as the instantaneous rate of change in the view of the limit, the difference can be set as follows. \begin{aligned} \cfrac{\Delta y}{\Delta x} = f'(x) + \epsilon \end{aligned}
Note that $\epsilon$ is smaller when $\Delta x$ gets smaller. It implies that the product of them is an infinitesimal number and it can be ignored. Therefore, the approximation of $\Delta y$ is $f'(x) \Delta x$. The limit of this approximation is called the derivative of $y$ and is written as \begin{aligned} dy = df(x) = f'(x) dx \end{aligned}
Another point of view of $dy$ is that is is the error of $y$. Moreover, $dy/y$ is called the relative error. Meanwhile, this concept leads to the approximation formula about $f(x + \Delta x)$. \begin{aligned} f(x + \Delta x) \approx f(x) + f'(x) \Delta x \end{aligned}
### 6. Asymptotes
Asymptotes make it easier to draw a graph of a function. Vertical asymptotes can be found when a function goes to infinity as $x$ approaches a value. Also, horizontal asymptotes can be found when a function goes to a value as $x$ approaches $\pm \infty$. Other than these, there exist oblique asymptotes, which can be found in the following steps.
• Divide both sides of the function by the highest order term and find a slope $m$ of the asymptote by evaluating the limits at infinity.
• Set $y = mx + n$ from the found $m$ and substitute this equation to the function. Again, divide both sides by the highest order term and find $n$ by evaluating the limits at infinity.
For example, the oblique asymptotes of $y^3 = (x - 1)(x - 3)^2$ can be found as follows. \begin{aligned} &\lim_{x \to \infty} \left( \cfrac{y}{x} \right)^3 = \lim_{x \to \infty} \cfrac{(x - 1)(x - 3)^2}{x^3} = 1 \\\\ &\implies y = x + n \\ &\implies (x + n)^3 = (x - 1)(x - 3)^2 = x^3 - 7x^2 + 15x - 9 \\ &\implies 3nx^2 + 3n^2x + n^3 = -7x^2 + 15x - 9 \\ &\implies \lim_{x \to \infty} \cfrac{3nx^2 + 3n^2x + n^3}{x^2} = \lim_{x \to \infty} \cfrac{-7x^2 + 15x - 9}{x^2} \\ &\implies 3n = -7 \\ &\implies y = x - \frac{7}{3} \end{aligned}
### 7. Curvature
Let $y = f(x)$ be a twice-differentiable curve on a plane. Now, consider $P(x, y)$ and $Q(x + \Delta x, y + \Delta y)$. When $\alpha$ is the angle between the tangent line on $P$ and $x$-axis and $\alpha + \Delta \alpha$ is the angle between the tangent line on $Q$ and $x$-axis, the tangent line is rotated by $\Delta \alpha$ moving the arc length $\Delta s$ on the curve. The mean curvature denotes the ratio of $\Delta \alpha$ to $\Delta s$. Based on this concept, curvature $K$ of $f(x)$ is computed as follows. \begin{aligned} K = \cfrac{\vert y'' \vert}{(1 + (y')^2)^{\frac{3}{2}}} \end{aligned}
If this curve is a parametric function, $K$ is \begin{aligned} \begin{cases} \; x = f(t) \\ \; y = g(t) \end{cases} \implies K = \cfrac{\vert x' \cdot y'' - x'' \cdot y' \vert}{((x')^2 + (y')^2)^{\frac{3}{2}}} \end{aligned}
Meanwhile, the radius of curvature $R$ can be calculated from $K$. \begin{aligned} K = \cfrac{1}{R} \end{aligned}
Furthermore, the center of curvature $(X, Y)$ on $P(x, y)$ can be calculated as follows. \begin{aligned} X = x - \cfrac{y' (1 + (y')^2)}{y''}, \quad Y = y + \cfrac{1 + (y')^2}{y''} \end{aligned}
The gradient has different definitions depending on whether the function is explicit or implicit. \begin{aligned} \begin{cases} \; \text{Explicit Function } z = f(x, y) \implies \nabla f = \cfrac{\partial f}{\partial x} i + \cfrac{\partial f}{\partial y} j \\\\ \; \text{Implicit Function } f(x, y, z) = 0 \implies \nabla f = \cfrac{\partial f}{\partial x} i + \cfrac{\partial f}{\partial y} j + \cfrac{\partial f}{\partial z} k \end{cases} \end{aligned}
The gradient of an explicit function means the tangent vector at the point, while the gradient of an implicit function means the normal vector at the point. The details about this can be found in this note. Therefore, the normal vector from an explicit function can be obtained after reforming to an implicit function as follows. \begin{aligned} z - f(x, y) = 0 \implies \left( -\cfrac{\partial f}{\partial x}, \; -\cfrac{\partial f}{\partial y}, \; 1 \right) \end{aligned}
### 9. Directional Derivative
The directional derivative of a multivariable differentiable function $f$ along a given vector $u$ at a given point $x$ represents the slope of the tangent at $x$ with the direction $u$. If $u$ is parallel with the coordinate axes, the directional derivative is the partial derivative. Accordingly, the directional derivative can be viewed as an orthogonal projection of the partial derivative on $u$. Consider $P(a, b)$ on the function $f(x, y)$. The directional derivative along the normalized direction $u$ at the point $P$ is \begin{aligned} D_u f(P) &= \lim_{h \to 0} \cfrac{f(a + hu_x, b + hu_y) - f(a, b)}{h} = \nabla f(P) \cdot u \\\\ &= \left( \cfrac{\partial f}{\partial x} i + \cfrac{\partial f}{\partial y} j \right) \cdot (u_x i + u_y j) = \cfrac{\partial f}{\partial x}u_x + \cfrac{\partial f}{\partial y}u_y \end{aligned}
Therefore, the maximum of $D_u f(P)$ is $\Vert \nabla f(P) \Vert$ and the minimum is $-\Vert \nabla f(P) \Vert$.
### 10. Derivative of an Implicit Function
The general formula for the derivative of an implicit function is as follows. \begin{aligned} \begin{cases} \; f(x, y) = 0 \implies \cfrac{dy}{dx} = -\cfrac{f_x}{f_y} \\\\ \; f(x, y, z) = 0 \implies \cfrac{\partial z}{\partial x} = -\cfrac{f_x}{f_z}, \quad \cfrac{\partial z}{\partial y} = -\cfrac{f_y}{f_z} \end{cases} \end{aligned}
### 11. Bivariate Extreme Value
To find extreme values of a function $f(x, y)$, the point $(x, y)$ such that $f_x = 0$ and $f_y = 0$ should be calculated. Let that point be $(a, b)$. If $D$ is defined as follows, $(a, b)$ is the extreme value when $D>0$. \begin{aligned} D &= f_{xx}(a, b) f_{yy}(a, b) - (f_{xy}(a, b))^2 \\\\ D > 0 &\implies \begin{cases} \; f_{xx}(a, b) > 0 & \text{local minimum} \\\\ \; f_{xx}(a, b) < 0 & \text{local maximum} \end{cases} \end{aligned}
If $D<0$, there are no extreme values. Besides, if $D = 0$, it is inconclusive.
### 12. Extrema of an Implicit Function
The extrema of $y$ of an implicit function $f(x, y) = 0$ can be calculated from $f_x = 0$ and $f(x, y) = 0$. Let such point be $(a, b)$. If $f_y \ne 0$, \begin{aligned} \begin{cases} \; -\cfrac{f_{xx}(a, b)}{f_y(a,b)} > 0 & \text{local minimum } b \\\\ \; -\cfrac{f_{xx}(a, b)}{f_y(a,b)} < 0 & \text{local maximum } b \end{cases} \end{aligned}
It is inconclusive if $f_{xx}(a, b) = 0$.
### 13. Lagrange Multiplier
When finding the maximum and minimum of $f(x)$, this problem becomes often easy if an unknown variable can be represented using other unknown variables under constraints. Even so, however, the problem can sometimes get complex, and A is the proper method for this case. First, if a constraint $g(x, y, z) = 0$ is the only one, \begin{aligned} \begin{cases} \; \nabla f = \lambda \nabla g \\\\ \; g(x, y, z) = 0 \end{cases} \end{aligned}
the maximum or minimum can be found by calculating $\lambda$ and $(x_0, y_0, z_0)$ satisfying the above conditions. Here $\lambda$ is called the Lagrange multiplier. If another constraint $h(x, y, z) = 0$ is added, \begin{aligned} \begin{cases} \; \nabla f = \lambda \nabla g + \mu \nabla h \\\\ \; g(x, y, z) = 0 \\\\ \; h(x, y, z) = 0 \end{cases} \end{aligned}
the maximum or minimum can be found by calculating $\lambda$, $\mu$, and $(x_0, y_0, z_0)$ satisfying the above conditions. For example, the intersection of a cylinder $y^2 + z^2 = 4$ and a plane $x + y + z = 1$ forms an ellipse. The Lagrange multiplier makes it simple to find the maximum and minimum of $f = x + 2y$ for a point on this ellipse as follows. \begin{aligned} f &= x + 2y, \quad \begin{cases} \; g = x + y + z - 1 = 0 \\ \; h = y^2 + z^2 - 4 = 0 \end{cases} \\\\ &\implies \nabla f = (1, 2, 0), \quad \begin{cases} \; \nabla g = (1, 1, 1) \\ \; \nabla h = (0, 2y, 2z) \end{cases} \\\\ &\implies \nabla f = \lambda \nabla g + \mu \nabla h \iff (1, 2, 0) = \lambda(1, 1, 1) + \mu(0, 2y, 2z) \\ &\implies \lambda = 1, \quad \lambda + 2y\mu = 2, \quad \lambda + 2z\mu = 0 \\ &\implies \lambda = 1, \quad y = \cfrac{1}{2\mu}, \quad z = -\cfrac{1}{2\mu} \\ &\implies h = y^2 + z^2 - 4 = \cfrac{1}{4\mu^2} + \cfrac{1}{4\mu^2} - 4 = 0 \\ &\implies \mu = \pm \cfrac{1}{2\sqrt{2}}, \quad y = \pm 2\sqrt{2}, \quad z = \mp 2\sqrt{2} \\ &\implies g = x + y + z - 1 = x - 1 = 0 \\ &\implies x = 1 \end{aligned}
Therefore, the maximum is $f(1, \sqrt{2}, -\sqrt{2}) = 1 + 2\sqrt{2}$, and the minimum is $f(1, -\sqrt{2}, \sqrt{2}) = 1 - 2\sqrt{2}$. | 4,642 | 12,503 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 143, "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} | 4.75 | 5 | CC-MAIN-2024-30 | latest | en | 0.576057 |
http://www.ask.com/web?q=Amount+of+Money+Subtracted+from+the+Sales+Price&o=2603&l=dir&qsrc=3139&gc=1 | 1,484,845,569,000,000,000 | text/html | crawl-data/CC-MAIN-2017-04/segments/1484560280723.5/warc/CC-MAIN-20170116095120-00436-ip-10-171-10-70.ec2.internal.warc.gz | 343,372,024 | 15,612 | Web Results
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Deflation. Inflation means to increase the amount of money in circulation, deflation means what you have up on top. Hope this helps! I'm assuming you are doing ...
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www.trulia.com/voices/Home_Selling/Roughly_as_a_seller_what_percentage_of_your_sale-39962
Jun 6, 2008 ... Julie's response below is right on the money. Take whatever your sales price is and times it by 9% and that's your TOTAL figure for your costs.
## How do I subtract a percentage from a price? | Reference.com
www.reference.com/math/subtract-percentage-price-a345114148a1842d
The answer is the amount to subtract from the original price. ... Subtracting 25 percent of the price is the same as subtracting \$5 from the price. ... How do you calculate discount and sale prices? ... Subtract Percentages from Money · Learn How to Add or Subtract Percentages · Formula to Subtract Percentage · Subt...
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The amount of money that is applied toward the rental credit is subtracted along with the option consideration from the total sales price of the home at the time of ...
## Personal Finance– Glossary | LifeSmarts
lifesmarts.org/lifesmarts-u/personal-finance/personal-finance-glossary/
Emergency Fund–An amount of money set aside to cover bills in case of emergency ... of a person or company, which can be calculated by subtracting liabilities from total assets ... POS Terminal– A point-of-sale terminal serves as a computerized cash ... Impulse purchase– Making a purchase without comparing prices and ...
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## Closing Costs Should Be Subtracted to Determine Adjusted Sale Price
articles.latimes.com/1990-12-23/realestate/re-9532_1_closing-costs
Dec 23, 1990 ... In addition to the realtor's sales commission of 6% of the sales price, we had to pay ... From this amount you subtract your adjusted cost basis, usually the ... a partial interest in the house to a man who loaned her some money.
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www.wyzant.com/resources/lessons/math/elementary_math/money/money_calculating_discounts
Now, please realize that \$5.00 is the amount that will be discounted; it is not the amount you will pay for the item. In order to figure out ... This would be the price you would pay for the sale item. There is ... Prev (Adding and Subtracting Money). | 726 | 3,133 | {"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} | 2.640625 | 3 | CC-MAIN-2017-04 | latest | en | 0.897509 |
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### Sven-Hendrik Loßin, s#6, Schach-Aktiv 10/2010
Although the judge did not mention it, I still like this problem of mine (P1191126):
(= 10+9 )
Solution:
1.Rf4-a4 (thr. 2.Rg3xf3) 1...Sg7-f5 2.Rg3-f3 Sf5-e3 3.Pf2-e3 (thr. 4.e3-e4) 3...Qf8-h6 4.Bh4-g5 (thr. 5.e3-e4) 4...Qh6-b6 (4...Qh6-g5 5.Qc8-f5 Qg5-f5 6.Pe3-e4 Bh1-f3/Qf5-f3#) 5.Qc8-c5 (thr. 6.Pe3-e4 Bh1-f3#) 5...Qb6-c5 6.Bb3-c2 Qc5-c2#
White wants to force Bh1-f3# but there is still a lot of work to be done. White forces the black queen first to interfere on h6-e3 and afterwards to go to either g5 or c5 which is decisive. Tries are for example 1.Rc4?, 1.Rg5?, 1.Rgg4? and 1.Rfg4?. The problem with 1.Rc4? is that after 4...Qh6-b6 the c-File is closed, while for example 1.Rgg4? has the following defect: After 1...Sg7-f5 2.Rf4-f3 Sf5-e3 3.Pf2-e3 there is no threat 4.Pe3-e4 because the black king has a flight square on d4.
Perhaps there should be an additional wPe5 so that there is another try with 1.Rgg4? Sg7-f5 2.Rf4-f5 Qf8-h6 3.Bh4-g5? Qh6-g5! 4.Qc8-f5?? (without wPe5 3...Qh6-b6 also works in this line). Just an idea.
This piece is about "gestaffelte Lenkung" (dont know the English word for that). I hope that some of you feel the same enjoyment for it like I did and still do. The only problem that I have with it is the existence of the a-file because the bPa5 tells too much about the solution. 7x8-board would be better. | 650 | 1,677 | {"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} | 2.84375 | 3 | CC-MAIN-2020-10 | latest | en | 0.851706 |
https://math.stackexchange.com/questions/4155600/poisson-random-measure-and-alpha-stable-processes | 1,653,360,997,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662562410.53/warc/CC-MAIN-20220524014636-20220524044636-00038.warc.gz | 443,945,692 | 66,147 | # Poisson random measure and $\alpha$-stable processes
Let $$N$$ be a Poisson random measure in $$(0,\infty)^2$$ with intensity $$\eta$$ given by $$\eta(ds, dx) = \mathbb{I}_{\{x>0\}} \dfrac{C}{x^{\alpha + 1}} ds \, dx.$$
• Find the values $$\alpha$$ for which $$\eta$$ is a $$\sigma$$-finite measure.
• Let $$X_t = N(f_t)$$ with $$f_t(s,x) = \mathbb{I}_{\{s\leq t\}}\cdot x.$$
Find the values $$\alpha$$ for which $$X_t<\infty$$ for all $$t\geq 0$$ a-s.
• Compute $$\mathbb{E}\left[e^{-\lambda X_t}\right]$$
• Prove $$X_t \overset{d}{=} t^{1/\alpha}X_1$$.
Here is the problem, I have read a bit about stable Levy $$\alpha$$-processes and I can realize that the $$\eta$$ measure is related to a $$\alpha$$-stable process, which means that $$\alpha$$ must be between 0 and 2 for the $$\eta$$ measure to be $$\sigma$$-finite.
My problem with the first question is that I don't know how to prove that alpha must take those values mentioned. I understand that for a measure to be sigma-finite we must find a succession of countable sets with finite measure covering all $$S = (0,\infty)^2$$ and that is easy considering the intervals $$A_n = \left(\dfrac{1}{n}, n\right)$$. So my reasoning (which I think is not correct) is the following
$$\eta\left(A_n\times A_n\right) = \int_{1/n}^{n} \int_{1/n}^{n}\mathbb{I}_{\{x>0\}} \dfrac{C}{x^{\alpha + 1}} \, ds \, dx$$
but here we can conclude that this measure is finite for any value $$\alpha \in \mathbb{R}$$, which I believe is incorrect to assume from what I have already said about $$\alpha -$$stable processes.
For questions 2 and 3 I am using Campbell's theorem: in question two I had no problem finding that $$\alpha \in (1,2]$$ (this is assuming I had concluded that $$\alpha \in (0,2]$$ in question 1).
For question 3, again using the result of Campbell's theorem and assuming $$\alpha \in (1,2]$$, I have the following calculation:
\begin{align*} \mathbb{E}\left[e^{-\lambda X_t}\right] &= \exp\{-\int_S (1-e^{-\lambda f_t(s,x)})\ \eta\left(ds,dx\right)\}\\[0.2cm] &= \exp\{-\int_0^\infty \int_0^\infty \left(1-e^{-\lambda f_t(s,x)}\right) \mathbb{I}_{\{x>0\}} \dfrac{C}{x^{\alpha+1}}\ ds\,dx\}\\[0.2cm] &= \exp\{-\int_0^\infty \int_0^\infty \left(1-e^{-\lambda \mathbb{I}_{\{s\leq t\}}x}\right) \mathbb{I}_{\{x>0\}} \dfrac{C}{x^{\alpha+1}}\ ds\,dx\}\\[0.2cm] &= \exp\{-\int_0^\infty \int_0^\infty \left(1-e^{-\lambda \mathbb{I}_{\{s\leq t\}} x}\right) \dfrac{C}{x^{\alpha+1}}\ ds\,dx\}\\[0.2cm] &= \exp\{-\int_0^\infty \dfrac{C}{x^{\alpha+1}}\left[ \int_0^\infty \left(1-e^{-\lambda \mathbb{I}_{\{s\leq t\}}x}\right)\ ds\right]\,dx\}\\[0.2cm] &= \exp\{-\int_0^\infty \dfrac{C}{x^{\alpha+1}}\left[ \int_0^t \left(1-e^{-\lambda x}\right)\ ds\right]\,dx\}\\[0.2cm] &= \exp\{-\int_0^\infty \dfrac{C}{x^{\alpha+1}}\left[ \left(1-e^{-\lambda x}\right) \int_0^t \ ds\right]\,dx\}\\[0.2cm] &= \exp\{-\int_0^\infty \dfrac{C}{x^{\alpha+1}} \left(1-e^{-\lambda x}\right) \cdot t\ dx\}\\[0.2cm] &= \exp\{-C\cdot t \int_0^\infty \left(1-e^{-\lambda x} \right) \dfrac{1}{x^{\alpha+1}} \, dx\} \end{align*}
However, once that point is reached, I am stuck, since I don't know how to calculate that remaining integral, I am almost sure that the development up to that point is correct, but I would appreciate some comment in case something is wrong or some alternative how to conclude the calculation.
Finally, for question 4, I think it is enough to find the expectation of the Laplace transform of $$t^{1/\alpha} X_1$$, and see that it matches the previous calculation, but again I run into a problem when developing the calculation
\begin{align*} \mathbb{E}\left[e^{-\lambda t^{1/\alpha} X_1}\right] &= \exp\{-\int_S \left(1-e^{-\lambda t^{1/\alpha} f_1(s,x)}\right)\ \eta\left(ds,dx\right)\}\\[0.2cm] &= \exp\{-\int_{0}^{\infty} \int_{0}^{\infty} \left(1-e^{-\lambda t^{1/\alpha} \mathbb{I}_{\{s\leq 1\}}x}\right)\ \mathbb{I}_{\{x>0\}} \dfrac{C}{x^{\alpha + 1}}\ ds \, dx\}\\[0.2cm] &= \exp\{-\int_0^\infty \int_0^1 \left(1-e^{-\lambda t^{1/\alpha} x}\right)\ \dfrac{C}{x^{\alpha + 1}}\ ds \, dx\}\\[0.2cm] &= \exp\{-\int_0^\infty \dfrac{C}{x^{\alpha + 1}} \left[\int_0^1 \left(1-e^{-\lambda t^{1/\alpha} x}\right)\ \ ds\right] dx\} \end{align*}
Again, I would appreciate any comments or observations that may help me to proceed.
1. For $$\alpha>0$$, the integral, $$\int_0^\infty(1-e^{-\lambda x}){dx\over x^{1+\alpha}}$$ converges if and only if $$0<\alpha<1$$, because the integrand is $$\sim \lambda x^{-\alpha}$$ for $$x\to 0+$$. This is precisely the range of $$\alpha$$s for which $$X_t<\infty$$ a.s.
For $$\alpha\in(0,1)$$, the change of variables $$u=\lambda x$$ results in $$\int_0^\infty(1-e^{-\lambda x}){dx\over x^{1+\alpha}}=\lambda^\alpha\int_0^\infty(1-e^{-u}){du\over u^{1+\alpha}}.$$ Now write $$1-e^{-u} =\int_0^u e^{-t} dt$$ and use Fubini: \eqalign{ \int_0^\infty(1-e^{-u}){du\over u^{1+\alpha}} &=\int_0^\infty\int_0^t e^{-t}dt {du\over u^{1+\alpha}} \cr &=\int_0^\infty\int_t^\infty {du\over u^{1+\alpha}} e^{-t} dt\cr &=\int_0^\infty t^{-\alpha}e^{-t} dt\cr &=\int_0^\infty t^{(1-\alpha)-1}e^{-t} dt\cr &=\Gamma(1-\alpha).\cr }
2. The measure $$\eta$$ is $$\sigma$$-finite for each $$\alpha>0$$. Indeed $$\eta\left( (0,n)\times (1/n,n)\right)=\lambda n\cdot{C\over\alpha}(n^\alpha-n^{-\alpha})<\infty$$ for each positive integer $$n$$.
• Thank you very much for your help! I was able to realize some mistakes in question 2 to find that $\alpha \in (0,1)$ and thanks to the development of the integral I was able to complete the whole exercise. May 30, 2021 at 23:32 | 2,115 | 5,547 | {"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": 48, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5625 | 4 | CC-MAIN-2022-21 | latest | en | 0.867197 |
https://community.wolfram.com/groups/-/m/t/2340643 | 1,631,939,727,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780056297.61/warc/CC-MAIN-20210918032926-20210918062926-00659.warc.gz | 227,079,642 | 24,669 | # [GIF] Perpetual 2 (Parallelograms with vertices traversing curves)
Posted 1 month ago
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Inspired by the neat post from @Clayton Shonkwiler, I thought it would be cool to generalize his approach. I started by defining three families of curves: (* A typical elliptic orbit. *) ellipticOrbit[t_,a_:1,b_:1]:={a Cos[t],b Sin[t]}; (* The circle of radius b rolls on the outside of the circle of radius a. Setting a=b results in a cardioid, setting a=2b results in a nephroid. *) epicycloidOrbit[t_,a_:1,b_:1]:={(a+b)Cos[t]-b Cos[(a/b+1)t],(a+b)Sin[t]-b Sin[(a/b+1)t]}; (* The circle of radius b rolls on the outside of the circle of radius a. The point P is at a distance c from the center of the circle of radius b. *) epitrochoidOrbit[t_,a_:1,b_:.25,c_:.5]:={(a+b)Cos[t]-c Cos[(a/b+1)t],(a+b)Sin[t]-c Sin[(a/b+1)t]}; This next bit of code lays out k copies of a curve f around a circle of radius r: compositeOrbits[t_,k_:4,f_:ellipticOrbit,r_:2,curvePars__]:=Table[ r ReIm[Exp[2 I i Pi/k]]+f[t-(2 i Pi/k),##]&@@{curvePars}, {i,0,k-1} ]; To generate an animation in your notebook evaluate: animatedCurves[n_:12,k_:4,f_:ellipticOrbit,r_:1,curvePars___]:=Animate[ Graphics[{ Thickness[.005], Table[ {FaceForm[{RGBColor["#E5F6C6"],Opacity[.2]}], EdgeForm[{RGBColor["#E5F6C6"],Thickness[.002]}], Polygon[compositeOrbits[t+i,k,f,r,curvePars]], RGBColor["#E5F6C6"], Point/@compositeOrbits[t+i,k,f,r,curvePars]}, {i,0,2Pi(1-1/n),2Pi/n} ]}, Background->RGBColor["#5D414D"], PlotRange->If[ Length@{curvePars}==0, {{-4r,4r},{-4r,4r}}, {{-1.5(r+Total[{curvePars}]),1.5(r+Total[{curvePars}])},{-1.5(r+Total[{curvePars}]),1.5(r+Total[{curvePars}])}} ], AspectRatio->Automatic ], {t,0,2Pi}, AnimationRate->Pi/30 ]; The first argument determines the number of polygons, the second argument determines the number of curves, the third argument sets the type of curve, the fourth determines their distance from the origin, and the remaining arguments are curve parameters.To export this as a GIF use the following function. After the first argument the argument structure is the same as before. Remember to set your directory before exporting to make it easier to locate your GIF: gifAnimatedCurves[name_String,n_:12,k_:4,f_:ellipticOrbit,r_:1,curvePars___]:=Export[ name<>".gif", Table[ Graphics[{ Thickness[.005], Table[ {FaceForm[{RGBColor["#E5F6C6"],Opacity[.2]}], EdgeForm[{RGBColor["#E5F6C6"],Thickness[.002]}], Polygon[compositeOrbits[t+i,k,f,r,curvePars]], RGBColor["#E5F6C6"], Point/@compositeOrbits[t+i,k,f,r,curvePars]}, {i,0,2Pi(1-1/n),2Pi/n} ]}, Background->RGBColor["#5D414D"], PlotRange->If[ Length@{curvePars}==0, {{-4r,4r},{-4r,4r}}, {{-1.5(r+Total[{curvePars}]),1.5(r+Total[{curvePars}])},{-1.5(r+Total[{curvePars}]),1.5(r+Total[{curvePars}])}} ], AspectRatio->Automatic ], {t,0,4Pi/n,.05} ] ]; For example, in the GIF at the top of this post each vertex of a triangle traverses a different cardioid. The animation was generated by: gifAnimatedCurves["animation", 12, 3, epicycloidOrbit, 1.5, .5, 0.5] With this approach any 2D parametric curve, with any number of parameters, can be used to generate animations of this type.
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A neat side effect of the GIF having a period of 2Pi/n is that the size of the GIF file will be more or less proportional to 1/n, at least for small values of n. This means you can generate quite "complex" animations with a tiny memory footprint. | 1,130 | 3,448 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.96875 | 3 | CC-MAIN-2021-39 | longest | en | 0.62162 |
https://www.gradesaver.com/textbooks/math/algebra/intermediate-algebra-for-college-students-7th-edition/chapter-1-review-exercises-page-98/39 | 1,680,201,483,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296949355.52/warc/CC-MAIN-20230330163823-20230330193823-00425.warc.gz | 870,224,309 | 12,972 | Intermediate Algebra for College Students (7th Edition)
$1$
Recall the acronym PEDMAS, which tells us the order of operations: P: Parenthesis and other grouping symbols (including fraction bars) E: Exponents D/M: Division and Multiplication (from left to right) A/S: Addition and Subtraction (from left to right) Use the rule above to obtain: $=\dfrac{16 + 9}{4-(-21)} \\=\dfrac{25}{4+21} \\=\dfrac{25}{25} \\=1$ | 118 | 413 | {"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-2023-14 | latest | en | 0.83337 |
http://www.mrexcel.com/learnexcel/tag/array-formula/ | 1,516,340,630,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084887746.35/warc/CC-MAIN-20180119045937-20180119065937-00156.warc.gz | 500,137,034 | 11,699 | # Dueling Excel – “Working on the Railroad”: Podcast #1883
Today’s question for Duel 158: “You have to rent rail tank car #123 for a 3 month period. The daily rate changes twice ; once in
# Dueling Excel – “Category and Settings in Same Cell”: Podcast #1867
Every once in a while, you run into ugly Data Sets and this one qualifies. In ‘Dueling Excel’ Episode #154, we find that someone decided it would be brilliant to Back
# Dueling Excel – “LOOKUP First Over 100”: Podcast #1855 / Duel #150
Wow. 150 Dueling Excel Podcasts and counting! Wouldn’t it be nice if you could do =VLOOKUP(>100,MyTable,-1,False)? Back
# Learn Excel – “Count Cells Containing Digits 423 in Any Order”: Podcast #1834
Can Excel count Cells that contain any sequence of the digits 423? My solution in today’s Episode #1843 seems a bit clunky. To me, it seems like the first step has to be The Formula:=SMALL(MID(TEXT(A2,’000′),{1;2;3},1)*1,1)*100+SMALL(MID(TEXT(A2,’000′),{1;2;3},1)*1,2)*10+SMALL(MID(TEXT(A2,’000′),{1;2;3},1)*1,3)*Make sure to Array-enter this formula by holding down Ctrl+Shift while pressing Enter. Although elusive, Array Formulas can be used to(…)
# Learn Excel – “3 Level Validation Using Slicers”: Podcast #1798
Ian wrote in looking for an Array Formula that would build a unique list of managers who Back | 377 | 1,314 | {"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} | 3.28125 | 3 | CC-MAIN-2018-05 | latest | en | 0.84549 |
http://en.wikipedia.org/wiki/Local_base | 1,412,103,424,000,000,000 | text/html | crawl-data/CC-MAIN-2014-41/segments/1412037663060.18/warc/CC-MAIN-20140930004103-00432-ip-10-234-18-248.ec2.internal.warc.gz | 101,692,772 | 10,393 | # Neighbourhood system
(Redirected from Local base)
In topology and related areas of mathematics, the neighbourhood system or neighbourhood filter $\mathcal{V}(x)$ for a point x is the collection of all neighbourhoods for the point x.
A neighbourhood basis or local basis for a point x is a filter base of the neighbourhood filter, i.e. a subset
$\mathcal{B}(x) \subset \mathcal{V}(x)$
such that
$\forall V \in \mathcal{V}(x) \quad \exists B \in \mathcal{B}(x) \mbox{ with } B \subset V$.
That is, for any neighbourhood $V$ we can find a neighbourhood $B$ in the neighbourhood basis that is contained in $V$.
Conversely, as with any filter base, the local basis allows to get back the corresponding neighbourhood filter as $\mathcal{V}(x) =\left\{ V \supset B~:~ B \in \mathcal{B}(x)\right\}$.[1]
## Examples
• Trivially the neighbourhood system for a point is also a neighbourhood basis for the point.
• Given a space X with the indiscrete topology the neighbourhood system for any point x only contains the whole space, $\mathcal{V}(x) = \{ X \}$
• In a metric space, for any point x, the sequence of open balls around x with radius 1/n form a countable neighbourhood basis $\mathcal{B}(x) = \{ B_{1/n}(x) ; n \in \mathbb N^* \}$. This means every metric space is first-countable.
• In the weak topology on the space of measures on a space E, a neighbourhood base about $\nu$ is given by
$\left \{ \mu \in \mathcal{M}(E) : | \mu f_i - \nu f_i | < \varepsilon_i , i=1,\ldots, n\right \}$
where $f_i$ are continuous bounded functions from E to the real numbers.
## Properties
In a semi normed space, that is a vector space with the topology induced by a semi norm, all neighbourhood systems can be constructed by translation of the neighbourhood system for the point 0,
$\mathcal{V}(x) = \mathcal{V}(0) + x .$
This is because, by assumption, vector addition is separate continuous in the induced topology. Therefore the topology is determined by its neighbourhood system at the origin. More generally, this remains true whenever the topology is defined by a translation invariant metric or pseudometric.
Every neighbourhood system for a non empty set A is a filter called the neighbourhood filter for A. | 580 | 2,218 | {"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": 13, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.65625 | 4 | CC-MAIN-2014-41 | latest | en | 0.906087 |
http://www.wikihow.com/Calculate-Dog-Years | 1,472,548,096,000,000,000 | text/html | crawl-data/CC-MAIN-2016-36/segments/1471982974985.86/warc/CC-MAIN-20160823200934-00204-ip-10-153-172-175.ec2.internal.warc.gz | 787,345,896 | 52,276 | Expert Reviewed
# How to Calculate Dog Years
Most people know that dogs age faster than people. Conventional wisdom suggests that 1 dog year is equivalent to 7 human years. You've probably heard, for example, that you can multiply your dog's age by 7 to find the equivalent in "dog years," essentially a dog's estimated age measured in human years. But this calculation is oversimplified. For example, dogs ages much more rapidly than human during the first two years of life, such that the first year of a dog's life is equal to about 15 human years. There are also other factors that affect canine aging and that are essential for understand how to calculate a dog's age.[1]
### Method 1 Determining Your Dog's Actual Age
1. 1
Learn your dog's age in actual years. Knowing your dog's actual biological age is necessary to calculate his age in dog years. It also helps you make decisions during his life that are usually based on age, such as when to spay or neuter and when to change to a senior diet.
• Finding out your dog's age is easy to do if you have the dog's records or got him when he was a puppy.
• However, if you don't have this information readily available, there are other ways to determine his age. Most of these methods involving an investigation of a number of physical characteristics, including the condition of the teeth, size, coat condition, and eye conditions.
2. 2
Inspect your dog's teeth. Look at the stage and condition of the teeth.
• Determine what stage the teeth are at. Puppies will usually have all of their baby teeth by the time they are 8 weeks old and should have all of their permanent teeth by the time they are 6 to 7 months old. They start getting their permanent canine teeth, the longest ones, usually right at 6 months old, which makes this the easiest time to accurately age a puppy. If they have all of their permanent teeth and they are white and clean, they could be up to 1 1/2 to 2 years, but probably around 1 year old.[2]
3. 3
Assess the condition of the teeth. Between the ages of 1 and 2 years old, many dogs will start to show some yellowing on their back teeth. Tartar buildup starts shortly after this.[3] Most dogs will start to show wear on their incisors, the teeth along the front of their mouth, between 3 and 5 years. The wear on the teeth will gradually increase as your dog ages. Significant tartar build-up with evidence of gum disease (look for red, inflamed gums) can be seen after they are 5 years old. Missing teeth usually means that the dog is a senior and could use some dental care.[4]
• The stage and condition of the teeth can help to give you a good estimate of your dog's age, but remember that there are many factors besides age that can affect their appearance including breed, genetics, chewing behavior, and history of dental care.
4. 4
Assess your dog's size and growth. If your dog is continuing to grow, then he is less than 1 1/2 to 2 years old, depending on overall size. Small dogs have reached their full height and length by around one year of age, but large and especially giant breed dogs may take until they are 18 months to 2 years old.[5][6]
5. 5
Assess your dog's muscle tone. Younger dogs are more likely to have more muscle definition because they generally have higher activity levels. In contrast, older dogs can be a little bonier or fatter from decreased activity.[7]
• Watch your dog when he moves around. You may find he moves with a little more stiffness; this is also a sign of aging.[8]
6. 6
Examine your dog's coat. Younger dogs usually have soft, fine coats. Older dogs tend to have thicker and coarser fur that can sometimes also be oilier. A senior dog may also have grays or patches of white, particularly around the snout.[9]
7. 7
Inspect your dog's eyes. Younger dogs tend to have bright, clear eyes without tearing or discharge, whereas older dogs' eyes can be cloudy or opaque.[10]
8. 8
Take your dog to the vet. Your vet can estimate your dog's age based on a comprehensive physical exam or tests that evaluate the bones, joints, muscles, and organs. This is the best way to get an accurate assessment of your dog's actual age.[11]
### Method 2 Calculating Dog Years Based on Size
1. 1
Understand how size impacts dog years. The most accurate dog year calculations consider the size of the dog. Smaller dogs do tend to live longer than larger dogs, although they may mature more quickly in the first few years of life. A large dog may mature more slowly initially but be considered a "senior" at age five. Medium-sized dogs are somewhere in the middle in terms of maturation and lifespan.[12]
• In other words, after they reach adulthood, larger dogs will be considered older in dog years than a smaller dog of the same age.[13]
2. 2
Determine the size of your adult dog. You can weigh your dog at home or, for a more accurate assessment, have your dog weighed at the vet. Here are general categories based on dog size:[14]
• Small dogs: 20 pounds or less
• Medium dogs: 21 to 50 pounds
• Large dogs: 51 to 100 pounds
• Giant breed dogs: 100 pounds or more.
• Note that most dogs are considered adults by the time they are two years old. For dogs of all sizes the relationship of dog age to human years is the same for the first two years of a dog's life. For example, irrespective of whether he is small, medium or large, a dog that is 1 year old is about 15 years old in dog years. At 2 years old, he is 24 in dog years. After they reach 2 years old, however, is when a dog's size begins to affect his age in dog years.[15]
3. 3
Calculate for your dog's age after 2 years old. For every year after your dog is 2 years old, add the following number of years to 24 (his age at 2 years), based on your dog's weight: for small and medium dogs add 5 years, for large dogs add 6 years, and for giant breed dogs add 7 years.[16]
• Consider the age of a pug at 10 years old. At 2 years old, the pug is 24 in dog years. Every year after that, he ages by 5 years. This means that by his 10th year, he is actually 64 years old in dog years.
• A giant-breed dog, such as a Great Dane, ages more rapidly after age 2. At 2 years old, he is 24 in dog years. Each year after that, he ages by 7 years. By his 10th year, he is in fact 80 years old in dog years.
### Method 3 Determining Dog Years with a Quick Calculation
1. 1
Take your dog's age and subtract by 2. Note that your dog needs to be at least two years old. This when dogs are considered to have reached adulthood.[17]
2. 2
Multiple that number by 4 and then add 21. Now you will have an estimate of your dog's age in dog years.[18]
• For example, if your dog is 10 years old, then his age in dogs years is 53.
3. 3
Be aware that this calculation is just a rough estimate. Because this method does not take your dog's size into consideration, it cannot be completely accurate.
• This calculation will be more accurate for smaller dogs than larger dogs. For example, as we calculated above, when factoring in size, a Great Dane at 10 years old is 80 in dog years, whereas this method would calculate his age in dog years to be 53.
• However, this method is an easy trick to calculate dog years quickly if you don't know your dog's size or are in a hurry.
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• My dog is 2 human months old; how do I convert it to dog years?
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You dog needs to be two human years old to calculate his age using this formula, but a rough estimate if that your dog is around three in dog years.
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• Remember that using these guidelines only gives you an estimate of your dog's age in dog years and actual years. It can also be helpful to get another opinion from your veterinarian.
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Categories: Dogs
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Thanks to all authors for creating a page that has been read 80,890 times. | 1,979 | 8,077 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.125 | 3 | CC-MAIN-2016-36 | longest | en | 0.978249 |
https://forums.nrel.gov/t/power-law-exponent-problem/1885 | 1,653,758,928,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652663016949.77/warc/CC-MAIN-20220528154416-20220528184416-00780.warc.gz | 334,957,826 | 6,776 | # Power law exponent problem
Hello
I hope everybody is fine
I have a wind field from a large eddy simulation (LES) that I want to make something similar to it in Turb sim. I was advised to use the power law exponent to calculate the wind speed profile but I am not sure what exactly should I do to choose the best power-law exponent that gives me results similar to the one I have in the LES.
I can export wind field data from the LES to post process it to calculate the best power exponent.my problem is that i am not sure how to post-process the data to get the most fitting component
What do I have in mind?
I thought if I know the average velocity at a certain height in the LES and the average velocity at hub height I can calculate the power exponent.
The idea is I can output the average velocity for many different heights so then I will be getting many power law exponent. How to pick the best one that fits? I am not sure what to do?
Also, I was wondering what wind field should I be using from the LES simulation? Should I output the 3d wind velocities hitting the rotor and average it? Do I output the wind fields positioned before my turbine how many rotor diameters before the turbine ?? What should I do? should I only care for wind speed hitting the rotor of my turbine in the les simulation?
Dear Mostafa,
I would suggest time-averaging the wind speed at multiple heights from your LES output and fitting a power lower through a least-squares regression.
If you are not familiar, you may want to consider using the TIMESR feature of TurbSim v2, whereby the wind time series derived from TurbSim can match to some prescribed time series (from LES) at specific points (but not all points) in space; other points in the field will be based on the spectra of the prescribed time series and standard coherence functions. See the TurbSim v2 archive for more information: nwtc.nrel.gov/alphas.
Another option if you want to use your LES field within FAST, is to save your LES data in e.g. TurbSim format.
Best regards,
so I read about the regression method and I wanted to ask you if I am implementing it correctly
I cannot save my les files in turbsim format and someone using the Les implemented a Matlab script to output the velocities out of a saved data in the les but there is something making it unable to work properly
Please correct me if I am wrong in the following concerning the regression method. I will output time series of different heights for the velocity U at position (x,y)(2D before the first wind turbine, the position of the centre of the rotor) (as my les is in a 3d simulation)I average the time series of the U component at different heights to get an average U velocity at every height.I output the average Uhub speed also. THen I use least square regression to calculate the best power coefficient that fits the different points. I have attached picture of the law I will be using . I wanted to know should i pick a different x y positions other than the ones i chose or what exactly?
Also,I wanted to ask how should i pick the turbulence intensity for the turb sim .
Can I calculate the turbulence intensity at a specific point in my les simulation and use it for the turbsim simulation? I know that the turbulence intensity is the standard deviation /mean value of a time series of U component at a specific point. My question is where should I pick to calculate the turbulence intensity?I mean i have many positions to calculate the turbulence intensity which should I choose? I was thinking of calculating the average turbulence intensity and inputting it to turb sim but I am not sure how to calculate the average turbulence intensity. I was thinking about the following method to calculate it but I am not sure. this is the method i was thinking about :
I have a 3d grid in the x y z with u components for different positions in the x y z. I was going to pick all u components that are situated inside of my rotor disk in the yz plane and cube each velocity component then sum them about the rotor plane then divided by the area of my rotor then cubic root the answer to getting an average rotor speed about the rotor at a certain x position. Keep repeating the previous thing for different x positions, then calculate the mean value of different x positions. So now I have one velocity value at a certain time instant for an average velocity U about the rotor disk for different X positions.I repeat the following for different time instances to have a time series of the average velocity about the rotor disk for different X positions and use this time series to calculate the average Turbulence intensity .Am i doing it correctly ?
Dear Mostafa,
Yes, I agree with your least-squares regression approach.
So, does your LES simulation include a wind turbine or does the LES simulation only consider the atmospheric boundary layer? If a wind turbine is modeled within your LES, I agree picking points in the ambient wind say 2D upstream of the rotor makes sense. If the wind turbine is not modeled within the LES, then I would pick points at the rotor plane.
Are you cubing the wind speed in your spatial averaging to derive some sort of power-equivalent average wind speed? I agree that there are many points that you could derive the turbulence intensity from and I’m not sure whether there is a best practice on the approach taken. I would first judge how different the turbulence intensity is across the domain. If you want to limit the number of points considered, I would use points at the hub height.
Best regards,
thank you so so so much
my wind turbine is modelled in my les . I am using palm if you heard about it .
I am cubing the wind speed to derive some sort of power equivalent average wind speed as it helps in what I am doing.I wil judge according how different the turbulance intensity at hub height.
I am having some hard time as my supervisor is on vacation and everybody in the research centre I am working in are very busy to help. I am very thankful for your kind help. | 1,258 | 6,050 | {"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} | 2.703125 | 3 | CC-MAIN-2022-21 | latest | en | 0.916504 |
http://ecoursesonline.iasri.res.in/mod/page/view.php?id=95439 | 1,725,745,191,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700650920.0/warc/CC-MAIN-20240907193650-20240907223650-00858.warc.gz | 10,716,547 | 12,538 | ## Lesson 3. RHEOLOGICAL METHODS
Module 1. Rheology of foods
Lesson 3
RHEOLOGICAL METHODS
3.1 Introduction
Generally rheological properties are judged by sensory panel, it has its advantages and disadvantages depending on the person selected for judging the products. To have unbiased scores as well as reproducibility of the values of rheological attributes, it is necessary to go for instrumental measurement. There are many instrumental methods are developed based on fundamental principle as well as experimental data. There are certain mathematical models developed by different scientist based on empirical methods, which are widely used for measurement of rheological properties of most of the food products.
## 3.2 Tests for Measurement of Rheological Properties
Instrumental methods for measurement of rheological properties are classified into two broad categories as follow:
• Fundamental tests which measure the properties that are inherent to the material and do not depend on geometry and shape of the sample, conditions of loading or type of apparatus used, e.g. relaxation time, Poisson’s ratio, shear modulus and bulk modulus;
• Empirical tests (because data are based on comparison with sensory) or imitative tests (because these imitate the chewing in mouth). e.g. properties like puncture force, extrusion energy, cutting force required, pressing/compression force required for juice extraction, etc. – where mass of sample, geometry and speed of test will decide the magnitude of parameter estimated.
Generally fundamental tests are applied on solid foods and these are further classified into quasi-static and dynamic tests
The tests conducted under conditions of static/quasi-static loading are known as quasi-static tests while those conducted under dynamic loading conditions are called dynamic tests.
The use of Instron in determining the modulus of elasticity under compression is an example of quasi-static test while if the determination is done using a vibrating device of certain frequency (generally 200 Hertz), then the test is dynamic. You can say that rate of loading can be used to determine whether test is dynamic/quasistatic.
3.3 Quasi-Static Testing of Solid Food Products
Two types of behaviour can be studied– elastic behaviour of solid and another is pure viscous flow in case of liquids. Pure elastic behaviour is defined such that when force is applied to the material, it will instantaneously and finitely deform and when the force is released, the material will instantaneously come to the original form. Such materials are called ‘Hooken solids’ i.e. which follow Hook’s law. The amount of deformation is proportional to the magnitude of the force. Rheological representation of this type of solids is a spring. The material of this nature can be given a rheological constant modulus of elasticity is ratio of stress/strain, where stress = force/area, and strain = deformation due to force applied/original dimension. There are 3 types of moduli depending on type of force applied.
• If force is applied perpendicular to area defined by stress and it is calculated as – modulus of elasticity(E)
• If modulus is calculated by applying force parallel to area defined by stress i.e. a shearing stress, then it is called a shear modulus or modulus of rigidity(G or n) and
• If force is applied from all directions (isotropic force) then change in volume over original volume is obtained that can be calculated by bulk modulus(B or K)
Creep: In an experiment if a constant stress is applied to sample and corresponding strain is followed as a function of time and results are expressed in terms of a parameter of compliance (J=strain/stress). The change in the strain of material can be measured, when stress is removed it known as creep curve. In short we can say that creep curve shows strain as a function of time at constant stress. Visco-elastic materials can often be characterized by a modulus and relaxation time, which can be determined by an analysis of strain curve with time.
Relaxation curve (stress relaxation) – It is the curve obtained when stress is applied as a function of time at a constant strain. That means that instead of applying constant force and measuring the change in strain with time, it is also possible to apply a constant strain and measure change in stress with time. This type of experiment is called relaxation stress and the curve is known as relaxation curve.
These relaxation and creep experiments are known as Transient experiments in which a constant force is applied to the material and resulting strain is measured as a function of time and vice-versa. | 912 | 4,654 | {"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} | 3.40625 | 3 | CC-MAIN-2024-38 | latest | en | 0.919372 |
https://socratic.org/questions/how-do-you-find-the-circumference-and-area-of-a-circle-with-radius-of-7-feet | 1,579,381,874,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579250593937.27/warc/CC-MAIN-20200118193018-20200118221018-00063.warc.gz | 648,911,975 | 6,362 | # How do you find the circumference and area of a circle with radius of 7 feet?
Mar 18, 2018
The formula for area is $\pi \times {r}^{2}$ and the circumference is $2 \times \pi \times r$
So if $r = 7$, let's find these values:
$\cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot$
Area
$\pi \times {7}^{2}$
$\pi \times 49$
$49 \pi$ feet
$\cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot$
Circumference
$2 \times \pi \times 7$
$14 \pi$ feet
Mar 18, 2018
Use these formulae
#### Explanation:
The circumference of a circle is $\pi d$
The area of a circle is $\pi {r}^{2}$
Remember that the diameter is double the radius | 244 | 748 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 12, "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} | 4.53125 | 5 | CC-MAIN-2020-05 | longest | en | 0.419207 |
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#### Resources tagged with Arcs, sectors and segments similar to Triangles' Triangle:
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### There are 17 results
Broad Topics > Measures and Mensuration > Arcs, sectors and segments
### Giant Holly Leaf
##### Stage: 4 Challenge Level:
Find the perimeter and area of a holly leaf that will not lie flat (it has negative curvature with 'circles' having circumference greater than 2πr).
### Arclets
##### Stage: 4 Challenge Level:
Each of the following shapes is made from arcs of a circle of radius r. What is the perimeter of a shape with 3, 4, 5 and n "nodes".
### Floored
##### Stage: 3 Challenge Level:
A floor is covered by a tessellation of equilateral triangles, each having three equal arcs inside it. What proportion of the area of the tessellation is shaded?
##### Stage: 4 Challenge Level:
Given a square ABCD of sides 10 cm, and using the corners as centres, construct four quadrants with radius 10 cm each inside the square. The four arcs intersect at P, Q, R and S. Find the. . . .
### Rolling Coins
##### Stage: 4 Challenge Level:
A blue coin rolls round two yellow coins which touch. The coins are the same size. How many revolutions does the blue coin make when it rolls all the way round the yellow coins? Investigate for a. . . .
### Pericut
##### Stage: 4 and 5 Challenge Level:
Two semicircle sit on the diameter of a semicircle centre O of twice their radius. Lines through O divide the perimeter into two parts. What can you say about the lengths of these two parts?
### Speeding Up, Slowing Down
##### Stage: 3 Challenge Level:
Experiment with the interactivity of "rolling" regular polygons, and explore how the different positions of the red dot affects its speed at each stage.
### Track Design
##### Stage: 4 Challenge Level:
Where should runners start the 200m race so that they have all run the same distance by the finish?
### How Far Does it Move?
##### Stage: 3 Challenge Level:
Experiment with the interactivity of "rolling" regular polygons, and explore how the different positions of the red dot affects the distance it travels at each stage.
### 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?
### Bound to Be
##### Stage: 4 Challenge Level:
Four quadrants are drawn centred at the vertices of a square . Find the area of the central region bounded by the four arcs.
### Two Circles
##### Stage: 4 Challenge Level:
Draw two circles, each of radius 1 unit, so that each circle goes through the centre of the other one. What is the area of the overlap?
### Holly
##### Stage: 4 Challenge Level:
The ten arcs forming the edges of the "holly leaf" are all arcs of circles of radius 1 cm. Find the length of the perimeter of the holly leaf and the area of its surface.
### Just Rolling Round
##### Stage: 4 Challenge Level:
P is a point on the circumference of a circle radius r which rolls, without slipping, inside a circle of radius 2r. What is the locus of P?
### Two Shapes & Printer Ink
##### Stage: 4 Challenge Level:
If I print this page which shape will require the more yellow ink?
### Far Horizon
##### Stage: 4 Challenge Level:
An observer is on top of a lighthouse. How far from the foot of the lighthouse is the horizon that the observer can see?
### Round and Round
##### Stage: 4 Challenge Level:
Prove that the shaded area of the semicircle is equal to the area of the inner circle. | 825 | 3,598 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.765625 | 4 | CC-MAIN-2017-39 | latest | en | 0.86721 |
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# Design Equation Grinding Mill India
• Grinding Millnance and reliable grinding system The structural design of the agitator mill has a number of advantages The horizontal orientation ensures homo geneous distribution of grinding media in the grinding chamber Thus start up of the mill with product is unproblematic The highest fl ow rates can be set without pressure build up
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• Optimization of mill performance by usingliner design such as the toe and shoulder regions Production is disrupted for the duration of the procedure grind out mill stoppage mill start up and the transition period to steady state Stresses generated in the ball charge increase which may result in
• Modelling SAG milling power and specific energy basic design of the mills for example using DW i in the specific energy equation Morrell 2004 but studying the inclusion of the feed size distribution not just the F80 as a model variable represented by an operation relevant size fraction 2 Operational data Operational data from 4 grinding circuits corresponding to 3
• CALCULATION OF BALL MILL GRINDING EFFICIENCY Page 1 of 108/03/2013· calculation of ball mill grinding efficiency dear experts please tell me how to calculate the grinding efficiency of a closed ckt open ckt ball mill in literatures it is written that the grinding efficiency of ball mill is very less [less than 10%] please expalin in
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• design equations of clay grindersDesign equations for undrained stability of opening in In recent years a concrete diaphragm wall is commonly employed as a permanent retaining structure for deep excavations in soft soils in various constructions including basements underpasses shaft stations of tunnels cutandcover tunnels and among others eg Ou 2006 Puller 2003This paper concerns with the study of undrained stability of an
• SIMPLIFIED BOND WORK INDEX DETERMINATION By R O To design grinding circuits or evaluate their performance the Bond Theory of comrninution is used Equation 1 is the mathematical statement of the Third Theory where W = the energy input to the mill kwh/t Wi Work Indéx kwh/t P square sieve opening which 80% of the product passes microns
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• Back to Basics Hammer Milling and Jet Milling FundamentalsA hammer mill is typically good for grinding softer materials with Mohs hard ness ranging from 1 to 5 while a jet mill can grind materials with Mohs hardness as high as 10 High hardness materi als become very abrasive therefore they are not suitable for high speed hammer mills In a typical jet mill grinding
• Design of Closed Circuit Grinding System with Tube Mill 01/05/2002· Article Views are the COUNTER compliant sum of full text article downloads since November 2008 both PDF and HTML across all institutions and individuals | 1,841 | 9,082 | {"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} | 2.53125 | 3 | CC-MAIN-2022-21 | latest | en | 0.894481 |
http://www.enotes.com/homework-help/factors-affecting-friction-experiment-179385 | 1,477,553,357,000,000,000 | text/html | crawl-data/CC-MAIN-2016-44/segments/1476988721142.99/warc/CC-MAIN-20161020183841-00303-ip-10-171-6-4.ec2.internal.warc.gz | 433,517,773 | 10,405 | # Factors Affecting Friction ExperimentI have to design an experiment to determine how certain factors (weight, surface area, and speed) affect the force of friction. I have taken a tissue box...
Factors Affecting Friction Experiment
I have to design an experiment to determine how certain factors (weight, surface area, and speed) affect the force of friction.
I have taken a tissue box weighing 122g and attached an elastic with a spring constant of 14.9N/m. The surface area is 688cm^2, but I'm pretty certain this has no effect, and neither would velocity.
Basically, I need some help understanding these friction equations. I need to figure out the force of friction, but I'm not sure which equations to use or how to use them. If someone could explain and/or offer some tips, I would be extremely grateful.
hnystrom | High School Teacher | (Level 1) Assistant Educator
Posted on
Simple friction is caused by the motion of one surface sliding against another. The amount of frictional force is proportional to the force pressing two surfaces together. For a book resting on the surface of the table the force is the weight of the book. If we press the book into the surface the force increases by the amount of press. If we lift up on the book the force is reduced by the amount of lift. The force that best represents the amount that is pressing the two surfaces together is what we call the Normal force.
Normal force = Weight + Press into Surface - Lift from Surface
The friction is proportional to the Normal force. The proportion deprends upon what the surfaces are, whether lubrication is present, the surface area and the speed.
For example, friction for wood sliding on wood is 30%-40%. So if the Normal force is 100 N then the friction is 30 N -40 N. The proportionality factor is called the co-efficient of friction. (The coefficient of wood on wood is .3 - .4) There are tables of approximate values of the co-efficient of friction.
Friction force = coefficient of friction x Normal force
There are two forces of friction: static friction and sliding or kinetic friction. Static friction is the force you need to overcome to get an object starting to slide from rest. Kinetic friction is the friction you need to overcome to keep it going. In general static friction is more than kinetice friction.
To measure friction have the box at rest. Pull slowly on the spring until it just starts to move watching all the time. You will see the force increase until it just starts to move and then ease up a bit. The maximum force you see on the spring at the moment it starts to move is the static friction. Once it gets going you need to pull just hard enough so that the box slides at a steady speed. At steady speed the spring force will be equal to the kinetic friction. | 613 | 2,812 | {"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-2016-44 | latest | en | 0.922801 |
https://www.slideserve.com/Gabriel/neural-heuristics | 1,591,161,531,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347432237.67/warc/CC-MAIN-20200603050448-20200603080448-00599.warc.gz | 892,147,484 | 14,186 | # Neural heuristics for Problem Solving - PowerPoint PPT Presentation
Neural heuristics for Problem Solving
1 / 18
Neural heuristics for Problem Solving
## Neural heuristics for Problem Solving
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##### Presentation Transcript
1. 10-09-2004 Siena Neural heuristics for Problem Solving Outline • Problem Solving • Heuristic Search • Optimal Search and Admissibility • Neural heuristics for Problem Solving • Architecture • Asymmetric Regression • Dataset Generation • Likely-admissibility • Multiple neural heuristics Marco Ernandes - email: ernandes@dii.unisi.it
2. 1 2 3 4 5 11 9 15 5 6 7 8 8 1 6 10 xo 9 10 11 12 3 7 14 g P 13 14 15 4 13 2 12 Problem Solving • PS is a decision-making process that aims to find a sequence of operators that takes the agent from a given state to a goal state. • We talk about: states, successor function (defines the operators available at each state), initial state, goal state, problem space. Marco Ernandes - email:ernandes@dii.unisi.it
3. Heuristic Search • Search algorithms (best-first, greedy, …): define a strategy to investigate the search-tree. • Heuristic information h(n): typically the distance from node n (of the search-tree)to goal • Heuristic usage policy : how to combine h(n) and g(n) to obtain f(n). n g(n) h(n) xo * g P Marco Ernandes - email:ernandes@dii.unisi.it
4. Optimal Heuristic Search • GOAL: retrieve the solution that minimizes the path cost (sum of all the operator costs) C=C* • Requires: • an optimal search algorithm (A*, IDA*, BS*) • an admissible heuristic, h(n)h*(n) (i.e. Manhattan) • an admissible heuristic usage (f(n) =h(n) + g(n) , ) • Complexity: • Optimal solving: any puzzle is NP-Hard. • Non-admissible search is polynomial. Marco Ernandes - email:ernandes@dii.unisi.it
5. Our idea about admissible search • The best performance in literature: memory-based heuristics (Disjoint Pattern DBs, <Korf&Taylor, 2002>) • Offline phase: resolution of all possibile subproblems and storage of all the results. • Online phase: decomposition of a node in subproblems and database querying. • Our idea: • memory-based heuristics have little future because they only shift NP-completeness from time to space. • ANN (as universal approx.) can provide effective non-memory-based, “nearly” admissible heuristics. Marco Ernandes - email:ernandes@dii.unisi.it
6. Outline: • Problem Solving • Heuristic Search • Optimal Search and Admissibility • Neural heuristics for Problem Solving • Architecture • Asymmetric Regression • Dataset Generation • Likely-admissibility • Multiple neural heuristics Marco Ernandes - email:ernandes@dii.unisi.it
7. heuristic evaluation error check 7 3 5 BP 2 4 6 01001000010 1 8 request an estimation extraction of an example Neural heuristics We used standard MLP networks. ONLINE PHASE OFFLINE PHASE h(n) Marco Ernandes - email:ernandes@dii.unisi.it
8. – Neural heuristics –Outputs, Targets & Entrances • It’s a regression problem, hence we used 1 linear output neuron (modified a posteriori exploiting information from Manhattan-like heuristics). • 2 possible targets: • A) “direct” target function o(x) = h*(x) • B) “gap” target o(x) = h*(x)-hM(x) (which takes advantage of Manhattan too) • Entrances coding: • we tried 3 different vector-valued codings • future work: represent configurations as graphs, in order to have non-dimension-dependent learning. (i.e. exploit learning from 15-puzzle, in the 24-puzzle). Marco Ernandes - email:ernandes@dii.unisi.it
9. – Neural heuristics –Learning Algorithm • Normal backpropagation algorithm, but … • Introducing a coefficient of asymmetry in the error function. This stresses admissibility: • Ed = (1-w) (od –td) if (od –td) < 0 • Ed = (1+w) (od –td) if (od –td) > 0 • We used a dynamic decreasing w, in order to stress underestimations when learning is simple and to ease it successively. Momentum a=0,8 helped smoothness. with 0 < w < 1 Marco Ernandes - email:ernandes@dii.unisi.it
10. – Neural heuristics –Asymmetric Regression • This is a general idea for backprop learning. • It can suit any regression problem where overestimations harm more than underestimations (or contrary). • Heuristic machine learning is an ideal application field. • We believe that totally admissible neural heuristics are theoretically impossible, or at least impracticable. Symmetric error Asymmetric error Marco Ernandes - email:ernandes@dii.unisi.it
11. – Neural heuristics –Dataset Generation • Examples are previously optimally solved configurations. It seems a big problem, but … • Few examples are sufficient for good learning. A few hundreds to have faster search than Manhattan. We used a training set of 25000 to (500 million times smaller than the problem space). • These examples are mainly “easy” ones, over 60% of 15-puzzle examples have d < 30, whereas only 0,1% of random cases have d < 30 [see 15-puzzle search tree distribution]. • All the process is fully parallelizable. • Further works: auto-feed learning. Marco Ernandes - email:ernandes@dii.unisi.it
12. – Neural heuristics –Are sub-symbolic heuristics “online”? • We believe so. Even that there is an offline learning phase. For 2 reasons: • 1. Nodes visited during search are generally UNSEEN. • Exactly like often humans do with learned heuristics: we don’t recover a heuristic value from a database, we compute it employing the inner rules that the heuristic provides. • 2. The learned heuristic should be dimension-independent: learning over small problems could be used for bigger problems (i.e. 8-puzzle 15-puzzle). This is not possible with mem-based heuristics. Marco Ernandes - email:ernandes@dii.unisi.it
13. Outline: • Problem Solving • Heuristic Search • Optimal Search and Admissibility • Neural heuristics for Problem Solving • Architecture • Asymmetric Regression • Dataset Generation • Likely-admissibility • Multiple neural heuristics Marco Ernandes - email:ernandes@dii.unisi.it
14. Likely-Admissible Search • We relax the optimality requirement in a probabilistic sense (not qualitatively like e-admissible search). • Why is it a better approach than e-admissibility? • It allows to retrieve TRULY OPTIMAL solutions. • It still allows to change the nature of search complexity. • Because search can rely on any heuristic, unlike e-admissible search that works only on already-proven-admissible ones. • Because we can better combine search with statistical machine learning techniques. Using universal approximators we can automatically generate heuristics. Marco Ernandes - email:ernandes@dii.unisi.it
15. – Likely-Admissible Search –A statistical framework • One requisite: to have a previous statistical analysis of overestimation frequencies of our h. • P(h\$) shall be the probability that heuristic h underestimates h* for any given state xX. • ph shall be the probability of optimally solving a problem using h and A*. • TO ESTIMATE OPTIMALITY FROM ADMISSIBILITY: Marco Ernandes - email:ernandes@dii.unisi.it
16. – Likely-Admissible Search –Multiple Heuristics • To enrich the heuristic information we can generate many heuristics and use them simultaneously, as: • Thus: • If we will consider for simplicity that all j heuristics have the same given P(h+2): j grows logarithmically with d and pH Marco Ernandes - email:ernandes@dii.unisi.it
17. – Likely-Admissible Search –Optimality prediction • Unfortunately the last equation is very optimistic since it assumes a total error independency among neural heuristics. • For predicions we have to use which is: • Extremely precise for optimality over 80%. • Imprecise for low predictions. • Predictions are much more accurate than e-admissible search predictions. Marco Ernandes - email:ernandes@dii.unisi.it
18. Experimental Results & Demo • Compared to Manhattan: • IDA* with 1 ANN (optimality 30%): 1/1000 execution time, 1/15000 nodes visited • IDA* with 2 ANN (opt. 50%): 1/500 time, 1/13000 nodes. • IDA* with 4 ANN-1 (opt. 90%): 1/70 time, 1/2800 nodes. • Compared to DPDBs: • IDA* with 1 ANN (optimality 30%): between -17% and +13% nodes visited, between 1,4 and 3,5 times slower • IDA* with 2 ANN (opt. 50%): -5% nodes visited, 5 times slower (but this could be parallelized completely!) Try the “demo” at: http://www.dii.unisi.it/ ~ernandes/samloyd/ Marco Ernandes - email:ernandes@dii.unisi.it | 2,242 | 8,437 | {"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} | 3.03125 | 3 | CC-MAIN-2020-24 | latest | en | 0.56784 |
http://sciencepantheism.com/worksheet/pearson-education-math-worksheets.php | 1,611,649,922,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610704799711.94/warc/CC-MAIN-20210126073722-20210126103722-00269.warc.gz | 88,511,256 | 11,983 | ## sciencepantheism.com - the pro math teacher
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# Pearson Education Math Worksheets
Public on 30 Oct, 2016 by Cyun Lee
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std 4 maths worksheets | 1,754 | 4,441 | {"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} | 2.890625 | 3 | CC-MAIN-2021-04 | latest | en | 0.161016 |
https://www.gradesaver.com/textbooks/math/trigonometry/trigonometry-10th-edition/chapter-4-graphs-of-the-circular-functions-section-4-1-graphs-of-the-sine-and-cosine-functions-4-1-exercises-page-143/17 | 1,544,750,800,000,000,000 | text/html | crawl-data/CC-MAIN-2018-51/segments/1544376825123.5/warc/CC-MAIN-20181214001053-20181214022553-00567.warc.gz | 890,849,305 | 12,396 | ## Trigonometry (10th Edition)
RECALL: The graph of the function $y=a\cdot \cos{x}$: (1) has a period of $2\pi$ and an amplitude of $|a|$; and (2) involves a reflection about the x-axis of the parent function $y=\cos{x}$ when $a \lt 0$. The given function has a period of $2\pi$ and since $a=-1$, its amplitude is $1$, and it involves a reflection about the x-axis of the parent function $y=\cos{x}$. To graph the given function, perform the following steps: (1) Create a table of values. (refer to the table below) (2) Plot each point then connect them using a sinusoidal curve with an amplitude of $1$ Refer to the graph in the answer part above. | 181 | 649 | {"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} | 3.5 | 4 | CC-MAIN-2018-51 | latest | en | 0.831367 |
http://www.chegg.com/homework-help/definitions/distance-formula-63?cp=CHEGGFREESHIP | 1,467,114,832,000,000,000 | text/html | crawl-data/CC-MAIN-2016-26/segments/1466783396875.58/warc/CC-MAIN-20160624154956-00003-ip-10-164-35-72.ec2.internal.warc.gz | 445,903,172 | 13,096 | # Distance Formula
The distance formula is used to determine the distance, d, between two points. If the coordinates of the two points are (x1, y1) and (x2, y2), the distance equals the square root of x2x1 squared + y2y1 squared.
The distance formula is derived by creating a triangle and using the Pythagorean theorem to find the length of the hypotenuse. The hypotenuse of the triangle is the distance between the two points.
See more Geometry topics
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if we are given two similar triangles, both with a 90 degree angle where they both span 6 units (both adjacent sides added together) what are the adjacent side lengths?
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what is tan 67 degrees? Have a rght triangle with 67 and 23 degrees. the side from 90 degrees to 67 degrees is 5. the side from 90 to 23 degrees is 12 and the side fron 23 to 67 degrees is 13.
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Get help on Geometry with Chegg Study | 283 | 1,146 | {"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} | 3.375 | 3 | CC-MAIN-2016-26 | latest | en | 0.910824 |
http://www.sawaal.com/odd-man-out-and-series-questions-and-answers/insert-the-missing-number--3-7-6-5-9-3-12-1-15-_6468 | 1,521,560,841,000,000,000 | text/html | crawl-data/CC-MAIN-2018-13/segments/1521257647498.68/warc/CC-MAIN-20180320150533-20180320170533-00096.warc.gz | 434,756,151 | 16,227 | 38
Q:
# Insert the missing number.3, 7, 6, 5, 9, 3, 12, 1, 15, (....)
A) 18 B) 13 C) -1 D) 3
Explanation:
There are two series, beginning respectively with 3 and 7. In one 3 is added and in another 2 is subtracted.
The next number is 1 - 2 = -1.
Q:
Find the missing number in the given number series?
5 11 21 44 ? 175
A) 86 B) 78 C) 94 D) 102
Explanation:
Here the given number series is 5 11 21 44 ? 175
Now the given series follows a pattern, that
x2 + 1, x2 - 1, x2 + 2, x2 - 2, x2 + 3, x2 - 3,...
Then the missing number is x2 - 2
= 44 x 2 - 2 = 88 - 2 = 86
Hence, the missing number in the number series is 86.
0 8
Q:
Find the Odd one out of the following Numbers?
A) 3 B) 19 C) 67 D) 133
Explanation:
Here among the given numbers, 3, 19, 67 are prime numbers. And 133 is not a prime number.
Hence, the odd number of the given numbers is 133.
0 150
Q:
Find the odd man out of the number series?
18 37 76 155 314 635
A) 37 B) 155 C) 314 D) 635
Explanation:
Given Number series is 18 37 76 155 314 635
follows a pattern that,
18
18 x 2 +1 = 37
37 x 2 +2 = 76
76 x 2 +3 = 155
155 x 2 +4 = 314
314 x 2 +5 = 633 not 635
Hence, the odd man is 635.
5 62
Q:
Please enter the Missing number: 4 8 14 22 ?
A) 28 B) 32 C) 46 D) 36
Explanation:
Here the given series follows a pattern that,
2- 0 = 4
3- 1 = 8
4- 2 = 14
5- 3 = 22
6- 4 = 32
Hence, the missing number is 32
Method 2 ::
The difference between 4 and 8 is 4, the difference between 8 and 14 is 6, the difference between 14 and 22 is 8. So the next difference between 22 and the next number will be 10 since it follows (4,6,8,10 - add +2 each time)
Hence, 22 + 10 = 32 is the missing number.
4 109
Q:
Find the next number in the given number series?
0, 3, 8, 15, 24, ?
A) 35 B) 37 C) 34 D) 36
Explanation:
Here the given number series is 0, 3, 8, 15, 24, ?
The next number in the given series can be found such a way that the number series follows a pattern,
2 97
Q:
Find the Missing number in the given number series?
5, 9, 15, 23, ?, 45
A) 33 B) 29 C) 31 D) 35
Explanation:
The given series follows a pattern that,
+4, +6, +8, +10, +12...
5
5 + 4 = 9
9 + 6 = 15
15 + 8 = 23
23 + 10 = 33
33 + 12 = 45
Hence, the missing number in the series is 33.
1 79
Q:
Find the next number in the number series?
1, 4, 12, 32, ?
A) 80 B) 100 C) 192 D) 169
Explanation:
Next number of the sequence 1, 4, 12, 32,...
1 = 2 x ${2}^{0}$
4 = 2 x ${2}^{1}$
12 = 3 x ${2}^{2}$
32 = 4 x ${2}^{3}$
80 = 5 x ${2}^{4}$.
2 187
Q:
Find the Odd One Out?
3 24 186 1008 5040 20160
A) 24 B) 20160 C) 186 D) 5040
Explanation:
The given number series is 3 24 186 1008 5040 20160
Here the given series follows a pattern that,
3
3 x 8 = 24
24 x 7 = 168 not 186
168 x 6 = 1008
1008 x 5 = 5040
5040 x 4 = 20160
Hence the Odd One in the number series is 186. | 1,176 | 2,929 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 5, "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-2018-13 | longest | en | 0.756333 |
https://exponentiations.com/2-to-the-98th-power | 1,696,089,929,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510697.51/warc/CC-MAIN-20230930145921-20230930175921-00230.warc.gz | 274,154,131 | 23,464 | Home » Powers of 2 » 2 to the 98th Power
# 2 to the 98th Power
Welcome to 2 to the 98th power, our post about the mathematical operation exponentiation of 2 to the power of 98.
If you have been looking for 2 to the ninety-eighth power, or if you have been wondering about 2 exponent 98, then you also have come to the right place. 🙂
The number 2 is called the base, and the number 98 is called the exponent. In this post we are going to answer the question what is 2 to the 98th power.
Keep reading to learn everything about two to the ninety-eighth power.
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## What is 2 to the 98th Power?
2 to the 98th power is conventionally written as 298, with superscript for the exponent, but the notation using the caret symbol ^ can also be seen frequently: 2^98.
298 stands for the mathematical operation exponentiation of two by the power of ninety-eight.
As the exponent is a positive integer, exponentiation means a repeated multiplication:
2 to the 98th power =
The exponent of the number 2, 98, also called index or power, denotes how many times to multiply the base (2).
Thus, we can answer what is 2 to the 98th power as
2 to the power of 98 = 298 = 3169126500 5705735037 4175801344.
If you have come here in search of an exponentiation different to 2 to the ninety-eighth power, or if you like to experiment with bases and indices, then use our calculator above.
To stick with 2 to the power of 98 as an example, insert 2 for the base and enter 98 as the index, also known as exponent or power.
2 to the 98th power is an exponentiation which belongs to the category powers of 2.
Similar exponentiations on our site in this category include, but are not limited, to:
Ahead is more info related to 2 to the 98 power, along with instructions how to use the search form, located in the sidebar or at the bottom, to obtain a number like 2^98.
## 2 to the Power of 98
Reading all of the above, you already know most about 2 to the power of 98, except for its inverse which is discussed a bit further below in this section.
Using the aforementioned search form you can look up many numbers, including, for instance, 2 to the power 98, and you will be taken to a result page with relevant posts.
Now, we would like to show you what the inverse operation of 2 to the 98th power, (298)−1, is. The inverse is the 98th root of 298, and the math goes as follows:
(298)−1
=
=
=
=
2
Because the index of 98 is a multiple of 2, which is even, in contrast to odd numbers, the operation produces two results: (298)−1
; the positive value is the principal root.
Make sure to understand that exponentiation is not commutative, which means that 298 ≠ 982, and also note that (298)-1 ≠ 2-98, the inverse and reciprocal of 298, respectively.
You already know what 2 to the power of 98 equals, but you may also be interested in learning what 2 to the negative 98th power stands for.
Next is the summary of our content.
## Two to the Ninety-eighth Power
You have reached the concluding section of two to the ninety-eighth power = 298. Two to the ninety-eighth power is, for example, the same as 2 to the power 98 or 2 to the 98 power.
Exponentiations like 298 make it easier to write multiplications and to conduct math operations as numbers get either big or small, such as in case of decimal fractions with lots of trailing zeroes.
If you have been looking for 2 power 98, what is 2 to the 98 power, 2 exponent 98 or 98 power of 2, then it’s safe to assume that you have found your answer as well.
If our explanations have been useful to you, then please hit the like button to let your friends know about our site and this post 2 to the 98th power.
And don’t forget to bookmark us.
## Conclusion
In summary,
If you like to learn more about exponentiation, the mathematical operation conducted in 298, then check out the articles which you can locate in the header menu of our site.
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Thanks for visiting 2 to the 98th power. | 1,063 | 4,249 | {"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-2023-40 | latest | en | 0.960488 |
http://forum.allaboutcircuits.com/threads/finding-phase-angle-of-circuit-using-phasors.107018/ | 1,485,177,652,000,000,000 | text/html | crawl-data/CC-MAIN-2017-04/segments/1484560282926.64/warc/CC-MAIN-20170116095122-00117-ip-10-171-10-70.ec2.internal.warc.gz | 101,376,210 | 14,603 | # Finding phase angle of circuit using phasors.
Discussion in 'Homework Help' started by danielb33, Feb 12, 2015.
1. ### danielb33 Thread Starter Member
Aug 20, 2012
105
0
See circuit in file attached (and problem statement).
I first found the voltage across the inductor - I * Z = 5.6*<-28.
Next I simply used V1 - delta V = V2. I found that 100 - 5.6*cos(-28)+ i*sin(-28)*5.6 = ... and simplified the expression and brought back to phasor notation. I did not get the right answer...at all!
Note that my answer in the image attached was my first attempt. The way I solved above gave a different answer.
Any help as to where I am going wrong? Thanks!
File size:
59.2 KB
Views:
31
2. ### WBahn Moderator
Mar 31, 2012
18,091
4,917
What's the angle of the impedance of the inductor?
3. ### shteii01 AAC Fanatic!
Feb 19, 2010
3,516
515
i would change all the values to the complex notation
sum all the voltages is equal to zero: (real v1+ imaginary v1) + (real v inductor+ imaginary v inductor) + (real v2+ imaginary v2)=0
solve for real v2 and imaginary v2
use trig to find magnitude and angle of v2
4. ### danielb33 Thread Starter Member
Aug 20, 2012
105
0
shteii01 - 100 + 0.5<90 + V2<theta = 0 --> 100 + 0.5*cos(90) + i*sin(90)0.5+V2*cos(theta)+i*sin(theta)*V2 = 0
How doe this help me?
Wbahn - impedence of inductor is 0.5<90
5. ### WBahn Moderator
Mar 31, 2012
18,091
4,917
So now find the voltage across the inductor a bit more carefully.
6. ### shteii01 AAC Fanatic!
Feb 19, 2010
3,516
515
Those are not complex numbers.
You can do your own thing... invent your own math... if you like. I aint gonna stop u. | 528 | 1,633 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.703125 | 4 | CC-MAIN-2017-04 | latest | en | 0.835311 |
https://discuss.leetcode.com/topic/47540/very-easy-to-understand-5-ms-o-n-solution-without-stack-or-dp | 1,513,184,604,000,000,000 | text/html | crawl-data/CC-MAIN-2017-51/segments/1512948529738.38/warc/CC-MAIN-20171213162804-20171213182804-00233.warc.gz | 530,332,449 | 8,676 | # Very Easy To Understand - 5 ms - O(n) solution without Stack or DP
• for input example:
"()()))((()(())(()))((())(()()()()()()(((((())))))())()())()(())(())())))))(())()((((((((()()(())))))())())))()(((()())()))(((()()((((("
I scan the string forward and backward change the string to:
()()##((()(())(()))((())(()()()()()()(((((())))))())()())()(())(())())####(())()((((((((()()(())))))())())))()(((()())()))###()()#####
the invalid ( or ) are removed.
Then you count the longest valid substring length.
``````public class Solution {
public int longestValidParentheses(String s) {
char[] c = s.toCharArray();
int len = c.length, t = 0, ans = 0;
for (int i = 0, y = 0; i < len; i++)
if (c[i] == '(')
y++;
else if (c[i] == ')' && --y < 0) {
y = 0;
c[i] = '#';
}
for (int i = len - 1, y = 0; i >= 0; i--) {
if (c[i] == ')')
y++;
else if (c[i] == '(' && --y < 0) {
y = 0;
c[i] = '#';
}
t = (c[i] == '#') ? 0 : t + 1;
ans = Math.max(ans, t);
}
System.out.println(new String(c));
return ans;
}
}``````
Looks like your connection to LeetCode Discuss was lost, please wait while we try to reconnect. | 376 | 1,104 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.859375 | 3 | CC-MAIN-2017-51 | latest | en | 0.378315 |
https://plainmath.org/algebra-ii/29146-solve-the-equation-frac-7-y-2-4-frac-4-y-plus-2-equal-frac-5-y-2 | 1,717,078,465,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971667627.93/warc/CC-MAIN-20240530114606-20240530144606-00336.warc.gz | 390,555,737 | 21,739 | preprekomW
2021-09-16
Solve the equation.
$\frac{7}{{y}^{2}-4}-\frac{4}{y+2}=\frac{5}{y-2}$
SchepperJ
Step 1
We have to solve the equation:
$\frac{7}{{y}^{2}-4}-\frac{4}{y+2}=\frac{5}{y-2}$
Rewriting the equation,
$\frac{7}{{y}^{2}-{2}^{2}}-\frac{4}{y+2}=\frac{5}{y-2}$
$\frac{7}{\left(y-2\right)\left(y+2\right)}-\frac{4}{y+2}=\frac{5}{y-2}$ (since ${a}^{2}-{b}^{2}=\left(a-b\right)\left(a+b\right)$)
We know for any rational function denominator cant
Do you have a similar question? | 219 | 489 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 22, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.09375 | 4 | CC-MAIN-2024-22 | latest | en | 0.548499 |
https://www.teacherspayteachers.com/Product/Busy-Bee-Math-Single-and-Double-Digit-Addition-and-Subtraction-662570 | 1,529,475,117,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267863463.3/warc/CC-MAIN-20180620050428-20180620070428-00026.warc.gz | 936,232,033 | 15,851 | Busy Bee Math: Single and Double Digit Addition and Subtraction
Subject
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1 MB|5 pages and 5 answer keys
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Product Description
Busy Bee Math includes 5 PRINTABLES and ANSWER KEYS.
In the first worksheet students will do a cryptogram to practice single and double digit addition. They will solve the riddle, "Why do bees have sticky hair?"
In the second worksheet students do another cryptogram to practice subtraction. They will solve another riddle, “Why did Busy Bee go to the doctor?”
In the third and fourth worksheets entitled “Busy Bee and the Hive” students will do a variety of addition and subtraction problems to see how fast they can help Busy Bee return to her hive.
In the fifth worksheet students will need to add all the numbers in the petals of each of the flowers. They will write their answers below each flower. Then, as a challenge they will need to add the numbers under each flower to find out the total number in the flower bouquet.
Total Pages
5 pages and 5 answer keys
Included
Teaching Duration
N/A
Report this Resource
\$2.00 | 253 | 1,134 | {"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} | 2.53125 | 3 | CC-MAIN-2018-26 | latest | en | 0.910285 |
https://www.reference.com/world-view/factors-8-2ec9cc9be8083878 | 1,618,413,313,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038077818.23/warc/CC-MAIN-20210414125133-20210414155133-00520.warc.gz | 1,067,429,108 | 23,510 | # What Are the Factors of 8?
By Staff WriterLast Updated Apr 2, 2020 7:34:13 AM ET
The factors of the number 8 are 1, 2, 4 and 8. Since the number is divisible by more than 1 and itself, it is not a prime number.
The number 8 is a rational, even and positive integer. It is also referred to as a composite number, or a nonprime, since it is divisible by factors other than 1 and itself. The number 8 also contains the factor pairs of 1 and 8 and 2 and 4. Another related characteristic of the factors of 8 is that its distinct prime factor is 2, which is the prime number that divides 8 evenly.
More From Reference | 167 | 618 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.515625 | 4 | CC-MAIN-2021-17 | longest | en | 0.969686 |
https://www-num.math.uni-wuppertal.de/de/amna/forschung/anwendungen/finance/ | 1,660,816,613,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882573193.35/warc/CC-MAIN-20220818094131-20220818124131-00463.warc.gz | 547,440,030 | 14,035 | Fakultät für Mathematik und Naturwissenschaften
## Finance
The famous Black-Scholes equation is an effective model for option pricing. It was named after the pioneers Black, Scholes and Merton who suggested it 1973.
In this research field our aim is the development of effective numerical schemes for solving linear and nonlinear problems arising in the mathematical theory of derivative pricing models.
An option is the right (not the duty) to buy (`call option') or to sell (`put option') an asset (typically a stock or a parcel of shares of a company) for a price E by the expiry date T. European options can only be exercised at the expiration date T. For American options exercise is permitted at any time until the expiry date. The standard approach for the scalar Black-Scholes equation for European (American) options results after a standard transformation in a diffusion equation posed on an bounded (unbounded) domain.
Another problem arises when considering American options (most of the options on stocks are American style). Then one has to compute numerically the solution on a semi-unbounded domain with a free boundary. Usually finite differences or finite elements are used to discretize the equation and artificial boundary conditions are introduced in order to confine the computational domain.
In this research field we want to design and analyze new efficient and robust numerical methods for the solution of highly nonlinear option pricing problems. Doing so, we have to solve adequately the problem of unbounded spatial domains by introducing artificial boundary conditions and show how to incorporate them in a high-order time splitting method.
Nonlinear Black-Scholes equations have been increasingly attracting interest over the last two decades, since they provide more accurate values than the classical linear model by taking into account more realistic assumptions, such as transaction costs, risks from an unprotected portfolio, large investor's preferences or illiquid markets, which may have an impact on the stock price, the volatility, the drift and the option price itself.
## Special Interests
2022
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Bohrmann-Linde, Claudia; Siehr, Ilona
CHEMIE Einführungsphase Nordrhein-Westfalen
Herausgeber: C.C.Buchner Verlag, Bamberg
August 2022
ISBN: 9783661060019
### 5628.
[german] Monique, Meier; Zeller, Diana; Stinken-Rösner, Lisa
Interaktive Videoformate für den naturwissenschaftlichen Unterricht. Vom Rezipieren zum Interagieren
Unterricht Biologie, 475 :44-47
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### 5627.
[german] Grandrath, Rebecca; Bohrmann-Linde, Claudia
Strom aus Bäckerhefe
Nachrichten aus der Chemie, 70 (7-8) :18-21
Juli 2022
### 5626.
[german] Zeller, Diana
Medialab – ein dreistufiges Modul zur Entwicklung digitalisierungsbezogener Kompetenzen im Studium des Chemie‐ und Sachunterrichtslehramts
CHEMKON, 29 (S1) :287-292
Juni 2022
### 5625.
[german] Grandrath, Rebecca; Bohrmann-Linde, Claudia
Entwicklung eines lowcost Experiments für den Chemieunterricht am Beispiel der enzymatischen Brennstoffzelle mit Lactase
CHEMKON, 29 (S1) :233-238
Juni 2022
### 5624.
[english] Bohrmann-Linde, Claudia; Zeller, Diana; Meuter, Nico; Tausch, Michael W.
Teaching Photochemistry: Experimental Approaches and Digital Media
ChemPhotoChem, 6 (6) :1-11
Juni 2022
### 5623.
[german] Zeller, Diana; Meier, Monique
Videos interaktiv erweitern - Forschendes Lernen vielseitig unterstützen
Digital Unterricht Biologie, 4 :10-11
Mai 2022
### 5622.
[german] Gökkus, Yasemin; Tausch, Michael W.
Explorative Studie zur partizipativen und nutzenorientierten Forschung in der Chemiedidaktik
CHEMKON, 29 (3) :117-124
April 2022
### 5621.
Frommer, Andreas; Kahl, Karsten; Schweitzer, Marcel; Tsolakis, Manuel
Krylov subspace restarting for matrix Laplace transforms
2022
### 5620.
Gerlach, Moritz; Glück, Jochen
On characteristics of the range of integral operators
2022
### 5619.
Bolten, M.; De Sturler, E.; Hahn, C.
Krylov Subspace Recycling for Evolving Structures
Comput. Methods Appl. Mech. Engrg., 391 :114222
2022
### 5618.
Klamroth, Kathrin; Stiglmayr, Michael; Sudhoff, Julia
Ordinal Optimization Through Multi-objective Reformulation
math.OC, arXiv:2204.02003
2022
Herausgeber: arXiv
### 5617.
Jacob, Birgit; Morris, Kirsten
On solvability of dissipative partial differential-algebraic equations
IEEE Control. Syst. Lett., 6 :3188-3193
2022
### 5616.
Farkas, Bálint; Jacob, Birgit; Schmitz, Merlin
On exponential splitting methods for semilinear abstract Cauchy problems
2022
### 5615.
Botchev, M. A.; Knizhnerman, L. A.; Schweitzer, M.
Krylov subspace residual and restarting for certain second order differential equations
2022
### 5614.
Klamroth, Kathrin; Stiglmayr, Michael; Sudhoff, Julia
Multi-objective Matroid Optimization with Ordinal Weights
Discrete Applied Mathematics
2022
### 5613.
Frommer, Andreas; Kahl, Karsten; Schweitzer, Marcel; Tsolakis, Manuel
Krylov subspace restarting for matrix Laplace transforms
2022
### 5612.
Botchev, M. A.; Knizhnerman, L. A.; Schweitzer, M.
Krylov subspace residual and restarting for certain second order differential equations
2022
### 5611.
Daners, Daniel; Glück, Jochen; Mui, Jonathan
Local uniform convergence and eventual positivity of solutions to biharmonic heat equations
2022
### 5610.
[german] Tausch, Michael W.
LED statt Gasbrenner - Mehr Licht für nachhaltigen Chemieunterricht
Chemie in unserer Zeit, 56 (3/2022) :188–196
2022
### 5609.
Frommer, Andreas; Kahl, Karsten; Schweitzer, Marcel; Tsolakis, Manuel
Krylov subspace restarting for matrix Laplace transforms
2022
### 5608.
[german] Banerji, Amitabh; Dörschelln, Jennifer; Schwarz, D.
Organische Leuchtdioden im Chemieunterricht
Chemie in unserer Zeit, 52 (1) :34-41
2022
### 5607.
[german] Zeller, Diana; Bohrmann-Linde, Claudia
#debunk YouTube-Videos - Ein didaktisches Konzept zum Einsatz von Videos im Chemieunterricht zur Stärkung der Digital Scientific Literacy
MNU journal, 75 (03) :197-201
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### 5606.
Ballaschk, Frederic; Kirsch, Stefan F.
Oxidations with Iodine(V) Compounds – From Stoichiometric Compounds to Catalysts
In Ishihara, Kazuaki and Muñiz, Kilian, Editor, Iodine Catalysis in Organic Synthesis
Seite 299–334
Herausgeber: Wiley
1 Edition
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299–334
### 5605.
Teng, L.; Ehrhardt, M.; Günther, M.
Stochastic Correlation: Modelling, Analysis and Numerical Simulation with Applications in Finance
Herausgeber: World Scientific
Weitere Infos über #UniWuppertal: | 1,818 | 6,472 | {"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} | 2.65625 | 3 | CC-MAIN-2022-33 | latest | en | 0.899016 |
https://frequentlyasked.net/what-happens-when-you-scratch-on-the-9-ball/ | 1,653,589,215,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662619221.81/warc/CC-MAIN-20220526162749-20220526192749-00276.warc.gz | 335,077,869 | 18,066 | # What happens when you scratch on the 9 ball?
The 9-ball is pocketed. This is a win unless the player scratches, in which case the 9-ball (any other available high numbered ball is adequate) is spotted and the turn passes to the opponent with ball-in-hand anywhere on the table. d.
## What happens if you sink the 9 ball on a break?
If a player sinks the 9 Ball on their break, they will automatically win. This is referred to as the “Golden Break.” If both the 9 Ball and the Cue Ball are sunk on break, the 9 Ball is spotted back to its original position and the other player is given Ball in Hand.
## Can you combo 9 ball?
You can combo into higher number balls, including the 9 ball, if you contact the lowest number ball first. For example, say you’re shooting for the 5 ball, but you see that you could “combo” the 5 into the 9 and pocket the 9. You cannot shoot directly at the nine ball until it is the only ball remaining on the table.
## What is a foul in 9 ball?
If no object ball is pocketed, failure to drive the cue ball or any numbered ball to a rail after the cue ball contacts the object ball on is a foul. 9.
## Is it ball in hand when you scratch in 9-ball?
The exception concerning scratching on the break does not apply to 9-Ball. Scratching on the break is ball- in-hand anywhere just as other fouls. 2. The 9-ball does not have to be called; therefore, the pocket the 9-ball is going to be pocketed in does not have to be called.
## Can you move the cue ball away from the bumper?
yes. If the ball needs to be repositioned (for instance after a foul) or sometimes in nine ball when the “ball in hand’ rule is in play. you can but unless it is the break, or you have ball in hand, because your opponent fouled. you will be giving your opponent ball in hand.
## How do you play cut throat?
Cut throat is a casual pool game played with three people. Each person is assigned five balls to protect (1 through 5, 6 through 10, or 11 through 15). The objective of the game is to pocket your opponents’ balls. The last person with one or more balls remaining on the table wins the game.
## What is an illegal break in 9 ball?
You must either pocket a ball or cause at least four object balls to contact the cushions, or it is an illegal break. If you pocket a ball you continue to shoot; if you do not pocket a ball or if you commit a foul, your inning ends.
## Can you put the cue ball anywhere after a scratch?
A scratch, foul, or illegal shot results in ball-in-hand, where your opponent can place the cue ball anywhere on the table in preparation for the next shot. The only exception to this is the break (see the 8-ball and 9-ball differences that follow).
## Where do you put the ball after scratching?
Tournament rules are “ball in hand”, which means that if you scratch the ball your opponent can place the cue ball anywhere on the table, without any restrictions. By contrast, bar rules require you to put the cue ball in the “kitchen”, which is the area behind the head spot where you break.
## Can you move the white ball if you have 2 shots?
When a player gets two shots, they can pick up the white ball and place it anywhere behind the line. 6.
## Can you intentionally scratch in pool?
A foul is called only if the player fouls while actually stroking the cue ball, meaning a double hit of the cue ball (sometimes called double clutching). Without this rule, a player could benefit by accidentally or purposely scratching or fouling.
## What happens if you scratch in 8ball?
If you pocket the 8 ball and scratch, you lose the game. Regardless of whether or not you hit the 8 ball with the cue ball, if you scratch, the opponent has ball-in-hand. Most places it’s a loss. Some places say the 8-ball must be made “clean”, yet the same place will allow a win on an 8 made on the break.
## How do you rack cutthroat?
The rack in Cutthroat is simple. Just use a standard triangle rack and place the 1-ball at the top of your rack (the apex) so it sits on the foot spot. Place the 6 and 11 balls in the corners of the triangle. The rest of the balls can be randomly placed in the rack.
## What does it mean to be a cutthroat?
1 : killer, murderer. 2 : a cruel unprincipled person. cutthroat. adjective. Definition of cutthroat (Entry 2 of 2)
## How do you assign balls on cutthroat?
When racking the fifteen balls for cutthroat, the 1 ball is placed on the foot spot and the 6 ball and 11 ball are placed on the other two corners of the billiards rack. In cutthroat, there are (usually) three sets of balls. Balls 1–5 are called the “low” balls, the “mid” balls are 6–10, and the “high” balls are 11–15.
## How do you stack in 9-ball pool?
Arrange them in a diamond shape, with the 9-ball in the center and the 1-ball closest to the shooter. Place balls in the rest of the diamond at random. The rack goes in the usual place, with the closest point on top of the foot spot. If you can find a diamond-shaped 9-ball rack, use it.
## When you scratch in pool Do you take the ball out?
Any scratch on a players turn, will end their turn as well as give them a one ball penalty. This means if a player has 3 balls in his pocket and he accidentally scratches, then one of his balls will come back out onto the table and have to be re-pocketed.
## What side do you shoot from when you scratch in pool?
Also known as gameplay scratches, these will often result in your opponent getting an immediate ball in hand. However, in some rule variations, your opponent will instead get to shoot from behind the headstring. These scratch rules are designed to prevent players from scratching intentionally during the game.
## What happens when white ball goes in pocket?
In pool, the white ball used to drive the other balls into the pockets is called the cue ball. When the cue ball goes into the pocket during your shot, this is considered a foul no matter what rules you’re playing by. Bar Rules: Your opponent can place the cue ball anywhere they would like behind the head string.
## Can you put the white ball anywhere in pool?
The white ball should be placed anywhere behind the service line on the table. If it is the first game in a match, a coin should be tossed to decide who gets to choose whether to break. After that, the break is taken in turns. | 1,483 | 6,309 | {"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} | 3.015625 | 3 | CC-MAIN-2022-21 | latest | en | 0.95045 |
http://www.jiskha.com/display.cgi?id=1382035799 | 1,493,263,863,000,000,000 | text/html | crawl-data/CC-MAIN-2017-17/segments/1492917121865.67/warc/CC-MAIN-20170423031201-00278-ip-10-145-167-34.ec2.internal.warc.gz | 564,958,834 | 3,943 | posted by on .
A man of mass 70 kg is in free fall in the air. He reaches a terminal speed of about v_term = 54 m/s. (Note that v_term >> v_crit and we are in the pressure dominated regime where F_fric is directly propotional to v^2 ; we have g = 9.8 m/s )
What is the force (in N) due to air-drag on the man?
at a constant speed, there is no acceleration, so no net force. The upward force = weight = 9.8*70 | 119 | 411 | {"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} | 2.84375 | 3 | CC-MAIN-2017-17 | latest | en | 0.943017 |
http://reference.wolfram.com/mathematica/ref/InverseZTransform.html | 1,398,048,568,000,000,000 | text/html | crawl-data/CC-MAIN-2014-15/segments/1397609539447.23/warc/CC-MAIN-20140416005219-00530-ip-10-147-4-33.ec2.internal.warc.gz | 200,503,934 | 7,582 | BUILT-IN MATHEMATICA SYMBOL
# InverseZTransform
InverseZTransform[expr, z, n]
gives the inverse Z transform of expr.
InverseZTransform[expr, {z1, z2, ...}, {n1, n2, ...}]
gives the multiple inverse Z transform of expr.
## Details and OptionsDetails and Options
• The inverse Z transform of a function is given by the contour integral .
• The multidimensional inverse Z transform is given by .
• The following options can be given:
• Assumptions \$Assumptions assumptions to make about parameters Method Automatic method to use
• In TraditionalForm, InverseZTransform is output using .
## ExamplesExamplesopen allclose all
### Basic Examples (2)Basic Examples (2)
Univariate inverse transforms:
Out[1]=
Out[2]=
Multivariate inverse transforms:
Out[1]=
Out[2]= | 194 | 774 | {"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} | 2.6875 | 3 | CC-MAIN-2014-15 | latest | en | 0.511128 |
https://www.physicsforums.com/threads/prove-that-a-sequence-is-eventually-decreasing.377837/ | 1,511,215,322,000,000,000 | text/html | crawl-data/CC-MAIN-2017-47/segments/1510934806225.78/warc/CC-MAIN-20171120203833-20171120223833-00773.warc.gz | 887,200,350 | 15,460 | # Prove that a sequence is eventually decreasing.
1. Feb 12, 2010
### The_Iceflash
1. The problem statement, all variables and given/known data
Prove that if a > 0, the sequence $$\frac{a^{n}}{n!}$$ is eventually decreasing for large n.
2. Relevant equations
N/A
3. The attempt at a solution
I know I'm to prove this by finding the index but I'm at a loss on how to find the index on such a sequence as I'm not used to using a variable that's in the sequence.
To warmup I was given a values 110 and 300 to find an index for but with 'a' being in the sequence itself I'm not sure how to do that.
2. Feb 12, 2010
### Gib Z
Try showing that $n! - a^n$ is eventually increasing.
3. Feb 13, 2010
### HallsofIvy
Staff Emeritus
The numerator and denominator are products of n terms but each term in the numerator is a while the terms in the denominator are increasing. The "average" value (geometric average since we are multiplying) in the denominator is $\sqrt[n]{n!}$. Can you show that, for fixed a, there exist an integer, n, such that $\sqrt[n]{n!}> a$?
4. Feb 14, 2010
### The_Iceflash
I don't really follow.
5. Feb 14, 2010
### rsa58
try subtracting the n+1 term of the sequence from the nth term. if this quantitiy is less that zero the the sequence is decreasing. however you can factor out a to the power n. this is greater than zero so it doesn't matter. look at the other factor, if you choose a <N+1 you get the desired result. | 405 | 1,454 | {"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} | 3.296875 | 3 | CC-MAIN-2017-47 | longest | en | 0.942461 |
https://www.cravencountryjamboree.com/personal-blog/what-does-cv-mean-for-a-valve/ | 1,701,492,346,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100327.70/warc/CC-MAIN-20231202042052-20231202072052-00163.warc.gz | 801,737,392 | 14,156 | # What does CV mean for a valve?
## What does CV mean for a valve?
Valve Flow Coefficient
What is a CV value?
What is Cv value? The valve flow coefficient according to the JIS standard, represents the flow capacity in US gallons/minute of 60 F pure water when it is flowing through the valve with a pressure difference of 1 psi at the specified travel (operation range).
### What is KV in control valve?
Kv-value The Kv-value is a parameter defining the flow rate of valves. It describes the amount of water from 5° to 30°C which flows through the valve at a pressure loss of 1 bar. The Kvs-value describes the Kv-value when the valve is 100% open.
What is the difference between KV and KVS?
The Kv value expresses the amount of flow in a regulating valve at a given valve position with a pressure loss of 1 bar. The Kvs value expresses the amount of flow in a regulating valve at a fully-open valve position and a pressure differential of 1 bar. …
#### How do you calculate valve Kv?
Flow Calculation – (Kv)Kv = m3/h – Flow coefficient.Q = m3/h – Flow.Qn = m3n/h – Normal flow (20°C 760mm Hg)P1 = bar – Inlet pressure – (Gauge pressure + 1)P2 = bar – Outlet pressure – (Gauge pressure – 1)DP = bar – Pressure drop – (Differential pressure between inlet and outlet pressure)
What is nominal valve size?
The nominal diameter is referred to as the pipe diameter or the mating dimension of a valve (with valves this is usually referred to as the nominal size). | 355 | 1,469 | {"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} | 2.921875 | 3 | CC-MAIN-2023-50 | latest | en | 0.90095 |
https://github.com/torch/torch.github.io/blob/master/docs/00-five-simple-examples.md | 1,537,846,322,000,000,000 | text/html | crawl-data/CC-MAIN-2018-39/segments/1537267160923.61/warc/CC-MAIN-20180925024239-20180925044639-00332.warc.gz | 540,189,148 | 16,635 | # torch/torch.github.io
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five-simple-examples
Five simple examples
docs
/docs/five-simple-examples.html
getting-started.html
Here are five simple hands-on steps, to get started with Torch! This tutorial supposes the package `torch` is already required via
`require 'torch'`
or that you are using the REPL `th` (which requires it automatically).
## 1. Define a positive definite quadratic form
We rely on a few torch functions here:
• `rand()` which creates tensor drawn from uniform distribution
• `t()` which transposes a tensor (note it returns a new view)
• `dot()` which performs a dot product between two tensors
• `eye()` which returns a identity matrix
• `*` operator over matrices (which performs a matrix-vector or matrix-matrix multiplication)
We first make sure the random seed is the same for everyone
`torch.manualSeed(1234)`
```-- choose a dimension
N = 5
-- create a random NxN matrix
A = torch.rand(N, N)
-- make it symmetric positive
A = A*A:t()
-- make it definite
b = torch.rand(N)
function J(x)
return 0.5*x:dot(A*x)-b:dot(x)
end```
Printing the function value (here on a random point) can be easily done with:
`print(J(torch.rand(N)))`
## 2. Find the exact minimum
We can inverse the matrix (which might not be numerically optimal)
```xs = torch.inverse(A)*b
print(string.format('J(x^*) = %g', J(xs)))```
## 3. Search the minimum by gradient descent
We first define the gradient w.r.t. `x` of `J(x)`:
```function dJ(x)
return A*x-b
end```
We then define some current solution:
`x = torch.rand(N)`
And then apply gradient descent (with a given learning rate `lr`) for a while:
```lr = 0.01
for i=1,20000 do
x = x - dJ(x)*lr
-- we print the value of the objective function at each iteration
print(string.format('at iter %d J(x) = %f', i, J(x)))
end```
You should see
``````...
at iter 19995 J(x) = -3.135664
at iter 19996 J(x) = -3.135664
at iter 19997 J(x) = -3.135665
at iter 19998 J(x) = -3.135665
at iter 19999 J(x) = -3.135665
at iter 20000 J(x) = -3.135666
``````
## 4. Using the optim package
Want to use more advanced optimization techniques, like conjugate gradient or LBFGS? The `optim` package is there for that purpose! First, we need to install it:
`luarocks install optim`
#### A word on local variables
In practice, it is never a good idea to use global variables. Use `local` at everywhere. In our examples, we have defined everything in global, such that they can be cut-and-pasted in the interpreter command line. Indeed, defining a local like:
`local A = torch.rand(N, N)`
will be only available to the current scope, which, when running the interpreter, is limited to the current input line. Subsequent lines would not have access to this local.
In lua one can define a scope with the `do...end` directives:
```do
local A = torch.rand(N, N)
print(A)
end
print(A)```
If you cut-and-paste this in the command line, the first print will be a 5x5 matrix (because the local `A` is defined for the duration of the scope `do...end`), but will be `nil` afterwards.
#### Defining a closure with an upvalue
We need to define a closure which returns both `J(x)` and `dJ(x)`. Here we define a scope with `do...end`, such that the local variable `neval` is an upvalue to `JdJ(x)`: only `JdJ(x)` will be aware of it. Note that in a script, one would not need to have the `do...end` scope, as the scope of `neval` would be until the end of the script file (and not the end of the line like the command line).
```do
local neval = 0
function JdJ(x)
local Jx = J(x)
neval = neval + 1
print(string.format('after %d evaluations J(x) = %f', neval, Jx))
return Jx, dJ(x)
end
end```
#### Training with optim
The package is not loaded by default, so let's require it:
`require 'optim'`
We first define a state for conjugate gradient:
```state = {
verbose = true,
maxIter = 100
}```
and now we train:
```x = torch.rand(N)
optim.cg(JdJ, x, state)```
You should see something like:
``````after 120 evaluation J(x) = -3.136835
after 121 evaluation J(x) = -3.136836
after 122 evaluation J(x) = -3.136837
after 123 evaluation J(x) = -3.136838
after 124 evaluation J(x) = -3.136840
after 125 evaluation J(x) = -3.136838
``````
## 5. Plot
Plotting can be achieved in various ways. For example, one could use the recent iTorch package. Here, we are going to use `gnuplot`.
`luarocks install gnuplot`
### Store intermediate function evaluations
We modify slightly the closure we had previously, such that it stores intermediate function evaluations (as well as the real time it took to train so far):
```evaluations = {}
time = {}
timer = torch.Timer()
neval = 0
function JdJ(x)
local Jx = J(x)
neval = neval + 1
print(string.format('after %d evaluations, J(x) = %f', neval, Jx))
table.insert(evaluations, Jx)
table.insert(time, timer:time().real)
return Jx, dJ(x)
end```
Now we can train it:
```state = {
verbose = true,
maxIter = 100
}
x0 = torch.rand(N)
cgx = x0:clone() -- make a copy of x0
timer:reset()
optim.cg(JdJ, cgx, state)
-- we convert the evaluations and time tables to tensors for plotting:
cgtime = torch.Tensor(time)
cgevaluations = torch.Tensor(evaluations)```
Let's add the training with stochastic gradient, using `optim`:
```evaluations = {}
time = {}
neval = 0
state = {
lr = 0.1
}
-- we start from the same starting point than for CG
x = x0:clone()
-- reset the timer!
timer:reset()
-- note that SGD optimizer requires us to do the loop
for i=1,1000 do
optim.sgd(JdJ, x, state)
table.insert(evaluations, Jx)
end
sgdtime = torch.Tensor(time)
sgdevaluations = torch.Tensor(evaluations)```
### Final plot
We can now plot our graphs. A first simple approach is to use `gnuplot.plot(x, y)`. Here we precede it with `gnuplot.figure()` to make sure plots are on different figures.
`require 'gnuplot'`
```gnuplot.figure(1)
gnuplot.title('CG loss minimisation over time')
gnuplot.plot(cgtime, cgevaluations)
gnuplot.figure(2)
gnuplot.title('SGD loss minimisation over time')
gnuplot.plot(sgdtime, sgdevaluations)```
A more advanced way, which plots everything on the same graph would be the following. Here we save everything in a PNG file.
```gnuplot.pngfigure('plot.png')
gnuplot.plot(
{'CG', cgtime, cgevaluations, '-'},
{'SGD', sgdtime, sgdevaluations, '-'})
gnuplot.xlabel('time (s)')
gnuplot.ylabel('J(x)')
gnuplot.plotflush()``` | 1,847 | 6,459 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.625 | 4 | CC-MAIN-2018-39 | latest | en | 0.719035 |
http://www.mywordsolution.com/question/write-down-any-four-properties-of-the-good/912597 | 1,490,569,389,000,000,000 | text/html | crawl-data/CC-MAIN-2017-13/segments/1490218189313.82/warc/CC-MAIN-20170322212949-00653-ip-10-233-31-227.ec2.internal.warc.gz | 622,130,066 | 8,507 | +1-415-315-9853
info@mywordsolution.com
## Engineering
Civil Engineering Chemical Engineering Electrical & Electronics Mechanical Engineering Computer Engineering Engineering Mathematics MATLAB Other Engineering Digital Electronics Biochemical & Biotechnology
1) describe the function of economizer?
2) describe what is meant by the water hammer?
3) Sketch the p-V diagram of the two stage reciprocating air compressor with the intercooler?
4) describe the benefits of the multi stage compression with the intercooler over the single stage compression for same pressure ratio?
5) prepare down any four properties of the good refrigerant?
6) describe the term relative humidity?
7) describe about the S.I and C.I engines.
8) describe what is meant by the compensation in the carburetor?
9) describe the Fourier’s law of heat Conduction.
10) Specify and describe Planck’s Distribution law?
Mechanical Engineering, Engineering
• Category:- Mechanical Engineering
• Reference No.:- M912597
Have any Question?
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Describe what you learned about the impact of economic, social, and demographic trends affecting the US labor environmen | 1,074 | 5,083 | {"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} | 2.6875 | 3 | CC-MAIN-2017-13 | longest | en | 0.860759 |
https://iffaustralia.com/how-to-divide-decimals | 1,709,266,692,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474948.91/warc/CC-MAIN-20240301030138-20240301060138-00770.warc.gz | 308,148,982 | 14,344 | # How to Divide Decimals: A Comprehensive Guide
Baca Cepat
## Understanding Decimal Numbers
Before we delve into the process of dividing decimals, let’s first understand what decimal numbers are. Decimal numbers are the numbers that have a decimal point, like 5.8, 3.25, or 0.73. Decimals are used to represent numbers that fall between two whole numbers. They are commonly used in everyday calculations, such as calculating prices, measurements, and percentages.
The decimal point separates the whole number part of the number from its fractional part. For example, in the number 4.25, 4 is the whole number part, and .25 is the fractional part. When we divide decimals, we divide the whole number parts and the fractional parts separately, and then combine the results.
## How to Divide Decimals Step by Step
Dividing decimals can be tricky, but once you understand the process, it becomes much simpler. Here’s a step-by-step guide on how to divide decimals:
### Step 1: Set Up the Problem
Write the problem in long division format, placing the dividend (the number being divided) inside the division bracket and the divisor (the number dividing the dividend) outside the bracket. Make sure the decimal points are in line with each other.
Dividend Divisor
25.35 5.7
### Step 2: Move the Decimal Point
Move the decimal point in the divisor to the right until it becomes a whole number. Then, move the decimal point in the dividend the same number of places to the right. This ensures that both the divisor and the dividend are whole numbers.
Dividend Divisor
253.5 57
### Step 3: Divide Whole Numbers
Divide the whole numbers just as you would with whole numbers. In our example, 253 ÷ 57 = 4.
Dividend Divisor Quotient
253.5 57 4
### Step 4: Divide Decimal Places
After dividing the whole numbers, subtract from the dividend the product of the quotient and the divisor. Write the decimal point in the quotient above the decimal point in the dividend. Then, bring down the next digit from the dividend.
Dividend Divisor Quotient
253.5 57 4.
196
In our example, 4 x 57 = 228. Subtracting 228 from 253.5 gives us 25.5. We then bring down the next digit (5) to get 255.
### Step 5: Repeat Step 4
Repeat Step 4 until you get the desired number of decimal places in the quotient. In our example, we can stop after one more iteration:
Dividend Divisor Quotient
253.5 57 4.45
196
190
4 x 57 = 228. Subtracting 228 from 253.5 gives us 25.5. We then bring down the next digit (5) to get 255. 45 x 57 = 2565. Subtracting 2565 from 2550 gives us 15. We then bring down the next digit (0) to get 150. 150 x 57 = 8550. Subtracting 8550 from 9150 gives us 600. We have now found our quotient, which is 4.45.
## Common Mistakes to Avoid When Dividing Decimals
Dividing decimals can be challenging, but with practice, it becomes more comfortable. Here are some common mistakes to avoid when dividing decimals:
### Dividing in the Wrong Order
When dividing decimals, it’s essential to divide in the right order. Start by dividing the whole numbers, then follow the steps to divide the decimal places. Failing to follow this order can lead to incorrect results.
### Forgetting to Line Up the Decimal Points
When dividing decimals, it’s crucial to line up the decimal points in the divisor and the dividend. Failing to do so can lead to incorrect results.
### Dividing by Zero
Dividing by zero is undefined, so make sure to avoid it when dividing decimals.
### Q: What do you do when the divisor is a decimal?
A: To divide decimals when the divisor is a decimal, move the decimal point in both the divisor and dividend to the right until the divisor becomes a whole number. Then, divide as usual.
### Q: What do you do when the dividend is smaller than the divisor?
A: When the dividend is smaller than the divisor, the quotient will be less than one. Add a zero after the decimal point in the dividend and continue to divide until you reach the desired number of decimal places in the quotient.
### Q: What is the shortcut to dividing decimals?
A: There is no shortcut to dividing decimals. The process must be followed precisely to get accurate results.
### Q: Can you round the quotient when dividing decimals?
A: Yes, you can round the quotient when dividing decimals. However, you must follow the rounding rules for decimals.
### Q: Can long division be used to divide decimals?
A: Yes, long division can be used to divide decimals. The same steps apply, but you will have to deal with decimal points instead of whole numbers.
### Q: Is it necessary to align the decimal points when dividing decimals?
A: Yes, aligning the decimal points is necessary when dividing decimals. This ensures that the place values match up correctly.
### Q: Can the dividend and divisor have different numbers of decimal places?
A: Yes, the dividend and divisor can have different numbers of decimal places. To divide decimals, move the decimal point in the divisor and the dividend until the divisor becomes a whole number.
## Conclusion
Dividing decimals can be challenging, but with practice and patience, it becomes more comfortable. Remember to follow the steps precisely and avoid common mistakes. Dividing decimals is a crucial skill in many real-world applications, so it’s essential to learn it well.
Now that you know how to divide decimals, try practicing with different problems to improve your skills. With enough practice, you’ll master the art of dividing decimals in no time!
## Closing Disclaimer
This article is intended for educational purposes only. The author and publisher are not liable for any damages that may result from the use or misuse of this information. Always consult a professional before making any financial, legal or medical decisions. | 1,306 | 5,788 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.84375 | 5 | CC-MAIN-2024-10 | longest | en | 0.819779 |
http://slideplayer.com/slide/4397149/ | 1,524,196,575,000,000,000 | text/html | crawl-data/CC-MAIN-2018-17/segments/1524125937113.3/warc/CC-MAIN-20180420022906-20180420042906-00271.warc.gz | 312,610,713 | 20,929 | # Geography Revision Notes Mapping Mind maps, practice exercises, key points to assist revision Recommended links to useful websites.
## Presentation on theme: "Geography Revision Notes Mapping Mind maps, practice exercises, key points to assist revision Recommended links to useful websites."— Presentation transcript:
Geography Revision Notes Mapping Mind maps, practice exercises, key points to assist revision Recommended links to useful websites
Mapping All maps should have the following: F – Frame A – Arrow indicating direction (North) K – Key T – Title S - Scale
Types of maps Physical maps- show physical features such as mountains, rivers, oceans, lakes Political maps –show countries and cities
Map of Europe
Lines of Latitude & Longitude Lines of latitude are parallel lines that run around the Earth. The centre line is the Equator which divides the Earth into the Northern and Southern Hemispheres Lines of longitude run from north to south. The main line of longitude is called the Prime Meridian or the Greenwich Meridian and runs through Greenwich in London
Mapping Symbols Church station Caravan site camping site picnic site bus station Windmill Golf Course View Point Non-coniferous wood Public Telephone Parking National Trust Light House Nature Reserve Windpump/generator Motorway
Four figure grid references Let’s find grid square 21 31 To read a 4 figure grid reference go along to the line number of the first 2 numbers and then go up to the line with the second 2 numbers.
In order to give something a six-figure grid reference, imagine the larger square split into one hundred smaller squares. 26 25 7778 Then add numbers 1 to 10 between the main lines.
To work out the six-figure grid reference, first go along the bottom line and calculate the first three figures. 26 25 7778 1 3 2 5 4 7 6 9 8 123456789 773 253 So, the six- figure grid reference for the church is..... 773253 Next, work out the last three figures by going up the side of the square.
Try to remember this phrase: ‘Go along the corridor, then up the stairs’. Can’t remember which lines to use first? c. R. Langley
Rural & Urban Settlements Rural – green fields, open spaces, lots of vegetations, farmlands. Small villages with shops, church, post office, houses, small roads Urban – built up, lots of tall skyscrapers and buildings close together, factories, houses, busy roads, shopping malls and centres, lots of traffic and congestion, pollution,
London Docklands – a changing settlement
Just by turning off the tap while you brush your teeth in the morning and before bedtime, you can save up to 8 gallons of water! Taking a shower uses much less water than filling up a bathtub. To save even more water, keep your shower under five minutes long Washing your bike or car with a bucket and sponge instead of a hose saves a lot of water. The best time to water your garden is in the early morning or late evening when it's cool outside. Did you know that less than 1% of all the water on Earth can be used by people? The rest is salt water (the kind you find in the ocean) or is permanently frozen and we can't drink it, wash with it, or use it to water plants. As our population grows, more and more people are using up this limited resource. Therefore, it is important that we use our water wisely and not waste it. WATER
Mountain Ranges of the world
The surface of our world is made up of huge plates of rock. These tectonic plates slowly move against each other… When two plates push against each other, earth and rocks are gradually raised up to make a range (or group) of mountains. Making mountains Plates pushing towards one another Land is pushed up between the plates and forms mountains
The lava rushes out (or erupts). Then it cools down to make solid rock. Over time, the rock builds up in huge piles to make a special kind of mountain called a volcano. A volcano that can still erupt is active. A volcano that has stopped erupting is extinct. Making mountains
The higher you go, the colder the air gets. And different plants and animals do well at different temperatures… Mountain weather & visitors Snowline – no plants can survive at this height. Only birds such as eagles at this level Warmest temperatures at base of mountain – lots of green plants, trees, animals & birds Temperature decreases as we go higher, fewer plants mostly heathers & moss, fewer animals
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Similar presentations | 970 | 4,567 | {"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} | 3.09375 | 3 | CC-MAIN-2018-17 | latest | en | 0.848031 |
https://hextobinary.com/unit/angularacc/from/rpw2/to/arcsecpmicros2 | 1,716,521,172,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058677.90/warc/CC-MAIN-20240524025815-20240524055815-00387.warc.gz | 250,280,023 | 17,058 | # Revolution/Square Week to Arcsec/Square Microsecond Converter
Angular Acceleration
Revolution/Square Week
Arcsec/Square Microsecond
1 Revolution/Square Week = 3.5430839002268e-18 Arcsec/Square Microsecond
## How many Arcsec/Square Microsecond are in a Revolution/Square Week?
The answer is one Revolution/Square Week is equal to 3.5430839002268e-18 Arcsec/Square Microsecond and that means we can also write it as 1 Revolution/Square Week = 3.5430839002268e-18 Arcsec/Square Microsecond. Feel free to use our online unit conversion calculator to convert the unit from Revolution/Square Week to Arcsec/Square Microsecond. Just simply enter value 1 in Revolution/Square Week and see the result in Arcsec/Square Microsecond.
## How to Convert Revolution/Square Week to Arcsec/Square Microsecond (r/week2 to arcsec/μs2)
By using our Revolution/Square Week to Arcsec/Square Microsecond conversion tool, you know that one Revolution/Square Week is equivalent to 3.5430839002268e-18 Arcsec/Square Microsecond. Hence, to convert Revolution/Square Week to Arcsec/Square Microsecond, we just need to multiply the number by 3.5430839002268e-18. We are going to use very simple Revolution/Square Week to Arcsec/Square Microsecond conversion formula for that. Pleas see the calculation example given below.
$$\text{1 Revolution/Square Week} = 1 \times 3.5430839002268e-18 = \text{3.5430839002268e-18 Arcsec/Square Microsecond}$$
## What is Revolution/Square Week Unit of Measure?
Revolution per square week is a unit of measurement for angular acceleration. By definition, if an object accelerates at one revolution per square week, its angular velocity is increasing by one revolution per week every week.
## What is the symbol of Revolution/Square Week?
The symbol of Revolution/Square Week is r/week2. This means you can also write one Revolution/Square Week as 1 r/week2.
## What is Arcsec/Square Microsecond Unit of Measure?
Arcsec per square microsecond is a unit of measurement for angular acceleration. By definition, if an object accelerates at one arcsec per square microsecond, its angular velocity is increasing by one arcsec per microsecond every microsecond.
## What is the symbol of Arcsec/Square Microsecond?
The symbol of Arcsec/Square Microsecond is arcsec/μs2. This means you can also write one Arcsec/Square Microsecond as 1 arcsec/μs2.
## Revolution/Square Week to Arcsec/Square Microsecond Conversion Table
Revolution/Square Week [r/week2]Arcsec/Square Microsecond [arcsec/μs2]
13.5430839002268e-18
27.0861678004535e-18
31.062925170068e-17
41.4172335600907e-17
51.7715419501134e-17
62.1258503401361e-17
72.4801587301587e-17
82.8344671201814e-17
93.1887755102041e-17
103.5430839002268e-17
1003.5430839002268e-16
10003.5430839002268e-15
## Revolution/Square Week to Other Units Conversion Table
Revolution/Square Week [r/week2]Output
1 revolution/square week in degree/square second is equal to9.8418997228521e-10
1 revolution/square week in degree/square millisecond is equal to9.8418997228521e-16
1 revolution/square week in degree/square microsecond is equal to9.8418997228521e-22
1 revolution/square week in degree/square nanosecond is equal to9.8418997228521e-28
1 revolution/square week in degree/square minute is equal to0.0000035430839002268
1 revolution/square week in degree/square hour is equal to0.012755102040816
1 revolution/square week in degree/square day is equal to7.35
1 revolution/square week in degree/square week is equal to360
1 revolution/square week in degree/square month is equal to6806.51
1 revolution/square week in degree/square year is equal to980137.19
1 revolution/square week in radian/square second is equal to1.7177355481489e-11
1 revolution/square week in radian/square millisecond is equal to1.7177355481489e-17
1 revolution/square week in radian/square microsecond is equal to1.7177355481489e-23
1 revolution/square week in radian/square nanosecond is equal to1.7177355481489e-29
1 revolution/square week in radian/square minute is equal to6.1838479733359e-8
1 revolution/square week in radian/square hour is equal to0.00022261852704009
1 revolution/square week in radian/square day is equal to0.12822827157509
1 revolution/square week in radian/square week is equal to6.28
1 revolution/square week in radian/square month is equal to118.8
1 revolution/square week in radian/square year is equal to17106.62
1 revolution/square week in gradian/square second is equal to1.0935444136502e-9
1 revolution/square week in gradian/square millisecond is equal to1.0935444136502e-15
1 revolution/square week in gradian/square microsecond is equal to1.0935444136502e-21
1 revolution/square week in gradian/square nanosecond is equal to1.0935444136502e-27
1 revolution/square week in gradian/square minute is equal to0.0000039367598891408
1 revolution/square week in gradian/square hour is equal to0.014172335600907
1 revolution/square week in gradian/square day is equal to8.16
1 revolution/square week in gradian/square week is equal to400
1 revolution/square week in gradian/square month is equal to7562.79
1 revolution/square week in gradian/square year is equal to1089041.33
1 revolution/square week in arcmin/square second is equal to5.9051398337113e-8
1 revolution/square week in arcmin/square millisecond is equal to5.9051398337113e-14
1 revolution/square week in arcmin/square microsecond is equal to5.9051398337113e-20
1 revolution/square week in arcmin/square nanosecond is equal to5.9051398337113e-26
1 revolution/square week in arcmin/square minute is equal to0.00021258503401361
1 revolution/square week in arcmin/square hour is equal to0.76530612244898
1 revolution/square week in arcmin/square day is equal to440.82
1 revolution/square week in arcmin/square week is equal to21600
1 revolution/square week in arcmin/square month is equal to408390.5
1 revolution/square week in arcmin/square year is equal to58808231.63
1 revolution/square week in arcsec/square second is equal to0.0000035430839002268
1 revolution/square week in arcsec/square millisecond is equal to3.5430839002268e-12
1 revolution/square week in arcsec/square microsecond is equal to3.5430839002268e-18
1 revolution/square week in arcsec/square nanosecond is equal to3.5430839002268e-24
1 revolution/square week in arcsec/square minute is equal to0.012755102040816
1 revolution/square week in arcsec/square hour is equal to45.92
1 revolution/square week in arcsec/square day is equal to26448.98
1 revolution/square week in arcsec/square week is equal to1296000
1 revolution/square week in arcsec/square month is equal to24503429.85
1 revolution/square week in arcsec/square year is equal to3528493897.96
1 revolution/square week in sign/square second is equal to3.2806332409507e-11
1 revolution/square week in sign/square millisecond is equal to3.2806332409507e-17
1 revolution/square week in sign/square microsecond is equal to3.2806332409507e-23
1 revolution/square week in sign/square nanosecond is equal to3.2806332409507e-29
1 revolution/square week in sign/square minute is equal to1.1810279667423e-7
1 revolution/square week in sign/square hour is equal to0.00042517006802721
1 revolution/square week in sign/square day is equal to0.24489795918367
1 revolution/square week in sign/square week is equal to12
1 revolution/square week in sign/square month is equal to226.88
1 revolution/square week in sign/square year is equal to32671.24
1 revolution/square week in turn/square second is equal to2.7338610341256e-12
1 revolution/square week in turn/square millisecond is equal to2.7338610341256e-18
1 revolution/square week in turn/square microsecond is equal to2.7338610341256e-24
1 revolution/square week in turn/square nanosecond is equal to2.7338610341256e-30
1 revolution/square week in turn/square minute is equal to9.8418997228521e-9
1 revolution/square week in turn/square hour is equal to0.000035430839002268
1 revolution/square week in turn/square day is equal to0.020408163265306
1 revolution/square week in turn/square week is equal to1
1 revolution/square week in turn/square month is equal to18.91
1 revolution/square week in turn/square year is equal to2722.6
1 revolution/square week in circle/square second is equal to2.7338610341256e-12
1 revolution/square week in circle/square millisecond is equal to2.7338610341256e-18
1 revolution/square week in circle/square microsecond is equal to2.7338610341256e-24
1 revolution/square week in circle/square nanosecond is equal to2.7338610341256e-30
1 revolution/square week in circle/square minute is equal to9.8418997228521e-9
1 revolution/square week in circle/square hour is equal to0.000035430839002268
1 revolution/square week in circle/square day is equal to0.020408163265306
1 revolution/square week in circle/square week is equal to1
1 revolution/square week in circle/square month is equal to18.91
1 revolution/square week in circle/square year is equal to2722.6
1 revolution/square week in mil/square second is equal to1.7496710618404e-8
1 revolution/square week in mil/square millisecond is equal to1.7496710618404e-14
1 revolution/square week in mil/square microsecond is equal to1.7496710618404e-20
1 revolution/square week in mil/square nanosecond is equal to1.7496710618404e-26
1 revolution/square week in mil/square minute is equal to0.000062988158226253
1 revolution/square week in mil/square hour is equal to0.22675736961451
1 revolution/square week in mil/square day is equal to130.61
1 revolution/square week in mil/square week is equal to6400
1 revolution/square week in mil/square month is equal to121004.59
1 revolution/square week in mil/square year is equal to17424661.22
1 revolution/square week in revolution/square second is equal to2.7338610341256e-12
1 revolution/square week in revolution/square millisecond is equal to2.7338610341256e-18
1 revolution/square week in revolution/square microsecond is equal to2.7338610341256e-24
1 revolution/square week in revolution/square nanosecond is equal to2.7338610341256e-30
1 revolution/square week in revolution/square minute is equal to9.8418997228521e-9
1 revolution/square week in revolution/square hour is equal to0.000035430839002268
1 revolution/square week in revolution/square day is equal to0.020408163265306
1 revolution/square week in revolution/square month is equal to18.91
1 revolution/square week in revolution/square year is equal to2722.6
Disclaimer:We make a great effort in making sure that conversion is as accurate as possible, but we cannot guarantee that. Before using any of the conversion tools or data, you must validate its correctness with an authority. | 2,999 | 10,557 | {"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} | 3.171875 | 3 | CC-MAIN-2024-22 | latest | en | 0.88594 |
https://financequestions4you.com/what-is-the-average-living-wage-in-the-us/ | 1,670,464,248,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446711232.54/warc/CC-MAIN-20221208014204-20221208044204-00210.warc.gz | 282,808,001 | 19,901 | What is the average living wage in the us?
What Is The Average Living Wage In The U.S.?
The average living wage in the United States is \$16.50 per hour, or \$68,000 per year. This wage is based on the cost of living in the United States, which includes the cost of food, housing, transportation, and healthcare. The living wage is the wage that a person needs to earn to cover their basic needs.
There is no one answer to this question because the average living wage in the United States can vary depending on where you are located and what type of job you have. However, according to The Daily Review, the national average living wage was \$27.06 per hour as of September 2017. This means that a full-time worker earning the national average living wage would need to work approximately 40 hours per week to make ends meet.
10 Different Average Living Wage Figures In The US
What is a living wage in the us 2020?
According to national data compiled by MIT, the U.S. minimum is less than half the living wage for a single adult. It’s less than a third of what a family of four needs to live, or almost \$90,000 a year.
1. The average living wage in the United States is \$51,939 per year.
2. The median household income in the United States is \$56,516 per year.
3. The poverty line in the United States is \$11,880 per year for a single person and \$24,300 for a family of four.
4. The federal minimum wage in the United States is \$7.25 per hour.
5. In some states, the minimum wage is as high as \$12 per hour.
The average living wage in the United States is \$27.73 per hour, according to the Bureau of Labor Statistics. That’s a little more than the national average of \$25.19, but less than the state average of \$28.21. California has the highest living wage at \$34.13 per hour, followed by New York (\$32.05) and Hawaii (\$31.78).
There is no one answer to this question, as the living wage in the US can vary significantly depending on where you are located and what type of job you have. However, according to The Daily Review, the average hourly wage for a full-time worker in the US was \$25.53 as of September 2017. This means that if you are working 40 hours per week, you would be earning an hourly wage of \$1022.60 per month— which is over \$30,000 annually! So while it might not be possible to live off of this amount alone, it should provide a decent baseline from which to start saving or taking other steps towards achieving financial independence.
1. The average living wage in the United States is \$16.50 per hour.
2. This wage is based on a 40-hour work week and does not include benefits.
3. The cost of living varies greatly across the country, with some areas requiring a much higher wage to maintain a decent standard of living.
4. Many jobs in the United States do not pay a living wage, leaving workers struggling to make ends meet.
5. There has been recent momentum towards raising the minimum wage to a livable wage, but progress has been slow.
What is The Average Wage in the US?
What is the living wage in america 2021?
The living wage for a household is \$13.47 per hour. Two working adults are paid \$11.09 per hour, while two working adults and two children are paid \$19.60 per hour.
Average wages and salaries in the united states are offered according to multiple factors. This information can help you figure out what impacts salaries and compensation. Financial data for the year is reflected in reports by the bureau of labor and statistics.
If a person works 52 weeks in a year, they would make a national annual salary of over \$50,000. There are 117.5 million full-time workers in the united states, according to the bureau of labor statistics. The average weekly earnings for a person working in the U.S. is \$1,037. Some people are compensated with an annual salary while others work for hourly wages.
Is 16 dollars an hour a living wage?
Two people who work full-time and earn \$16 dollars an hour have a net take- home of \$4157, which is a decent amount for living well in most places!
1. What is the average living wage in the United States?
2. How does the cost of living affect the average living wage in the United States?
3. What are some of the factors that contribute to the cost of living in the United States?
4. How does the average living wage in the United States compare to other developed countries?
5. What are some ways that people can reduce their costs of living in the United States?
Which state has the highest minimum wage?
Each state has its own average salary, and geographic location and cost of living can dictate the parameters of individual compensation and wages. The five states with the highest average salaries are Massachusetts, New York, California, and Washington.
1. The living wage in the United States is \$15.77 per hour.2. The average living wage varies by state, but is generally higher in states with higher costs of living.3. The living wage is an important measure of workplace and economic justice, and helps to ensure that all workers can afford to live decently while working.
The living wage in maine is \$80,336.
The living wage in delaware is \$71,254.
Natural or man-made geographic regions are areas with distinct features. These areas reflect unique economic statistics. Highly populated metropolitan cities are often industry hub for jobs with high earning potential.
The level of education a person has contributes to the national average salary amounts. College graduates earn an average of \$79,300 per year, while people with a high school diploma earn an average of \$43,004. | 1,236 | 5,615 | {"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} | 2.53125 | 3 | CC-MAIN-2022-49 | latest | en | 0.968801 |
http://www.emathzone.com/tutorials/basic-statistics/definition-of-probability.html | 1,493,563,501,000,000,000 | text/html | crawl-data/CC-MAIN-2017-17/segments/1492917125654.80/warc/CC-MAIN-20170423031205-00412-ip-10-145-167-34.ec2.internal.warc.gz | 501,145,714 | 12,582 | # Definition of Probability
Probability is something strange and it has been defined in different manners. We can define probability in objective or subjective manner. Let us first use objective approach to define probability.
The Classical Definition of Probability:
This definition is for equally likely outcomes. If an experiment can produced $N$ mutually exclusive and equally likely outcomes out of which $n$ outcomes are favorable to the occurrence of event $A$, then the probability of $A$is denoted by $P\left( A \right)$and is defined as the ratio $\frac{n}{N}$. Thus the probability of $A$ is given by
This definition can be applied in a situation in which all possible outcomes and the outcomes in the events $A$ can be counted. This definition is due to P.S. Laplace (1749 – 1827). The classical definition is also called the priori definition of probability. The word priori is from prior and is used because the definition of base on the previous knowledge that the outcomes are equally likely. When a coin is tossed, the probability of head is assumed to be $\frac{1}{2}$. This probability of $\frac{1}{2}$ is based on this classical definition of probability. It is assumed that the two faces of the coin are equally likely. In practical life the people do believe that a coin will do justice when it is tossed. In the playgrounds, the participating teams toss the coin to start the match. A coin in which probability of head is assumed to be equal to the probability of tail is called a true or uniform or an unbiased coin. But it is an all assumption. The probability of a certain event is a number which lies between$0$ and$1$. If the event does not contain any outcome, it is called impossible event and its probability is zero. If the event is as big as the sample space, the probability of the event is one because
When probability of an event is one, it is called a “Sure” or “Certain”, event.
Criticism:
The classical definition of probability has always been criticized for the following reasons:
1. This definition assumes that the outcomes are equally likely. The term equally likely is almost as difficult as the word probability itself. Thus the definition uses the circular reasoning.
2. The definition is not applicable when the numbers of outcomes are not equally likely.
3. The definition is also not applicable when the total number of outcomes is infinite or it is difficult to count the total outcomes or the outcomes favorable to the event. It is difficult to count the fish in the ocean. Thus it is difficult to find the probability of catching a fish of some weight say more than one kilogram.
Example:
One day 20 files were presented to an income tax officer for disposal. Five files contained bogus entries. All the files were thoroughly mixed and there was no indication about bogus files. What is the probability that one file with bogus entries is selected.
Solution:
Here all possible outcomes $= 20$
Let $A$ be the event that the file has bogus entries.
Thus, number of favorable outcomes $= 5$
Here we shall apply the classical definition of probability. All the $20$ files are assumed to be equally likely for the purpose of selecting a file.
Probability of selecting a file with bogus entries is written as $P\left( A \right)$
Example:
A fair die is thrown. Find the probabilities that the face on the die is (1) Maximum (2) Prime (3) Multiple of $3$ (4) Multiple of $7$
Solution:
There are $6$ possible outcomes when a die is tossed. We assumed that all the $6$ faces are equally likely. The classical definition of probability is to be applied here
The sample space is $S = \left\{ {1,2,3,4,5,6} \right\}$, $n\left( S \right) = 6$
(1) Let $A$ be the event that the face is maximum
Thus,
$A = \left\{ 6 \right\}$, $n\left( A \right) = 1$
Therefore, $P\left( A \right) = \frac{{n\left( A \right)}}{{n\left( S \right)}} = \frac{1}{6}$
(2) Let $B$ be the event that the face is maximum
Thus,
$B = \left\{ {2,3,5} \right\}$, $n\left( B \right) = 3$
Therefore, $P\left( B \right) = \frac{{n\left( B \right)}}{{n\left( S \right)}} = \frac{3}{6} = \frac{1}{2}$
(3) Let $C$ be the event that the face is maximum
Thus,
$C = \left\{ {3,6} \right\}$$n\left( C \right) = 2$
Therefore, $P\left( C \right) = \frac{{n\left( C \right)}}{{n\left( S \right)}} = \frac{2}{6} = \frac{1}{3}$
(4) Let $D$ be the event that the face is maximum
Thus,
$D = \phi$, $n\left( D \right) = 0$
Therefore, $P\left( D \right) = \frac{{n\left( D \right)}}{{n\left( S \right)}} = \frac{0}{6} = 0$ (not possible) | 1,181 | 4,570 | {"found_math": true, "script_math_tex": 39, "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": 43, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2017-17 | longest | en | 0.961434 |
http://www.caraudio.com/forums/archive/index.php/t-416367.html?s=0ebed208eaf2c5430f98e82e02b732ea | 1,448,457,662,000,000,000 | text/html | crawl-data/CC-MAIN-2015-48/segments/1448398445142.9/warc/CC-MAIN-20151124205405-00104-ip-10-71-132-137.ec2.internal.warc.gz | 334,929,128 | 2,677 | PDA
View Full Version : Areo Port Placement Question
TheMan007
07-03-2009, 12:02 AM
I am trying to figure out what I need to do while build my box to put my port firing out the side of my box and my subs forward. I am building a box for my CC pickup and the subs will be forward firing and the ports will fire out the side of the box into the doors. The box needs to be 1.36 and that is including sub displacement and it needs a 3'' port tuned to 38Hz and the length needs to be 8.51''. One question is does the box need to be bigger to add the port or can I just put in the box. And does it matter if I put the port in the bottom corner of the box or does it have to be a certain place?
**Example Of How I Want My Box, Except With An Aero Port**
hockey512546
07-03-2009, 02:15 AM
you will have to account for the port displacement, though that port would only take up .0348ft^3, which probably won't matter
Did you calculate the port length with the 1.36 cubic foot box or the 1.36 - sub displacement?
If you calculated it as 1.36, redo it and use the 1.36-sub number as the box volume
TheMan007
07-03-2009, 02:39 AM
I did 1.36 that is sub displacement included, so does it matter how I put it in the box? And do I need to redo it with the port displacement added?
hockey512546
07-03-2009, 02:48 AM
I don't think it will matter where you put it in the box, I'd say as far away from the sub as possible, but let someone else answer that...
and to clarify, your box is bigger than 1.36 and you subtracted the sub displacement number from your total volume
eq: Box (1.5) - sub (.14)=1.36? like that?
I don't think you will need to subtract the port displacement, as it is really small.
TheMan007
07-03-2009, 03:01 AM
No the manual says the box needed is 1.36 and this is with the sub displacement added
hockey512546
07-03-2009, 03:35 AM
then that should be good | 527 | 1,870 | {"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} | 2.96875 | 3 | CC-MAIN-2015-48 | latest | en | 0.940751 |
https://www.shakuhachi.net/what-is-the-amp-rating-of-8-gauge-wire/ | 1,679,533,861,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296944606.5/warc/CC-MAIN-20230323003026-20230323033026-00184.warc.gz | 1,093,376,949 | 11,107 | # Shakuhachi.net
Best place for relax your brain
# What is the amp rating of 8 gauge wire?
## What is the amp rating of 8 gauge wire?
#8 Gauge THHN Copper Stranded Wire is the most popular single conductor copper wire we carry and is RATED FOR 55 AMPS @ 90°C.
Is 8 gauge wire enough for AMP?
8 gauge wire is recommended to be used with amps that produce 200 – 400 watts RMS. So for example, let’s say you have a 4 channel amp, putting out 50 watts RMS per channel. 4 x 50 = 200, so you’ll want to use 8 gauge wire with this amp.
What gauge wire is good for 50 amps?
6
For a maximum of 50 amps, you’ll need a wire gauge of 6. Fifty amp breakers are most often used to power many different appliances. However, a kitchen oven can alone require 50 amps. Many electric dryers also require a 50 amp breaker.
### What is the size of American Wire Gauge?
American Wire Gauge Conductor Size Table. American wire gauge (AWG) is a standardized wire gauge system for the diameters of round, solid, nonferrous, electrically conducting wire. The larger the AWG number or wire guage, the smaller the physical size of the wire. The smallest AWG size is 40 and the largest is 0000 (4/0).
What does AWG mean in wire gauge?
American wire gauge (AWG) is a standardized wire gauge system for the diameters of round, solid, nonferrous, electrically conducting wire. The larger the AWG number or wire guage, the smaller the physical size of the wire.
Where can I find information about DC wire size selection?
The Circuit Wizard, at circuitwizard.bluesea.com, is a resource for a more detailed treatment of wire size selection for DC circuits. It allows you to input detailed information including wire insulation temperature rating and other derating factors. The Circuit Wizard is easy to use, and is accessible from any computer with an Internet connection.
## What is the difference between 6 gauge and 3 gauge wire?
The general rule of thumb is for every 6 gauge decrease the wire diameter doubles and every 3 gauge decrease doubles the cross sectional area. Diameter Notes: A mil is a unit of length equal to 0.001 inch (a “milli-inch” or a “thousandth of one inch”) ie. 1 mil = 0.001″. | 529 | 2,186 | {"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} | 2.53125 | 3 | CC-MAIN-2023-14 | longest | en | 0.881261 |
https://socratic.org/questions/how-do-you-solve-x-5-2x-1-20 | 1,601,151,894,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400245109.69/warc/CC-MAIN-20200926200523-20200926230523-00486.warc.gz | 602,146,357 | 5,806 | # How do you solve x/5 = (2x+1)/20?
Mar 26, 2018
$x = \frac{1}{2}$
#### Explanation:
$\frac{x}{5} = \frac{2 x + 1}{20}$
= x×20=5×(2x+1)
$= 20 x = 10 x + 5$
$= 20 x - 10 x = 5$
$= 10 x = 5$
$= x = \frac{5}{10}$
$= x = \frac{1}{2}$ | 129 | 239 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 8, "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} | 4.25 | 4 | CC-MAIN-2020-40 | latest | en | 0.402912 |
https://statsmaths.github.io/stat209-s19/assets/solutions/lab08-solutions.html | 1,618,644,100,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038118762.49/warc/CC-MAIN-20210417071833-20210417101833-00078.warc.gz | 632,609,905 | 241,825 | Start by reading in all of the R packages that we will need for today:
library(dplyr)
library(ggplot2)
library(tmodels)
library(readxl)
## Class data collection
We are going to build a dataset as a class. Download the datafile and save it as “class08.csv”. Read in the dataset and call it simply class using the read_csv function:
class <- read_csv
Run all three correlation tests on the data:
Are there any large differences between the p-values in the tests? What do you conclude from the analysis about the relationship between the variables?
## Graduate pay information by major
Now, read in the pay by major dataset that we used last time:
pay <- read_csv("https://raw.githubusercontent.com/statsmaths/stat_data/gh-pages/grad_info.csv")
Run Pearson’s product-moment correlation test on this dataset with median_pay as the response and unemployment as the independent variable. Note the correlation point estimate and the p-value:
tmod_pearson_correlation_test(median_pay ~ unemployment, data = pay)
##
## Pearson's product-moment correlation test
##
## H0: True correlation is zero
## HA: True correlation is non-zero
##
## Test statistic: t(171) = -1.4317
## P-value: 0.1541
##
## Parameter: (Pearson) correlation coefficient
## Point estimate: -0.10883
## Confidence interval: [-0.253910, 0.041033]
Re-run the analysis with median_pay as the independent variable and unemployment as the response.
tmod_pearson_correlation_test(unemployment ~ median_pay, data = pay)
##
## Pearson's product-moment correlation test
##
## H0: True correlation is zero
## HA: True correlation is non-zero
##
## Test statistic: t(171) = -1.4317
## P-value: 0.1541
##
## Parameter: (Pearson) correlation coefficient
## Point estimate: -0.10883
## Confidence interval: [-0.253910, 0.041033]
The results should be exactly the same due to symmetry in the test.
## Graduate pay by major in three categories
The following code creates a new variable in the dataset. Run it and look at the dataset to see what it does
pay$mcategory <- "non-science" pay$mcategory[pay$sciences == "yes"] <- "other science" pay$mcategory[grep("ENGINEERING", pay$major)] <- "engineering" The variable mcategory now has three different values: non-science, other science, and engineering. Use the one-way ANOVA function to test a relationship between median pay and the variable mcategory: tmod_one_way_anova_test(median_pay ~ mcategory, data=pay) ## ## One-way Analysis of Variance (ANOVA) ## ## H0: True means are the same in each group. ## HA: True means are the same in each group. ## ## Test statistic: F(2, 170) = 72.95 ## P-value: < 2.2e-16 Now, run the same test with Kruskal-Wallis rank sum test: tmod_kruskal_wallis_test(median_pay ~ mcategory, data=pay) ## ## Kruskal-Wallis rank sum test ## ## H0: Location parameters are the same in each group. ## HA: Location parameters are not the same in each group. ## ## Test statistic: chi-squared(2) = 59.129 ## P-value: 1.447e-13 Finally, make a copy of the dataset and introduce a bad data point as we did last time: pay_bad <- pay pay_bad$median_pay[173] <- 22000000
Verify that the one-way ANOVA function is sensitive to this outlier but the Kruskal-Wallis test is not:
tmod_one_way_anova_test(median_pay ~ mcategory, data=pay_bad)
##
## One-way Analysis of Variance (ANOVA)
##
## H0: True means are the same in each group.
## HA: True means are the same in each group.
##
## Test statistic: F(2, 170) = 0.19071
## P-value: 0.8265
tmod_kruskal_wallis_test(median_pay ~ mcategory, data=pay_bad)
##
## Kruskal-Wallis rank sum test
##
## H0: Location parameters are the same in each group.
## HA: Location parameters are not the same in each group.
##
## Test statistic: chi-squared(2) = 56.896
## P-value: 4.418e-13 | 1,011 | 3,772 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.765625 | 3 | CC-MAIN-2021-17 | latest | en | 0.788144 |
https://handwiki.org/wiki/Astronomy:Gibbons%E2%80%93Hawking%E2%80%93York_boundary_term | 1,674,792,376,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764494936.89/warc/CC-MAIN-20230127033656-20230127063656-00629.warc.gz | 321,299,825 | 21,164 | Astronomy:Gibbons–Hawking–York boundary term
In general relativity, the Gibbons–Hawking–York boundary term is a term that needs to be added to the Einstein–Hilbert action when the underlying spacetime manifold has a boundary. The Einstein–Hilbert action is the basis for the most elementary variational principle from which the field equations of general relativity can be defined. However, the use of the Einstein–Hilbert action is appropriate only when the underlying spacetime manifold $\displaystyle{ \mathcal{M} }$ is closed, i.e., a manifold which is both compact and without boundary. In the event that the manifold has a boundary $\displaystyle{ \partial\mathcal{M} }$, the action should be supplemented by a boundary term so that the variational principle is well-defined.
The necessity of such a boundary term was first realised by York and later refined in a minor way by Gibbons and Hawking.
For a manifold that is not closed, the appropriate action is
$\displaystyle{ \mathcal{S}_\mathrm{EH} + \mathcal{S}_\mathrm{GHY} = \frac{1}{16 \pi} \int_\mathcal{M} \mathrm{d}^4 x \, \sqrt{-g} R + \frac{1}{8 \pi} \int_{\partial \mathcal{M}} \mathrm{d}^3 y \, \epsilon \sqrt{h}K, }$
where $\displaystyle{ \mathcal{S}_\mathrm{EH} }$ is the Einstein–Hilbert action, $\displaystyle{ \mathcal{S}_\mathrm{GHY} }$ is the Gibbons–Hawking–York boundary term, $\displaystyle{ h_{ab} }$ is the induced metric (see section below on definitions) on the boundary, $\displaystyle{ h }$ its determinant, $\displaystyle{ K }$ is the trace of the second fundamental form, $\displaystyle{ \epsilon }$ is equal to $\displaystyle{ +1 }$ where the normal to $\displaystyle{ \partial \mathcal{M} }$ is spacelike and $\displaystyle{ -1 }$ where the normal to $\displaystyle{ \partial \mathcal{M} }$ is timelike, and $\displaystyle{ y^a }$ are the coordinates on the boundary. Varying the action with respect to the metric $\displaystyle{ g_{\alpha\beta} }$, subject to the condition
$\displaystyle{ \delta g_{\alpha \beta} \big|_{\partial \mathcal{M}} = 0, }$
gives the Einstein equations; the addition of the boundary term means that in performing the variation, the geometry of the boundary encoded in the transverse metric $\displaystyle{ h_{ab} }$ is fixed (see section below). There remains ambiguity in the action up to an arbitrary functional of the induced metric $\displaystyle{ h_{ab} }$.
That a boundary term is needed in the gravitational case is because $\displaystyle{ R }$, the gravitational Lagrangian density, contains second derivatives of the metric tensor. This is a non-typical feature of field theories, which are usually formulated in terms of Lagrangians that involve first derivatives of fields to be varied over only.
The GHY term is desirable, as it possesses a number of other key features. When passing to the Hamiltonian formalism, it is necessary to include the GHY term in order to reproduce the correct Arnowitt–Deser–Misner energy (ADM energy). The term is required to ensure the path integral (a la Hawking) for quantum gravity has the correct composition properties. When calculating black hole entropy using the Euclidean semiclassical approach, the entire contribution comes from the GHY term. This term has had more recent applications in loop quantum gravity in calculating transition amplitudes and background-independent scattering amplitudes.
In order to determine a finite value for the action, one may have to subtract off a surface term for flat spacetime:
$\displaystyle{ S_{EH} + S_{GHY,0} = \frac{1}{16 \pi} \int_\mathcal{M} \mathrm{d}^4 x \, \sqrt{-g} R + \frac{1}{8 \pi} \int_{\partial \mathcal{M}} \mathrm{d}^3 y \, \epsilon \sqrt{h} K - {1 \over 8 \pi} \int_{\partial \mathcal{M}} \mathrm{d}^3 y \, \epsilon \sqrt{h} K_0, }$
where $\displaystyle{ K_0 }$ is the extrinsic curvature of the boundary imbedded flat spacetime. As $\displaystyle{ \sqrt{h} }$ is invariant under variations of $\displaystyle{ g_{\alpha \beta} }$, this addition term does not affect the field equations; as such, this is referred to as the non-dynamical term.
Introduction to hyper-surfaces
Defining hyper-surfaces
In a four-dimensional spacetime manifold, a hypersurface is a three-dimensional submanifold that can be either timelike, spacelike, or null.
A particular hyper-surface $\displaystyle{ \Sigma }$ can be selected either by imposing a constraint on the coordinates
$\displaystyle{ f (x^\alpha) = 0, }$
or by giving parametric equations,
$\displaystyle{ x^\alpha = x^\alpha (y^a), }$
where $\displaystyle{ y^a (a=1,2,3) }$ are coordinates intrinsic to the hyper-surface.
For example, a two-sphere in three-dimensional Euclidean space can be described either by
$\displaystyle{ f (x^\alpha) = x^2 + y^2 + z^2 - r^2 = 0, }$
where $\displaystyle{ r }$ is the radius of the sphere, or by
$\displaystyle{ x = r \sin \theta \cos \phi, \quad y = r \sin \theta \sin \phi, \quad z = r \cos \theta, }$
where $\displaystyle{ \theta }$ and $\displaystyle{ \phi }$ are intrinsic coordinates.
Hyper-surface orthogonal vector fields
We take the metric convention (-,+,...,+). We start with the family of hyper-surfaces given by
$\displaystyle{ f (x^\alpha) = C }$
where different members of the family correspond to different values of the constant $\displaystyle{ C }$. Consider two neighbouring points $\displaystyle{ P }$ and $\displaystyle{ Q }$ with coordinates $\displaystyle{ x^\alpha }$ and $\displaystyle{ x^\alpha + d x^\alpha }$, respectively, lying in the same hyper-surface. We then have to first order
$\displaystyle{ C = f (x^\alpha + d x^\alpha) = f (x^\alpha) + {\partial f \over \partial x^\alpha} d x^\alpha. }$
Subtracting off $\displaystyle{ C = f (x^\alpha) }$ from this equation gives
$\displaystyle{ {\partial f \over \partial x^\alpha} d x^\alpha = 0 }$
at $\displaystyle{ P }$. This implies that $\displaystyle{ f_{, \alpha} }$ is normal to the hyper-surface. A unit normal $\displaystyle{ n_\alpha }$ can be introduced in the case where the hyper-surface is not null. This is defined by
$\displaystyle{ n^\alpha n_\alpha \equiv \epsilon =\begin{cases} -1 & \text{if } \Sigma \text{ is spacelike} \\ +1 & \text{if } \Sigma \text{ is timelike} \end{cases} }$
and we require that $\displaystyle{ n^\alpha }$ point in the direction of increasing $\displaystyle{ f : n^\alpha f_{, \alpha} \gt 0 }$. It can then easily be checked that $\displaystyle{ n_\alpha }$ is given by
$\displaystyle{ n_\alpha = {\epsilon f_{, \alpha} \over |g^{\alpha \beta} f_{, \alpha} f_{, \beta}|^{1 \over 2}} }$
if the hyper-surface either spacelike or timelike.
Induced and transverse metric
The three vectors
$\displaystyle{ e^\alpha_a = \left( {\partial x^\alpha \over \partial y^a} \right)_{\partial \mathcal{M}} \quad a=1,2,3 }$
are tangential to the hyper-surface.
The induced metric is the three-tensor $\displaystyle{ h_{ab} }$ defined by
$\displaystyle{ h_{ab} = g_{\alpha \beta} e^\alpha_a e^\beta_b . }$
This acts as a metric tensor on the hyper-surface in the $\displaystyle{ y^a }$ coordinates. For displacements confined to the hyper-surface (so that $\displaystyle{ x^\alpha = x^\alpha (y^a) }$)
\displaystyle{ \begin{align} ds^2 &= g_{\alpha \beta} dx^\alpha dx^\beta \\ &= g_{\alpha \beta} \left(\frac{\partial x^\alpha}{\partial y^a} dy^a \right) \left(\frac{\partial x^\beta}{\partial y^b} dy^b \right) \\ &= \left( g_{\alpha \beta} e^\alpha_a e^\beta_b \right) dy^a dy^b \\ &= h_{ab} dy^a dy^b \end{align} }
Because the three vectors $\displaystyle{ e^\alpha_1, e^\alpha_2, e^\alpha_3 }$ are tangential to the hyper-surface,
$\displaystyle{ n_\alpha e^\alpha_a = 0 }$
where $\displaystyle{ n_\alpha }$ is the unit vector ($\displaystyle{ n_\alpha n^\alpha = \pm 1 }$) normal to the hyper-surface.
We introduce what is called the transverse metric
$\displaystyle{ h_{\alpha \beta} = g_{\alpha \beta} - \epsilon n_\alpha n_\beta. }$
It isolates the part of the metric that is transverse to the normal $\displaystyle{ n^\alpha }$.
It is easily seen that this four-tensor
$\displaystyle{ {h^\alpha}_{\beta} = {\delta^\alpha}_{\beta} - \epsilon n^\alpha n_\beta }$
projects out the part of a four-vector transverse to the normal $\displaystyle{ n^\alpha }$ as
$\displaystyle{ {h^\alpha}_{\beta} n^\beta = ({\delta^\alpha}_{\beta} - \epsilon n^\alpha n_\beta) n^\beta = (n^\alpha - \epsilon^2 n^\alpha) = 0 \quad \text{and } \; \mathrm{if} \quad w^\alpha n_\alpha = 0 \quad \mathrm{then} \quad {h^\alpha}_{\beta} w^\beta = w^\alpha. }$
We have
$\displaystyle{ h_{ab} = h_{\alpha \beta} e^\alpha_a e^\beta_b. }$
If we define $\displaystyle{ h^{ab} }$ to be the inverse of $\displaystyle{ h_{ab} }$, it is easy to check
$\displaystyle{ h^{\alpha \beta} = h^{ab} e^\alpha_a e^\beta_b }$
where
$\displaystyle{ h^{\alpha \beta} = g^{\alpha \beta} - \epsilon n^\alpha n^\beta. }$
Note that variation subject to the condition
$\displaystyle{ \delta g_{\alpha \beta} \big|_{\partial \mathcal{M}} = 0, }$
implies that $\displaystyle{ h_{ab} = g_{\alpha \beta} e^\alpha_a e^\beta_b }$, the induced metric on $\displaystyle{ \partial \mathcal{M} }$, is held fixed during the variation. See also [1] for clarification on $\displaystyle{ \delta h_{\alpha \beta} }$ and $\displaystyle{ \delta n_{\alpha} }$ etc.
On proving the main result
In the following subsections we will first compute the variation of the Einstein–Hilbert term and then the variation of the boundary term, and show that their sum results in
$\displaystyle{ \delta S_{TOTAL} = \delta S_{EH} + \delta S_{GHY} = \frac{1}{16 \pi} \int_\mathcal{M} G_{\alpha \beta} \delta g^{\alpha \beta} \sqrt{-g} d^4x }$
where $\displaystyle{ G_{\alpha \beta} = R_{\alpha \beta} - {1 \over 2} g_{\alpha \beta} R }$ is the Einstein tensor, which produces the correct left-hand side to the Einstein field equations, without the cosmological term, which however is trivial to include by replacing $\displaystyle{ S_{EH} }$ with
$\displaystyle{ {1 \over 16 \pi} \int_\mathcal{M} (R - 2 \Lambda) \sqrt{-g} d^4x }$
where $\displaystyle{ \Lambda }$ is the cosmological constant.
In the third subsection we elaborate on the meaning of the non-dynamical term.
Variation of the Einstein–Hilbert term
We will use the identity
$\displaystyle{ \delta \sqrt{-g} \equiv - {1 \over 2}\sqrt{-g} g_{\alpha \beta} \delta g^{\alpha \beta}, }$
and the Palatini identity:
$\displaystyle{ \delta R_{\alpha \beta} \equiv \nabla_\mu (\delta \Gamma^\mu_{\alpha \beta}) - \nabla_\beta (\delta \Gamma^\mu_{\alpha \mu}), }$
which are both obtained in the article Einstein–Hilbert action.
We consider the variation of the Einstein–Hilbert term:
\displaystyle{ \begin{align} (16 \pi) \delta S_{EH} & = \int_\mathcal{M} \delta \left ( g^{\alpha \beta} R_{\alpha \beta} \sqrt{-g} \right ) d^4x \\ & = \int_\mathcal{M} \left( R_{\alpha \beta} \sqrt{-g} \delta g^{\alpha \beta} + g^{\alpha \beta} R_{\alpha \beta} \delta \sqrt{-g} + \sqrt{-g} g^{\alpha \beta} \delta R_{\alpha \beta} \right) d^4x \\ & = \int_\mathcal{M} \left( R_{\alpha \beta} - {1 \over 2} g_{\alpha \beta} R \right ) \delta g^{\alpha \beta} \sqrt{-g} d^4x + \int_\mathcal{M} g^{\alpha \beta} \delta R_{\alpha \beta} \sqrt{-g} d^4x . \end{align} }
The first term gives us what we need for the left-hand side of the Einstein field equations. We must account for the second term.
By the Palatini identity
$\displaystyle{ g^{\alpha \beta} \delta R_{\alpha \beta} = \delta {V^\mu}_{; \mu}, \qquad \delta V^\mu = g^{\alpha \beta} \delta \Gamma^\mu_{\alpha \beta} - g^{\alpha \mu} \delta \Gamma^\beta_{\alpha \beta} . }$
We will need Stokes theorem in the form:
\displaystyle{ \begin{align} \int_\mathcal{M} {A^\mu}_{; \mu} \sqrt{-g} d^4x & = \int_\mathcal{M} (\sqrt{-g} A^\mu)_{, \mu} d^4x \\ & = \oint_{\partial \mathcal{M}} A^\mu d \Sigma_\mu \\ & = \oint_{\partial \mathcal{M}} \epsilon A^\mu n_\mu \sqrt{|h|} d^3y \end{align} }
where $\displaystyle{ n_\mu }$ is the unit normal to $\displaystyle{ \partial_\mathcal{M} }$ and $\displaystyle{ \epsilon \equiv n^\mu n_\mu = \pm 1 }$, and $\displaystyle{ y^a }$ are coordinates on the boundary. And $\displaystyle{ d \Sigma_\mu = \epsilon n_\mu d \Sigma }$ where $\displaystyle{ d \Sigma = |h|^{1 \over 2} d^3 y }$ where $\displaystyle{ h = \det [h_{ab}] }$, is an invariant three-dimensional volume element on the hyper-surface. In our particular case we take $\displaystyle{ A^\mu = \delta V^\mu }$.
We now evaluate $\displaystyle{ \delta V^\mu n_\mu }$ on the boundary $\displaystyle{ \partial \mathcal{M} }$, keeping in mind that on $\displaystyle{ \partial \mathcal{M}, \delta g_{\alpha \beta} = 0 = \delta g^{\alpha \beta} }$. Taking this into account we have
$\displaystyle{ \delta \Gamma^\mu_{\alpha \beta} \big|_{\partial \mathcal{M}} = \frac{1}{2} g^{\mu \nu} (\delta g_{\nu \alpha, \beta} + \delta g_{\nu \beta, \alpha} - \delta g_{\alpha \beta, \nu}). }$
It is useful to note that
\displaystyle{ \begin{align} g^{\alpha \mu} \delta \Gamma^\beta_{\alpha \beta} \big|_{\partial \mathcal{M}} & = {1 \over 2} g^{\alpha \mu} g^{\beta \nu} (\delta g_{\nu \alpha, \beta} + \delta g_{\nu \beta, \alpha} - \delta g_{\alpha \beta, \nu}) \\ & = {1 \over 2} g^{\mu \nu} g^{\alpha \beta} (\delta g_{\nu \alpha, \beta} + \delta g_{\alpha \beta, \nu} - \delta g_{\nu \beta, \alpha}) \end{align} }
where in the second line we have swapped around $\displaystyle{ \alpha }$ and $\displaystyle{ \nu }$ and used that the metric is symmetric. It is then not difficult to work out $\displaystyle{ \delta V^\mu = g^{\mu \nu} g^{\alpha \beta} (\delta g_{\nu \beta, \alpha} - \delta g_{\alpha \beta, \nu}) }$.
So now
\displaystyle{ \begin{align} \delta V^\mu n_\mu \big|_{\partial \mathcal{M}} & = n^\mu g^{\alpha \beta} (\delta g_{\mu \beta, \alpha} - \delta g_{\alpha \beta, \mu}) \\ & = n^\mu (\epsilon n^\alpha n^\beta + h^{\alpha \beta}) (\delta g_{\mu \beta, \alpha} - \delta g_{\alpha \beta, \mu}) \\ & = n^\mu h^{\alpha \beta} (\delta g_{\mu \beta, \alpha} - \delta g_{\alpha \beta, \mu}) \end{align} }
where in the second line we used the identity $\displaystyle{ g^{\alpha \beta} = \epsilon n^\alpha n^\beta + h^{\alpha \beta} }$, and in the third line we have used the anti-symmetry in $\displaystyle{ \alpha }$ and $\displaystyle{ \mu }$. As $\displaystyle{ \delta g_{\alpha \beta} }$ vanishes everywhere on the boundary $\displaystyle{ \partial \mathcal{M} , }$ its tangential derivatives must also vanish: $\displaystyle{ \delta g_{\alpha \beta, \gamma} e^\gamma_c = 0 }$. It follows that $\displaystyle{ h^{\alpha \beta} \delta g_{\mu \beta, \alpha} = h^{ab} e^\alpha_a e^\beta_b \delta g_{\mu \beta, \alpha} = 0 }$. So finally we have
$\displaystyle{ n^\mu \delta V_\mu \big|_{\partial \mathcal{M}} = - h^{\alpha \beta} \delta g_{\alpha \beta, \mu} n^\mu. }$
Gathering the results we obtain
$\displaystyle{ (16 \pi) \delta S_{EH} = \int_\mathcal{M} G_{\alpha \beta} \delta g^{\alpha \beta} \sqrt{-g} d^4x - \oint_{\partial \mathcal{M}} \epsilon h^{\alpha \beta} \delta g_{\alpha \beta, \mu} n^\mu \sqrt{h} d^3 y \quad Eq 1. }$
We next show that the above boundary term will be cancelled by the variation of $\displaystyle{ S_{GHY} }$.
Variation of the boundary term
We now turn to the variation of the $\displaystyle{ S_{GHY} }$ term. Because the induced metric is fixed on $\displaystyle{ \partial \mathcal{M}, }$ the only quantity to be varied is $\displaystyle{ K }$ is the trace of the extrinsic curvature.
We have
\displaystyle{ \begin{align} K & = {n^\alpha}_{; \alpha} \\ & = g^{\alpha \beta} n_{\alpha ; \beta} \\ & = \left (\epsilon n^\alpha n^\beta + h^{\alpha \beta} \right ) n_{\alpha ; \beta} \\ & = h^{\alpha \beta} n_{\alpha ; \beta} \\ & = h^{\alpha \beta} (n_{\alpha, \beta} - \Gamma^\gamma_{\alpha \beta} n_\gamma) \end{align} }
where we have used that $\displaystyle{ 0 = (n^\alpha n_\alpha)_{; \beta} }$ implies $\displaystyle{ n^\alpha n_{\alpha; \beta} = 0. }$ So the variation of $\displaystyle{ K }$ is
\displaystyle{ \begin{align} \delta K &= -h^{\alpha \beta} \delta \Gamma^\gamma_{\alpha \beta} n_\gamma \\ &= -h^{\alpha \beta} n_\gamma \frac{1}{2} g^{\gamma \sigma} \left (\delta g_{\sigma \alpha, \beta} + \delta g_{\sigma \beta, \alpha} - \delta g_{\alpha \beta, \sigma} \right ) \\ &= -{1 \over 2} h^{\alpha \beta} \left( \delta g_{\mu \alpha, \beta} + \delta g_{\mu \beta, \alpha} - \delta g_{\alpha \beta, \mu} \right ) n^\mu \\ &= \frac{1}{2} h^{\alpha \beta} \delta g_{\alpha \beta, \mu} n^\mu \end{align} }
where we have use the fact that the tangential derivatives of $\displaystyle{ \delta g_{\alpha \beta} }$ vanish on $\displaystyle{ \partial \mathcal{M}. }$ We have obtained
$\displaystyle{ (16 \pi) \delta S_{GHY} = \oint_{\partial \mathcal{M}} \epsilon h^{\alpha \beta} \delta g_{\alpha \beta, \mu} n^\mu \sqrt{h} d^3 y }$
which cancels the second integral on the right-hand side of Eq. 1. The total variation of the gravitational action is:
$\displaystyle{ \delta S_{TOTAL} = {1 \over 16 \pi} \int_\mathcal{M} G_{\alpha \beta} \delta g^{\alpha \beta} \sqrt{-g} d^4x . }$
This produces the correct left-hand side of the Einstein equations. This proves the main result.
This result was generalised to fourth-order theories of gravity on manifolds with boundaries in 1983[2] and published in 1985.[3]
The non-dynamical term
We elaborate on the role of
$\displaystyle{ S_0 = {1 \over 8 \pi} \oint_{\partial \mathcal{M}} \epsilon K_0 |h|^{1 \over 2} d^3y }$
in the gravitational action. As already mentioned above, because this term only depends on $\displaystyle{ h_{ab} }$, its variation with respect to $\displaystyle{ g_{\alpha \beta} }$ gives zero and so does not effect the field equations, its purpose is to change the numerical value of the action. As such we will refer to it as the non-dynamical term.
Let us assume that $\displaystyle{ g_{\alpha \beta} }$ is a solution of the vacuum field equations, in which case the Ricci scalar $\displaystyle{ R }$ vanishes. The numerical value of the gravitational action is then
$\displaystyle{ S = {1 \over 8 \pi} \oint_{\partial \mathcal{M}} \epsilon K |h|^{1 \over 2} d^3y , }$
where we are ignoring the non-dynamical term for the moment. Let us evaluate this for flat spacetime. Choose the boundary $\displaystyle{ \partial \mathcal{M} }$ to consist of two hyper-surfaces of constant time value $\displaystyle{ t= t_1, t_2 }$ and a large three-cylinder at $\displaystyle{ r=r_0 }$ (that is, the product of a finite interval and a three-sphere of radius $\displaystyle{ r_0 }$). We have $\displaystyle{ K=0 }$ on the hyper-surfaces of constant time. On the three cylinder, in coordinates intrinsic to the hyper-surface, the line element is
\displaystyle{ \begin{align} ds^2 & = - dt^2 + r_0^2 d \Omega^2 \\ & = - dt^2 + r_0^2 (d \theta^2 + \sin^2 \theta d \phi^2) \end{align} }
meaning the induced metric is
$\displaystyle{ h_{ab} = \begin{bmatrix} -1 & 0 & 0 \\ 0 & r_0^2 & 0 \\ 0 & 0 & r_0^2 \sin^2 \theta \end{bmatrix}. }$
so that $\displaystyle{ |h|^{1 \over 2} = r_0^2 \sin \theta }$. The unit normal is $\displaystyle{ n_\alpha = \partial_\alpha r }$, so $\displaystyle{ K = {n^\alpha}_{; \alpha} = 2/r_0 }$. Then
$\displaystyle{ \oint_{\partial \mathcal{M}} \epsilon K |h|^{1 \over 2} d^3y = \int_{t_1}^{t_2} dt \int_0^{2 \pi} d \varphi \int_0^\pi d \theta \left( {2 \over r_0} \right) (r_0^2 \sin \theta) = 8 \pi r_0 (t_2 - t_1) }$
and diverges as $\displaystyle{ r_0 \to \infty }$, that is, when the spatial boundary is pushed to infinity, even when the $\displaystyle{ \mathcal{M} }$ is bounded by two hyper-surfaces of constant time. One would expect the same problem for curved spacetimes that are asymptotically flat (there is no problem if the spacetime is compact). This problem is remedied by the non-dynamical term. The difference $\displaystyle{ S_{GHY} - S_0 }$ will be well defined in the limit $\displaystyle{ r_0 \to \infty }$.
Variation of modified gravity terms
Main page: Physics:Alternatives to general relativity
There are many theories which attempt to modify General Relativity in different ways, for example f(R) gravity replaces R, the Ricci scalar in the Einstein–Hilbert action with a function f(R). Guarnizo et al. found the boundary term for a general f(R) theory.[4] They found that the "modified action in the metric formalism of f(R) gravity plus a Gibbons–York–Hawking like boundary term must be written as:"
$\displaystyle{ S_{mod} = \frac{1}{2\kappa} \int_V d^4x\sqrt{-g} f(R) +2 \int_{\partial V} d^3y \epsilon |h| f'(R) K }$
where $\displaystyle{ f'(R) \equiv \frac{d f(R)}{d R} }$.
By using the ADM decomposition and introducing extra auxiliary fields, in 2009 Deruelle et al. found a method to find the boundary term for "gravity theories whose Lagrangian is an arbitrary function of the Riemann tensor."[5] This method can be used to find the GHY boundary terms for Infinite derivative gravity.[6]
A path-integral approach to quantum gravity
As mentioned at the beginning, the GHY term is required to ensure the path integral (a la Hawking et al.) for quantum gravity has the correct composition properties.
This older approach to path-integral quantum gravity had a number of difficulties and unsolved problems. The starting point in this approach is Feynman's idea that one can represent the amplitude
$\displaystyle{ \langle g_2, \phi_2, \Sigma_2 | g_1, \phi_1, \Sigma_1 \rangle }$
to go from the state with metric $\displaystyle{ g_1 }$ and matter fields $\displaystyle{ \phi_1 }$ on a surface $\displaystyle{ \Sigma_1 }$ to a state with metric $\displaystyle{ g_2 }$ and matter fields $\displaystyle{ \phi_2 }$ on a surface $\displaystyle{ \Sigma_2 }$, as a sum over all field configurations $\displaystyle{ g }$ and $\displaystyle{ \phi }$ which take the boundary values of the fields on the surfaces $\displaystyle{ \Sigma_1 }$ and $\displaystyle{ \Sigma_2 }$. We write
$\displaystyle{ \langle g_2, \phi_2, \Sigma_2 | g_1, \phi_1, \Sigma_1 \rangle = \int \mathcal{D} [g,\phi] \exp (i S [g,\phi]) }$
where $\displaystyle{ \mathcal{D} [g,\phi] }$ is a measure on the space of all field configurations $\displaystyle{ g }$ and $\displaystyle{ \phi }$, $\displaystyle{ S [g,\phi] }$ is the action of the fields, and the integral is taken over all fields which have the given values on $\displaystyle{ \Sigma_1 }$ and $\displaystyle{ \Sigma_2 }$.
It is argued that one need only specify the three-dimensional induced metric $\displaystyle{ h }$ on the boundary.
Now consider the situation where one makes the transition from metric $\displaystyle{ h_1 }$, on a surface $\displaystyle{ \Sigma_1 }$, to a metric $\displaystyle{ h_2 }$, on a surface $\displaystyle{ \Sigma_2 }$ and then on to a metric $\displaystyle{ h_3 }$ on a later surface $\displaystyle{ \Sigma_3 }$
One would like to have the usual composition rule
$\displaystyle{ \langle h_3, \Sigma_3 | h_1, \Sigma_1 \rangle = \sum_{h_2} \langle h_3, \Sigma_3 | h_2, \Sigma_2 \rangle \langle h_2, \Sigma_2 | h_1, \Sigma_1 \rangle }$
expressing that the amplitude to go from the initial to final state to be obtained by summing over all states on the intermediate surface $\displaystyle{ \Sigma_2 }$.
Let $\displaystyle{ g_1 }$ be the metric between $\displaystyle{ \Sigma_1 }$ and $\displaystyle{ \Sigma_2 }$ and $\displaystyle{ g_2 }$ be the metric between $\displaystyle{ \Sigma_2 }$ and $\displaystyle{ \Sigma_3 }$. Although the induced metric of $\displaystyle{ g_1 }$ and $\displaystyle{ g_2 }$ will agree on $\displaystyle{ \Sigma_2 }$, the normal derivative of $\displaystyle{ g_1 }$ at $\displaystyle{ \Sigma_2 }$ will not in general be equal to that of $\displaystyle{ g_2 }$ at $\displaystyle{ \Sigma_2 }$. Taking the implications of this into account, it can then be shown that the composition rule will hold if and only if we include the GHY boundary term.[7]
In the next section it is demonstrated how this path integral approach to quantum gravity leads to the concept of black hole temperature and intrinsic quantum mechanical entropy.
Calculating black-hole entropy using the Euclidean semi-classical approach
Main page: Physics:Euclidean quantum gravity
Application in loop quantum gravity
Main page: Physics:Loop quantum gravity
Transition amplitudes and the Hamilton's principal function
In the quantum theory, the object that corresponds to the Hamilton's principal function is the transition amplitude. Consider gravity defined on a compact region of spacetime, with the topology of a four dimensional ball. The boundary of this region is a three-dimensional space with the topology of a three-sphere, which we call $\displaystyle{ \Sigma }$. In pure gravity without cosmological constant, since the Ricci scalar vanishes on solutions of Einstein's equations, the bulk action vanishes and the Hamilton's principal function is given entirely in terms of the boundary term,
$\displaystyle{ S [q] = \int_\Sigma K^{ab} [q] q_{ab} \sqrt{q} \; d^3 \sigma }$
where $\displaystyle{ K^{ab} }$ is the extrinsic curvature of the boundary, $\displaystyle{ q_{ab} }$ is the three-metric induced on the boundary, and $\displaystyle{ \sigma }$ are coordinates on the boundary.
The functional $\displaystyle{ S [q] }$ is a highly non-trivial functional to compute; this is because the extrinsic curvature $\displaystyle{ K^{ab} [q] }$ is determined by the bulk solution singled out by the boundary intrinsic geometry. As such $\displaystyle{ K^{ab} [q] }$ is non-local. Knowing the general dependence of $\displaystyle{ K^{ab} }$ from $\displaystyle{ q_{ab} }$ is equivalent to knowing the general solution of the Einstein equations.
Background-independent scattering amplitudes
Loop quantum gravity is formulated in a background-independent language. No spacetime is assumed a priori, but rather it is built up by the states of theory themselves – however scattering amplitudes are derived from $\displaystyle{ n }$-point functions (Correlation function (quantum field theory)) and these, formulated in conventional quantum field theory, are functions of points of a background space-time. The relation between the background-independent formalism and the conventional formalism of quantum field theory on a given spacetime is far from obvious, and it is far from obvious how to recover low-energy quantities from the full background-independent theory. One would like to derive the $\displaystyle{ n }$-point functions of the theory from the background-independent formalism, in order to compare them with the standard perturbative expansion of quantum general relativity and therefore check that loop quantum gravity yields the correct low-energy limit.
A strategy for addressing this problem has been suggested;[8] the idea is to study the boundary amplitude, or transition amplitude of a compact region of spacetime, namely a path integral over a finite space-time region, seen as a function of the boundary value of the field.[9][10] In conventional quantum field theory, this boundary amplitude is well-defined[11][12] and codes the physical information of the theory; it does so in quantum gravity as well, but in a fully background-independent manner.[13] A generally covariant definition of $\displaystyle{ n }$-point functions can then be based on the idea that the distance between physical points – arguments of the $\displaystyle{ n }$-point function is determined by the state of the gravitational field on the boundary of the spacetime region considered.
The key observation is that in gravity the boundary data include the gravitational field, hence the geometry of the boundary, hence all relevant relative distances and time separations. In other words, the boundary formulation realizes very elegantly in the quantum context the complete identification between spacetime geometry and dynamical fields.
Notes
1. Feng, J. C., Matzner R. A. The Weiss variation of the gravitational action. Theory Group, Department of Physics, University of Texas at Austin. arXiv:1708.04489v3 [gr-qc]. 24 July 2018 https://arxiv.org/pdf/1708.04489
2. Barth, N H (1985-07-01). "The fourth-order gravitational action for manifolds with boundaries". Classical and Quantum Gravity (IOP Publishing) 2 (4): 497–513. doi:10.1088/0264-9381/2/4/015. ISSN 0264-9381. Bibcode1985CQGra...2..497B.
3. Guarnizo, Alejandro; Castaneda, Leonardo; Tejeiro, Juan M. (2010). "Boundary Term in Metric f(R) Gravity: Field Equations in the Metric Formalism". General Relativity and Gravitation 42 (11): 2713–2728. doi:10.1007/s10714-010-1012-6. Bibcode2010GReGr..42.2713G.
4. Deruelle, Nathalie; Sasaki, Misao; Sendouda, Yuuiti; Yamauchi, Daisuke (2010). "Hamiltonian formulation of f(Riemann) theories of gravity". Progress of Theoretical Physics 123: 169–185. doi:10.1143/PTP.123.169. Bibcode2010PThPh.123..169D.
5. Teimouri, Ali; Talaganis, Spyridon; Edholm, James; Mazumdar, Anupam (2016). "Generalised Boundary Terms for Higher Derivative Theories of Gravity". Journal of High Energy Physics 2016 (8): 144. doi:10.1007/JHEP08(2016)144. Bibcode2016JHEP...08..144T.
6. For example see the book "Hawking on the big bang and black holes" by Stephen Hawking, chapter 15.
7. Modesto, Leonardo; Rovelli, Carlo (2005-11-01). "Particle Scattering in Loop Quantum Gravity". Physical Review Letters 95 (19): 191301. doi:10.1103/physrevlett.95.191301. ISSN 0031-9007. PMID 16383970. Bibcode2005PhRvL..95s1301M.
8. Oeckl, Robert (2003). "A "general boundary" formulation for quantum mechanics and quantum gravity". Physics Letters B (Elsevier BV) 575 (3–4): 318–324. doi:10.1016/j.physletb.2003.08.043. ISSN 0370-2693. Bibcode2003PhLB..575..318O.
9. Oeckl, Robert (2003-11-03). "Schrödinger's cat and the clock: lessons for quantum gravity". Classical and Quantum Gravity 20 (24): 5371–5380. doi:10.1088/0264-9381/20/24/009. ISSN 0264-9381. Bibcode2003CQGra..20.5371O.
10. Conrady, Florian; Rovelli, Carlo (2004-09-30). "Generalized Schrödinger equation in Euclidean field theory". International Journal of Modern Physics A (World Scientific Pub Co Pte Lt) 19 (24): 4037–4068. doi:10.1142/s0217751x04019445. ISSN 0217-751X. Bibcode2004IJMPA..19.4037C.
11. Doplicher, Luisa (2004-09-24). "Generalized Tomonaga-Schwinger equation from the Hadamard formula". Physical Review D (American Physical Society (APS)) 70 (6): 064037. doi:10.1103/physrevd.70.064037. ISSN 1550-7998. Bibcode2004PhRvD..70f4037D.
12. Conrady, Florian; Doplicher, Luisa; Oeckl, Robert; Rovelli, Carlo; Testa, Massimo (2004-03-18). "Minkowski vacuum in background independent quantum gravity". Physical Review D (American Physical Society (APS)) 69 (6): 064019. doi:10.1103/physrevd.69.064019. ISSN 1550-7998. Bibcode2004PhRvD..69f4019C. | 9,271 | 30,671 | {"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": 8, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.828125 | 3 | CC-MAIN-2023-06 | latest | en | 0.835595 |
https://www.nuprl.org/wip/Mathematics/cubical!sets/cubical-type-ap-rename-one-equal.html | 1,696,081,869,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510676.40/warc/CC-MAIN-20230930113949-20230930143949-00849.warc.gz | 1,017,233,490 | 2,595 | ### Nuprl Lemma : cubical-type-ap-rename-one-equal
`∀[X:CubicalSet]. ∀[A:{X ⊢ _}]. ∀[I:Cname List]. ∀[x,y:Cname]. ∀[a:X([x / I])]. ∀[u,v:A(a)].`
` u = v ∈ A(a) `⇐⇒` (u a rename-one-name(x;y)) = (v a rename-one-name(x;y)) ∈ A(rename-one-name(x;y)(a)) `
` supposing (¬(y ∈ I)) ∧ (¬(x ∈ I))`
Proof
Definitions occuring in Statement : cubical-type-ap-morph: `(u a f)` cubical-type-at: `A(a)` cubical-type: `{X ⊢ _}` cube-set-restriction: `f(s)` I-cube: `X(I)` cubical-set: `CubicalSet` rename-one-name: `rename-one-name(z1;z2)` coordinate_name: `Cname` l_member: `(x ∈ l)` cons: `[a / b]` list: `T List` uimplies: `b supposing a` uall: `∀[x:A]. B[x]` iff: `P `⇐⇒` Q` not: `¬A` and: `P ∧ Q` equal: `s = t ∈ T`
Definitions unfolded in proof : uall: `∀[x:A]. B[x]` member: `t ∈ T` uimplies: `b supposing a` and: `P ∧ Q` iff: `P `⇐⇒` Q` implies: `P `` Q` squash: `↓T` cand: `A c∧ B` true: `True` prop: `ℙ` rev_implies: `P `` Q` all: `∀x:A. B[x]` subtype_rel: `A ⊆r B` guard: `{T}`
Lemmas referenced : cubical-type-ap-morph_wf cons_wf coordinate_name_wf rename-one-name_wf equal_wf cubical-type-at_wf cube-set-restriction_wf not_wf l_member_wf I-cube_wf rename-one-comp name-morph_wf id-morph_wf rename-one-same squash_wf true_wf iff_weakening_equal cube-set-restriction-comp cube-set-restriction-when-id list_wf cubical-type_wf cubical-set_wf cubical-type-ap-morph-id cubical-type-ap-morph-comp
Rules used in proof : sqequalSubstitution sqequalTransitivity computationStep sqequalReflexivity isect_memberFormation introduction cut sqequalHypSubstitution productElimination thin independent_pairFormation lambdaFormation applyEquality lambdaEquality imageElimination extract_by_obid isectElimination because_Cache hypothesis hypothesisEquality independent_isectElimination equalitySymmetry natural_numberEquality sqequalRule imageMemberEquality baseClosed independent_pairEquality dependent_functionElimination axiomEquality productEquality isect_memberEquality equalityTransitivity universeEquality independent_functionElimination applyLambdaEquality hyp_replacement instantiate
Latex:
\mforall{}[X:CubicalSet]. \mforall{}[A:\{X \mvdash{} \_\}]. \mforall{}[I:Cname List]. \mforall{}[x,y:Cname]. \mforall{}[a:X([x / I])]. \mforall{}[u,v:A(a)].
u = v \mLeftarrow{}{}\mRightarrow{} (u a rename-one-name(x;y)) = (v a rename-one-name(x;y))
supposing (\mneg{}(y \mmember{} I)) \mwedge{} (\mneg{}(x \mmember{} I))
Date html generated: 2017_10_05-AM-10_12_39
Last ObjectModification: 2017_07_28-AM-11_18_26
Theory : cubical!sets
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http://math.stackexchange.com/questions/257001/spectral-radius-for-linear-compact-maps | 1,464,422,570,000,000,000 | text/html | crawl-data/CC-MAIN-2016-22/segments/1464049277475.33/warc/CC-MAIN-20160524002117-00095-ip-10-185-217-139.ec2.internal.warc.gz | 188,829,845 | 17,086 | # Spectral radius for linear compact maps
Prove or disprove the following assertions for a linear map $C$ from a Banach space $X$ into itself:
a) If C is compact then its spectral radius equals the maximum of the absolute value of $C$
I'm not sure what the definition of absolute value of an operator is. But the spectral value is defined as $|\sigma(M)| = max_{\lambda \in \sigma(M)} |\lambda|$
edit: I split it up to two questions
-
I can't make sense of the question, and you also don't know what it means. Where did it come from? – Jonas Meyer Dec 12 '12 at 15:23
An old exam, maybe there was a miss print. – Johan Dec 12 '12 at 16:09
If the absolute value of $C$ is the norm of $C$, then the answer is no. It becomes YES though, assuming further that $C$ is self-adjoint. – Yiorgos S. Smyrlis Dec 30 '15 at 10:40
This is false, consider Volterra operator on $C[0,1]$. Radius $0$ but $\mid Cf \mid \ge 1$, just take $\chi_{[0,1]}$
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If your Banach space is $\mathbb R^2$ and $C=\left(\begin{matrix}0&1\\ 0&0 \end{matrix}\right)$, then $\lvert C\rvert=1$, while its spectral radius is $0$.
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# Robert Bosch - 24th Aug. 2004 by BOSCH
Details of Robert Bosch - 24th Aug. 2004 by BOSCH conducted by BOSCH for job interview.
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SVNIT Surat CAMPUS RECRUITEMENT PROGRAMME
ON 24th Aug. 2004
BOSCH REXROTH AG , Vatva , Ahmedabad
1. Critical stress at which material will start to
flow
(a)yield
(b)ultimate tensile stress
(c) proof stress
(d) none o the above
2. maximum degree of freedom in space
Ans. 6
(a) thrust bearing
(b) deep groove
(c) taper roller
(d) ---
4. volume of water per cubic metre of air.
(a)specific volume
(b) specific gravity
(c) vapour pressure
(d) none of above
5. fluid is flowing through a frustum of cone. what is
the nature of graph of velocity
Vs. c/s area. {flow = velocity X c/s area}
(a) parabola
(b) hyperbola
(c) y=mx+c
(d) x=x.
6. Fluid in a pipe having no change in profile with
change in length of pipe ...explain type of flow.
a)Laminar
b)Turbulent
c)fully developed profile <ans>
7. Which are correct
I) Navier Stokes Eq. is .... of momentum
conservation eq.
II)bernoulis eq. is for viscous flow
III) REYNOLD no. >3000 always for turbulent flow.
a)I and II correct
b)I and II not correct
c)II and III correct
d)II and III not correct
8. Which can,t be removed during alloying from Fe
alloys
a) co
b) N
c) Si
d) As
9.There are two ropes burning non uniformly (rate) and
clock half an hour each. which can,t be clocked ?
a) 1 Hr.
b) 45 min
c)15 min
d)none
10. Lim n->0 (1+ 1/n)^n =
a) 1
b)0
c)e
d)infinity
11.time dependent increase in length at steady temp.
is called
a) superplasticity
b) creep
c) fatigue
d) none
12. Vicosity change with increasing temp.
a) decreases
b) increases
c) same
d) first decreases and then increases.
13. Direction of friction in bicycle tyres
a) along motion
b) opp. motion
c) Front opp. motion and vice versa
d) rear opp. motion and vice versa
14. Sp. gravity of a fluid
a) density of fluid at 0`c / density of water at o`c.
b) density of fluid at 0`c / density of water at T`c.
c) density of fluid at T`c / density of water at o`c.
d) density of fluid at T`c / density of water at T`c.
15. lateral strain / longitudinal strain
a) Poission,s Ratio
b) Bulk Modulus
c) Modulus Of Elasticity
d) None
16. 10 coins & 1 defected coin and 2 weighs Find min
weighs req. to detect faulty coin
a) 2
b)3
c)4
d)5
17. 9 members in meeting scheluled at 10:00 A.M. , all
reaches at 9:48 A.M. . one member req. 2 min to intro
with other
find meeting delay with scheduled time of 10:00 A.M..
a) 1 Hr.
b) 6 min
c) ...
d)At scheduled time
18. A completes job in 10 hr. and B completes in 15
hr.. Find when both together works
a) 6 hr.
b) 8 hr
c) 10 hr
d) none
19. A ball dropped from H height and moves 80% of
height each time.Total dist. covered
a) 4H
b) 5H
c) 7H
d) 9H
20. A cantilever beam loaded at free end find
reinforcement place
a) AT neutral axis
b) above neutral axis
c) below neutral axis
d) above and below neutral axis
21. punching a sheet of D dia and T thickness and S
ultimate tensile stress
a) 3.14 DTS
b) 3.14TS X D^2
c) 6.28 DTS
d)6.28 TS x D^2
22. & 23. two easy ques. of string mass equilibrium
system...
24. Microstructure after quenching
a) Mertensite
b) pearlite
c) ........
d).......... | 1,099 | 3,301 | {"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} | 2.515625 | 3 | CC-MAIN-2018-17 | longest | en | 0.769269 |
https://www.mathcelebrity.com/community/threads/frank-is-a-plumber-who-charges-a-35-service-charge-and-15-per-hour-for-his-plumbing-services-find.1091/ | 1,719,314,350,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198865972.21/warc/CC-MAIN-20240625104040-20240625134040-00051.warc.gz | 790,231,344 | 8,871 | # Frank is a plumber who charges a \$35 service charge and \$15 per hour for his plumbing services. Find
Discussion in 'Calculator Requests' started by math_celebrity, May 9, 2017.
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1. ### math_celebrityAdministratorStaff Member
Frank is a plumber who charges a \$35 service charge and \$15 per hour for his plumbing services. Find a linear function that expresses the total cost C for plumbing services for h hours.
Cost functions include a flat rate and a variable rate. The flat rate is \$35 and the variable rate per hour is 15. The cost function C(h) where h is the number of hours Frank works is:
C(h) = 15h + 35 | 153 | 627 | {"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} | 2.859375 | 3 | CC-MAIN-2024-26 | latest | en | 0.913139 |
http://caml.inria.fr/pub/docs/manual-ocaml-311/libref/ListLabels.html | 1,500,591,429,000,000,000 | text/html | crawl-data/CC-MAIN-2017-30/segments/1500549423512.93/warc/CC-MAIN-20170720222017-20170721002017-00711.warc.gz | 54,512,391 | 4,883 | # Module ListLabels
`module ListLabels: `sig` .. `end``
List operations.
Some functions are flagged as not tail-recursive. A tail-recursive function uses constant stack space, while a non-tail-recursive function uses stack space proportional to the length of its list argument, which can be a problem with very long lists. When the function takes several list arguments, an approximate formula giving stack usage (in some unspecified constant unit) is shown in parentheses.
The above considerations can usually be ignored if your lists are not longer than about 10000 elements.
`val length : `'a list -> int``
Return the length (number of elements) of the given list.
`val hd : `'a list -> 'a``
Return the first element of the given list. Raise `Failure "hd"` if the list is empty.
`val tl : `'a list -> 'a list``
Return the given list without its first element. Raise `Failure "tl"` if the list is empty.
`val nth : `'a list -> int -> 'a``
Return the `n`-th element of the given list. The first element (head of the list) is at position 0. Raise `Failure "nth"` if the list is too short. Raise `Invalid_argument "List.nth"` if `n` is negative.
`val rev : `'a list -> 'a list``
List reversal.
`val append : `'a list -> 'a list -> 'a list``
Catenate two lists. Same function as the infix operator `@`. Not tail-recursive (length of the first argument). The `@` operator is not tail-recursive either.
`val rev_append : `'a list -> 'a list -> 'a list``
`List.rev_append l1 l2` reverses `l1` and concatenates it to `l2`. This is equivalent to `ListLabels.rev`` l1 @ l2`, but `rev_append` is tail-recursive and more efficient.
`val concat : `'a list list -> 'a list``
Concatenate a list of lists. The elements of the argument are all concatenated together (in the same order) to give the result. Not tail-recursive (length of the argument + length of the longest sub-list).
`val flatten : `'a list list -> 'a list``
Same as `concat`. Not tail-recursive (length of the argument + length of the longest sub-list).
###### Iterators
`val iter : `f:('a -> unit) -> 'a list -> unit``
`List.iter f [a1; ...; an]` applies function `f` in turn to `a1; ...; an`. It is equivalent to `begin f a1; f a2; ...; f an; () end`.
`val map : `f:('a -> 'b) -> 'a list -> 'b list``
`List.map f [a1; ...; an]` applies function `f` to `a1, ..., an`, and builds the list `[f a1; ...; f an]` with the results returned by `f`. Not tail-recursive.
`val rev_map : `f:('a -> 'b) -> 'a list -> 'b list``
`List.rev_map f l` gives the same result as `ListLabels.rev`` (``ListLabels.map`` f l)`, but is tail-recursive and more efficient.
`val fold_left : `f:('a -> 'b -> 'a) -> init:'a -> 'b list -> 'a``
`List.fold_left f a [b1; ...; bn]` is `f (... (f (f a b1) b2) ...) bn`.
`val fold_right : `f:('a -> 'b -> 'b) -> 'a list -> init:'b -> 'b``
`List.fold_right f [a1; ...; an] b` is `f a1 (f a2 (... (f an b) ...))`. Not tail-recursive.
###### Iterators on two lists
`val iter2 : `f:('a -> 'b -> unit) -> 'a list -> 'b list -> unit``
`List.iter2 f [a1; ...; an] [b1; ...; bn]` calls in turn `f a1 b1; ...; f an bn`. Raise `Invalid_argument` if the two lists have different lengths.
`val map2 : `f:('a -> 'b -> 'c) -> 'a list -> 'b list -> 'c list``
`List.map2 f [a1; ...; an] [b1; ...; bn]` is `[f a1 b1; ...; f an bn]`. Raise `Invalid_argument` if the two lists have different lengths. Not tail-recursive.
`val rev_map2 : `f:('a -> 'b -> 'c) -> 'a list -> 'b list -> 'c list``
`List.rev_map2 f l1 l2` gives the same result as `ListLabels.rev`` (``ListLabels.map2`` f l1 l2)`, but is tail-recursive and more efficient.
`val fold_left2 : `f:('a -> 'b -> 'c -> 'a) -> init:'a -> 'b list -> 'c list -> 'a``
`List.fold_left2 f a [b1; ...; bn] [c1; ...; cn]` is `f (... (f (f a b1 c1) b2 c2) ...) bn cn`. Raise `Invalid_argument` if the two lists have different lengths.
`val fold_right2 : `f:('a -> 'b -> 'c -> 'c) -> 'a list -> 'b list -> init:'c -> 'c``
`List.fold_right2 f [a1; ...; an] [b1; ...; bn] c` is `f a1 b1 (f a2 b2 (... (f an bn c) ...))`. Raise `Invalid_argument` if the two lists have different lengths. Not tail-recursive.
###### List scanning
`val for_all : `f:('a -> bool) -> 'a list -> bool``
`for_all p [a1; ...; an]` checks if all elements of the list satisfy the predicate `p`. That is, it returns `(p a1) && (p a2) && ... && (p an)`.
`val exists : `f:('a -> bool) -> 'a list -> bool``
`exists p [a1; ...; an]` checks if at least one element of the list satisfies the predicate `p`. That is, it returns `(p a1) || (p a2) || ... || (p an)`.
`val for_all2 : `f:('a -> 'b -> bool) -> 'a list -> 'b list -> bool``
Same as `ListLabels.for_all`, but for a two-argument predicate. Raise `Invalid_argument` if the two lists have different lengths.
`val exists2 : `f:('a -> 'b -> bool) -> 'a list -> 'b list -> bool``
Same as `ListLabels.exists`, but for a two-argument predicate. Raise `Invalid_argument` if the two lists have different lengths.
`val mem : `'a -> set:'a list -> bool``
`mem a l` is true if and only if `a` is equal to an element of `l`.
`val memq : `'a -> set:'a list -> bool``
Same as `ListLabels.mem`, but uses physical equality instead of structural equality to compare list elements.
###### List searching
`val find : `f:('a -> bool) -> 'a list -> 'a``
`find p l` returns the first element of the list `l` that satisfies the predicate `p`. Raise `Not_found` if there is no value that satisfies `p` in the list `l`.
`val filter : `f:('a -> bool) -> 'a list -> 'a list``
`filter p l` returns all the elements of the list `l` that satisfy the predicate `p`. The order of the elements in the input list is preserved.
`val find_all : `f:('a -> bool) -> 'a list -> 'a list``
`find_all` is another name for `ListLabels.filter`.
`val partition : `f:('a -> bool) -> 'a list -> 'a list * 'a list``
`partition p l` returns a pair of lists `(l1, l2)`, where `l1` is the list of all the elements of `l` that satisfy the predicate `p`, and `l2` is the list of all the elements of `l` that do not satisfy `p`. The order of the elements in the input list is preserved.
###### Association lists
`val assoc : `'a -> ('a * 'b) list -> 'b``
`assoc a l` returns the value associated with key `a` in the list of pairs `l`. That is, `assoc a [ ...; (a,b); ...] = b` if `(a,b)` is the leftmost binding of `a` in list `l`. Raise `Not_found` if there is no value associated with `a` in the list `l`.
`val assq : `'a -> ('a * 'b) list -> 'b``
Same as `ListLabels.assoc`, but uses physical equality instead of structural equality to compare keys.
`val mem_assoc : `'a -> map:('a * 'b) list -> bool``
Same as `ListLabels.assoc`, but simply return true if a binding exists, and false if no bindings exist for the given key.
`val mem_assq : `'a -> map:('a * 'b) list -> bool``
Same as `ListLabels.mem_assoc`, but uses physical equality instead of structural equality to compare keys.
`val remove_assoc : `'a -> ('a * 'b) list -> ('a * 'b) list``
`remove_assoc a l` returns the list of pairs `l` without the first pair with key `a`, if any. Not tail-recursive.
`val remove_assq : `'a -> ('a * 'b) list -> ('a * 'b) list``
Same as `ListLabels.remove_assoc`, but uses physical equality instead of structural equality to compare keys. Not tail-recursive.
###### Lists of pairs
`val split : `('a * 'b) list -> 'a list * 'b list``
Transform a list of pairs into a pair of lists: `split [(a1,b1); ...; (an,bn)]` is `([a1; ...; an], [b1; ...; bn])`. Not tail-recursive.
`val combine : `'a list -> 'b list -> ('a * 'b) list``
Transform a pair of lists into a list of pairs: `combine [a1; ...; an] [b1; ...; bn]` is `[(a1,b1); ...; (an,bn)]`. Raise `Invalid_argument` if the two lists have different lengths. Not tail-recursive.
###### Sorting
`val sort : `cmp:('a -> 'a -> int) -> 'a list -> 'a list``
Sort a list in increasing order according to a comparison function. The comparison function must return 0 if its arguments compare as equal, a positive integer if the first is greater, and a negative integer if the first is smaller (see Array.sort for a complete specification). For example, `compare` is a suitable comparison function. The resulting list is sorted in increasing order. `List.sort` is guaranteed to run in constant heap space (in addition to the size of the result list) and logarithmic stack space.
The current implementation uses Merge Sort. It runs in constant heap space and logarithmic stack space.
`val stable_sort : `cmp:('a -> 'a -> int) -> 'a list -> 'a list``
Same as `ListLabels.sort`, but the sorting algorithm is guaranteed to be stable (i.e. elements that compare equal are kept in their original order) .
The current implementation uses Merge Sort. It runs in constant heap space and logarithmic stack space.
`val fast_sort : `cmp:('a -> 'a -> int) -> 'a list -> 'a list``
Same as `List.sort` or `List.stable_sort`, whichever is faster on typical input.
`val merge : `cmp:('a -> 'a -> int) -> 'a list -> 'a list -> 'a list``
Merge two lists: Assuming that `l1` and `l2` are sorted according to the comparison function `cmp`, `merge cmp l1 l2` will return a sorted list containting all the elements of `l1` and `l2`. If several elements compare equal, the elements of `l1` will be before the elements of `l2`. Not tail-recursive (sum of the lengths of the arguments). | 2,756 | 9,276 | {"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} | 2.546875 | 3 | CC-MAIN-2017-30 | latest | en | 0.697803 |
https://www.teachoo.com/164/52/Free-Club-Membership-for-Employees/category/Taxability-of-Perquisites/ | 1,726,574,134,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651773.64/warc/CC-MAIN-20240917104423-20240917134423-00441.warc.gz | 962,311,110 | 22,531 | Taxability of Perquisites
Income Tax
Income from Salaries
----------------------------------------------------------------------------------------------------------------------------------------------------------
QUESTIONS
Q1
Basic salary = 40000 p.m.
Company owns club where Employees get free use
Cost of running the club per member is Rs.2000 P.M and fees charged from members is Rs.3000
Calculate income from salary
Value of club Perquisite Rs.2000x12 = 24000
Q2 Suppose club facility is not free Employee have to pay a nominal charge of Rs.500 per month
Q3 Suppose Employee and his wife has joined a health club and they pay Rs.3000 each (6,000 for both)
As per company policy they claim reimbursement of the same of maximum Rs.5,000
What is the value of Perquisite. | 164 | 785 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.75 | 3 | CC-MAIN-2024-38 | latest | en | 0.898437 |
http://www.theswamp.org/index.php?topic=49235.0 | 1,713,797,839,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296818312.80/warc/CC-MAIN-20240422144517-20240422174517-00370.warc.gz | 54,538,909 | 8,778 | ### Author Topic: Need related level (Read 6839 times)
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• Swamp Rat
• Posts: 1422
##### Need related level
« on: April 07, 2015, 12:37:23 PM »
Hi all
I do not know where exactly should I post this question, I hope this place can give an answer
this is road levels (Z Level) at spacific points. we have to get the point on CYAN and RED lines but by mistake landscape man got these points,
So is there any way to get points at cyan and red lines.
Thaks all
#### mjfarrell
• Seagull
• Posts: 14444
• Every Student their own Lesson
##### Re: Need related level
« Reply #1 on: April 07, 2015, 12:47:02 PM »
if those 'points' are really surface labels - just move them they will update....
if they are points on a surface - just move them they should update
care should be taken NOT to snap to anything that does or doe not have an elevation assigned or they will grab the wrong level (elevation)
Michael Farrell
http://primeservicesglobal.com/
• Swamp Rat
• Posts: 1422
##### Re: Need related level
« Reply #2 on: April 07, 2015, 12:50:37 PM »
string is Z level and as you see the road has a slop so i want to copy the new point at the same slop.
#### mjfarrell
• Seagull
• Posts: 14444
• Every Student their own Lesson
##### Re: Need related level
« Reply #3 on: April 07, 2015, 12:53:51 PM »
string is Z level and as you see the road has a slop so i want to copy the new point at the same slop.
YES...but what are they? are they labels on a surface
or are they points on a surface
either case they should update correctly IF you simply move them XX in that direction
Michael Farrell
http://primeservicesglobal.com/
• Swamp Rat
• Posts: 1152
##### Re: Need related level
« Reply #4 on: April 07, 2015, 12:54:38 PM »
they are just blocks, inserted with Z values between 9 and 11, with 0=10.
So, you'll be interpolating new locations and elevations
The only thing more dangerous to the liberty of a free people than big government is big business
• Swamp Rat
• Posts: 1422
##### Re: Need related level
« Reply #5 on: April 07, 2015, 01:01:02 PM »
I am doing it manualy
draw 3DPOLY then eatend then get new elevation by inserting a bloch has attribute with field for Z value.
But It will take long time, I am seraching for faster and easyer way.
#### mjfarrell
• Seagull
• Posts: 14444
• Every Student their own Lesson
##### Re: Need related level
« Reply #6 on: April 07, 2015, 02:26:15 PM »
give me a copy of your file...post it here I will test a method that should be faster
Michael Farrell
http://primeservicesglobal.com/
#### Matt__W
• Seagull
• Posts: 12955
• I like my water diluted.
##### Re: Need related level
« Reply #7 on: April 07, 2015, 02:31:34 PM »
What about using the ordinate dimension command?? Would that work?
Autodesk Expert Elite
Revit Subject Matter Expert (SME)
Owner/FAA sUAS Pilot @ http://skyviz.io
#### mjfarrell
• Seagull
• Posts: 14444
• Every Student their own Lesson
##### Re: Need related level
« Reply #8 on: April 07, 2015, 02:44:43 PM »
What about using the ordinate dimension command?? Would that work?
not sure that would solve or resolve this particular issue ?
Knowing what or how those 'points' were generated would...
Feature Line(s) and Stepped offset(s) are now coming to me as a potential solution...as the offset distance to the two reference lines in NOT equal....
but a stepped offset would allow that to be compensated for
And or IF this is a road....just use generic links build a corridor and get the elevations from the point codes/points from corridor command...done in under five minutes
Michael Farrell
http://primeservicesglobal.com/
#### Matt__W
• Seagull
• Posts: 12955
• I like my water diluted.
##### Re: Need related level
« Reply #9 on: April 07, 2015, 02:49:33 PM »
After looking at the image again, the ordinate command wouldn't work.
*goes back to his Revit world*
Autodesk Expert Elite
Revit Subject Matter Expert (SME)
Owner/FAA sUAS Pilot @ http://skyviz.io
• Swamp Rat
• Posts: 1422
##### Re: Need related level
« Reply #10 on: April 07, 2015, 02:53:59 PM »
give me a copy of your file...post it here I will test a method that should be faster
posted in first post
#### mjfarrell
• Seagull
• Posts: 14444
• Every Student their own Lesson
##### Re: Need related level
« Reply #11 on: April 07, 2015, 03:37:35 PM »
I dont think anything can 'save' you on this one...
two things prevent me from going further...
ONE
the point data grid is NOT aligned parallel or perpendicular to the Cyan or the RED lines so they are skewed to the lines of points.
TWO
and probably the biggest concern. a random sampling of the roadway cross slopes...shows that they are inconsistent
so there is no way to simply assume that the slope remains consistent....
The next question..is this DESIGN data? Or is this SURVEY data?
Not sure there is a solution either way
Michael Farrell
http://primeservicesglobal.com/
#### mjfarrell
• Seagull
• Posts: 14444
• Every Student their own Lesson
##### Re: Need related level
« Reply #12 on: April 07, 2015, 03:44:35 PM »
see attached
Michael Farrell
http://primeservicesglobal.com/
• Swamp Rat
• Posts: 1422
##### Re: Need related level
« Reply #13 on: April 08, 2015, 05:13:42 AM »
Thanks all for help
#### mjfarrell
• Seagull
• Posts: 14444
• Every Student their own Lesson
##### Re: Need related level
« Reply #14 on: April 08, 2015, 08:53:47 AM »
I am not saying that the elevations can NOT be transferred over to those lines.
They can, however a lot of assumption and understanding of those assumptions would be needed prior to doing so.
For one..transferring the exact elevation farther left or right will change the cross slope.
Secondly, assuming that the slope as it exists is constant over to the line, is no guarantee that the slope is
constant or the project elevation is correct.
From a surveying or engineering perspective that is a little too much to leave to chance, and would be
safer to simply resurvey the area at the proper locations. From what I saw in the file most of
those elevations are NOT at the locations you want(ed) them to be at. I presume
that was supposed to be Center, right and left edges. And they are not at those locations. | 1,712 | 6,254 | {"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} | 2.625 | 3 | CC-MAIN-2024-18 | latest | en | 0.867609 |
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# Answer to Question #42366 in Calculus for shasha
Question #42366
f(x) = x2 + 3; g(x)= square root x-2
Find f(g(x)).
Let g(x)=sqrt(x-2)=y. (1)
We need to find f(y). So, we substitute (1) in f(y)=y^2+3:
f(g(x))=(sqrt(x-2))^2+3=x-2+3=x+1.
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for any assignment or question with DETAILED EXPLANATIONS! | 161 | 462 | {"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} | 3.140625 | 3 | CC-MAIN-2018-39 | latest | en | 0.831652 |
https://itunes.apple.com/us/book/hands-on-calculus/id585497623?mt=13 | 1,542,272,941,000,000,000 | text/html | crawl-data/CC-MAIN-2018-47/segments/1542039742569.45/warc/CC-MAIN-20181115075207-20181115101207-00338.warc.gz | 650,455,648 | 15,480 | Opening the iTunes Store.If iTunes doesn't open, click the iTunes application icon in your Dock or on your Windows desktop.Progress Indicator
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# Hands-On Calculus
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#### Description
Learn calculus the fun way, with dozens of hands-on interactives that let you explore the subject and over 10 hours of animated video explanations. In Hands-On Calculus, you'll never go more than a minute without interacting with our Virtual Tutor, and you can even log in to track your progress.
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All 3 chapters from Hands-On Precalculus (functions, limits, and series) Definition of the derivative Derivatives of common functions Addition, product, and quotient rules Chain rule and implicit differentiation Second and higher derivatives Linear approximations L'Hopital's rule Integrals and Riemann sums Fundamental theorem of calculus Integration by parts, u-substitution, and partial fractions Why all of this matters!
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• Free | 554 | 2,701 | {"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} | 2.578125 | 3 | CC-MAIN-2018-47 | latest | en | 0.882016 |
https://www.physicsforums.com/threads/rotational-inertia-globes-connected-by-a-thin-rod.975569/ | 1,624,529,279,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623488552937.93/warc/CC-MAIN-20210624075940-20210624105940-00241.warc.gz | 818,238,219 | 17,466 | # Rotational inertia - globes connected by a thin rod
Homework Statement:
Finding the rotational inertia
Relevant Equations:
I=∑m*r^2
Problem Statement: Finding the rotational inertia
Relevant Equations: I=∑m*r^2
A rigid body of 2 massive globes with homogenous mass distribution and a thin rod is connecting the 2 globes. The globes has radius R1 = 0.18 m and R2 =0.28 m and masses m1=193 kg and m2=726 kg. The thin rod has the mass ms=10kg and the length L=0.88 m. THe body can rotate frictionless around the rotation axis in the middle of the thin rod.
a) Calculate the the body's overall inertiamoment and show that the inertiamoment is I=476 kgm^2
my calculations are m1*(1/2*L+r1)^2+m2*(1/2*L+r2)^2 and i doubt if this is even correct and cant tell how the inertiamoment of the rod would look. Ignore the F(force)
Doc Al
Mentor
Hint: Don't treat the spheres as if they were point masses. (Think parallel axis theorem.)
haruspex
Homework Helper
Gold Member
2020 Award
cant tell how the inertiamoment of the rod would look
You should have a standard formula for the MoI of a uniform stick about its centre. Similarly for a uniform solid sphere.
okay so for a solid sphere i use 2/5*m*r^2 and for the rod it is 1/12*m*L^2?
Doc Al
Mentor
okay so for a solid sphere i use 2/5*m*r^2 and for the rod it is 1/12*m*L^2?
Okay so you use the formula I=Icm+Md^2 and therefore:
2/5*m
1*r1^2+m1*(r1+1/2*L)^2 + 2/5*m2*r2^2+m2*(-r2-1/2*L)^2 + 1/12*m3*L+m3*1
haruspex
Homework Helper
Gold Member
2020 Award
m3*1
Where does this term come from?
The rest is fine.
Where does this term come from?
The rest is fine.
honestly i am
Where does this term come from?
The rest is fine.
I am not sure how to calculate the rod when the rotation axis is right on the middle of it
haruspex | 547 | 1,777 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.8125 | 4 | CC-MAIN-2021-25 | latest | en | 0.865024 |
https://physics.stackexchange.com/questions/707217/number-of-fringes-obtained-on-interference-of-light-of-different-wavelengths | 1,720,826,099,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514459.28/warc/CC-MAIN-20240712224556-20240713014556-00711.warc.gz | 380,120,867 | 37,171 | # Number of fringes obtained on interference of light of different wavelengths [closed]
The Question goes as follows:
I am stuck at the part where we obtain general equation for path difference of two interfering lights. Since wavelength of those interfering lights are different, so I cannot apply the equation $$\Delta \psi = 2\pi\Delta x/\lambda$$, and thus cannot find the general condition for the position of dark fringes.
Would be really grateful if someone suggests me how to tackle these kind of problem.
• I think the question has no definite answer. Since each wavelength generates a (formally) infinite number of dark fringes, the only thing that reduces that number is either 1) screen size or 2) diffraction by non-zero width slits. Commented May 6, 2022 at 11:27
• I think the idea is you need to count how many places are there where both banks produce a dark fine at the same place. Commented May 6, 2022 at 14:30
• @ThePhoton Even then, there's an infinite number of those (they're evenly spaced). Commented May 6, 2022 at 19:00
• @Miyase You do actually obtain some sort of Limiting Condition by ∆x = d sinθ, where d is the Distance btw the slits (since sinθ cannot be >1), and then you can relate ∆x=(2n+1)λ/2 for dark fringes, and thus you get Limiting Condition for n (Number of Dark Fringes). But since this equation can be used for Interference of Lights of same Wavelength only, I am having some difficulty using that, which is basically my doubt. Commented May 7, 2022 at 15:15
• Since these are different wavelength beams, they are incoherent (assuming no extreme measures were taken to produce coherent beams). Therefore for practical purposes the intensities of the beams add, rather than the electric fields (which is usually what we mean when we talk about the superposition principle in optics). Commented May 7, 2022 at 15:21
The condition for a maximum is $$n \lambda = d\sin \theta$$ where $$d$$ Is the slit separation and $$n$$ is the order of the fringe. Using this formula with $$\theta = \pm \pi/2$$ will give you the number of bright fringes either side of the central maximum and this eliminates two of the answers.
The wavelengths have been chosen so that they are in a simple ratio $$(9/7)$$ which makes finding the position of the minima relatively easy.
For coincidence of minima at a certain position you must have $$(n+1/2)420 = (m+1/2)540 = d\sin \theta$$ where $$m$$ and $$n$$ are integers which you need to find. | 611 | 2,467 | {"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": 9, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.34375 | 3 | CC-MAIN-2024-30 | latest | en | 0.938195 |
http://www.personal.psu.edu/dpl14/java/geometry/euclidean/inversion.html | 1,656,329,133,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103331729.20/warc/CC-MAIN-20220627103810-20220627133810-00014.warc.gz | 102,568,565 | 2,151 | David Little
Mathematics Department
Penn State University
Eberly College of Science
University Park, PA 16802
Office: 403 McAllister
Phone: (814) 865-3329
Fax: (814) 865-3735
e-mail:dlittle@psu.edu
Inversion of a Point across a Circle
Our goal will be to construct the inversion of the blue point, P, with respect to the circle centered at the green point.
To do so, first pick any point, Q, inside the circle. Next, draw a circle centered at P that goes through the point Q. This circle must intersect the original circle in exactly two points. Mark these two points of intersection.
Next, draw two lines. Each line should go through P and one of the points of intersection you just constructed. These two lines should each intersect the original circle in two places. Mark the points of intersection of these two lines and the original circle. These four points of intersection form the vertices of a trapezoid.
And finally, draw two line segments that join opposite vertices of the trapezoid. These two line segments intersect at a point P', which is called the inversion of P. The inversion point P' has the following property. Let D represent the distance from P to the green point, let d represent the distance from P' to the green point and let r represent the radius of the circle centered at the green point. Then
Dd = r2.
Can you explain why?
Once you have performed the above construction... consider a circle and a point on a line. move the point on the line, what curve is traced out by it's inversion do the same but with another circle | 355 | 1,556 | {"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-2022-27 | latest | en | 0.910516 |
https://uphillathlete.com/forums/topic/help-with-heart-rate-drift-test/ | 1,726,302,327,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651559.58/warc/CC-MAIN-20240914061427-20240914091427-00630.warc.gz | 558,101,413 | 30,416 | Help with heart rate drift test | Uphill Athlete
• Creator
Topic
• #34689
MMK
Participant
I fear I need a little help interpreting the results of my recent test. Here’s what I have for the hour long run, after a 10-15min warmup (listed as pace / hr):
1. 8:21 / 154
2. 8:14 / 158
3. 8:09 / 160
4. 8:06 / 161
5. 8:14 / 160
6. 8:09 / 161
7. 8:11 / 161
.4 8:03 / 161
Training peaks says the Pa:Hr is .84%, which from what I understand, suggests that my AeT is substantially higher. However, about 6 months ago, I ran a half at ~1:30 with an average heart rate of 172-173 (which, I was told means my AnT is probably around 180), so, I’m not sure how much higher the AeT can go.
So, main question about how to interpret these results: Does it make sense that my AeT is higher, maybe somewhere around 165? (Which would mean I don’t suffer from ADS?). Or might I be missing something / reading this wrong somehow?
Thanks!
• Participant
MMK on #34841
Giving this a light boost, for I fear it may have gotten lost amidst all the other requests for help interpreting the test.
I also have a little more data: during an hour long run today (minus ~ 2mile warmup), I averaged ~8:21/mi pace and a flat 155 heart rate all the way through.
Any and all suggestions are welcome!
Participant
MMK on #35316
At the risk of annoyance, a last request for some interpretive help, along with a final data point.
9.6 mile run with the following averages:
First half (minus warmup): HR 160, pace 8:31
Second half: HR 159, pace 8:41
Total Pa:Hr = 1.23%
Any help interpreting all these results would be much appreciated, as I feel a little lost.
Participant
Dada on #35337
Hi,
What I would do:
1. Redo the test on a treadmill to rule out the course as a matter of error
2. Start with 165 for the AeT test
3. Do an AnT test to determine your real threshold to see if you have ADS
Dada
Participant
MMK on #35379
Thanks! I’ll go ahead and give it a shot. 165 is much higher than what I usually run for the majority of my miles, but I guess it’s possible for it to be around my AeT.
My only hesitation with the treadmill is that for some reason I’m usually much less efficient on it (higher heart rate, lower pace) but for the purposes of this test, perhaps that won’t matter too much.
Participant
Dada on #35392
Reasons I can think off why a treadmill could be different:
– slope
– room temperature (Missing heat acclimatization)
– house dust allergy (asthma)
Can you rule out some reasons?
Participant
MMK on #35438
-Slope — not sure how this would effect anything, if I’m on the treadmill it’s at max 1%
-room temperature — yes probably, it’s definitely hotter than running outside in Chicago right now
-allergy — no
I’ll try the test this week and report back!
Thanks for the suggestions
Participant
Dada on #35439
I opened the window during my last AeT test and my heartrate dropped by 5 beats. And in the Caribbean, the heat forced me to walk to stay below my AeT.
Viewing 7 replies - 1 through 7 (of 7 total)
• The forum ‘General Training Discussion’ is closed to new topics and replies. | 842 | 3,096 | {"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} | 2.59375 | 3 | CC-MAIN-2024-38 | latest | en | 0.923227 |
https://instrumentationtools.com/flow-meter-k-factor-and-calculations/ | 1,726,638,032,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651836.76/warc/CC-MAIN-20240918032902-20240918062902-00651.warc.gz | 290,308,019 | 59,216 | # Flow Meter K-factor and Calculations
The axial design of Turbine Flow Meter is inherently linear within a known turndown range, typically 15:1, based on velocity of the measured fluid.
## Flow Meter K Factor
The device has unmatched capability of precise and repeatable K-Factor generation based on the turning of the balanced rotor and the subsequently generated frequency pulse signal via the magnetic coil assembly, providing accuracy of +/-.15% with special calibration applied.
Each pulse generated represents a discrete amount of volumetric throughput. Dividing the total number of pulses generated by the specific amount of liquid product that passed through the PTF determines the K-Factor.
Put in the simplest of terms, K-Factor is no more complicated than understanding that the meter will generate a specific number of pulses for every unit of product passing through it. If the user can detect the pulses, then it is a straightforward task to determine flow rate and totals.
The K-Factor, expressed in pulses per unit volume, may be used to electronically provide an indication of volumetric throughput directly in engineering units.
Mass flow can be accomplished by the addition of pressure and temperature sensors that are factored electronically. The electronic device continuously divides the incoming pulses by the K-Factor, or multiplies them with the inverse of the K-Factor, to provide factored totalization, rate indication, and various outputs.
The frequency of the pulse output, or number of pulses per unit time, is directly proportional to the rotational rate of the turbine rotor.
Therefore, this frequency of the pulse output is proportional to the rate of the flow.
By dividing the pulse rate by the K-Factor, the volumetric throughput per unit time of the rate of flow can be determined.
Standard electronic devices are commonly used to provide instantaneous totalization and flow rate indication.
Plotting the electrical signal output versus flow rate provides the characteristics profile or calibration curves for the PTF.
Although the concept of K-Factor is applied widely with other types of meters, it should be noted that this value is calculated from analog values in other meters where a magnetic coil assembly reading a turning rotor is not used.
This calculation has built-in error factors based on a relatively imprecise primary input coupled with conversions from analog frequency and sometimes back to analog for transmission purposes.
Also See: Turbine Flow Meter Animation
## What is a K-factor
Simply stated a K-factor is a dividing factor. The term is usually encountered when dealing with pulse signals although analog K-factors are sometimes used.
## Pulse Signal K-factors
All pulse output type flow meters when they are dispatched by their manufacturer will have a calibration certificate.
The calibration certificate will show that the meter has been calibrated over its flow range and noted on the certificate will be the average K-factor for the meter.
This K-factor will be given in terms of the number of pulses produced by the meter for a given volumetric flow. (e.g.) 200 pulses per U.S. gallon, 150 pulses per liter etc.
This K-factor is the value that is entered into a batch meter or indicator/totalizer in order to give a readout in engineering units.
## Calculations Using K-Factors
Example 1
If the display on a rate meter is required in U.S. gallons per second, and the K-factor of the flow meter is 210 pulses per U.S. gallon, then the K-factor entered into the rate meter would be 210.
If a totalizer associated with the same flow meter was to be set up so as to totalize in U.S. gallons the totalizer K-factor would be 210.
If the totalizer was to be set to totalize in tenths of a gallon the K-factor would be 210/10 = 21
Example 2
If the display on a rate meter is required in U.S. gallons per minute, and the K-factor of the flow meter is 210 pulses per U.S. gallon, then the K-factor entered into the rate meter would be: 210/60 = 3.5
When batching, indication or totalization has to be carried out using an analog input signal a flow meter first converts the 4 to 20 mA signal into a 0 to 10000 Hz. signal.
The K-factor is then calculated by relating the engineering unit equivalent of 20 mA to the 10000 Hz. signal.
Example 3
A vortex meter outputs 20 mA when the flow is 2000 U.S. gallons per minute, we wish to display the rate in gallons per minute. The rate K-factor is = 10000/2000 = 5
The value of the totalizer K-factor will depend on whether the flow rate was given in units per second, minute or hour and whether it is desired to totalize in whole units, tenths, hundredths etc.
If the flow rate was given in units per second the totalizer K-factor (for whole units) is obtained by multiplying the rate K-factor by 1.
If the flow rate was given in units per minute the totalizer K-factor (for whole units) is obtained by multiplying the rate K-factor by 60.
If the flow rate was given in units per hour the totalizer K-factor (for whole units) is obtained by multiplying the rate K-factor by 3600.
The totalizer K-factor in example 3 will be = 5 x 60 = 300 in order to totalize in gallons.
If we wished to totalize in tenths of a gallon the K-factor would be 5 x 60/10 = 30
Example 4
An electromagnetic flow meter outputs 20 mA when the flow is 20 liters per second, we wish to display the flow rate in liters per second and totalize in M³.
The rate K-factor is 10000/20 = 500
The totalizer K-factor will be 500 x 1/0.001 = 500000
## Multi-Point K-factor
Some flow applications dictate that multiple K-factors are used. Two applications that require multiple K-factors are:
• flow meters with non linear outputs
• wide turndown flow applications
Some Flow meters have an option available that allows the user to input from 3 to 16 K-factors.
This multi-point K-factor option is available for both pulse and analog inputs.
## Multi-Point K-factor for Pulse inputs
The first step is to calculate K-factors to cover each flow range. This is done by taking the information on the flow meter manufacturer’s calibration sheet and calculating the K-factors.
An alternative to using the manufacturers’ data is to conduct tests on site against a calibrated standard.
The second step is to relate an incoming frequency range from the flow meter to a given calculated K-factor.
The final step is to program these values into the instrument.
Example 5
A turbine flow meter has the following calibration data.
From the above calibration sheet data we can relate the incoming frequency to the K-factor as follows.
The 16 point K-factor would be programmed as follows.
Note that because the K-factor for the last two points are the same, any frequency higher than 43.67 Hz will be modified by a K-factor of 52.4 The set up is now complete.
The procedure for analog inputs is essentially the same as for pulse inputs.
The first step is to calculate K-factors to cover each flow range. This is done by taking the information on the flow meter manufacturer’s calibration sheet and calculating the K-factors.
An alternative to using the manufacturer’s data is to conduct tests on site against a calibrated standard.
The second step is to relate an incoming flow value from the flow meter to a given calculated K-factor.
The final step is to program these values into the instrument.
Example 6
A vortex flow meter has the following calibration data.
Base K-factor 10000/100 = 100
Using the MASStrol as an example the K-factors would be programmed as follows.
Note that the point after the final one should have a flow value entered that is very much higher than the true maximum flow rate of the meter.
Note also that as the last two K-factors are the same any flow above 100 gpm will be modified by a K-factor of 104. The setup is now complete.
Credits: Sponsler & Kessler-Ellis (KEP)
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## 5 thoughts on “Flow Meter K-factor and Calculations”
1. Dear Manager
We have on the same GAS LINE 2 flowmeters for Gas Turbine.
The first near the Gas turbine is Corriolis Micro motion (as reference) and is measuring the flow of gas on kg/h.
The Second is little bit far to the gas Turbine and is an Fluidwell E018-P associated with turbinemeter BARTON 7403.
The K-factor for the turbinemeter is (on calibration certificate) : 25.52 pul/Gal (UK).
linearity : 0.30 %
I need to introduce the K-factor on the Fluidwell E018-P
Wich value I have to us……. 25.52 pulse/Gal or ……?
I need the same unit on the totalizer fluidwell for flowrate : kg/h
but for totalizer : I need to use : cubic meter
2. could you explain about k-factor in other flow meter?(Coriolis, ultrasonic, orifice)
3. if we have a liquid having density 801 kg/m3, what will be the k factor for flow meter (i-tech) to measure correct flow rate and total flow…
4. Dear
I am very grateful for your wonderful cooperation and this shows that you are very wonderful people I will send you a picture of the report and the way in which the values of (KF) are calculated Is this method correct? According to international references API.
calculated K-factor= factory meter K-factor/calculated meter factor
with friendliness. | 2,067 | 9,336 | {"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} | 3.109375 | 3 | CC-MAIN-2024-38 | latest | en | 0.906562 |
https://cran.microsoft.com/snapshot/2018-09-26/web/packages/agridat/vignettes/agridat_papadakis.html | 1,675,540,438,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764500151.93/warc/CC-MAIN-20230204173912-20230204203912-00406.warc.gz | 210,221,173 | 27,526 | # R setup
Papadakis (1937) believed that traditional blocking in field experiments did not adequately represent the patchiness of soil fertility patterns and he instead proposed adjusting the yield of each plot by the performance of the neighboring plots.
If there is heterogeneity in the field that is of a scale smaller than the block (but larger than the individual plots) then adjacent plots will be positively correlated and this information about the neighboring plots can be used to reduce the effect of spatial heterogeneity and increase the accuracy of the treatment effects.
The Papadakis method is a nearest neighbor method that uses a residual covariate in the analysis. In essence, the method follows the following steps.
1. Fit a treatment model and calculate the residuals from the model.
2. Calculate covariates that are the average of the neighboring residuals.
3. Fit a model with additional covariate terms for the residuals.
The left-right (LR) covariate for the (i,j)th plot is the average of the residuals for the plots immediately to the left and right of the (i,j)th plot. If one of these neighbors is missing, then the covariate is constructed from the single remaining neighboring residual. Border plots use only one neighboring residual. The up-down (UD) covariate is similarly constructed from residuals for plots immediately up or down from the (i,j)th plot.
papcov <- function(resid,x,y){
# Make sure x and y are numeric
if(is.factor(x)) x <- as.numeric(as.character(x))
if(is.factor(y)) y <- as.numeric(as.character(y))
xy <- paste(x,y,sep=":")
# Average neighboring residuals in up/down direction
xym1 <- paste(x,y-1,sep=":")
xyp1 <- paste(x,y+1,sep=":")
rm1 <- resid[match(xym1,xy,NA)]
rp1 <- resid[match(xyp1,xy,NA)]
ud <- (rm1+rp1)/2
# If only one neighboring residual is available, then just use it
ud <- ifelse(is.na(ud) & !is.na(rm1),rm1,ud)
ud <- ifelse(is.na(ud) & !is.na(rp1),rp1,ud)
# Average neighboring residuals in left/right direction
xm1y <- paste(x-1,y,sep=":")
xp1y <- paste(x+1,y,sep=":")
cm1 <- resid[match(xm1y,xy,NA)]
cp1 <- resid[match(xp1y,xy,NA)]
lr <- (cm1+cp1)/2
# If only one neighboring residual is available, then just use it
lr <- ifelse(is.na(lr) & !is.na(cm1),cm1,lr)
lr <- ifelse(is.na(lr) & !is.na(cp1),cp1,lr)
return(list(LR=lr, UD=ud))
}
# Reproduce Hinz 1987 case 2
Hinz (1987) used the Papadakis technique to analzye a field experiment of tobacco.
library("agridat")
data(federer.tobacco)
dat <- federer.tobacco
dat <- transform(dat, height=height-600) # For simplicity
# Model 1 - RCB
m1 <- aov(height ~ factor(block) + factor(dose), dat)
anova(m1)
## Analysis of Variance Table
##
## Response: height
## Df Sum Sq Mean Sq F value Pr(>F)
## factor(block) 7 388315 55474 1.8352 0.1056
## factor(dose) 6 273875 45646 1.5100 0.1985
## Residuals 42 1269586 30228
# Model 2 - Row/Col as class variables
m2 <- aov(height ~ factor(block) + factor(dose) + factor(row), dat)
anova(m2)
## Analysis of Variance Table
##
## Response: height
## Df Sum Sq Mean Sq F value Pr(>F)
## factor(block) 7 388315 55474 7.5455 1.355e-05 ***
## factor(dose) 6 273875 45646 6.2088 0.0001521 ***
## factor(row) 6 1004920 167487 22.7816 6.767e-11 ***
## Residuals 36 264666 7352
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
# Model 3 - Two-step Papadakis
m3 <- aov(height ~ factor(dose), dat)
dat <- cbind(dat, papcov(m3$resid, dat$block, dat$row)) m4 <- aov(height ~ factor(dose) + LR + UD, data=dat) anova(m4) ## Analysis of Variance Table ## ## Response: height ## Df Sum Sq Mean Sq F value Pr(>F) ## factor(dose) 6 273875 45646 3.6857 0.004407 ** ## LR 1 1061352 1061352 85.6998 3.636e-12 *** ## UD 1 14477 14477 1.1689 0.285136 ## Residuals 47 582073 12385 ## --- ## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1 # Resid MS uses 1 df less to account for covariates. Matches Hinz. 582073 / 46 ## [1] 12653.76 # Iterated example as given in Stroup et al, Table 2 Stroup, Baenziger, and Mulitze (1994) used the Papadakis tecnique in an iterative manner. library("agridat") data(stroup.nin) dat2 <- stroup.nin dat2 <- subset(dat2,!is.na(yield)) n.gen <- nlevels(dat2$gen)
# RCB model, ranks match Stroup Table 2, RCB Alliance
m5 <- lm(yield ~ gen -1 + rep, data=dat2)
pred.rcb <- coef(m5)[1:n.gen] # RCB adj means
rev(57-sort(rank(pred.rcb)))
## genNE86503 genNE87619 genNE86501 genRedland genCenturk78 genNE83498
## 1 2 3 4 5 6
## genSiouxland genNE86606 genArapahoe genNE87613 genNE86607 genLancer
## 7 8 9 10 11 12
## genTAM107 genCheyenne genNE87446 genHomestead genScout66 genNE83404
## 13 14 15 16 17 18
## genColt genNE86509 genNE87513 genLancota genNE85556 genNE87408
## 19 20 21 22 23 24
## genBrule genNE87463 genNE87615 genBuckskin genNE87403 genNE87522
## 25 26 27 28 29 30
## genNE87451 genNE86582 genGage genNorkan genNE86482 genNE83406
## 31 32 33 34 35 36
## genKS831374 genNE87457 genNE86507 genVona genNE87512 genNE87627
## 37 38 39 40 41 42
## genNE83407 genNE86527 genNE87612 genNE85623 genCentura genNE83T12
## 43 44 45 46 47 48
## genNE86T666 genNE87409 genTAM200 genCody genRoughrider genNE84557
## 49 50 51 52 53 54
## genNE87499 genNE83432
## 55 56
# Initial genotype model (no blocks)
m6 <- lm(yield ~ gen -1, data=dat2)
pp <- papcov(resid(m6), dat2$col, dat2$row)
dat2$LR <- pp$LR
dat2$UD <- pp$UD
# Single iteration of Papadakis model
m7 <- lm(yield ~ gen - 1 + LR + UD, data=dat2)
# Residual = observed - adjusted mean
resid <- dat2$yield - adjmn[match(paste0("gen",dat2$gen),names(adjmn))]
# Now iterate Papadakis method to convergence
iter <- 0
notConv <- TRUE
while(notConv){
iter <- iter + 1
# Covariates based on residuals
pp <- papcov(resid, dat2$col, dat2$row)
dat2$LR <- pp$LR
dat2$UD <- pp$UD
m8 <- lm(yield ~ gen - 1 + LR + UD, data=dat2)
# Check convergence of adjusted means
cat("Iteration: ",iter," tol: ",tol,"\n")
notConv <- tol > .001
resid <- dat2$yield - adjmn[match(paste0("gen",dat2$gen),names(adjmn))]
}
## Iteration: 1 tol: 51.60518
## Iteration: 2 tol: 7.823055
## Iteration: 3 tol: 1.445812
## Iteration: 4 tol: 0.3619413
## Iteration: 5 tol: 0.09555693
## Iteration: 6 tol: 0.02840853
## Iteration: 7 tol: 0.008783593
## Iteration: 8 tol: 0.003297561
## Iteration: 9 tol: 0.001650757
## Iteration: 10 tol: 0.001157637
## Iteration: 11 tol: 0.001000046
## Iteration: 12 tol: 0.0009401906
pred.pap <- adjmn
# Ranks almost match Stroup et al, Table 2, Alliance, RCB+NNA-PAP
all <- data.frame(rcb=57-rank(pred.rcb), nna=57-rank(pred.pap))
all[order(all\$rcb),]
## rcb nna
## genNE86503 1 14
## genNE87619 2 4
## genNE86501 3 24
## genRedland 4 6
## genCenturk78 5 15
## genNE83498 6 2
## genSiouxland 7 28
## genNE86606 8 11
## genArapahoe 9 18
## genNE87613 10 9
## genNE86607 11 16
## genLancer 12 35
## genTAM107 13 37
## genCheyenne 14 21
## genNE87446 15 46
## genScout66 17 13
## genNE83404 18 23
## genColt 19 17
## genNE86509 20 42
## genNE87513 21 45
## genLancota 22 49
## genNE85556 23 3
## genNE87408 24 36
## genBrule 25 7
## genNE87463 26 39
## genNE87615 27 32
## genBuckskin 28 1
## genNE87403 29 53
## genNE87522 30 52
## genNE87451 31 33
## genNE86582 32 48
## genGage 33 29
## genNorkan 34 51
## genNE86482 35 25
## genNE83406 36 19
## genKS831374 37 5
## genNE87457 38 34
## genNE86507 39 10
## genVona 40 30
## genNE87512 41 47
## genNE87627 42 55
## genNE83407 43 27
## genNE86527 44 8
## genNE87612 45 20
## genNE85623 46 31
## genCentura 47 26
## genNE83T12 48 40
## genNE86T666 49 56
## genNE87409 50 12
## genTAM200 51 54
## genCody 52 43
## genRoughrider 53 22
## genNE84557 54 38
## genNE87499 55 41
## genNE83432 56 50
# Visually compare the coefficients from the two methods
lims=range(c(pred.rcb,pred.pap))
plot(pred.rcb, pred.pap, xlim=lims, ylim=lims,
abline(0,1) | 3,170 | 8,976 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.265625 | 3 | CC-MAIN-2023-06 | latest | en | 0.847969 |
foldable-robotics.github.io | 1,695,530,846,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233506559.11/warc/CC-MAIN-20230924023050-20230924053050-00186.warc.gz | 286,557,916 | 5,984 | # Least Squares Optimization
%matplotlib inline
import numpy
import numpy.random
import matplotlib.pyplot as plt
import numpy.linalg
x = numpy.r_[-10:10:.5]
x
array([-10. , -9.5, -9. , -8.5, -8. , -7.5, -7. , -6.5, -6. ,
-5.5, -5. , -4.5, -4. , -3.5, -3. , -2.5, -2. , -1.5,
-1. , -0.5, 0. , 0.5, 1. , 1.5, 2. , 2.5, 3. ,
3.5, 4. , 4.5, 5. , 5.5, 6. , 6.5, 7. , 7.5,
8. , 8.5, 9. , 9.5])
Define y as a function of x. This can be anything
y = 3*x
#y = x**2
y = 2*x**3
y += 2*numpy.sin(x)
y /= y.max()
y
array([-1.16581845e+00, -1.00000000e+00, -8.50824970e-01, -7.17279360e-01,
-5.98377987e-01, -4.93191502e-01, -4.00859731e-01, -3.20588089e-01,
-2.51627928e-01, -1.93245642e-01, -1.44688088e-01, -1.05152789e-01,
-7.37702189e-02, -4.96025012e-02, -3.16588415e-02, -1.89239181e-02,
-1.03922769e-02, -5.10030999e-03, -2.14798940e-03, -7.05033996e-04,
0.00000000e+00, 7.05033996e-04, 2.14798940e-03, 5.10030999e-03,
1.03922769e-02, 1.89239181e-02, 3.16588415e-02, 4.96025012e-02,
7.37702189e-02, 1.05152789e-01, 1.44688088e-01, 1.93245642e-01,
2.51627928e-01, 3.20588089e-01, 4.00859731e-01, 4.93191502e-01,
5.98377987e-01, 7.17279360e-01, 8.50824970e-01, 1.00000000e+00])
Now create an array of normally distributed noise
rand = numpy.random.randn(*y.shape)/10
y_rand = y+rand
y_rand
array([-1.03966184, -1.11103693, -0.89738405, -0.7897717 , -0.61606444,
-0.58075587, -0.32366094, -0.41986849, -0.2252123 , -0.27872011,
-0.06530092, -0.05975687, -0.0774088 , -0.1278407 , 0.08502485,
-0.06129926, -0.12204749, 0.14838957, 0.02424667, 0.00377575,
0.11386753, 0.00182104, 0.01021305, -0.00858236, 0.14212112,
0.18574359, 0.03174494, -0.01499514, 0.12695571, 0.08401802,
0.21914263, 0.24385571, 0.1560743 , 0.38365295, 0.59747615,
0.46985472, 0.55933946, 0.61241949, 0.9059338 , 0.79946197])
Plot y against the random vector
plt.plot(x,y)
plt.plot(x,y_rand,'o')
[<matplotlib.lines.Line2D at 0x7ff170474310>]
Now assume you create a model of the form $y(k)^$ where k are your model coefficients. You want to pick your model in such a way that the error from $y-y^$ is minimized. The residual error, or just residual can be expressed as
$$r=y-y^*$$
thus the sum of squared error is
$$||r||^2 = ||y-y^||^2 = y^T y - 2y^Ty^* + {y^}^T y^$$
Now in matrix form, $y^$ takes the form $y^=Ak$, where $k$ is your set of model weights in the form
$$k=\left[\begin{matrix}k_1&k_2&\ldots &k_m\end{matrix}\right]^T,$$
and $A$ is your set of models applied to the input variable $x$ $$A = A(x) = \left[\begin{matrix} a_1(x_0) & a_2(x_0)& \ldots& a_m(x_0)\ a_1(x_1) & a_2(x_1)& \ldots &a_m(x_1)\ \vdots & \vdots & \ddots& \vdots\ a_1(x_n) & a_2(x_n)& \ldots& a_m(x_n)\ \end{matrix} \right]$$
### Example
In our case we will try to model our data with $A(x) = \left[\begin{matrix}x&x^2&x^3&\sin(x)\end{matrix}\right]$
A = numpy.array([x,x**2,x**3,x**4,x**5,x**6]).T
A.shape
(40, 6)
With the model stated, we may now expand the sum of squared error with our model:
$$||r||^2 = y^T y - 2y^T(Ak) + k^TA^TAk$$
But when optimizing, you are not optimizing for $x$ or even $A(x)$, which is either given or selected by you, but for the $k$ weighting parameters, which you may select in order to minimize the above equation. Thus the error will be minimized at the point where $$\frac{d(||r||^2)}{dk}=0,$$ or
$$- 2y^T A+ 2k^T A^T A=0$$
Solving for k, $$k^T A^T A= y^T A$$ $$(k^T A^T A)^T= (y^T A)^T$$ $$(A^T A)^Tk= A^Ty$$ $$(A^T A)k= A^Ty$$ $$(A^T A)^{-1}(A^T A)k= (A^T A)^{-1}A^Ty$$
The optimimum value for k is thus
$$k= (A^T A)^{-1}A^Ty$$
### Example
B = numpy.linalg.inv(A.T.dot(A))
coeff = B.dot(A.T.dot(y_rand))
coeff
array([-2.18923289e-03, 2.77108020e-03, 1.55743851e-03, -8.14782525e-05,
-4.86516898e-06, 4.69654727e-07])
Plotting the coefficients, we see that the weights for $x^3$ are near 1 while all other weights are quite small.
xx = numpy.r_[:6]
labels = '$x$','$x^2$','$x^3$','$x^4$','$x^5$','$x^6$'
f = plt.figure()
ax.bar(xx,coeff)
ax.set_xticks(xx)
ax.set_xticklabels(labels)
[Text(0, 0, '$x$'),
Text(1, 0, '$x^2$'),
Text(2, 0, '$x^3$'),
Text(3, 0, '$x^4$'),
Text(4, 0, '$x^5$'),
Text(5, 0, '$x^6$')]
To return $y^*$,
y_model = A.dot(coeff)
Plotting the noisy data against the model, we get
fig = plt.figure()
a = ax.plot(x,y_rand,'o')
b = ax.plot(x,y_model)
ax.legend(a+b,['data','model'])
<matplotlib.legend.Legend at 0x7ff17008f490>
And finally, to plot the residual
plt.figure()
plt.plot(x,y_model-y_rand)
[<matplotlib.lines.Line2D at 0x7ff15d69c6d0>]
Now try other models, higher resolution data, and different domains | 2,117 | 4,639 | {"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": 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} | 2.53125 | 3 | CC-MAIN-2023-40 | latest | en | 0.292834 |
http://www.profr.org/homework-help/view/273-the-circumference-of-a-circle-whose-diameter-is-7-inches-is-approximately | 1,576,288,805,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575540579703.26/warc/CC-MAIN-20191214014220-20191214042220-00232.warc.gz | 219,904,040 | 22,293 | Connect faster with
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# The circumference of a circle whose diameter is 7 inches is approximately:
The circumference of a circle whose diameter is 7 inches is approximately:
Posted in Math, asked by Furqan Ali, 6 years ago. 30236 hits.
## 10
Diameter = 7 inches therefore, Radius = Diameter/2 = 3.5 inches
Circumference of Circle = 2∏r (2∏ = Constant, r = Radius)
= 2 x 22/7 x 3.5 (∏ = 22/7)
= 22 inches
Ans: Circumference of circle = 22 inches.
Gurudas Kadam - 6 years ago
- Just now
## 0
Just now
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/homework-help | 197 | 665 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.203125 | 3 | CC-MAIN-2019-51 | latest | en | 0.828876 |
https://www.techwhiff.com/issue/a-tennis-ball-rolls-horizontally-to-the-left-for-30--589826 | 1,675,115,551,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764499829.29/warc/CC-MAIN-20230130201044-20230130231044-00453.warc.gz | 1,049,931,992 | 12,093 | # A tennis ball rolls horizontally to the left for 30 m at a constant speed. what is the velocity after 15 seconds? what is the speed? **PLEASE HELP - for physics, list variables
###### Question:
a tennis ball rolls horizontally to the left for 30 m at a constant speed. what is the velocity after 15 seconds? what is the speed?
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### The histogram below shows the distribution of times, in minutes, required for 25 rats in an animal behavior experiment to navigate a maze successfully.The histogram has 8 intervals, each with a width of 1 minute and with midpoints at 2, 3, 4, ..., and 9 minutes. The frequencies are 3, 8, 6, 4, 2, 1, 0, and 1.To describe the center and spread of the above distribution, the appropriate numerical measures are:a. the mean and the standard deviationb. the mean and the medianc. the median and the IQRd
The histogram below shows the distribution of times, in minutes, required for 25 rats in an animal behavior experiment to navigate a maze successfully.The histogram has 8 intervals, each with a width of 1 minute and with midpoints at 2, 3, 4, ..., and 9 minutes. The frequencies are 3, 8, 6, 4, 2, 1,... | 839 | 3,494 | {"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} | 2.578125 | 3 | CC-MAIN-2023-06 | latest | en | 0.878789 |
http://maskray.me/leetcode/paint-house.cc | 1,721,144,340,000,000,000 | text/plain | crawl-data/CC-MAIN-2024-30/segments/1720763514759.37/warc/CC-MAIN-20240716142214-20240716172214-00053.warc.gz | 16,475,068 | 844 | // Paint House #define REP(i, n) FOR(i, 0, n) #define FOR(i, a, b) for (int i = (a); i < (b); i++) class Solution { public: int minCost(vector>& costs) { int n = costs.size(); if (! n) return 0; FOR(i, 1, n) REP(j, 3) costs[i][j] += min(costs[i-1][(j+1)%3], costs[i-1][(j+2)%3]); return *min_element(costs[n-1].begin(), costs[n-1].end()); } }; | 135 | 343 | {"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} | 2.625 | 3 | CC-MAIN-2024-30 | latest | en | 0.192319 |
https://erkaman.github.io/posts/ray_trace_inverse_ray.html | 1,566,276,903,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027315222.56/warc/CC-MAIN-20190820045314-20190820071314-00350.warc.gz | 453,848,395 | 3,868 | # How to ray-intersect a transformed geometry, without actually transforming it: a geometric illustration
$\newcommand{\mvec}[1]{\mathbf{#1}}\newcommand{\gvec}[1]{\boldsymbol{#1}}\definecolor{eqcol2}{RGB}{114,0,172}\definecolor{eqcol1}{RGB}{172,0,114}\definecolor{eqcol3}{RGB}{0,172,114}\definecolor{eqcol4}{RGB}{230,190,120}$The goal of this article, is to describe how we can transform some geometry with a transformation matrix, and then compute the intersection between the transformed geometry, and some ray. However, we want to do this without actually transforming the geometry, since this can be expensive.
Let us say we have a ray $\mvec{p}(t) = \mvec{o} + \mvec{d}t$, with origin $\mvec{o}$ and direction $\mvec{d}$, and we want to find its intersection with some geometry, by finding the value of $t$ for this intersection.
For an example, consider the intersection between above ray, with the given values of $\mvec{o}$ and $\mvec{d}$. and the green geometry. The ray certainly intersects this geometry, and the value of $t$ is $t=2$, since the ray is heading straight towards the wall of the geometry, and the distance between $\mvec{o}$ and the wall of the geometry is 2. Let us now translate this geometry, 4 units upwards on the $y$-axis, by applying the following matrix: \begin{equation*} T = \begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 4 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix}, \end{equation*} then we get this geometry:
This translated geometry is intersected by the ray with origin $\mvec{o}=(-4,4)$ and direction $\mvec{d}=(+1,0)$, with again $t=2$. However, is there a way we can calculate this exact intersection, without having to translate the geometry at all. The geometry might be very complex, and have millions upon millions of triangles, so transforming the whole geometry might be expensive. Also, the geometry might also be storing some kind of spatial data structure for fast ray-intersection queries(like a bounding-volume hierarchy), which would also have to be updated if we transformed and updated the geometry vertex data. If we found a way to intersect the translated geometry, without actually having to translate it all, these costs would become zero.
So, is there a way to intersect the geometry with the ray, without having to transform the geometry at all? Yes, and it is actually rather simple: instead of transforming the geometry, we apply the inverse of the transform on the ray. And then, we intersect this inverse transformed ray, with the original geometry. A small example should convince the reader that this is a valid approach. The inverse of the translation matrix $T$ we just used, is \begin{equation*} T^{-1} = \begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & -4 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix}, \end{equation*} If we apply this matrix to the ray with properties $\mvec{o}=(-4,4)$, $\mvec{d}=(+1,0)$, we arrive at the following situation:
Now as we can observe, we can intersect this inverse transformed ray, with the untransformed geometry, and by doing so, we arrive at our desired result $t=2$.
One subtle detail here, is that we must make sure to apply this inverse transformation on both the ray origin, and direction. To understand this, let us go through a slightly more subtle example. In this example, the transformation matrix is \begin{equation*} M = \begin{bmatrix} 0 & -1 & 0 & 0 \\ 1 & 0 & 0 & 4 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix} = \begin{bmatrix} 0 & -1 & 0 & 0 \\ 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix} \begin{bmatrix} 1 & 0 & 0 & 4 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix} , \end{equation*} This is a transform that first translates +4 units on the $x$-axis, and then does a 90 degrees counter-clockwise rotation about the origin $(0,0)$. If we apply this transform to a geometry, we arrive at the below situation
We want to intersect the above ray with the original geometry instead of the transformed geometry, and so we begin by inverse transforming the ray origin. The inverse of the transform is \begin{equation*} M^{-1} = \begin{bmatrix} 0 & 1 & 0 & -4 \\ -1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix} \end{equation*} now apply this to the ray origin: \begin{equation*} \begin{bmatrix} 0 & 1 & 0 & -4 \\ -1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix} \begin{bmatrix} -3 \\ +4 \\ 0 \\ 1 \end{bmatrix} = \begin{bmatrix} 0 \\ +3 \\ 0 \\ 1 \end{bmatrix} \end{equation*} And after doing this transform, we obtain:
And we can see that the ray obviously will not intersect the original geometry in this configuration. Because we also need to inverse transform the ray direction, so let us do this: \begin{equation*} \begin{bmatrix} 0 & 1 & 0 & -4 \\ -1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix} \begin{bmatrix} +1 \\ 0 \\ 0 \\ 0 \end{bmatrix} = \begin{bmatrix} 0 \\ -1 \\ 0 \\ 0 \end{bmatrix} \end{equation*} And by doing this, we end up with
Now the ray does intersect the geometry, and the point it does intersect, is the same point it would have intersected, on the transformed geometry. We highlight these two intersection points in blue for clarity
As long as we remember to inverse transform both the ray origin and direction, we can intersect a transformed geometry, without actually having to transform it. It even works if the transform matrix has a scaling component to it. We must just make sure to not normalize the ray direction after inverse transforming, and we will get the correct result. The one requirement to using the technique is that the transform matrix is invertible. If it is not invertible, then we are not able to inverse transform the ray to begin with. | 1,688 | 5,702 | {"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} | 4.3125 | 4 | CC-MAIN-2019-35 | latest | en | 0.859846 |
https://pro.arcgis.com/en/pro-app/2.7/help/analysis/raster-functions/vector-field-function.htm | 1,719,269,590,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198865490.6/warc/CC-MAIN-20240624214047-20240625004047-00277.warc.gz | 391,911,693 | 6,451 | # Vector Field function
## Overview
The Vector Field function is used to composite and convert two rasters into a two-band raster that is of either data type Magnitude-Direction or data type U-V.
Once you specify the parameters, you can use the Vector Field Renderer function to display your raster.
## Notes
If the Input Data Type parameter is Magnitude-Direction, you also need to specify the Angular Reference System. parameter.
The U value is sometimes referred to as the zonal velocity or velocity along latitudinal lines. The V value is sometimes referred to as the meridional velocity or velocity along longitudinal lines.
### Conversion between U-V and Magnitude-Direction
The two subsections below contain the formulas for the appropriate conversions. The variables are as follows:
• u = vector-U
• v = vector-V
• m = magnitude
• d = direction
#### Magnitude-Direction to U-V
``u = m * cos (pi *d / 180)``
``v = m * sin (pi * d /180)``
#### U-V to Magnitude-Direction
``m = sqrt (u * u + v * v)``
``d = atan2 (v / u) * 180 / p``
## Parameters
ParameterDescription
Raster1
Input Raster 1
The first input raster.
This will be either the U value or the magnitude.
Raster2
Input Raster 2
The second input raster.
This will be either the V value or the direction.
Input Data Type
The type of vector field your inputs represent.
• Unknown—It is unknown whether the inputs represent U-V or Magnitude-Direction.
• U-V—Input data represents U and V components.
• Magnitude-Direction—Input data represents magnitude and direction.
Angular Reference System
Specifies how the direction component was measured.
• Geographic—0° points due north, and 90° points due east
• Arithmetic—0° points due east, and 90° points due north
Output Data Type
Specifies the type of vector field your output will represent.
• Magnitude-Direction—Output data represents magnitude and direction.
• U-V—Output data represents U and V components. | 459 | 1,955 | {"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} | 2.6875 | 3 | CC-MAIN-2024-26 | latest | en | 0.723518 |
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# For Speed Of Ball Mill India
• Ball Mills an overview ScienceDirect Topics Ball mills The ball mill is a tumbling mill that uses steel balls as the grinding media The length of the cylindrical shell is usually 1 times the shell diameter Figure The feed can be dry with less than 3% moisture to minimize ball coating or slurry containing 20 40% water by weight
• Speeds and Feeds 101 In The Loupe02 10 2017· Speeds and feeds are the cutting variables used in every milling operation and vary for each tool based on cutter diameter operation material etc Understanding the right speeds and feeds for your tool and operation before you start machining is
• End Mills The Essential Beginners Guide Ball Nose End Mills Ball nose mills have a radius at the bottom which makes for a nicer surface finish in your workpiece meaning less work for you as the piece won t need to be finished any further They are used for contour milling shallow slotting pocketing and contouring applications
• Ball Mill Critical Speed Mineral Processing Metallurgy19 06 2015· A Ball Mill Critical Speed actually ball rod AG or SAG is the speed at which the centrifugal forces equal gravitational forces at the mill shell s inside surface and no balls will fall from its position onto the shell The imagery below helps explain what goes on inside a mill as speed varies Use our online formula The mill speed is typically defined as the percent of the Theoretical
• Tips for Selecting a Lubricant for Ball Screws14 01 2010· Speeds that are high for ball screws are no problem for grease so speed is not a criterion for selection For example grease is the accepted lubricant for machine tool spindles with dn bearing bore mm x rpm values as high as 1 000 000
• SPEEDS AND FEEDS FOR CARBIDE ENDMILLS Internal Toolnote all speeds and feeds listed here are provided for reference only formula for computing table speed x number of teeth in cutter x rpm example cutting aluminum with a 3/4 key x 10 x 1600 = 16 tiplier number of teeth rpm formula for computing spindle speed is
• TECHNICAL How to Spec a Mill Gear Power TransmissionThe critical speed of rotation is the speed in rpm at which an infinite ly small particle will cling to the inside of the liners of the mill for a complete revolution 1 CS = √Mill Diameter where CS is the theoretical critical speed of rotation and is the mill speed rpm Mill diameter is the nominal inside diameter of the mill m
• Lab Ball MillsLab Jar Mill Lab Roll Ball Mill Dual Planetary Ball Mill Cryogenic Planetary Ball Mill Vertical Planetary Ball Mill for Glove Box Use Heavy duty Full directional Planetary Ball Mill Laboratory Full Directional Planetary Ball Mill Laboratory Horizontal Planetary Ball Mill Mini Vertical Planetary Ball Mill
• Optimization of mill performance by usingOptimization of mill performance by using online ball and pulp measurements by B Clermont and B de Haas Synopsis Ball mills are usually the largest consumers of energy within a mineral concentrator Comminution is responsible for 50% of the total mineral processing cost In today s global markets expanding mining groups are trying
• Speeds and Feeds 101 In The Loupe02 10 2017· Speeds and feeds are the cutting variables used in every milling operation and vary for each tool based on cutter diameter operation material etc Understanding the right speeds and feeds for your tool and operation before you start machining is
• EFFECT OF BALL SIZE DISTRIBUTION ON MILLING PARAMETERS Ball size distribution in tumbling mills 37 Milling performance of a ball size distribution 40 Summary 41 Chapter 3 Experimental equipment and programme 43 Laboratory grinding mill configuration 43 Preparation of mono size grinding media 44 Feed material preparation 46
• What is the critical rotation speed in revolutions per What is the critical rotation speed in revolutions per second for a ball mill of m diameter charged with 70 mm dia balls a b c d
• Ball mill WikipediaA ball mill a type of grinder is a cylindrical device used in grinding or mixing materials like ores chemicals ceramic raw materials and mills rotate around a horizontal axis partially filled with the material to be ground plus the grinding medium Different materials are used as media including ceramic balls flint pebbles and stainless steel balls
• Feed Rate Calculator DaycounterWhen milling or drilling or creating a tool path for a CNC machine the feed rate must be determined Materials have rated surface speeds for a given type of cutter The harder the material the slower the speed Given the diameter of the tool and the surface speed the RPMs of the spindle can be calculated
• Speeds And Feeds For Milling With End MillsMilling Speeds and Feeds Charts The most important aspect of milling with carbide end mills is to run the tool at the proper rpm and feed rate We have broken these recommendations down into material categories so you can make better decisions with how to productively run your end mills
• Rpm Speed Of Ball Mill Rpm Speed Of Ball Mill Critical rotation speed of dry ball mill was studied by experiments and by numerical simulation using discrete element method demhe results carried out by both methods showed good agreementt has been commonly accepted that the critical rotation speed
• Operating Speed Of Ball Mill Operating Speed Of Ball Mill 2019 8 29higher mill speed should always be considered when set ting gear ratios service factors and motor power ratingsor gearless drives motor torque output at the maximum desired operating mill speed and ball charge volume has to be determinedrimary mills torque output by motor vs torque demand by mill
• Factors Affecting Ball Mill Grinding Efficiency25 10 2017· 25 10 2017· The following are factors that have been investigated and applied in conventional ball milling in order to maximize grinding efficiency a Mill Geometry and Speed Bond 1954 observed grinding efficiency to be a function of ball mill diameter and established empirical relationships for recommended media size and mill speed that take this factor into account
• End Mills The Essential Beginners Guide Ball Nose End Mills Ball nose mills have a radius at the bottom which makes for a nicer surface finish in your workpiece meaning less work for you as the piece won t need to be finished any further They are used for contour milling shallow slotting pocketing and contouring applications
• EFFECT OF BALL SIZE DISTRIBUTION ON MILLING PARAMETERS Ball size distribution in tumbling mills 37 Milling performance of a ball size distribution 40 Summary 41 Chapter 3 Experimental equipment and programme 43 Laboratory grinding mill configuration 43 Preparation of mono size grinding media 44 Feed material preparation 46
• Ball Mill Used in Minerals Processing Plant Prominer This ball mill is typically designed to grind mineral ores and other materials with different hardness and it is widely used in different fields such as ore dressing building material field chemical industry etc Due to the difference of its slurry discharging method it is divided to two types grid type ball mill and overflow type ball mill
• Ball mill factorsSpeed of rotation of mill at low speeds the balls simply roll over one another and little grinding is obtained while at very high speeds the balls are simply carried along the walls of the shells and little or no grinding takes place so for an effective grinding the ball mill should be operated at a speed that is optimum speed equal to 50 to 75 percent of the critical speed the minimum
• Planetary Mills Planetary Mills are ideally suited for fine grinding of hard medium hard soft brittle tough and moist materials The comminution of the material to be ground takes place primarily through the high energy impact of grinding balls in rotating grinding bowls
• accupro ball end mill feeds and speeds Ball nose end mills are ideal for machining 3 dimensional contour shapes typically found in the mold and die industry the manufacturing of turbine blades and fulfilling general part radius properly employ a ball nose end mill with no tilt angle and gain the optimal tool life and part finish follow the 2 step process below see Figure 1
• THE EFFECT OF BALL MILL OPERATING PARAMETERS ON MINERAL Mill speed mill charge ball size and wet grinding are the parameters which have been selected for the present study It is hoped that the analysis of the data acquired will allow insight as to which parameters will require a functional form if they were to be integrated into
• Speed Of Ball Mill Ball Nose Precision Contouring End Mills Guhring Browse our offering of inch and metric ball nose precision contouring end website uses use cookies to make our site cookies enable core functionality like security network management and full line end mill catalog rf 100 speed end mills rf 100 90 degree end rf 100 speed
• Ballnose Cut Depth Musings for Feeds and Speeds in G Ballnose Cut Depth Musings for Feeds and Speeds in G Wizard Calculator by Bob Warfield Blog CNC Basics Feeds and Speeds but just the diameter out near the end of the ball Yet he had a situation where the ball was down in a slot and this is what was worrying
• Size reduction Solids and powder milling Ball mills as described above are also a machine of choice for milling solids in wet phase The design applied here is mainly a ball mill with an agitator the movement of the agitator making the beads move and impact or friction the particles to be milled
• High Performance Solid Carbide End MillsBall Nose BI Ball 47 BI4T Ball 47 BI220 220˚ 2 Flute Ball 47 BI2 2 Flute Long Series Ball 48 BI4 4 Flute Long Millstar s new High Performance and ultra precise solid carbide end mills were designed for high speed high velocity and hard steel milling Designed with specially selected
• Steel Ball Mills and Milling Equipment Paul O AbbéSteel Ball Mills Benefits of Tumble Milling 1 HIGH EFFICIENCY Due to the relatively slow rotational speed but large mass of media more of the energy goes into milling and less wasted as heat 2 NARROW PARTICLE DISTRIBUTION Solids milled in tumble mills are normally so fine and consistent in size that it s rare to require | 2,059 | 10,259 | {"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} | 2.625 | 3 | CC-MAIN-2022-27 | latest | en | 0.885077 |
http://stacks.math.columbia.edu/tag/0ABI | 1,503,216,326,000,000,000 | text/html | crawl-data/CC-MAIN-2017-34/segments/1502886106358.80/warc/CC-MAIN-20170820073631-20170820093631-00510.warc.gz | 372,780,154 | 6,309 | # The Stacks Project
## Tag 0ABI
Lemma 39.14.10. Let $(U, R, s, t, c)$ be a groupoid scheme. Let $u \in U$. Assume
1. $s, t$ are finite morphisms,
2. $U$ is separated and locally Noetherian,
3. $\dim(\mathcal{O}_{U, u'}) \leq 1$ for every point $u'$ in the orbit of $u$.
Then $u$ is contained in an $R$-invariant affine open of $U$.
Proof. The $R$-orbit of $u$ is finite. By conditions (2) and (3) it is contained in an affine open $U'$ of $U$, see Varieties, Proposition 32.41.7. Then $t(s^{-1}(U \setminus U'))$ is an $R$-invariant closed subset of $U$ which does not contain $u$. Thus $U \setminus t(s^{-1}(U \setminus U'))$ is an $R$-invariant open of $U'$ containing $u$. Replacing $U$ by this open we may assume $U$ is quasi-affine.
By Lemma 39.14.6 we may replace $U$ by its reduction and assume $U$ is reduced. This means $R$-invariant subschemes $W' \subset W \subset U$ of Lemma 39.14.2 are equal $W' = W$. As $U = t(s^{-1}(\overline{W}))$ some point $u'$ of the $R$-orbit of $u$ is contained in $\overline{W}$ and by Lemma 39.14.6 we may replace $U$ by $\overline{W}$ and $u$ by $u'$. Hence we may assume there is a dense open $R$-invariant subscheme $W \subset U$ such that the morphisms $s_W, t_W$ of the restriction $(W, R_W, s_W, t_W, c_W)$ are finite locally free.
If $u \in W$ then we are done by Groupoids, Lemma 38.24.1 (because $W$ is quasi-affine so any finite set of points of $W$ is contained in an affine open, see Properties, Lemma 27.29.5). Thus we assume $u \not \in W$ and hence none of the points of the orbit of $u$ is in $W$. Let $\xi \in U$ be a point with a nontrivial specialization to a point $u'$ in the orbit of $u$. Since there are no specializations among the points in the orbit of $u$ (Lemma 39.14.8) we see that $\xi$ is not in the orbit. By assumption (3) we see that $\xi$ is a generic point of $U$ and hence $\xi \in W$. As $U$ is Noetherian there are finitely many of these points $\xi_1, \ldots, \xi_m \in W$. Because $s_W, t_W$ are flat the orbit of each $\xi_j$ consists of generic points of irreducible components of $W$ (and hence $U$).
Let $j : U \to \mathop{\rm Spec}(A)$ be an immersion of $U$ into an affine scheme (this is possible as $U$ is quasi-affine). Let $J \subset A$ be an ideal such that $V(J) \cap j(W) = \emptyset$ and $V(J) \cup j(W)$ is closed. Apply Lemma 39.14.7 to the groupoid scheme $(W, R_W, s_W, t_W, c_W)$, the morphism $j|_W : W \to \mathop{\rm Spec}(A)$, the points $\xi_j$, and the ideal $J$ to find an $f \in J$ such that $(j|_W)^{-1}D(f)$ is an $R_W$-invariant affine open containing $\xi_j$ for all $j$. Since $f \in J$ we see that $j^{-1}D(f) \subset W$, i.e., $j^{-1}D(f)$ is an $R$-invariant affine open of $U$ contained in $W$ containing all $\xi_j$.
Let $Z$ be the reduced induced closed subscheme structure on $$U \setminus j^{-1}D(f) = j^{-1}V(f).$$ Then $Z$ is set theoretically $R$-invariant (but it may not be scheme theoretically $R$-invariant). Let $(Z, R_Z, s_Z, t_Z, c_Z)$ be the restriction of $R$ to $Z$. Since $Z \to U$ is finite, it follows that $s_Z$ and $t_Z$ are finite. Since $u \in Z$ the orbit of $u$ is in $Z$ and agrees with the $R_Z$-orbit of $u$ viewed as a point of $Z$. Since $\dim(\mathcal{O}_{U, u'}) \leq 1$ and since $\xi_j \not \in Z$ for all $j$, we see that $\dim(\mathcal{O}_{Z, u'}) \leq 0$ for all $u'$ in the orbit of $u$. In other words, the $R_Z$-orbit of $u$ consists of generic points of irreducible components of $Z$.
Let $I \subset A$ be an ideal such that $V(I) \cap j(U) =\emptyset$ and $V(I) \cup j(U)$ is closed. Apply Lemma 39.14.7 to the groupoid scheme $(Z, R_Z, s_Z, t_Z, c_Z)$, the restriction $j|_Z$, the ideal $I$, and the point $u \in Z$ to obtain $h \in I$ such that $j^{-1}D(h) \cap Z$ is an $R_Z$-invariant open affine containing $u$.
Consider the $R_W$-invariant (Groupoids, Lemma 38.23.2) function $$g = \text{Norm}_{s_W}(t_W^\sharp(j^\sharp(h)|_W)) \in \Gamma(W, \mathcal{O}_W)$$ (In the following we only need the restriction of $g$ to $j^{-1}D(f)$ and in this case the norm is along a finite locally free morphism of affines.) We claim that $$V = (W_g \cap j^{-1}D(f)) \cup (j^{-1}D(h) \cap Z)$$ is an $R$-invariant affine open of $U$ which finishes the proof of the lemma. It is set theoretically $R$-invariant by construction. As $V$ is a constuctible set, to see that it is open it suffices to show it is closed under generalization in $U$ (Topology, Lemma 5.19.9 or the more general Topology, Lemma 5.23.5). Since $W_g \cap j^{-1}D(f)$ is open in $U$, it suffices to consider a specialization $u_1 \leadsto u_2$ of $U$ with $u_2 \in j^{-1}D(h) \cap Z$. This means that $h$ is nonzero in $j(u_2)$ and $u_2 \in Z$. If $u_1 \in Z$, then $j(u_1) \leadsto j(u_2)$ and since $h$ is nonzero in $j(u_2)$ it is nonzero in $j(u_1)$ which implies $u_1 \in V$. If $u_1 \not \in Z$ and also not in $W_g \cap j^{-1}D(f)$, then $u_1 \in W$, $u_1 \not \in W_g$ because the complement of $Z = j^{-1}V(f)$ is contained in $W \cap j^{-1}D(f)$. Hence there exists a point $r_1 \in R$ with $s(r_1) = u_1$ such that $h$ is zero in $t(r_1)$. Since $s$ is finite we can find a specialization $r_1 \leadsto r_2$ with $s(r_2) = u_2$. However, then we conclude that $f$ is zero in $u'_2 = t(r_2)$ which contradicts the fact that $j^{-1}D(h) \cap Z$ is $R$-invariant and $u_2$ is in it. Thus $V$ is open.
Observe that $V \subset j^{-1}D(h)$ for our function $h \in I$. Thus we obtain an immersion $$j' : V \longrightarrow \mathop{\rm Spec}(A_h)$$ Let $f' \in A_h$ be the image of $f$. Then $(j')^{-1}D(f')$ is the principal open determined by $g$ in the affine open $j^{-1}D(f)$ of $U$. Hence $(j')^{-1}D(f)$ is affine. Finally, $j'(V) \cap V(f') = j'(j^{-1}D(h) \cap Z)$ is closed in $\mathop{\rm Spec}(A_h/(f')) = \mathop{\rm Spec}((A/f)_h) = D(h) \cap V(f)$ by our choice of $h \in I$ and the ideal $I$. Hence we can apply Lemma 39.14.9 to conclude that $V$ is affine as claimed above. $\square$
The code snippet corresponding to this tag is a part of the file more-groupoids.tex and is located in lines 2805–2815 (see updates for more information).
\begin{lemma}
\label{lemma-find-affine-codimension-1}
Let $(U, R, s, t, c)$ be a groupoid scheme. Let $u \in U$. Assume
\begin{enumerate}
\item $s, t$ are finite morphisms,
\item $U$ is separated and locally Noetherian,
\item $\dim(\mathcal{O}_{U, u'}) \leq 1$ for every point $u'$
in the orbit of $u$.
\end{enumerate}
Then $u$ is contained in an $R$-invariant affine open of $U$.
\end{lemma}
\begin{proof}
The $R$-orbit of $u$ is finite. By conditions (2) and (3) it is contained
in an affine open $U'$ of $U$, see
Varieties, Proposition
\ref{varieties-proposition-finite-set-of-points-of-codim-1-in-affine}.
Then $t(s^{-1}(U \setminus U'))$ is an $R$-invariant
closed subset of $U$ which does not contain $u$. Thus
$U \setminus t(s^{-1}(U \setminus U'))$ is an $R$-invariant open
of $U'$ containing $u$.
Replacing $U$ by this open we may assume $U$ is quasi-affine.
\medskip\noindent
By Lemma \ref{lemma-find-affine-integral} we may replace $U$ by its reduction
and assume $U$ is reduced. This means $R$-invariant subschemes
$W' \subset W \subset U$ of
Lemma \ref{lemma-finite-flat-over-almost-dense-subscheme}
are equal $W' = W$. As $U = t(s^{-1}(\overline{W}))$ some point
$u'$ of the $R$-orbit of $u$ is contained in $\overline{W}$
and by Lemma \ref{lemma-find-affine-integral}
we may replace $U$ by $\overline{W}$ and $u$ by $u'$.
Hence we may assume there is
a dense open $R$-invariant subscheme $W \subset U$ such that
the morphisms $s_W, t_W$ of the restriction $(W, R_W, s_W, t_W, c_W)$ are
finite locally free.
\medskip\noindent
If $u \in W$ then we are done by
Groupoids, Lemma \ref{groupoids-lemma-find-invariant-affine}
(because $W$ is quasi-affine so any finite set of points
of $W$ is contained in an affine open, see
Properties, Lemma \ref{properties-lemma-ample-finite-set-in-affine}).
Thus we assume $u \not \in W$ and hence none of the points of the
orbit of $u$ is in $W$. Let $\xi \in U$
be a point with a nontrivial specialization to a point $u'$ in the orbit
of $u$. Since there are no specializations among the points in the
orbit of $u$ (Lemma \ref{lemma-no-specializations-map-to-same-point})
we see that $\xi$ is not in the orbit.
By assumption (3) we see that $\xi$ is a generic point of $U$
and hence $\xi \in W$.
As $U$ is Noetherian there are finitely many of these
points $\xi_1, \ldots, \xi_m \in W$. Because $s_W, t_W$ are flat the orbit
of each $\xi_j$ consists of generic points of irreducible components
of $W$ (and hence $U$).
\medskip\noindent
Let $j : U \to \Spec(A)$ be an immersion of $U$ into an affine scheme
(this is possible as $U$ is quasi-affine). Let $J \subset A$
be an ideal such that $V(J) \cap j(W) = \emptyset$ and $V(J) \cup j(W)$
is closed. Apply Lemma \ref{lemma-find-almost-invariant-function}
to the groupoid scheme $(W, R_W, s_W, t_W, c_W)$, the morphism
$j|_W : W \to \Spec(A)$, the points $\xi_j$, and the ideal $J$
to find an $f \in J$ such that $(j|_W)^{-1}D(f)$ is an $R_W$-invariant
affine open containing $\xi_j$ for all $j$. Since $f \in J$
we see that $j^{-1}D(f) \subset W$, i.e., $j^{-1}D(f)$ is
an $R$-invariant affine open of $U$ contained in $W$
containing all $\xi_j$.
\medskip\noindent
Let $Z$ be the reduced induced closed subscheme structure on
$$U \setminus j^{-1}D(f) = j^{-1}V(f).$$
Then $Z$ is set theoretically
$R$-invariant (but it may not be scheme theoretically $R$-invariant).
Let $(Z, R_Z, s_Z, t_Z, c_Z)$ be the restriction of $R$ to $Z$.
Since $Z \to U$ is finite, it follows that $s_Z$ and $t_Z$ are finite.
Since $u \in Z$ the orbit of $u$ is in $Z$ and agrees with the
$R_Z$-orbit of $u$ viewed as a point of $Z$. Since
$\dim(\mathcal{O}_{U, u'}) \leq 1$ and since $\xi_j \not \in Z$
for all $j$, we see that $\dim(\mathcal{O}_{Z, u'}) \leq 0$ for
all $u'$ in the orbit of $u$. In other words, the $R_Z$-orbit of $u$
consists of generic points of irreducible components of $Z$.
\medskip\noindent
Let $I \subset A$ be an ideal such that $V(I) \cap j(U) =\emptyset$
and $V(I) \cup j(U)$ is closed. Apply
Lemma \ref{lemma-find-almost-invariant-function} to
the groupoid scheme $(Z, R_Z, s_Z, t_Z, c_Z)$, the restriction $j|_Z$,
the ideal $I$, and the point $u \in Z$ to obtain $h \in I$ such that
$j^{-1}D(h) \cap Z$ is an $R_Z$-invariant open affine containing $u$.
\medskip\noindent
Consider the $R_W$-invariant (Groupoids, Lemma
\ref{groupoids-lemma-determinant-trick}) function
$$g = \text{Norm}_{s_W}(t_W^\sharp(j^\sharp(h)|_W)) \in \Gamma(W, \mathcal{O}_W)$$
(In the following we only need the restriction of $g$ to $j^{-1}D(f)$ and
in this case the norm is along a finite locally free morphism of affines.)
We claim that
$$V = (W_g \cap j^{-1}D(f)) \cup (j^{-1}D(h) \cap Z)$$
is an $R$-invariant affine open of $U$ which finishes the proof of the lemma.
It is set theoretically $R$-invariant by construction. As $V$ is a
constuctible set, to see that it is open it suffices to show it is
closed under generalization in $U$ (Topology, Lemma
\ref{topology-lemma-characterize-closed-Noetherian}
or the more general
Topology, Lemma
\ref{topology-lemma-constructible-stable-specialization-closed}).
Since $W_g \cap j^{-1}D(f)$ is open in $U$, it suffices to consider
a specialization $u_1 \leadsto u_2$ of $U$ with
$u_2 \in j^{-1}D(h) \cap Z$.
This means that $h$ is nonzero in $j(u_2)$ and $u_2 \in Z$.
If $u_1 \in Z$, then $j(u_1) \leadsto j(u_2)$ and since
$h$ is nonzero in $j(u_2)$ it is nonzero in $j(u_1)$ which
implies $u_1 \in V$. If $u_1 \not \in Z$ and
also not in $W_g \cap j^{-1}D(f)$, then $u_1 \in W$, $u_1 \not \in W_g$
because the complement of $Z = j^{-1}V(f)$ is contained in $W \cap j^{-1}D(f)$.
Hence there exists a point $r_1 \in R$ with $s(r_1) = u_1$
such that $h$ is zero in $t(r_1)$. Since $s$ is finite we
can find a specialization $r_1 \leadsto r_2$ with $s(r_2) = u_2$.
However, then we conclude that $f$ is zero in $u'_2 = t(r_2)$
which contradicts the fact that $j^{-1}D(h) \cap Z$
is $R$-invariant and $u_2$ is in it. Thus $V$ is open.
\medskip\noindent
Observe that $V \subset j^{-1}D(h)$ for our function $h \in I$.
Thus we obtain an immersion
$$j' : V \longrightarrow \Spec(A_h)$$
Let $f' \in A_h$ be the image of $f$. Then $(j')^{-1}D(f')$
is the principal open determined by $g$ in the affine
open $j^{-1}D(f)$ of $U$.
Hence $(j')^{-1}D(f)$ is affine. Finally,
$j'(V) \cap V(f') = j'(j^{-1}D(h) \cap Z)$
is closed in $\Spec(A_h/(f')) = \Spec((A/f)_h) = D(h) \cap V(f)$
by our choice of $h \in I$ and the ideal $I$. Hence we can apply
Lemma \ref{lemma-get-affine}
to conclude that $V$ is affine as claimed above.
\end{proof}
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## Add a comment on tag 0ABI
In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the lower-right corner). | 4,564 | 12,933 | {"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} | 2.59375 | 3 | CC-MAIN-2017-34 | latest | en | 0.839544 |
https://mlmadesimple.com/2014/04/29/the-mighty-distance-function-part-1/ | 1,618,405,314,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038077818.23/warc/CC-MAIN-20210414125133-20210414155133-00017.warc.gz | 490,853,810 | 24,092 | # The Mighty Distance Function – Part 1
I’ll start my posts with the distance function, as I believe a complete understanding of it, provides the basic idea behind most of the ML algorithms. Some of the approaches which come next are closely related to some known algorithms in ML, but I am not going to name any of them, as I believe that a complete understanding of the intuition behind any of them is more important than their names.
Assume we have two points ( a and b) in a 2-d space, and our goal is to measure the distance between these two points. So we are looking for a function in the form d(a,b) which gives a smaller value if points are closer and gives a large value if they are far apart. There are different possibilities for the distance function we can use, but the simplest and most widely used is the Euclidean distance. In other words the length of the straight line segment connecting the two points:
Assuming we have two points (a and b), the length of the straight line segment connecting the two points can be used as a measure of distance between them.
Looks very simple, so now let’s make the problem a bit more challenging. Assume that instead of a single point b, we have several points B, and we want to calculate the distance between a single point a and a set of points B.
Unlike the distance between two points, the distance between a single point (a) and a set of points (B) does not have a single solution.
There are several possible approaches we can take:
One possibility is to use the distance between point a and its closest neighbor in set B.
To measure the distance between point (a) and the set (B), the closest neighbor to (a) in the set ($b_c$) can be used as a representative of the set.
This is often a good strategy, but as shown below, sometimes the result is not a good estimate for the distance between point a and the whole set B.
Sometimes the closest neighbor is not a good representative of the whole set.
The opposite approach is to use the distance between point a and its farthest neighbor in set B:
To measure the distance between point (a) and the set (B), the farthest neighbor ($b_f$) in the set can be used as a representative of the set.
This also could be a good estimation, but it suffers from the shortcoming similar to that of the closest neighbor approach.
Farthest neighbor is not a good representative of the whole set.
The approach which makes more sense is to use the average distance between a and all points in B. In other words we take the Euclidean distance between point a and all points in set B and then take the average of the calculated distances. For example, for the points in the below figure we have:
$d(a,B) = \frac{d(a,b_1)+d(a,b_2)+d(a,b_3)}{3}$
To measure the distance between point a and set B, we can measure the distance between point a and each point in set B and then take the average.
This process is computationally expensive, as calculating the distance between set B and any new given point a’ needs calculation of the distance between point a’ and all the points in the set. A more computationally efficient approach is to calculate the average point $\mu(B)$ of set B beforehand, and then calculate the distance between point a and the average point. This way for any new given point a’ distance calculation would take constant time.
we can calculate the average point of the set $\mu(B)$ and then calculate the distance between point a and $\mu(B)$
Well that was it for my first post, it was simple (wasn’t it?). That is the main idea anyway (remember the simplicity in the title of this page). But I promise that it will be more interesting for the next part. | 802 | 3,681 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 6, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.984375 | 4 | CC-MAIN-2021-17 | latest | en | 0.963839 |
https://www.br.freelancer.com/projects/php-windows/option-project-extension-repost/ | 1,532,022,384,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676591150.71/warc/CC-MAIN-20180719164439-20180719184439-00598.warc.gz | 816,534,906 | 27,139 | # Option Project Extension(repost)
This is an extension to the previous project and will act as a bridge to the next project.
## Deliverables
This is an extension to the previous project and will act as a bridge to the next project. The following things should be implemented:
1. Regarding the PP / SD calc where we take the 10 values and break them up to calculate the PP by denoting 1 or 0. I want to utilize 40 values for each SD instead of 10.
2. Regarding the Date to Expiration, Mid Point and Today calculations, I want to triple the amount of readings we take at a particular stock value. This would also affect the chart as the chart is built off this data.
3. Regarding the main grid where we display the options data. I want to use the individual options IV to perform theoretical pricing for D,G,T,V and option values.
4. In the current iteration of the program some of the options cannot price, meaning we cannot receive a IV, D,G,T,V for these values. When the strike price of options become far away from the stock price they become difficult to price, but there is a work around. Let’s say the current stock price is 30 and we are trying to price a 25.00 option but we cannot (program returns 0.00 for Th. Pr., D,G,T,V). This happens because the volatility value (SV) we are using to price that option is too low. So, what we need to do is use the volatility of the option closest to the option we cannot price. So, let’s say the SV was 20.00% and the IV of the 26.00 Strike was 29.78%. If we had no values available for the 25.00 Strike we would try to use 29.78% for the volatility input to price the 25 Strike option. We would do this for all prices below it. So if a the 25 option got a price and a volatility we would use that 25.00 Strike vol to price the 24.00 and the 24.00 for the 23.00 and so on.
There are a few other small things which can address on an hourly basis? | 460 | 1,902 | {"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} | 2.578125 | 3 | CC-MAIN-2018-30 | latest | en | 0.926842 |
http://openstudy.com/updates/5589b39de4b03f41bbd94125 | 1,516,342,717,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084887746.35/warc/CC-MAIN-20180119045937-20180119065937-00690.warc.gz | 266,589,590 | 8,483 | • anonymous
Maria looks at the architectural plan of a four-walled room in which the walls meet each other at right angles. The length of one wall in the plan is 13 inches. The length of the diagonal of the floor of the room in the plan is approximately 15.26 inches. Is the room in the shape of a square? Explain how you determined your answer
Mathematics
• Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
Looking for something else?
Not the answer you are looking for? Search for more explanations. | 322 | 1,274 | {"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} | 2.59375 | 3 | CC-MAIN-2018-05 | latest | en | 0.462456 |
https://mrkirkmath.com/tag/polar-form/ | 1,585,547,877,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585370496669.0/warc/CC-MAIN-20200330054217-20200330084217-00255.warc.gz | 599,615,569 | 14,837 | # Fractals & Chaos Recap for 12/11
Yesterday, we identified that adding complex numbers is equivalent to translating a point and multiplying complex numbers is equivalent to dilating and rotating that point. Furthermore, we noted that while the conventional rectangular coordinates for identifying a complex number are useful for identifying the direction and distance of a translation, to properly understand the dilation/rotation we need the polar coordinates.
Today, we quickly reviewed converting between rectangular and polar form at the start of class, then spent a considerable amount of time in class practicing this conversion and identifying the transformations they refer to. This work was on the Complex Numbers Transformations sheet and together we finished some of the front and about half of the back. Your homework is to finish the front.
# Fractals & Chaos Recap for 12/10
We saw yesterday that a sequence of geometric transformations can illustrate the same “fixed point attractor” behavior that we’ve seen when iterating a mathematical expression. We explained this by illustrating that such geometric transformations are analogous to iterating a linear function in the Complex number system. Specifically, adding two complex numbers is analogous to translating, and multiplying two complex numbers is analogous to a combined dilation/rotation.
The conventional rectangular form of a complex number a + bi tells us the horizontal and vertical components of the translation achieved by adding the complex number, but this form is not useful for when the number is being used as a multiplication factor. For that, we need the number’s Polar Form. Once finding it, the value of r tells us the dilation factor and the value of θ tell us the rotation factor.
We’ll dig in to this some more tomorrow, but for now please watch this video recapping polar vs rectangular coordinates (which the video refers to as “Cartesian coordinates”) and how to convert between the two. | 375 | 1,989 | {"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} | 3.84375 | 4 | CC-MAIN-2020-16 | latest | en | 0.897518 |
https://math.stackexchange.com/questions/4506816/on-the-rank-of-a-submatrix | 1,660,558,889,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882572163.61/warc/CC-MAIN-20220815085006-20220815115006-00088.warc.gz | 361,389,323 | 63,356 | # On the rank of a submatrix
The following is an exercise.
Suppose that $$A$$ is an $$n\times n$$ matrix and that $$m$$ rows of $$A$$ are selected to form an $$m\times n$$ submatrix $$B$$. By considering the number of zero rows in the normal form, prove that $$\text{rank}B \ge m - n + \text{rank} A$$.
My solution showed that $$A$$ could also be a non-square matrix and the result would still follow. Is my observation correct?
• Yes. In general, if $B$ is chosen in such a fashion from the $n \times k$ matrix $A$, then $\operatorname{rank}(B) \geq m-n+ \operatorname{rank}(A)$ Aug 5 at 21:26
• @BenGrossmann thanks Aug 5 at 21:29 | 194 | 636 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 8, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.765625 | 3 | CC-MAIN-2022-33 | latest | en | 0.876383 |
https://www.physicsforums.com/threads/problem-when-calculating-the-center-of-mass-of-a-triangle.964906/ | 1,722,814,767,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640417235.15/warc/CC-MAIN-20240804230158-20240805020158-00823.warc.gz | 745,561,211 | 22,109 | # Problem when calculating the center of mass of a triangle
• Kampret
In summary, the center of mass for a triangle is found by solving the median equation for each of the three vertices.
Kampret
## Homework Statement
if the black dot is assumed by (0,0).find the center of mass coordinate of this triangle
[/B]
i'm sorry but since the pic won't show ill attach the link here
https://ibb.co/4Ptw5T7
## Homework Equations
centroid is 2/3 of median [/B]
## The Attempt at a Solution
the problem is when i try solve this problem by both method it always give me different answer first I'll use the 2/3 theorem first
~ by seeing the picture it shows us that the length of EF is 2√2(2/sin45) and length from D to midpoint(H) is √2 ,[/B]
EH √2 then the length of center of mass from point D, ⅔DH or ⅔√2 from here i turn this into x-axis by multiply it by cos 45(since 90 on point D split into half) so i get ⅔√2.½√2 it become ⅔ this is the length from right side (point D) but the problem demand it from left side (black point) so it become 2-⅔=4/3(x axis)
for y-axis is using same method and it will give also 4/3result
~and the second method is using vector addition and have form like (x1m1+x2m2)/(m1+m2) in my case (x1l1+x2l2)/(l1+l2) since it doesn't show any mass and just length , and I am sure most you already know about this method
so ill straight to my attempt
for x-axis (x1l1+x2l2+x3l3)/(l1+l2+l3)
where 1= ED 2=EF 3=DF
(1.2+1.2√2+2.2)/(2+2√2+2)=1.292
and exact same method for y axis
and give same result, and from here the problem is arises , now if we take a look 1.292 is length from left side to right (from black point 0,0) so it length from point D is 2-1.292=½√2 so for y-axis it length from point D is 2-1.292=½√2 then if we use phytagorean theorem on both result we should able to determine the length from point D to the centrium ( i already calculated it on method 1) so it turned out like this √((½√2)²+(½√2)²)=1 and the answer should be ⅔√2 and if we try further more the length from D to mid point is 1/(2/3)=1,5 and how can there become 1,5??
sorry if i use weird term since I am not native english speaker so I'm not familiar with it i , ill really glad if someone can explain this to me since both formula is valid at least so i think
#### Attachments
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Do you know where the center of mass of any triangle should be at, just with basic geometry?
Kampret said:
since it doesn't show any mass and just length
There's the problem. Lengths are not areas.
YoungPhysicist said:
Do you know where the center of mass of any triangle should be at, just with basic geometry?
YoungPhysicist said:
Do you know where the center of mass of any triangle should be at, just with basic geometry?
i don't say any type of triangle. if you open the link you will see it only ordinary triangle
Kampret said:
i don't say any type of triangle. if you open the link you will see it only ordinary triangle
Well, I mean this:
https://en.m.wikipedia.org/wiki/Triangle_center
It just seems like this problem isn’t that complicated. The medians can be considered as linear equations. Then solve the system.
For instance:
The median from point D to line EF is y=x
YoungPhysicist said:
Well, I mean this:
https://en.m.wikipedia.org/wiki/Triangle_center
It just seems like this problem isn’t that complicated. The medians can be considered as linear equations. Then solve the system.
For instance:
The median from point D to line EF is y=x
yes it just when i try both method it doesn't display same result so it leave me restless since i think both are valid equation but I'm not sure about the second since so far in problem lIke this is always give either mass of area just like what haruspex stated above . frankly it first time i learn about center of mass so the density is always same and it basically give me the generic equation that also applicable to area and when i try to match it with my knowledge about triangle centrium it just give more contradiction . but it also perhaps caused by my shallowness and mistakes, so if i the one that make mistake please show me the right things to do but if both method really doesn't show same result then I'm not in the position to talking about it
## 1. What is the center of mass of a triangle?
The center of mass of a triangle is the point at which the triangle would balance if it were placed on a pin. It is the average position of all the mass in the triangle.
## 2. Why is calculating the center of mass of a triangle important?
Calculating the center of mass of a triangle is important in physics and engineering because it helps to determine the stability and balance of objects. It is also used in many real-world applications, such as in construction and architecture.
## 3. How do you calculate the center of mass of a triangle?
To calculate the center of mass of a triangle, you need to find the average of the x-coordinates and the average of the y-coordinates of the triangle's vertices. These averages will give you the coordinates of the center of mass.
## 4. What are some common problems that may arise when calculating the center of mass of a triangle?
One common problem is not considering the density of the triangle. The center of mass is affected by the distribution of mass within the triangle, so if the density is not uniform, the center of mass calculation will be incorrect. Another problem can arise if the triangle is not a simple shape, as it may be difficult to determine the exact coordinates of the vertices.
## 5. How can you ensure accuracy when calculating the center of mass of a triangle?
To ensure accuracy, it is important to use the correct formula for calculating the center of mass and to carefully consider the distribution of mass within the triangle. It may also be helpful to break the triangle into smaller, simpler shapes and calculate the center of mass for each of those shapes before finding the overall center of mass for the entire triangle.
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213 | 1,666 | 6,608 | {"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-2024-33 | latest | en | 0.914674 |
https://metanumbers.com/18943 | 1,722,660,838,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640356078.1/warc/CC-MAIN-20240803025153-20240803055153-00562.warc.gz | 324,327,973 | 7,442 | # 18943 (number)
18943 is an odd five-digits composite number following 18942 and preceding 18944. In scientific notation, it is written as 1.8943 × 104. The sum of its digits is 25. It has a total of 2 prime factors and 4 positive divisors. There are 17,928 positive integers (up to 18943) that are relatively prime to 18943.
## Basic properties
• Is Prime? no
• Number parity odd
• Number length 5
• Sum of Digits 25
• Digital Root 7
## Name
Name eighteen thousand nine hundred forty-three
## Notation
Scientific notation 1.8943 × 104 18.943 × 103
## Prime Factorization of 18943
Prime Factorization 19 × 997
Composite number
Distinct Factors Total Factors Radical ω 2 Total number of distinct prime factors Ω 2 Total number of prime factors rad 18943 Product of the distinct prime numbers λ 1 Returns the parity of Ω(n), such that λ(n) = (-1)Ω(n) μ 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 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 18943 is 19 × 997. Since it has a total of 2 prime factors, 18943 is a composite number.
## Divisors of 18943
1, 19, 997, 18943
4 divisors
Even divisors 0 4 2 2
Total Divisors Sum of Divisors Aliquot Sum τ 4 Total number of the positive divisors of n σ 19960 Sum of all the positive divisors of n s 1017 Sum of the proper positive divisors of n A 4990 Returns the sum of divisors (σ(n)) divided by the total number of divisors (τ(n)) G 137.634 Returns the nth root of the product of n divisors H 3.79619 Returns the total number of divisors (τ(n)) divided by the sum of the reciprocal of each divisors
The number 18943 can be divided by 4 positive divisors (out of which none is even, and 4 are odd). The sum of these divisors (counting 18943) is 19960, the average is 4990.
## Other Arithmetic Functions (n = 18943)
1 φ(n) n
Euler Totient Carmichael Lambda Prime Pi φ 17928 Total number of positive integers not greater than n that are coprime to n λ 2988 Smallest positive number such that aλ(n) ≡ 1 (mod n) for all a coprime to n π ≈ 2160 Total number of primes less than or equal to n r2 0 The number of ways n can be represented as the sum of 2 squares
There are 17,928 positive integers (less than 18943) that are coprime with 18943. And there are approximately 2,160 prime numbers less than or equal to 18943.
## Divisibility of 18943
m n mod m
2 1
3 1
4 3
5 3
6 1
7 1
8 7
9 7
18943 is not divisible by any number less than or equal to 9.
## Classification of 18943
• Arithmetic
• Semiprime
• Deficient
• Polite
• Square Free
### Other numbers
• LucasCarmichael
## Base conversion 18943
Base System Value
2 Binary 100100111111111
3 Ternary 221222121
4 Quaternary 10213333
5 Quinary 1101233
6 Senary 223411
8 Octal 44777
10 Decimal 18943
12 Duodecimal ab67
20 Vigesimal 2773
36 Base36 em7
## Basic calculations (n = 18943)
### Multiplication
n×y
n×2 37886 56829 75772 94715
### Division
n÷y
n÷2 9471.5 6314.33 4735.75 3788.6
### Exponentiation
ny
n2 358837249 6797454007807 128764171269888001 2439179696365488402943
### Nth Root
y√n
2√n 137.634 26.6573 11.7317 7.16951
## 18943 as geometric shapes
### Circle
Diameter 37886 119022 1.12732e+09
### Sphere
Volume 2.84731e+13 4.50928e+09 119022
### Square
Length = n
Perimeter 75772 3.58837e+08 26789.4
### Cube
Length = n
Surface area 2.15302e+09 6.79745e+12 32810.2
### Equilateral Triangle
Length = n
Perimeter 56829 1.55381e+08 16405.1
### Triangular Pyramid
Length = n
Surface area 6.21524e+08 8.01088e+11 15466.9
## Cryptographic Hash Functions
md5 09dbda0ec297f8e1fb8fa397efd0f70a 6550254133733f32c5db10dac35273d06c36c7c4 3166bfbd51d01c91e88bfd0dd53693aa4b4c1aa807fd45970ab375d282e1be65 d2af8c92bd84531d1f19b27de09e235025301494547f6185e35e50bc16447f008245e96447405f45f04028f94417b06fa5099dadfcee747c8d2a53bd544cd1c7 40d0170ee9b906e9e4f98f8123a6879ef7bf34ed | 1,387 | 4,022 | {"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-2024-33 | latest | en | 0.825334 |
http://poj.org/problem?id=3972 | 1,726,013,245,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651323.1/warc/CC-MAIN-20240910224659-20240911014659-00644.warc.gz | 29,411,530 | 3,851 | Online JudgeProblem SetAuthorsOnline ContestsUser
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Language:
BF
Time Limit: 5000MS Memory Limit: 65536K Total Submissions: 268 Accepted: 71
Description
BF is a language that was designed to challenge and amuse programmers. BF uses a simple machine model consisting, besides the program, of:
• An array of 30,000 byte cells – byte value ranges from 0 to 255 - initialized to zeroes.
• A movable pointer into a circular array (initialized to point to location 0)
• Two streams of bytes for input and output (most often connected to a keyboard and a monitor respectively, and using the ASCII character encoding).
In this problem you are asked to implement a BF interpreter, your program should read a BF program and an input stream and should print the output stream.
BF syntax is easy, only the following commands are available:
Character Spelled as Meaning > Greater than Increment the pointer (to point to the next cell to the right). < Less than Decrement the pointer (to point to the next cell to the left). + Plus sign Increment (increase by one) the byte at the pointer. - Minus sign Decrement (decrease by one) the byte at the pointer. . Dot (period) Puts the value of the byte at the pointer into the output stream. , Comma Accept one byte of input - from the input stream - storing its value in the byte at the pointer. [ Left square bracket Jump forward to the command after the corresponding ‘]’ if the byte at the pointer is zero. ] Right square bracket Jump back to the command after the corresponding ‘[‘ if the byte at the pointer is nonzero.
The given BF programs will be syntactically valid also the ‘>’ and ‘<’ operators will never lead to a value outside range [0-30000].
Input
Input starts with the number of test cases on a line by itself. Then a number of test cases; each test case is formatted as follows.
1. A BF program (possibly with extra letters and/or white space characters embedded anywhere in the text).
2. A white space and a dollar sign – ‘\$’ – indicating end of BF program.
3. A space, the input stream, another space and a dollar sign terminator indicating the end of the input stream.
4. A new line.
A test case may span on multiple lines and each two consecutive test cases are separated by a blank line.
The embedded text could be any characters except the dollar sign terminator, you should ONLY process the 8 commands and detect the end of program/input by the dollar signs, discard any other characters when processing. In the sample input only the relevant input is underlined, however in the real test cases there will be no signs of input termination except for the dollar signs.
Output
Output should be “Case “ then test case number (starting with 1) , a colon, a space followed by the contents of the BF program output stream and a new line after each test case.
Sample Input
```3
,>,>,>,.<.<.<. \$ abcde \$
This is a sample BF program ,>++++++[<-------->-],[<+>-]<. That
Prints the summation of 2 numbers \$ 23 \$
every cell 8*4=32 and prints 'em in reverse order \$ AB \$```
Sample Output
```Case 1: dcba
Case 2: 5
Case 3: ba```
Source
[Submit] [Go Back] [Status] [Discuss] | 783 | 3,337 | {"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} | 2.828125 | 3 | CC-MAIN-2024-38 | latest | en | 0.857478 |
http://digital.library.unt.edu/explore/partners/UNT/browse/?sort=date_a&fq=str_degree_level%3ADoctoral&fq=str_degree_discipline%3AComputer+Science | 1,477,110,873,000,000,000 | text/html | crawl-data/CC-MAIN-2016-44/segments/1476988718423.65/warc/CC-MAIN-20161020183838-00066-ip-10-171-6-4.ec2.internal.warc.gz | 67,288,373 | 17,267 | ## You limited your search to:
Partner: UNT Libraries
Degree Discipline: Computer Science
Degree Level: Doctoral
Results 1 - 24 of 75
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Date: December 1986
Creator: Atwood, Larry D. (Larry Dale)
Description: This dissertation deals with the problem of manipulating and storing an image using quadtrees. A quadtree is a tree in which each node has four ordered children or is a leaf. It can be used to represent an image via hierarchical decomposition. The image is broken into four regions. A region can be a solid color (homogeneous) or a mixture of colors (heterogeneous). If a region is heterogeneous it is broken into four subregions, and the process continues recursively until all subregions are homogeneous. The traditional quadtree suffers from dependence on the underlying grid. The grid coordinate system is implicit, and therefore fixed. The fixed coordinate system implies a rigid tree. A rigid tree cannot be translated, scaled, or rotated. Instead, a new tree must be built which is the result of one of these transformations. This dissertation introduces the independent quadtree. The independent quadtree is free of any underlying coordinate system. The tree is no longer rigid and can be easily translated, scaled, or rotated. Algorithms to perform these operations axe presented. The translation and rotation algorithms take constant time. The scaling algorithm has linear time in the number nodes in the tree. The disadvantage of independent quadtrees is ...
Contributing Partner: UNT Libraries
### A Comparative Analysis of Guided vs. Query-Based Intelligent Tutoring Systems (ITS) Using a Class-Entity-Relationship-Attribute (CERA) Knowledge Base
Date: August 1987
Creator: Hall, Douglas Lee
Description: One of the greatest problems facing researchers in the sub field of Artificial Intelligence known as Intelligent Tutoring Systems (ITS) is the selection of a knowledge base designs that will facilitate the modification of the knowledge base. The Class-Entity-Relationship-Attribute (CERA), proposed by R. P. Brazile, holds certain promise as a more generic knowledge base design framework upon which can be built robust and efficient ITS. This study has a twofold purpose. The first is to demonstrate that a CERA knowledge base can be constructed for an ITS on a subset of the domain of Cretaceous paleontology and function as the "expert module" of the ITS. The second is to test the validity of the ideas that students guided through a lesson learn more factual knowledge, while those who explore the knowledge base that underlies the lesson through query at their own pace will be able to formulate their own integrative knowledge from the knowledge gained in their explorations and spend more time on the system. This study concludes that a CERA-based system can be constructed as an effective teaching tool. However, while an ITS - treatment provides for statistically significant gains in achievement test scores, the type of treatment seems ...
Contributing Partner: UNT Libraries
### A Timescale Estimating Model for Rule-Based Systems
Date: December 1987
Creator: Moseley, Charles Warren
Description: The purpose of this study was to explore the subject of timescale estimating for rule-based systems. A model for estimating the timescale necessary to build rule-based systems was built and then tested in a controlled environment.
Contributing Partner: UNT Libraries
### Computer Realization of Human Music Cognition
Date: August 1988
Creator: Albright, Larry E. (Larry Eugene)
Description: This study models the human process of music cognition on the digital computer. The definition of music cognition is derived from the work in music cognition done by the researchers Carol Krumhansl and Edward Kessler, and by Mari Jones, as well as from the music theories of Heinrich Schenker. The computer implementation functions in three stages. First, it translates a musical "performance" in the form of MIDI (Musical Instrument Digital Interface) messages into LISP structures. Second, the various parameters of the performance are examined separately a la Jones's joint accent structure, quantified according to psychological findings, and adjusted to a common scale. The findings of Krumhansl and Kessler are used to evaluate the consonance of each note with respect to the key of the piece and with respect to the immediately sounding harmony. This process yields a multidimensional set of points, each of which is a cognitive evaluation of a single musical event within the context of the piece of music within which it occurred. This set of points forms a metric space in multi-dimensional Euclidean space. The third phase of the analysis maps the set of points into a topology-preserving data structure for a Schenkerian-like middleground structural analysis. This ...
Contributing Partner: UNT Libraries
### Semaphore Solutions for General Mutual Exclusion Problems
Date: August 1988
Creator: Yue, Kwok B. (Kwok Bun)
Description: Automatic generation of starvation-free semaphore solutions to general mutual exclusion problems is discussed. A reduction approach is introduced for recognizing edge-solvable problems, together with an O(N^2) algorithm for graph reduction, where N is the number of nodes. An algorithm for the automatic generation of starvation-free edge-solvable solutions is presented. The solutions are proved to be very efficient. For general problems, there are two ways to generate efficient solutions. One associates a semaphore with every node, the other with every edge. They are both better than the standard monitor—like solutions. Besides strong semaphores, solutions using weak semaphores, weaker semaphores and generalized semaphores are also considered. Basic properties of semaphore solutions are also discussed. Tools describing the dynamic behavior of parallel systems, as well as performance criteria for evaluating semaphore solutions are elaborated.
Contributing Partner: UNT Libraries
### Speech Recognition Using a Synthesized Codebook
Date: August 1988
Creator: Smith, Lloyd A. (Lloyd Allen)
Description: Speech sounds generated by a simple waveform synthesizer were used to create a vector quantization codebook for use in speech recognition. Recognition was tested over the TI-20 isolated word data base using a conventional DTW matching algorithm. Input speech was band limited to 300 - 3300 Hz, then passed through the Scott Instruments Corp. Coretechs process, implemented on a VET3 speech terminal, to create the speech representation for matching. Synthesized sounds were processed in software by a VET3 signal processing emulation program. Emulation and recognition were performed on a DEC VAX 11/750. The experiments were organized in 2 series. A preliminary experiment, using no vector quantization, provided a baseline for comparison. The original codebook contained 109 vectors, all derived from 2 formant synthesized sounds. This codebook was decimated through the course of the first series of experiments, based on the number of times each vector was used in quantizing the training data for the previous experiment, in order to determine the smallest subset of vectors suitable for coding the speech data base. The second series of experiments altered several test conditions in order to evaluate the applicability of the minimal synthesized codebook to conventional codebook training. The baseline recognition rate ...
Contributing Partner: UNT Libraries
### Inheritance Problems in Object-Oriented Database
Date: May 1989
Creator: Auepanwiriyakul, Raweewan
Description: This research is concerned with inheritance as used in object-oriented database. More specifically, partial bi-directional inheritance among classes is examined. In partial inheritance, a class can inherit a proper subset of instance variables from another class. Two subclasses of the same superclass do not need to inherit the same proper subset of instance variables from their superclass. Bi-directional partial inheritance allows a class to inherit instance variables from its subclass. The prototype of an object-oriented database that supports both full and partial bi-directional inheritance among classes was developed on top of an existing relational database management system. The prototype was tested with two database applications. One database application needs full and partial inheritance. The second database application required bi-directional inheritance. The result of this testing suggests both advantages and disadvantages of partial bi-directional inheritance. Future areas of research are also suggested.
Contributing Partner: UNT Libraries
### A Mechanism for Facilitating Temporal Reasoning in Discrete Event Simulation
Date: May 1992
Creator: Legge, Gaynor W.
Description: This research establishes the feasibility and potential utility of a software mechanism which employs artificial intelligence techniques to enhance the capabilities of standard discrete event simulators. As background, current methods of integrating artificial intelligence with simulation and relevant research are briefly reviewed.
Contributing Partner: UNT Libraries
### Using Extended Logic Programs to Formalize Commonsense Reasoning
Date: May 1992
Creator: Horng, Wen-Bing
Description: In this dissertation, we investigate how commonsense reasoning can be formalized by using extended logic programs. In this investigation, we first use extended logic programs to formalize inheritance hierarchies with exceptions by adopting McCarthy's simple abnormality formalism to express uncertain knowledge. In our representation, not only credulous reasoning can be performed but also the ambiguity-blocking inheritance and the ambiguity-propagating inheritance in skeptical reasoning are simulated. In response to the anomalous extension problem, we explore and discover that the intuition underlying commonsense reasoning is a kind of forward reasoning. The unidirectional nature of this reasoning is applied by many reformulations of the Yale shooting problem to exclude the undesired conclusion. We then identify defeasible conclusions in our representation based on the syntax of extended logic programs. A similar idea is also applied to other formalizations of commonsense reasoning to achieve such a purpose.
Contributing Partner: UNT Libraries
### Using Normal Deduction Graphs in Common Sense Reasoning
Date: May 1992
Creator: Munoz, Ricardo A. (Ricardo Alberto)
Description: This investigation proposes a powerful formalization of common sense knowledge based on function-free normal deduction graphs (NDGs) which form a powerful tool for deriving Horn and non-Horn clauses without functions. Such formalization allows common sense reasoning since it has the ability to handle not only negative but also incomplete information.
Contributing Partner: UNT Libraries
### Efficient Linked List Ranking Algorithms and Parentheses Matching as a New Strategy for Parallel Algorithm Design
Date: December 1993
Creator: Halverson, Ranette Hudson
Description: The goal of a parallel algorithm is to solve a single problem using multiple processors working together and to do so in an efficient manner. In this regard, there is a need to categorize strategies in order to solve broad classes of problems with similar structures and requirements. In this dissertation, two parallel algorithm design strategies are considered: linked list ranking and parentheses matching.
Contributing Partner: UNT Libraries
### Multiresolutional/Fractal Compression of Still and Moving Pictures
Date: December 1993
Creator: Kiselyov, Oleg E.
Description: The scope of the present dissertation is a deep lossy compression of still and moving grayscale pictures while maintaining their fidelity, with a specific goal of creating a working prototype of a software system for use in low bandwidth transmission of still satellite imagery and weather briefings with the best preservation of features considered important by the end user.
Contributing Partner: UNT Libraries
### Temporal Connectionist Expert Systems Using a Temporal Backpropagation Algorithm
Date: December 1993
Creator: Civelek, Ferda N. (Ferda Nur)
Description: Representing time has been considered a general problem for artificial intelligence research for many years. More recently, the question of representing time has become increasingly important in representing human decision making process through connectionist expert systems. Because most human behaviors unfold over time, any attempt to represent expert performance, without considering its temporal nature, can often lead to incorrect results. A temporal feedforward neural network model that can be applied to a number of neural network application areas, including connectionist expert systems, has been introduced. The neural network model has a multi-layer structure, i.e. the number of layers is not limited. Also, the model has the flexibility of defining output nodes in any layer. This is especially important for connectionist expert system applications. A temporal backpropagation algorithm which supports the model has been developed. The model along with the temporal backpropagation algorithm makes it extremely practical to define any artificial neural network application. Also, an approach that can be followed to decrease the memory space used by weight matrix has been introduced. The algorithm was tested using a medical connectionist expert system to show how best we describe not only the disease but also the entire course of the disease. ...
Contributing Partner: UNT Libraries
### A Multi-Time Scale Learning Mechanism for Neuromimic Processing
Date: August 1994
Creator: Mobus, George E. (George Edward)
Description: Learning and representing and reasoning about temporal relations, particularly causal relations, is a deep problem in artificial intelligence (AI). Learning such representations in the real world is complicated by the fact that phenomena are subject to multiple time scale influences and may operate with a strange attractor dynamic. This dissertation proposes a new computational learning mechanism, the adaptrode, which, used in a neuromimic processing architecture may help to solve some of these problems. The adaptrode is shown to emulate the dynamics of real biological synapses and represents a significant departure from the classical weighted input scheme of conventional artificial neural networks. Indeed the adaptrode is shown, by analysis of the deep structure of real synapses, to have a strong structural correspondence with the latter in terms of multi-time scale biophysical processes. Simulations of an adaptrode-based neuron and a small network of neurons are shown to have the same learning capabilities as invertebrate animals in classical conditioning. Classical conditioning is considered a fundamental learning task in animals. Furthermore, it is subject to temporal ordering constraints that fulfill the criteria of causal relations in natural systems. It may offer clues to the learning of causal relations and mechanisms for causal reasoning. The ...
Contributing Partner: UNT Libraries
### A Highly Fault-Tolerant Distributed Database System with Replicated Data
Date: December 1994
Creator: Lin, Tsai S. (Tsai Shooumeei)
Description: Because of the high cost and impracticality of a high connectivity network, most recent research in transaction processing has focused on a distributed replicated database system. In such a system, multiple copies of a data item are created and stored at several sites in the network, so that the system is able to tolerate more crash and communication failures and attain higher data availability. However, the multiple copies also introduce a global inconsistency problem, especially in a partitioned network. In this dissertation a tree quorum algorithm is proposed to solve this problem, imposing a logical tree structure along with dynamic system reconfiguration on all the copies of each data item. The proposed algorithm can be viewed as a dynamic voting technique which, with the help of an appropriate concurrency control algorithm, exhibits the major advantages of quorum-based replica control algorithms and of the available copies algorithm, so that a single copy is read for a read operation and a quorum of copies is written for a write operation. In addition, read and write quorums are computed dynamically and independently. As a result expensive read operations, like those that require several copies of a data item to be read in most ...
Contributing Partner: UNT Libraries
### Recognition of Face Images
Date: December 1994
Creator: Pershits, Edward
Description: The focus of this dissertation is a methodology that enables computer systems to classify different up-front images of human faces as belonging to one of the individuals to which the system has been exposed previously. The images can present variance in size, location of the face, orientation, facial expressions, and overall illumination. The approach to the problem taken in this dissertation can be classified as analytic as the shapes of individual features of human faces are examined separately, as opposed to holistic approaches to face recognition. The outline of the features is used to construct signature functions. These functions are then magnitude-, period-, and phase-normalized to form a translation-, size-, and rotation-invariant representation of the features. Vectors of a limited number of the Fourier decomposition coefficients of these functions are taken to form the feature vectors representing the features in the corresponding vector space. With this approach no computation is necessary to enforce the translational, size, and rotational invariance at the stage of recognition thus reducing the problem of recognition to the k-dimensional clustering problem. A recognizer is specified that can reliably classify the vectors of the feature space into object classes. The recognizer made use of the following principle: ...
Contributing Partner: UNT Libraries
### A Theoretical Network Model and the Incremental Hypercube-Based Networks
Date: May 1995
Creator: Mao, Ai-sheng
Description: The study of multicomputer interconnection networks is an important area of research in parallel processing. We introduce vertex-symmetric Hamming-group graphs as a model to design a wide variety of network topologies including the hypercube network.
Contributing Partner: UNT Libraries
### Practical Cursive Script Recognition
Date: August 1995
Creator: Carroll, Johnny Glen, 1953-
Description: This research focused on the off-line cursive script recognition application. The problem is very large and difficult and there is much room for improvement in every aspect of the problem. Many different aspects of this problem were explored in pursuit of solutions to create a more practical and usable off-line cursive script recognizer than is currently available.
Contributing Partner: UNT Libraries
### An Algorithm for the PLA Equivalence Problem
Date: December 1995
Creator: Moon, Gyo Sik
Description: The Programmable Logic Array (PLA) has been widely used in the design of VLSI circuits and systems because of its regularity, flexibility, and simplicity. The equivalence problem is typically to verify that the final description of a circuit is functionally equivalent to its initial description. Verifying the functional equivalence of two descriptions is equivalent to proving their logical equivalence. This problem of pure logic is essential to circuit design. The most widely used technique to solve the problem is based on Binary Decision Diagram or BDD, proposed by Bryant in 1986. Unfortunately, BDD requires too much time and space to represent moderately large circuits for equivalence testing. We design and implement a new algorithm called the Cover-Merge Algorithm for the equivalence problem based on a divide-and-conquer strategy using the concept of cover and a derivational method. We prove that the algorithm is sound and complete. Because of the NP-completeness of the problem, we emphasize simplifications to reduce the search space or to avoid redundant computations. Simplification techniques are incorporated into the algorithm as an essential part to speed up the the derivation process. Two different sets of heuristics are developed for two opposite goals: one for the proof of equivalence ...
Contributing Partner: UNT Libraries
### Convexity-Preserving Scattered Data Interpolation
Date: December 1995
Creator: Leung, Nim Keung
Description: Surface fitting methods play an important role in many scientific fields as well as in computer aided geometric design. The problem treated here is that of constructing a smooth surface that interpolates data values associated with scattered nodes in the plane. The data is said to be convex if there exists a convex interpolant. The problem of convexity-preserving interpolation is to determine if the data is convex, and construct a convex interpolant if it exists.
Contributing Partner: UNT Libraries
### Quantifying Design Principles in Reusable Software Components
Date: December 1995
Creator: Moore, Freeman Leroy
Description: Software reuse can occur in various places during the software development cycle. Reuse of existing source code is the most commonly practiced form of software reuse. One of the key requirements for software reuse is readability, thus the interest in the use of data abstraction, inheritance, modularity, and aspects of the visible portion of module specifications. This research analyzed the contents of software reuse libraries to answer the basic question of what makes a good reusable software component. The approach taken was to measure and analyze various software metrics as mapped to design characteristics. A related research question investigated the change in the design principles over time. This was measured by comparing sets of Ada reuse libraries categorized into two time periods. It was discovered that recently developed Ada reuse components scored better on readability than earlier developed components. A benefit of this research has been the development of a set of "design for reuse" guidelines. These guidelines address coding practices as well as design principles for an Ada implementation. C++ software reuse libraries were also analyzed to determine if design principles can be applied in a language independent fashion. This research used cyclomatic complexity metrics, software science metrics, and ...
Contributing Partner: UNT Libraries
### Rollback Reduction Techniques Through Load Balancing in Optimistic Parallel Discrete Event Simulation
Date: May 1996
Creator: Sarkar, Falguni
Description: Discrete event simulation is an important tool for modeling and analysis. Some of the simulation applications such as telecommunication network performance, VLSI logic circuits design, battlefield simulation, require enormous amount of computing resources. One way to satisfy this demand for computing power is to decompose the simulation system into several logical processes (Ip) and run them concurrently. In any parallel discrete event simulation (PDES) system, the events are ordered according to their time of occurrence. In order for the simulation to be correct, this ordering has to be preserved. There are three approaches to maintain this ordering. In a conservative system, no lp executes an event unless it is certain that all events with earlier time-stamps have been executed. Such systems are prone to deadlock. In an optimistic system on the other hand, simulation progresses disregarding this ordering and saves the system states regularly. Whenever a causality violation is detected, the system rolls back to a state saved earlier and restarts processing after correcting the error. There is another approach in which all the lps participate in the computation of a safe time-window and all events with time-stamps within this window are processed concurrently. In optimistic simulation systems, there is ...
Contributing Partner: UNT Libraries
### A Machine Learning Method Suitable for Dynamic Domains
Date: July 1996
Creator: Rowe, Michael C. (Michael Charles)
Description: The efficacy of a machine learning technique is domain dependent. Some machine learning techniques work very well for certain domains but are ill-suited for other domains. One area that is of real-world concern is the flexibility with which machine learning techniques can adapt to dynamic domains. Currently, there are no known reports of any system that can learn dynamic domains, short of starting over (i.e., re-running the program). Starting over is neither time nor cost efficient for real-world production environments. This dissertation studied a method, referred to as Experience Based Learning (EBL), that attempts to deal with conditions related to learning dynamic domains. EBL is an extension of Instance Based Learning methods. The hypothesis of the study related to this research was that the EBL method would automatically adjust to domain changes and still provide classification accuracy similar to methods that require starting over. To test this hypothesis, twelve widely studied machine learning datasets were used. A dynamic domain was simulated by presenting these datasets in an uninterrupted cycle of train, test, and retrain. The order of the twelve datasets and the order of records within each dataset were randomized to control for order biases in each of ten runs. ...
Contributing Partner: UNT Libraries
### Automatic Speech Recognition Using Finite Inductive Sequences
Date: August 1996
Creator: Cherri, Mona Youssef, 1956-
Description: This dissertation addresses the general problem of recognition of acoustic signals which may be derived from speech, sonar, or acoustic phenomena. The specific problem of recognizing speech is the main focus of this research. The intention is to design a recognition system for a definite number of discrete words. For this purpose specifically, eight isolated words from the T1MIT database are selected. Four medium length words "greasy," "dark," "wash," and "water" are used. In addition, four short words are considered "she," "had," "in," and "all." The recognition system addresses the following issues: filtering or preprocessing, training, and decision-making. The preprocessing phase uses linear predictive coding of order 12. Following the filtering process, a vector quantization method is used to further reduce the input data and generate a finite inductive sequence of symbols representative of each input signal. The sequences generated by the vector quantization process of the same word are factored, and a single ruling or reference template is generated and stored in a codebook. This system introduces a new modeling technique which relies heavily on the basic concept that all finite sequences are finitely inductive. This technique is used in the training stage. In order to accommodate the variabilities ...
Contributing Partner: UNT Libraries
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#### Resources tagged with Generalising similar to Where Can We Visit?:
Filter by: Content type:
Age range:
Challenge level:
### There are 123 results
Broad Topics > Using, Applying and Reasoning about Mathematics > Generalising
### Where Can We Visit?
##### Age 11 to 14 Challenge Level:
Charlie and Abi put a counter on 42. They wondered if they could visit all the other numbers on their 1-100 board, moving the counter using just these two operations: x2 and -5. What do you think?
### How Much Can We Spend?
##### Age 11 to 14 Challenge Level:
A country has decided to have just two different coins, 3z and 5z coins. Which totals can be made? Is there a largest total that cannot be made? How do you know?
### What Numbers Can We Make?
##### Age 11 to 14 Challenge Level:
Imagine we have four bags containing a large number of 1s, 4s, 7s and 10s. What numbers can we make?
### Frogs
##### Age 11 to 14 Challenge Level:
How many moves does it take to swap over some red and blue frogs? Do you have a method?
### Tower of Hanoi
##### Age 11 to 14 Challenge Level:
The Tower of Hanoi is an ancient mathematical challenge. Working on the building blocks may help you to explain the patterns you notice.
### Searching for Mean(ing)
##### Age 11 to 14 Challenge Level:
Imagine you have a large supply of 3kg and 8kg weights. How many of each weight would you need for the average (mean) of the weights to be 6kg? What other averages could you have?
### Number Pyramids
##### Age 11 to 14 Challenge Level:
Try entering different sets of numbers in the number pyramids. How does the total at the top change?
### Summing Consecutive Numbers
##### Age 11 to 14 Challenge Level:
15 = 7 + 8 and 10 = 1 + 2 + 3 + 4. Can you say which numbers can be expressed as the sum of two or more consecutive integers?
### What Numbers Can We Make Now?
##### Age 11 to 14 Challenge Level:
Imagine we have four bags containing numbers from a sequence. What numbers can we make now?
### Litov's Mean Value Theorem
##### Age 11 to 14 Challenge Level:
Start with two numbers and generate a sequence where the next number is the mean of the last two numbers...
### Elevenses
##### Age 11 to 14 Challenge Level:
How many pairs of numbers can you find that add up to a multiple of 11? Do you notice anything interesting about your results?
### Card Trick 2
##### Age 11 to 14 Challenge Level:
Can you explain how this card trick works?
### Tourism
##### Age 11 to 14 Challenge Level:
If you can copy a network without lifting your pen off the paper and without drawing any line twice, then it is traversable. Decide which of these diagrams are traversable.
### Nim-7 for Two
##### Age 5 to 14 Challenge Level:
Nim-7 game for an adult and child. Who will be the one to take the last counter?
### Arithmagons
##### Age 14 to 16 Challenge Level:
Can you find the values at the vertices when you know the values on the edges?
### More Number Pyramids
##### Age 11 to 14 Challenge Level:
When number pyramids have a sequence on the bottom layer, some interesting patterns emerge...
### Cunning Card Trick
##### Age 11 to 14 Challenge Level:
Delight your friends with this cunning trick! Can you explain how it works?
### Cubes Within Cubes Revisited
##### Age 11 to 14 Challenge Level:
Imagine starting with one yellow cube and covering it all over with a single layer of red cubes, and then covering that cube with a layer of blue cubes. How many red and blue cubes would you need?
### Hidden Rectangles
##### Age 11 to 14 Challenge Level:
Rectangles are considered different if they vary in size or have different locations. How many different rectangles can be drawn on a chessboard?
### Maths Trails
##### Age 7 to 14
The NRICH team are always looking for new ways to engage teachers and pupils in problem solving. Here we explain the thinking behind maths trails.
### Consecutive Negative Numbers
##### Age 11 to 14 Challenge Level:
Do you notice anything about the solutions when you add and/or subtract consecutive negative numbers?
### Nim-7
##### Age 5 to 14 Challenge Level:
Can you work out how to win this game of Nim? Does it matter if you go first or second?
### Cuboid Challenge
##### Age 11 to 16 Challenge Level:
What's the largest volume of box you can make from a square of paper?
### Chess
##### Age 11 to 14 Challenge Level:
What would be the smallest number of moves needed to move a Knight from a chess set from one corner to the opposite corner of a 99 by 99 square board?
### Squares in Rectangles
##### Age 11 to 14 Challenge Level:
A 2 by 3 rectangle contains 8 squares and a 3 by 4 rectangle contains 20 squares. What size rectangle(s) contain(s) exactly 100 squares? Can you find them all?
### Picturing Square Numbers
##### Age 11 to 14 Challenge Level:
Square numbers can be represented as the sum of consecutive odd numbers. What is the sum of 1 + 3 + ..... + 149 + 151 + 153?
### Picturing Triangular Numbers
##### Age 11 to 14 Challenge Level:
Triangular numbers can be represented by a triangular array of squares. What do you notice about the sum of identical triangle numbers?
### Route to Infinity
##### Age 11 to 14 Challenge Level:
Can you describe this route to infinity? Where will the arrows take you next?
### Window Frames
##### Age 5 to 14 Challenge Level:
This task encourages you to investigate the number of edging pieces and panes in different sized windows.
### Christmas Chocolates
##### Age 11 to 14 Challenge Level:
How could Penny, Tom and Matthew work out how many chocolates there are in different sized boxes?
### AMGM
##### Age 14 to 16 Challenge Level:
Can you use the diagram to prove the AM-GM inequality?
### Mini-max
##### Age 11 to 14 Challenge Level:
Consider all two digit numbers (10, 11, . . . ,99). In writing down all these numbers, which digits occur least often, and which occur most often ? What about three digit numbers, four digit numbers. . . .
### Go Forth and Generalise
##### Age 11 to 14
Spotting patterns can be an important first step - explaining why it is appropriate to generalise is the next step, and often the most interesting and important.
### Special Sums and Products
##### Age 11 to 14 Challenge Level:
Find some examples of pairs of numbers such that their sum is a factor of their product. eg. 4 + 12 = 16 and 4 × 12 = 48 and 16 is a factor of 48.
### Shear Magic
##### Age 11 to 14 Challenge Level:
What are the areas of these triangles? What do you notice? Can you generalise to other "families" of triangles?
### Sum Equals Product
##### Age 11 to 14 Challenge Level:
The sum of the numbers 4 and 1 [1/3] is the same as the product of 4 and 1 [1/3]; that is to say 4 + 1 [1/3] = 4 × 1 [1/3]. What other numbers have the sum equal to the product and can this be so for. . . .
### Games Related to Nim
##### Age 5 to 16
This article for teachers describes several games, found on the site, all of which have a related structure that can be used to develop the skills of strategic planning.
### ...on the Wall
##### Age 11 to 14 Challenge Level:
Explore the effect of reflecting in two intersecting mirror lines.
### Enclosing Squares
##### Age 11 to 14 Challenge Level:
Can you find sets of sloping lines that enclose a square?
##### Age 7 to 14 Challenge Level:
I added together some of my neighbours' house numbers. Can you explain the patterns I noticed?
### Multiplication Square
##### Age 14 to 16 Challenge Level:
Pick a square within a multiplication square and add the numbers on each diagonal. What do you notice?
### Winning Lines
##### Age 7 to 16
An article for teachers and pupils that encourages you to look at the mathematical properties of similar games.
### Mystic Rose
##### Age 14 to 16 Challenge Level:
Use the animation to help you work out how many lines are needed to draw mystic roses of different sizes.
### Got it for Two
##### Age 7 to 14 Challenge Level:
Got It game for an adult and child. How can you play so that you know you will always win?
### Steps to the Podium
##### Age 7 to 14 Challenge Level:
It starts quite simple but great opportunities for number discoveries and patterns!
### Magic Letters
##### Age 11 to 14 Challenge Level:
Charlie has made a Magic V. Can you use his example to make some more? And how about Magic Ls, Ns and Ws?
### Multiplication Arithmagons
##### Age 14 to 16 Challenge Level:
Can you find the values at the vertices when you know the values on the edges of these multiplication arithmagons?
### Partly Painted Cube
##### Age 14 to 16 Challenge Level:
Jo made a cube from some smaller cubes, painted some of the faces of the large cube, and then took it apart again. 45 small cubes had no paint on them at all. How many small cubes did Jo use?
### Handshakes
##### Age 11 to 14 Challenge Level:
Can you find an efficient method to work out how many handshakes there would be if hundreds of people met?
### Egyptian Fractions
##### Age 11 to 14 Challenge Level:
The Egyptians expressed all fractions as the sum of different unit fractions. Here is a chance to explore how they could have written different fractions. | 2,197 | 9,219 | {"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-2019-43 | latest | en | 0.896155 |
http://nrich.maths.org/public/leg.php?code=31&cl=2&cldcmpid=7280 | 1,503,373,314,000,000,000 | text/html | crawl-data/CC-MAIN-2017-34/segments/1502886109893.47/warc/CC-MAIN-20170822031111-20170822051111-00676.warc.gz | 310,868,907 | 10,122 | # Search by Topic
#### Resources tagged with Addition & subtraction similar to Area and Perimeter:
Filter by: Content type:
Stage:
Challenge level:
### Oddly
##### Stage: 2 Challenge Level:
Find the sum of all three-digit numbers each of whose digits is odd.
### 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?
### The Pied Piper of Hamelin
##### Stage: 2 Challenge Level:
This problem is based on the story of the Pied Piper of Hamelin. Investigate the different numbers of people and rats there could have been if you know how many legs there are 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.
### Rabbits in the Pen
##### Stage: 2 Challenge Level:
Using the statements, can you work out how many of each type of rabbit there are in these pens?
### How Old?
##### Stage: 2 Challenge Level:
Cherri, Saxon, Mel and Paul are friends. They are all different ages. Can you find out the age of each friend using the information?
### Sitting Round the Party Tables
##### Stage: 1 and 2 Challenge Level:
Sweets are given out to party-goers in a particular way. Investigate the total number of sweets received by people sitting in different positions.
### 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?
### 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?
### Hubble, Bubble
##### Stage: 2 Challenge Level:
Winifred Wytsh bought a box each of jelly babies, milk jelly bears, yellow jelly bees and jelly belly beans. In how many different ways could she make a jolly jelly feast with 32 legs?
### Pouring the Punch Drink
##### Stage: 2 Challenge Level:
There are 4 jugs which hold 9 litres, 7 litres, 4 litres and 2 litres. Find a way to pour 9 litres of drink from one jug to another until you are left with exactly 3 litres in three of the jugs.
### A-magical Number Maze
##### Stage: 2 Challenge Level:
This magic square has operations written in it, to make it into a maze. Start wherever you like, go through every cell and go out a total of 15!
##### 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?
### Calendar Calculations
##### Stage: 2 Challenge Level:
Try adding together the dates of all the days in one week. Now multiply the first date by 7 and add 21. Can you explain what happens?
### Arranging the Tables
##### Stage: 2 Challenge Level:
There are 44 people coming to a dinner party. There are 15 square tables that seat 4 people. Find a way to seat the 44 people using all 15 tables, with no empty places.
### On Target
##### Stage: 2 Challenge Level:
You have 5 darts and your target score is 44. How many different ways could you score 44?
### Rod Measures
##### Stage: 2 Challenge Level:
Using 3 rods of integer lengths, none longer than 10 units and not using any rod more than once, you can measure all the lengths in whole units from 1 to 10 units. How many ways can you do this?
### Magic Circles
##### Stage: 2 Challenge Level:
Put the numbers 1, 2, 3, 4, 5, 6 into the squares so that the numbers on each circle add up to the same amount. Can you find the rule for giving another set of six numbers?
### Super Value Shapes
##### Stage: 2 Challenge Level:
If each of these three shapes has a value, can you find the totals of the combinations? Perhaps you can use the shapes to make the given totals?
### 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.
### 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?
### Doplication
##### Stage: 2 Challenge Level:
We can arrange dots in a similar way to the 5 on a dice and they usually sit quite well into a rectangular shape. How many altogether in this 3 by 5? What happens for other sizes?
### Number Squares
##### Stage: 1 and 2 Challenge Level:
Start with four numbers at the corners of a square and put the total of two corners in the middle of that side. Keep going... Can you estimate what the size of the last four numbers will be?
### Build it Up
##### Stage: 2 Challenge Level:
Can you find all the ways to get 15 at the top of this triangle of numbers?
### Sums and Differences 1
##### Stage: 2 Challenge Level:
This challenge focuses on finding the sum and difference of pairs of two-digit numbers.
### 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.
### X Is 5 Squares
##### Stage: 2 Challenge Level:
Can you arrange 5 different digits (from 0 - 9) in the cross in the way described?
### Magic Constants
##### Stage: 2 Challenge Level:
In a Magic Square all the rows, columns and diagonals add to the 'Magic Constant'. How would you change the magic constant of this square?
### The Dice Train
##### Stage: 2 Challenge Level:
This dice train has been made using specific rules. How many different trains can you make?
##### Stage: 2 Challenge Level:
Lolla bought a balloon at the circus. She gave the clown six coins to pay for it. What could Lolla have paid for the balloon?
### Spell by Numbers
##### Stage: 2 Challenge Level:
Can you substitute numbers for the letters in these sums?
### Strike it Out for Two
##### Stage: 1 and 2 Challenge Level:
Strike it Out game for an adult and child. Can you stop your partner from being able to go?
### Throw a 100
##### Stage: 2 Challenge Level:
Can you score 100 by throwing rings on this board? Is there more than way to do it?
### Christmas Shopping
##### Stage: 2 Challenge Level:
Vera is shopping at a market with these coins in her purse. Which things could she give exactly the right amount for?
### Machines
##### Stage: 2 Challenge Level:
What is happening at each box in these machines?
### The Puzzling Sweet Shop
##### Stage: 2 Challenge Level:
There were chews for 2p, mini eggs for 3p, Chocko bars for 5p and lollypops for 7p in the sweet shop. What could each of the children buy with their money?
### A Rod and a Pole
##### Stage: 2 Challenge Level:
A lady has a steel rod and a wooden pole and she knows the length of each. How can she measure out an 8 unit piece of pole?
### Six Ten Total
##### Stage: 2 Challenge Level:
This challenge combines addition, multiplication, perseverance and even proof.
### One Wasn't Square
##### Stage: 2 Challenge Level:
Mrs Morgan, the class's teacher, pinned numbers onto the backs of three children. Use the information to find out what the three numbers were.
### Clock Face
##### Stage: 2 Challenge Level:
Where can you draw a line on a clock face so that the numbers on both sides have the same total?
### A Mixed-up Clock
##### Stage: 2 Challenge Level:
There is a clock-face where the numbers have become all mixed up. Can you find out where all the numbers have got to from these ten statements?
### Bean Bags for Bernard's Bag
##### Stage: 2 Challenge Level:
How could you put eight beanbags in the hoops so that there are four in the blue hoop, five in the red and six in the yellow? Can you find all the ways of doing this?
### Sealed Solution
##### Stage: 2 Challenge Level:
Ten cards are put into five envelopes so that there are two cards in each envelope. The sum of the numbers inside it is written on each envelope. What numbers could be inside the envelopes?
### 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?
### Times
##### Stage: 2 Challenge Level:
Which times on a digital clock have a line of symmetry? Which look the same upside-down? You might like to try this investigation and find out!
### Code Breaker
##### Stage: 2 Challenge Level:
This problem is based on a code using two different prime numbers less than 10. You'll need to multiply them together and shift the alphabet forwards by the result. Can you decipher the code?
### The Clockmaker's Birthday Cake
##### Stage: 2 Challenge Level:
The clockmaker's wife cut up his birthday cake to look like a clock face. Can you work out who received each piece?
### 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.
### Two Primes Make One Square
##### Stage: 2 Challenge Level:
Can you make square numbers by adding two prime numbers together? | 2,230 | 9,503 | {"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-2017-34 | latest | en | 0.906785 |
https://www.fxsolver.com/browse/?like=1117&p=3 | 1,685,758,703,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224648911.0/warc/CC-MAIN-20230603000901-20230603030901-00322.warc.gz | 857,031,328 | 41,755 | '
# Search results
Found 1552 matches
Euler's theorem (excircles)
The circumscribed circle or circumcircle of a triangle is a circle which passes through all the vertices of the triangle. The center of this circle is ... more
Area of a deltoid
In geometry, a deltoid, also known as a tricuspoid or Steiner curve, is a hypocycloid of three cusps. In other words, it is the roulette created by a point ... more
Perimeter of a deltoid
In geometry, a deltoid, also known as a tricuspoid or Steiner curve, is a hypocycloid of three cusps. In other words, it is the roulette created by a point ... more
Epicyclic gearing (overal gear ratio)
An epicyclic gear train consists of two gears mounted so that the center of one gear revolves around the center of the other. A carrier connects the ... more
Klein bagel ( "figure 8" immersion x-coordinate)
In mathematics, the Klein bottle is an example of a non-orientable surface, informally, it is a surface (a two-dimensional manifold) in which notions of ... more
Klein bagel ( "figure 8" immersion y-coordinate)
In mathematics, the Klein bottle is an example of a non-orientable surface, informally, it is a surface (a two-dimensional manifold) in which notions of ... more
Astroid (Perimeter)
An astroid is a particular mathematical curve: a hypocycloid with four cusps. The astroid is a real locus of a plane algebraic curve of genus zero. The ... more
Astroid (Area)
An astroid is a particular mathematical curve: a hypocycloid with four cusps. The astroid is a real locus of a plane algebraic curve of genus zero. The ... more
Length of an arc of a circle (central angle in radians)
Circular arc is a segment of a circle, or of its circumference (boundary) if the circle is considered to be a disc. Central angle is an angle whose apex ... more
Area of an arbitrary triangle (incircle and excircles)
The incircle or inscribed circle of a triangle is the largest circle contained in the triangle; it touches (is tangent to) the three sides. The center of ... more
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https://www.sophia.org/tutorials/sampling-with-or-without-replacement--3 | 1,508,299,267,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187822739.16/warc/CC-MAIN-20171018032625-20171018052625-00447.warc.gz | 1,043,354,701 | 19,316 | ### Free Educational Resources
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4 Tutorials that teach Sampling With or Without Replacement
# Sampling With or Without Replacement
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Author: Jonathan Osters
##### Description:
This lesson will explain sampling with and without replacement
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Tutorial
Source: Playing Cards; Public Domain: http://www.jfitz.com/cards/
## Video Transcription
This tutorial is going to cover sampling, both with and without replacement. With replacement means that you put everything back once you've selected it. And without replacement means that each observation is not put back once it's selected. And so once it's selected, it's out. It can't be selected again.
So typically, one big requirement for statistical inference is that the individuals, the values from the sample, are independent. That is that one doesn't affect any of the others. And ideally, this would mean sampling with replacement.
Let's go through an example real quick involving cards. So the probability that you select a spade, something from this bottom row, on the first draw is one fourth. Now suppose that you draw it and you don't put it back. So I took the 10 of spades, and I pulled it out.
Now the probability of a spade, there's only 12 left out of 51 cards. That's not one fourth. That's a different number. So the first draw and the second draw are dependent. The probability of a spade changed after knowing that we got a spade on the first draw.
Now consider if the card got replaced though. The probability of a spade on the first draw is one fourth. And then you pull the 10 of spades. And then you put it back.
Now what's the probability of a spade on the second draw? Well it's one fourth again. It's the same 52 cards. And so you have the same likelihood of selecting a spade.
Now in real life, we often don't sample with replacement. And this is a huge deal because typically sampling with replacement will lead to independence, which is a requirement for a lot of statistical analysis. But you wouldn't call a person for their opinion in a poll twice. So we don't put someone back into the population and see if we can sample them again. It just doesn't make sense to do in real life.
So we need a little bit of a workaround. So what we're going to do is something, even though the sampling done in real life doesn't technically fit the definition for independent observations, there's going to be a workaround. There's a big but here.
So suppose that our population was very large. So suppose we had, instead of the 52 cards, four decks of cards, so 208 different cards. Now suppose the worst case scenario happened in terms of independence. And maybe every card we picked was the same suit. So we'll take five diamonds from the group. So we've selected five cards, all of which were diamonds.
The probability of a diamond on the first draw-- there were 52 diamonds here total listed out of 208 cards-- and so that's one fourth probability, same as if there were one deck. But the larger population actually has an effect now. Look at the probability of a diamond on the last draw. It's 48 remaining diamonds out of 204 remaining cards.
The probability is about 0.24, which is different than 0.25, but not hugely. And this is even after five draws. And so we're going to say that the probability of a diamond didn't change particularly that much from the first to the last draw.
So we're going to have sort of a catch for independence. So when we sample without replacement, if the population is large enough, then the probabilities don't shift very much as we sample. And so the sampling without replacement becomes almost independent because the probabilities don't change very much.
Now the question is, it says when the population is large enough, and that's not very well defined term. How large is considered a large population? What we're going to do is we're going to institute a rule. And a large population is going to be at least 10 times larger than the sample. So the population is greater than or equal to 10 times n, the sample size.
And if that's the case, then we're going to say that the probabilities don't shift very much when you sample several items, n items from the population. And so therefore, we can treat the sampling as being almost independent.
So to recap, sampling with replacement is kind of the gold standard. It always creates independent trials. So the probability of particular events don't change at all-- at all-- from trial to trial. However, in real life, when we sample without replacement, the probabilities do necessarily change. However, our workaround is that if the population from which we're sampling is at least 10 times larger than the sample that we're drawing, the trials can be considered nearly independent.
And so we talked about sampling with replacement and sampling without replacement and how independence relates to those two. Sampling with replacement will guarantee independence. Sampling without replacement can guarantee independence almost if certain conditions are met.
Good luck. And we'll see you next time.
## Notes on "Sampling With or Without Replacement"
.
Terms to Know
Sampling with Replacement
A sampling plan where each observation that is sampled is replaced after each time it is sampled, resulting in an observation being able to be selected more than once.
Sampling without Replacement
A sampling plan where each observation that is sampled is kept out of subsequent selections, resulting in a sample where each observation can be selected no more than one time. | 1,297 | 6,226 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.625 | 4 | CC-MAIN-2017-43 | longest | en | 0.94124 |
http://www.ck12.org/book/Basic-Geometry/r1/section/8.6/ | 1,493,197,119,000,000,000 | text/html | crawl-data/CC-MAIN-2017-17/segments/1492917121216.64/warc/CC-MAIN-20170423031201-00569-ip-10-145-167-34.ec2.internal.warc.gz | 478,143,630 | 35,710 | # 8.6: Inverse Trigonometric Ratios
Difficulty Level: At Grade Created by: CK-12
## Learning Objectives
• Use the inverse trigonometric ratios to find an angle in a right triangle.
• Solve a right triangle.
## Review Queue
Find the lengths of the missing sides. Round your answer to the nearest tenth.
1. Draw an isosceles right triangle with legs of length 3. What is the hypotenuse?
2. Use the triangle from #3, to find the sine, cosine, and tangent of \begin{align*}45^\circ\end{align*}.
Know What? The longest escalator in North America is at the Wheaton Metro Station in Maryland. It is 230 feet long and is 115 ft. high. What is the angle of elevation, \begin{align*}x^\circ\end{align*}, of this escalator?
## Inverse Trigonometric Ratios
In mathematics, the word inverse means “undo.” For example, addition and subtraction are inverses of each other because one undoes the other. When we apply inverses to the trigonometric ratios, we can find acute angle measures as long as we are given two sides.
Inverse Tangent: Labeled \begin{align*}\tan^{-1}\end{align*}, the “-1” means inverse.
\begin{align*}\tan^{-1} \left (\frac{b}{a} \right ) = m \angle B && \tan^{-1} \left( \frac{a}{b} \right) = m \angle A\end{align*}
Inverse Sine: Labeled \begin{align*}\sin^{-1}\end{align*}.
\begin{align*}\sin^{-1} \left( \frac{b}{c} \right) = m \angle B && \sin^{-1} \left(\frac{a}{c} \right) = m \angle A\end{align*}
Inverse Cosine: Labeled \begin{align*}\cos^{-1}\end{align*}.
\begin{align*}\cos^{-1} \left(\frac{a}{c} \right) = m \angle B && \cos^{-1} \left( \frac{b}{c} \right) = m \angle A\end{align*}
In order to find the measure of the angles, you will need you use your calculator. On most scientific and graphing calculators, the buttons look like \begin{align*}[ \text{SIN}^{-1}], [ \text{COS}^{-1}]\end{align*}, and \begin{align*}[\text{TAN}^{-1}]\end{align*}. You might also have to hit a shift or \begin{align*}2^{nd}\end{align*} button to access these functions.
Example 1: Use the sides of the triangle and your calculator to find the value of \begin{align*}\angle A\end{align*}. Round your answer to the nearest tenth of a degree.
Solution: In reference to \begin{align*}\angle A\end{align*}, we are given the opposite leg and the adjacent leg. This means we should use the tangent ratio.
\begin{align*}\tan A = \frac{20}{25} = \frac{4}{5}\end{align*}. So, \begin{align*}\tan^{-1} \frac{4}{5} = m \angle A\end{align*}. Now, use your calculator.
If you are using a TI-83 or 84, the keystrokes would be: \begin{align*}[2^{nd}]\end{align*}[TAN]\begin{align*}\left( \frac{4}{5} \right)\end{align*} [ENTER] and the screen looks like:
\begin{align*}m \angle A = 38.7^\circ\end{align*}
Example 2: \begin{align*}\angle A\end{align*} is an acute angle in a right triangle. Find \begin{align*}m \angle A\end{align*} to the nearest tenth of a degree.
a) \begin{align*}\sin A = 0.68\end{align*}
b) \begin{align*}\cos A = 0.85\end{align*}
c) \begin{align*}\tan A = 0.34\end{align*}
Solution:
a) \begin{align*}m \angle A = \sin^{-1} 0.68=42.8^\circ\end{align*}
b) \begin{align*}m \angle A = \cos^{-1} 0.85=31.8^\circ\end{align*}
c) \begin{align*}m \angle A = \tan^{-1} 0.34=18.8^\circ\end{align*}
## Solving Triangles
To solve a right triangle, you need to find all sides and angles in a right triangle, using sine, cosine or tangent, inverse sine, inverse cosine, or inverse tangent, or the Pythagorean Theorem.
Example 3: Solve the right triangle.
Solution: To solve this right triangle, we need to find \begin{align*}AB, m \angle C\end{align*} and \begin{align*}m \angle B\end{align*}. Only use the values you are given.
\begin{align*}\underline{AB}\end{align*}: Use the Pythagorean Theorem.
\begin{align*}24^2 + AB^2 &= 30^2\\ 576 + AB^2 &= 900\\ AB^2 &= 324\\ AB &= \sqrt{324} = 18\end{align*}
\begin{align*}\underline{m \angle B}\end{align*}: Use the inverse sine ratio.
\begin{align*}\sin B &= \frac{24}{30} = \frac{4}{5}\\ \sin^{-1} \left( \frac{4}{5} \right) &= 53.1^\circ = m \angle B\end{align*}
\begin{align*}\underline{m \angle C}\end{align*}: Use the inverse cosine ratio.
\begin{align*}\cos C = \frac{24}{30} = \frac{4}{5} \longrightarrow \cos^{-1}\left( \frac{4}{5} \right) = 36.9^\circ = m \angle C\end{align*}
Example 4: Solve the right triangle.
Solution: To solve this right triangle, we need to find \begin{align*}AB, BC\end{align*} and \begin{align*}m \angle A\end{align*}.
\begin{align*}\underline{AB}\end{align*}: Use sine ratio.
\begin{align*}\sin 62^\circ &= \frac{25}{AB}\\ AB &= \frac{25}{\sin 62^\circ}\\ AB & \approx 28.31\end{align*}
\begin{align*}\underline{BC}\end{align*}: Use tangent ratio.
\begin{align*}\tan 62^\circ &= \frac{25}{BC}\\ BC &= \frac{25}{\tan 62^\circ}\\ BC & \approx 13.30\end{align*}
\begin{align*}\underline{m \angle A}\end{align*}: Use Triangle Sum Theorem
\begin{align*}62^\circ + 90^\circ + m \angle A &= 180^\circ\\ m \angle A &= 28^\circ\end{align*}
Example 5: Solve the right triangle.
Solution: The two acute angles are congruent, making them both \begin{align*}45^\circ\end{align*}. This is a 45-45-90 triangle. You can use the trigonometric ratios or the special right triangle ratios.
Trigonometric Ratios
\begin{align*}\tan 45^\circ &= \frac{15}{BC} && \sin 45^\circ = \frac{15}{AC}\\ BC &= \frac{15}{\tan 45^\circ} = 15 && \quad \ AC = \frac{15}{\sin 45^\circ} \approx 21.21\end{align*}
45-45-90 Triangle Ratios
\begin{align*}BC = AB = 15, AC = 15 \sqrt{2} \approx 21.21\end{align*}
## Real-Life Situations
Example 6: A 25 foot tall flagpole casts a 42 feet shadow. What is the angle that the sun hits the flagpole?
Solution: Draw a picture. The angle that the sun hits the flagpole is \begin{align*}x^\circ\end{align*}. We need to use the inverse tangent ratio.
\begin{align*}\tan x &= \frac{42}{25}\\ \tan^{-1} \frac{42}{25} & \approx 59.2^\circ = x\end{align*}
Example 7: Elise is standing on top of a 50 foot building and sees her friend, Molly. If Molly is 35 feet away from the base of the building, what is the angle of depression from Elise to Molly? Elise’s eye height is 4.5 feet.
Solution: Because of parallel lines, the angle of depression is equal to the angle at Molly, or \begin{align*}x^\circ\end{align*}. We can use the inverse tangent ratio.
\begin{align*}\tan^{-1} \left( \frac{54.5}{30} \right) = 61.2^\circ = x\end{align*}
Know What? Revisited To find the escalator’s angle of elevation, use the inverse sine.
\begin{align*}\sin^{-1} \left( \frac{115}{230} \right)= 30^\circ \qquad \quad \text{The angle of elevation is} \ 30^\circ.\end{align*}
## Review Questions
• Questions 1-6 are similar to Example 1.
• Questions 7-12 are similar to Example 2.
• Questions13-21 are similar to Examples 3 and 4.
• Questions 22-24 are similar to Examples 6 and 7.
• Questions 25-30 are a review of the trigonometric ratios.
Use your calculator to find \begin{align*}m \angle A\end{align*} to the nearest tenth of a degree.
Let \begin{align*}\angle A\end{align*} be an acute angle in a right triangle. Find \begin{align*}m \angle A\end{align*} to the nearest tenth of a degree.
1. \begin{align*}\sin A = 0.5684\end{align*}
2. \begin{align*}\cos A = 0.1234\end{align*}
3. \begin{align*}\tan A = 2.78\end{align*}
4. \begin{align*}\cos^{-1} 0.9845\end{align*}
5. \begin{align*}\tan^{-1} 15.93\end{align*}
6. \begin{align*}\sin^{-1} 0.7851\end{align*}
Solving the following right triangles. Find all missing sides and angles. Round any decimal answers to the nearest tenth.
Real-Life Situations Use what you know about right triangles to solve for the missing angle. If needed, draw a picture. Round all answers to the nearest tenth of a degree.
1. A 75 foot building casts an 82 foot shadow. What is the angle that the sun hits the building?
2. Over 2 miles (horizontal), a road rises 300 feet (vertical). What is the angle of elevation?
3. A boat is sailing and spots a shipwreck 650 feet below the water. A diver jumps from the boat and swims 935 feet to reach the wreck. What is the angle of depression from the boat to the shipwreck?
Examining Patterns Below is a table that shows the sine, cosine, and tangent values for eight different angle measures. Answer the following questions.
\begin{align*}10^\circ\end{align*} \begin{align*}20^\circ\end{align*} \begin{align*}30^\circ\end{align*} \begin{align*}40^\circ\end{align*} \begin{align*}50^\circ\end{align*} \begin{align*}60^\circ\end{align*} \begin{align*}70^\circ\end{align*} \begin{align*}80^\circ\end{align*}
Sine 0.1736 0.3420 0.5 0.6428 0.7660 0.8660 0.9397 0.9848
Cosine 0.9848 0.9397 0.8660 0.7660 0.6428 0.5 0.3420 0.1736
Tangent 0.1763 0.3640 0.5774 0.8391 1.1918 1.7321 2.7475 5.6713
1. What value is equal to \begin{align*}\sin 40^\circ\end{align*}?
2. What value is equal to \begin{align*}\cos 70^\circ\end{align*}?
3. Describe what happens to the sine values as the angle measures increase.
4. Describe what happens to the cosine values as the angle measures increase.
5. What two numbers are the sine and cosine values between?
6. Find \begin{align*}\tan85^\circ, \tan 89^\circ\end{align*}, and \begin{align*}\tan 89.5^\circ\end{align*} using your calculator. Now, describe what happens to the tangent values as the angle measures increase.
## Review Queue Answers
1. \begin{align*}\sin 36^\circ = \frac{y}{7} \qquad \cos 36^\circ = \frac{x}{7}\!\\ {\;} \qquad y = 4.11 \qquad \quad \ \ x = 5.66\end{align*}
2. \begin{align*}\cos 12.7^\circ = \frac{40}{x} \qquad \ \tan 12.7^\circ = \frac{y}{40}\!\\ {\;} \qquad \ \ \ x = 41.00 \qquad \qquad \ \ y = 9.01\end{align*}
3. \begin{align*}{\;} \ \sin 45^\circ = \frac{3}{3 \sqrt{2}} = \frac{\sqrt{2}}{2}\!\\ {\;} \ \cos 45^\circ = \frac{3}{3 \sqrt{2}} = \frac{\sqrt{2}}{2}\!\\ {\;} \ \tan 45^\circ = \frac{3}{3} = 1\end{align*}
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Math Central Quandaries & Queries
Question from Savanna, a student: How would put .12 with 2 repeating in fraction? and if you do figure it out (which im sure u will) could you please explain why so i can know how to do it in my future assignments!!! Thanks a ton! Savanna!
Savanna,
I am going to convert 0.13777... into a fraction and you can use the same technique for your problem.
Let d = 0.13777... and then perform the subtraction
10d = 1.37777... - d = 0.13777... ______________ 9d = 1.24
Multiply both sides by 100 to eliminate the decimal point
900d = 124
and hence
d = 124/900 = 31/225
You can check this by having your calculator find 31/225.
I multiplied both sides of d = 0.13777. by 101 = 10 because the repeat is 1 digit long. If the repeating fraction were 0.13757575... so that the repeat (75) is two digits long I would let d = 0.13757575... and multiply both sides by 102 = 100 before subtracting. Try it and use your calculator to check your answer.
If the repeat were 3 digits long I would multiply by 103 = 1000 and so on.
I hope this helps,
Penny
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## Oscar Corporation is planning to construct an elliptical gate at its headquarters. The width of the ellipse will be 5 feet across and its maximum height along the center will be 3 feet. The company wants to place two bright spots at the foci of the ellipse. How far from the center of the ellipse will the spots be located?
Draw the elliptical gate and show the foci, the width, and height as in the figure below.
Note that
a = 5 ft, the width
b = 3 ft, the height
The center is at (0,0).
The foci are at (c, 0) and at (-c, 0).
The foci are related to a and b by
c² = a² – b²
= 5² – 3² = 16
c = +4 or -4
The foci are located 4 ft horizontally from the center of the ellipse.
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## Ead the passage from “The Caged Bird.” The free bird thinks of another breeze and the trade winds soft through the sighing trees and the fat worms waiting on a dawn bright lawn and he names the sky his own What are the connotative meanings of sighing, as used in the poem? Check all that apply. longing sadness relaxation exhaling peacefulness breathing
Ead the passage from “The Caged Bird.” The free bird thinks of another breeze and the trade winds soft through the sighing trees and the fat worms waiting on a dawn bright lawn and he names the sky his own What are the connotative meanings of sighing, as used in the poem? Check all that apply. longing sadness relaxation exhaling peacefulness breathing
Categories
## Read the excerpt from Act I, scene i of Romeo and Juliet. Yet tell me not, for I have heard it all. Here’s much to do with hate, but more with love: Why then, O brawling love! O loving hate! O any thing! of nothing first create. O heavy lightness! serious vanity! Mis-shapen chaos of well-seeming forms! Feather of lead, bright smoke, cold fire, sick health! Still-waking sleep, that is not what it is! This love feel I, that feel no love in this. Dost thou not laugh? The oxymorons in Romeo’s dialogue emphasize a) the anger he feels toward a certain woman. b) his certainty about his romantic fate. c) the extreme emotions that he is feeling. d) his confusion about Benvolio’s advice.
Read the excerpt from Act I, scene i of Romeo and Juliet. Yet tell me not, for I have heard it all. Here’s much to do with hate, but more with love: Why then, O brawling love! O loving hate! O any thing! of nothing first create. O heavy lightness! serious vanity! Mis-shapen chaos of well-seeming forms! Feather of lead, bright smoke, cold fire, sick health! Still-waking sleep, that is not what it is! This love feel I, that feel no love in this. Dost thou not laugh? The oxymorons in Romeo’s dialogue emphasize a) the anger he feels toward a certain woman. b) his certainty about his romantic fate. c) the extreme emotions that he is feeling. d) his confusion about Benvolio’s advice.
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## Marie had a bright mind and a __personality
Marie had a bright mind and a __personality
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## Match the words in the left-hand column to the appropriate blank in the sentences in the right-hand column. use each word only once. hints helpreset coma plasma tail oort cloud kuiper belt meteor shower nucleus 1. about a trillion comets are thought to be located far, far beyond pluto in the . 2. the bright spherical part of a comet observed when it is close to the sun is the . 3. a comet’s stretches directly away from the sun. 4. a comet’s is the frozen portion of a comet. 5. particles ejected from a comet can cause a(n) on earth. 6. the extends from about beyond the orbit of neptune to about twice the distance of neptune from the sun.
1. About a trillion comets are thought to be located far, far beyond pluto in the Oort cloud. Oort cloud is theoretical cloud of predominantly icy planetesimals and. Many scientists have a belief that it surrounds the Sun at the distance of approximately between 50,000 and 200,000 AU. Such forces as passing stars and of the Milky Way usually easily affect the outer Oort cloud and dislodge comets right from their orbits through the Oort the cloud and then they send them to the the inner Solar System.
2. The bright spherical part of a comet observed when it is close to the sun is the coma. When coma is viewed by a telescope it looks quite fuzzy and, unlike the stars, it does not have legible form. Such a phenomenon as the coma is created when a comet comes too close to the sun, then a comet warms and sublimes its parts. These parts are placed around the nucleus of a comet forming sort of envelope.
3. A comet’s plasma tail stretches directly away from the sun. Both coma and comet’s tail are visible parts of a comet, and when the comet passes through the inner Solar System we it becomes visible right from Earth. Usually comets have two tails :the blue plasma tail and the red dust tail. The plasma tail is formed by an interaction between the solar wind and the cometary plasma. And the dust tail is caused due to the activity of solar radiation pressure directed to the cometary dust.
4. A comet’s nucleus is the frozen portion of a comet. The conet nucleus is the solid center of the head of a comet that is also called ”dirty snowball” or an ”icy dirtball” among astronomists. It consists of rock, dust, and frozen gases and when they are heated by the Sun they form a coma (which is an atmosphere that surrounds nucleus) by sublimating of the gases.
5. Particles ejected from a comet can cause a meteor shower on Earth. A meteor shower is a celestial phenomenon during which we can see how meteors radiate. The meteors which we can observe are formed by meteoroids that enter Earth’s atmosphere at extremely high speeds and they usually move in parallel directions. Some of the meteors are too small and often don’t even reach the surface of Earth as they disintegrate during the entering process.
6. The Kuiper Belt extends from about beyond the orbit of neptune to about twice the distance of neptune from the sun. It is the circumstellar disc that uccurs in the Solar System beyond the already known planets. Its formreminds of asteroid belt, but it is far larger, to be more exact in 20 times. I
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## Blue light, which has a wavelength of about 475 nm, is made to pass through a slit of a diffraction grating that has 425 lines per mm and forms a first-order bright band. What is the angle of diffraction? Round answer to the nearest whole number.
a) The time the police officer required to reach the motorist was 15 s.
b) The speed of the officer at the moment she overtakes the motorist is 30 m/s
c) The total distance traveled by the officer was 225 m.
Explanation:
The equations for the position and velocity of an object moving in a straight line are as follows:
x = x0 + v0 · t + 1/2 · a · t²
v = v0 + a · t
Where:
x = position at time t
x0 = initial position
v0 = initial velocity
t = time
a = acceleration
v = velocity at time t
a)When the officer reaches the motorist, the position of the motorist is the same as the position of the officer:
x motorist = x officer
Using the equation for the position:
x motirist = x0 + v · t (since a = 0).
x officer = x0 + v0 · t + 1/2 · a · t²
Let´s place our frame of reference at the point where the officer starts following the motorist so that x0 = 0 for both:
x motorist = x officer
x0 + v · t = x0 + v0 · t + 1/2 · a · t² (the officer starts form rest, then, v0 = 0)
v · t = 1/2 · a · t²
Solving for t:
2 v/a = t
t = 2 · 15.0 m/s/ 2.00 m/s² = 15 s
The time the police officer required to reach the motorist was 15 s.
b) Now, we can calculate the speed of the officer using the time calculated in a) and the equation for velocity:
v = v0 + a · t
v = 0 m/s + 2.00 m/s² · 15 s
v = 30 m/s
The speed of the officer at the moment she overtakes the motorist is 30 m/s
c) Using the equation for the position, we can find the traveled distance in 15 s:
x = x0 + v0 · t + 1/2 · a · t²
x = 1/2 · 2.00 m/s² · (15s)² = 225 m
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## I have travelled the length and breadth of Alabama, Mississippi and all the other southern states. On sweltering summer days and crisp autumn mornings I have looked at her beautiful churches with their spires pointing heavenward. I have beheld the impressive outlay of her massive religious education buildings. Over and over again I have found myself asking: “Who worships here? Who is their God? Where were their voices when the lips of Governor Barnett dripped with words of interposition and nullification? Where were their voices of support when tired, bruised, and weary Negro men and women decided to rise from the dark dungeons of complacency to the bright hills of creative protest?” What is the main idea of the paragraph?
Here are a few pointers; hope this is useful)
Ovation-by definition- is show of appreciation from an audience, for a person’s accomplishments or flaw.
“Everyone deserves a standing ovation because we all overcometh the world.”
A person’s accomplishment could be how they made a positive change in this world, strong leadership- that makes them a effective leader or simply helping others. A person’s flaw- mistakes in life, sin or even guilt should also be considered an appreciation- an ovation for representing mankind’s flaw and that humanity makes mistakes, fulfilling at least one deadly sin such as greed, lust, selfishness etc.
Thus, regardless of a person’s achievement or flaw- a person deserves an applause for, not the least, living in this society and this world that we are all living together and dying together.
That was just the introduction.. the best part is yet to come.. now it’s your turn!!
Here are other pointers to talk about in your essay:
Shakespeare’s famous line “All the World’s a stage. That agrees with your line: “Everybody deserves a standing ovation…”
Shakespeare explains that men and women are like players: they live, and die, some being celebrated and some forever living in solitude till their death. Shakespeare states the world is a “stage” which symbolizes that mankind is in its peak. The world is changing everyday: little by little and humanity is falling behind.
Due to our world turning into machinery: factories, an automotive future: where humans only job to live (entrance) and to die (exit) the famous humans remembered and the flawed not recalled.
This is according to Shakespeare’s imagery.
I don’t know what grade your in, but I think simplifying Shakespeare’s word of mouth in your essay would be handy and useful as it has strong references of your quote, and agrees strongly in your essay.
Hope this helps 🙂
Categories
## Read the excerpt from "Annabel Lee," by Edgar Allan Poe. And the stars never rise, but I feel the bright eyes Of the beautiful Annabel Lee; How does Poe use a sound device in the excerpt? He uses euphony to create the sounds of nature at night. He uses alliteration to link images of the night with images of his loss. He uses cacophony to proclaim the depth of his emotion. He uses internal rhyme to enhance the rhythm of his lyrical poem.
Read the excerpt from “Annabel Lee,” by Edgar Allan Poe. And the stars never rise, but I feel the bright eyes Of the beautiful Annabel Lee; How does Poe use a sound device in the excerpt? He uses euphony to create the sounds of nature at night. He uses alliteration to link images of the night with images of his loss. He uses cacophony to proclaim the depth of his emotion. He uses internal rhyme to enhance the rhythm of his lyrical poem.
Categories
## Hector has written this topic sentence. Dr. King’s “I Have a Dream” speech includes many metaphors to show the suffering of African Americans. Which excerpt from the speech best supports this topic sentence? One hundred years later, the Negro is still languished in the corners of American society and finds himself an exile in his own land. And they have come to realize that their freedom is inextricably bound to our freedom. We cannot walk alone. The whirlwinds of revolt will continue to shake the foundations of our nation until the bright day of justice emerges. And those who hope that the Negro needed to blow off steam and will now be content will have a rude awakening if the nation returns to business as usual.
C) A certain cool-headedness had come to him
E) [D]oggedly he swam in that direction, swimming with slow, deliberate strokes, conserving his strength
Both of these sentences show that Rainsford can easily handle stress. The first sentence says that he was overcome by “cool-headedness”. In the second stanza the speaker uses words like “doggedly, slow, deliberate and conserving his strength.” In all of these instances Rainsford was in danger,; however, he was able to keep calm and think clearly about what he should do next.
Categories
## A 39-year-old male sustained a stab wound to the groin during an altercation at a bar. as you approach the patient, you note that he is conscious, is screaming in pain, and is attempting to control the bleeding, which is bright red and spurting from his groin area. you should: a. apply direct pressure to the wound. b. elevate his legs and keep him warm. c. ensure that his airway is patent. d. administer 100% supplemental oxygen.
Answer : i) Drinking alcohol and smoking cigarettes – It will have direct harm on individual’s body and the health will deteriorate.
ii) Vegetable Hater – The major chunk of proteins, fibers, vitamins comes from veggies. If they are skipped it will cause a health issue in a person.
iii) Hate socializing – Mental health is also important. One needs to socialize and make new friends to release stress and anxiety.
iv) Avoiding Traveling – Being a homebound has its own side effects. There will be lot of respiratory issues if one cannot breathe in the fresh air of the surrounding nature.
v) Not playing sports – A physical activity keeps an individual’s body fit. Lacking of this may give invitation to many diseases.
vi) Not drinking enough water – Health is largely influenced by the water that we drink, as our bodies are composed of 70% water. It is essential to take the correct amount of water in our body.
Categories
## What causes the bright lines in the emission spectrum of an element to occur?
What causes the bright lines in the emission spectrum of an element to occur?
Categories
c. The discoloration of ages had been great.
Explanation:
The Fall of the House of Usher (The Fall of the House of Usher) is a short story by Edgar Allan Poe first published in 1839. The story begins with an unnamed narrator who arrives at the home of his friend, Roderick Usher, having received a letter from him in a distant part of the country complaining about an illness, and asking for his help. When he arrives, the narrator observes a thin crack that extends from the roof at the front of the building to the lake.
Although Poe wrote this tale before the invention of modern psychological science, Roderick’s condition can be described according to his terminology. It includes a form of sensory overload known as hyperesthesia (hypersensitivity to textures, light, sounds, smells and tastes), hypochondria (an excessive worry or temerity in having a serious illness), and acute anxiety. It is revealed that Roderick’s twin sister, Madeline, is also ill and falls into cataleptic trances, as if in a state of death. The narrator is impressed by Roderick’s paintings, and tries to cheer him up by reading with him and listening to his improvised musical compositions on guitar. Roderick sings “The Haunted Palace,” then tells the narrator that he believes that the house in which he lives may be alive, and that this sensitivity stems from the arrangement of the house and the surrounding vegetation.
Categories
## The author chose to tell Lizzie Bright and the Buckminster Boy from the perspective of children rather than adults. Which best describes the effectiveness of this strategy in recounting the story? It is ineffective because the reader is not able to understand who is involved in the racial conflict occurring between Phippsburg and Malaga. It is ineffective because the reader is not able to understand the reasons behind the racial conflict occurring between Phippsburg and Malaga. It is effective because the reader is able to better understand how young people are experiencing the racial conflict between Phippsburg and Malaga. It is effective because the reader is able to better understand what occurs as a result of the racial conflict between Phippsburg and Malaga.
The example of deductive reasoning is A. In my community, it’s a law that all dogs out on walks must be on leashes, so most dogs I see on walks today will be on leashes.
Explanation:
Deductive reasoning goes from a broad premise to a more specific/limited conclusion. That means we first have information that reaches everyone or a large group of people/things and, from that information, we can reach a logical conclusion about one or more individuals.
Letter A is the only option that presents deductive reasoning. Our broad premise is the existence of a law that all dogs on walks must be on leashes. If that premise is true, it is only logical to conclude that, when there is a law, most people will obey it. If most people obey it, then most of the dogs I see on walks today will be on leashes.
Categories
## In “I Wandered Lonely as a Cloud,” the speaker is a human, who experiences the startling beauty of nature through the unexpected discovery of an entire sea of daffodils by the water. This poem is pensive and calm, using light, frivolous vocabulary: the daffodils are “fluttering and dancing in the breeze,” and “tossing their heads in sprightly dance.” The waves in the bay, as well, dance and sparkle, and yet the daffodils are more captivating even than the ocean, multitudinous as they are, as the stars in the sky. In Wordsworth’s poem nature is powerful and inviting, exhibiting forces of healing in the form of bright colors and gentle vibes. It is recounted from a comfortable, safe perspective; when the speaker is resting on his safe, warm couch, the memories of his solo walk along the bay …flash upon that inward eye Which is the bliss of solitude; And then my heart with pleasure fills, And dances with the daffodils. These recollections serve as a comfort and pleasure to him, even when he is comfortable in a pleasant environment. Such was the power of the scene. De la Mare’s poem also presents nature as a powerful force, but an impersonal, destructive one. The poem is told from the perspective of sea birds in a storm, and the vocabulary is a violent as Wordsworth’s is serene: “And the wind rose, and the sea rose,/To the angry billows’ roar,” and in the second verse, And the yeasty surf curdled over the sands, The gaunt grey rocks between; And the tempest raved, and the lightning’s fire Struck blue on the spindrift hoar – Here the birds have lost control, and the storm is forcing them onto the shore, waves tossing and wind howling, a wholly different scene than Wordsworth’s happy spring day. Even in the end, when the storm breaks and the sun comes out, we see the lingering effects of the chaos – “the bright green headlands shone/As they’d never shone before,” and yet within this setting we have vast hoards of sea birds breaking this lovely post-storm calm with their “screeching, scolding, [and] scrabbling.” But in the final two lines of the poem, we see also “A snowy, silent, sun-washed drift/Of sea-birds on the shore.” And herein lies the true destruction: while a whole host of birds are tumbling through the sky, another host of birds has been killed by the violence of the storm. Both poems depict the unpredictability of nature, and yet because Wordsworth’s poem is from the point of view of a man, on a bright spring day, his poem is more domestic and simple than that of de la Mare. The latter presents the point of view of nature itself, only to switch to a third person, withdrawn perspective at the end of the poem; humans have no role in the events that unfold. Any humans that exist in the area would have been safely indoors during the storm, away from any danger. We therefore get the rawness of nature where we would normally escape it for our fires and our beds; here is the flip-side of natural beauty – natural destruction. This poem is no walk in the garden, but a story of the wildness of natural processes. I NEED HELP WRITING THIS IN MY OWN WORDS PLEASE HELP
In “I Wandered Lonely as a Cloud,” the speaker is a human, who experiences the startling beauty of nature through the unexpected discovery of an entire sea of daffodils by the water. This poem is pensive and calm, using light, frivolous vocabulary: the daffodils are “fluttering and dancing in the breeze,” and “tossing their heads in sprightly dance.” The waves in the bay, as well, dance and sparkle, and yet the daffodils are more captivating even than the ocean, multitudinous as they are, as the stars in the sky.
In Wordsworth’s poem nature is powerful and inviting, exhibiting forces of healing in the form of bright colors and gentle vibes. It is recounted from a comfortable, safe perspective; when the speaker is resting on his safe, warm couch, the memories of his solo walk along the bay
…flash upon that inward eye
Which is the bliss of solitude;
And then my heart with pleasure fills,
And dances with the daffodils.
These recollections serve as a comfort and pleasure to him, even when he is comfortable in a pleasant environment. Such was the power of the scene.
De la Mare’s poem also presents nature as a powerful force, but an impersonal, destructive one. The poem is told from the perspective of sea birds in a storm, and the vocabulary is a violent as Wordsworth’s is serene: “And the wind rose, and the sea rose,/To the angry billows’ roar,” and in the second verse,
And the yeasty surf curdled over the sands,
The gaunt grey rocks between;
And the tempest raved, and the lightning’s fire
Struck blue on the spindrift hoar –
Here the birds have lost control, and the storm is forcing them onto the shore, waves tossing and wind howling, a wholly different scene than Wordsworth’s happy spring day. Even in the end, when the storm breaks and the sun comes out, we see the lingering effects of the chaos – “the bright green headlands shone/As they’d never shone before,” and yet within this setting we have vast hoards of sea birds breaking this lovely post-storm calm with their “screeching, scolding, [and] scrabbling.” But in the final two lines of the poem, we see also “A snowy, silent, sun-washed drift/Of sea-birds on the shore.” And herein lies the true destruction: while a whole host of birds are tumbling through the sky, another host of birds has been killed by the violence of the storm.
Both poems depict the unpredictability of nature, and yet because Wordsworth’s poem is from the point of view of a man, on a bright spring day, his poem is more domestic and simple than that of de la Mare. The latter presents the point of view of nature itself, only to switch to a third person, withdrawn perspective at the end of the poem; humans have no role in the events that unfold. Any humans that exist in the area would have been safely indoors during the storm, away from any danger. We therefore get the rawness of nature where we would normally escape it for our fires and our beds; here is the flip-side of natural beauty – natural destruction. This poem is no walk in the garden, but a story of the wildness of natural processes. | 5,430 | 23,484 | {"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} | 3.046875 | 3 | CC-MAIN-2021-25 | latest | en | 0.941312 |
https://support.sas.com/documentation/cdl/en/statug/66103/HTML/default/statug_phreg_details15.htm | 1,627,318,137,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046152144.81/warc/CC-MAIN-20210726152107-20210726182107-00002.warc.gz | 555,097,227 | 6,988 | ### The Frailty Model
You can use the frailty model to model correlations between failures of the same cluster. The hazard rate for the jth individual in the ith cluster is
where is an arbitrary baseline hazard rate, is the vector of (fixed-effect) covariates, is the vector of regression coefficients, and is the random effect for cluster i.
Frailties are the exponential transformations of the random components, and the frailty model can be written as
The random components (alternatively, the frailties ) are assumed to be independent and identically distributed. Modeling is based on the random effects rather than on the frailties.
Two frailty distributions are available in PROC PHREG: gamma and lognormal. Use the DIST= option in the RANDOM statement to choose the distribution. Let be an unknown parameter. The frailty distributions are listed in Table 67.11.
Table 67.11: Frailty Distributions
Frailty Option Distribution Density f( Mean and Variance DIST=GAMMA E()=1 V()= DIST=LOGNORMAL E()=0 V()=
The unknown parameter is a dispersion parameter. Each frailty distribution has a central tendency of 1 (the gamma frailty has a mean of 1, and the lognormal frailty has a median of 1). Thus, you can infer that individuals in cluster i with frailty (or ) tend to fail at a faster (or slower) rate than they fail under an independence model.
PROC PHREG estimates the regression coefficients , the random effects , and the dispersion parameter . The RANDOM statement in PROC PHREG enables you to fit a shared frailty model by a penalized partial likelihood approach. If you also specify the BAYES statement, PROC PHREG performs a Bayesian analysis of the shared frailty model.
If the RANDOM statement is specified, any ASSESS, BASELINE, and OUTPUT statements are ignored. Also ignored are the COVS options in the PROC PHREG statement and the following options in the MODEL statement: BEST=, DETAILS, HIERARCHY=, INCLUDE=, NOFIT, PLCONV=, SELECTION=, SEQUENTIAL, SLENTRY=, SLSTAY=, TYPE1, and TYPE3(ALL, LR, SCORE). Profile likelihood confidence intervals for the hazard ratios are not available for the frailty model analysis. | 487 | 2,148 | {"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} | 2.703125 | 3 | CC-MAIN-2021-31 | longest | en | 0.833946 |
https://www.doorsteptutor.com/Exams/KVPY/Stream-SA/Physics/Questions/Part-54.html | 1,529,607,705,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267864256.26/warc/CC-MAIN-20180621172638-20180621192638-00183.warc.gz | 785,183,284 | 20,647 | # KVPY (Kishore Vaigyanik Protsahan Yojana) Stream-SA (Class 11) Physics: Questions 199 - 203 of 253
Get 1 year subscription: Access detailed explanations (illustrated with images and videos) to 253 questions. Access all new questions we will add tracking exam-pattern and syllabus changes. View Sample Explanation or View Features.
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## Question number: 199
MCQ▾
### Question
The activity of a radioactive sample reduces to of its original value in 2 hours. What will be its activity in 6 hours?
### Choices
Choice (4) Response
a.
b.
c.
d.
## Question number: 200
MCQ▾
### Question
The energy required to accelerate a car from rest to 12 ms-1 is E. What energy will be required to accelerate the car from 12 ms-1 to 24 ms-1?
### Choices
Choice (4) Response
a.
E
b.
7E
c.
5E
d.
3E
## Question number: 201
MCQ▾
### Question
A train of length l (180 m) is going towards north directional constant speed v1=16 m/s. A parrot flies at a constant speed of v2 = 4 m/s towards south parallel to be railway track. The time taken by the parrot to cross the train is –
### Choices
Choice (4) Response
a.
7 s
b.
3 s
c.
9 s
d.
5 s
## Question number: 202
MCQ▾
### Question
The isotopes 238U and 235U occur in nature in the ratio 150: 1. Assuming that at the time of earth formation they were present in equal amount; make an estimation of the age of the earth. The half-lives of 238U and 235U are and respectively.
### Choices
Choice (4) Response
a.
b.
c.
d.
## Question number: 203
MCQ▾
### Question
The height of a waterfall is 75 m, If the water at the bottom of the fall is at rest. Calculate the difference between temperature of water at the top and bottom of the fall (specific heat of water and ).
### Choices
Choice (4) Response
a.
0.175 0C
b.
0.97 0C
c.
0.126 0C
d.
0.117 0C
f Page | 539 | 1,855 | {"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} | 3.0625 | 3 | CC-MAIN-2018-26 | latest | en | 0.790228 |
https://brainmass.com/math/basic-algebra/absolute-value-distributive-property-92309 | 1,721,457,752,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763515020.57/warc/CC-MAIN-20240720052626-20240720082626-00345.warc.gz | 130,532,817 | 6,681 | Purchase Solution
# Absolute Value and Distributive Property
Not what you're looking for?
This is a True or False problem and I need to explain the answer.
1. If we add the absolute values of -3 and -5, we get 8.
using the distributive property to remove the parentheses.
2. -3(6-p)
Can you help me with this problems and explain them.
Thank you, your help is GREATLY appeciated.
##### Solution Summary
Absolute Value and Distributive Property are investigated. The solution is detailed and well presented. The response received a rating of "5/5" from the student who originally posted the question.
##### Solution Preview
1. True.
First, we should understand the meaning of absolute value. We use |x| to represent the absolute value of x. The definition is: |x|=x if x>=0; |x|=-x if x<0. For example, ...
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This quiz helps you easily identify a function and test your understanding of ranges, domains , function inverses and transformations.
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Some questions on probability
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# Garrett Knapp
## Mat 275 Lab
W 1:30
MAT 275 MATLAB Assingment #1
%Problem 1
theta=linspace(0,5*pi/4,6); %Creates row vector for theta
x=2*cos(theta); %Creates x-component for circle
y=2*sin(theta); %Creates y-compononent for circle
r=sqrt(x.^2+y.^2) %Confirms x&y create circle, r=2 for all cases
%Problem 2
function ex2
%This function solves the function in excercise 2
e=2.718281828;
t=linspace(1,10,91)
output=((e^(0.1.*t))*sin(t'))/(t.^2+1)
end
%Problem 3
t=linspace(0,20,250); %Creating x,y, and z values as defined by problem
statement
x=sin(t);
y=cos(t);
z=t;
plot3(x,y,z) %Plotting all variables in three dimensions
%Problem 4
b=linspace(0,pi,2000);
y=cos(b);
z=1-(b.^2)/2+(b.^4)/24;
plot(b,y,'r',b,z,'--')
grid on
%Problem 5
function ex5
x=linspace(0,4,200);
y1=x.^2/2+2*x-1;
y2=x.^2/2+2*x;
y3=x.^2/2+2*x+1;
plot(x,y1,'-',x,y2,':',x,y3,'--')
title('Solutions to dy/dx=x+2')
legend('C=-1','C=0','C=1')
end
function y=f(x,C)
y=(x.^2)/2+2x+C
end | 408 | 981 | {"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} | 3.046875 | 3 | CC-MAIN-2020-10 | latest | en | 0.404444 |
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# I don't understand my Kaplan score, can somebody help me?
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I don't understand my Kaplan score, can somebody help me? [#permalink]
### Show Tags
26 Feb 2013, 17:14
Hello!
After 3 months of preparation, I just took my 1st Kaplan CAT and I am pleasantly surprise by the score: 690, with 92 percentile for Quant and 53 for Verbal. However, I have 2 questions that I am unable to solve:
1- I failed to find the "real scores" of each part (the ones scaled on 51) in the Kaplan report... Can anyone tell me how to find them please?
2- I don't understand how can I score 690 with a 53 percentile for Verbal... Does anyone have an explanation for this?
Thanks everyone.
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Joined: 05 Apr 2011
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Re: I don't understand my Kaplan score, can somebody help me? [#permalink]
### Show Tags
06 Mar 2013, 01:15
roughtell wrote:
Hello!
After 3 months of preparation, I just took my 1st Kaplan CAT and I am pleasantly surprise by the score: 690, with 92 percentile for Quant and 53 for Verbal. However, I have 2 questions that I am unable to solve:
1- I failed to find the "real scores" of each part (the ones scaled on 51) in the Kaplan report... Can anyone tell me how to find them please?
2- I don't understand how can I score 690 with a 53 percentile for Verbal... Does anyone have an explanation for this?
Thanks everyone.
Am not sure about the first point as i have never taken Kaplan test.
Regarding the second point:
Yes you can get a 690 with 53 percentile in Verbal
My GMAT score was 700 with a 31 in Verbal which was 58 percentile!
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Re: I don't understand my Kaplan score, can somebody help me? [#permalink]
### Show Tags
06 Mar 2013, 07:40
roughtell wrote:
Hello!
After 3 months of preparation, I just took my 1st Kaplan CAT and I am pleasantly surprise by the score: 690, with 92 percentile for Quant and 53 for Verbal. However, I have 2 questions that I am unable to solve:
1- I failed to find the "real scores" of each part (the ones scaled on 51) in the Kaplan report... Can anyone tell me how to find them please?
2- I don't understand how can I score 690 with a 53 percentile for Verbal... Does anyone have an explanation for this?
Thanks everyone.
Dont rely on Kaplan scores. Use them for practice. Recently few folks are seeing their scores inflated. Kaplan tests are tough, so you get more questions wrong, but they adjust your score accordingly. What are your GMATPREP/MGMAT scores?
_________________
Thanks,
Prashant Ponde
Tough 700+ Level RCs: Passage1 | Passage2 | Passage3 | Passage4 | Passage5 | Passage6 | Passage7
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Kudos [?]: 930 [0], given: 184
Re: I don't understand my Kaplan score, can somebody help me? [#permalink] 06 Mar 2013, 07:40
Display posts from previous: Sort by | 1,125 | 4,269 | {"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} | 2.984375 | 3 | CC-MAIN-2017-43 | latest | en | 0.923925 |
https://www.excelforum.com/excel-formulas-and-functions/354057-formula-for-calculating-interest-payments.html | 1,642,456,526,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320300624.10/warc/CC-MAIN-20220117212242-20220118002242-00578.warc.gz | 828,998,259 | 13,993 | # Formula for calculating interest payments?
1. ## Formula for calculating interest payments?
Hello
My first post to this news group.
formula I should use to calculate the monthly/fortnightly payments which
would need to be made on an interest only loan.
I have used PMT and IPMT but my results don't quite correspond with other
online calculators. I really would like to be able to develop my own
calculations in Excel and be confident the resultrs are reasonably accurate.
Jack
2. ## Re: Formula for calculating interest payments?
On Mon, 14 Mar 2005 14:17:31 GMT, "Jack Shearer" <aylict@msn.com> wrote:
>Hello
>
>My first post to this news group.
>
>formula I should use to calculate the monthly/fortnightly payments which
>would need to be made on an interest only loan.
>
>I have used PMT and IPMT but my results don't quite correspond with other
>online calculators. I really would like to be able to develop my own
>calculations in Excel and be confident the resultrs are reasonably accurate.
>
>
>Jack
>
if it is an interest-only loan, then the payments would be given by the
formula:
=Interest_rate_per_period * Principal
So if your interest is quoted as yearly but your payments are monthly, you
would divide that number by 12.
--ron
3. ## Re: Formula for calculating interest payments?
Interest only means you are not paying down any principal. You are paying
only the interest charged in a period. Therefore you can't use the financial
functions, because they all assume you will have the loan paid off by the
end of the term.
In your case, to calculate the monthly interest charge, use:
=AnnualRate * Principal / 12
To calculate the fortnightly charge:
=AnnualRate * Principal / 26
--
Regards,
Fred
"Jack Shearer" <aylict@msn.com> wrote in message
news:%1hZd.197651\$K7.130392@news-server.bigpond.net.au...
> Hello
>
> My first post to this news group.
>
> formula I should use to calculate the monthly/fortnightly payments which
> would need to be made on an interest only loan.
>
> I have used PMT and IPMT but my results don't quite correspond with other
> online calculators. I really would like to be able to develop my own
> calculations in Excel and be confident the resultrs are reasonably
> accurate.
>
>
> Jack
>
>
4. ## Re: Formula for calculating interest payments?
Thanks guys.
Of course this is the method. Benn quite a while since I worked with
figures and was looking for a complicated process. Well that's my excuse
and I'm sticking to it.
Jack
"Jack Shearer" <aylict@msn.com> wrote in message
news:%1hZd.197651\$K7.130392@news-server.bigpond.net.au...
> Hello
>
> My first post to this news group.
>
> formula I should use to calculate the monthly/fortnightly payments which
> would need to be made on an interest only loan.
>
> I have used PMT and IPMT but my results don't quite correspond with other
> online calculators. I really would like to be able to develop my own
> calculations in Excel and be confident the resultrs are reasonably
accurate.
>
>
> Jack
>
>
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Search Engine Friendly URLs by vBSEO 3.6.0 RC 1 | 815 | 3,296 | {"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} | 3.25 | 3 | CC-MAIN-2022-05 | latest | en | 0.946284 |
https://studydaddy.com/question/he-latin-club-charted-a-bus-to-travel-to-and-participate-in-a-state-competition | 1,550,357,044,000,000,000 | text/html | crawl-data/CC-MAIN-2019-09/segments/1550247481122.31/warc/CC-MAIN-20190216210606-20190216232606-00174.warc.gz | 686,020,033 | 10,087 | Answered You can buy a ready-made answer or pick a professional tutor to order an original one.
QUESTION
# he Latin Club charted a bus to travel to and participate in a state competition. The cost for each adult to ride the bus was $25 and the cost for each student to ride the bus was$15.A total of 17 pe
he Latin Club charted a bus to travel to and participate in a state competition. The cost for each adult to ride the bus was $25 and the cost for each student to ride the bus was$15.
A total of 17 people rode the bus. The total paid for the bus was $305. Let a represent the number of adults and s represent the number of students. • @ ANSWER Tutor has posted answer for$15.00. See answer's preview
\$15.00
**** ** equation *** ******** ****** *** = 25and *** each ******* **** ***** then equation ** **** 15sfor ***** ****** ***** ******
or | 212 | 856 | {"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} | 3.09375 | 3 | CC-MAIN-2019-09 | longest | en | 0.931804 |
http://www.grundfos.com/service-support/encyclopedia-search/calculating-heatloss.html | 1,511,489,411,000,000,000 | text/html | crawl-data/CC-MAIN-2017-47/segments/1510934807056.67/warc/CC-MAIN-20171124012912-20171124032912-00608.warc.gz | 401,222,516 | 11,660 | # Calculating heat loss
The calculated transmissions loss is the sum of the heat losses from each individual section of the building
The calculated transmissions loss is the sum of the heat losses from each individual section of the building.
These can be calculated using the following formula:
Φ =U x A x (tout- tin)
where
Φ= Heat loss for the building section [W]
U=Coefficient of heat transmission for the building section [W/m²K]
A= Area of the building section [m²]
tout= Outdoor temperature at design conditions [°C]
tin = Indoor temperature at design conditions[°C]
The calculated heat loss is used to size the heating system of the building according to design conditions for the temperatures outside and inside of the building. Approximately 10-20% is typically added to the calculated heat loss, in order to ensure extra capacity following night set-back or similar situations. | 189 | 894 | {"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} | 2.84375 | 3 | CC-MAIN-2017-47 | longest | en | 0.918005 |
http://kingyojima.net/pje/015.html | 1,723,347,137,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640843545.63/warc/CC-MAIN-20240811013030-20240811043030-00100.warc.gz | 14,861,516 | 2,732 | # Project Euler 解答
## Project Euler Problem 015
Project Euler Problem 015
```ans1 :: Integer
ans1 = nckp 40 20
ans2 :: Integer
ans2 = ncki 40 20
-- 137846528820
main :: IO ()
main = print ans1
nckp :: Integral a => a -> a -> a
nckp n k = product [n,n-1..(n-k+1)] `quot` product [1..k]
ncki :: (Enum a, Num a) => Int -> Int -> a
ncki n k = (iterate(scanl1(+))[1,1..]) !! (n-k) !! k
-- ナイーブに実装するならこんな感じ。でも遅くて本問は解けない
nck :: Integral a => a -> a -> a
nck _ 0 = 1
nck n 1 = n
nck n k | n == k = 1
nck n k | n < k = 0
nck n k | k > (n`quot`2) = nck n (n - k)
nck n k = nck (n-1) k + nck (n-1) (k-1)
```
• はじめに
• プロジェクトオイラー問題
• リンク等 | 298 | 630 | {"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-2024-33 | latest | en | 0.350554 |
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