url stringlengths 6 1.61k | fetch_time int64 1,368,856,904B 1,726,893,854B | content_mime_type stringclasses 3
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http://clay6.com/qa/25620/to-what-series-does-the-spectral-lines-of-atomic-hydrogen-belong-if-its-wav | 1,524,658,839,000,000,000 | text/html | crawl-data/CC-MAIN-2018-17/segments/1524125947803.66/warc/CC-MAIN-20180425115743-20180425135743-00133.warc.gz | 64,425,582 | 26,744 | # To what series does the spectral lines of atomic hydrogen belong if its wave number is equal to the difference between the wave numbers of the following two lines of the Balmer series 486.1 and 410.2 nm?
$(a)\;2^{nd} to 3^{rd} level\qquad(b)\;4^{th} to 2^{nd} level\qquad(c)\;5^{th} to 1^{st} level\qquad(d)\;6^{th} to 4^{th} level$
Given
$\lambda_1 = 486.1\times10^{-9}$m = 486.1$\times10^{-7}$ cm
$\lambda_2 = 410.2\times10^{-9}$m = 410.2$\times10^{-7}$ cm
$\overline V = \overline V_2 - \overline V_1$ = $\large\frac{1}{\lambda_2}$ - $\large\frac{1}{\lambda_1}$
$=R_H[\large\frac{1}{2^2}$ -$\large\frac{1}{n_2^2}-R_H[\large\frac{1}{2^2} -\large\frac{1}{n_1^2}]$
$\overline V = R_H[\large\frac{1}{n_1^2} -\large\frac{1}{n_2^2}]$
For I case of Balmer Series :
$\large\frac{1}{\lambda_1}= R_H[\large\frac{1}{2^2} -\large\frac{1}{n_1^2}]$
= 109678 [$\large\frac{1}{2^2} - \large\frac{1}{n_1^2} ]$
Or [$\large\frac{1}{486.1\times10^{-7}}= 109678 [\large\frac{1}{2^2}-\large\frac{1}{n_1^2}]$
For II case of Balmer series:
$\large\frac{1}{\lambda_2}= R_H[\large\frac{1}{2^2} -\large\frac{1}{n_2^2}]$
= 109678 [$\large\frac{1}{2^2} - \large\frac{1}{n_2^2} ]$
Or [$\large\frac{1}{410.2\times10^{-7}}= 109678 [\large\frac{1}{2^2}-\large\frac{1}{n_2^2}]$
$\therefore n_2$ = 6
$\therefore$ given transition occurs from $6^{th}$ level to $4^{th}$ level | 598 | 1,346 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.53125 | 4 | CC-MAIN-2018-17 | latest | en | 0.466998 |
http://www.briddles.com/2014/01/famous-probability-problem.html | 1,503,229,213,000,000,000 | text/html | crawl-data/CC-MAIN-2017-34/segments/1502886106465.71/warc/CC-MAIN-20170820112115-20170820132115-00577.warc.gz | 484,942,012 | 19,942 | ## Search This Blog
### Famous Probability Problem
Famous Probability Problem - 23 January
You are offered an opportunity of winning a fortune. There are 100 precious black stones and 100 unworthy white stones. There are two different sacks labelled as 'Heads' and 'Tails'. You can distribute the stones as per you wish. Then a coin will be flipped. Then you will have to choose a stone from the corresponding sack. If you pick up a black stone, all the fortune of black stones is yours but if you pick up white, you get nothing.
How will you distribute the stones so that you can maximize your chances of winning?
For Solution : Click Here
#### 3 comments:
1. One black stone in one of the sacks, and the rest 199 in the other sack.
2. don't put any white stones in any of the bags?
1. Best Answer! The puzzle should specify that all stones must be placed into one of the sacks, but it does not specify this. Therefore, your answer gives a 100% probability of choosing a black stone. You win. | 227 | 1,002 | {"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} | 3.484375 | 3 | CC-MAIN-2017-34 | longest | en | 0.891505 |
https://www.convertunits.com/from/reed+%5BIsrael%5D/to/fall | 1,653,360,196,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662562410.53/warc/CC-MAIN-20220524014636-20220524044636-00783.warc.gz | 791,890,552 | 17,164 | ## ››Convert reed [Israel] to fall [English]
reed [Israel] fall
Did you mean to convert reed [Israel] to fall [English] fall [Scotland]
How many reed [Israel] in 1 fall? The answer is 2.5599104143337.
We assume you are converting between reed [Israel] and fall [English].
You can view more details on each measurement unit:
reed [Israel] or fall
The SI base unit for length is the metre.
1 metre is equal to 0.3732736095558 reed [Israel], or 0.14581510644503 fall.
Note that rounding errors may occur, so always check the results.
Use this page to learn how to convert between reed [Israel] and falls.
Type in your own numbers in the form to convert the units!
## ››Quick conversion chart of reed [Israel] to fall
1 reed [Israel] to fall = 0.39064 fall
5 reed [Israel] to fall = 1.95319 fall
10 reed [Israel] to fall = 3.90639 fall
20 reed [Israel] to fall = 7.81277 fall
30 reed [Israel] to fall = 11.71916 fall
40 reed [Israel] to fall = 15.62555 fall
50 reed [Israel] to fall = 19.53193 fall
75 reed [Israel] to fall = 29.2979 fall
100 reed [Israel] to fall = 39.06387 fall
## ››Want other units?
You can do the reverse unit conversion from fall to reed [Israel], or enter any two units below:
## Enter two units to convert
From: To:
## ››Metric conversions and more
ConvertUnits.com provides an online conversion calculator for all types of measurement units. You can find metric conversion tables for SI units, as well as English units, currency, and other data. Type in unit symbols, abbreviations, or full names for units of length, area, mass, pressure, and other types. Examples include mm, inch, 100 kg, US fluid ounce, 6'3", 10 stone 4, cubic cm, metres squared, grams, moles, feet per second, and many more! | 494 | 1,741 | {"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 | 0.789837 |
https://ask.sagemath.org/answers/12284/revisions/ | 1,597,039,599,000,000,000 | text/html | crawl-data/CC-MAIN-2020-34/segments/1596439738609.73/warc/CC-MAIN-20200810042140-20200810072140-00443.warc.gz | 208,754,837 | 6,747 | # Revision history [back]
The original scenario involves two variables and only one equation. That is what the "shape mismatch" error is saying. ff needs to return a list. Here is a working example (with two equations):
import scipy.optimize
var('x,y')
f = x^2 + y^2 - 1
g = x - y^2
def ff(v):
x, y = map(float, v)
return [f(x=x, y=y), g(x=x, y=y)]
print scipy.optimize.fsolve(ff, [1, 1])
The code could be slightly more readable if f and g are defined with explicit arguments:
import scipy.optimize
var('x,y')
f(x,y) = x^2 + y^2 - 1
g(x,y) = x - y^2
def ff(v):
x, y = map(float, v)
return [f(x, y), g(x, y)]
print scipy.optimize.fsolve(ff, [1, 1])
In both cases, the result should look like: [ 0.61803399 0.78615138]
The original scenario involves two variables and only one equation. That is what Because ff accepts two variables but does not return a list (or tuple or array, ...) of length 2, the "shape mismatch" error is saying. ff needs to return a list. thrown. Here is a working example (with two equations):
import scipy.optimize
var('x,y')
f = x^2 + y^2 - 1
g = x - y^2
def ff(v):
x, y = map(float, v)
return [f(x=x, y=y), g(x=x, y=y)]
print scipy.optimize.fsolve(ff, [1, 1])
The code could be slightly more readable if f and g are defined with explicit arguments:
import scipy.optimize
var('x,y')
f(x,y) = x^2 + y^2 - 1
g(x,y) = x - y^2
def ff(v):
x, y = map(float, v)
return [f(x, y), g(x, y)]
print scipy.optimize.fsolve(ff, [1, 1])
In both cases, the result should look like: [ 0.61803399 0.78615138]
The original scenario involves two variables and only one equation. Because ff accepts two variables but does not return a list (or tuple tuple, or array, ...) of length 2, the "shape mismatch" error is thrown. Here is a working example (with two equations):
import scipy.optimize
var('x,y')
f = x^2 + y^2 - 1
g = x - y^2
def ff(v):
x, y = map(float, v)
return [f(x=x, y=y), g(x=x, y=y)]
print scipy.optimize.fsolve(ff, [1, 1])
The code could be slightly more readable if f and g are defined with explicit arguments:
import scipy.optimize
var('x,y')
f(x,y) = x^2 + y^2 - 1
g(x,y) = x - y^2
def ff(v):
x, y = map(float, v)
return [f(x, y), g(x, y)]
print scipy.optimize.fsolve(ff, [1, 1])
In both cases, the result should look like: [ 0.61803399 0.78615138]
The original scenario involves two variables and only one equation. Because ff accepts two variables but does not return a list (or tuple, or array, ...) of length 2, the "shape mismatch" error is thrown. Here is a working example (with two equations):equations and minimal modifications):
import scipy.optimize
var('x,y')
f = x^2 + y^2 - 1
g = x - y^2
def ff(v):
x, y = map(float, v)
return [f(x=x, y=y), g(x=x, y=y)]
print scipy.optimize.fsolve(ff, [1, 1])
The code could be Here is a more compact version using lambda functions and the trick described link:here
import scipy.optimize
var('x,y')
f(x,y) = (x^2 + y^2 - 1, x - y^2)
print scipy.optimize.fsolve(lambda v: f(*map(float, v)), (1, 1))
Since you said you have a large number of variables and equations, here is a slightly more readable if f and g are defined with explicit arguments:"scale-friendly" approach (but somewhat kludgy):
import scipy.optimize
var('x,y')
f(x,y) = x^2 # Create symbolic variables "x_0" and "x_1".
# This is a pretty messy kludge
x = tuple(
SR.var(
"".join(['x_', str(t)])
)
for t in range(2)
)
# Represent system of equations as a vector over callable symbolic ring.
# I use the intermediate "exprs" variable to allow breaking up the tuple
# over multiple lines.
exprs = (
x[0]^2 + y^2 - 1
g(x,y) = x - y^2
def ff(v):
x, y = map(float, v)
return [f(x, y), g(x, y)]
x[1]^2 - 1,
x[0] - x[1]^2
)
f(*x) = exprs
# Guess values
guess = (1, 1)
print scipy.optimize.fsolve(ff, [1, 1])
scipy.optimize.fsolve(lambda v: f(*map(float, v)), guess)
In both all cases, the result should look like: [ 0.61803399 0.78615138]
The original scenario involves two variables and only one equation. Because ff accepts two variables but does not return a list (or tuple, or array, ...) of length 2, the "shape mismatch" error is thrown. Here is a working example (with two equations and minimal modifications):
import scipy.optimize
var('x,y')
f = x^2 + y^2 - 1
g = x - y^2
def ff(v):
x, y = map(float, v)
return [f(x=x, y=y), g(x=x, y=y)]
print scipy.optimize.fsolve(ff, [1, 1])
Here is a more compact version using lambda functions and the trick described link:here
import scipy.optimize
var('x,y')
f(x,y) = (x^2 + y^2 - 1, x - y^2)
print scipy.optimize.fsolve(lambda v: f(*map(float, v)), (1, 1))
Since you said you have a large number of variables and equations, here is a slightly more "scale-friendly" approach (but somewhat kludgy):
import scipy.optimize
# Create symbolic variables "x_0" and "x_1".
# This is a pretty messy kludge
x = tuple(
SR.var(
"".join(['x_', str(t)])
)
for t in range(2)
)
# Represent system of equations using "x" as a vector over callable symbolic ring.
defined above.
# I use the intermediate "exprs" variable to allow breaking up the tuple
# over multiple lines.
exprs = (
x[0]^2 + x[1]^2 - 1,
x[0] - x[1]^2
)
f(*x) = exprs
# Guess values
guess = (1, 1)
print scipy.optimize.fsolve(lambda v: f(*map(float, v)), guess)
In all cases, the result should look like: [ 0.61803399 0.78615138] | 1,638 | 5,336 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.140625 | 3 | CC-MAIN-2020-34 | latest | en | 0.776044 |
https://www.symbolab.com/study-guides/collegealgebra1/section-exercises-36.html | 1,721,586,529,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763517768.40/warc/CC-MAIN-20240721182625-20240721212625-00353.warc.gz | 872,815,712 | 46,566 | We've updated our
1. How does the power rule for logarithms help when solving logarithms with the form ${\mathrm{log}}_{b}\left(\sqrt[n]{x}\right)$? 2. What does the change-of-base formula do? Why is it useful when using a calculator? For the following exercises, expand each logarithm as much as possible. Rewrite each expression as a sum, difference, or product of logs. 3. ${\mathrm{log}}_{b}\left(7x\cdot 2y\right)$ 4. $\mathrm{ln}\left(3ab\cdot 5c\right)$ 5. ${\mathrm{log}}_{b}\left(\frac{13}{17}\right)$ 6. ${\mathrm{log}}_{4}\left(\frac{\text{ }\frac{x}{z}\text{ }}{w}\right)$ 7. $\mathrm{ln}\left(\frac{1}{{4}^{k}}\right)$ 8. ${\mathrm{log}}_{2}\left({y}^{x}\right)$ For the following exercises, condense to a single logarithm if possible. 9. $\mathrm{ln}\left(7\right)+\mathrm{ln}\left(x\right)+\mathrm{ln}\left(y\right)$ 10. ${\mathrm{log}}_{3}\left(2\right)+{\mathrm{log}}_{3}\left(a\right)+{\mathrm{log}}_{3}\left(11\right)+{\mathrm{log}}_{3}\left(b\right)$ 11. ${\mathrm{log}}_{b}\left(28\right)-{\mathrm{log}}_{b}\left(7\right)$ 12. $\mathrm{ln}\left(a\right)-\mathrm{ln}\left(d\right)-\mathrm{ln}\left(c\right)$ 13. $-{\mathrm{log}}_{b}\left(\frac{1}{7}\right)$ 14. $\frac{1}{3}\mathrm{ln}\left(8\right)$ For the following exercises, use the properties of logarithms to expand each logarithm as much as possible. Rewrite each expression as a sum, difference, or product of logs. 15. $\mathrm{log}\left(\frac{{x}^{15}{y}^{13}}{{z}^{19}}\right)$ 16. $\mathrm{ln}\left(\frac{{a}^{-2}}{{b}^{-4}{c}^{5}}\right)$ 17. $\mathrm{log}\left(\sqrt{{x}^{3}{y}^{-4}}\right)$ 18. $\mathrm{ln}\left(y\sqrt{\frac{y}{1-y}}\right)$ 19. $\mathrm{log}\left({x}^{2}{y}^{3}\sqrt[3]{{x}^{2}{y}^{5}}\right)$ For the following exercises, condense each expression to a single logarithm using the properties of logarithms. 20. $\mathrm{log}\left(2{x}^{4}\right)+\mathrm{log}\left(3{x}^{5}\right)$ 21. $\mathrm{ln}\left(6{x}^{9}\right)-\mathrm{ln}\left(3{x}^{2}\right)$ 22. $2\mathrm{log}\left(x\right)+3\mathrm{log}\left(x+1\right)$ 23. $\mathrm{log}\left(x\right)-\frac{1}{2}\mathrm{log}\left(y\right)+3\mathrm{log}\left(z\right)$ 24. $4{\mathrm{log}}_{7}\left(c\right)+\frac{{\mathrm{log}}_{7}\left(a\right)}{3}+\frac{{\mathrm{log}}_{7}\left(b\right)}{3}$ For the following exercises, rewrite each expression as an equivalent ratio of logs using the indicated base. 25. ${\mathrm{log}}_{7}\left(15\right)$ to base e 26. ${\mathrm{log}}_{14}\left(55.875\right)$ to base 10 For the following exercises, suppose ${\mathrm{log}}_{5}\left(6\right)=a$ and ${\mathrm{log}}_{5}\left(11\right)=b$. Use the change-of-base formula along with properties of logarithms to rewrite each expression in terms of a and b. Show the steps for solving. 27. ${\mathrm{log}}_{11}\left(5\right)$ 28. ${\mathrm{log}}_{6}\left(55\right)$ 29. ${\mathrm{log}}_{11}\left(\frac{6}{11}\right)$ For the following exercises, use properties of logarithms to evaluate without using a calculator. 30. ${\mathrm{log}}_{3}\left(\frac{1}{9}\right)-3{\mathrm{log}}_{3}\left(3\right)$ 31. $6{\mathrm{log}}_{8}\left(2\right)+\frac{{\mathrm{log}}_{8}\left(64\right)}{3{\mathrm{log}}_{8}\left(4\right)}$ 32. $2{\mathrm{log}}_{9}\left(3\right)-4{\mathrm{log}}_{9}\left(3\right)+{\mathrm{log}}_{9}\left(\frac{1}{729}\right)$ For the following exercises, use the change-of-base formula to evaluate each expression as a quotient of natural logs. Use a calculator to approximate each to five decimal places. 33. ${\mathrm{log}}_{3}\left(22\right)$ 34. ${\mathrm{log}}_{8}\left(65\right)$ 35. ${\mathrm{log}}_{6}\left(5.38\right)$ 36. ${\mathrm{log}}_{4}\left(\frac{15}{2}\right)$ 37. ${\mathrm{log}}_{\frac{1}{2}}\left(4.7\right)$ 38. Use the product rule for logarithms to find all x values such that ${\mathrm{log}}_{12}\left(2x+6\right)+{\mathrm{log}}_{12}\left(x+2\right)=2$. Show the steps for solving. 39. Use the quotient rule for logarithms to find all x values such that ${\mathrm{log}}_{6}\left(x+2\right)-{\mathrm{log}}_{6}\left(x - 3\right)=1$. Show the steps for solving. 40. Can the power property of logarithms be derived from the power property of exponents using the equation ${b}^{x}=m?$ If not, explain why. If so, show the derivation. 41. Prove that ${\mathrm{log}}_{b}\left(n\right)=\frac{1}{{\mathrm{log}}_{n}\left(b\right)}$ for any positive integers > 1 and > 1. 42. Does ${\mathrm{log}}_{81}\left(2401\right)={\mathrm{log}}_{3}\left(7\right)$? Verify the claim algebraically. | 1,603 | 4,441 | {"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-2024-30 | latest | en | 0.514287 |
https://windrawwin-prediction.com/betting/is-rolling-two-dice-a-normal-distribution.html | 1,652,740,227,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662512249.16/warc/CC-MAIN-20220516204516-20220516234516-00497.warc.gz | 686,825,399 | 17,989 | Is rolling two dice a normal distribution?
Contents
What is the distribution of rolling 2 dice?
Since there are six possible outcomes, the probability of obtaining any side of the die is 1/6. The probability of rolling a 1 is 1/6, the probability of rolling a 2 is 1/6, and so on.
Is rolling a die a probability distribution?
Probability Distribution
Take rolling a die, for example. We can let the random variable D represent the number showing on the die when rolling the die. Then, D equals either 1, 2, 3, 4, 5, or 6. A function that puts together a probability with its outcome in an experiment is known as a probability distribution.
Is dice roll a uniform distribution?
For instance, while any one roll of a dice has a uniform distribution, summing up the totals of rolling a dice lots of time, or taking their average, does not have a uniform distribution, but approximates a Gaussian distribution, which we will discuss later.
Is rolling 2 dice uniform probability?
The sum of two dice rolls will not have uniform distribution.
Is rolling two dice independent or dependent?
Sample Problem
If we roll two dice, the event of rolling 5 on the first die and the event of the numbers on the two dice summing to 8 are dependent.
IT IS SURPRISING: What is a soft total blackjack?
What are the outcomes of rolling 2 dice?
Note that there are 36 possibilities for (a,b). This total number of possibilities can be obtained from the multiplication principle: there are 6 possibilities for a, and for each outcome for a, there are 6 possibilities for b. So, the total number of joint outcomes (a,b) is 6 times 6 which is 36.
When you roll a die what type of distribution would you expect to see in the output of values?
Understanding Uniform Distribution
Therefore, the roll of a die generates a discrete distribution with p = 1/6 for each outcome. There are only 6 possible values to return and nothing in between.
What is normal probability distribution?
Normal distribution, also known as the Gaussian distribution, is a probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean. In graph form, normal distribution will appear as a bell curve. | 493 | 2,258 | {"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} | 4.65625 | 5 | CC-MAIN-2022-21 | latest | en | 0.89888 |
https://fr.mathworks.com/matlabcentral/profile/authors/7931408 | 1,719,071,776,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198862404.32/warc/CC-MAIN-20240622144011-20240622174011-00568.warc.gz | 228,594,447 | 21,818 | # Mallouli Marwa
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Réponse apportée
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réponse | 1,954 | 6,336 | {"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} | 2.90625 | 3 | CC-MAIN-2024-26 | latest | en | 0.584687 |
https://www.geeksforgeeks.org/count-unordered-pairs-ij-such-that-product-of-ai-and-aj-is-power-of-two/ | 1,576,272,017,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575540569146.17/warc/CC-MAIN-20191213202639-20191213230639-00377.warc.gz | 724,768,434 | 29,780 | Count unordered pairs (i,j) such that product of a[i] and a[j] is power of two
Given an array of N elements. The task is to count unordered pairs (i, j) in the array such that the product of a[i] and a[j] can be expressed as a power of two.
Examples:
```Input : arr[] = {2, 3, 4, 8, 10}
Output : 3
Explanation: The pair of array element will be
(2, 4), (2, 8), (4, 8) whose product are
8, 16, 32 respectively which can be expressed
as power of 2, like 2^3, 2^4, 2^5.
Input : arr[] = { 2, 5, 8, 16, 128 }
Output : 6
```
Recommended: Please try your approach on {IDE} first, before moving on to the solution.
If you multiply and and their product become , then z=x*y, now if it’s possible to express as power of two then it can be proved that both and can be expressed as power of two. Basically z= 2a = 2(b+c) = 2b * 2c = x * y, where and both
can hold a minimum value 0.
So now we have to count the number of elements in the array which can be expressed as a power of two. If the count is k, then answer will be kC2 = k*(k-1)/2, as we need the count of unordered pairs.
Below is the implementation of above approach:
C++
`// C++ program to Count unordered pairs (i, j) ` `// in array such that product of a[i] and a[j] ` `// can be expressed as power of two ` `#include ` `using` `namespace` `std; ` ` ` `/* Function to check if x is power of 2*/` `bool` `isPowerOfTwo(``int` `x) ` `{ ` ` ``/* First x in the below expression is ` ` ``for the case when x is 0 */` ` ``return` `x && (!(x&(x-1))); ` `} ` ` ` `// Function to Count unordered pairs ` `void` `Count_pairs(``int` `a[], ``int` `n) ` `{ ` ` ``int` `count = 0; ` ` ` ` ``for` `(``int` `i = 0; i < n; i++) { ` ` ` ` ``// is a number can be expressed ` ` ``// as power of two ` ` ``if` `(isPowerOfTwo(a[i])) ` ` ``count++; ` ` ``} ` ` ` ` ``// count total number ` ` ``// of unordered pairs ` ` ``int` `ans = (count * (count - 1)) / 2; ` ` ` ` ``cout << ans << ``"\n"``; ` `} ` ` ` `// Driver code ` `int` `main() ` `{ ` ` ``int` `a[] = { 2, 5, 8, 16, 128 }; ` ` ` ` ``int` `n = ``sizeof``(a) / ``sizeof``(a[0]); ` ` ` ` ``Count_pairs(a, n); ` ` ` ` ``return` `0; ` `} `
Java
`// Java program to Count unordered pairs (i, j) ` `// in array such that product of a[i] and a[j] ` `// can be expressed as power of two ` ` ` `import` `java.io.*; ` ` ` `class` `GFG { ` ` ` ` ` `/* Function to check if x is power of 2*/` `static` `boolean` `isPowerOfTwo(``int` `x) ` `{ ` `/* First x in the below expression is ` ` ``for the case when x is 0 */` `return` `(x >``0``&& (!((x&(x-``1``))>``0``))); ` `} ` ` ` `// Function to Count unordered pairs ` `static` `void` `Count_pairs(``int` `a[], ``int` `n) ` `{ ` ` ``int` `count = ``0``; ` ` ` ` ``for` `(``int` `i = ``0``; i < n; i++) { ` ` ` ` ``// is a number can be expressed ` ` ``// as power of two ` ` ``if` `(isPowerOfTwo(a[i])) ` ` ``count++; ` ` ``} ` ` ` ` ``// count total number ` ` ``// of unordered pairs ` ` ``int` `ans = (count * (count - ``1``)) / ``2``; ` ` ` ` ``System.out.println( ans); ` `} ` ` ` `// Driver code ` ` ` ` ``public` `static` `void` `main (String[] args) { ` ` ``int` `a[] = { ``2``, ``5``, ``8``, ``16``, ``128` `}; ` ` ` ` ``int` `n = a.length; ` ` ``Count_pairs(a, n); ` ` ` ` ``} ` `} ` ` ` `// This code is contributed ` `// by shs `
Python 3
`# Python3 program to Count unordered pairs ` `# (i, j) in array such that product of a[i] ` `# and a[j] can be expressed as power of two ` ` ` `# Function to check if x is power of 2 ` `def` `isPowerOfTwo(x) : ` ` ` ` ``# First x in the below expression ` ` ``# is for the case when x is 0 ` ` ``return` `(x ``and``(``not``(x & (x ``-` `1``)))) ` ` ` `# Function to Count unordered pairs ` `def` `Count_pairs(a, n) : ` ` ` ` ``count ``=` `0` ` ` ` ``for` `i ``in` `range``(n) : ` ` ` ` ``# is a number can be expressed ` ` ``# as power of two ` ` ``if` `isPowerOfTwo(a[i]) : ` ` ``count ``+``=` `1` ` ` ` ``# count total number ` ` ``# of unordered pairs ` ` ``ans ``=` `(count ``*` `(count ``-` `1``)) ``/` `2` ` ` ` ``print``(ans) ` ` ` `# Driver code ` `if` `__name__ ``=``=` `"__main__"` `: ` ` ` ` ``a ``=` `[ ``2``, ``5``, ``8``, ``16``, ``128``] ` ` ` ` ``n ``=` `len``(a) ` ` ` ` ``Count_pairs(a, n) ` ` ` `# This code is contributed by ANKITRAI1 `
C#
`// C# program to Count unordered pairs (i, j) ` `// in array such that product of a[i] and a[j] ` `// can be expressed as power of two ` ` ` `using` `System; ` ` ` `public` `class` `GFG{ ` ` ` ` ` `/* Function to check if x is power of 2*/` `static` `bool` `isPowerOfTwo(``int` `x) ` `{ ` `/* First x in the below expression is ` ` ``for the case when x is 0 */` `return` `(x >0&& (!((x&(x-1))>0))); ` `} ` ` ` `// Function to Count unordered pairs ` `static` `void` `Count_pairs(``int` `[]a, ``int` `n) ` `{ ` ` ``int` `count = 0; ` ` ` ` ``for` `(``int` `i = 0; i < n; i++) { ` ` ` ` ``// is a number can be expressed ` ` ``// as power of two ` ` ``if` `(isPowerOfTwo(a[i])) ` ` ``count++; ` ` ``} ` ` ` ` ``// count total number ` ` ``// of unordered pairs ` ` ``int` `ans = (count * (count - 1)) / 2; ` ` ` ` ``Console.WriteLine( ans); ` `} ` ` ` `// Driver code ` ` ` ` ``static` `public` `void` `Main (){ ` ` ``int` `[]a = { 2, 5, 8, 16, 128 }; ` ` ` ` ``int` `n = a.Length; ` ` ``Count_pairs(a, n); ` ` ` ` ``} ` `} ` ` ` `// This code is contributed ` `// by Sach_Code `
PHP
` `
Output:
```6
```
Time Complexity: O(N), where N is the number of elements in the array.
My Personal Notes arrow_drop_up
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Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below. | 2,340 | 6,312 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5625 | 4 | CC-MAIN-2019-51 | longest | en | 0.813278 |
https://forums.airlinesim.aero/t/no-benefit-of-a-quick-change-plane/5776 | 1,679,727,788,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296945317.85/warc/CC-MAIN-20230325064253-20230325094253-00401.warc.gz | 310,577,816 | 6,024 | # no benefit of a quick change plane
you can get a quick change aircraft but it has no benefit. it could give you an option for quick change aircraft cargo or passangers.
like this
passenger flights between 11 am to 9pm
1 hour change over (10pm ready)
cargo flights 10pm to 4pm
1 hour change over (5am ready)
5am to 11 am maintenance
I’m afraid, I do not completely understand the relevance of such a plane.
Two things coming to my mind immediately:
1. Why not operate 2 seperate planes, one for cargo and the other for pax? This way you wouldn’t waste time converting the aircrafts for pax/cargo missions which would result in a higher possible profit margin.
2. Many planes do carry pax as well as cargo at the same time. For me this looks a lot more profitable again.
Cheers
look at it like this
you pay twice every thing ie lease
You have to think about it a little bit different.
Imagine you have 1 quick change aircraft on the one hand and 1 normal aircraft with pax and cargo transport at the same time. Both make 2 Flights.
Also: One flight with the quick change with pax + one flight with the quick change with cargo = Two flights with a normal aircraft with pax and cargo.
(lets’s say: Flight 1: 200 Pax, Flight 2: 100 Cargo = Flight 1: 100 Pax + 50 Cargo, Flight 2: 100 Pax + 50 Cargo)
So in the end the number of passengers and the amount of cargo transported is the same in both cases. If now both types of aircraft also have the same leasing rate, then the normal aircrafts are more profitable than quick change aircrafts.
That is, because you will have more flight time available on the normal aircrafts - namely: the 2 hours of changing time for the conversion of the quick change aircrafts. The daily conversion would make these aircrafts unprofitable.
If you imagine a second scenario in which you have the quick change aircraft of above and on the other hand you have one pax aircraft and one cargo aircraft. The quick change aircraft makes 2 flights, the normal ones make one each, so that we have 2 flights on each side. Of course the leasing rate of one quick change aircraft is the same as the rate for one normal aircraft.
You may pay twice the leasing rate in this scenario for the normal aircrafts, since you have 2 of them, BUT you can fly twice as much, which means that on average you pay the same leasing rate per flight hour on each side.
(this time: Flight 1: 200 Pax, Flight 2: 100 Cargo = Pax Aircraft: 200 Pax, Cargo Aircraft: 100 Cargo)
Also in this case the normal aircrafts will be better, since you will lose valuable opertating time due to the conversion process of the quick change aircraft.
Do these planes exist in real life? It takes month to convert an aircraft
Yes the boeing 737-300qc
Bae 146-200/300QT to name another…
Had the pleasure of getting seats in and out of these birds, and the 737-300QC for real…man its a pain in the ~@!.
So I guess you don’t need them in the game also?
Too many engines, and not enough capacity!
I seem to remember reading about an ATR operated like this, seats taken out at night, put back in the morning.
other airlines has made it work
Do these planes exist in real life? It takes month to convert an aircraft
They do not convert them totally they just unfasten the seats and lift them out and take out the floor panels. For example Jet 2 has 4 Boeing 737-300qc aircraft based in EDI, everyday they make a couple of flights in the morning to afternoon whit passagers and then they return to EDI get converted in an hour and fly whit all Scottish Royal Mail to London Stansted and then fly back to EDI and get converted to passagers again and start a new day.
They do not convert them totally they just unfasten the seats and lift them out and take out the floor panels. For example Jet 2 has 4 Boeing 737-300qc aircraft based in EDI, everyday they make a couple of flights in the morning to afternoon whit passagers and then they return to EDI get converted in an hour and fly whit all Scottish Royal Mail to London Stansted and then fly back to EDI and get converted to passagers again and start a new day.
Give me the Q400QC NOOOOOOOW!!!!! :) | 976 | 4,161 | {"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-2023-14 | latest | en | 0.942764 |
https://learn.careers360.com/school/question-provide-solution-for-rd-sharma-maths-class-12-chapter19-definite-integrals-exercise-19-point-5-question-7/?question_number=7.0 | 1,717,066,545,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971059632.19/warc/CC-MAIN-20240530083640-20240530113640-00849.warc.gz | 308,951,957 | 39,761 | #### Provide solution for RD sharma maths class 12 chapter19 Definite Integrals exercise 19.5 question 7
-4
Hint:
To solve the given statement, we have to use the formula of addition limits.
Given:
$\int_{3}^{5}(2-x) d x$
Solution:
We have,
$\int_{a}^{b}(f x) d x=\lim _{h \rightarrow 0} h[f(a)+f(a+h)+f(a+2 h)+\ldots f(a+(n-1) h)]$
Here,
\begin{aligned} &a=3, b=5, f(x)=2-x \\ &h=\frac{2}{n} \end{aligned}
Thus, we have
\begin{aligned} &I=\int_{3}^{5}(2-x) d x \\ &I=\lim _{h \rightarrow 0} h[f(3)+f(3+h)+f(3+2 h)+\ldots f(3+(n-1)) h] \end{aligned}
\begin{aligned} &=\lim _{h \rightarrow 0} h[(2-3)+\{2-(3+h)\}+(2-(3+2 h)) \ldots 2-(3+(n-1) h)\rfloor \\ &=\lim _{h \rightarrow 0} h[-n-h(1+2+3+\ldots(n-1) h)] \end{aligned}
$=\lim _{n \rightarrow \infty} \frac{2}{n}\left[-n-\frac{2}{n} \cdot \frac{n(n-1)}{2}\right] \quad\left[\begin{array}{l} \mathrm{Q} h \rightarrow 0 \& h=\frac{2}{n} \\ n \rightarrow \infty \end{array}\right]$
\begin{aligned} &=\lim _{n \rightarrow \infty}\left[-2-\frac{2}{n^{2}} \cdot n^{2}\left(1-\frac{1}{n}\right)\right] \\ &=\lim _{n \rightarrow \infty}-2-2\left(1-\frac{1}{n}\right) \\ &=-2-2=-4 \end{aligned}
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• Faculty Support | 516 | 1,260 | {"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": 7, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-22 | latest | en | 0.36239 |
http://mathhelpforum.com/calculus/127803-calculus-tangent-line.html | 1,527,249,672,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794867085.95/warc/CC-MAIN-20180525102302-20180525122302-00274.warc.gz | 187,588,104 | 10,102 | 1. Calculus/tangent line
Find all values of x at which the tangent line to the given curve passes through the origin for y = 1/(x + 4)
2. Originally Posted by pantera
Find all values of x at which the tangent line to the given curve passes through the origin for y = 1/(x + 4)
The tangent is a line of slope "m".
The derivative of the function gives this slope.
As the tangent passes through the origin, it's equation is y=mx, as c=0.
Hence, the point of intersection of this tangent and the curve gives us x.
$\displaystyle f'(x)=\frac{(x+4)(0)-1(1)}{(x+4)^2}$
The equation of the tangent(s) through (0,0) is $\displaystyle y=mx=\frac{-x}{(x+4)^2}$
Solving f(x) for the curve = mx for the line finds x.
$\displaystyle \frac{1}{x+4}=\frac{-x}{(x+4)^2}$
Solving this, bearing in mind that x=-4 is not part of the domain of f(x) or f'(x) discovers x.
Hence multiply both fractions by (x+4) before solving for x.
3. Lines look like this: $\displaystyle (y-y_{0}) = m(x-x_{0})$
We know we need the Origin: $\displaystyle y_{0} = m \cdot x_{0}$
The derivative gives the slope: $\displaystyle m = -\frac{1}{(x_{0}+4)^{2}}$
We know: $\displaystyle y_{0} = \frac{1}{x_{0}+4}$
I'm a little surprised to find only one solution. I expected two solutions when I started the problem. I was wrong.
Let's see what you get.
4. Here's why there is only one solution | 421 | 1,366 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.28125 | 4 | CC-MAIN-2018-22 | latest | en | 0.87372 |
http://mathhelpforum.com/advanced-algebra/208384-roots-cubic.html | 1,481,251,512,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698542668.98/warc/CC-MAIN-20161202170902-00070-ip-10-31-129-80.ec2.internal.warc.gz | 172,974,586 | 10,792 | 1. ## Roots of Cubic
Hi , I don't know what's going on here with the sigma notation (attached question and answer) .
A simplified explanation would be greatly appreciated .
Thank you
2. ## Re: Roots of Cubic
There are a many sigmas/ Which one are you referring to?
3. ## Re: Roots of Cubic
Hi , alpha=2 the first one and then why gamma is omitted in the second .
4. ## Re: Roots of Cubic
Originally Posted by minicooper58
Hi , I don't know what's going on here with the sigma notation (attached question and answer) .
A simplified explanation would be greatly appreciated .
The cubic is $z^3-2z^2+0z+k=0$
The sum of the roots; $\alpha+\beta+\gamma=2$
The product of roots two at a time: $\alpha\beta+\alpha\gamma+\beta\gamma=0$
The products of the roots: $\alpha\beta\gamma=-k$ | 215 | 789 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 4, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.8125 | 3 | CC-MAIN-2016-50 | longest | en | 0.911642 |
https://docs.ncot.uk/computing-maths/fixed-point-arithmetic/fixed-point-concept.html | 1,680,267,658,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296949642.35/warc/CC-MAIN-20230331113819-20230331143819-00305.warc.gz | 256,920,388 | 4,052 | # Fixed Point Place Values¶
Before we begin, we are going to switch to using 16 bits, 8 bits doesn’t give us enough range for our numbers as you’ll see in a moment. Also to make this easier to explain we are working with unsigned integers. This works equally fine with 2s complement signed values, and in fact they are what you should use. With signed 2s complement 8 bit values the range will be +127 / -128
Instead of deciding the bits in our bytes follow this pattern:
32768
16384
8192
4096
2048
1024
512
256
128
64
32
16
8
4
2
1
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
What if we chose to use the first 8 bits as the whole number, and the last 8 bits as our fractional part?
The headings on the columns would now be this:
(The point column doesn’t exist, it is just here to make it more obvious what is going on. This is still a 16 bit number)
128
64
32
16
8
4
2
1
.
1/2
1/4
1/8
1/16
1/32
1/64
1/128
1/256
x
x
x
x
x
x
x
x
.
x
x
x
x
x
x
x
x
Now we can represent certain fractional numbers by working out which bits to turn on. We’ve turned our 16 bits into an 8 bit number that has 256 smaller values.
A half would be
128
64
32
16
8
4
2
1
.
1/2
1/4
1/8
1/16
1/32
1/64
1/128
1/256
0
0
0
0
0
0
0
0
.
1
0
0
0
0
0
0
0
12.0625 would be
128
64
32
16
8
4
2
1
.
1/2
1/4
1/8
1/16
1/32
1/64
1/128
1/256
0
0
0
0
1
1
0
0
.
0
0
0
1
0
0
0
0
What about values where the fractional part is not an even power of two? What about a simple number like 10.1?
10 decimal in binary is 00001010, but how do we convert the fractional part? Read on to find out how to convert floating point values to their fixed point representations. | 604 | 1,748 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.703125 | 4 | CC-MAIN-2023-14 | latest | en | 0.845601 |
http://www.numbersaplenty.com/73533 | 1,590,726,890,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347401260.16/warc/CC-MAIN-20200529023731-20200529053731-00070.warc.gz | 195,216,479 | 3,424 | Search a number
73533 = 3127193
BaseRepresentation
bin10001111100111101
310201212110
4101330331
54323113
61324233
7424245
oct217475
9121773
1073533
1150279
1236679
1327615
141cb25
1516bc3
hex11f3d
73533 has 8 divisors (see below), whose sum is σ = 99328. Its totient is φ = 48384.
The previous prime is 73529. The next prime is 73547. The reversal of 73533 is 33537.
It can be divided in two parts, 73 and 533, that added together give a palindrome (606).
It is a sphenic number, since it is the product of 3 distinct primes.
It is not a de Polignac number, because 73533 - 22 = 73529 is a prime.
It is a Duffinian number.
It is a plaindrome in base 12.
It is a congruent number.
It is not an unprimeable number, because it can be changed into a prime (73523) by changing a digit.
It is a pernicious number, because its binary representation contains a prime number (11) of ones.
It is a polite number, since it can be written in 7 ways as a sum of consecutive naturals, for example, 285 + ... + 477.
It is an arithmetic number, because the mean of its divisors is an integer number (12416).
273533 is an apocalyptic number.
It is an amenable number.
73533 is a deficient number, since it is larger than the sum of its proper divisors (25795).
73533 is a wasteful number, since it uses less digits than its factorization.
73533 is an odious number, because the sum of its binary digits is odd.
The sum of its prime factors is 323.
The product of its digits is 945, while the sum is 21.
The square root of 73533 is about 271.1696885716. The cubic root of 73533 is about 41.8948617444.
The spelling of 73533 in words is "seventy-three thousand, five hundred thirty-three".
Divisors: 1 3 127 193 381 579 24511 73533 | 515 | 1,734 | {"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} | 3.453125 | 3 | CC-MAIN-2020-24 | latest | en | 0.897005 |
https://maniacs.info/does-my-vote-count/how-much-does-my-vote-count-in-a-presidential-election-in-mississippi.html | 1,638,578,444,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964362923.11/warc/CC-MAIN-20211204003045-20211204033045-00007.warc.gz | 443,009,650 | 2,684 | How much does my vote count in a Presidential Election in Mississippi?
Are you going to vote in the next Presidential Election? Do you want to find out how much your vote counts if you live in Mississippi? Here we will determine how much your vote counts in Mississippi based on Electoral College information, how many people vote in Mississippi, and how many electoral votes Mississippi has.
Our calculations are based on the United States Census Bureau data from the previous election and information about the U.S. Electoral College from the National Archives and Records Administration:
Total Electoral College: 538
Electoral College to win: 270
People who vote in Mississippi: 1,470,000
Because Mississippi has 6 out of 538 electoral votes, we can calculate (6/538) x 100 = 1.12. This means that Mississippi's total influence over the Presidential Election is 1.12%
If we divide 1.12% by the Mississippi voting population of 1,470,000 we get how much one person's Mississippi vote influences the Presidential Election. The answer is 0.000000758667779379408%
Since a Presidential Candiate only needs 270 Electoral votes to win, we could do our math like this: (6/270) x 100 = 2.22 which means that Mississippi can contribute 2.22% of the Electoral College to get to 270.
Furthermore, 2.22 divided by 1,470,000 is 0.00000151171579743008. This means that your vote in Mississippi will be 0.00000151171579743008% of what is needed for the Presidential Candidate to win the White House.
How much does my vote count in a Presidential Election?
Go here to find out how much your vote would count on a national level or to look up a different state.
How much does my vote count in a Presidential Election in Missouri?
Go here to find out what your vote is worth in the next state on our list.
Disclaimer: There are many factors that influence voting numbers and we made multiple assumptions. This page is for entertainment only. | 433 | 1,935 | {"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} | 2.765625 | 3 | CC-MAIN-2021-49 | longest | en | 0.933568 |
http://mooshak.in/diagram/ge-dishwasher-schematic-diagram | 1,606,584,248,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141195687.51/warc/CC-MAIN-20201128155305-20201128185305-00504.warc.gz | 60,850,231 | 8,107 | # Ge Dishwasher Schematic Diagram
• Schematic Diagram
• Date : November 28, 2020
## Ge Dishwasher Schematic Diagram
Dishwasher
Downloads Ge Dishwasher Schematic Diagram dishwasher dishwashers canada dishwasher cleaning dishwasher reviews dishwasher manual dishwasher parts dishwasher safe dishwasher sale dishwasher parts canada dishwasher not cleaning dishes properly dishwashers for sale dishwasher sale ontario
Ge Dishwasher Schematic Diagram - What Is Illustrated at a UML Sequence Diagram? ? In this piece I will be looking at what is illustrated in a UML sequence diagram. The UML diagrams are built around the notion of two countries, which are at different stages of transition from one another. These arrows represent the flow of the items to which they point to. What's exemplified in a UML sequence diagram is that there are ways of linking objects together which then can make relations between the objects. In the tree structure you have the origin that has the widest level of branches and leaves. Each branch is joined to another through a range of leaves which branches off to the left and right and are occasionally called vertices. At a tree structure the objects are arranged depending on the properties that they have within their states, which can be either the very first or the previous state. It may be the case that objects in a tree structure have exactly the same number of leaves. In this case there would be no need for connecting the items in a tree structure at all. When you look at a tree structure in your mind you may envision that objects which can be found closer to the border of the trees are going to have fewer leaves. Similarly, when you take a look at the tree structure in your thoughts you may imagine that objects which can be found closer to the centre of the shrub will have fewer branches, and so on. All things in a tree structure is going to have a general property that they share with the rest of the items they are connected to. A good instance of that is that all objects in a tree structure share a common property because they may be dropped to one kind of thing. This is referred to as the origin and if you examine the origins of this tree you will see that they are closed types of items, i.e. All the branches is not the same type of branch of an item. When you have a look at a branch, then it might be that it is below the rest of the branch. The next illustration of how things can be linked is that when you examine a tree structure that you might envision that every branch has a different shape. As, well as the normal rounded shape the branches of the tree structure may also be shaped in various ways, such as square or rectangle feet. | 547 | 2,710 | {"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 | 3 | CC-MAIN-2020-50 | latest | en | 0.949863 |
https://mathinfocusanswerkey.com/math-in-focus-grade-7-chapter-1-lesson-1-3-answer-key/ | 1,726,241,927,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651523.40/warc/CC-MAIN-20240913133933-20240913163933-00122.warc.gz | 359,915,587 | 45,280 | # Math in Focus Grade 7 Chapter 1 Lesson 1.3 Answer Key Introducing Irrational Numbers
Go through the Math in Focus Grade 7 Workbook Answer Key Chapter 1 Lesson 1.3 Introducing Irrational Numbers to finish your assignments.
## Math in Focus Grade 7 Course 2 A Chapter 1 Lesson 1.3 Answer Key Introducing Irrational Numbers
### Math in Focus Grade 7 Chapter 1 Lesson 1.3 Guided Practice Answer Key
Hands-On Activity
Materials:
• paper
• ruler
• scissors
Find the value of $$\sqrt{2}$$ using a square.
Work in pairs.
Step 1.
Draw a square that has a length of 2 inches on a piece of paper. Then cut out the square.
Step 2.
Find the area of the square (square A).
Step 3.
Fold the four vertices of square A towards the center to form square B as shown below.
Step 4.
State how the areas of square A and square B are related. State the area of square B. How can you represent the length of a side of square B?
Step 5.
Using your answer in step 4, find the length of a side of square B with a calculator. Round your answer to 2 decimal places.
Math Journal
Place an edge of square B alongsi& a ruler to measure its length. Explain why the reading from the ruler is different from the answer in step 5.
Copy and complete.
Question 1.
Graph $$\sqrt{5}$$ on the number line using rational approximations.
Which two whole numbers is $$\sqrt{5}$$ between? Justify your reasoning. Using a calculator, $$\sqrt{5}$$ = .
Graph an interval where $$\sqrt{5}$$ is located.
The value of $$\sqrt{5}$$ with two decimal places is
is closer to than to . So, $$\sqrt{5}$$ is located closer to .
By using an approximate value of $$\sqrt{5}$$, locate $$\sqrt{5}$$ on the number line.
The $$\sqrt{5}$$ is between two whole numbers. The two whole numbers are 2 and 3.
By using calculator $$\sqrt{5}$$ = 2.236067977….
The value of $$\sqrt{5}$$ with two decimal places is 2.24.
The decimal 2.24 is closer to 2.2 than to 2.3.
So, $$\sqrt{5}$$ is located closer to 2.2.
By using an approximate value of $$\sqrt{5}$$, located $$\sqrt{5}$$ on the number line as we can observe in the below image.
Copy and complete.
Question 2.
Graph –$$\sqrt{2}$$ on the number line using rational approximations.
Which two integers is –$$\sqrt{2}$$ between? Justify your reasoning. Using a calculator, –$$\sqrt{2}$$ = .
Graph an interval where –$$\sqrt{2}$$ is located.
The value of –$$\sqrt{2}$$ with two decimal places is ? .
is closer to than to . So, –$$\sqrt{2}$$ is located closer to ?
By using an approximate value of –$$\sqrt{2}$$, locate –$$\sqrt{2}$$ on the number line.
The –$$\sqrt{2}$$ is between two whole numbers. The two whole numbers are -1 and -2.
By using calculator –$$\sqrt{2}$$ = -1.414213562…..
The value of –$$\sqrt{2}$$ with two decimal places is -1.41.
The decimal -1.41 is closer to -1.4 than to -1.5.
So, –$$\sqrt{2}$$ is located closer to -1.4.
By using an approximate value of –$$\sqrt{2}$$, located –$$\sqrt{2}$$ on the number line as we can observe in the below image.
Solve.
Question 3.
Graph –$$\sqrt{7}$$ on the number line using rational approximations.
The –$$\sqrt{7}$$ is between two whole numbers. The two whole numbers are -2 and -3.
By using calculator –$$\sqrt{7}$$ = -2.645751311…..
The value of –$$\sqrt{7}$$ with two decimal places is -2.64.
The decimal -2.64 is closer to -2.6 than to -2.7.
So, –$$\sqrt{7}$$ is located closer to -2.6.
By using an approximate value of –$$\sqrt{7}$$, located –$$\sqrt{7}$$ on the number line as we can observe in the below image.
### Math in Focus Course 2A Practice 1.3 Answer Key
Locate each positive irrational number on the number line using rational approximations. First tell which two whole numbers the square root is between.
Question 1.
$$\sqrt{3}$$
The $$\sqrt{3}$$ is between two whole numbers. The two whole numbers are 1 and 2.
By using calculator $$\sqrt{3}$$ = 1.732050807….
The value of $$\sqrt{3}$$ with two decimal places is 1.73.
The decimal 1.73 is closer to 1.7 than to 1.8.
So, $$\sqrt{3}$$ is located closer to 1.7.
By using an approximate value of $$\sqrt{3}$$, the positive irrational number $$\sqrt{3}$$ is located on the number line as we can observe in the above image.
Question 2.
$$\sqrt{7}$$
The $$\sqrt{7}$$ is between two whole numbers. The two whole numbers are 2 and 3.
By using calculator $$\sqrt{7}$$ = 2.645751311….
The value of $$\sqrt{7}$$ with two decimal places is 2.64.
The decimal 2.64 is closer to 2.6 than to 2.7.
So, $$\sqrt{7}$$ is located closer to 2.6.
By using an approximate value of $$\sqrt{7}$$, the positive irrational number $$\sqrt{7}$$ is located on the number line as we can observe in the above image.
Question 3.
$$\sqrt{11}$$
The $$\sqrt{11}$$ is between two whole numbers. The two whole numbers are 3 and 4.
By using calculator $$\sqrt{11}$$ = 3.316624790….
The value of $$\sqrt{11}$$ with two decimal places is 3.31.
The decimal 3.31 is closer to 3.3 than to 3.4.
So, $$\sqrt{11}$$ is located closer to 3.3.
By using an approximate value of $$\sqrt{11}$$, the positive irrational number $$\sqrt{11}$$ is located on the number line as we can observe in the above image.
Question 4.
$$\sqrt{26}$$
The $$\sqrt{26}$$ is between two whole numbers. The two whole numbers are 5 and 6.
By using calculator $$\sqrt{26}$$ = 5.099019513….
The value of $$\sqrt{26}$$ with two decimal places is 5.09.
The decimal 5.09 is closer to 5.1 than to 5.0.
So, $$\sqrt{26}$$ is located closer to 5.1.
By using an approximate value of $$\sqrt{26}$$, the positive irrational number $$\sqrt{26}$$ is located on the number line as we can observe in the above image.
Question 5.
$$\sqrt{34}$$
The $$\sqrt{34}$$ is between two whole numbers. The two whole numbers are 5 and 6.
By using calculator $$\sqrt{34}$$ = 5.830951894….
The value of $$\sqrt{34}$$ with two decimal places is 5.83.
The decimal 5.83 is closer to 5.8 than to 5.9.
So, $$\sqrt{34}$$ is located closer to 5.8.
By using an approximate value of $$\sqrt{34}$$, the positive irrational number $$\sqrt{34}$$ is located on the number line as we can observe in the above image.
Question 6.
$$\sqrt{48}$$
The $$\sqrt{48}$$ is between two whole numbers. The two whole numbers are 6 and 7.
By using calculator $$\sqrt{48}$$ = 6.928203230….
The value of $$\sqrt{48}$$ with two decimal places is 6.92.
The decimal 6.92 is closer to 6.9 than to 7.0.
So, $$\sqrt{48}$$ is located closer to 6.9.
By using an approximate value of $$\sqrt{48}$$, the positive irrational number $$\sqrt{48}$$ is located on the number line as we can observe in the above image.
Locate each negative irrational number on the number line using rational approximations. First tell which two integers the square root is between.
Question 7.
–$$\sqrt{5}$$
The –$$\sqrt{5}$$ is between two integers. The two integers are -2 and -3.
By using calculator –$$\sqrt{5}$$ = -2.236067977…..
The value of –$$\sqrt{5}$$ with two decimal places is -2.23.
The decimal -2.23 is closer to -2.2 than to -2.3.
So, –$$\sqrt{5}$$ is located closer to -2.2.
By using an approximate value of –$$\sqrt{5}$$, the negative irrational number –$$\sqrt{5}$$ is located on the number line as we can observe in the above image.
Question 8.
–$$\sqrt{6}$$
The –$$\sqrt{6}$$ is between two integers. The two integers are -2 and -3.
By using calculator –$$\sqrt{6}$$ = -2.449489742…..
The value of –$$\sqrt{6}$$ with two decimal places is -2.44.
The decimal -2.44 is closer to -2.4 than to -2.5.
So, –$$\sqrt{6}$$ is located closer to -2.4.
By using an approximate value of –$$\sqrt{6}$$, the negative irrational number –$$\sqrt{6}$$ is located on the number line as we can observe in the above image.
Question 9.
–$$\sqrt{17}$$
The –$$\sqrt{17}$$ is between two integers. The two integers are -4 and -5.
By using calculator –$$\sqrt{17}$$ = -4.123105625…..
The value of –$$\sqrt{17}$$ with two decimal places is -4.12.
The decimal -4.12 is closer to -4.1 than to -4.2.
So, –$$\sqrt{17}$$ is located closer to -4.1.
By using an approximate value of –$$\sqrt{17}$$, the negative irrational number –$$\sqrt{17}$$ is located on the number line as we can observe in the above image.
Question 10.
–$$\sqrt{26}$$
The –$$\sqrt{26}$$ is between two integers. The two integers are -5 and -6.
By using calculator –$$\sqrt{26}$$ = -5.099019513….
The value of –$$\sqrt{26}$$ with two decimal places is -5.09.
The decimal -5.09 is closer to -5.1 than to -5.0.
So, –$$\sqrt{26}$$ is located closer to -5.1.
By using an approximate value of –$$\sqrt{26}$$, the negative irrational number –$$\sqrt{26}$$ is located on the number line as we can observe in the above image.
Question 11.
–$$\sqrt{53}$$
The –$$\sqrt{53}$$ is between two integers. The two integers are -7 and -8.
By using calculator –$$\sqrt{53}$$ = -7.280109889….
The value of –$$\sqrt{53}$$ with two decimal places is -7.28.
The decimal -7.28 is closer to -7.3 than to -7.2.
So, –$$\sqrt{53}$$ is located closer to -7.3.
By using an approximate value of –$$\sqrt{53}$$, the negative irrational number –$$\sqrt{53}$$ is located on the number line as we can observe in the above image.
Question 12.
–$$\sqrt{80}$$
The –$$\sqrt{80}$$ is between two integers. The two integers are -8 and -9.
By using calculator –$$\sqrt{80}$$ = -8.944271909….
The value of –$$\sqrt{80}$$ with two decimal places is -8.94.
The decimal -8.94 is closer to -8.9 than to -9.0.
So, –$$\sqrt{80}$$ is located closer to -8.9.
By using an approximate value of –$$\sqrt{80}$$, the negative irrational number –$$\sqrt{80}$$ is located on the number line as we can observe in the above image.
Use a calculator. Locate each irrational number to 3 decimal places on the number line using rational approximations.
Question 13.
$$\sqrt{47}$$
By using calculator $$\sqrt{47}$$ = 6.855654600….
The value of $$\sqrt{47}$$ with three decimal places is 6.855.
The decimal 6.855 is closer to 6.86 than to 6.85.
So, $$\sqrt{47}$$ is located closer to 6.86.
The given irrational number $$\sqrt{47}$$ is located on the number line by using rational approximation as we can observe in the above image.
Question 14.
–$$\sqrt{15}$$
By using calculator –$$\sqrt{15}$$ = –3.872983346….
The value of –$$\sqrt{15}$$ with three decimal places is –3.872.
The decimal -3.872 is closer to –3.87 than to –3.88.
So, –$$\sqrt{15}$$ is located closer to -3.87.
The given irrational number –$$\sqrt{15}$$ is located on the number line by using rational approximation as we can observe in the above image.
Question 15.
By using calculator = 4.54683594….
The value of with three decimal places is 4.546.
The decimal 4.546 is in between 4.5 and 4.6.
So, is located on 4.55.
The given irrational number is located on the number line by using rational approximation as we can observe in the above image.
Locate each irrational number on the number line using rational approximations.
Question 16.
$$\sqrt{101}$$
By using calculator $$\sqrt{101}$$ = 10.049875….
The value of $$\sqrt{101}$$ with two decimal places is 10.04.
The decimal 10.04 is closer to 10 than to 10.1.
So, $$\sqrt{101}$$ is located closer to 10.
The given irrational number $$\sqrt{101}$$ is located on the number line by using rational approximation as we can observe in the above image.
Question 17.
–$$\sqrt{132}$$
By using calculator –$$\sqrt{132}$$ = – 11.489125….
The value of –$$\sqrt{132}$$ with two decimal places is –11.48.
The decimal -11.48 is closer to –11.5 than to –11.4.
So, –$$\sqrt{132}$$ is located closer to –11.5.
The given irrational number –$$\sqrt{132}$$ is located on the number line by using rational approximation as we can observe in the above image.
Question 18.
$$\sqrt{2,255}$$
By using calculator $$\sqrt{2,255}$$ = 47.4868….
The value of $$\sqrt{2,2551}$$ with two decimal places is 47.48.
The decimal 47.48 is closer to 47.5 than to 47.4.
So, $$\sqrt{2,255}$$ is located closer to 47.5.
The given irrational number $$\sqrt{2,255}$$ is located on the number line by using rational approximation as we can observe in the above image.
Solve.
Question 19.
Locate the value of the constant, π, on the number line using rational numbers.
We know that π = 3.14159265
In the above image we can observe the value of the constant, π, is located on the number line using rational numbers.
Question 20.
3.1416 and $$\frac{22}{7}$$ are two rational approximate values of π.
a) Graph 3.1416, $$\frac{22}{7}$$, and π on the number line.
b) Which of the two rational approximate values is closer to π?
a)We know that 22/7 = 3.1428, π = 3.14159
In the above image we can observe 3.1416, 22/7 and π on the number line.
b) The two rational approximate values closer to π are 3.1416 and 22/7.
Question 21.
A triangle is cut from a square as shown in the diagram. The area of the square is 59 square inches. Approximate the height of the triangle to 3 decimal places.
Question 22.
Math Journal When do you need to approximate an irrational number with a rational value? Explain and illustrate with an example. | 3,988 | 12,926 | {"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} | 5.03125 | 5 | CC-MAIN-2024-38 | latest | en | 0.867294 |
https://discourse.julialang.org/t/modelingtoolkit-external-base-excitation/105983 | 1,702,159,676,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100972.58/warc/CC-MAIN-20231209202131-20231209232131-00819.warc.gz | 238,269,421 | 6,570 | # ModelingToolkit external base excitation
hello
I’m trying to simulate a TMD system using ModelingToolkit.
This is the code i’m runing:
``````using ModelingToolkit
using DifferentialEquations
using LinearAlgebra
using Symbolics: scalarize
@variables t
D = Differential(t)
function Mass(; name, m = 1.0, xy = [0.0, 0.0], u = [0.0, 0.0])
ps = @parameters m=m
@variables pos(t)[1:2]=xy v(t)[1:2]=u
eqs = scalarize(D.(pos) .~ v)
ODESystem(eqs, t, [pos..., v...], ps; name)
end
function Spring(; name, k = 1e4, l = 1.0)
ps = @parameters k=k l=l
@variables x(t), dir(t)[1:2]
ODESystem(Equation[], t, [x, dir...], ps; name)
end
function Damper(; name, c = 2)
ps = @parameters c=c
@variables x(t), dir(t)[1:2]
ODESystem(Equation[], t, [x, dir...], ps; name)
end
function connect_element(element, x, y)
[element.x ~ norm(scalarize(x .- y))
scalarize(element.dir .~ scalarize(x .- y))]
end
function damper_force(damper)
-damper.c .* scalarize(damper.dir) .* D.(damper.x)
end
function spring_force(spring)
-spring.k .* scalarize(spring.dir) .* (spring.x - spring.l) ./ spring.x
end
# main mass
m1 = 1.0
xy1 = [0.0, 1.0]
k1 = 1e6
l1 = 1.0
c1 = 50.0
@named mass1 = Mass(m = m1, xy = xy1)
@named damper1 = Damper(c = c1)
@named spring1 = Spring(k = k1, l = l1)
# tmd mass
m2 = m1 * 0.05 # 5% of the main mass
xy2 = [0.0, 1.5]
k2 = (main_mass_fn*2π)^2*m2
l2 = 0.5
c2 = 50.0
@named mass2 = Mass(m = m2, xy = xy2)
@named damper2 = Damper(c = c2)
@named spring2 = Spring(k = k2, l = l2)
base_pos = [0.0, 0.1]
g = [0.0, -9.81]
eqs = [connect_element(damper1, mass1.pos, base_pos)
connect_element(spring1, mass1.pos, base_pos)
connect_element(damper2, mass2.pos, mass1.pos)
connect_element(spring2, mass2.pos, mass1.pos)
scalarize(D.(mass2.v) .~ damper_force(damper2) / mass2.m .+ spring_force(spring2) / mass2.m .+ g)
scalarize(D.(mass1.v) .~ damper_force(damper1) / mass1.m .+ spring_force(spring1) / mass1.m .- damper_force(damper2) / mass1.m .- spring_force(spring2) / mass1.m .+ g)]
@named _model = ODESystem(eqs, t, [ damper1.x; damper1.dir; spring1.x; spring1.dir; mass1.pos;
damper2.x; damper2.dir; spring2.x; spring2.dir; mass2.pos], [])
@named model = compose(_model, mass1, damper1, spring1, mass2, damper2, spring2)
sys = structural_simplify(model)
prob = ODEProblem(sys, [], (0.0, 1.0))
solve(prob, Rosenbrock23())
``````
my issue is, how to replace the static base position, `base_pos = [0.0, 0.1]`,
to a function in time, `base_pos(t)=sin(2π*f*t)` for example.
thanks
Have you tried
``````base_pos = [0.0, sin(2π*f*t)]
``````
?
More generally, you could make use of ModelingToolkitStandardLibrary.jl: A Standard Library for ModelingToolkit · ModelingToolkitStandardLibrary.jl in order to avoid having to define your own components for a simple system like this. Using the stdlib, you could use a `Sine` source as input.
You find a relevant tutorial here
In this tutorial, a `Step` is used as input rather than a `Sine`. For a system closer to a mass-spring-damper, see this tutorial instead
1 Like
thanks a lot, I’ll look into it. | 1,080 | 3,054 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.703125 | 3 | CC-MAIN-2023-50 | latest | en | 0.357477 |
https://webwork.libretexts.org/webwork2/html2xml?answersSubmitted=0&sourceFilePath=Library/UMN/calculusStewartCCC/s_2_2_5.pg&problemSeed=1234567&courseID=anonymous&userID=anonymous&course_password=anonymous&showSummary=1&displayMode=MathJax&problemIdentifierPrefix=102&language=en&outputformat=libretexts | 1,653,759,395,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652663016949.77/warc/CC-MAIN-20220528154416-20220528184416-00563.warc.gz | 682,966,971 | 4,325 | Use the given graph of the function $g$ to find the following limits:
1. $\displaystyle \lim_{x \to 2^{-}}g(x) =$ help (limits)
2. $\displaystyle \lim_{x \to 2^{+}}g(x) =$
3. $\displaystyle \lim_{x \to 2}g(x) =$
4. $\displaystyle \lim_{x \to 0}g(x) =$
5. $\displaystyle g(2) =$
Note: You can click on the graph to enlarge the image. | 123 | 338 | {"found_math": true, "script_math_tex": 6, "script_math_asciimath": 0, "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-2022-21 | latest | en | 0.524918 |
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# A pump started filling an empty pool with water and continue
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A pump started filling an empty pool with water and continued at a constant rate until the pool was full. At noon the pool was 1/3 full, and $$1\frac{1}{4}$$ hours later it was 3/4 full. What was the total number of hours that it took the pump to fill the pool?
A. $$2\frac{1}{3}$$
B. $$2\frac{2}{3}$$
C. 3
D. $$3\frac{1}{2}$$
E. $$3\frac{2}{3}$$
[Reveal] Spoiler: OA
Last edited by Bunuel on 11 Apr 2014, 02:29, edited 1 time in total.
Renamed the topic, edited the question, added the OA and moved to PS forum.
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20 Nov 2008, 22:18
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Jcpenny wrote:
A pump started filling an empty pool with water and continued at
a constant rate until the pool was full. At noon the pool was 1/3 full,
and 1+1/4 hours later it was 3/4 full. What was the total number of
hours that it took the pump to fill the pool?
A. 2+1/3 B. 2+2/3 C. 3
D. 3+1/2 E. 3+2/3
in 1+1/4 =5/4 hrs the pump filled 3/4-1/3 = 5/12th of the pool
in 1 hr it fills 1/3 of the pool
so in 3 hrs it will fill the entire pool.
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### Show Tags
10 Apr 2014, 16:22
alpha_plus_gamma wrote:
Jcpenny wrote:
A pump started filling an empty pool with water and continued at
a constant rate until the pool was full. At noon the pool was 1/3 full,
and 1+1/4 hours later it was 3/4 full. What was the total number of
hours that it took the pump to fill the pool?
A. 2+1/3 B. 2+2/3 C. 3
D. 3+1/2 E. 3+2/3
in 1+1/4 =5/4 hrs the pump filled 3/4-1/3 = 5/12th of the pool
in 1 hr it fills 1/3 of the pool
so in 3 hrs it will fill the entire pool.
I was with you up until 5/12th of the pool. How did you get the in 1 hour it fills 1/3 of the poor?
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Re: A pump started filling an empty pool with water and [#permalink]
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10 Apr 2014, 19:13
Expert's post
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BOOKMARKED
Jcpenny wrote:
A pump started filling an empty pool with water and continued at
a constant rate until the pool was full. At noon the pool was 1/3 full,
and 1+1/4 hours later it was 3/4 full. What was the total number of
hours that it took the pump to fill the pool?
A. 2+1/3 B. 2+2/3 C. 3
D. 3+1/2 E. 3+2/3
The time elapsed between 1/3 full pool and 3/4 full pool is 5/4 hrs. This means 3/4 - 1/3 = 5/12 of the pool was filled in 5/4 hrs.
Again, (5/12)th of the pool will be filled in 5/4 hrs
1 pool will be filled in $$\frac{5/4}{5/12} * 1 = 3$$ hrs
_________________
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Get started with Veritas Prep GMAT On Demand for $199 Veritas Prep Reviews GMAT Club Legend Joined: 09 Sep 2013 Posts: 13459 Followers: 575 Kudos [?]: 163 [0], given: 0 Re: A pump started filling an empty pool with water and continue [#permalink] ### Show Tags 12 Apr 2015, 21:07 Hello from the GMAT Club BumpBot! Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos). Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email. _________________ Manager Status: I am not a product of my circumstances. I am a product of my decisions Joined: 20 Jan 2013 Posts: 132 Location: India Concentration: Operations, General Management GPA: 3.92 WE: Operations (Energy and Utilities) Followers: 4 Kudos [?]: 93 [1] , given: 68 A pump started filling an empty pool with water and continue [#permalink] ### Show Tags 02 Jun 2015, 06:04 1 This post received KUDOS Jcpenny wrote: A pump started filling an empty pool with water and continued at a constant rate until the pool was full. At noon the pool was 1/3 full, and $$1\frac{1}{4}$$ hours later it was 3/4 full. What was the total number of hours that it took the pump to fill the pool? A. $$2\frac{1}{3}$$ B. $$2\frac{2}{3}$$ C. 3 D. $$3\frac{1}{2}$$ E. $$3\frac{2}{3}$$ Taking a smart number. Let the capacity of the pool be 12 Liters At 12 PM = $$\frac{1}{3}*12$$ = 4 Liters of water was in the pool. After $$1\frac{1}{4}$$ Hours i.e. at 13:15 Hrs = $$\frac{3}{4}*12$$ = 9 liters of water was in the pool. So in 75 mins the pump fills 9-4 = 5 liters of water in the pool. setting up ratio and proportion $$\frac{X}{75}=\frac{12}{5}$$ $$X = 180 mins = 3 Hours$$ Answer is C Optimus Prep Instructor Joined: 06 Nov 2014 Posts: 1782 Followers: 51 Kudos [?]: 393 [0], given: 21 Re: A pump started filling an empty pool with water and continue [#permalink] ### Show Tags 02 Jun 2015, 07:54 Expert's post 1 This post was BOOKMARKED A pump started filling an empty pool with water and continued at a constant rate until the pool was full. At noon the pool was 1/3 full, and 1 1/4 hours later it was 3/4 full. What was the total number of hours that it took the pump to fill the pool? (3/4) - (1/3) = 5/12. 5/12 of the pool was full in 1 1/4 hours time (5/4 hours). To determine a rate of filling we divide how much is full by the time. That is (5/12) ÷ (5/4) = 1/3. This means it fills a pool in 3 hours. A. 2 1/3 B. 2 2/3 C. 3 D. 3 1/2 E. 3 2/3 _________________ # Janielle Williams Customer Support Special Offer:$80-100/hr. Online Private Tutoring
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Hello from the GMAT Club BumpBot!
Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).
Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
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Re: A pump started filling an empty pool with water and continue [#permalink]
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24 Sep 2016, 04:14
at noon or 12:00PM it is filled 1/3 of total, say x = 1/3x
1hour 15 mins later it is 3/4x filled
so amount it filled in 75mins = (3/4)x - (1/3)x
=> 15*12 mins => 3 hours
C is the answer
Re: A pump started filling an empty pool with water and continue [#permalink] 24 Sep 2016, 04:14
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# A pump started filling an empty pool with water and continue
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Powered by phpBB © phpBB Group and phpBB SEO Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®. | 2,795 | 8,883 | {"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} | 3.984375 | 4 | CC-MAIN-2017-04 | latest | en | 0.904617 |
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51
##### Chapter 3 / Re: Chapter 3.1 Theorem 4
« Last post by Victor Ivrii on February 27, 2022, 07:52:52 PM »
Indeed, to be corrected. Thanks
52
##### Chapter 3 / Chapter 3.1 Theorem 4
« Last post by Zicheng Ding on February 26, 2022, 07:43:05 PM »
In the online text book, chapter 3.1 theorem 4, we have the formula for inhomogeneous heat equations. For the second integral in the formula, since it represents the homogeneous part of the heat equation and the heat equation is linear, should the range be from $-\infty$ to $\infty$ instead of $0$ to $\infty$? The equation we are solving have the range of $-\infty < x < \infty$ and $t > 0$ as usual.
53
##### Test 1 / 2020 Test 1 Q2
« Last post by kevin on February 23, 2022, 08:45:59 PM »
how did we derive 2 from 1 and 4 from 3
54
##### Chapter 2 / Textbook Chapter 2.5 Example 2
« Last post by Yifei Hu on February 21, 2022, 11:29:41 AM »
Hi Professor,
I think one coefficient is wrong in the solution.
In this problem the solution to the inhomogeneous equation with homogeneous initial condition should start with coefficient $\frac{1}{2c} = \frac{1}{6}$. After cancelling the 36 in the $f(x,t)$, we should arrive at $6\int_0^t \int_{x-3(t-t')}^{x+3(t-t')}\frac{1}{t^2+1}dx'dt'$ instead of $3\int_0^t \int_{x-3(t-t')}^{x+3(t-t')}\frac{1}{t^2+1}dx'dt'$.
Can you help me confirm if I missed something or the textbook has a typo?
55
##### Chapter 2 / Re: Derivation of D' Alembert formula under Characteristic Coordinate
« Last post by Victor Ivrii on February 21, 2022, 04:34:15 AM »
Quote
But how does this qualify us to replace the indefinite integral with the definite one?
Did you take Calculus I? Then you must know that if the preimitive (indefinite integral) is a set of definite integrals which differ by an arbitrary constant.
56
##### Chapter 2 / Re: Derivation of D' Alembert formula under Characteristic Coordinate
« Last post by Yifei Hu on February 20, 2022, 08:12:28 PM »
I understand that I can show $\phi'(\xi)=0$ by:
1) when t = 0, $\xi = x+ct = x = x-ct = \eta$
2) $u_\xi = u_t \frac{dt}{d\xi} + u_x \frac{dx}{d\xi}$ by chain rule.
3)By initial condition: $u_t|_{t=0} = u_x|_{t=0}$ = 0, we must have $u_\xi=0$
4) $u_\xi = \phi'(x)$ hence $\phi'(\xi)=0$
But how does this qualify us to replace the indefinite integral with the definite one?
57
##### Chapter 2 / Re: Text Book 2.4 Example 2.1
« Last post by Victor Ivrii on February 20, 2022, 10:04:48 AM »
We integrate from $t=0$ because for $t=0$ initial conditions are done. $-1<t$ is a domain where $f(x,t)$ is defined
58
##### Chapter 2 / Re: Derivation of D' Alembert formula under Characteristic Coordinate
« Last post by Victor Ivrii on February 20, 2022, 10:01:28 AM »
Reproduce formula correctly (there are several errors) and think about explanation why $\phi'(\xi)=0$.
59
##### Chapter 2 / Text Book 2.4 Example 2.1
« Last post by Yifei Hu on February 19, 2022, 09:16:55 PM »
In this problem, $-1 < t < -\infty$, when applying D' Alembert's formula, why we integrate from 0 to t instead of from -1 to t?
In text book 2.4, we had one line in the derivation: $\tilde{u_{\xi}}=-\frac{1}{4c^2} \int^\xi f(\xi,\eta')d\eta'=-\frac{1}{4c^2} \int_\xi^\eta\tilde f(\xi,\eta')d\eta' + \phi'(\xi)$.
Why can we replace the definite integral with the indefinite integral? Why we choose $\xi$ as lower limit and $\eta$ as upper limit? | 1,161 | 3,410 | {"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-2023-23 | longest | en | 0.823719 |
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Question
Updated 4/28/2014 7:51:23 PM
This conversation has been flagged as incorrect.
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Original conversation
User: solve problem for x. x - 3 > - 2
Weegy: x=3.
wedekind|Points 1012|
User: solve the problem for x. 3/5x < 2/3
User: solve problem for x. - 7x > 35
User: solve problem for x. -7x > 35
Weegy: -6x>35 -x>35/6 x<-35/6
Question
Updated 4/28/2014 7:51:23 PM
This conversation has been flagged as incorrect.
Flagged by yeswey [4/28/2014 7:50:48 PM]
Rating
3
x - 3 > - 2
x > - 2 + 3
x > 1
Confirmed by alfred123 [4/28/2014 8:05:23 PM]
3
3/5x < 2/3
multiple 5/3 on both sides:
(3/5)(5/3)x < (2/3)(5/3)
x < 10/9
Confirmed by andrewpallarca [4/29/2014 2:02:52 PM]
3
-7x > 35
Divided by -7 on both sides:
x < -5
Confirmed by alfred123 [4/28/2014 8:05:38 PM]
Questions asked by the same visitor
solve inequality for x . 4 ( x + 3 ) > 6 ( x -5 ) User: Graph this inequality on a number line x User: I don't understand it
Weegy: what dont you understand User: forget it Weegy: Are you sure? If you need anything answered, just ask! User: ok Weegy: I'm happy you are ok. Is there anything else you want answering today? User: Write the inequality that corresponds to this graph -1 (More)
Question
Updated 5/5/2014 4:00:52 PM
4 ( x + 3 ) > 6 ( x -5 )
4x + 12 > 6x - 30
42 > 2x
x < 21
Confirmed by andrewpallarca [5/5/2014 4:00:46 PM]
Solve for x. 3 ( x -2 )
Weegy: 4 ? -2 x (2)3 = 4 ? -2 x 8 = -2 * 8 = -16 (More)
Question
Updated 5/5/2014 4:00:30 PM
3 ( x -2 ) 18/-3;
x>-6
Unconfirmed by andrewpallarca [5/5/2014 4:00:05 PM]
It's strange that greater or less than symbols are not showing up again.
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Home | Contact | Blog | About | Terms | Privacy | © Purple Inc. | 1,218 | 3,439 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.609375 | 4 | CC-MAIN-2018-34 | latest | en | 0.905621 |
http://stackoverflow.com/questions/tagged/plot+statistics | 1,412,236,749,000,000,000 | text/html | crawl-data/CC-MAIN-2014-41/segments/1412037663718.7/warc/CC-MAIN-20140930004103-00284-ip-10-234-18-248.ec2.internal.warc.gz | 299,254,722 | 26,258 | # Tagged Questions
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http://courtneygibbons.org/category/prof-life/page/4/ | 1,544,458,087,000,000,000 | text/html | crawl-data/CC-MAIN-2018-51/segments/1544376823348.23/warc/CC-MAIN-20181210144632-20181210170132-00026.warc.gz | 75,069,181 | 14,592 | # Prof Life
## WMC REU, day 2
My group decided to meet daily at 10 am for the first week, which gives me plenty of time each morning to do some research work (that’s when my brain is at its freshest). I started at 7:30 at The Gov Cup where I made some progress on merging several drafts of The Paper. Some ideas for streamlining some of the mathematics also occurred to me. Why use cases if you can help it, right? When I reached the point where I would need a new a hard copy of The Paper, I turned to outlining a different paper. (This one is work from the past year with my senior research fellow.) It was hard to shift gears that quickly between projects, though, and I was disappointed with my work on the outline.
During the meeting with my REU students, we talked about how the research reading went. This discussion led us to create a list of terms that we should define. We started with one they all knew as a baseline (“ring”) and ramped up to a special kind of ring (“standard graded k-algebra”, where k is any field). The classic example of a standard graded k-algebra is a polynomial ring over k where each variable has degree 1. We also defined “module,” “graded module,” “Hilbert function,” and we did some examples. The research project possibilities all include working with a quotient of a polynomial ring by a homogeneous regular sequence (also known as a “complete intersection”). Thus, I focused my examples on quotients of the polynomial ring by homogeneous ideals. We started to talk a bit about free resolutions and Betti tables, too. Based on our morning session, I assigned some homework.
By afternoon, my triple latte had worn off, so I spent some time on the general bureaucratic professor stuff I alluded to yesterday. I’m about to start my third year at Hamilton College, and that means I have to start putting together my reappointment materials. Instead of creating any of these materials, I made a list. Then I made a more precise list. Then I added due dates. Then I decided that it’s summer, so that was enough for one day. Back to the fun stuff!
To round out the work day, I read a bit ahead of where my students were in their research papers with an eye toward what we might talk about next time (free resolutions, Betti diagrams), and I prepared a few examples to clarify some definitions (I hope!).
## Greetings from the Willamette Mathematics Consortium
Today marks the first day of the WMC REU. I never went to a formal REU as an undergraduate, so I’m a first-timer just like many of the students. My first task today finding a suitable coffee shop for my summer caffeinated home base. (I settled on The Governor’s Cup, which sports an exposed brick wall, generously-sized table tops, and coffee that’s roasted in-house).
Although I’d corresponded with my three research students via email, it was a treat to meet them in person. All nine of the students seem like good people, in fact, and so do the other research mentors. It’s rare to find a group where there isn’t a single unlikeable person (which leads me to wonder if, in reality, I’m the unlikable one? Nah…).
Besides having found a coffee shop that will inevitably see many hours of work on my part, I have an office in the mathematics “hearth” at Willamette. I haven’t had much experience visiting places for long enough to have an office; I find myself distracted by trying to calculate the number of math books I have in common with the usual owner. In my professional opinion, it’s a big number. We also have several climbing guides in common. What is it about math and climbers? (One of the REU students is a climber, too.)
In my first meeting with my group, we talked about the basics — scheduling, their goals for the summer, my goals for the summer, project possibilities, and, of course, homework. Since the homework required starting to read research papers, we talked about how to do that.
They had plenty to do for the day, so we broke and I spent the afternoon working on revising a paper that I’m writing with my graduate advisors. If I can get the paper out of my hands by next week, I’ll be a jubilant mathematician. After a productive session, I outlined some of my other summer goals. I’ve got other research work to do, and some general administrative professor “stuff” that I have been, well, lazy about getting done. While I feel well-prepared for my role as an REU mentor, I would like to wrangle some resources together in case they’re helpful.
It’s been a fun day!
## What the wha? Bernoulli’s theorem, applied.1
Today, as part of the AGAM camp, the girls had a mini course on aerodynamics. And then we flew in tiny airplanes.
I wish I could say that this post was going to contain some math, but really, I’m just going to post pictures.
I got to steer the plane, to everyone’s dismay.
But despite my lack off subtlety and grace, no high schoolers were harmed this evening. And now, I think I’ll look into flight schools!
## All Girls/All Math
All Girls/All Math 2014
This week, I’m teaching one of the parallel “Codes and Cryptography” classes at the All Girls/All Math camp at University of Nebraska-Lincoln. It’s so much fun to be in a room full of budding mathematicians from all over the country, and it reminds me of what it felt like to love the idea of math before getting too deep into any particular specialization.
I’ve made a few resources for the class, which I’ll drop in this post as the week continues.
Obviously, the rock climbing links come first…:
Rock climbing and teaching math
And here are some relevant math links:
The Bletchley Circle on PBS:
http://www.pbs.org/program/bletchley-circle/
A quick review of the symmetric cryptography systems we talked about on Day 1:
http://animoto.com/play/x7rvb7VPhZZWbFKdzH3J0g
Here’s an article about a new kind of space race — the race to bounce perfectly secret codes off low-orbit satellites using quantum cryptography:
http://www.technologyreview.com/view/528671/the-space-based-quantum-cryptography-race/
For encrypting and decrypting messages, you can play with the web apps at http://www.cryptoclub.org.
And after the jump, there’s an except lifted from a blog post by Cathy O’Neil at mathbabe.org:
## Undercover at an Algebraic Combinatorics Conference
Before I went to graduate school and became a commutative algebraist, I did a little work in graph theory. It was fun! I proved theorems! My mentor/coauthor Josh Laison provided a wonderful introduction to the fun inherent in doing research.
Well, roughly a decade later, I found myself at an Algebraic Combinatorics conference in honor of Chris Godsil’s 65th birthday (whose book with Royle was my introduction to Algebraic Graph Theory almost *exactly* 10 years ago). It’s exciting to see that some of the work we did is still relevant.
It’s also a lovely drive to get to U. Waterloo from Hamilton College. In fact, on the way home, I spent a night camping on Nairne Island at the Upper Canada Migratory Bird Sanctuary. It was the maiden voyage for my new solo tent, and I had plenty of math to think about in the peace and quiet!
Sometimes when I say I’m going to work at a coffee shop, I realize that it sounds like I’m changing careers. But today is the first time I’ve been mistaken for an actual employee at an actual coffee shop. My latest, greatest thought was just interrupted by someone earnestly claiming to be one of my customers. I was very confused until we both realized what was happening.
I guess I’m a natural. Good to know there are options if I don’t like being a math prof! | 1,758 | 7,594 | {"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-2018-51 | latest | en | 0.961116 |
https://hackage.haskell.org/package/hledger-irr-0.1 | 1,579,563,001,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579250601040.47/warc/CC-MAIN-20200120224950-20200121013950-00021.warc.gz | 476,729,544 | 7,458 | # hledger-irr: computes the internal rate of return of an investment
[ bsd3, finance, program ] [ Propose Tags ]
hledger-irr is a small command-line utility based on Simon Michael's hleder library. Its purpose is to compute the internal rate of return, also known as the effective interest rate, of a given investment. After specifying what account holds the investment, and what account stores the gains (or losses, or fees, or cost), it calculates the hypothetical annualy rate of fixed rate investment that would have provided the exact same cash flow.
As an example, consider the following irregular investment recorded in a file called speculation.ledger. The account “Speculation” holds the investment which could be, for example, a stock. Regularly, we make sure that the value of the account matches the value of the stock, by moving money from or to the account “Rate Gain”. It does not really matter when we adjust the price, as long as it is correct at the end of our reporting period.
2011-01-01 Some wild speculation – I wonder if it pays off
Speculation €100.00
Cash
2011-02-01 More speculation (and adjustment of value)
Cash -€10.00
Rate Gain -€1.00
Speculation
2011-03-01 Lets pull out some money (and adjustment of value)
Cash €30.00
Rate Gain -€3.00
Speculation
2011-04-01 More speculation (and it lost some money!)
Cash -€50.00
Rate Gain € 5.00
Speculation
2011-05-01 Getting some money out (and adjustment of value)
Speculation -€44.00
Rate Gain -€ 3.00
Cash
2011-06-01 Emptying the account (after adjusting the value)
Speculation -€85.00
Cash €90.00
Rate Gain -€ 5.00
We can now calculate the rate of return for the whole time or just for parts of it (and be freaked out by the volatility of the investment):
$hledger-irr -f speculation.ledger -t "Rate Gain" -i Speculation -c 2011/01/01: €-100.00 2011/02/01: €-10.00 2011/03/01: €30.00 2011/04/01: €-50.00 2011/05/01: €47.00 2011/06/01: €90.00 2011/01/01 - 2012/12/07: 17.72%$ hledger-irr -f speculation.ledger -t "Rate Gain" -i Speculation -e 2011-03-01
2011/01/01 - 2011/03/01: 26.11%
$hledger-irr -f speculation.ledger -t "Rate Gain" -i Speculation -b 2011-03-01 2011/03/01 - 2012/12/07: 12.28%$ hledger-irr -f speculation.ledger -t "Rate Gain" -i Speculation --monthly
2011/01/01 - 2011/02/01: 12.43%
2011/02/01 - 2011/03/01: 41.57%
2011/03/01 - 2011/04/01: -51.45%
2011/04/01 - 2011/05/01: 32.27%
2011/05/01 - 2011/06/01: 96.01%
Running the utility with --help gives a brief overview over the available options:
\$ hledger-irr --help
Usage: hledger-irr [OPTION...]
-h --help print this message and exit
-V --version show version number and exit
-c --cashflow also show all revant transactions
-f FILE --file=FILE input ledger file (pass '-' for stdin)
-i ACCOUNT --investment-account=ACCOUNT investment account
-t ACCOUNT --interest-account=ACCOUNT interest/gain/fees/losses account
-b DATE --begin=DATE calculate interest from this date
-e DATE --end=DATE calculate interest until this date
-D --daily calculate intereste for each day
-W --weekly calculate intereste for each week
-M --monthly calculate intereste for each month
-Y --yearly calculate intereste for each year
Known bugs and issues:
• Currenlty, hledger-irr does not cope well with multiple commodities (e.g. Euro and Dollar, or shares).
• Also, interest or fees that do not pass through the account selected by --investment-account are not taken into consideration.
Versions [faq] 0.1, 0.1.1, 0.1.1.1, 0.1.1.2, 0.1.1.3, 0.1.1.4, 0.1.1.5, 0.1.1.6, 0.1.1.7, 0.1.1.8, 0.1.1.9, 0.1.1.10, 0.1.1.11, 0.1.1.12, 0.1.1.13, 0.1.1.14, 0.1.1.15, 0.1.1.16 base (>=3 && <5), Cabal, hledger-lib (>=0.19.3), statistics (>=0.10), time [details] BSD-3-Clause Joachim Breitner Joachim Breitner Finance head: darcs get http://darcs.nomeata.de/hledger-irr by JoachimBreitner at Fri Dec 7 22:20:40 UTC 2012 NixOS:0.1.1.16 hledger-irr 7799 total (579 in the last 30 days) (no votes yet) [estimated by Bayesian average] λ λ λ Docs not available All reported builds failed as of 2016-12-23 | 1,360 | 4,365 | {"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} | 3.078125 | 3 | CC-MAIN-2020-05 | latest | en | 0.85428 |
https://techcommunity.microsoft.com/t5/sharepoint/adding-many-conditions-in-calculated-value-sp/m-p/1705749/highlight/true | 1,606,660,971,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141198409.43/warc/CC-MAIN-20201129123729-20201129153729-00523.warc.gz | 504,752,657 | 81,509 | Highlighted
New Contributor
# Adding many conditions in calculated value SP
Hello everyone,
I have a calculated value field and i want to look in a column1 and check for values and return OK or NOK, example :
IF(column1 contains 0% or 25% or 50% or 75%) if yes return "OK" if not "NOK.
Any ideas ?
3 Replies
Highlighted
# Re: Adding many conditions in calculated value SP
Hi @GeorgesElias - try something like this:
=IF([COLUMN 1]=0,"ok",IF([COLUMN 1]=25,"ok", IF([COLUMN 1]=50,"ok",IF([COLUMN 1]=75,"ok","nok"))))
Highlighted
# Re: Adding many conditions in calculated value SP
Hello @Kelly E
I did this formula :
=SI([Avancement :]=0%;"OK";SI([Avancement :]=25%;"OK";SI([Avancement :]=50%;"OK";SI([Avancement :]=75%;"OK";SI([Avancement :]=100%;"OK";[Avancement :])))))
and the result is the opposite.
I want to look in the Column "Avancement" and the result in "Statut simplifié".
Thank you
# Re: Adding many conditions in calculated value SP
@GeorgesElias it doesn't look like your last IF statement isn't complete, your example shows the formula ending with ;[Avancement :] and it should end with the value to display if the statement is not true.
I also see semi-colons instead of commas. Try this:
=IF([Avancement :]=0,"ok",IF([Avancement :]=25,"ok", IF([Avancement :]=50,"ok",IF([Avancement :]=75,"ok","nok")))) | 383 | 1,338 | {"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} | 2.796875 | 3 | CC-MAIN-2020-50 | latest | en | 0.801643 |
https://solutionsadda.in/2023/10/03/process-synchronization/ | 1,716,948,907,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971059167.30/warc/CC-MAIN-20240529010927-20240529040927-00347.warc.gz | 458,935,278 | 69,041 | October 3, 2023
October 3, 2023
October 3, 2023
###### Algorithms
October 3, 2023
Question 21
Consider the following threads, T 1 , T 2 , and T 3 executing on a single processor, synchronized using three binary semaphore variables, S 1 , S 2 , and S 3 , operated upon using standard wait() and signal(). The threads can be context switched in any order and at any time.
Which initialization of the semaphores would print the sequence BCABCABCA….?
A S 1 = 1; S 2 = 1; S 3 = 1 B S 1 = 1; S 2 = 1; S 3 = 0 C S 1 = 1; S 2 = 0; S 3 = 0 D S 1 = 0; S 2 = 1; S 3 = 1
Question 21 Explanation:
Explanation: Since B is to be printed first semaphore S1 has to be set to 1, so that the thread enters its critical section and prints B. Then C needs to be printed, this happens when Thread 1 is blocked on semaphore s3 (blocked, s3=0). Then Thread1 invokes Thread 3 to print A.
Question 21 Explanation:
Explanation: Since B is to be printed first semaphore S1 has to be set to 1, so that the thread enters its critical section and prints B. Then C needs to be printed, this happens when Thread 1 is blocked on semaphore s3 (blocked, s3=0). Then Thread1 invokes Thread 3 to print A. | 358 | 1,169 | {"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-2024-22 | latest | en | 0.914999 |
https://www.physicsforums.com/threads/speed-of-light-vs-the-universes-inflation.520625/ | 1,550,414,089,000,000,000 | text/html | crawl-data/CC-MAIN-2019-09/segments/1550247481994.42/warc/CC-MAIN-20190217132048-20190217154048-00017.warc.gz | 930,382,783 | 13,629 | # Speed of light vs. the universe's inflation
1. Aug 10, 2011
### Trailblzn
I understand that the Speed of light is the universal speed limit (I do not believe this is in question). However, I watched Steven Hawking's show this weekend on TV where they stated "10 minutes after the big bang the universe was thousands of light years across."
How is this possible if the speed of light is the "speed limit?" How could anything be light years across if it has only been 10 minutes? I would think that at most it could be 20 "light minutes" across. I am sure there is an answer, I am just now aware of one.
2. Aug 10, 2011
### Staff: Mentor
Space is not a thing that moves, so the speed of light does not apply to its expansion.
3. Aug 10, 2011
### Trailblzn
Russ, thanks for the reply. I understand that space does not move but the things in the "space" of the early universe did move namely radiation. Was there empty universe expanding faster than radiation? The expanding radiation is what made up the universe as there was no space yet. I appreciate your thoughts.
4. Aug 10, 2011
### Drakkith
Staff Emeritus
Spacetime existed at the beginning of the universe. All radiation, matter, and everything occupies spacetime. The expansion between points in spacetime has no limit and as such it can, and does, cause objects to move faster than c. The key is that the objects don't move WITHIN spacetime greater than c, but that spacetime itself is expanding.
5. Aug 10, 2011
### Trailblzn
Drakkith ~
Thank you VERY much that makes a lot of sense. Thank you so much for your help.
6. Aug 12, 2011
### Lobezno
If you had a Cartesian planes that you "placed" upon the Universe the size that it is right now, from an external vantage point, then yes, however, matter would have traveled a fair bit faster than c. But the speed is only in reference to an already expanding reference point, and so the example is not valid.
I generally think of it as an expanding polka dot balloon, but how it works in my head would probably confuse most. | 495 | 2,050 | {"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} | 2.859375 | 3 | CC-MAIN-2019-09 | latest | en | 0.96082 |
https://cynohub.com/jntua-b-tech-r-20-2-3-syllabus-for-electrical-circuit-analysis-pdf-2022/ | 1,660,151,078,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882571198.57/warc/CC-MAIN-20220810161541-20220810191541-00446.warc.gz | 199,308,856 | 21,897 | # JNTUA B.TECH R 20 2-3 Syllabus For Electrical circuit analysis PDF 2022
### Get Complete Lecture Notes for Electrical circuit analysis on Cynohub APP
You will be able to find information about Electrical circuit analysis along with its Course Objectives and Course outcomes and also a list of textbook and reference books in this blog.You will get to learn a lot of new stuff and resolve a lot of questions you may have regarding Electrical circuit analysis after reading this blog. Electrical circuit analysis has 5 units altogether and you will be able to find notes for every unit on the CynoHub app. Electrical circuit analysis can be learnt easily as long as you have a well planned study schedule and practice all the previous question papers, which are also available on the CynoHub app.
All of the Topic and subtopics related to Electrical circuit analysis are mentioned below in detail. If you are having a hard time understanding Electrical circuit analysis or any other Engineering Subject of any semester or year then please watch the video lectures on the official CynoHub app as it has detailed explanations of each and every topic making your engineering experience easy and fun.
### Electrical circuit analysis Unit One
#### Locus Diagrams & Resonance
HrsSeries R-L, R-C, R-L-C and Parallel Combination with Variation of Various Parameters -Resonance-Series, Parallel Circuits, Frequency Response, Concept of Bandwidth and Q Factor.
### Electrical circuit analysis Unit Two
#### Two PortNetworks
Two Port Network Parameters –Impedance –Admittance -Transmission and Hybrid Parameters and their Relations -Concept of Transformed Network -Two Port Network Parameters Using Transformed Variables.
### Electrical circuit analysis Unit Three
#### Transient Analysis
D.C Transient Analysis: Transient Response of R-L, R-C, R-L-C Series Circuits for D.C Excitation -Initial Conditions in network -Initial Conditions in elements -Solution Method Using Differential Equation and Laplace Transforms -Response of R-L & R-C Networks to Pulse Excitation. A.C Transient Analysis: Transient Response of R-L, R-C, R-L-C Series Circuits for Sinusoidal Excitations -Solution Method Using Differential Equations and Laplace Transforms.
### Electrical circuit analysis Unit Four
#### Fourier Transforms
Fourier Theorem -Trigonometric Form and Exponential Form of Fourier series –Conditions of Symmetry -Line Spectra and Phase Angle Spectra -Analysis of Electrical Circuits to Non Sinusoidal Periodic Waveforms. Fourier Integrals and Fourier Transforms –Properties of Fourier Transforms and Application to Electrical Circuits.
### Electrical circuit analysis Unit Five
#### Filters
Filters –Low Pass –High Pass, Band Pass and Band Stop–RC, RL filters–derived filters and composite filters design –Attenuators –Principle of Equalizers –Series and Shunt Equalizers –L Type -T type and Bridged –T and Lattice Equalizers.
### Electrical circuit analysis Course Objectives
To know the analysis of three phase balanced and unbalanced circuits and to measure active and reactive powers in three phase circuits. Knowing how to determine the transient response of R-L, R-C, R-L-C series circuits for D.C and A.C excitations. To know the applications of Fourier transforms to electrical circuits excited by non sinusoidal sources. Study of Different types of filters, equalizers.
### Electrical circuit analysis Course Outcomes
Understand the analysis of three phase balanced and unbalanced circuits and to measure active and reactive powers in three phase circuits. To get knowledge about how to determine the transient response of R-L, R-C, R-L-C series circuits for D.C and A.C excitations. Applications of Fourier transforms to electrical circuits excited by non-sinusoidal sources are known. To design filtersand equalizers.
### Electrical circuit analysis Text Books
1. William Hayt, Jack E. Kemmerly and Jamie Phillips, “Engineering Circuit Analysis”, Mc Graw Hill, 9th Edition, 2019. 2. A. Chakrabarti, “Circuit Theory: Analysis & Synthesis”, Dhanpat Rai & Sons, 2008.cs, by Erwin Kreyszig, Wiley India
### Electrical circuit analysis Reference Books
1. M.E. Van Valkenberg, “Network Analysis”, 3rd Edition, Prentice Hall (India), 1980. 2. V. Del Toro, “Electrical Engineering Fundamentals”, Prentice Hall International, 2009. 3. Charles K. Alexander and Matthew. N. O. Sadiku, “Fundamentals of Electric Circuits” Mc Graw Hill, 5th Edition, 2013. 4. MahamoodNahvi and Joseph Edminister, “Electric Circuits” Schaum’s Series, 6th Edition, 2013. 5. John Bird, Routledge, “Electrical Circuit Theory and Technology”, Taylor & Francis, 5th Edition, 2014.
### Scoring Marks in Electrical circuit analysis
Information about JNTUA B.Tech R 20 Electrical circuit analysis was provided in detail in this article. To know more about the syllabus of other Engineering Subjects of JNTUH check out the official CynoHub application. Click below to download the CynoHub application.
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English is the Most used Language This Course will help... | 1,177 | 5,254 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.25 | 3 | CC-MAIN-2022-33 | latest | en | 0.840462 |
https://grist.org/article/how-hard-is-it-to-integrate-renewables-into-the-grid/ | 1,721,502,384,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763517515.18/warc/CC-MAIN-20240720174732-20240720204732-00680.warc.gz | 251,482,919 | 49,302 | In January, EEI said [PDF] that the incentives paid to renewable energy are going to jeopardize grid reliability and electricity costs. Many enviros have responded that this represents nothing more than the utility industry’s naked self interest. Who’s right?
I’d suggest neither. The true problem with the EEI report, as I pointed out here, is that they are confusing (perhaps intentionally) a pricing structure problem with a technology problem. There are technical issues, but they result from the structure of clean-energy incentives, not the technologies or the total volume of the incentive.
So there’s an easy solution: change the structure by which non-utility generators are paid. This is as logical as it is politically impossible. Utilities (like all businesses) are loath to set prices for the benefit of their competitors, and enviros are loath to concede that tax credits and renewable portfolio standards, won after hard-fought battles, might have structural flaws.
But it’s gotta happen. We have to decarbonize the grid as fast as possible without compromising cost or reliability — and the current status quo won’t get us there.
Here, then, is my effort to outline the key technical issues and how better economic signals might work. First, a few electricity basics you may remember from high-school physics:
1. Power = energy divided by time. A watt-hour is a unit of energy. A watt-hour per hour is a watt, which is a unit of power.
2. Electric devices need power and energy in precise volumes. Lighting a 60 watt bulb for 8 hours requires 60 x 8 = 480 watt hours — but if you try to zap the bulb with 480 watts for one hour, or dribble in one watt for 480 hours, it’s going to be dark.
3. Power (watts) = voltage x current.
4. Electric devices also need voltage and current at precise levels.
5. Voltage = current x resistance.
6. Electric power systems can be designed with “direct” or “alternating” current (this refers to whether current moves in one direction or oscillates). The modern electric grid uses alternating current (“AC”).
Now, let’s look at what follows from these six points.
First, energy storage and generation are complementary, not competitive, technologies. They can work well together, but neither is particularly good at doing the other’s job.
As a general rule, energy storage is a cheap source of power and an expensive source of energy, while energy generation is a cheap source of energy and an expensive source of power. This makes storage very good at providing short bursts and generation good at keeping the lights on. That’s why you use a battery to start your car but a gas engine to keep it going.
The time-dependent nature of energy and power makes this true even for technologies that haven’t been invented yet. The physical size of an energy storage technology is a function of how much energy it stores while the physical size of a generator is a function of its peak power output. Increasing physical size = investing more capital which is always expensive in time and dollars. By contrast, incremental decisions to open up a throttle a little wider are cheap, operating-level decisions. For energy storage, opening the throttle increases the energy extracted per unit time = power. For energy generation, opening the throttle brings in a little more fuel and generates more energy.
As a result, the lowest cost way to run a power grid will always be to maximally use energy storage technologies to meet peak power needs and maximally use generation technologies to meet energy needs. Energy storage can play a bigger role on the grid, but it won’t eliminate the need for a “just in time” linkage between generation and load.
This leads to a second issue: the grid must maintain and control generation that can instantaneously ramp up and down in response to changes in load. This informs generator technology selection and contract structure, in the sense that both are necessary, but independently insufficient.
Historically, this led grid managers to monitor load and direct generators to ramp up or down in response to load variation. As wind and solar penetration has risen, grid managers now also need to take into account sudden increases or decreases in power output from these weather-dependent generators. The result is to add complexity to grid operation in excess of that contemplated by current control/contracting schemes. With its high penetration of wind and hydro, the Pacific Northwest is at the cutting edge of these challenges. So far, they’ve developed lots of workarounds, from dumping “excess” power into resistor banks to filing lawsuits that challenge existing contracts, but they’ve not yet found a long-term, viable solution.
Third, maintaining grid reliability requires precise synchronization of voltage and current. Since power = volts x amps and since current (amps) oscillates in an AC system, voltage has to oscillate in precisely the same way. To see this visually, first consider a 60 amp, 120 volt circuit operating in perfect synchrony.
Watts = volts x amps, so power peaks at (120 V x 60 A) = 7,200 W (7.2 kW).
Now look at the same circuit, but with the current slightly out of phase with the voltage:
The “real” power that comes out of this circuit is still the product of volts and amps, but since the volts and amps now peak at different times, we’re getting less power than we were before (6.7 kW in this example). The ratio of actual power to theoretical power is the “power factor,” and typically runs between 85-95%. As power factor falls, generators still make the same amount of power and burn the same amount of fuel, but less gets to the load, so the effect is to lower system-wide fuel efficiency.
Motors, capacitors, and other electrical devices cause current to shift out of phase with voltage, so power factor degradation is unavoidable and grid managers must take actions to correct. The most effective way to correct is with power plants that are sited near the load and use spinning generators that can maintain constant frequency but independently shift current and voltage to offset grid degradation. This is pretty easy with any power plant that naturally spins at 2000–7000 rpm. As it turns out, this is exactly the speed that steam turbines, reciprocating engines, and gas turbines normally operate.
Unfortunately, lots of emerging generation technologies would prefer to run at lower speeds (windmills), higher speeds (microturbines), or don’t spin at all (solar panels, fuel cells). That’s not a particularly big deal, except that as these new technologies serve an ever greater portion of the load, it gets relatively harder to maintain high system power factors.
Finally, location matters. The resistance of a wire is a direct function of the length of the wire. Since voltage is the product of current and resistance, the more wires that separate a generator and the load, the greater the current (and therefore, energy) loss through that wire for any given voltage. These line losses typically run 3-5% on average, but increase dramatically during peak periods when wires are congested, often exceeding 20%. This means not only that we have to burn more fuel to generate the same amount of useful energy, but also that we must over-invest in the power generation capacity of any system with a preponderance of remote generators.
Implications
The bottom line is that not all MWh are equal:
• It takes fewer MWh of generation to serve a MWh of load if that MWh is generated near the load.
• The ability to produce (or curtail) peak power output at a moment’s notice is valuable regardless of actual MWh generated.
• 1 MWh from a generator that can boost system power factor is worth more than 1 MWh to the system. The reverse is also true.
A key point is that none of these values depend on MWh output, nor do they depend on the fuel used or power plant ownership. They depend solely on location, dispatchability, and generator technology. And yet most of the ways the rules reward generators are with \$/MWh payments (PTCs, RECs, etc.) that are a function of fuel use and whether or not your ownership structure allows you to monetize tax attributes.
Which means that EEI is right when they say that current incentives for renewable energy are leading to sub-optimal capital allocation. But that’s not because renewable energy is a dangerous thing — it’s because it’s compensated in the wrong way. In the course of putting a premium on clean energy to try and monetize externalities, we’ve created a set of economic incentives that don’t map very well against the economics of grid operation. That’s a fixable problem — but only if we first admit that there are legitimate technical issues that can be addressed with better economic signals. Price it right, and they will come. This would be an ideal first initiative for new Chairman Binz. | 1,856 | 8,902 | {"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} | 2.53125 | 3 | CC-MAIN-2024-30 | latest | en | 0.930953 |
http://www.qacollections.com/Help-in-Math-for-Simplifying-Fractions-to-Obtain-a-Single-Fraction | 1,524,436,673,000,000,000 | text/html | crawl-data/CC-MAIN-2018-17/segments/1524125945660.53/warc/CC-MAIN-20180422212935-20180422232935-00167.warc.gz | 491,939,481 | 5,702 | # Help in Math for Simplifying Fractions to Obtain a Single Fraction?
Simplifying fractions is an integral concept in math that continues to be important throughout upper-level classes. There are several pointers that can help you to quickly and efficiently simplify ... Read More »
http://www.ehow.com/info_8358664_math-fractions-obtain-single-fraction.html
Top Q&A For: Help in Math for Simplifying Fractions to Obtai...
## How to Use Simplifying for Mixed Fractions?
Mixed fractions (also known as mixed numbers) tend to throw a curve at students who are still struggling to master proper fractions. A mixed fraction is a number consisting of a whole number and a ... Read More »
http://www.ehow.com/how_8295042_use-simplifying-mixed-fractions.html
## Hands-on Activities for Simplifying Fractions?
Fractions have been the bane of many people's math learning process. There are many activities that can be done with students to help them get a better understanding of how fractions work and why t... Read More »
http://www.ehow.com/info_8099182_handson-activities-simplifying-fractions.html
## How to Find a Fraction Between Two Fractions?
Some rules of fractional math were meant to be broken. A fraction is a value in two parts. Each part, the numerator or denominator, is an integer. The numerator is the fraction's top number, while ... Read More »
http://www.ehow.com/how_8485547_fraction-between-two-fractions.html
## How To Find Two Fractions That Are Equivalent to Another Fraction?
Equivalent fractions are fractions are equal to each other but have different numerators and denominators. The fractions 1/2 and 2/4, for example, are equivalent fractions. To illustrate this point... Read More »
http://www.ehow.com/how_8369168_two-fractions-equivalent-another-fraction.html
Related Questions | 404 | 1,814 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.09375 | 4 | CC-MAIN-2018-17 | latest | en | 0.904492 |
https://www.tutorialkart.com/python/python-divmod/ | 1,713,614,277,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817576.41/warc/CC-MAIN-20240420091126-20240420121126-00125.warc.gz | 951,245,898 | 18,944 | ## Python divmod()
Python divmod() builtin function takes two numbers as arguments, computes division operation, and returns the quotient and reminder.
In this tutorial, we will learn about the syntax of Python divmod() function, and learn how to use this function with the help of examples.
## Syntax
The syntax of divmod() function is
`divmod(a, b)`
where
Note: `a` or `b` cannot be complex numbers.
Returns
The function returns two numbers of type integer/float based on the arguments.
## Examples
### 1. divmod(22, 4)
In this example, we will find the quotient and reminder for the dividend 22 and divisor 4.
Python Program
```a = 22
b = 4
q, r = divmod(a, b)
print(f'Quotient is {q}')
print(f'Reminder is {r}')```
Try Online
Output
```Quotient is 5
Reminder is 2```
### 2. divmod(22, 0) – ZeroDivisionError
In this example, we will find the quotient and reminder for the dividend 22 and divisor 0. Since, the divisor is zero, divmod() throws ZeroDivisionError.
Python Program
```a = 22
b = 0
q, r = divmod(a, b)
print(f'Quotient is {q}')
print(f'Reminder is {r}')```
Try Online
Output
```Traceback (most recent call last):
File "d:/workspace/python/example.py", line 3, in <module>
q, r = divmod(a, b)
ZeroDivisionError: integer division or modulo by zero```
### 3. divmod() with Floating Point Number
In this example, we will find the quotient and reminder for the dividend 2.2 and divisor 0.4.
Python Program
```a = 2.2
b = 0.4
q, r = divmod(a, b)
print(f'Quotient is {q}')
print(f'Reminder is {r}')```
Try Online
Output
```Quotient is 5.0
Reminder is 0.20000000000000007```
## Conclusion
In this Python Tutorial, we have learnt the syntax of Python divmod() builtin function, and also learned how to use this function, with the help of Python example programs. | 515 | 1,798 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.984375 | 3 | CC-MAIN-2024-18 | latest | en | 0.728739 |
http://electronics2electrical.com/tag/frequency | 1,563,538,185,000,000,000 | text/html | crawl-data/CC-MAIN-2019-30/segments/1563195526237.47/warc/CC-MAIN-20190719115720-20190719141720-00479.warc.gz | 49,822,210 | 18,621 | # Recent questions tagged frequency
Draw the diagram and explain the working of ferrodynamic type frequency meter.
List the different types of frequency meters.
Explain working of Weston type frequency meter.
State the application of phase sequence indicator; clip on ammeter, frequency meter & P.F. meter.
List two types of oscillator, which generates frequency in RF range.
State the need of multistage amplifier. Draw frequency response of R-C coupled amplifier.
Classify the transformers based on the construction, frequency and power.
Define: 1) frequency, 2) cycle, 3) time period, 4) amplitude
A pulse radar has a pulse repetition frequency (PRF) of 3 KHz. If the radar is emitting at 10 GHz, what is the maximum unambiguous range for this radar? A) 5 Km B) 5 m C) 50 Km D) 15 m
A pulse compression radar transmits an encoded pulse having a frequency chirp from 990 MHz to 1010 MHz centred around 1 GHz. The range resolution capability of this radar is A) 7.5 m B) 7.5 Km C) 75 m D) 750 m
The clock frequency of an 8085 microprocessor based system is 3 MHz. What should be the minimum pulse width of the INTR signal so that it is recognized successfully? A) 5.6 µs B) 5.7 µs C) 5.8 µs D) 5.9 µs
What is the frequency range of VHF electromagnetic waves? A) 30-300 kHz B) 30-3000 kHz C) 30-300 MHz D) 30-3000 MHz
The SCR would be turned OFF by voltage reversal of applied anode-cathode ac supply frequency of A) 10 kHz B) 10 Hz C) 5 kHz D) 5 Hz
Which of the following thyristors can be used for high frequency of operation? A) TRIAC B) DIAC C) SCR D) GTO
In a series R-C circuit, with the increase in frequency, current (A) Reduces (B) Increases (C) Does not change (D) Becomes zero
A series R-L-C Circuit has a resonant frequency of 1000 Hz. If the values of R, L and C are doubled, the new resonant frequency is
What would happen if a power transformer designed for operation on 50 Hz (frequency) were connected to a 500 Hz (frequency) source of the same voltage: (A) Eddy Current and Hysteresis losses will be excessive (B) Current will be too much high (C) Transformer may start to smoke and burn (D) No effect
The speed of a three-phase induction motor is controlled by variable voltage variable frequency control (i.e. keeping V/f constant). As the frequency is reduced, the slip at maximum torque (1) Decreases (2) Increases (3) Remains constant (4) None of the above
For measurement of radio frequency ac signals __________ type meter is used
In a series RLC high Q circuit, the current peaks at a frequency (1) equal to the resonant frequency (2) greater than the resonant frequency (3) less than the resonant frequency (4) equal to half the resonant frequency
In a self-controlled synchronous motor fed from a variable frequency inverter (1) the rotor poles invariably have damper windings (2) there are stability problems (3) the speed of the rotor decides stator frequency (4) the frequency of the stator decides the rotor speed
For a 500 Hz frequency excitation, a 50 km long power line will be modeled as
What is the frequency of an alternator which has 2 poles and makes 3600 revolutions per minute? A) 3600 HZ B) 60 Hz C) 120 Hz D) None of these
What happens to the voltage in the secondary when the frequency of the supply in the primary is decreased? A) It increases B) It decreases C) It remains the same D) Cannot say unless the actual percentage increase in the frequency is known
The device which allows reverse power flow and withstands highest switch frequency is (A) GTO (B) MOSFET (C) IGBT (D) Inverter grade SCR
The time period of third harmonic of power supply frequency, in milliseconds, is (A) 20 (B) 3.33 (C) 6.66 (D) 60
Two waves have frequency of 500 Hz. One is set at its maximum value whereas the other at zero. The phase angle between them will be
A 4 GHz carrier is amplitude modulated by a low-pass signal of maximum cut off frequency 1 MHz. lf this signal is to be ideally sampled, the minimum sampling frequency should be nearly
In a 741 Op-Amp, there is 20 dB/decade fall-off starting at a relatively low frequency. This is due to the (A) applied load (B) internal compensation (C) impedance of the source (D) power dissipation in the chip
An accelerometer has input range of 0 to 10g, natural frequency 30 Hz and mass 0.001 kg. The range of the secondary displacement transducer in mm required to cover the input range is (A) 0 to 2.76 (B) 0 to 9.81 (C) 0 to 11.20 (D) 0 to 52.10
The frequency of an alternating current is?
The effective resistance of an iron-cored choke working on ordinary supply frequency is more than its true resistance because of?
In a series L-C circuit at the resonant frequency the?
In R-L-C series resonant circuit magnitude of resonance frequency can be changed by changing the value of?
For a frequency of 200 Hz, the time period will be?
Two waves of the same frequency have opposite phase when the phase angle between them is?
The maximum constant load torque under which a synchronous motor will pull into synchronism at rated rotor supply voltage and frequency is known as?
The maximum torque which a synchronous motor will develop at rest for any angular position of the rotor, at rated stator supply voltage and frequency, is known as?
For given applied voltage, with the increase in frequency of the applied voltage?
Is the speed of alternator is directly proportional to frequency?
Which capacitor is used for high frequency? | 1,366 | 5,460 | {"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-2019-30 | latest | en | 0.865416 |
https://en.wikibooks.org/wiki/Principles_of_Finance/Section_1/Chapter_7/Port/Correlation | 1,524,588,205,000,000,000 | text/html | crawl-data/CC-MAIN-2018-17/segments/1524125946807.67/warc/CC-MAIN-20180424154911-20180424174911-00598.warc.gz | 603,584,364 | 10,771 | # Principles of Finance/Section 1/Chapter 7/Port/Correlation
## A two security portfolio
When building a portfolio, it is important not only to consider the individual securities in a portfolio, but also how they interact with one another. For example, consider a portfolio that contains two stocks, XYZ and ABC.
The following formula finds the variance of a portfolio using the correlation (rho) between two stocks in a portfolio:
${\displaystyle \sigma _{portfolio}^{2}=x_{1}^{2}\sigma _{1}^{2}+x_{2}^{2}\sigma _{2}^{2}+2(x_{1}x_{2}\rho _{12}\sigma _{1}\sigma _{2})}$
Where x represents the weight of each security in the portfolio.
So, let us suppose the following information:
XYZ ABC $50$60 20 15 0.02 0.04 0.1414 0.2
And the correlation between the two is 0.25
Find the variance of this portfolio.
The first thing we must do is determine the relative weights of each stock. We see that we have \$1000 in XYZ, and 900 in ABC. Therefore x1=0.53 and x2=0.47. The other relevant information is listed, so we simply plug it into the equation:
${\displaystyle \sigma _{p}^{2}=0.53^{2}(0.02)+.47^{2}(0.04)+2(0.53*0.47*0.25*0.1414*0.2)=0.01798}$
Now let us say we were trying to find the Beta of the XYZ company, and we know that the market has a variance of 0.04. The following formula will allow us to do so:
${\displaystyle \beta _{i}={\frac {\sigma _{im}}{\sigma _{m}^{2}}}}$
First, we will need to find the covariance (${\displaystyle \sigma _{12}}$) between the market and XYZ. For that, we will use the following formula:
${\displaystyle \sigma _{12}=\rho _{12}\sigma _{1}\sigma _{2}}$
For this exercise we will need to know the correlation between the market and XYZ, which we will say is 0.75. Therefore:
${\displaystyle \sigma _{12}=0.75*0.1414*0.2=0.02121}$
Applying our covariance of 0.02121 to the formula for beta, we find that:
${\displaystyle \beta _{i}={\frac {0.02121}{0.04}}=0.53}$ | 576 | 1,917 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 7, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.1875 | 4 | CC-MAIN-2018-17 | longest | en | 0.873791 |
https://everything2.com/user/SpudTater/writeups/best-first+search | 1,685,349,128,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224644817.32/warc/CC-MAIN-20230529074001-20230529104001-00180.warc.gz | 290,299,445 | 6,958 | A method of searching a tree that uses a heuristic to estimate how close to the goal it is at any one moment. It is known as an informed search, as opposed to the uninformed searches. (depth-first, breadth-first and iterative deepening searches.)
Of course, this requires that the tree is sorted into some semblence of order, otherwise best-first search will be useless. Best-first is often used in problem solving; e.g. finding the best move in a chess game, or finding the way through a maze.
In agenda-based searching, best-first search is performed by the following steps:
• Add the first (or root) node to the agenda.
• Examine the first node on the agenda to see if it is the required item.
• Otherwise, expand the node (find all its children), and add those children to the agenda.
• Sort the agenda by heuristic value so that the most promising nodes come first.
• Repeat the second to fourth steps until the goal is reached.
A best-first search often finds solutions more quickly than a breadth-first or depth-first search, since the algorithm has some idea of the direction to head in (thanks to the sorting of the agenda). However, it will not always find the shortest solution. Of course, the effectiveness of best-first search depends entirely on the suitability of its heuristic; if you have a bad heuristic, you might as well not take the time sorting the agenda.
More complex versions of the best-first search are known as Algorithm A and A*. | 305 | 1,462 | {"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} | 2.515625 | 3 | CC-MAIN-2023-23 | latest | en | 0.934725 |
http://www.pfhub.com/understanding-earnings-per-share-eps/ | 1,394,573,096,000,000,000 | text/html | crawl-data/CC-MAIN-2014-10/segments/1394011278661/warc/CC-MAIN-20140305092118-00031-ip-10-183-142-35.ec2.internal.warc.gz | 459,058,984 | 23,042 | # Understanding Earnings Per Share (EPS)
The earnings per share metric is a valuable tool for investors to use when investigating the profitability of a company against its outstanding equity. From there, the components that come together to make this metric are as useful to investors as the output that they create, because it allows us to understand how it is that different earnings and dilution scenarios would change a single share’s claim against the company’s overall earnings, depending on the nature of the securities that are being used to finance this dilution.
Calculating earnings per share is simple, you take a company’s net incomes, you subtract out preferred share dividends, and then you divide the result by the average number of shares outstanding over the course of the year. The preferred dividends are subtracted out from the profits to show only those earnings that are available to common shareholders, and therefore give us an understanding of what an investor’s nominal claim to the company is. However, if the company changes the volume of shares it has issued throughout the year, it will also change the way in which we calculate actual number of shares used in the equation.
There are three main ways for a company to change the average number of shares outstanding during the year: a buyback, an issuance, or a split/reverse-split. In a split/reverse-split situation, the net effect on proportionate shareholder ownership remains constant, because everyone still owns the same ratio of shares to one-another. This means that a split doesn’t change an individual investor’s EPS claim, but instead impacts the EPS that an incremental purchase would have access to (because you’d need to buy a greater volume of shares to have a similar claim).
In those instances where a company chooses to issue or buy-back shares part-way through a year, it is important to take into account the weighted impact of such an event. This is accomplished by multiplying the change in share volume by the proportionate date of the change. For example, if a company bought back 1,000 shares in March, we’d multiply 1,000x(3/12), because March is the third month out of 12. This shows us the weighted number of shares bought back over the course of the year, and ensures that we are calculating a final result that accurately represents the time value of shares outstanding. | 467 | 2,387 | {"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-2014-10 | longest | en | 0.951091 |
https://quizgecko.com/q/1-14-divided-by-4-yeozp | 1,726,884,729,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725701427996.97/warc/CC-MAIN-20240921015054-20240921045054-00652.warc.gz | 439,191,271 | 33,339 | # 1 1/4 divided by 4
#### Understand the Problem
The question is asking us to perform a division operation involving a mixed number and a whole number. We will convert the mixed number into an improper fraction and then divide it by the whole number.
$$\frac{5}{16}$$
The final answer is ( \frac{5}{16} )
#### Steps to Solve
1. Convert the mixed number to an improper fraction
Take the mixed number $1 \frac{1}{4}$ and convert it to an improper fraction:
$$1 \frac{1}{4} = \frac{1 \cdot 4 + 1}{4} = \frac{4 + 1}{4} = \frac{5}{4}$$
1. Write the division as a multiplication by the reciprocal
Dividing by 4 is the same as multiplying by the reciprocal of 4, which is $\frac{1}{4}$. So, we rewrite the division:
$$\frac{5}{4} \div 4 = \frac{5}{4} \times \frac{1}{4}$$
1. Multiply the fractions
Multiply the numerators and the denominators:
$$\frac{5 \cdot 1}{4 \cdot 4} = \frac{5}{16}$$
1. Simplify the fraction if needed
The fraction $\frac{5}{16}$ is already in its simplest form.
The final answer is $\frac{5}{16}$
The final answer is ( \frac{5}{16} ) | 329 | 1,069 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 2, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.8125 | 5 | CC-MAIN-2024-38 | latest | en | 0.815618 |
https://justaaa.com/finance/155641-different-cash-flowgiven-the-following-cash | 1,721,578,627,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763517747.98/warc/CC-MAIN-20240721152016-20240721182016-00412.warc.gz | 296,558,122 | 10,093 | Question
# Different cash flow. Given the following cash inflow at the end of each year, what is...
Different cash flow. Given the following cash inflow at the end of each year, what is the future value of this cash flow at 4 %, 8 %, and 17% interest rates at the end of year 7? Year 1: \$12,000 Year 2: \$22,000 Year 3: \$30,000 Years 4 through 6: \$0 Year 7: \$150,000
Future value. A speculator has purchased land along the southern Oregon coast. He has taken a loan with the end-of-year payments of \$8,000 for 8 years. The loan rate is 5%. At the end of 8 years, he believes that he can sell the land for \$60,000. If he is correct on the future price, did he make a wise investment?
#### Homework Answers
Answer #1
future value of cashflows @4% at the end of 7 year ,required equation will be
=12000(1.04)^6 +22000(1.04)^5 +30000(1.04)^4 + 150000
=227045.94
future value of cashflow @8% at the end of 7 year
= 1200091.08)^6 +22000(1.08)^5 +30000(1.08)^4 + 150000
=242182.3784
future value of cashflows@17% at the end of 7 year
= 12000(1.17)^6 + 22000(1.17)^5 +30000(1.17)^4 +150000
=285232.4435
(b) as he take the loan with the end year payments the present value of the of the stream @5% will be
=8000*(1-(1.05)^-8/0.05)
=51705.70 this is the loan amount for the land and he sold the land for 60000 after 8 years this is not a wise investment as we find the future value of the loans amount after 8 years and it will be 51705.70(1.05)^8=76392.86 but he sold the land for 60000
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Ask your own homework help question | 566 | 1,747 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.59375 | 4 | CC-MAIN-2024-30 | latest | en | 0.900917 |
https://forum.mrmoneymustache.com/antimustachian-wall-of-shame-and-comedy/instead-of-a-regular-battery-here-have-a-highly-inefficient-battery!!/ | 1,606,795,024,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141542358.71/warc/CC-MAIN-20201201013119-20201201043119-00605.warc.gz | 299,476,558 | 9,534 | ### Author Topic: Instead of a regular battery here have a highly inefficient battery!! (Read 3346 times)
#### jj20051
• Posts: 51
##### Instead of a regular battery here have a highly inefficient battery!!
« on: March 10, 2015, 08:52:23 PM »
I saw this and thought it was pretty anti-mustachian:
http://www.aravua.com/2015/03/kraftwerk-highly-innovative-portable.html
The math:
The retail price is going to be \$149.00. (Versus a \$50 rechargeable battery pack on amazon).
Each canister costs \$4 - \$8 dependent on shipping and who you get it from and charges your phone up to 11 times. (Versus that battery pack which can charge your phone the same number of times and do it for as little as \$0.0428 (Took the cost to charge an iPhone over a year, divided it out by 365 days and multiplied it by 11, then took that number and multiplied it by 10 to account for battery inefficiency. If my math is wrong please correct me. ))
Summary:
Even if you're going to be in the middle of nowhere for weeks you can just buy extra battery packs, get a hand crank power station or solar station for MUCH cheaper. Devices like this are 100% pure waste.
This is just a quick synopsis of my full rant here.
« Last Edit: March 10, 2015, 09:55:12 PM by jj20051 »
#### EricL
• Guest
##### Re: Instead of a regular battery here have a highly inefficient battery!!
« Reply #1 on: March 11, 2015, 12:23:17 AM »
Agreed. It's pointless. Plus if you're off the grid there's no internet so half the usefulness of those gadgets is gone.
#### TheAnonOne
• Handlebar Stache
• Posts: 1622
##### Re: Instead of a regular battery here have a highly inefficient battery!!
« Reply #2 on: March 11, 2015, 11:03:15 AM »
At least from the article you linked, you can just put gas into this and it will charge your device 11 times?? Where are the \$4-\$8 canisters coming from?
Looks cool to me...
#### Cromacster
• Handlebar Stache
• Posts: 1697
• Location: Minnesnowta
##### Re: Instead of a regular battery here have a highly inefficient battery!!
« Reply #3 on: March 11, 2015, 11:08:58 AM »
Things like these can be a game changer in developing countries and countries with poor electrical infrastructure.
#### Syonyk
• Magnum Stache
• Posts: 3999
##### Re: Instead of a regular battery here have a highly inefficient battery!!
« Reply #4 on: March 11, 2015, 11:43:49 AM »
So, I understood the "11 charges" as the capacity of the charger alone, before you needed to refill it from an external canister.
https://www.kickstarter.com/projects/265641170/kraftwerk-highly-innovative-portable-power-plant indicates 11 charges "per filling," so you'd have many, many charges per external canister.
According to the weight, it holds 40 grams of fluid (for the 11 charges).
Elsewhere on the page, you can get 12 external canisters (1kg of fuel). So one of the small external canisters is ~80g, or 22 charges.
And if you don't care about "renewably produced gas," you can get refills a LOT cheaper. I found a 10oz can of butane for \$5 (http://www.cubancrafters.com/butane-refill-for-lighters-ultra-5x-refined-fuel-for-gas-lighters-universal-adapters/, just the first thing I happened across) - that's 283 grams of butane, or ~80 "charges" for \$5. Again, not amazing, but given that you don't have to carry solar with you and butane/propane is easily available, that's not horrible.
I entirely agree with the OP that it's nonsensical for most people in developed countries (and I have no idea why so many people want this when a \$40 battery pack does the same job in 99.9% of use cases), but it is a genuinely useful device in terms of pushing mobile devices out into developing countries that may not have power. A lot of the Kickstarter stuff tends this way (the Biolite stoves are a good example), so I don't have a huge problem with it.
... that said, I looked at it for a while, did some math, and concluded that it wasn't of any value to me, so I didn't order one, for much the same reasons listed. And I play a lot of Ingress, so mobile power in desolate areas is a very useful thing to me. Just, my battery pack is a better solution, for me.
#### v10viperbox
• Posts: 48
##### Re: Instead of a regular battery here have a highly inefficient battery!!
« Reply #5 on: March 11, 2015, 11:57:53 AM »
Agreed. It's pointless. Plus if you're off the grid there's no internet so half the usefulness of those gadgets is gone.
Sat phones, camera while traveling, my GPS. I have been in places where you absolutely need juice for safety sake and not been around a functioning electric grid and something like this would be a godsend. Batteries are insanely heavy when you are backpacking as well, third world countries with different phase power.
#### jj20051
• Posts: 51
##### Re: Instead of a regular battery here have a highly inefficient battery!!
« Reply #6 on: March 11, 2015, 03:30:33 PM »
@v10viperbox - I mentioned in my article that you could use a solar panel instead. Light weight ones (2 Lbs or less) are just as useful and don't require refilling.
#### Syonyk
• Magnum Stache
• Posts: 3999
##### Re: Instead of a regular battery here have a highly inefficient battery!!
« Reply #7 on: March 11, 2015, 03:50:18 PM »
A solar panel requires sun and either a battery pack to store it, or to hang out in the sun charging for a while. Also, it depends on where you're traveling. I've tried various solar phone chargers in the Seattle area, and they're junk 9 months of the year.
#### jj20051
• Posts: 51
##### Re: Instead of a regular battery here have a highly inefficient battery!!
« Reply #8 on: March 11, 2015, 04:29:53 PM »
@Syonyk - There are decent solar chargers which will charge a phone in 6 - 8 hours even in bad weather. In reality even if you're camping out you can set up the charger to recharge the battery pack and then use the battery to recharge your phone at night... Either way the device is perfectly fine for people who live in the middle of nowhere most of the time... the average person? not so much.
The ad they created shows some dude sitting in a coffee shop charging his iPaid with the pack. Three inches away is a socket... good lord.
#### Syonyk
• Magnum Stache
• Posts: 3999
##### Re: Instead of a regular battery here have a highly inefficient battery!!
« Reply #9 on: March 11, 2015, 04:34:28 PM »
That's fair. The solar I've worked with simply doesn't work unless it's in pretty direct sun. Do you have links to the ones that will charge in bad weather?
And, yes, I agree, coffee shops are a silly place for this. | 1,711 | 6,567 | {"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} | 2.640625 | 3 | CC-MAIN-2020-50 | latest | en | 0.894987 |
https://math.stackexchange.com/questions/222907/primes-in-gaussian-integers | 1,566,464,975,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027317037.24/warc/CC-MAIN-20190822084513-20190822110513-00448.warc.gz | 557,351,795 | 32,651 | # Primes in Gaussian Integers
Let $p$ be a rational prime. It is is well known that if $p\equiv 3\;\;mod\;4$, then $p$ is inert in the ring of gaussian integers $G$, that is, $p$ is a gaussian prime. If $p\equiv 1\;mod\;4$ then $p$ is decomposed in $G$, that is, $p=\pi_1\pi_2$ where $\pi_1$ and $pi_2$ are gaussian primes not associated. The rational prime $2$ ramifies in $G$, that is $2=u\pi^2$, where $u$ is a unit in $G$ and $\pi$ a prime in $G$.
where can I find a proof of this fact? I want a direct proof, not a proof for the quadratic integers and then deduce this as a particular case.
Define the norm $N(a+bi) = a^2 + b^2$. It is straightfoward to check it is multiplicative. Now, consider a prime $p \equiv 3 \pmod{4}$. Suppose it is not inert, i.e. it factors as $p = \alpha \beta$ for some $\alpha, \beta \in \mathbb{Z}[i]$ and neither of them are units.. Then it is easy to see $N(\alpha) = N(\beta) = p$. However, the equation $a^2 + b^2 = p$ has no solutions modulo $4$, therefore we have reached a contradiction so primes $3 \pmod{4}$ remain inert.
Now to prove primes $1 \pmod{4}$ split. It suffices to show $p = a^2 + b^2$ has a solution where $a,b$ are integers. Let $z$ denote the least value of $\sqrt{-1} \pmod{p}$. Now define $\mathcal L = \{(a,b) \in \mathbb{Z}^2 | a \equiv zb \pmod{p}\}$. It is straightfoward to check $\mathcal L$ is a lattice whose fundamental parallelogram has area $p$. Now by Minkowski's theorem one has $\mathcal L$ contains a nontrivial lattice point inside the circle $x^2 + y^2 < 2p$. Call this point $(a,b)$. But then $a^2 + b^2 \equiv 0 \pmod{p}$ based on the definition of the lattice, thus it must be $a^2 + b^2 = p$. But then $p = (a+bi)(a-bi)$, proving it factors so we are done.
Proving $2$ ramifies is trivial, since it's just $2 = -i(1+i)^2$. | 609 | 1,808 | {"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} | 3.953125 | 4 | CC-MAIN-2019-35 | latest | en | 0.857511 |
https://www.insideaiml.com/blog/Python---Hash-Table-438 | 1,627,116,348,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046150134.86/warc/CC-MAIN-20210724063259-20210724093259-00657.warc.gz | 838,425,628 | 30,822 | Machine Learning with Python & Statistics
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Python - Hash Table
Jessi Day
10 months ago
Python - Hash Table | Insideaiml
Hash tables are a type of data structure in which the address or the index value of the data element is generated from a hash function. That makes accessing the data faster as the index value behaves as a key for the data value. In other words, Hash table stores key-value pairs but the key is generated through a hashing function.
So the search and insertion function of a data element becomes much faster as the key values themselves become the index of the array which stores the data.
In Python, the Dictionary data types represent the implementation of hash tables. The Keys in the dictionary satisfy the following requirements.
• The keys of the dictionary are hashable i.e. they are generated by hashing function which generates unique result for each unique value supplied to the hash function.
• The order of data elements in a dictionary is not fixed.
So we see the implementation of hash table by using the dictionary data types as below.
Accessing Values in Dictionary
To access dictionary elements, you can use the familiar square brackets along with the key to obtaining its value.
``````
# Declare a dictionary
dict = {'Name': 'Zara', 'Age': 7, 'Class': 'First'}
# Accessing the dictionary with its key
print("dict['Name']: ", dict['Name'])
print("dict['Age']: ", dict['Age'])
``````
When the above code is executed, it produces the following result
``````dict['Name']: Zara
dict['Age']: 7``````
Updating Dictionary
You can update a dictionary by adding a new entry or a key-value pair, modifying an existing entry, or deleting an existing entry as shown below in the simple example −
``````
# Declare a dictionary
dict = {'Name': 'Zara', 'Age': 7, 'Class': 'First'}
dict['Age'] = 8; # update existing entry
dict['School'] = "DPS School"; # Add new entry
print("dict['Age']: ", dict['Age'])
print("dict['School']: ", dict['School'])
``````
When the above code is executed, it produces the following result
``````dict['Age']: 8
dict['School']: DPS School
``````
Delete Dictionary Elements
You can either remove individual dictionary elements or clear the entire contents of a dictionary. You can also delete the entire dictionary in a single operation. To explicitly remove an entire dictionary, just use the del statement. −
``````
dict = {'Name': 'Zara', 'Age': 7, 'Class': 'First'}
del dict['Name']; # remove entry with key 'Name'
dict.clear(); # remove all entries in dict
del dict ; # delete entire dictionary
print("dict['Age']: ", dict['Age'])
print("dict['School']: ", dict['School'])
``````
This produces the following result. Note that an exception is raised because after del dict dictionary does not exist any more −
``````
dict['Age']:
Traceback (most recent call last):
File "test.py", line 8, in
print "dict['Age']: ", dict['Age'];
TypeError: 'type' object is unsubscriptable
`````` | 714 | 3,016 | {"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} | 2.71875 | 3 | CC-MAIN-2021-31 | longest | en | 0.773538 |
https://de.mathworks.com/matlabcentral/cody/problems/1239-decoding-find-the-value/solutions/201882 | 1,582,612,160,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875146033.50/warc/CC-MAIN-20200225045438-20200225075438-00179.warc.gz | 333,406,053 | 15,306 | Cody
Problem 1239. Decoding : Find the value
Solution 201882
Submitted on 7 Feb 2013 by Freddy
This solution is locked. To view this solution, you need to provide a solution of the same size or smaller.
Test Suite
Test Status Code Input and Output
1 Pass
%% Test 1 u = 3; v= 0.05; y_correct = 1020; assert(isequal(your_fcn_name(u,v),y_correct));
2 Pass
%% Test 2 u=30; v = 0.01; y_correct = [5100 4800]; assert(isequal(your_fcn_name(u,v),y_correct));
3 Pass
%% Test 3 u=3; v = 0.03; y_correct = 1700; assert(isequal(your_fcn_name(u,v),y_correct)); | 192 | 560 | {"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} | 2.5625 | 3 | CC-MAIN-2020-10 | latest | en | 0.615206 |
https://scholar.archive.org/search?q=A+basis+for+the+implicit+representation+of+planar+rational+cubic+B%C3%A9zier+curves. | 1,660,617,999,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882572215.27/warc/CC-MAIN-20220815235954-20220816025954-00547.warc.gz | 446,264,595 | 14,900 | Filters
244 Hits in 4.5 sec
### A basis for the implicit representation of planar rational cubic Bézier curves
Oliver J.D. Barrowclough
2014 Computer Aided Geometric Design
We present an approach to finding the implicit equation of a planar rational parametric cubic curve, by defining a new basis for the representation. ... The basis, which contains only four cubic bivariate polynomials, is defined in terms of the B\'ezier control points of the curve. ... I would like to thank Tor Dokken for reading through the manuscript and for helpful suggestions. I would also like to thank the anonymous reviewers for their comments and corrections. ...
### Geometric Modeling [chapter]
Shashi Shekhar, Hui Xiong
2008 Encyclopedia of GIS
For example, a 90 degrees arc as a rational Bezier curve: P1 w0 P0 B02 (t ) + w1P1B12 (t ) + w2 P2 B22 (t ) The B-spline functions are then ... by introducing a more powerful representation functions is non-zero of rational curves ...
### A Review on Approaches for Handling Bezier Curves in CAD for Manufacturing
Hetal N. Fitter, Akash B. Pandey, Divyang D. Patel, Jitendra M. Mistry
2014 Procedia Engineering
A review is done on various approaches for handling Bezier curve in Computer Aided Design for the purpose of manufacturing for example tool path optimization, profile design, reverse engineering, etc. ... Various techniques and methodologies like curve fitting, curve manipulation, blending and merging of curves have been proposed over the years for better handling and enhancing Bezier curve use with every ... This made the entire Voronoi edge consisting of implicit bisectors to be converted into rational quadratic Bezier curve for unified representation. ...
### Page 4321 of Mathematical Reviews Vol. , Issue 96g [page]
1996 Mathematical Reviews
Summary: “An explicit expression for the unique implicitization of any planar rational cubic Bézier curve is presented, the only exception being when the tangents at the ends of the curve are parallel. ... The representation is in the form of a cubic Bernstein- Bézier triangle and is invariant under affine transformations of the curve. ...
### PHYSICS BASED GEOMETRIC DESIGN
HONG QIN
1996 International journal of shape modeling
We demonstrate that D-NURBS can not only serve as a basis for the future research o f physics-based geometric design but also become readily appropriate for a large variety of important applications in ... In this paper, we s u r v ey the diversity of shape representations ranging from the primitive polynomial to the sophisticated Non-Uniform Rational B-Spline (NURBS). ... Chou and Piegl used piecewise cubic rational Bezier splines to t a set of data points and their tangent directions in the least-square sense 19]. ...
### Resolution independent curve rendering using programmable graphics hardware
Charles Loop, Jim Blinn
2005 ACM Transactions on Graphics
A simple implicit equation for a parametric curve is found in a space that can be thought of as an analog to texture space. ... When the triangle(s) corresponding to the Bézier curve control hull are rendered, a pixel shader program evaluates the implicit equation for a pixel's interpolated texture coordinates to determine an inside ... Integral Cubics Solving for the roots of I(t, s) will, for arbitrary rational cubic curves, require solving a cubic equation. ...
### Resolution independent curve rendering using programmable graphics hardware
Charles Loop, Jim Blinn
2005 ACM SIGGRAPH 2005 Papers on - SIGGRAPH '05
A simple implicit equation for a parametric curve is found in a space that can be thought of as an analog to texture space. ... When the triangle(s) corresponding to the Bézier curve control hull are rendered, a pixel shader program evaluates the implicit equation for a pixel's interpolated texture coordinates to determine an inside ... Integral Cubics Solving for the roots of I(t, s) will, for arbitrary rational cubic curves, require solving a cubic equation. ...
### A geometric characterization of parametric cubic curves
Maureen C. Stone, Tony D. DeRose
1989 ACM Transactions on Graphics
In this paper, we analyze planar parametric cubic curves to determine conditions for loops, cusps, or inflection points. ... Such a characterization forms the basis for an easy and efficient solution to this problem. ... Su and Liu [ll] have presented a specific geometric solution for the Bezier representation, and Forrest [6] has studied rational cubic curves. ...
### Page 4587 of Mathematical Reviews Vol. , Issue 2003f [page]
2003 Mathematical Reviews
In particular, a family of si- nusoidal spirals, including the cubic curves of Tschirnhausen, are identified as polar Bézier curves. ... The derivation formulae, basis function properties, envelope theorems of the L-surface and an equivalent representation of the L-surface as a Bezier surface are presented. In particular, a C? ...
### Geometries for CAGD [chapter]
Helmut Pottmann, Stefan Leopoldseder
2002 Handbook of Computer Aided Geometric Design
For the use of algebraic geometry in geometric design, the reader is referred to chapter ?? on implicit surfaces. ... This holds especially for subdivision curves and surfaces (chapter ??) and multiresolution techniques (chapter ??), where discrete models of curves and surfaces play a fundamental role. ... Planar cubic Bézier curve via projection of a cubic normal curve frame point F i (see Figure 2 ). ...
### Envelope computation in the plane by approximate implicitization
Tino Schulz, Bert Jüttler
2011 Applicable Algebra in Engineering, Communication and Computing
Given a rational family of planar rational curves in a certain region of interest, we are interested in computing an implicit representation of the envelope. ... This connection enables us to use approximate implicitization for computing the (exact or approximate) implicit representation of the envelope. ... This research has been supported by the Austrian Science Fund (FWF) in the frame of the FSP S092 "Industrial Geometry", subproject 2. ...
### An Efficient Trim Structure for Rendering Large B-Rep Models [article]
Frédéric Claux, David Vanderhaeghe, Loïc Barthe, Mathias Paulin, Jean-Pierre Jessel, David Croenne
2012 International Symposium on Vision, Modeling, and Visualization
To get a good tradeoff between performance and visual accuracy, we propose to use a vectorial but approximated representation of the model that allows efficient, real-time GPU exploitation. ... We get interactive frame rates for models that consists of hundreds of thousands of B-Rep faces, regardless of the zoom level. ... The Cubic Bézier line counts the number of cubic Bézier surface patches that are created as a result of a 0.1mm approximation of the original basis surfaces. ...
### Interrogation of spline surfaces with application to isogeometric design and analysis of lattice-skin structures
Xiao Xiao, Malcolm Sabin, Fehmi Cirak
2019 Computer Methods in Applied Mechanics and Engineering
Subsequently, the intersections of the identified B\'ezier patches with the curve segment are determined with a matrix-based implicit representation leading to the computation of a sequence of small singular ... For mechanical analysis, the skin is modelled as a Kirchhoff-Love thin-shell and the lattice as a pin-jointed truss. ... For displacements there is usually no need for rational splines as the exact representation of quadric sections is not relevant. ...
### Approximate algebraic methods for curves and surfaces and their applications
Bert Jüttler, Pavel Chalmovianský, Mohamed Shalaby, Elmar Wurm
2005 Proceedings of the 21st spring conference on Computer graphics - SCCG '05
We report on approximate techniques for conversion between the implicit and the parametric representation of curves and surfaces, i.e., implicitization and parameterization. ... It is shown that these techniques are able to handle general free-form surfaces, and they can therefore be used to exploit the duality of implicit and parametric representations. ... and Emerging Technologies" arm of the Fifth Framework Programme. ...
### Rational Bézier Formulas with Quaternion and Clifford Algebra Weights [chapter]
Rimvydas Krasauskas, Severinas Zubė
2014 Geometry and Computing
We consider Bézier-like formulas with weights in quaternion and geometric (Clifford) algebra for parametrizing rational curves and surfaces. ... Applications include Bézier curves and surfaces in the conformal model of Euclidean space, bilinear Clifford-Bézier patches on isotropic cyclides, and rational offset surface modeling. ... The majority of the numerical experiments and symbolic computations were made for this paper using the software package CLU-Calc/CLUViz described in [16] and the MAPLE package Clifford [1] . ...
« Previous Showing results 1 — 15 out of 244 results | 1,994 | 8,933 | {"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-2022-33 | latest | en | 0.851335 |
http://philosophy.hku.hk/think/sci/causal-fallacies.php | 1,490,583,201,000,000,000 | text/html | crawl-data/CC-MAIN-2017-13/segments/1490218189377.63/warc/CC-MAIN-20170322212949-00402-ip-10-233-31-227.ec2.internal.warc.gz | 287,976,985 | 4,571 | [S08] Causal fallacies
Here are some typical mistakes in causal reasoning:
• Post hoc fallacy - Inferring that X causes Y just because X is followed by Y. Example: "Last time I wore these red pants I got hit by a car. It must be because they bring bad luck."
• Mistaking correlation as causation - "Whenever I take this pill my cough clears up within a week, so this pill is very effective in curing coughs." But perhaps mild coughs go away eventually even without taking medicine?
• Reversing causal direction - Assuming that X causes Y without considering the possibility that Y is the cause of X - "Children who like violent video games are more likely to show violent behavior. This must be because they are copying the games." But can it be that children who are more prone to violence are more fond of such video games?
• Genetic fallacy - Thinking that if some item X is associated with a source with a certain property, then X must have the same property as well. But of course this might not be the case. Example: "Eugenics was practised by the Nazis so it is obviously disgusting and unacceptable."
• Fallacy of the single cause - Wrongly presupposing that an event has a single cause when there are many causally relevant factors involved. This is a fallacy where causal interactions are being over-simplified. For example, after tragedy such as a student committing suicide, people and the news media might start looking for "the cause", and blame it on either the parents, the amount of school work, the society, etc. But there need not be a single cause that led to the suicide. Many factors might be at work.
• Confusing good causal consequences with reasons for belief - Thinking that a claim C must be true because believing in C brings about some benefit. Example: "God exists because after I have become a believer I am a lot happier and is now a better person."
Read the following research about the use of Facebook and longevity: https://www.sciencedaily.com/releases/2016/10/161031165135.htm. Should you use Facebook more and post more photos in order to live longer? | 450 | 2,094 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2017-13 | longest | en | 0.955994 |
https://dsollberger.netlify.app/post/vote-by-mail-data/ | 1,632,812,846,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780060538.11/warc/CC-MAIN-20210928062408-20210928092408-00249.warc.gz | 261,467,605 | 5,358 | # Vote by Mail data
library("dplyr")
library("ggplot2")
# Data
Today I am going to try to make a choropleth (map + data) of the percentage of registered voters that sent in ballots for the 2018 elections in the USA. I gathered the data from and by the following sites:
states <- tolower( sort(c(state.name, "District of Columbia")))
voted2018 <- c(2,5,62,1,54,88,4,3,2,25,4,38,9,6,17,16,10,1,2,18,4,2,15,21,3,6,63,15,6,5,10,36,3,2,NA,13,4,104,3,4,2,15,1,4,89,6,2,92,1,5,22)
vote_data <- data.frame(states, voted2018)
The 538 article’s table caption included the following notes:
• “It’s not possible to calculate these numbers for North Dakota, as it doesn’t have voter registration. However, it does have a relatively high rate of mail voting — 23 percent of its votes were cast by mail in 2016, and 29 percent in 2018. Republicans control both the governorship and state legislature in North Dakota.”
• “When asked how Oregon mailed ballots to 104 percent of voters in 2018, state election officials responded that the EAC number was “off” but did not provide further details."
• Sources: Ballotpedia, U.S. Election Assistance Commission
# Choropleth
Adapted from Choropleth Map in ggplot2 by ‘data technik’
states_map <- map_data("state")
fact_join <- left_join(states_map, vote_data, by = c("region" = "states"))
fact_join %>%
ggplot(aes(long, lat, group = group)) +
geom_polygon(aes(fill = voted2018), color = "black") +
scale_fill_viridis_c(option = "C") +
labs(title = "A View of Voting by Mail",
subtitle = "Percentage of registered voters that sent in ballots in 2018",
caption = "Alaska: 2%, Hawaii: 38%, North Dakota: NA \n Data Source: FiveThirtyEight",
fill = "% voted by mail",
x = "",
y = "") +
theme(axis.text.x = element_blank(),
axis.text.y = element_blank(),
axis.ticks.x = element_blank(),
axis.ticks.y = element_blank(),
panel.background = element_rect(fill = "white")) | 549 | 1,899 | {"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} | 2.796875 | 3 | CC-MAIN-2021-39 | latest | en | 0.853859 |
https://romannumeralsconverter.net/dccxlvii-roman-numerals | 1,708,651,380,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947473871.23/warc/CC-MAIN-20240222225655-20240223015655-00214.warc.gz | 511,831,450 | 6,712 | # DCCXLVII Roman Numerals
Result
DCCXLVII = 747
DCCXLVII Roman Numerals in a number represented as:
Roman Numeral Number DCCXLVII 747
747 meaning in roman numeral is “DCCXLVII” number.
## How to write DCCXLVII Roman Numerals?
You can write DCCXLVII roman numeral by combining each alphabet of roman numeral i.e,
• DCCXLVII = DCC + XL + VII = 700 + 40 + 7 = 747.
According to basic roman numerals rule when a higher letter comes before the smaller letter then the letter will be added.
For example:
DCCXLVII is a roman numeral:
DCC = 700. XL = 40. VII = 7.
The letters will be added according to the rule.
Result:
DCCXLVII numbers are simply translated to 747 considering the rule.
## How to Convert DCCXLVII in Numbers?
Step 1:
Add DCCXLVII into the given area of tool,
Step 2:
Click “Convert To Number”
Step 3:
Get your results in translated form which is 747. | 256 | 884 | {"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} | 2.734375 | 3 | CC-MAIN-2024-10 | latest | en | 0.79341 |
https://www.jiskha.com/questions/13779/derivatives-differentiate-y-x-7-3-So-far-I-have-Y-7-3X-Is-this-all-there-is-to-this | 1,553,554,882,000,000,000 | text/html | crawl-data/CC-MAIN-2019-13/segments/1552912204461.23/warc/CC-MAIN-20190325214331-20190326000331-00446.warc.gz | 791,430,076 | 5,423 | calculus
derivatives
differentiate
y=x^7/3
So far I have
Y=7/3X
Is this all there is to this problem?
Using the power rule, you bring the exponent down, just like you did. Then, you subtract one from the given exponent to get the new exponent. 7/3 - 1 (which is 3/3) gives you a new exponent of 4/3. Your answer should be:
7/3 x^4/3
Hope that helped!
Amy :)
Thanks...
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3. 👁 93
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asked by I'm stumped on June 19, 2009
4. Calculus
Hi~ Thank you for your help! I was trying to work on a problem about Taylor series, but i don't think im approaching the problem the right way. I have to find the fifth order Taylor polynomial and Taylor series for the function
asked by KIKSY on March 7, 2011
5. math
I studied derivatives before, but kind of forgot them all, is there a website that gives me an overview of derivatives. What is first and second derivatives, how to find it, all the derivative rules and so on... thank you
asked by Sam on October 9, 2008
6. convert derivatives of x,y,z, to r, Ɵ,z
convert derivatives of x,y,z, to r, Ɵ,z with chain rule? I want provment rules of converting rectangular derivatives to cylindrical derivatives and and also cylindrical to spherical.I know the rules but I cant prove EQUATIONS.thx
asked by parsa on December 1, 2011
7. Calculus
I'm having problems with this one. Can't get the right answer. Differentiate the following using the chain rule. f(x) = squareroot(x^2+1)/(3x+1) I know that I can take the whole thing and put it to the (1/2) and differentiate that
asked by Angie on September 22, 2012
8. Calculus: Differentiation
I am having trouble solving these. Could you explain how to do these 3 questions, step by step? That would be helpful! 1. Differentiate y=sin ^-1 (x^2) at x=0.5 . Round to 3 decimal points. 3. Differentiate y= (x^2 +1) tan ^-1 (x)
asked by Paige on December 6, 2018
9. Math - Calculus 1
Approximate using derivatives. (Must show work using derivatives!) Do not use a calculator: (1.1)^4 - (1.1)^2
asked by James on October 7, 2016 | 869 | 2,848 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4 | 4 | CC-MAIN-2019-13 | latest | en | 0.956431 |
https://www.emathhelp.net/en/calculators/calculus-2/integral-calculator/?f=ln%28x%5E2%29 | 1,719,332,453,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198866143.18/warc/CC-MAIN-20240625135622-20240625165622-00843.warc.gz | 645,447,049 | 6,241 | Integral of $\ln\left(x^{2}\right)$
The calculator will find the integral/antiderivative of $\ln\left(x^{2}\right)$, with steps shown.
Related calculator: Definite and Improper Integral Calculator
Please write without any differentials such as $dx$, $dy$ etc.
Leave empty for autodetection.
If the calculator did not compute something or you have identified an error, or you have a suggestion/feedback, please write it in the comments below.
Find $\int 2 \ln\left(x\right)\, dx$.
Solution
The input is rewritten: $\int{\ln{\left(x^{2} \right)} d x}=\int{2 \ln{\left(x \right)} d x}$
Apply the constant multiple rule $\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$ with $c=2$ and $f{\left(x \right)} = \ln{\left(x \right)}$:
$${\color{red}{\int{2 \ln{\left(x \right)} d x}}} = {\color{red}{\left(2 \int{\ln{\left(x \right)} d x}\right)}}$$
For the integral $\int{\ln{\left(x \right)} d x}$, use integration by parts $\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$.
Let $\operatorname{u}=\ln{\left(x \right)}$ and $\operatorname{dv}=dx$.
Then $\operatorname{du}=\left(\ln{\left(x \right)}\right)^{\prime }dx=\frac{dx}{x}$ (steps can be seen here) and $\operatorname{v}=\int{1 d x}=x$ (steps can be seen here).
The integral can be rewritten as
$$2 {\color{red}{\int{\ln{\left(x \right)} d x}}}=2 {\color{red}{\left(\ln{\left(x \right)} \cdot x-\int{x \cdot \frac{1}{x} d x}\right)}}=2 {\color{red}{\left(x \ln{\left(x \right)} - \int{1 d x}\right)}}$$
Apply the constant rule $\int c\, dx = c x$ with $c=1$:
$$2 x \ln{\left(x \right)} - 2 {\color{red}{\int{1 d x}}} = 2 x \ln{\left(x \right)} - 2 {\color{red}{x}}$$
Therefore,
$$\int{2 \ln{\left(x \right)} d x} = 2 x \ln{\left(x \right)} - 2 x$$
Simplify:
$$\int{2 \ln{\left(x \right)} d x} = 2 x \left(\ln{\left(x \right)} - 1\right)$$
Add the constant of integration:
$$\int{2 \ln{\left(x \right)} d x} = 2 x \left(\ln{\left(x \right)} - 1\right)+C$$
Answer: $\int{2 \ln{\left(x \right)} d x}=2 x \left(\ln{\left(x \right)} - 1\right)+C$ | 742 | 2,099 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.34375 | 4 | CC-MAIN-2024-26 | latest | en | 0.574222 |
https://gitlab.doc.ic.ac.uk/vpalladi/HgcTpgSim/-/commit/94cfffcbaf1ca462ed9e217ea02257bc443ca878 | 1,642,838,622,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320303779.65/warc/CC-MAIN-20220122073422-20220122103422-00100.warc.gz | 340,681,950 | 55,551 | Commit 94cfffcb by dauncey
### Rewriting AnalysisPhoton
parent 1ea3f838
No preview for this file type
... ... @@ -75,8 +75,9 @@ The resulting RMS using the best fit values is given by \end{eqnarray*} But since the solution is defined by $Ma=v$, then $a^T M a = a^T v$. Hence \mathrm{RMS}^2_\mathrm{min} = \frac{1}{N} \sum_e T_e^2 - a^T M a = \frac{1}{N} \sum_e T_e^2 - v^T M^{-1} v \mathrm{RMS}^2_\mathrm{min} = \frac{1}{N} \left(\sum_e T_e^2\right) - a^T M a = \frac{1}{N} \left(\sum_e T_e^2\right) - v^T M^{-1} v = \frac{1}{N} \left(\sum_e T_e^2\right) - a^T v The above can be extended slightly, which may improve the energy response linearity as well as the RMS. The energy estimation for the event (i.e. the ... ...
... ... @@ -10,12 +10,15 @@ #include "TH1F.h" #include "TH2F.h" #include "TProfile.h" #include "TVectorD.h" #include "TMatrixDSym.h" #include "Random.hh" #include "TE1F.hh" #include "TE2F.hh" #include "AnalysisBase.hh" #include "SubAnalysisPhotonSimHit.hh" #include "SubAnalysisPhotonTrgHit.hh" #include "Event.hh" #include "ErrorMatrix.hh" ... ... @@ -26,7 +29,113 @@ public: enum { kNumberOfTests=AdcReading::fMaximumNumberOfMethods+26 }; /* class Coefficients { public: enum { kNumberOfCoefficients=Geometry::kNumberOfLayers }; Coefficients() : fSum(kNumberOfCoefficients), fSSq(kNumberOfCoefficients) { fNum=0.0; for(unsigned i(0);i0.5); if(fNum==1.0) return; for(unsigned i(0);i> n; assert(n==kNumberOfCoefficients); fin >> fNum; double d; for(unsigned i(0);i> d; fSum(i)=d; } for(unsigned i(0);i> d; fSSq(i,j)=d; } } } void write(const std::string &fName) { std::ofstream fout(fName.c_str()); if(!fout) return; normalise(); fout << kNumberOfCoefficients; fout << fNum << std::endl; for(unsigned i(0);iFill(i,fErrorMatrix.fSum[i]); std::cout << "fErrMean[" << i << "]=" << fErrorMatrix.fSum[i] ... ... @@ -2425,7 +2547,8 @@ public: hErrMatrix->Fill(i,j,em[i][j]);//fErrorMatrix.fSSq[i][j]); hCovMatrix->Fill(i,j,em[i][j]/sqrt(em[i][i]*em[j][j]));//fErrorMatrix.fSSq[i][j]/sqrt(fErrorMatrix.fSSq[i][i]*fErrorMatrix.fSSq[j][j])); } } } */ } bool fitC3dShape(const double *a, double &et) { ... ... @@ -2621,10 +2744,15 @@ public: //double stInv(1.0/proj.position().sinTheta()); /* if(proj.position().rho()> 30.0 && proj.position().rho()<145.0 //&& stInv>=2.5 && stInv<3.0 ) good=true; */ if(fabs(proj.position().eta())>=1.7 && fabs(proj.position().eta())< 2.7) good=true; } } ... ... @@ -2709,8 +2837,14 @@ public: double tanMax[2]; double phiMax[2]; //simEvent(event); //return true; if(fSimEvent) fSAPSimHit.event(event); //event->digHitNoiseCreator(); //event->digHitProcessor(); //event->trgHitProcessor(); if(fTrgEvent) fSAPTrgHit.event(event); return true; for(unsigned e(0);e
#ifndef Coefficients_HH #define Coefficients_HH #include "TVectorD.h" #include "TMatrixDSym.h" class Coefficients { public: enum { kNumberOfCoefficients=Geometry::kNumberOfHGCLayers+Geometry::kNumberOfBHLayers }; Coefficients(bool b=false) : fPlainRms(b), fSum(kNumberOfCoefficients), fCoe(kNumberOfCoefficients), fSSq(kNumberOfCoefficients) { fNum=0.0; fErg=0.0; for(unsigned i(0);i0.5); assert(lm!=0); uint64_t one(1); TVectorD sum(fSum); for(unsigned i(0);i> n; if(n!=kNumberOfCoefficients) return false; fin >> fNum >> fErg; double d; for(unsigned i(0);i> d; fSum(i)=d; } for(unsigned i(0);i> d; fSSq(i,j)=d; } } if(!fin) return false; return true; } bool write(const std::string &fName) { std::ofstream fout(fName.c_str()); if(!fout) return false; fout << kNumberOfCoefficients << " " << std::setprecision(12) << fNum << " " << std::setprecision(12) << fErg << std::endl; for(unsigned i(0);i
... ... @@ -602,6 +602,8 @@ public: } void trgHitProcessor() { std::cout << "Event::trgHitProcessor called" << std::endl; for(unsigned e(0);e
... ... @@ -49,11 +49,13 @@ class ReadSimHitTTree { fTTree->SetBranchAddress( "SimHit_E", &_SimHit_E ); fTTree->SetBranchAddress( "SimHit_Id_BH", &_SimHit_Id_BH ); fTTree->SetBranchAddress( "SimHit_E_BH", &_SimHit_E_BH ); fTTree->SetBranchAddress( "StringTestVector", &_SimHit_str ); _SimHit_Id=0; _SimHit_E=0; _SimHit_Id_BH=0; _SimHit_E_BH=0; _SimHit_str=0; if(PrintLevel::print(2)) std::cout << "ReadSimHitTTree opening file " << fFileName <<" successful" << std::endl; ... ... @@ -115,7 +117,15 @@ class ReadSimHitTTree { vHbh.push_back(s); } } if(_SimHit_str!=0) { std::cout << "_SimHit_str size = " << _SimHit_str->size() << std::endl; std::vector &vStr(*_SimHit_str); for(unsigned j(0);j<_SimHit_str->size() && j<100;j++){ if(vStr[j]!="abc0") std::cout << "Line " << j << " = " << vStr[j] << std::endl; } } return true; } ... ... @@ -163,6 +173,8 @@ class ReadSimHitTTree { std::vector *_SimHit_Id_BH; std::vector *_SimHit_E_BH; std::vector *_SimHit_str; }; #endif
... ... @@ -47,6 +47,12 @@ public: return fEvent; } // Position Point point() const { return Geometry::point(*this); } // Deposited energy in GeV double deposit() const { return fScale*fDeposit; ... ... @@ -64,6 +70,11 @@ public: return Constants::fMipPerGev[Geometry::numberOfSublayers(layer(),wafer())-1]*deposit(); } // Equivalent transverse number of MIPs double transverseMips() const { return mips()*point().sinTheta(); } // Equivalent incoming energy in GeV double energy() const { return mips()*Calibration::layerWeight(layer()); ... ... @@ -71,7 +82,7 @@ public: // Equivalent incoming transverse energy in GeV double transverseEnergy() const { return energy()*Geometry::point(*this).sinTheta(); return energy()*point().sinTheta(); } bool read(std::istream &fin) { ... ...
#ifndef SubAnalysisPhotonSimHit_HH #define SubAnalysisPhotonSimHit_HH #include #include #include #include #include #include #include "TH1F.h" #include "TH2F.h" #include "TProfile.h" #include "TH1A.hh" #include "Average.hh" #include "Coefficients.hh" #include "utilities.cc" class SubAnalysisPhotonSimHit { public: SubAnalysisPhotonSimHit(const std::string &sRoot="", const std::string &sOut="") : fName("SubAnalysisPhotonSimHit"), fFile((sOut==""?"":sOut+"/")+fName+(sRoot==""?sRoot:"_"+sRoot)) { nEvt=0; hEvent=new TH1F((fName+"_Event").c_str(), ";Layer;Number of trgHits",100,0.0,1000.0); for(unsigned l(0);lrmsSq) { rmsSqMin=rmsSq; std::cout << "New minimum " << rmsSqMin << " found for lm = " << printBin(lm) << std::endl; } } */ uint64_t lm(0x000ffffffaaaaaaa); std::cout << "k = " << k << " RMS^2 for trigger coeffs = " << vCoefficients[k][4].solve(lm) << std::endl; vCoefficients[k][4].print(); std::cout << "k = " << k << " RMS^2 for all coeffs = " << vCoefficients[k][4].solve() << std::endl; vCoefficients[k][4].print(); TVectorD c4(vCoefficients[k][4].coefficients()); assert(vTVectorD[k].size()==vTruthE.size()); double rmsSq(0.0),rmsSqOverE(0.0); for(unsigned i(0);i | 2,244 | 6,686 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.796875 | 3 | CC-MAIN-2022-05 | latest | en | 0.335166 |
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for(int i=0; i
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It is from Fortran. Integers were i to n, everything else was real. i happened to be the very first integer letter. Everyone unknowingly just followed Fortran.
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# Nonparametric Hypothesis Testing for baseball data
This content was STOLEN from BrainMass.com - View the original, and get the already-completed solution here!
Set up a variable that divides the teams into two groups, those that had a winning season and those that did not. There are 162 games in the season, so define a winning season as having won 81 or more games. Next, divide the teams into two salary groups. Let the 15 teams with the largest salaries be in one group and the 15 teams with the smallest salaries in the other. At the .05 significance level is there a relationship between salaries and winning?
Formulate a hypothesis statement regarding your research issue.
Perform the five-step hypothesis test on the data using a nonparametric hypothesis test. Idenitfy what type of test you used
1. State the null hypothesis, state the alternate hypothesis
2. Select the significance level- 0.05 selected for this
3. Identify the test statistic
4. State the decision rule
5. Take a sample and arrive at a decision
Please include your raw data tables and the results of your computations using both graphical and tabular methods of displaying data and results.
https://brainmass.com/statistics/nonparametric-tests/nonparametric-hypothesis-testing-for-baseball-data-236516
#### Solution Summary
The solution provides step by step method for the Nonparametric Hypothesis Testing . Formula for the calculation and Interpretations of the results are also included.
\$2.19
## Nonparametric Testing in Megastat for Baseball Data
Non parametric Hypothesis Testing Questions:
Using the data from the Excel file (see attachment), conduct the equivalent, nonparametric test of hypothesis using the 5-step process.
Describing the results of the nonparametric hypothesis test. Be sure to include the following:
1. A clear and logical statement of your research question.
2. From your research question, formulate both a numerical and verbal hypothesis statement regarding your research issue.
3. Perform the five-step hypothesis test on the data.
4. Explain which nonparametric test you used to analyze your data and why you chose that test.
5. Interpret the results of your test,
6.Be sure to include your raw data tables and the results of the computations of your Z-test or t-test in your paper, using both graphical and tabular methods of displaying data and results.
Note: Methods of computation could include the usage of Excel®, or MegaStat.
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# Although it claims to delve into political issues,
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Although it claims to delve into political issues, [#permalink]
### Show Tags
16 Jun 2006, 21:11
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Although it claims to delve into political issues, television can be superficial such as when each of the three major networks broadcast exactly the same statement from a political candidate.
(A) superficial such as when each of the three major networks
(B) superficial, as can sometimes occur if all of the three major networks
(C) superficial if the three major networks all
(D) superficial whenever each of the three major networks
(E) superficial, as when the three major networks each
Please explain
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Re: SC - political issues [#permalink]
### Show Tags
16 Jun 2006, 21:45
mahesh004 wrote:
Although it claims to delve into political issues, television can be superficial such as when each of the three major networks broadcast exactly the same statement from a political candidate.
(A) superficial such as when each of the three major networks
(B) superficial, as can sometimes occur if all of the three major networks
(C) superficial if the three major networks all
(D) superficial whenever each of the three major networks
(E) superficial, as when the three major networks each
Please explain
By POE I find E as the correct option.
Since "broadcast" is a plural verb, we need plural subject.
A Incorrect
- "such as" is awkward
- "each of the three major networks" is singular subject
B Incorrect
- "can sometimes occur if" is also not proper.
- We need "when" instead of "if"
C Incorrect
- We need "when" instead of "if"
D Incorrect
- "each of the three major networks" is singular subject
E Correct
- "as when" is properly used.
- IMO "three major networks each" is a plural subject.
Regards,
Brajesh
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### Show Tags
16 Jun 2006, 22:21
Picking E over C.
Tough one for me.
C also looked right but preferred ' when ' over 'if'.
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17 Jun 2006, 08:44
employ the lesson provided by buddy willget800 about "each after verb" , i get E in this one
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17 Jun 2006, 09:48
OA is E.
17 Jun 2006, 09:48
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# Although it claims to delve into political issues,
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Powered by phpBB © phpBB Group and phpBB SEO Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®. | 1,162 | 4,445 | {"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} | 3.34375 | 3 | CC-MAIN-2017-04 | latest | en | 0.900411 |
https://engineeringtutorial.com/combining-independent-voltage-sources-series/ | 1,709,623,413,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707948223038.94/warc/CC-MAIN-20240305060427-20240305090427-00605.warc.gz | 229,811,112 | 13,709 | ### Combining Independent Voltage Sources in Series
Combining Independent Sources
An inspection of the KVL equations for a series circuit shows that the order in which elements are placed in a series circuit makes no difference. An inspection of the KCL equations for a parallel circuit shows that the order in which elements are placed in a parallel circuit makes no difference. We can use these facts to simplify voltage sources in series and current sources in parallel.
Combining Independent Voltage Sources in Series
It is not possible to combine independent voltage sources in parallel, since this would violate KVL. However, consider the series connection of two ideal voltage sources shown in (a) below:
From KVL we know that v v1 + v2 , and by the definition of an ideal voltage source, this must be the voltage between nodes a and b, regardless of what is connected to them. Thus, the series connection of two ideal voltage sources is equivalent to a single independent voltage source given by:
Clearly, the obvious generalization to N voltage sources in series holds.
Example
In a previous example we determined the current i in the one-loop circuit shown below:
By rearranging the order in this one loop circuit (of course this does not affect i), we obtain the circuit shown below:
We can now combine the series independent voltage sources and the series resistors into single equivalent elements:
By Ohm’s Law:
i = – 24/12 = -2 A
#### Engineering Tutorial Keywords:
• combining voltage sources
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http://www.chegg.com/homework-help/questions-and-answers/the-capacitor-c-in-fig-h31-is-initially-uncharged-at-t-0-the-switch-k-is-closed-determine--q3145022 | 1,369,320,595,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368703334458/warc/CC-MAIN-20130516112214-00075-ip-10-60-113-184.ec2.internal.warc.gz | 385,541,861 | 7,679 | ## Direct current circuits
The capacitor C in Fig H.31 is initially uncharged. At t-0, the switch K is closed. Determine an expression for potential difference V and the current i of the circuit. E= 42V r=150 ohm R=450 ohm C=50 uF | 65 | 231 | {"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} | 2.78125 | 3 | CC-MAIN-2013-20 | latest | en | 0.824248 |
https://www.easyelimu.com/qa/3100/product-first-three-terms-geometric-progression-common-ratio | 1,686,416,701,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224657735.85/warc/CC-MAIN-20230610164417-20230610194417-00292.warc.gz | 847,663,972 | 7,262 | # The product of the first three terms of geometric progression is 64.If the first term is a and the common ratio is r;
1.1k views
The product of the first three terms of geometric progression is 64.If the first term is a and the common ratio is r;
1. Express r in terms of a
2. Given that the sum of the three terms is 14
• Find the values of a and r ,hence write down two possible sequences each upto 4th term.
• Find the sum of the first 5 terms of each sequences.
1. nth= arn-1
a x ar x ar2=64
a3r3=64
r3 =64/a3
r=4/a
2.
• a + ar + ar2=14
a+a(4/a) + a(4/a)2=14
a+ 4+16/a=14
a2 + 4a + 16 =149 a=2 r=2
a2-10a + 16 =0 a=8 r=½
a2 -8a -2a + 16=0
a(a-8) -2(a-8)=0
a=2 or 8
r=4/8 or 4/2 ½or2
r=½=8,4,2,1,or
r=2=2,4,8,18
• S50= a(1-rn) r<1
1-r
(a(rn-1) r)/1-r> 1
(2(ss-1))/2-1 = 62 | 350 | 782 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.25 | 4 | CC-MAIN-2023-23 | latest | en | 0.782444 |
https://www.univerkov.com/15-g-of-a-mixture-of-zinc-and-copper-was-treated-with-an-excess-of-hydrochloric-acid-while-4-48-liters/ | 1,653,496,684,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662588661.65/warc/CC-MAIN-20220525151311-20220525181311-00009.warc.gz | 1,227,705,185 | 6,301 | # 15 g of a mixture of zinc and copper was treated with an excess of hydrochloric acid while 4.48 liters
15 g of a mixture of zinc and copper was treated with an excess of hydrochloric acid while 4.48 liters of gas were released, determine the composition of the mixture.
Given:
m mixture (Zn, Cu) = 15 g
V (gas) = 4.48 l
Find:
ω (Zn) -?
ω (Cu) -?
Solution:
1) Zn + 2HCl => ZnCl2 + H2 ↑;
Cu + HCl – the reaction will not proceed;
2) M (Zn) = Mr (Zn) = Ar (Zn) = 65 g / mol;
3) n (H2) = V (H2) / Vm = 4.48 / 22.4 = 0.2 mol;
4) n (Zn) = n (H2) = 0.2 mol;
5) m (Zn) = n (Zn) * M (Zn) = 0.2 * 65 = 13 g;
6) ω (Zn) = m (Zn) * 100% / m mixture (Zn, Cu) = 13 * 100% / 15 = 86.67%;
7) ω (Cu) = 100% – ω (Zn) = 100% – 86.67% = 13.33%.
Answer: The mass fraction of Zn is 86.67%; Cu – 13.33%.
One of the components of a person's success in our time is receiving modern high-quality education, mastering the knowledge, skills and abilities necessary for life in society. A person today needs to study almost all his life, mastering everything new and new, acquiring the necessary professional qualities. | 397 | 1,097 | {"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-2022-21 | longest | en | 0.896555 |
https://hoven-in.appspot.com/Home/Aptitude/Averages/maths-aptitude-mock-test-on-averages-015.html | 1,660,100,002,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882571097.39/warc/CC-MAIN-20220810010059-20220810040059-00072.warc.gz | 288,096,301 | 6,380 | # Averages Quiz Set 015
### Question 1
A, B and C are three numbers. The average of A, B and C is 30. The average of A and B is 17, and the average of B and C is 78, what is the value of number B?
A
100.
B
101.
C
99.
D
98.
Soln.
Ans: a
By the given conditions, A + B = 2 × 17 = 34. Similarly, B + C = 2 × 78 = 156. Adding we get A + 2B + C = 190. We have also been given that A + B + C = 3 × 30 = 90. Subtracting, we get B = 190 - 90 = 100.
### Question 2
What is the average of first 383 natural numbers?
A
192.
B
193.
C
191.
D
190.
Soln.
Ans: a
The sum of first n natural numbers is \$(n × (n + 1))/2\$. The average is \${(n × (n + 1))/2}/n\$ which is \$(n + 1)/2.\$ Putting n = 383, we get average = 192.
### Question 3
Two of the 22 numbers are 30 and 54, If these two are excluded the average of the remaining 20 numbers is 1 less than the average of all the 22 numbers. What is the average of all the 22 numbers?
A
32.
B
33.
C
31.
D
34.
Soln.
Ans: a
Let the required average be x. Then \$22x - (30 + 54) = 20 × (x - 1).\$ which gives \$22x - 84 = 20 × (x - 1).\$ Solving for x we get x = 32.
### Question 4
What is the average of first 527 natural numbers?
A
264.
B
265.
C
263.
D
262.
Soln.
Ans: a
The sum of first n natural numbers is \$(n × (n + 1))/2\$. The average is \${(n × (n + 1))/2}/n\$ which is \$(n + 1)/2.\$ Putting n = 527, we get average = 264.
### Question 5
The sales(in rupees) of a gift shop for six consecutive days is 3318, 3690, 2466, 5346, 1692 and 1470. What is the overall average sale for these six days?
A
Rs.2997.
B
Rs.3003.
C
Rs.2991.
D
Rs.3009.
Soln.
Ans: a
The total sale on first six days is 3318 + 3690 + 2466 + 5346 + 1692 + 1470 = 17982. The average for 6 days is: 17982/6, which gives Rs. 2997. | 663 | 1,795 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.3125 | 4 | CC-MAIN-2022-33 | latest | en | 0.85137 |
https://drawabox.com/community/sketchbook/Taelia/replies | 1,713,640,292,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817674.12/warc/CC-MAIN-20240420184033-20240420214033-00827.warc.gz | 195,264,083 | 17,588 | # Taelia
## taelia's Sketchbook
##### 5:47 PM, Wednesday September 6th 2023
https://imgur.com/a/JjhNfpB
Hereby the revisions, one page of organic forms and one with contour ellipses. Hope this is up to standards, I attempted to make the sausages less pointy at the ends.
Cheers!
##### 4:27 PM, Thursday April 27th 2023
Good going :) you definitely saw what I meant and corrected it. The arrowheads are a little bit weird, but luckily they're not part of the exercise!
Next Steps:
Hope you like nature. Onto lesson 3!
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##### 11:51 AM, Friday April 7th 2023
Good going! Skimming through the pages, everything looks good so consider that a pass. Let's review everything anyway to see if we can get you some useful feedback.
Arrows:
Lots of creativity! You made all sorts of curves and ribbons, I appreciate it. A few of your ribbons even have full twists instead of curves, you drew them well but it wasn't quite part of the exercise.
You understood the foreshortening part quite well, differing the width of the ribbon depending on how far in front it is, as well as changing up the space between the curves.
Do pay attention to your lineweight though, you overdid it. Just a gentle line is enough to be able to tell overlap, but yours are clearly visible, I'd even call them thick. Too subtle is better than too noticeable, in this case!
Sausages:
You only uploaded one page, did you forget the second one?
The shape of the sausages look good. Consistent in width with a slight bend, not too complex a shape. Your ellipses follow the shape of the sausage nicely, and I can tell you've tried to switch up the degree of the ellipses as well, though I'm afraid you were holding back a bit too much still. If you look at the example, you can tell the ellipse width really changes from thin to wide: https://drawabox.com/lesson/2/5/ellipses
Textures:
You did not quite make full gradients in your texture analysis page, there is no transition from dark to light. Imagine a very bright light on the right-side of the boxes, shining a light to the left. The goal is to imagine which shapes are there on the page, and draw only the hard shadows of the shapes that stick up. Because the light is located on the right, the further you go left, the more shadow you'll find, ending up in a full black shape.
For the dissections similarly, you ended up with mostly the same texture leaving a gap of white, as opposed to making a proper gradient from light to dark: https://drawabox.com/lesson/2/6/notransition
Also, your dissections don't follow the curves of the sausage properly, causing a rather flat look: https://drawabox.com/lesson/2/7/curvature
Form intersections:
Jeez, you went full ham on this exercise. I don't know if this is A3-format paper or you just made small shapes, but you made plenty of them! Awesome. They look good too, you draw nice spheres and ellipses, and most everything is drawn in proper perspective. I'm impressed.
However, you did draw with a large amount of foreshadowing, which results in a somewhat messy look. The idea of this exercise is to draw shallow perspective, nearly isometric even. When the amount of foreshortening differs per object, it becomes really hard to think about intersection.
As for the intersection lines themselves, you've drawn them a bit arbitrarily, but that's okay because that wasn't the goal of this exercise anyway. No worries about that.
Organic intersections:
Look fine, though be careful of drawing complex shapes. You should avoid draping your beans in this exercise, they have to retain their original shape and just lean against/over eachother. A bunch of your shadows have a bit of a weird shape. A few of them hug the form of the sausages they are cast from: https://drawabox.com/lesson/2/9/shadows
Other shadows are drawn properly on the body they're on, but don't follow the shape of the body they're drawn onto. If you cast a shadow on a curved surface, the shadow will follow the curve of the surface it's cast on.
That said and done, the sausages are otherwise well drawn and you have plenty creativity in your composition.
Overall:
I'm doubting. I'm gonna let you pass without asking for some additional pages, but should you find the time, consider re-doing one Texture Analysis and one Form Intersections page anyway. You draw well and your concept of perspective is solid enough to just continue.
The textures were quite messy, however, and you didn't quite grasp the idea behind gradients.
As for the intersections, try and draw a few shapes with similar foreshadowing, as opposed to mixing dramatic and shallow ones.
Next Steps:
On to Lesson 3 you go!
(but take into account the feedback you got!)
This community member feels the lesson should be marked as complete, and 2 others agree. The student has earned their completion badge for this lesson and should feel confident in moving onto the next lesson.
2 users agree
##### 11:26 AM, Friday April 7th 2023
Good going on finishing Lesson 2! Giving everything a quick glance, I can tell I'm gonna mark this as complete :) but let's review it anyway to see if we got some good feedback.
Arrows:
You have a solid understanding on how to foreshorten your ribbons. Large in front, small in back. One thing you didn't quite get right is that this also goes for the spaces inbetween the arrows. That is to say, if you look at this curve, you'll see that in the back the curve is also much closer to eachother than at the front: https://drawabox.com/lesson/2/4/step1
All in all, no worries though. You put all the shading at the right spots too, you have a fine understanding of depth.
Sausages:
Your sausage shapes are well done. Especially your second page is smooth anf confident, doesn't taper, is nice and consistent in width. Very good!
The ellipses are less confident, they're rather wobbly and you didn't quite draw all of the ellipses through twice. I recommend spending more time on that for warmups! Something you also missed is to shift the degree of the ellipse. This is required to give the sausages the illusion of "bending", as it stands they are quite flat. Check out this bit: https://drawabox.com/lesson/2/5/ellipses
Textures:
You did decently well on the textures exercises. Be careful to not draw the shapes themselves, you have quite a few forms drawn out fully: https://drawabox.com/lesson/2/6/drawingforms
Be sure to imagine one strong light source, the point of the exercise is to only draw the strong shadows that come from that single point of light shining onto your object.
On most of your dissections you did it fine, but there's a significant amount of shapes where you either didn't transition from dense to sparse as you went around the shape: https://drawabox.com/lesson/2/6/notransition
or where you forgot to actually follow the curve of the sausages themselves: https://drawabox.com/lesson/2/7/curvature
You did break the silhouette properly everywhere, which is great!
Form Intersections:
You've drawn all your objects in roughly the same perspective, and kept all the shapes equilateral, both perfect! A solid amount of objects too, with plenty of overlap. The intersection lines are a bit awkward, but that is fully understandable and not the point of this exercise anyway.
However, your linework is sloppy. Lots of unconfident wobbly marks have been made, the hatching isn't spaced properly, there has been a clear lack of ghosting and just overall.. hastiness. It's not enough to make you redo this part, but please take it into account! Clean, confident linework is at the core of this entire course, and this really isn't done properly.
Organic Intersections:
Well.. there is a certain lack of creativity here. But honestly, otherwise well done. The shapes look proper, the shadows follow the curves of the object below it nicely, you have a good understanding of how shapes touch eachother in a three-dimension space and how to adjust shapes to give a good illusion of solidity. (there's one sausage in the middle somewhere that's really off, but we'll ignore that). All in all, you did this exercise well. Consider drawing another page where you don't just stack like a jenga tower but get a bit more creative like in the example homework: https://drawabox.com/lesson/2/9/example
Overall:
Awesome, you did great. Continue on to lesson 3.
However, do take extra care of your linework, the form intersections exercise wasn't well done. Consider really slowing down the pace you complete the lessons at, and pay more attention to the marks you place.
Next Steps:
Onwards, lesson 3!
This community member feels the lesson should be marked as complete, and 2 others agree. The student has earned their completion badge for this lesson and should feel confident in moving onto the next lesson.
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##### 11:43 AM, Thursday April 6th 2023
Oh dear, I don't think you did it well either, I'm sorry :(
Arrows:
You made exactly one arrow that is passable, the rest you kept flat. Be sure to re-read the exercise properly, especially the "Common Mistakes" section. Your lines are wobbly and unconfident, you did not take perspective into account at all, you didn't draw shading inside the bends either.
Sausages:
Your sausages are oddly shaped. They bulk and bend at weird places, and are not consistent in width at all, tapering at the ends. Your ellipses aren't changing in degree either, and they don't follow the angle of the center lines you drew. On the page where the ellipses aren't fully drawn through you did take care to get the "ellipse-shape" in, so that's good. You did draw the wrong half of the ellipses, though.
Texture:
I agree that you didn't capture the textures well. The goal of this exercise isn't to draw the "shape" of the paper, it's to draw the "shadows". Imagine there being a bright lamp on the right-side of the paper, shining all its light to the left. What kind of shadows would then appear? That is what you're supposed to draw :)
I fear you've failed to submit the texture dissections entirely.
Form intersections:
These forms aren't drawn with proper perspective, so there is very little to give in terms of feedback. Please refer back to the 250 boxes exercise to learn about perspective and foreshortening before doing this part again!
Also, the idea was to draw lines where the forms are intersecting, but most of your forms aren't even overlapping. Please re-read the exercise before doing this again.
Organic intersections:
For this exercise, I.. you made floating shapes. I'm afraid you failed to read the exercise. The end result should look like a pile of sausages all lying on a solid floor. After you've done that, someone can give feedback on the shadows you were supposed to draw but didn't!
Overall:
I'm afraid I can't let this pass at all. Looking at your profile, I think noone gave you a passing grade for 250 boxes either, which explains why this lesson went so bad. The next lessons will only be harder, so please return to the 250 boxes challenge and make sure you fully understand it before you continue!
Next Steps:
2 users agree
##### 11:27 AM, Thursday April 6th 2023
Good job finishing Lesson 2. However, I'm afraid I can't mark this lesson as complete.
I will be going over some of the mistakes you have made, but none of the exercises have been done properly.
I'm sorry in advance for being harsh :(
Arrows:
Some of the arrows are finished, most of them are submitted in an unfinished state. You haven't taken perspective into account, the ribbons are the same width all the way through. Your shading has been placed at random spots on the ribbon instead of in the inside of the bends as was asked.
Sausages:
Your sausages are mostly ellipses instead of slightly bent shapes. I will admit, the ellipses inside the sausages look rather good, if your entire submission had the quality of these ellipses you'd pass. But I'm afraid that's where the compliments end. All your ellipses are the same degree, you'll want to vary them in width to give the illusion of bending in three-dimensional space, as it stands they are rather flat. Your sausages are tapering towards the end, please keep them a consistent width. Also the linework is somewhat wobbly and unconfident. Be sure to practice your ghosting before placing marks on paper!
Texture:
The idea of this exercise is to draw just cast shadows of actual photos of objects. I'm afraid you came up with some fantasy objects to draw, and have no done a proper study of them. In the dissections exercise, you pasted the random shapes as if they were a flat surface, not taking any light source into account. Also, be sure to avoid "scratching" the page to make texture, it is never required!
Form intersections:
None of the shapes are drawn in proper perspective, and none of the shapes are overlapping. This is the opposite of what was asked in the exercise, so I'm not even sure how to critique this.
Organic intersections:
Be sure to keep your shapes simple, these are elongated and draped. Be sure to draw the full shape of the sausages even when they would otherwise hide behind other objects. The shadows seem to have been placed arbitrarily, and on most drawings no shadows have been placed at all. Be sure to re-read the exercise to understand where to place the shadows.
Overall:
I don't think any of the exercises are passable, I recommend you re-read the exercises and re-submit the homework in full if you would like a proper grade. Sorry :(
However, looking at your profile I can see that you haven't received a review for your first lesson either, and you have not done the 250 boxes challenge yet. It is mandatory before starting this lesson, because it will get you the required amount of practice in drawing perspective needed to be able to do this lesson properly.
Next Steps:
Do the 250 boxes challenge first!
2 users agree
##### 11:04 AM, Thursday April 6th 2023
Hey hey, good job on finishing Lesson 2. I'll be splitting this in five sections to review separately, here we go.
Arrows:
You've drawn plenty of arrows! You made a few critical mistakes though, so I'm gonna ask you to do another page for me.
First of all, you have placed your "shading" on the wrong side of the curves. If you look at the examples, you'll see that the shading has been added on the inside of the bends, while you have placed them on the outside. This gives your arrows the impression of impossible perspective: https://drawabox.com/lesson/2/4/step4
A large amount of arrows aren't actually folding onto themselves. Don't be afraid to let lines cross! A few arrows have been drawn well, most the majority remains flat: https://drawabox.com/lesson/2/4/overlap
Also try and create more dramatic foreshortening. That is to say, in the distance, the arrow will be both a lot smaller than in the front, and the space between the curves will be shorter as well.
Check out this section of the lesson for a bit of an explanation: https://drawabox.com/lesson/2/4/overlap
Sausages:
Your sausages are mostly either too simple or too complex. Please be sure to make a gentle sausage-shape. Some of your shapes are near ellipses, and others are full C-shapes.
You did well on making the ellipses follow the shape of the sausage! And you've also drawn through your ellipses, making multiple loops which is great. Definitely keep practicing these though, your ellipses are somewhat wobbly, which is a result of drawing them too slow.
Most of your ellipses are the same shape, though. It's important to change the degree of your ellipses to give the illusion of curvature, otherwise the shapes will remain looking flat. See: https://drawabox.com/lesson/2/5/degree
Also be sure to doublecheck the angle of your ellipses. Some of them do not align with the center line of the sausage at all: https://drawabox.com/lesson/2/5/alignment
Lastly, I am missing a full page of sausages, did you forget to upload one?
Textures:
Well done on the textures! You might want to experiment a bit more with stronger shadow sections, but overall you did well on this exercise. Especially the dissection textures exercise is well done, you also work with giving the silhouette a good shape, making it seem nice and three-dimensional. If anything, try to keep in mind exactly from what angle the light is coming from, so you can place the cast shadows consistently in the same direction.
Form intersections:
You did well on these too. The intersections themselves are off, but that's to be expected, and that is also not the goal of this exercise. You drew most of the shapes with a similar rate of foreshadowing, which makes all the shapes feel like they fit in the same space well. You're also drawing confident straight lines, so I assume the work in the 250 boxes challenge has paid off!
Organic intersections:
These look fine as well, but you did make some important mistakes. I won't have you redraw them but it's good to take a note.
First of all, these shapes are draped, you still want to make sure to make solid sausage-shapes. For this exercise its important you don't draw complex forms: https://drawabox.com/lesson/2/9/complicated
You did well on not making the shadows hug the shape they're coming from! The next step is to make the shadows actually follow the shape it is attached to, however, which you sometimes did, but sometimes failed.
Also please be sure to draw the entire shape. I see you've only drawn a few sausages partially (because they would be behind other sausages anyway). For the purpose of Drawabox, always be sure to draw the full shape, front to back.
Overall:
I was expecting a pretty low quality submission, but halfway through it picked up immensely, and you actually did a really good job!! Going by the organic intersections, I'm gonna assume you have a decent grasp of the sausages exercise as well, so no worries about finding that missing page to upload.
However, I can't let those arrows slide, they're messy. Would you draw and upload another page for me, making sure to re-read the exercise properly?
Focus on two things mainly:
• stronger foreshortening (smaller arrows/empty space in the back, wider arrows/larger spaces in the front)
• drawing the shading in the proper place. Doublecheck the exercise and pay close attention to where you put it.
Good luck!
Next Steps:
One more page of arrows.
##### 2:11 PM, Thursday March 30th 2023
Ha, i wasn't expecting a comment in here, but I'll gladly take the compliment! Thank you!
##### 4:50 PM, Friday March 24th 2023
Very cool, those are no problem to you. Onwards!
Next Steps:
Lesson 2!
This community member feels the lesson should be marked as complete, and 4 others agree. The student has earned their completion badge for this lesson and should feel confident in moving onto the next lesson.
2 users agree
##### 2:04 PM, Tuesday March 21st 2023
You're in luck, you're getting a heckin' speedy review!
Awesome job on completing the 250 boxes, it's quite a feat. Taking a quick sneak peak at the 250th box, I think you did just fine, I could mark this lesson as complete right now.
Instead, let me go through the boxes 50 by 50, see if we can find you some nice feedback and see if you made some good progression while at it.
Boxes 1-50
The small size of the pictures you took make it somewhat harder to zoom in on imgur. Not a bad thing, just makes it a bit rougher for me here. Your first set of boxes look great already. I was going to make a remark on how similar all your boxes are looking, but within the first 50 boxes you already went full creative with different shapes like pizza boxes and stretched bars. Good going! Your lines are sharp and straight, and you are even hatching your boxes properly. Even if its optional, it is great that you are doing so and I'm sure I'll see the fruits of that effort down the line. I particularly enjoy your experiments with boxes 25-29.
I can't tell if you have been adding lineweight to the outside of the boxes though, I don't think I see any. I do recommend doing that as well, as it is an invaluable bit of practice ghosting lines, and it will greatly benefit the rest of the course to do so.
Boxes 51-100
You're experimenting with even more box shapes and even sizes now. It is clear that you have properly understood the goal of this exercise.
You infrequently have the odd box like number 92 and 94 where your lines are converging "in pairs" instead of all to the same point. Most of your boxes properly have all four lines converging to the same point, which is indeed how it's supposed to go. I'm noting this mainly cause it seems to be a bit of a recurring trend, so it's something you might want to keep an extra eye on.
Also, at box 91 you have redrawn a few of your lines. Please be careful not to do that! There's plenty boxes to be drawn still, if you make the wrong mark just accept the error and move on. There is no need to make boxes look better in hindsight, just make the next box better instead.
Boxes 101-150
I see some scratching in this set of boxes, avoid doing so! Bad, bad! Just leave the errors in and move on. Boxes 119, 125, 129, 130, 131.
I also see some dotted lines in 106 and 108. I'm not entirely sure what to make of these. You're not intended to try and draw additional lines over your already drawn boxes, but on the other hand.. this is clearly showing that you are fully understanding the mistakes you made and you know how to fix them. Which is good :)
Boxes 151-200
I think something clicked for you around this point, I'm seeing a large amount of great boxes. Of course there's still odd angles and plenty small mistakes, but the amount of boxes that are drawn correctly are no coincidence. It's clear that you have the right idea, the rest is a matter of a thousand more drawn boxes.
Boxes 201-250
Box 201 is oddly off. But its the only one really. You're even doing well on the hard long boxes like 208. I am particularly impressed by how you manage to keep all your "middle two" lines properly converging as well. At worst you draw them parallel (which results in the "converging in pairs of 2" issue), but that is so much better than diverging which you are solidly avoiding.
Overall
Honestly, great job. The main takeaway here is that all lines of a box should converge to the same point, and you understood that perfectly fine. Your lines are decently straight, even on the larger boxes, you hatched diligently, I have very little feedback for you.
Except for that I didn't see that line weight. I'm afraid that is a solid 1500 lines you could've practice ghosting on but didn't, you're just gonna have to do that in hindsight now! Also, don't redraw lines to fix things! Adjust and move on :) no redoing or chicken scratching.
On to Lesson 2.
Next Steps:
Lesson 2! Finally some non-straight lines to draw!
This community member feels the lesson should be marked as complete, and 2 others agree. The student has earned their completion badge for this lesson and should feel confident in moving onto the next lesson.
The recommendation below is an advertisement. Most of the links here are part of Amazon's affiliate program (unless otherwise stated), which helps support this website. It's also more than that - it's a hand-picked recommendation of something I've used myself. If you're interested, here is a full list.
### Pentel Pocket Brush Pen
This is a remarkable little pen. Technically speaking, any brush pen of reasonable quality will do, but I'm especially fond of this one. It's incredibly difficult to draw with (especially at first) due to how much your stroke varies based on how much pressure you apply, and how you use it - but at the same time despite this frustration, it's also incredibly fun.
Moreover, due to the challenge of its use, it teaches you a lot about the nuances of one's stroke. These are the kinds of skills that one can carry over to standard felt tip pens, as well as to digital media. Really great for doodling and just enjoying yourself. | 5,515 | 24,287 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.640625 | 3 | CC-MAIN-2024-18 | latest | en | 0.969743 |
https://philosophyofscienceportal.blogspot.com/2008/04/puzzles-and-like.html | 1,723,746,387,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641311225.98/warc/CC-MAIN-20240815173031-20240815203031-00112.warc.gz | 353,837,803 | 22,478 | ## Monday, April 21, 2008
### Puzzles and the like
Uncle Roy's Jigsaw Puzzle
Birthday:
This is one of my favorite tricks--guaranteed to astound your friends. Select an individual and tell this person that you can tell them the month and date of their birthday. They will need to follow your instructions and do some calculations.
1.) Ask the individual to multiply the number of the month of their birthday by 5.
3.) Multiply by 4.
5.) Multiply by 5.
6.) Add the day of the month.
Now, have the individual show or tell you the result. Mentally, subtract 205 from the given number.
Example January 29th
1.) 1x5=5.
2.) 5+7=12.
3.) 12x4=48.
4.) 48+13=61.
5.) 61x5=305.
6.) 305+29=334.
7.) 334-205=129 or 1 [January] 29th
Clocks:
A home has three clocks. On January 1st at midnight all three show the correct time. Only the first clock keeps perfect time. The second clock loses a minute every day. The third clock gains a minute every day. If the clocks were to continue in this way, how long would it be before all of them show the correct time again at the same time?
Cubes:
You have four solid cubes that are all made of the same material but have different heights. The first cube is 6 centimeters tall, the second is 8 centimeters tall, the third is 10 centimeters tall, and the fourth is 12 centimeters tall. If you were to place the cubes on the two pans of a balance, which cubes would you place in which pan so that the weights on both sides would be equal?
Clue: You would need to consider the volume of a cube rather than its height to determine which cubes go in which pan.
Einstein's Riddle:
"There are 5 houses in 5 different colors. In each house lives a person with a different nationality. The 5 owners drink a certain type of beverage, smoke a certain brand of cigar, and keep a certain pet. No owners have the same pet, smoke the same brand of cigar, or drink the same beverage.
The question is: Who owns the fish?
Hints:
The Brit lives in the red house.
The Swede keeps dogs as pets.
The Dane drinks tea.
The green house is on the left of the white house.
The green homeowner drinks coffee.
The person who smokes Pall Mall rears birds.
The owner of the yellow house smokes Dunhill.
The man living in the center house drinks milk.
The Norwegian lives in the first house.
The man who smokes Blend lives next to the one who keeps cats.
The man who keeps the horse lives next to the man who smokes Dunhill.
The owner who smokes Bluemaster drinks beer.
The German smokes Prince.
The Norwegian lives next to the blue house.
The man who smokes Blend has a neighbor who drinks water.
Einstein wrote this riddle early during the 19th century. He said 98% of the world could not solve it. Its not hard, you just need to pay attention and be patient.
Lone Electron:
Help the "lone electron" at the top left make its way to the atom at the center. Bounce in a straight line from atom to atom following the arrows.
Oddball:
There are twelve balls.
One weighs more or less than the others.
Using a simple balance scale, discover the "oddball" in only three weighings.
Tongue Twisters:
Chop shops stock chops.
Fat frogs flying past fast.
Friendly Frank flips fine flapjacks.
Give papa a cup of proper coffee in a copper coffee cup.
Knapsack straps.
Mrs. Smith's Fish Sauce Shop.
Peggy Babcock.
Penny's pretty pink piggy bank.
Plague-bearing prairie dogs.
She sifted thistles through her thistle-sifter.
Red lorry, yellow lorry, red lorry, yellow lorry.
Roger weaves red rugs.
Rubber baby buggy bumpers.
She sifted thistles through her thistle-sifter.
The sheik's sixth sheep's sick.
Silly Sally slid down a slippery slide.
Six sick slick slim sycamore saplings.
Six small slick seals.
Sly Sam slurps Sally's soup.
Sunshine City.
Thieves seize skis.
Three free throws.
Toy Boat, toy boat, toy boat.
We surely shall see the sun shine soon.
Which witch wished which wicked wish?
Which wrist watch is a Swiss wrist watch?
Wishy-washy Wanda watched her watch get washed.
Water Levels:
Say you are out fishing, in a boat on a lake, and you take the boat's anchor and drop it over the side, into the lake: Does the water level of the lake rise (since the anchor is now taking up space in the lake)? Or does the water level fall (because the boat is now floating higher in the water)? Or does the water level stay the same (because these two things cancel each other out)? Or is it impossible to determine what the water level will do without more information? | 1,108 | 4,515 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.125 | 4 | CC-MAIN-2024-33 | latest | en | 0.955688 |
https://minorfs.wordpress.com/tag/password-capabilities/ | 1,553,634,425,000,000,000 | text/html | crawl-data/CC-MAIN-2019-13/segments/1552912206016.98/warc/CC-MAIN-20190326200359-20190326222359-00203.warc.gz | 557,339,237 | 17,236 | ## Rumpelstiltskin and his children
When communicating about my efforts on the upcoming version of my least authority file-system collection, I often seem unable to fully communicate the base design of the core capability-based file-system, even to many cryptographically educated people. In this post I’ll try to elaborate on the algorithm that lies at the center of all my efforts: The Rumpelstiltskin hash-tree algorithm.
Rumpelstilskin
In a fairy tale by the Grimm brothers that most of you will probably know by heart, a girl, as a result of her fathers big mouth, found herself in a rather problematic situation . The king had locked her in a tower room filled with straw and had declared that she be decapitated unless she would spin the straw into gold by morning, An imp showed up and offered to help her out in exchange for her necklace, and the girl gladly accepted the offer. Next day, larger room with more straw, same story now in exchange for her ring. third and final day, even a larger room with even more straw, but as the girl has nothing left to give in exchange, the imp makes the girl promise to give up her firstborn child in exchange for his services.
Later in the story, the girl, now a queen married to the king is visited again by the imp who comes to claim his price. After pleading with the imp, the imp agrees that he will give up his claim if the queen guesses his true name within 3 days. When by chanche the queens messenger overhears the imp chanting:
“Today I bake, tomorrow I brew, then the Queen’s child I shall stew. For nobody knows my little game, for Rumpelstiltskin is my name.”
Learning his name, the queen is able to wield authority over the evil imp, forcing him to give up his claim to her firstborn.
You could say that the unguessable name “Rumpelstiltskin” is a token of authority not unlike what in information security theory we refer to as a capability, or more precisely a password-capability.
Directional tree graph
The CapFs file-system I’m working on, like most file-systems constitutes a tree. Unlike many file-systems though, the CapFs tree structure aims to be purely directional. That is, while in DOS, Unix, OSX or Linux we are used to navigate ‘up’ using a command like “cd ..”, the CapFs ‘directional’ tree adheres to the rule that:
“Authority to a node implies authority to its children but NOT to its parent.”
Names and true names
In our fairy tale, the imp its true name was Rumpelstiltskin (cap). This was the name that implied authority over the imp. The imp might have had an other non authoritative name used by others to address him, but if he had it was not relevant for the story. Maybe Rumpelstiltskin himself did not have an other non authoritative name, but if he had any children, and those children had true names like him that could be used to wield authority over them, Rumpelstiltskin would probably often address his children by their non authoritative names.
Now talking of authority, if Rumpelstiltskin (cap) would want to delegate the authority to one of his children to an other imp, what options would there be? Lets assume Rumpelstiltskin (cap) had a kid named Slartibartfast (cap) for who he used the casual (non authoritative) name Bob. There would be two ways someone could come to wield authority over Bob, either by using his true name Slartibartfast or by having authority over his parent Rumpelstiltskin that by proxy would give authority over Bob. There are thus two authoritive ways to designate Bob:
1. Slartibartfast
2. Rumpelstiltskin::Bob
Back to our directional tree
So if we project Rumpelstiltskin and Bob to our directional tree graph, we could say that Rumpelstiltskin might be our tree’s root node. Rumpelstilskin being the root node has no name but his true authority carrying name. Bob and any other of Rumpelstiltskin’s descendants would have two names:
• A normal name that implies no direct authority (name)
• An unguessable name that implies authority (capability)
Attenuation
Where using a name and a capability (true name) for each non-root node in our tree takes care of making authority to branches and leafs decomposable, the authority to any sub branch that is implied by a capability is still absolute. If we look back at our file-system example, one might want to delegate the authority to read without delegating the authority to write. We could provide non-root nodes with a second capability that would imply attenuated authority to a node, for example read only. So now we have:
• A normal name that implies no direct authority (name)
• An unguessable name that implies FULL authority (capability)
• An unguessable name that implies attenuated authority (capability)
The algorithm
Given that we have only a single type of attenuation (for example read-only), there is an interesting algorithm we can use to securely calculate new unguessable names for non root node’s given its name and the unguesable name of its parent.
In the above diagram Key1 would be the equivalent of ‘Rumpelstiltskin’, Key2 would be the equivalent of a read-only capability to Rumpelstiltskin. The subnode name would for example be ‘Bob’, and Subnode Key1 would be the equivalent of Slartibartfast.
Going from Rumpelstiltskin to a read-only attenuated capability to Rumpelstiltskin would be straight forward. We basically take an HMAC hash of the unattenuated authority capability (Rumpelstiltskin) and some static string, and use a representation of the resulting key2 as attenuated authority capability. There is no secret salt needed in this step, so a shared static string is used instead.
For going from a parent capability to a child capability, there are two scenario’s:
• From unattenuated parent capability to unattenuated child capability.
• From attenuated parent capability to attenuated child capability.
To allow both scenario’s to be used, the calculation of the unattenuated child capability is done by taking an HMAC hash of the parent attenuated authority capability and the child’s name, together with a secret salt.
The secret salt makes sure that we can create a system where a client holding an attenuated authority capability can ask a server for an attenuated authority child capability, but can’t itself derive an unattenuated authority child capability from an attenuated authority parent capability.
Secure persistence
When looking at the diagram of the Rumpelstiltskin hash-tree algorithm, we see there is is yet a HMAC operation and a key3 that we did not yet discus. The problem is that if we want to implement something of persistence using relatively public storage, for example with a cloud storage provide we don’t fully trust to keep our big bag of secrets secret, we will want to:
• encrypt physical file’s
• store our files in a way that does not disclose capabilities.
In order to address the first issue, we shall let our attenuated authority capability double as file encryption key for our file. We now use a third HMAC hash operation to create a 4th name that we use to determine the location where the encrypted file is stored.
High-security versus high-scalability
Regarding the final HMAC hashing steps there are two possible implementation mode’s for the Rumpelstiltskin hash-tree algoritm: high-security and high-scalability. In the high scalability version, encryption and decryption could potentially be client side operations, and the server takes care only of enforcing read-only attenuation. In this mode, the storage path (key3) is calculated using a HMAC hash of the attenuated authority capability together with a static string. If in contrast we value security so much that we are willing to sacrifice the scalability that comes with client side en/de-cryption in order to prevent a cloud storage provider with access to an attenuated authority (read-only) capability to combine his explicit and his implicit authority into an unattenuated authority capability, than high-security mode would include the servers secret salt in the HMAC calculation of key4, mandating server side encryption and decryption.
In my own project I shall be using the Rumpelstiltskin hash-tree algorithm in high-scalability mode for now. Maybe in the future switching to high-security may turn out to be desired, but for now the hole plugged by giving up the potential for adding scalability seems to small to justify the price of using high-security mode.
Conclusion
I hope reading this blog post somewhat clarifies the Rumpelstiltskin hash-tree algorithm, and I hope others may find the algoritm usefull for other projects as well. I’ll be using this algorithm in MinorFs2, but I feel it might have much broader usability than just user space file-systems. Please use this algorithm as you please, and comment on this post if you think this information is usefull or still needs clarifications somewhere. | 1,967 | 8,887 | {"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-2019-13 | latest | en | 0.967543 |
https://numberworld.info/53423 | 1,628,210,460,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046152085.13/warc/CC-MAIN-20210805224801-20210806014801-00379.warc.gz | 435,288,558 | 3,789 | # Number 53423
### Properties of number 53423
Cross Sum:
Factorization:
Divisors:
1, 41, 1303, 53423
Count of divisors:
Sum of divisors:
Prime number?
No
Fibonacci number?
No
Bell Number?
No
Catalan Number?
No
Base 2 (Binary):
Base 3 (Ternary):
Base 4 (Quaternary):
Base 5 (Quintal):
Base 8 (Octal):
Base 16 (Hexadecimal):
d0af
Base 32:
1k5f
sin(53423)
-0.21522840043566
cos(53423)
-0.97656373864992
tan(53423)
0.22039360250383
ln(53423)
10.885996643829
lg(53423)
4.7277282724323
sqrt(53423)
231.13416017543
Square(53423)
### Number Look Up
Look Up
53423 (fifty-three thousand four hundred twenty-three) is a amazing figure. The cross sum of 53423 is 17. If you factorisate 53423 you will get these result 41 * 1303. The figure 53423 has 4 divisors ( 1, 41, 1303, 53423 ) whith a sum of 54768. 53423 is not a prime number. The number 53423 is not a fibonacci number. 53423 is not a Bell Number. The number 53423 is not a Catalan Number. The convertion of 53423 to base 2 (Binary) is 1101000010101111. The convertion of 53423 to base 3 (Ternary) is 2201021122. The convertion of 53423 to base 4 (Quaternary) is 31002233. The convertion of 53423 to base 5 (Quintal) is 3202143. The convertion of 53423 to base 8 (Octal) is 150257. The convertion of 53423 to base 16 (Hexadecimal) is d0af. The convertion of 53423 to base 32 is 1k5f. The sine of 53423 is -0.21522840043566. The cosine of the figure 53423 is -0.97656373864992. The tangent of the figure 53423 is 0.22039360250383. The root of 53423 is 231.13416017543.
If you square 53423 you will get the following result 2854016929. The natural logarithm of 53423 is 10.885996643829 and the decimal logarithm is 4.7277282724323. that 53423 is very impressive number! | 614 | 1,719 | {"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} | 3.296875 | 3 | CC-MAIN-2021-31 | latest | en | 0.753464 |
https://codecap.org/typeerror-cant-multiply-sequence-by-non-int-of-type-float/ | 1,670,013,718,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710916.40/warc/CC-MAIN-20221202183117-20221202213117-00507.warc.gz | 214,613,535 | 70,555 | 13.7 C
New York
Saturday, November 26, 2022
TypeError can’t multiply sequence by non-int of type ‘float’
TypeError can’t multiply sequence by non-int of type ‘float’
You can multiply an integer value and a float value by converting the integer into a float. But if you are trying to multiply a float and a value that is not an integer or string, you will encounter an error. An error called “TypeError can’t multiply sequence by non-int of type ‘float’” will be raised.
The easiest way to resolve this is by converting the string into a float or integer and then multiplying it.
The following code will throw the error:
``````# Declare variables
val1 = '10'
val2 = 1.2
# Multiply variables
result = (val1*val2)
# Print Output
print('Multiply of val1 and val2: ',result);``````
Output
``````Traceback (most recent call last):
File "multiply.py", line 3, in <module>
result = (val1*val2)
TypeError: can't multiply sequence by non-int of type 'float'``````
In the above example, we have declared an integer variable as a string
``val1 = '10' # Declare variables as string``
Due to which while doing the multiplication between string and float variable it raised an error.
To solve this error, you need to change the code where you are multiplying “val1” with “val2″ Here is the fix:
``result = (float(val1)*val2)``
This converts the “val1” variable into a float and then multiplies it with “val2”
Correct Code:
``````# Declare variables
val1 = '10'
val2 = 1.2
# Multiply variables
result = (float(val1)*val2)
# Print Output
print('Multiply of val1 and val2: ',result);`````` | 423 | 1,591 | {"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-2022-49 | longest | en | 0.812367 |
http://ken.duisenberg.com/potw/archive/arch03/031124sol.html | 1,637,982,900,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964358078.2/warc/CC-MAIN-20211127013935-20211127043935-00255.warc.gz | 35,301,586 | 3,009 | ## Some Chessboard Problems
1. What is the smallest number of moves needed by a chess rook to visit every square on a chessboard? (A rook can only move horizontally. A "move" can be any length in a single direction.)
2. What is the smallest number of moves needed by a chess queen to visit every square on a chessboard? (A queen can move both horizontally and diagonally.)
3. A chess bishop can only move diagonally. What is the maximum number of squares a chess bishop can visit, if it is only allowed to visit each square once?
4. Cut a chessboard into two pieces along the grid lines, then rearrange the two pieces (with rotation possibly, but not reflection) to try to achieve the largest number of same-color neighboring squares. (For example, if you cut off one column, and rotate it, then there will be eight pairs of adjacent squares of the same color.)
5. Repeat problem 4, with the added restriction that the two pieces be identical (through rotation.)
Source: Original.
Solutions were received from Joseph DeVincentis, Jeremy Galvagni and Alan O'Donnell.
Joseph's solutions. He interpreted the cutting problem differently than I'd intended, creating continuously connected sets of squares - an interesting solution:
```You have some problems with the directions (horizontally, diagonally) in the
first two parts.
1. I'm guessing that you mean by "visit" that squares passed over in a move
are also visited. Otherwise it clearly takes 64 moves to visit all the
squares, regardless of how far the rook moves between squares, and the tour
can be completed trivially.
Vith this interpretation of "visit", it takes 15 moves for a rook to visit
every square. To see this, consider that in order for the rook to visit
every square in a row or column, it must either travel along some part of
the length of that row or column once, or else cross perpendicular to it 8
times in different positions. To do the latter requires 15 moves, 8
crossings and 7 other moves between the crossings to get the rook into a new
position for each subsequent move. To avoid having to do this 15-move
solution, you have to have a move along every row and every column, or thus
at least 16 moves. A 15 move solution can be done in a very simple way,
crossing each column from end to end and moving one step horizontally
between columns.
http://www.mathpuzzle.com/dots.html
In that problem, you have more freedom to go in directions other than
those that a queen moves (though only exact hits on the centers of
squares would get counted) AND you have the freedom to go outside the
edges of the board. But still, you can only ever save one line relative
to the boring rook solution, or at least, that is all anybody managed to
find after a lot of work on the problem.
So the answer is 14 moves, and an example that fits our queen problem:
http://www.mathpuzzle.com/conn8x8.GIF
3. With the same interpretation of "visit" as above, the answer seems to be
29. It's possible to do this in four zig-zag crossings across the board, but
at each turning you miss a square on the edge. other patterns Only seem to
move the three missed squares around.
01 03 05 07
02 04 06 08
XX 13 11 09
14 12 10 XX
15 17 19 21
16 18 20 22
29 27 25 23
28 26 24 XX
If you don't count the passed-over squares, then a complete tour is
possible.
32 06 23 25
05 22 24 26
04 21 07 13
20 02 12 14
19 03 31 15
28 11 16 08
29 27 17 09
30 18 10 01
4. Use this cutting pattern:
oooooooo
XXXXXXXo
Xooooooo
XXXXXXXo
Xooooooo
XXXXXXXo
Xooooooo
XXXXXXXX
When you move the pieces apart by one space, gives this pattern of black and
white squares:
WBWBWBWB
BWBWBWB W
W BWBWBWB
BWBWBWB W
W BWBWBWB
BWBWBWB W
W BWBWBWB
BWBWBWBW
Two pairs of adjacent white squares, and five sets of 8 same-color squares.
But if you want a single large group, you can do better. Cut this way:
oooooooo
oXoXoXoX
oXXXXXoX
oXoXXooX
oXooXXoX
oXoXXXXX
oXoXoXoX
oooooooo
When you rotate the piece, it gives this pattern of black and white squares:
WBWBWBWB
B B B B
W W
B B WB
W WB W
B B
W W W W
BWBWBWBW
B B B B
WBWBW W
B BW B
W BW W
B BWBWB
W W W W
WBWBWBWB
BBBBBBBB
WWBWBWWW
BBBBWWBB
WWWBBWWW
BBBBWBWB
WWWWWWWW
BWBWBWBW
24 connected black squares, and 22 connected white squares. I'm not sure
this is the best possible.
5. With both pieces identical:
Xooooooo
XXXXXXXo
XoXoXoXo
Xooooooo
XXXXXXXo
XoXoXoXo
Xooooooo
XXXXXXXo
Shift the pieces vertically by 3 positions, to give this pattern of black
and white squares:
BWBWBWB
W
B B B B
WWBWBWBW
BWBWBWBB
WWWWWWWW
BBWBWBWB
WBWBWBWW
B B B B
W
BWBWBWB
22 connected white squares.
Joseph DeVincentis
```
Jeremy's cutting solutions: Each has 48 pairs:
```11111111
12121212
12121212
12122222
12222212
12121212
12121212
11111111 rotate 180
11111111
12212121
11212221
12222111
11122221
12221211
12121221
11111111 rotate 90```
```Mail to Ken
``` | 1,473 | 4,990 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.34375 | 4 | CC-MAIN-2021-49 | latest | en | 0.933777 |
http://news.datascience.org.ua/2019/04/23/how-much-extra-for-one-square-metre/ | 1,571,487,881,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570986693979.65/warc/CC-MAIN-20191019114429-20191019141929-00166.warc.gz | 139,359,093 | 8,974 | # How much extra for one square metre?
A linear relationship.
There looks to be a positive correlation between Richmond house prices and their building area.
This gives us the go ahead to begin building our linear regression model.
On to the next one!2.
Applying a linear regression modelIn this section we attempt to fit a linear regression model to the observed relationship between Richmond house prices and their building area.
We will clean our data for any outliers or erroneous points and then make sense of the regression output.
First, let’s visualise our data with a box plot.
Figure 4: Box plot of building area for Richmond housesWe can establish the following points from Figure 4:There looks to be a minimum value of 0 m²There are 5 outliers from 200 m² onwardLet tackle point 1 first, it’s quite peculiar that that there is a house in Richmond with a Building Area of 0 m², so let’s have a look.
Table 5: House with Building Area of 1 m²The Building Area is in fact 1 m² but has a Land size of 0 m².
After investigating this property online it is clear this isn’t a pantry, but much larger than that.
Hence this data point is incorrect, so let’s filter it out.
Next we have the 5 outliers from 200 m² onward.
To identify these outliers we use the 1.
5 x IQR rule.
Table 6: Richmond house outliers based on Building AreaAlthough these houses are deemed high outliers based on the IQR computation, I am going to keep these in the data set.
Reason being, is if we return to the scatter plot in Figure 3, these values are important in capturing the linear relationship between house price and building area.
Now that we have cleaned our data and assessed all the outliers, we can build our regression model using least squares regression!Figure 5: Linear Regression PlotWe can see our regression line fits quite nicely through the data, so let’s see the results:Table 7: OLS Regression ResultsOur regression model is as follows:Price (\$) = 14140 + 10620*BuildingAreaKey takeaways from OLS regression results:Based off the adjusted R-squared, our regression model accounts for 70.
6% of the variability in Price around its mean.
We have an intercept of 14,140, which according to a significance level of 0.
05 is not statistically significant (p-value = 0.
817).
The coefficient of Building Area is 10,620, which is statistically significant (p-value is very small).
Let’s elaborate on each point in detail.
Takeaway 1:Although we have identified that our model explains 70.
6% of the total variance in Price around it’s mean, we cannot simply leave it at that.
It is important we check the residual plot, to examine if any of the explanatory power of our model exists in our residuals, as these are meant to be random and unpredictable.
With regression there exists two components, a deterministic and stochastic component.
For the stochastic component, this means the difference between our observed prices and actual prices should be random.
Figure 6: Standardised Residual PlotWe can see that the points on the residual plot are scattered randomly around zero, and we are unable to discern any pattern.
This means we cannot use one price residual to determine the value of the next price residual, hence there exists no explanatory power.
(this is good!)Takeaway 2:We have a statistically insignificant intercept of 14,140, how do we interpret this?Well for our model, the intercept is the price of a house in Richmond when we have a building area of 0 m².
This means for a house that has no floor area, it’s price will be \$14,140.
My belief is that whether or not the intercept is significant, it should be neglected, as a house with no floor area selling for \$14,140 is unrealistic.
However, for scenarios when the building area becomes more realistic e.
g.
30 m², the intercept will be important in establishing a baseline house price.
Takeaway 3:We find that the coefficient of Building Area is 10,620 and is statistically significant.
How do we interpret this?Firstly, as the coefficient is greater than zero and is statistically significant, there is a positive relationship between Richmond house prices and their building area.
Secondly, with all other variables held constant, a one unit increase in Building Area, increases the price of a house in Richmond by \$10,620.
However, it is important we take into account the uncertainty of the regression coefficient (i.
e.
the 95% confidence interval), so we could express it as:Holding all other variables constant, increasing the floor area by 1 m², will approximately increase the price of a house in Richmond by \$9,524 to \$11,700.
Finally…I hope this article has given you an understanding of the processes involved in preparing the data for linear regression, and also how to make sense of the results.
This example was used to illustrate an application of simple linear regression, but in future articles I plan to apply different methods such as multiple linear regression.
If you have any feedback on my work, how I can improve or anything you feel is incorrect, please let me know!.Thank you for reading!. | 1,107 | 5,134 | {"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} | 4.125 | 4 | CC-MAIN-2019-43 | latest | en | 0.907476 |
https://help.scilab.org/docs/2023.0.0/ja_JP/mesh2d.html | 1,723,375,694,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640997721.66/warc/CC-MAIN-20240811110531-20240811140531-00769.warc.gz | 228,894,637 | 5,119 | # mesh2d
Triangulation of n points in the plane
### Calling Sequence
```triEdges = mesh2d(x, y)
[triEdges, bdy] = mesh2d(x, y)
triEdges = mesh2d(x, y, bdy)```
x
real vector
y
real vector
bdy
integer vector
triEdges
integer matrix
### Description
`mesh2d` computes a triangulation of `n` points in the plane with coordinates given by vectors`x,y`. It returns a matrix `triEdges` of size `[3,nbt]` where `triEdges(:,i)` gives the vertices numbers of triangle `#i` and `nbt` is the number of triangles.
When `bdy` is given as an input parameter this vector defines the boundary and contains the indices of edges belonging to it, grouped by successive connected components. Each component is positively oriented, i.e. successive `bdy(i:i+1)` segments have the interior of the domain to their left. Hence, for a simply connected domain, the boundary is given counterclockwise, and eventual holes are always given clockwise. Each connected component must be closed and is represented by the vector `[i1,..,i_nc]` such that `i1 == i_nc`.
When `bdy` is given as an output parameter the boundary is computed prior to the triangulation as the convex hull of input points `x,y` and is returned in `bdy` with the same convention as above, i.e. counterclockwise sucessive vertices numbers.
Possible error cases are the following:
all nodes are collinear,
some points are identical,
wrong boundary array,
crossed boundary,
wrong orientation of the boundary,
size limitation,
an interior point is too close to the boundary,
an interior point is on the boundary,
### Warning
The triangulation computed by `mesh2d` is not guaranteed to be a Delaunay triangulation of points `(x,y)`.
### Examples
```function displayTri(X, Y, Tr)
plot(0,0,rect=[-1 -1 2 2])
[m, n] = size(Tr);
xpols = matrix(X(Tr), m, n);
ypols = matrix(Y(Tr), m, n);
xpolys(xpols, ypols, color("blue")*ones(n,1));
endfunction
r1 = 1;
n1 = 20;
u = linspace(2*%pi, 0, n1);
xc1 = r1*cos(u(1:\$-1));
yc1 = r1*sin(u(1:\$-1));
bdy1 = [1:n1-1, 1];
r2 = 2;
n2 = 40;
v = linspace(0, 2*%pi, n2);
xc2 = r2*cos(v(1:\$-1));
yc2 = r2*sin(v(1:\$-1));
bdy2 = n1-1+[1:n2-1, 1];
xr = (rand(1, 100)-.5)*2*r2;
yr = (rand(1, 100)-.5)*2*r2;
r = sqrt(xr.^2+yr.^2);
clf
gcf().position(4)=300
// [t, bdy] = mesh2d(x, y) syntax
subplot(1, 2, 1)
k = find(r <= r2);
[t, bdy] = mesh2d(xr(k), yr(k));
displayTri(xr(k), yr(k), t)
plot(xr(k(bdy)), yr(k(bdy)),"r-o")
xtitle("[triEdges, bdy] = mesh2d(x, y)")
isoview
// t = mesh2d(x, y, bdy) syntax
subplot(1, 2, 2)
k = find((r >= r1) & (r <= r2));
x = [xc1 xc2 xr(k)];
y = [yc1 yc2 yr(k)];
t = mesh2d(x, y, [bdy1 bdy2]);
displayTri(x, y, t)
plot(x(bdy1), y(bdy1),"r-o")
plot(x(bdy2), y(bdy2),"r-o")
xtitle("triEdges = mesh2d(x, y, bdy)")
isoview```
### References
`mesh2d` was previously part of the `metanet` ATOMS module.
Report an issue << lsq_splin Interpolation smooth >> | 973 | 2,875 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.71875 | 3 | CC-MAIN-2024-33 | latest | en | 0.847473 |
https://www.hackmath.net/en/example/2371 | 1,561,391,867,000,000,000 | text/html | crawl-data/CC-MAIN-2019-26/segments/1560627999615.68/warc/CC-MAIN-20190624150939-20190624172939-00278.warc.gz | 757,860,501 | 6,767 | # Divide money
Divide 1200 USD at a ratio of 1:2:3:4:5:6:9:10
Result
a = 30
b = 60
c = 90
d = 120
e = 150
f = 180
g = 270
h = 300
#### Solution:
Leave us a comment of example and its solution (i.e. if it is still somewhat unclear...):
Be the first to comment!
#### To solve this example are needed these knowledge from mathematics:
Need help calculate sum, simplify or multiply fractions? Try our fraction calculator.
## Next similar examples:
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Simplify this ratio 10 : 1/4
4. Andre
Andre, Thomas, and Ivan split 88 postage stamps in a 2:5:4 ratio. How much did Thomas get?
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Write the fraction 3/22 as a decimal.
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Write as a fraction and a decimal. One and two plus three and five hundredths
12. In fractions
An ant climbs 2/5 of the pole on the first hour and climbs 1/4 of the pole on the next hour. What part of the pole does the ant climb in two hours?
13. Cupcakes 2
Susi has 25 cupcakes. She gives 4/5. How much does she have left?
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16. Write 2
Write 791 thousandths as fraction in expanded form.
17. Mixed2improper
Write the mixed number as an improper fraction. 166 2/3 | 725 | 2,540 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.921875 | 4 | CC-MAIN-2019-26 | latest | en | 0.951264 |
http://julierankin.co.uk/essays/Cvp-Vs-Stocastic-Analysis-1640025.html | 1,548,121,918,000,000,000 | text/html | crawl-data/CC-MAIN-2019-04/segments/1547583823140.78/warc/CC-MAIN-20190122013923-20190122035923-00611.warc.gz | 124,657,737 | 25,027 | Cvp vs. Stocastic Analysis
Topics: Decision theory, Decision making, Risk Pages: 2 (565 words) Published: April 27, 2013
The subject of cost-volume profit analysis under uncertainty has had extensive literature collected in recent years even though the topic has been ignored in most textbooks. In many cases, entire chapters are devoted to cost-volume profit analysis but they ignore the possibility that one or more parameters of the problem are completely random and therefore the future values are unknown at the time the decision is made. The reluctance of textbook authors to discuss stochastic CVP models can be attributed to the diversity of the literature published. Since numerous models have been proposed and examined, such as single product versus multi-product, it doesn’t really matter the model you use because it is likely to be complicated because it involves various concepts from mathematical statistics. It is a well known fact that real world decision making takes place under conditions of uncertainty and the basic cost-volume profit model helps to create a clearer picture of the variables by generalizing the model to any uncertain situation. In this paper I will present, analyze and apply a stochastic CVP model that can be used as a gateway to decision-making under uncertainty. While
the full mathematical derivations of the results shown herein are probably too complicated for most undergraduates, the results themselves are fairly straightforward, and they facilitate a precise focus on such fundamental concepts in decision-making under uncertainty as the tradeoff between expected profits and breakeven probability. There is an inevitable tradeoff between the comprehensiveness and realism of a model (which tend to generate mathematical complexity) and its practicality and ease-of-use (the extent to which it can readily provide definite answers to specific questions). The model presented here attempts to strike an appropriate balance between these two competing criteria. It is a much simpler model than many of those found in the research literature. For example, the model...
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Perfect Girl Evolution | 8 Suttogások 44:14 ➤ | Casse-noisette et les quatre royaumes [HDRIP] – telecharger streaming | 441 | 2,330 | {"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-2019-04 | longest | en | 0.950819 |
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Report a problem | 529 | 2,104 | {"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} | 2.625 | 3 | CC-MAIN-2019-04 | latest | en | 0.840006 |
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Before selecting lottery numbers it is recommended to spread numbers over the given stretch. The numbers that are closer, are unsuitable to get. You may have to pop up with unique numbers, also it is the key behind the lottery system. Experts have usually selected favourite numbers. It will be better to depend on these numbers to get yourself a top photo. A well balanced combination will enough to play with your luck. Random numbers in order to be picked, together with their sum should lie between 121 and 186. Helps be the best combination potential. | 721 | 3,475 | {"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-2022-21 | latest | en | 0.963183 |
https://cstheory.stackexchange.com/questions/31669/embedding-a-graph-in-the-euclidean-space/31843 | 1,718,952,423,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198862040.7/warc/CC-MAIN-20240621062300-20240621092300-00898.warc.gz | 158,684,045 | 40,012 | # Embedding a graph in the euclidean space
Given a graph $G=(V,E)$, find a mapping $f\colon V \rightarrow \mathbb R^d$ such that for every edge $(u,v) \in E$ we have that $||f(u)-f(v)|| \leq r$; and for every $(u,v) \not \in E$, we have the opposite $||f(u)-f(v)|| > r$ for some given $r$. One could also view this problem as embedding a unit disc graph back into the euclidean space.
Now I have been told that - while the problem is NP-complete - the usual way to do it is as follows. This is where the hearsay starts, so excuse me if I am a little vague.
The proposed approach places an arbitrary node at $0$. Then, run a linear (or quadratic?) program that maximizes $\sum_{v \in V} g(f(v))$ under the side condition stated above, i.e. for all edges $(u,v) \in E$, $(f(u) - f(v))^2 \leq r^2$. $g$ is probably some norm, maybe $L_1$ or $L_2$.
This is how I more or less understood it. The general gist was to fix the position of an initial node and then "spread" the graph apart using the linear program.
I tried to find some sources on that, maybe a paper that introduced this approach, but so far my efforts failed - there are just too many ways to embed a graph in the $\mathbb R^d$. In particular, I am interested in how this approach deals with noisy data. For example, consider a grid graph of size $n \times n$ which should be perfectly embeddable in the euclidean plane. Now, add an edge from $(0,0)$ to $(n,n)$. Intuitively, in an embedding I would like to ignore the edge since the rest of the graph is a perfect grid. The approach above, however, would most likely produce a weird embedding.
Can anyone give me some pointers to work describing this approach?
• So what is your question?
– cody
Commented Jun 2, 2015 at 20:16
• I'm mainly looking for some paper introducing this approach. Any literature would be appreciated.
– HdM
Commented Jun 2, 2015 at 20:23
• Commented Jun 3, 2015 at 20:16
• Could you expand on that? I know of MDS, do you see a way to incorporate the above approach in some multi dimensional scaling?
– HdM
Commented Jun 3, 2015 at 21:42 | 576 | 2,080 | {"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} | 3.4375 | 3 | CC-MAIN-2024-26 | latest | en | 0.921786 |
https://www.pokerstrategy.com/forum/thread.php?postid=1140653 | 1,561,023,811,000,000,000 | text/html | crawl-data/CC-MAIN-2019-26/segments/1560627999200.89/warc/CC-MAIN-20190620085246-20190620111246-00156.warc.gz | 868,413,812 | 19,397 | # Calculating EV of a semi-bluff
• Super Moderator
Super Moderator
Joined: 02.09.2010
Part 1:
In the article "Mathematical Concepts for No-Limit Holdem (3)" is the following:
As bluffs and semi-bluffs play an important role in this article, it is important to introduce the possibility of an opponent's fold to the EV calculation. This formula:
EV = Equity * (Win + Investment) - Investment
will not be sufficient. There is no variable for an opponent's fold in this formula. We therefore need a more general formula to calculate the EV:
EV = Pfold * Pot + (1 - Pfold) * (Equity * (Win + Investment) - Investment)
The probability that you win the pot is Pfold; if you don't win the pot immediately we use the EV. Pfold is the probability that your opponent will fold. Under the assumption Equity = 0 the formula can be simplified to this:
EV=Pfold * Pot - (1 - Pfold) * Investment
Loss in this case is your continuation bet. This is the case for all pure bluffs. Given the definition of a pure bluff, you can only win if your opponent gives up.
The two bold statements seem to attribute two different values to Pfold.
The probability that you win is more than the probability of villain folding, because he could call and you suck out -- even on a pure bluff.
Part 2:
I was actually looking for the calculation that tells you how many times opponent has to fold for your bluff raise to be break-even.
Can someone please point me to it?
Thanks,
--VS
• 6 replies
• Silver
Joined: 14.09.2009
Originally posted by VorpalF2F
Part 1:
In the article "Mathematical Concepts for No-Limit Holdem (3)" is the following:
As bluffs and semi-bluffs play an important role in this article, it is important to introduce the possibility of an opponent's fold to the EV calculation. This formula:
EV = Equity * (Win + Investment) - Investment
will not be sufficient. There is no variable for an opponent's fold in this formula. We therefore need a more general formula to calculate the EV:
EV = Pfold * Pot + (1 - Pfold) * (Equity * (Win + Investment) - Investment)
The probability that you win the pot is Pfold; if you don't win the pot immediately we use the EV. Pfold is the probability that your opponent will fold. Under the assumption Equity = 0 the formula can be simplified to this:
EV=Pfold * Pot - (1 - Pfold) * Investment
Loss in this case is your continuation bet. This is the case for all pure bluffs. Given the definition of a pure bluff, you can only win if your opponent gives up.
The two bold statements seem to attribute two different values to Pfold.
The probability that you win is more than the probability of villain folding, because he could call and you suck out -- even on a pure bluff.
Part 2:
I was actually looking for the calculation that tells you how many times opponent has to fold for your bluff raise to be break-even.
Can someone please point me to it?
Thanks,
--VS
1) Under the assumption Equity = 0 the formula can be simplified to this:
under this assumption both Pfold should be the same. ie villain WILL NOT call you with worse.
2) if equity = 0
EV=Pfold * Pot - (1 - Pfold) * Investment
for break even (EV = 0)
Pfold * Pot - (1 - Pfold) * Investment = 0
gives you Pfold = Investment / ( Pot - Investment )
if equity > 0 ( and if I didn't make mistake in my equations)
Pfold * Pot + (1 - Pfold) * (Equity * (Win + Investment) - Investment) =0
I got
Pfold = (Investment - Equity * (Win + Investment) ) / (Pot + Investment - Equity * (Win + Investment))
• Bronze
Joined: 28.01.2012
The second bolded one is right.
Part 2:
I'm just thinking out loud here to get the formula
Say we have a draw with 40% equity.
We then have a current pot of \$300, and semibluff shove all in for \$1400. when we win we gain \$1700, and when we lose, we lose \$1400.
Then our EV if Pfold=0 is 1700*0.4-1400*0.6, or -160
We need the fold equity to equal this -ev sooo
300xPfold=160 Pfold must equal 53.333
To put this in a formula,
(-1x ( (profit*equity%) - (raisesize*(1-equity%)) ) ) /potsize = Pfold needed for breakeven move.
The -1x is there because the value that we divide by potsize must be negative (otherwise move is automatically above breakeven because this number is the quity and if its positive and we have fold equity obviously the play is +EV) and note here Profit = raisesize+current pot.
Also the factor of multiple streets complicates things in terms of changing equity etc. so it only really works for all ins.
EDIT: To simplify this, work out EV equity without folding as an option, THEN if this value is positive you can already semibluff profitably regardless of pfold, but if your EV is negative, balance this negative value with Pfold*Pot, which should be exactly identical to the negative ev (negative ev only when the opponent calls 100% of the time) to balance it.
Interestingly, if there is enough in the pot, even with a Pfold of 0 and an equity of less than 50%, say 40%, a shove can still be profitable. Eg. 700 pot, you shove for 1200, when you win you make 1900 and when you lose you lose 1200 so you actually are +EV even without Pfold factored in.
• Bronze
Joined: 19.11.2010
Originally posted by metza
Interestingly, if there is enough in the pot, even with a Pfold of 0 and an equity of less than 50%, say 40%, a shove can still be profitable. Eg. 700 pot, you shove for 1200, when you win you make 1900 and when you lose you lose 1200 so you actually are +EV even without Pfold factored in.
Just to put metza's last point in more concrete terms. With a Pfold of 0 and 40% equity, shoving 1200 into a 700 pot gives you an EV of:
40% * 1900 - 60%*1200 = 760 - 720 = +40
that is your equity wins you 1900 40% of the time (the 700 in the pot and 1200 from your opponent calling) and the other 60% of the time you lose the 1200 that you shoved.
• Basic
Joined: 09.07.2012
[quote]Originally posted by VorpalF2F
Part 1:
In the article "Mathematical Concepts for No-Limit Holdem (3)" is the following:
As bluffs and semi-bluffs play an important role in this article, it is important to introduce the possibility of an opponent's fold to the EV calculation. This formula:
EV = Equity * (Win + Investment) - Investment
will not be sufficient. There is no variable for an opponent's fold in this formula. We therefore need a more general formula to calculate the EV:
EV = Pfold * Pot + (1 - Pfold) * (Equity * (Win + Investment) - Investment)
The probability that you win the pot is Pfold; if you don't win the pot immediately we use the EV. Pfold is the probability that your opponent will fold. Under the assumption Equity = 0 the formula can be simplified to this:
EV=Pfold * Pot - (1 - Pfold) * Investment
Loss in this case is your continuation bet. This is the case for all pure bluffs. Given the definition of a pure bluff, you can only win if your opponent gives up.
Both bolded parts seem like they are saying the same thing, they just worded it kind of weird which makes it a little confusing. If your opponent folds then your equity does not matter because the hand is over, but if your opponent does not fold then you need to calculate equity since it is very rare that you are getting your money in drawing dead, and this part of the equation accounts for the times that you suck out on your opponent.
For part 2, what exactly do you mean by bluff raise? My interpretation is that you want to bluff raise a continuation bet or something of that nature.
So lets just do a quick example and show how I would calculate this. I am going to make the assumption that your raise is a pure bluff and that your equity is 0.
Since equity does not matter, the best equation to use is
EV=Pfold * Pot - (1 - Pfold) * Investment and we are going to let
x = Pfold just so it looks a little nicer.
So if the pot is 6bbs, and your opponent cbets 5bbs and you raise to 15bbs the eqaution looks like this
Ev = 0 Pot = 11 Investment = 15 Pfold =x
0 = 11*x - (1-x)(15)
0 = 11x - (15-15x)
0 =11x -15 +15x
15 = 26x
(15/26) = x = 0.5769 = 57.69% = Pfold
So in this example you would need your opponent to fold 57.69% of the time for your bluff raise to be break even. Considering your hand probably has some equity, this is most likely a high estimate but trying to calculate your equity when it is not an all-in situation is rather difficult. If you had a flush draw with 9 outs, you would have about 36% equity IF you got to see the river, but only about a 18% equity to improve by next card. And then you can always try to bluff the turn again so the calculation gets even more complex.
• Bronze
Joined: 28.01.2012
Originally posted by fdel15
For part 2, what exactly do you mean by bluff raise? My interpretation is that you want to bluff raise a continuation bet or something of that nature.
So lets just do a quick example and show how I would calculate this. I am going to make the assumption that your raise is a pure bluff and that your equity is 0.
.
Thread title is EV of a semi-bluff though, so equity is obviously not 0.
• Basic
Joined: 09.07.2012
Originally posted by metza
Originally posted by fdel15
For part 2, what exactly do you mean by bluff raise? My interpretation is that you want to bluff raise a continuation bet or something of that nature.
So lets just do a quick example and show how I would calculate this. I am going to make the assumption that your raise is a pure bluff and that your equity is 0.
.
Thread title is EV of a semi-bluff though, so equity is obviously not 0.
Very true, and at the end of my post I tried describe some of the difficulties of calculating your equity when it is not an all-in situation. If it is an All-in situation and you know your exact equity then you can use the equation
EV = Pfold*Pot + (1-Pfold)*((Equity*(Win+Investment)-Investment))
One benefit of assuming your equity is 0, besides making the math a little easier, is that it gives you a margin of safety. Calculating your opponents folding percentage and even your exact equity vs their hand is not an exact science as you are weighing a bunch of different variables. By assuming you never suck out on your opponent, your break even FE is going to be higher then it actually needs to be. In my example, if our opponent folds exactly 57.69% then instead of breaking even I am actually making a profit depending on how often I suck out. The bigger your draw, and the more you are relying on your showdown equity, the less helpful this assumption is but it is still not completely useless. | 2,655 | 10,496 | {"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-2019-26 | latest | en | 0.917998 |
https://www.marketplace.org/2014/07/21/breaking-down-fees-and-taxes-plane-ticket/ | 1,719,200,483,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198864986.57/warc/CC-MAIN-20240624021134-20240624051134-00558.warc.gz | 757,660,627 | 33,574 | 0:00
0% played
# Breaking down the fees and taxes in a plane ticket
Sabri Ben-Achour Jul 21, 2014
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# Breaking down the fees and taxes in a plane ticket
Sabri Ben-Achour Jul 21, 2014
###### HTML EMBED:
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The Transportation Security Administration is increasing the fees it assesses travelers to \$5.60 per leg of your flight.
What’s that mean? If a layover is more than four hours, then the TSA considers that to be two trips and assesses the fee again. So, that could mean about \$22 extra on your layover flight. That made us wonder what are we paying for air travel these days…. besides the actual airfare, that is.
Let’s say you’ve got a \$500 round trip ticket from New York City to Los Angeles with a long layover in each direction, here’s what you’ll be paying.
(Numbers courtesy of Airlines for America.)
Fare: \$500
TSA 9/11 Fee: \$5.6 per segment x 4 = \$22.40
Federal Aviation Excise Tax: 7.5% = \$37.50
Flight Segment Tax: \$4 per segment x 4 = \$16.00
Airport passenger facilty charge: \$4.50 per segment x 4 = \$18.00
Effective Tax Rate: 18%. This can vary depending on how many layovers and how expensive your fare is. If this example fare were for a direct flight, for example, it would be taxed at an effective rate of 9 percent.
Flying international? International flights have a whole additional set of fees depending on the country and the airport, and these fees can range into the hundreds of dollars.
And let us not forget the fees for some kind of basic rudimentary comfort, says airline analyst Robert Mann, Jr:
Want to pick your own seat? \$0 to \$25
What about a seat with legroom? \$25 up to hundreds of dollars
Check a bag? \$0 to \$80
Oh, you want to carry on that bag? \$0 to \$100
Hungry? How does \$5 for a snack and \$15 for a sandwich sound?
The TSA has said historically, it spends significantly more money on aviation safety than it receives from airlines or passengers. However, the revenues raised from government fees and excise taxes do not directly go to their supposed purpose, says George Hobica, founder of AirfareWatchDog.com.
“A lot of the money actually ends up in the general fund to reduce the deficit and never sees its way as was intended to improve air travel.”
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2:26 | 726 | 2,766 | {"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-2024-26 | latest | en | 0.907477 |
http://www.gadycosteff.com/cql-6-1/line.html | 1,718,647,525,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861733.59/warc/CC-MAIN-20240617153913-20240617183913-00582.warc.gz | 40,268,903 | 9,877 | # line
The line filter lets CQL look ahead or look backwards to determine what follows or precedes the current position.
## Regular expressions
The syntax for line is based on the familiar regular expression syntax.
Regular expressions are usually used as a way to search for strings of characters inside text. For example, the regular expression
That was a (good )+game
could be used to find instances of "That was a good game" or "That was a good good game" or "That was a good good good game" and so on.
To convert the concept of ordinary regular expressions to CQL and apply it on chess positions, two changes need to be made.
### Regular expressions on trees
Usually regular expressions are applied to a linear string of characters, like a text file. However, we can imagine a tree each node of which is labeled with a single characters. Each path from the root to a leaf of the tree denotes a particular linear string. We can match a regular expression to the tree by matching the expression to one of the strings from the root to the leaf.
### Regular expressions in chess
In normal regular expressions, we match a string made up of characters. In chess, instead each character can be any chess position. So a "string" of chess positions is just a linear sequence of chess positions, arranged to make a legal game.
The regular expression itself, now uses filters to match a particular linear sequence of chess positions. For example, the regular expression
check*
will match any sequence of 0 or more positions in which one side is in check. Likewise
check+ --> stalemate
would match any sequence of 1 or more positions of check, followed by a stalemate. (The --> is used to separate different regular expressions)
### Regular expressions in chess trees
Now we combine the concepts of regular expressions on trees and of regular expressions in chess to get what we want: regular expressions on chess trees. A chess tree is just a game tree: each node is a position, and a node is connected to its children by chess moves, according to the PGN file.
Thus, a chess regular expression matches a chess tree if there is a string of chess positions from the root of the tree to a leaf of the tree that is matched by the regular expression.
### Regular expression chess syntax
CQL supports most of the usual character regular expression syntax, applied to filters instead of to other character regular expressions:
regexp meaning character example chess example
. any .z . -->
* zero or more repetitions z* check*
+ one or more repetitions z+ check+
? optional z? check?
{} repetition z{2,3} check{2,3}
() grouping (yz)* (check --> move from k)*
## line syntax
A line filter consists of
1. line
2. optional parameters
3. <-- or -->
4. constituents separated by <-- or -->
The optional parameters are:
• a range indicating that the length of the sequence of positions found must be in the range to match;
• firstmatch indicating search should stop after the first sequence of positions matching the constituents is found;
• lastposition indicating that the position at the end of the sequence of positions that is found is to be returned (rather the length of the sequence)
• singlecolor indicating that only positions whose side to move is the same as to side to move of the position at the start of the line are to be considered
• quiet indicating that comments automatically generated by line are suppressed (these comments are also suppressed if quiet is set in the CQL header)
• nestban indicating that a position cannot start a line if that position was already part of a matching sequence of the same line filter (from an earlier position)
• primarywhen moving to the next position, only consider positions resulting from primary moves
• secondary when moving to the next position, only consider positions resulting from secondary moves
line --> check //current position is a check
line --> check*>5 // at least 5 checks
line 5 100 nestban --> check*
line --> check
--> move from k
--> R attacks _
line --> (move from k-->
move from K)+
## <-- and -->
The arrows <-- and --> are used inside of a line filter to denote the direction of motion through the game tree and to separate individual constituents.
A right arrow --> denotes that the direction is towards future moves: the line is looking forward. The --> comes after the line keyboard and also between entries in parentheses.
If the --> is replaced by the left arrows <-- then the line will look backwards, towards the past and previous moves.
A given line filter can use only one type of arrow: either <-- or -->.
A constituent is a either a filter, or it is formed from another constituent by using the special regular expression characters: +, *, (), {}, or ?.
## line filter semantics
We begin by discussing semantics when all the arguments are filters.
line --> filter1
--> filter2
--> filter3
...
--> filtern
matches the current position if:
• filter1 matches the current position;
• filter2 matches the next position following the current position
• filter3 matches the next position after the next position following the current position
• ... and so on
Before an argument filter is evaluated to determine whether it matches a position, the current position is set to that position.
If variations is set, then line will also consider positions that move into variations. Otherwise, only mainline positions are considered.
For example, the diagrammed position:
D. Gurgenidze 1985, start of sequence
(found from CQL file: checkcheckcheckstalemate.cql)
is matched by the CQL file checkcheckcheckstalemate.cql (when run on sample database). The relevant line filter:
line --> .
--> check
--> check
--> check
--> stalemate
## regular expression special characters
A regular expression always matches the longest possible sequence of positions. The specific characters in detail:
### The '*' symbol
'*' means "repeat 0 or more times".
Thus, consider the filter
line --> not check
--> check*
The last 'check' is modified by the '*' so it is repeated 0 or more times. Thus, the expression is equivalent to:
line --> check // repeat 0 times
or line --> not check
--> check // repeat 1 times
or line --> not check
--> check // repeat 1 times
--> check // repeat 2 times
or line --> not check
--> check // repeat 1 times
--> check // repeat 2 times
--> check // repeat 3 times
....; and so on forever
Therefore, the line filter matches a position if either:
• The current position is not a check, or
• The current position is not a check and the next position is a check, or
• The current position is not a check and the next position is a check and the following position is a check... and so on.
Because of the rule above about matching longest sequence of position, the actual filter will match the longest possible sequence of checks that it can starting from a not check in the current position.
### The '+' symbol
The '+' symbol following a constituent means "repeat 1 or more times". Thus,
line --> move from Q
--> move from [Kk]+
--> mate
will match any position from which the next move is a move by the White queen; following which is a sequence of one or more positions from which a King moves; following which there is a position that is mate.
### { range } repetition symbol
If n is a literal number, like 2, then { n } denotes repetition of the preceding constituent exactly n times. Likewise, {, n} denotes repetition between 0 and n times and {n , } denotes repetition at least n times. If m is also a literal number, then { m , n} denotes repetition between m and n times inclusive.
For example,
line --> move from Q
--> not move from Q {10,20}
--> move from Q
will match a sequence of moves that begins with a move of the a white queen, then has between 10 and 20 consecutive moves of any piece other than a white queen, and ends with a move of a white queen.
Version Notes: the CQL 6.0 syntax allowed a space character instead of the comma as a separator. This syntax is now supported but deprecated.
In CQL 6.0, the regular expression signal {..} was inefficient for a large number of repetitions. That is no longer the case: range-repetition in CQL 6.1 has efficiency comparable to * and +.
### The '?' symbol
The '?' following a constituent means "repeat 0 or 1 times". For example,
line --> move from Q
--> move from k?
--> mate
means
line --> move from Q
--> mate
or
line --> move from Q
--> move from k
--> mate
That is, either White delivers mate with the Queen, or after White's Queen move, black delivers mate by a King move.
### The '()' wildcard symbol
A sequence of filters separated by arrows inside parentheses matches a sequence of consecutive positions that match the filters respectively. This construct is used exclusively with wildcards.
For example, suppose you want to match a white queen move followed by a black move followed by a sequence of White checks by a pawn followed by black king moves followed by mate. You can use this:
line --> move from Q
--> btm
--> (move from P
--> check)+
--> wtm
--> mate
## value of line filter
By default the line filter has as value the number of positions in the longest sequence of positions that was matched. Thus, a line filter can be sorted by putting it in a sort filter:
sort
{line --> check*}
>= 5
or equivalently
sort line 5 1000 --> check*
### {+} and {*}
Sometimes it can be confusing to clearly distinguish between '+' and '*' as arithmetic operators and as wildcards. If you want to be absolutely sure these symbols are interpreted as wildcards, enclose them in braces.
## Linearization: using move inside line when variations is set
When variations is set in the CQL header, line evaluates its constituent move filters differently from usual. (The rules below seem complicated, but they give the intuitive behavior).
The problem these rules are designed to address is that the move filter matches a position X based on every possible move that arises from the position. However, as the line filter goes down the game tree, it only selects a single move at a time from X to evaluate. This can mean that the line filter winds up traversing a move from line that the move filter already rejected.
Suppose for example that a user wants to match positions that from which a bishop promotes, and from which the game lasts at least 10 moves. A natural way to write this is this:
cql (input test.pgn variations)
line 10 1000
--> move promote B
--> .*
Here, the user saying that in the current position, one side promotes to a bishop, and that following that move, there are 0 or more other moves. The range 10 1000 in the next says there must be at least 10 positions contained in the matched sequence.
Now conside the following excerpt from a PGN game:
23. e8Q (23. e8B? Rf4+! =)
23. ...Ng1
24. {many moves omitted}
...
44. Rf3 #
Call the position before white's 23'd move above X. The user does not want the CQL file to match X, because there is only one move following the bishop promotion. However, the above CQL would match the position without the rule specified below:
• When X is evaluated, the move filter evaluates to true, because there is a matching move (a bishop promotion) from X.
• From X, the line filter then evaluates the position after 23. e8Q. This position certainly matches the next constituent of the line filter, namely .*. The line filter goes on to match the position after 23...Ng1 and so on, all the way until 44. Rf3#
• Since this represents at least 40 position, the whole line filter matches X.
To prevent this behavior CQL uses a rule called linearization. Under linearization, before evaluating any filters, the line filter chooses a single "line" in the game tree, from the root to a terminal position. All the move filters are then evaluated as if that "line" gave the only valid moves from a position. The remaining moves are discarded.
For example, in the PGN excerpt above, there are two lines:
• the line 23. e8Q Ng1 ... 44. Rf3# is one line. When the next filter tries to choose the this line, the bishop underpromotion disappears from the game. Thus, the move filter in the CQL file, move promote B no longer matches the position X
• the second line is 23. e8B Rf4+ . When the line filter selects this line, then the position X now does match the move filter, since there is bishop promotion at move 23. However, this line only has a length of three positions (X, and the positions after the next two moves). Since the length of this line is not long enough to fall within the 10 1000 range of the line filter, the line filter does not match X
### Disabling linearization
Suppose a line filter L contains a filter F that contains a move filter M. If F is a : filter, another line filter, a move previous filter, a find filter, or an echo filter, then linearization of L does not affect M.
Thus, linearization may be disabled for a particular move filter inside a line filter by preceding the move filter with currentposition:, for example
line
--> move from q //linearization applies
--> currentposition:
{move from R and move from B} // no linearization
--> mate
To turn off linearization completely, use the -nolinearize option to cql (this option is unsupported).
## Using wildcards to look for maneuvers
Wildcards are useful in isolating the particular pieces you're interested in in a maneuver. For example, in turton.cql, there is a line filter with these subfilters:
line --> ...
--> {not move from Front or Side}*
--> move from Side to Criticalsquare
...
In this particular theme, without going into details, we are particularly interested in the movements of the pieces Front and Side. We want to track where they go. So the first wildcard operator ensures that that we ignore movements of pieces other than those, and just focus on those pieces.
# Examples
The line filter is used through the examples: bristol-universal.cql, bristol1.cql, bristol2.cql, clearance-delayed.cql, consecutive-checks-by-one-color.cql, consecutive-checks.cql, forced-moves-both-sides.cql, forced-moves-either-side.cql, forced-moves-white.cql, lacny.cql, parallelpaths-simple.cql, queentriangulation.cql, staircase-sort.cql, staircase.cql, two-piece-cycle.cql, turton.cql, white-try.cql.
Let's look in detail at how the line filter is used in turton.cql, which expresses the Turton theme. In the Turton theme, two line pieces switch places so that one can support the other. The first line piece has to cross a "critical square" to allow the second line piece to reach that critical square so that the pieces are correctly positioned.
The following study illustrates the theme. In the position below:
Costeff 2015, after 4...Rf4
(found from CQL file: turton.cql)
The first line piece (called Front in the CQL code) is the Qe3. The second line piece (called Side in the CQL code) is the Ra2. The critical square is e2.
The obvious thing for White to do is to move 4.Re2?, preparing a back rank mate:
after 4. Re2? (variation)
(found from CQL file: turton.cql)
Unfortunately, this maneuver fails to 4...R:f7 (all these lines can be seen in the comments in the turton-out.pgn of course).
White needs to get the queen behind the rook. White does this using the Turton maneuver.
White first executes the move 4. Qe7!, preparing the Turton but allowing the queen check 5...Qd1+ leading to the following position:
Costeff 2015, after 4...Qd1+
(found from CQL file: turton.cql)
This is where the Turton maneuver comes into play. The line filter of turton.cql begins
line --> move from Front
comment ("Front moves to " Front
" crossing critical square "
CriticalSquare)
The comment here is not a filter: it is parameter to the move filter. Thus, this line constituent consists of a single move filter with two parameters: from and comment.
Note that this particular constituent will be evaluated for many possible different CriticalSquare values. Due to smart comments however, the actual comment is output only if all the constituents of the line match (and are the longest possible match).
It turns out that when the current position is the above diagram, that Front is Qe7, that Side is Ra2 and that CriticalSquare is e2, all the rest of the line constituents match. That is why the comment is output to the critical move
5. Qe1! Front moves to Qe1 crossing critical square e2
reaching
after 5.Qe1
(found from CQL file: turton.cql)
Now, rather surprisingly the white cannot be taken, as white wins after 5...Q:e1+ 6. K:g2. Thus, the black queen retreat 5...Qd8 in response (glossing over the xray constituent, which eliminates certain pathological cases), we see an idiom that comes up a lot in this kinds of themes:
-->not move from (Front|Side)*
The subfilter Front|Side are the squares on which the pieces Front and Side currently stand (namely a2 and e1 in the current position, but that can change as the current position changes). Thus, not move from (Front|Side) matches a move from some piece other than Front or Side. Finally, the wildcard specifier * means that the filter is repeated 0 or more times. In other words, this constituent has the effect of ignoring all the moves other than those of the pieces we are interested in, Front and Side.
We now have matched (0 moves of) the wildcard constituent, and we reach the position before the rook move:
-> move from Side to CriticalSquare
comment ("Side moves to the critical square: " CriticalSquare)
As before, the comment here is a parameter to the move filter, and due to smart comments is only output when it is part of a successful Turton maneuver. Here, the output is the comment shown to 6.Re2!, namely Side moves to the critical square e2. (Recall that we are assuming CriticalSquare is bound to e2. Otherwise the move filter will not match, and so neither will the line filter, and so a new CriticalSquare value will be tried.)
We then reach the position:
after 6. Re2
(found from CQL file: turton.cql)
which is the same as the earlier diagram showing the position after 4. Re2? in the variation. Thus, White has successfully switched around the rook (Side) and the queen (Front).
The line filter concludes by ignoring any sequence of moves not from from Side or Front as before:
--> not move from (Front | Side) *
and reaches the final move of the Turton: the rook moves to e8, winning:
-->move from Side
reaching the position
after 7. Re8+ 1-0
(found from CQL file: turton.cql)
The final constituent of the filter is the point of the Turton: the queen supports the rook through the critical square:
--> xray (Front CriticalSquare Side) | 4,235 | 18,591 | {"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} | 2.5625 | 3 | CC-MAIN-2024-26 | latest | en | 0.927472 |
http://mizar.uwb.edu.pl/version/current/html/proofs/int_5/34 | 1,568,920,857,000,000,000 | text/plain | crawl-data/CC-MAIN-2019-39/segments/1568514573570.6/warc/CC-MAIN-20190919183843-20190919205843-00209.warc.gz | 120,062,486 | 2,180 | let a be Integer; :: thesis: for p being Prime st p > 2 & a gcd p = 1 holds
Lege (a,p),a |^ ((p -' 1) div 2) are_congruent_mod p
let p be Prime; :: thesis: ( p > 2 & a gcd p = 1 implies Lege (a,p),a |^ ((p -' 1) div 2) are_congruent_mod p )
assume that
A1: p > 2 and
A2: a gcd p = 1 ; :: thesis: Lege (a,p),a |^ ((p -' 1) div 2) are_congruent_mod p
not p divides a by ;
then A3: a mod p <> 0 by INT_1:62;
A4: p > 1 by INT_2:def 4;
then - p < - 1 by XREAL_1:24;
then A5: (- 1) mod p = p + (- 1) by NAT_D:63;
per cases ;
suppose A6: a is_quadratic_residue_mod p ; :: thesis: Lege (a,p),a |^ ((p -' 1) div 2) are_congruent_mod p
then (a |^ ((p -' 1) div 2)) mod p = 1 by A1, A2, Th17;
then (a |^ ((p -' 1) div 2)) mod p = 1 mod p by ;
then (a |^ ((p -' 1) div 2)) mod p = (Lege (a,p)) mod p by A6, Def3, A3;
hence Lege (a,p),a |^ ((p -' 1) div 2) are_congruent_mod p by NAT_D:64; :: thesis: verum
end;
suppose A7: not a is_quadratic_residue_mod p ; :: thesis: Lege (a,p),a |^ ((p -' 1) div 2) are_congruent_mod p
then (a |^ ((p -' 1) div 2)) mod p = p - 1 by A1, A2, Th19
.= (Lege (a,p)) mod p by A5, A7, Def3 ;
hence Lege (a,p),a |^ ((p -' 1) div 2) are_congruent_mod p by NAT_D:64; :: thesis: verum
end;
end; | 555 | 1,208 | {"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} | 3.453125 | 3 | CC-MAIN-2019-39 | latest | en | 0.598738 |
https://yutsumura.com/mathematics-about-the-number-2017/triangle2017/ | 1,603,399,160,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107880038.27/warc/CC-MAIN-20201022195658-20201022225658-00398.warc.gz | 996,576,090 | 22,527 | # Pythagorean triple 2017
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This site uses Akismet to reduce spam. Learn how your comment data is processed. | 875 | 2,710 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.84375 | 4 | CC-MAIN-2020-45 | latest | en | 0.717762 |
https://code.djangoproject.com/ticket/361 | 1,709,494,215,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947476397.24/warc/CC-MAIN-20240303174631-20240303204631-00310.warc.gz | 177,381,565 | 18,993 | Opened 19 years ago
Closed 13 years ago
# Some Basic Math Filters
Reported by: Owned by: ilikeprivacy@… Adrian Holovaty Template system normal filter math Danilo Bargen Unreviewed no no no no no no
## Description
This may be the wrong place to submit this...
I wanted some basic math filters (eg. add multiply divide subtract), there is already add so I created the others, no doubt probably the easiest filters I could create, though someone else may find them useful.
in django/core/defaultfilters.py
```def mult(value, arg):
"Multiplies the arg and the value"
return int(value) * int(arg)
def sub(value, arg):
"Subtracts the arg from the value"
return int(value) - int(arg)
def div(value, arg):
"Divides the value by the arg"
return int(value) / int(arg)
template.register_filter('mult', mult, True)
template.register_filter('sub', sub, True)
template.register_filter('div', div, True)
```
### comment:1 by Jacob, 19 years ago
Resolution: → invalid new → closed
The ticket system isn't the rigt place for this; feel free to create a wiki page for it for now; in the future we're going to have some sort of contributed apps repository for stuff like this.
### comment:2 by till@…, 16 years ago
Resolution: invalid closed → reopened
this is a very simple enhancement that would nicely blend into the set of django-supplied template filters.
maybe someone could reconsider inclusion of this trivial patch.
today it would look like this:
in django/template/defaultfilters.py
```def mult(value, arg):
"Multiplies the arg and the value"
return int(value) * int(arg)
mult.is_safe = False
def sub(value, arg):
"Subtracts the arg from the value"
return int(value) - int(arg)
sub.is_safe = False
def div(value, arg):
"Divides the value by the arg"
return int(value) / int(arg)
div.is_safe = False
template.register_filter('mult', mult, True)
template.register_filter('sub', sub, True)
template.register_filter('div', div, True)
```
### comment:3 by Jacob, 16 years ago
Resolution: → wontfix reopened → closed
Please don't reopen tickets marked closed by a committer. Take this to djangosnippets - it doesn't need to be part of Django.
### comment:4 by Dan Loewenherz, 13 years ago
"add" is a filter, yet none of the other basic math operations are. This seems a little inconsistent to me.
it doesn't need to be part of Django.
And yet something like "ipsum lorem dolor sit amet" does? I think this deserves a 2nd look.
### comment:5 by diramazioni, 13 years ago
As newcomer to django I find really weird there is no math operations or the elseif operator in the template system. They are so common in others... Nobody willing to explain why there are such limitation?
### follow-up: 7 comment:6 by anonymous, 13 years ago
Resolution: wontfix closed → reopened
Django runs it's shop like the Third Reich... ADD MATH FILTERS
### in reply to: 6 comment:7 by Łukasz Rekucki, 13 years ago
Django runs it's shop like the Third Reich... ADD MATH FILTERS
Your behavior doesn't really deserve a comment, especially that you were brave enough not to sign it with your name. Normally people discuss this kind of stuff on the developers list, but I'm not sure if I want to discuss with this kind of offensive comments.
### comment:8 by Łukasz Rekucki, 13 years ago
Resolution: → wontfix reopened → closed
### comment:9 by anonymous, 13 years ago
Easy pickings: unset unset
it doesn't need to be part of Django.
May I know why it according to you doesn't need to be part of Django? Every time I find myself writing these filters yet again I have the oposite impression. There are situations where not having the ability to do basic math in templates results in code that is more complicated (thus error prone) and longer to write then it would otherwise be necessary. I find this unwillingness to include something as basic and harmless as these filters are rather absurd.
### follow-up: 11 comment:10 by Russell Keith-Magee, 13 years ago
This topic has been discussed repeatedly on django-dev and django-users. The reasoning *isn't* absurd -- it's part of the fundamental design intent of the template language. The reasoning is that Django's template language shouldn't encourage putting business logic in templates. Providing extensive tools for in-template math would encourage the placement of business logic in templates.
### in reply to: 10 comment:11 by anonymous, 13 years ago
This topic has been discussed repeatedly on django-dev and django-users. The reasoning *isn't* absurd -- it's part of the fundamental design intent of the template language. The reasoning is that Django's template language shouldn't encourage putting business logic in templates. Providing extensive tools for in-template math would encourage the placement of business logic in templates.
Django templates is the most inconsistent and horrible templating language available. Can you not just switch to Jinja2 and be done with it?
I want to add two datetimes together in my template. Nice, now I have to do it in several places in my code. Not good.
### comment:12 by anonymous, 13 years ago
Unfortunately, your criticism lacks details that would allow us to understand your problem and actually improve the template language.
I think template tags are appropriate for the use case you're describing.
### comment:13 by adam.g.pullen@…, 13 years ago
Here is sample code where simple math in the template engine may make sense.
It is totally presentation related.
This post should not be taken as critizam, more as developers trying to improve the framework.
We have a list of items that need to be displayed in a "3*n" table
There are 2 issues
1) Primary Issue: The forloop.counter starts at 1 which, depending on how you look at it, is wrong. It should, to remain consistant with python and most other languages, start at 0.
2) Due to the lack of the simple math functions (requirement driving from incorrect forloop.counter value) I will have to try to implement the code listed above.
```{% comment %}
Counter is used to perform the "row grouping" when mod 3 == 0 then a new row is started
When counter is > 0 then the row is ended
{% endcomment %}
{% for item in items %}
{% if counter|divisibleby:3 or forloop.counter == 1 %}
{% if forloop.counter > 1 %}
</div>
{% endif %}
<div class="creator_gallery_row">
{% endif %}
{% include "terra_creator/gallery_item.html" %}
{% endfor %}
```
Just a note for reference: divisibleby "DOES need to be part of Django" but multiplication not?
Ok durring writing this i found that
will solve my problem
I still think that you should listen to your users. It up to them to decide if they want to "bastardise" their application by putting application logic into the view, not the framework.
### comment:16 by Robert Wallner, 12 years ago
I think this is a common misconception. Business logic code doesn't belong in the template, but at the same time, template logic doesn't belong in the code. So, not having arithmetic operations in the template do exactly the opposite. It encourages users to put the template logic inside their code, which is as bad as placing business logic code inside templates. Reminds me of Ocam's razor :)
### comment:17 by anonymous, 12 years ago
I know this is not the place, but I have to say it. It would be nice to be able to do things like:
```<div style="padding-top: {{ 200 - thumb.heigth }}">
<img src="{{ thumb }}">
</div>
```
### comment:18 by Cal Leeming, 12 years ago
Today I found the need for having math filters in the template, specifically for visualization (changing 0.01PPM into 1p) which is not business logic but indeed template logic.
Just as I was about to add a comment saying "omfg why doesn't Django support this", I realized why the core devs are rejecting this. And the simple reason is, it would get abused like crazy and it's would be a slippery path to failure. Sure it would be useful for sane developers on specific use cases, but it would be abused by the majority.
Although Adam makes a good point with this comment:
```It up to them to decide if they want to "bastardise" their application by putting application logic into the view, not the framework.
```
dloewenherz also makes another good point:
```"add" is a filter, yet none of the other basic math operations are. This seems a little inconsistent to me.
```
Could a core developer perhaps explain what the reasoning was behind having add, but not any of the others?? I had a look around the archives, but couldn't seem to find any justification for this. It seems to me that we should either have all of them, or none at all??
It seems this topic hits a nerve with a lot of people, so if a core dev is able to give some answers on the above, we will finally be able to put this issue to rest with a concrete justification.
### comment:19 by Cal Leeming, 12 years ago
As a side note, if math filters were to ever get accepted into the core, it would be so much better if they weren't as template tag filters, but instead as native expressions with nested support:
For example:
```{{ (someval * 10) + 5 }}
```
### follow-up: 21 comment:20 by will@…, 11 years ago
I would like to have a "subtract" filter for this use case:
I want to only show the first 10 items in a list, then have a final line saying how many other items there were,
E.g.
"And 3 other items..."
This is purely a display (template) logic, so it should be supported by the template language (same as 'add').
```{% for l in list_items %}
{% if forloop.counter < 10 %}
<li>l </li>
{% else %}
{% ifequal forloop.counter 10 %}
<li>
And {{ list_items|length|subtract:10 }} other objects...
</li>
{% endifequal %}
{% endif %}
{% endfor %}
```
### in reply to: 20 comment:21 by Preston Holmes, 11 years ago
I would like to have a "subtract" filter for this use case:
If you feel you have a display related argument here - you are better off opening a new ticket than posting a comment on a very old wontfix ticket.
If you choose to do so, please explicitly reference this ticket, and explain why you think it is a different circumstance.
### comment:22 by Danilo Bargen, 11 years ago
will@..., you could also add -10 instead of subtracting 10 in your case: `list_items|length|add:-10` | 2,405 | 10,266 | {"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-2024-10 | latest | en | 0.753099 |
https://www.exactlywhatistime.com/timezone-converter/cst-to-cst/1-25-am | 1,696,095,746,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510697.51/warc/CC-MAIN-20230930145921-20230930175921-00361.warc.gz | 798,350,039 | 5,026 | # 1:25 am CST is 1:25 am CST
Here, we have calculated the local time in Central Standard Time based on the given time of 1:25 am in Central Standard Time. Our precise algorithm takes into account various factors, such as daylight saving time and the specific offset between the two time zones, to provide you with an accurate conversion.
## How to calculate 1:25 am CST to CST yourself:
1. Identify the time zones: Determine the time zones for both the starting location (CST) and the destination location (CST).
2. Calculate the offset: Find the difference in hours (and sometimes minutes) between the two time zones. This is usually expressed as a positive or negative offset from Coordinated Universal Time (UTC). In this case, the offset between Central Standard Time and Central Standard Time is 0 hours.
3. Consider daylight saving time: Check if either the starting or destination time zone is currently observing daylight saving time. If so, make sure to adjust your calculation by adding or subtracting one hour as needed.
4. Apply the offset: Add or subtract the calculated offset to the given time in the starting time zone (1:25 am). Be mindful of any potential "wrap-around" when crossing the International Date Line or switching between AM and PM.
By following these steps, you can convert the time between Central Standard Time and Central Standard Time accurately. However, if you prefer a hassle-free way to find the time in another part of the world, feel free to use our Time Zone Conversion Calculator on the directory page.
### Central Standard Time
Central Standard Time (CST) is used in the central part of the United States, including cities like Chicago and Dallas. It is 6 hours behind UTC during standard time and 5 hours behind during daylight saving time.
• Chicago
• Dallas
• Houston
• New Orleans
### Central Standard Time
Central Standard Time (CST) is used in the central part of the United States, including cities like Chicago and Dallas. It is 6 hours behind UTC during standard time and 5 hours behind during daylight saving time.
• Chicago
• Dallas
• Houston
• New Orleans | 447 | 2,120 | {"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} | 2.78125 | 3 | CC-MAIN-2023-40 | latest | en | 0.897088 |
https://www.widelands.org/forum/topic/2910/?page=1 | 1,580,029,772,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579251687958.71/warc/CC-MAIN-20200126074227-20200126104227-00322.warc.gz | 1,157,590,690 | 6,670 | # Topic: New soldiers fights program
Posted at: 2017-08-23, 00:50
### Hi everyone!
My last thoughts were around soldiers problem: my fight simulation was doing great job, but it wasn't exact. My target was to create complete and exact simulation of fights. It isn't an easy job. So I was thinking about this for a long time and finished with (another) program .
### The code:
You can see the code here:
https://github.com/einstein13/wl_soldiers_exact
Old program:
https://github.com/einstein13/wl_soldiers
The old one is checked many times and should not contain any bug. The new one is rather "new" and I don't know if there is any problem. Hope not!
### Comparison between tables:
Old one:
vs. bar_10 emp_10 atl_10
bar_10 54.2% 52.2% 50.5%
emp_10 55.4% 53.3% 48.9%
atl_10 56.7% 57.3% 53.1%
##### (taken from: https://wl.widelands.org/forum/post/19573/)
New ones:
vs. bar_10 emp_10 atl_10
bar_10 54.8% 52.2% 50.6%
emp_10 55.3% 53.3% 49.5%
atl_10 56.7% 56.8% 53.1%
### How I calculated all the stuff?
I have splitted calculations into three parts:
1. Probability of killing any soldier by any other soldier in N hits (with no miss!)
2. Probability of killing any soldier at exactly M tries and N no-miss hits
3. Probability of not-killing any soldier at exactly M-1 tries and up to N' no-miss hits
So for example if we have soldier A and B, A is an attacker (attacks first).
Then we calculate how many hits are needed to kill A -> B and B -> A, and all probabilities for that. So we will have a matrix of all possibilities:
vs. N_a1 N_a2 ...
N_b1 p(N_a1)*p(N_b1) p(N_a2)*p(N_b1) ...
N_b2 p(N_a1)*p(N_b2) p(N_a2)*p(N_b2) ...
... ... ... ...
To calculate that probabilities I was trying to calculate convolution (https://en.wikipedia.org/wiki/Convolution) of continuous uniform distribution (https://en.wikipedia.org/wiki/Uniform_distribution_(continuous)) and then calculate cumulative distribution function (https://en.wikipedia.org/wiki/Cumulative_distribution_function) for each result to get probabilities p(N_aX) & p(N_bX). But I failed. I've noticed that for N hits you can just draw N-dimensional cube and cut N-dimensional pyramid based on one apex of this cube. The probability is a ration between two volumes. OK, not exactly, but after a few modifications you will get it!
I can't prove this, so if there is a problem, please prove it. If not, please prove that too!
After that we have to calculate how possible is to kill B with N_a hits while he can kill A with N_b hits. That contained some steps inside:
Possibility of killing B: we know exactly how many are miss-tries and a possibility of that, also how many are hit-tries and a possibility of that. The only missing factor is how many times we can "pick" those tries between all tries. That is binomial coefficient (https://en.wikipedia.org/wiki/Binomial_coefficient).
Possibility of not-killing A: we have to calculate all the possibilities of correct miss-hit correct combination. After that we sum all them and get the result. With binomial coefficient in use we can simplify that a bit and make it faster for the computer.
Last part is to make the calculations working. Because we have infinite sum of all possibilities of 5-tries fights, 6-tries fights, 7, 8, ... and so on fights (with N_a and N_b kill-wins), we have to pick a border. Let it be 0.000...1 (10^-10) of "best" probability. So approximately after 15 cases, the calculations stopped and you can sum all the possibilities.
After all those problems solved I've spotted another one: computer calculations aren't exact (about 15-digits exact for standard floating point numbers). So I've researched a bit and found that Python has rational numbers and calculations on them are exact (you can't lose information with infinite numerator and denominator precision).
In short version that is whole my engine and hope it has no mistakes. Probably in a few weeks I will provide a pdf with all the equations I've used, but I am not sure if anybody is interested on that .
If you have any questions, please write them! Maybe I've made terrible mistake or you proved something else - check that!
einstein13
calculations & maps packages: http://wuatek.no-ip.org/~rak/widelands/ | 1,079 | 4,225 | {"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-2020-05 | latest | en | 0.897584 |
https://www.jiskha.com/display.cgi?id=1254957401 | 1,516,781,345,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084893530.89/warc/CC-MAIN-20180124070239-20180124090239-00593.warc.gz | 916,130,757 | 4,062 | # Algebra
posted by .
(-6+b)^2
• Algebra -
distribute first which will give you
• Algebra -
WRONG!
(-6+b)^2
= (-6+b)(-6+b)
= 36 - 6b - 6b + b^
= b^2 - 12b + 36
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## an unpublished tip for pyradev users
Started by gef, 02:12, 03 April 10
0 Members and 1 Guest are viewing this topic.
#### gef
Hi,
this is a flashback from my CPC activities 25 years ago...
...how to hack well protected games.
if anyone of you is still playing with Pyradev, trying to hack games that use the R register,
XOR the hell out of them, play funny tricks with interrupts etc, there is an easy way out:
you need a small patch which I can't supply but, could describe in good detail, since I invented it ;-)
The patch allows you to trace through any Z80 code and is based on two important facts:
* The R register is changing deterministically among subsequent commands (say, by +1 or +2).
* The state of the IFF (Interrupt flag) is actually possible to be found with a "LD I,A" or so.
This latter tip is very crucial since some games/hacking protections require a certain state
of the interrupt, at any given point of the code execution. The funny aspect is that I owned
both a Zilog and a Mostek (=z80 clone manufacturer) manual and the important information
was *only* referenced in the Mostek documentation, though it has been working fine in any
z80 implementation I was ever able to put my hands on. Bingo!
So, the rough sketch of the code patching Pyradev's breakpoint logic was:
a) Store the value of A accumulator immediately in a memory byte "X"
b) Do LD I,A and then branch according the Overflow flag (AFAI remember that's the one),
effectively storing either of "EI" or "DI" commands, for later use upon exiting - major deal for step (g)
c) Store the value of R in A, then store in memory "Y".
d) Jump back in Pyradev (after having forcing it to exit back to next step "f")
e) Perform your regular Pyradev tasks, as normal (disassembly, breakpoints etc)
f) Put "Y" value in A, and perform subtract in 7-bit logic of a constant "Z" (read how R functions in z80!)
g) do EI or DI, in accordance with step (b) above
h) Store A to R
i) Get value for A from memory "X"
j) jump back to original (game) code as if nothing happened
All the trick was tuning "Z" to the correct value for getting code transparency!
But, that should be a constant number and it's easy to track down with minimal effort.
Needless to say, no hacking-protection attempt was able to resist the patch...
...the first game that fall back then was "Live and Let Die", then "Double Dragon", then many others
... if Internet existed, this patch would have been extremely popular; effective and just <50 bytes to patch!
You are the first guys to know: I had a great time when I was using this ;-)
cheers,
F.
#### gef
#1
So,
Quote from: gef on 02:12, 03 April 10
[...]
The patch allows you to trace through any Z80 code and is based on two important facts:
* The R register is changing deterministically among subsequent commands.
* The state of the IFF (Interrupt flag) is actually possible to be found with a "LD I,A" or so.
[...]
b) Do LD I,A and then branch according the Overflow flag (AFAI remember),
[...]
OK, the assembly command is actually written LD A,I and the z80 undocumented trick
is explained here in sections 6 an 8 (it is even buggy in certain implementations, wtf!):
http://www.worldofspectrum.org/z88forever/dn327/z80undoc.htm
I have obviously been influenced by the Intel mnemonics rather than AT&T over the years!
http://en.wikipedia.org/wiki/X86_assembly_language#Syntax
btw.
The clone company' name was indeed Mostek, not Mustek; I've got gray hair since those times ;-)
ps.
Someone might ask why this has been posted under Emulators...
...the reason is that I'd expect people having used Pyradev to be doing this kind of stuff these days;
if you know better how to navigate in the forum, please supply advice.
Also, if people reading this are close to Lugano, Switzerland, please get in contact.
enjoy,
F.
#### gef
Quote from: gef on 12:18, 03 April 10
I have obviously been influenced by the Intel mnemonics rather than AT&T over the years!
http://en.wikipedia.org/wiki/X86_assembly_language#Syntax
Actually, it seems that the common assembly style for CPCs *was* called Intel style... whatever.
Anyway, I will be strictly talking about Emulators now:
* Has anyone managed to get an CPC6128 emulation running at good state in recent smartphone?
I am specifically interested for compatibility in the iPhone vs Android debate
and I am willing to make my buy based solely on this answer
It has been a great outcome to use my current Nokia N95 with s60cpc/caprice emulators,
hooked on my HD projector along with a bluetooth keyboard: highly recommended experience!
Also, if you are Spindizzy fans, please report here under what platforms you have revived the game
(including PCs etc).
tia,
F.
#### Devilmarkus
Sorry, I don't own any cellphone or something like that running an emulator.
I only play games with my real CPC or with the emulator I am coding...
About spindizzy:
Yes, was funny in the 80s, but today I prefer other CPC games
I often have been asked why I don't adapt my emulator for Java-cellphones.
The answer is quite simple:
- in 2004 I found JEMU (http://jemu.winape.net .... yes, JavaCPC has JEMU core inside!)
- I begun modifing it's sources and also adding improvements, new apps, etc...
- I really don't know much about Java. All I did was "doing by learning".
- Also I don't know anything (or maybe 10%) of emulating things... I'm testing around with source modifications.
- Java for cellphones is much different than JDK for PC's
- JavaCPC/WebCPC is also open source. (Like JEMU) So maybe it will be found a person who can adapt it for phones, too
When you put your ear on a hot stove, you can smell how stupid you are ...
Amstrad CPC games in your webbrowser
JavaCPC Desktop Full Release
#### Cholo
Thanks for posting. This pretty much answers one of my oldest still lasting questions:
1. how did people hack those tape protections before the multifaces showed up and
2. how did the programmers compile a protection like that.
Back in the 90's i did manage to read 2 books on machine code and did get fairly fluid in assembler. Even tho as a teenager i had a hard time seen the "big picture" in mc/assem.
Did try to dissassemble and later singlestep into a couple of tape protections, but came to a loss on how to continue. Knew there had to be a trick to it of cause. If it was just lack of knowledge and/or/of additional hardware i didnt know.
Seems like i know at least the theory how to get into a protection and i assume a similar trick can be used to compile one too.
#### arnoldemu
Quote from: Cholo on 20:48, 04 April 10
Thanks for posting. This pretty much answers one of my oldest still lasting questions:
1. how did people hack those tape protections before the multifaces showed up and
2. how did the programmers compile a protection like that.
Back in the 90's i did manage to read 2 books on machine code and did get fairly fluid in assembler. Even tho as a teenager i had a hard time seen the "big picture" in mc/assem.
Did try to dissassemble and later singlestep into a couple of tape protections, but came to a loss on how to continue. Knew there had to be a trick to it of cause. If it was just lack of knowledge and/or/of additional hardware i didnt know.
Seems like i know at least the theory how to get into a protection and i assume a similar trick can be used to compile one too.
I used another trick for hacking tapes to disc but this only works well for 128k machines and you need a reset switch.
You setup the 2nd 64k ram to be a mirror of the 1st.
You switch to it and load the game.
Many times the loader will run and the computer will effectively reset, but the loader is now decoded in the 2nd bank of ram.
Then you could look at the loader to find where the data is loaded and where the program is called.
Then patch the call to the program so that it loops forever.
Again load in 2nd bank. wait for it to finish loading. reset the computer using the switch.
Dump the data from the 2nd bank and convert into files.
Make a loader. job done.
My games. My Games
My website with coding examples: Unofficial Amstrad WWW Resource
#### Gryzor
Quote from: arnoldemu on 07:10, 05 April 10
I used another trick for hacking tapes to disc but this only works well for 128k machines and you need a reset switch.
But how are you sure the loader won't overwrite your cloning code?
#### arnoldemu
Quote from: Gryzor on 08:12, 05 April 10
But how are you sure the loader won't overwrite your cloning code?
The code copies the lower and upper jumpblocks and firmware data.
So that it is almost like having a virtual machine in the 2nd 64k.
But the good thing about the 2nd 64k is that it is not cleared by basic.
If you reset the computer using a reset switch the data remains good.
If you turn the computer off then the data will be corrupted because there will not be power to refresh it.
The reason that some loaders will reset the computer is that they change the memory configuration.
So you can get the state of the memory at that point (when it changes memory configuration it then starts to execute code in the first 64k).
But you are also free to wait until a time of your choosing and press the reset button.
This is how I hacked a lot of games from tape to disc.
The main program was often converted in this way. But of course you have to look at the loader to work out the memory range it actually loads. So that you can save the correct size of data. Also you have to join the bits of memory together because you access the extra ram in 16k pages.
Once the main program had been converted I looked through it to find the level loader code.
Then I wrote another program which called this, loaded each block and then saved this to disc.
When that was done, the main program had to be patched to load the files in.
Often there wasn't enough room to enable the firmware so you have to copy the data in the firmware area into the screen, enable firmware, load files, turn off firmware and copy the data back.
My games. My Games
My website with coding examples: Unofficial Amstrad WWW Resource
#### OCT
Quote from: arnoldemu on 07:10, 05 April 10
I used another trick for hacking tapes to disc but this only works well for 128k machines and you need a reset switch.
You setup the 2nd 64k ram to be a mirror of the 1st.
You switch to it and load the game.
Many times the loader will run and the computer will effectively reset, but the loader is now decoded in the 2nd bank of ram.
Quote from: arnoldemu on 08:31, 05 April 10
The code copies the lower and upper jumpblocks and firmware data.
So that it is almost like having a virtual machine in the 2nd 64k.
I understand you use a program to copy over the RAM to the second bank.
There's yet another trick that involves a little hardware hack, which has the advantage of being undetectable by software:
On a 6128, you use a toggle switch to exchange the two bank-select lines, plus a button to ground READY (a.k.a. /WAIT) on the Expansion Port.
Once the loading and decoding is done, hit the button to temporarily freeze the Z80 in its tracks, switch the memory banks and then release the freeze button. If the machine won't already reset on its own, hit the reset button (conveniently also located on the expansion port, as in my "CPC Brake" which also featured slow-motion. There you are in the other bank, ready to switch in and save the 4*16k pages via the usual MEMORY &3FFF:OUT &7F00,&C4 etc. (leaving you just the task of figuring out the jump point and maybe writing a loader to run on 64k machines as well).
#### arnoldemu
Quote from: OCT on 13:40, 05 April 10
I understand you use a program to copy over the RAM to the second bank.
There's yet another trick that involves a little hardware hack, which has the advantage of being undetectable by software:
On a 6128, you use a toggle switch to exchange the two bank-select lines, plus a button to ground READY (a.k.a. /WAIT) on the Expansion Port.
Once the loading and decoding is done, hit the button to temporarily freeze the Z80 in its tracks, switch the memory banks and then release the freeze button. If the machine won't already reset on its own, hit the reset button (conveniently also located on the expansion port, as in my "CPC Brake" which also featured slow-motion. There you are in the other bank, ready to switch in and save the 4*16k pages via the usual MEMORY &3FFF:OUT &7F00,&C4 etc. (leaving you just the task of figuring out the jump point and maybe writing a loader to run on 64k machines as well).
Nice.
I never thought to use hardware to help.
Is this info on the wiki? because it would be useful to have it there.
My games. My Games
My website with coding examples: Unofficial Amstrad WWW Resource
#### OCT
Quote from: arnoldemu on 17:14, 05 April 10Nice.
I never thought to use hardware to help.
Is this info on the wiki? because it would be useful to have it there.
To the best of my knowledge, the bank exchanger was first contrived (and instructions with a photo -which I have dug up again indeed- sold for a nominal fee) by a German scener changing his pseudo from Khany to Cherry-T in 1993.
Since copyright subsists in these materials as a matter of law, I cannot republish them, but I hope that someone is still in contact with the original author (not a member of this site nor CPC-Forum.de under any of the above user names I'm afraid) to obtain the required permission.
What I may say at this time is that it works by unsoldering the right-hand pins of R160 and R161 and connecting them via pull-ups to the nearest source of +5V which happens to be the top leftmost pin of IC207, and wiring a double-pole double-throw toggle switch to exchange the bank-select lines.
While I wouldn't know how to best place new articles on each in http://www.cpcwiki.eu/index.php/Peripherals, I attach an image showing the "CPC Brake" a.k.a. "CPC Slow Motion", switchable between slowdown and stop for both the "Centronics type" (Plus and German 6128, left, with a slide switch) and "PCB/card type" connectors (continental 464/664 and UK CPCs except Plus, right on Multiface II, with a jumper), which conveniently feature both the brake/stop (car-like foot pedal) and reset buttons.
The banks are of course exchanged by one of the toggles to the right of the hood (which BTW features the era's Amstrad commercial sticker "computers for people with brains" - as we have always known ).
#### redbox
These methods are really cool and it's nice to see how people did it.
One of the (simple) methods I used was if there was a poke for the game (like in 'Cheat Mode' in Amstrad Action) then I knew this had to break the game's copy protection to work. Therefore I would disassemble the cheat and use it to transfer the game to disc.
Nowhere near as neat as the above, but it worked for me for a few games
#### Cholo
Nice, didnt know there was so many ways, to get past a tape protection.
Didnt even get the idea to look at tape pokes as a way to look at the protections, lol. Even tho later on i did look at some pokes to find the right spot to poke games in a emulator.
#### gef
Quote from: Cholo on 20:48, 04 April 10
Thanks for posting. This pretty much answers one of my oldest still lasting questions:
1. how did people hack those tape protections before the multifaces showed up and
2. how did the programmers compile a protection like that.
[...]
I can answer #2 quite well for the cases I have encountered:
a) Some trivial cases were (XOR) code transformation trickery, _without_ using R register.
b) The most commonly found titles had one or two invocations of loops using the R register.
c) The titles that were bound to become commercial success would use some annoying tricks:
c1. using features of the magnetic media, that were characteristic of that particular game
c2. multiple nested XOR-decoding loops using R register in a back-to-back fashion
a) and b) would fall easily with some mild application of Pyradev, time and patience.
category c) was a different story altogether, since they required fairly good understanding
of many aspects of the system and fiddling with the game in non-trivial ways. examples:
c1. double dragon was becoming very annoying on our cpc6128, since it would force
disk drive to make worrysome noises and would not always actually start. Eventual
reverse engineering on both soft-n-hard fronts, forced me to learn the ins and outs
of NEC765 floppy disk driver (which was a big deal to occupy a single chip back then),
write a sector reader for it *from scratch* (aka device driver nowdays), understand
the mechanical properties of my drive and that track #42 was not really available
until the moment I unscrewed and adjusted a physical limiter inside that drive...
there were also some other disk tricks with oddly formatted sectors etc and I guess
tape protections were of similar style with timing-sensitive write/read operations.
c2. OK this was a category on its own. The amount of work involved here was non-trivial.
Even if you mastered Pyradev and assembly very well, it was not easy at all to push
breakpoints at interesting code points at will, because that would imply mangling the
R register which was part of the decoding-soft-logic itself. If the number of loops was
small (say, <5) with stepwise patience and craft the game could fall. The problem here
was though that the guys protected the game had time with their side, since they could
easily automate the creation of multiple nested tough-to-break R-based XOR loops.
R was difficult because it increased by 1 or 2 with each and every processor operation!
So, if you run Pyradev or another monitor in between, your breakpoint would ruin it all.
The antidote to this technique was to patch Pyradev in the way described earlier;
I think the first game that eventually got hacked like this was quite more >20 loops,
so there's no easy way someone could circumvent it by manual code-relocation tricks...
...or use the trick with bank-shifting, which I didn't know and was ingenious indeed!
For the record: when I was doing all these operations I was maximum 14 y.o.
(and my feeling since then is, that I just decline over time! LOL)
so next time you see a teenager busy with some weird computer problem,
better not underestimate the crap with which he might be fiddling with ;-)
enjoy,
F.
#### arnoldemu
Quote from: gef on 01:56, 06 April 10
I can answer #2 quite well for the cases I have encountered:
a) Some trivial cases were (XOR) code transformation trickery, _without_ using R register.
b) The most commonly found titles had one or two invocations of loops using the R register.
c) The titles that were bound to become commercial success would use some annoying tricks:
c1. using features of the magnetic media, that were characteristic of that particular game
c2. multiple nested XOR-decoding loops using R register in a back-to-back fashion
Yes indeed.
Speedlock protection had about 80 of these XOR type loops. Then it tried to detect the multiface. It was a beast of a protection and always changing.
I too was about 13/14 when I was converting tapes to discs. I did this mostly because the disc games were so expensive.
I initially started learning assembler and I was very lucky because I got a SOFT968 manual and my father could obtain the chip datasheets easily from his company library.
I never tried using Pyradev. First I used the very poor Amsoft assembler (MONA etc). With it's terrible input and line numbers. Then I moved onto Maxam (I had a strange chip on a PCB which connected to the back of the computer. I've not seen another since and I've since sold this.)
I never really found a debugger that I could get on with on the CPC.
It is interesting to read how others hacked and looked at code.
My games. My Games
My website with coding examples: Unofficial Amstrad WWW Resource
#### Gryzor
Quote from: OCT on 20:11, 05 April 10
Since copyright subsists in these materials as a matter of law, I cannot republish them,
Copyright issues? What are you talking about? I can publish it if you want, I don't think we should be exactly scared someone will come after us for hacking (again) a game from 1985...
#### OCT
Quote from: arnoldemu on 08:31, 05 April 10
The code copies the lower and upper jumpblocks and firmware data.
So that it is almost like having a virtual machine in the 2nd 64k.
I wonder what the code was that allowed you to page in the whole 4*16k of second-bank memory without hardware help.
AFAIK the OUT &7F00,&C4+page would only show 16k at a time at &4000 to &7FFF, which is an uncommon location for loaders.
#### OCT
#17
Quote from: Gryzor on 07:54, 06 April 10
Copyright issues? What are you talking about? I can publish it if you want, I don't think we should be exactly scared someone will come after us for hacking (again) a game from 1985...
I'm not into cracking protections at all, nor scared for that matter - it's just that the author and photographer of a (really nice BTW) writeup on the CPC6128 hardware modification does hold rights to it (and one day even his heirs will, for 70 years post mortem auctoris in most jurisdictions), just like the Heise publishers do for the other piece which first showed how to tamper with bank selection and expand the internal memory, "Aus David wird Goliath: 512 KB RAM für Schneider CPC", c't 10/1987, 156-162, as referenced in http://www.cpcwiki.eu/index.php/C't_512_KB_internal_RAM_expansion and much later picked up again/reinvented for up to 4MB by Yarek.
As for the games themselves, in rare cases there is some commercial interest indeed (in particular in the level maps, audio and artwork) whenever something like the http://en.wikipedia.org/wiki/C64_Direct-to-TV or a port to a mobile phone comes up, but (as for out-of-print books/records) most of the time waiting for infringements (rather than distribution) is the only viable option to still extract money out of these "ancient" rights - so there are firms specialized in monitoring these and sending out nastygrams (with bills attached, or bringing suits right away where they can't) when they occur.
Interestingly in the earliest decisions on video games, courts did have their doubts whether the excessive duration afforded to literary and musical works provided an adequate regime for the short-lived IT products (as we know now it only makes hobbyists' conservation efforts a target indeed), but the political consensus has been to keep applying it (and extend various aspects of overprotection ever since).
#### arnoldemu
Quote from: OCT on 08:28, 06 April 10
I wonder what the code was that allowed you to page in the whole 4*16k of second-bank memory without hardware help.
AFAIK the OUT &7F00,&C4+page would only show 16k at a time at &4000 to &7FFF, which is an uncommon location for loaders.
`10 addr=&c00020 READ a\$:IF a\$="*" then call &c000 elsepoke addr,val("&"+a\$):addr=addr+1:goto 2030 data 01,c4,7f,ed,49,21,00,00,11,00,40,01,00,40,ed,b0,01,c6,7f,ed,49,21,00,80,11,00,40,01,00,40,ed,b0,01,c0,7f,ed,49,c9,*`
Asm code:
`ld bc,&7fc4out (c),cld hl,0ld de,&4000ld bc,&4000ldirld bc,&7fc6out (c),cld hl,&8000ld de,&4000ld bc,&4000ldirld bc,&7fc0out (c),cret`
Run the basic program then type:
OUT &7F00,&C2
now press return a few times and the screen scrolls, you can type basic commands but you can't see results (because video hardware is reading from first 64k).
So now the z80 is executing code entirely from 2nd 64k bank
My games. My Games
My website with coding examples: Unofficial Amstrad WWW Resource
#### OCT
#19
Quote from: arnoldemu on 08:55, 06 April 10type:
OUT &7F00,&C2
[...] now the z80 is executing code entirely from 2nd 64k bank
OIC, thanks, I hadn't realized this could be done in software for an entire 64k bank (the hardware switch still has the advantage of being able to see the video and toggle to another bank to work from at a moment of one's choice, of course - and immune to loaders sending C0 just to prevent such tricks being played on them ).
Apparently the use of the 2 least significant bits had been documented, albeit in French:
http://cpcrulez.fr/coding_asm34b.htm (not entirely clear about the use for C2 yet, though)
http://quasar.cpcscene.com/doku.php?id=assem:gate_array#le_schema_general_du_gate_array (bottom right)
#### Gryzor
Quote from: OCT on 08:52, 06 April 10
I'm not into cracking protections at all, nor scared for that matter - it's just that the author and photographer of a (really nice BTW) writeup on the CPC6128 hardware modification does hold rights to it
Hahaha! Thanks - copyright issues are a very dear subject for me, but it never does harm discussing it
What I meant was, I thought you mentioned copyright because if the hardware's uses, not of the schematics' copyright itself... But even so, it does sound a bit far-fetched. Personally, I wouldn't have an issue with publishing it - but of course, if the owner objected I'd withdraw it at once... but what are the chances, really???
#### villain
I was in regular contact with Khany/Cherry-T back in the 90s. But I don´t remember his real name in case anybody wants to contact him... Maybe he published his bank exchanger in the paperfanz Rundschlag. So if someone could take a look through the issues... I can´t to this because my collection is still stored at the home of my parents.
#### OCT
#22
While all I have available here is a Linux machine, I dug up a disk from 1993 on which -besides his real name- he does give permission for unaltered distribution of his article in German and sketches (RUN"#HI), which I converted using `dskdump -itype floppy -otype dsk -iside 0 -idstep /dev/fd0 BankSwap.dsk` (libdsk-1.3.0) and successfully tested against http://java.cpc-live.com/jemu/JavaCPC.jnlp (not sure which compression would be universally accepted, or how I should link it to launch in a JavaCPC right away).
HTH...
#### gef
#23
Quote from: arnoldemu on 07:19, 06 April 10
Yes indeed.
Speedlock protection had about 80 of these XOR type loops. Then it tried to detect the multiface. It was a beast of a protection and always changing.
I too was about 13/14, when I was converting tapes to discs. I did this mostly because the disc games were so expensive.
So, Speedlock was the name of it, hm? Nice to know after all this time. :-)
I don't know if the number of loops was close to 80, but it could easily be,
since I remember I got bored trying loop after loop; I kept pressing forward,
mostly as a matter of testing the functionality of the patch on Pyradev!
btw.
A good motive for spending time on doing all this reverse engineering, was indeed cost issues:
The teenager budget of me and my brother was easily overwhelmed by the cost of floppy disks
at that time, and protected games tended to occupy a complete 180KB side of a 3" disk. Insulting!
The "Live and let die" I think was the first to get factorized in files & a small nice custom bootloader
and the size of it became 1/6th of what it was before, a testament to its programmers' efficiency!
Some other times it was just a matter of being efficient, eg. Discology's sector editor was
a mere 22KBs (why do I still remember these numbers??? ;-), and was handy to move around,
so that it was not necessary to swap floppies in and out, a heavy tax time-wise, while hacking.
(along with the regularly needed reboot which needed a mere handful of seconds)
Quote from: arnoldemu on 07:19, 06 April 10
I never tried using Pyradev. First I used the very poor Amsoft assembler (MONA etc). With it's terrible input and line numbers. Then I moved onto Maxam (I had a strange chip on a PCB which connected to the back of the computer. I've not seen another since and I've since sold this.)
I never really found a debugger that I could get on with on the CPC.
Pyradev suite was -and probably still is!- the best name in the field.
People in this emulator forum, should at least be able to call it by name ,
so that in case of funny emulation interactions they can resort to it for testing.
I recommend you check for its "monitor" screenshot to get an idea.
With its help, It was possible to disassemble and understand good parts of the BASIC ROM
of the cpc6128 and even load basic programs in memory at will, many of them at once,
calling them *from* assembly code and in effect using them as subroutines.
This can be quite important at moments, because that was the only substitute available
for a high level language... assembly is good, but you need tools around it.
Also, for any people around here studying Computer Science and such, I highly recommend
to disassemble/study the routines on mathematical operations (sqrt, sin, cos ...) of the lower ROM.
That was very educating stuff, involving both theoretical and practical aspects (eg. premade matrices).
> It is interesting to read how others hacked and looked at code.
Indeed. We urge others to step forward and let be known what they were doing at the time!
There were so many games hacked, that there must be dozens of tricks like the above!
If you have references to such discussions/articles, please provide them.
cheers,
F.
#### Cholo
Was looking thru some AA mags and happened to notice that no. 24 & no. 25 actually had something about hacking on the "Hackers only" (usually this page had very little with actually hacking i recall .. usually was just some random machine code math).
No. 24 about Xor and such:
http://img64.imageshack.us/img64/1589/aa24p33.jpg
No. 25 about the bank switching:
http://img189.imageshack.us/img189/8905/aa25p29.jpg
Never would have guessed any of that (especially the bank switching, cant recall any book speaking of it at all).
Ah yes, disc space. Had a friend back then who had a 6128 and during the perhaps 5-6 years he had the 6128 he managed to get about 15-17 "empty" discs in all. As a kid those discs was way to expensive, so had to squeezing ever byte. Cant complain about speed of cause, but i still wonder if the amstrad had had 5.25" drives like C64, then i may have been even more popular (at least kids friendly).
Powered by SMFPacks Menu Editor Mod | 7,357 | 30,216 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2024-33 | latest | en | 0.941807 |
https://www.jiskha.com/questions/595431/over-seven-days-a-plant-grew-from-12-cm-to-33-express-its-average-growth-rate-in-cm-day | 1,607,145,747,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141746320.91/warc/CC-MAIN-20201205044004-20201205074004-00529.warc.gz | 644,140,764 | 5,623 | # science
over seven days, a plant grew from 12 cm to 33 express its average growth rate in cm/day
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1. 33 - 12 = ?
Divide the difference by 7.
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Ms. Sue
2. thanks
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3. i got 3 cm
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4. Right. The plant grew an average of 3 cm a day.
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Ms. Sue
5. thank you so much
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Ms. Sue
7. a plant grew from 12 cm to 33 cm.express its average growth rate in cm/day
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A Bacteria has a doubling period of 8 days. If there are 2250 bacteria present now, how many will there be in 33 days? find the growth rate (Round this to 4 decimal places). Growth Rate___________. There will be _________ | 946 | 3,378 | {"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-2020-50 | latest | en | 0.914439 |
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• Feb 22nd 2007, 03:49 AM
related rates help
WAAAAAAHHHHHH!!!! please move this to calculus section... am very sorry for this....
A water tank in the form of an inverted cone is being emptied at the rate of 6m^3 / min. The altitude of the cone is 24m, and the radius is 12m. Find how fast the water level is lowering when the water is 10m deep.
• Feb 22nd 2007, 05:06 AM
Jhevon
We first want to draw a diagram to see what's going on. And fill in everything we know, and want to know. Notice the rate dV/dt is negative, since the volume is decreasing.
We realize from the second diagram (the cross section of the tank) that we have a similar triangle relationship between the radius and the height. We will rewrite the height in terms of the radius so we can use it, since we are not given the rate the radius is changing. The formula for the volume of the tank is V=(1/3)pi*r^2*h, so here goes.
From similar triangles we realize that r/h = 12/24 => r = h/2
Now V=(1/3)pi*r^2*h
=> V=(1/3)pi*(h/2)^2*h ..............plugged in the formula for r from above
V=(1/3)pi*[(h^2)/4]*h
V = (1/12)pi*h^3
=> dV/dt = (pi/12)*3h^2 dh/dt
=> dV/dt = (pi*h^2/4)dh/dt
=> dh/dt = (4/pi*h^2) dV/dt
plugging in the values we want: h=10, dV/dt= -6
so dh/dt = - 6/25pi m/min................which is the rate of change of the height of the water when h=10 m, note it is negative since the height is decreasing
So that's the way to do it, at least the only way I know. Will one of you fine mathematicians check my calculations, I'm late for school, so i was in a rush typing this. | 495 | 1,583 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.25 | 4 | CC-MAIN-2017-26 | longest | en | 0.926195 |
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Written by justan brandt
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Whether you’re looking to install a larger or smaller wheel on your vehicle, selecting the right size tire to mount on your wheels is important. Most professionals suggest that your new tire and wheel combination should be within a 3 per cent, positive or negative, variance of the Original Equipment (O.E.).
Skill level:
Easy
### Things you need
• Original Equipment Tire and Wheel Information
## Instructions
1. 1
Locate the first number of your O.E. tire size, this can be found on the side of your tire and should look like 225/45R17.
225 represents the width across the tire, in millimetres, from the widest point.
2. 2
Locate the second number of your O.E. tire size, this can be found on the side of your tire and should look like 225/45R17.
45 represents the height of the tire’s sidewall, from rim to tread, as a percentage of the tire width. In this case the tire’s height would be 90mm or 3.54 inches, which is 40 per cent of 225 millimetres. This is also known as the tire’s series.
3. 3
Locate the third number of your O.E. tire size, this can be found on the side of your tire and should look like 225/45R17
17 represents the wheel diameter, in inches, that the tire is intended to be mounted on.
1. 1
Calculate the overall tire height, in inches, using the following formula ((Width/25.4)Sidewall Percentage)2
This means a 225/45R17 tire would have an overall tire height of 7.09 inches, ((225/25.4)40%)2).
2. 2
Combine the overall tire height with the wheel diameter to reach the final diameter of the wheel and tire combination.
Using the previous example of, 225/45R17, you would add 7.09 to 17 for a final diameter of 24.09 inches.
3. 3
Calculate your acceptable tire and wheel diameter using the generally accepted rule of a 3 per cent variance.
Using the examples from above you would be able to install a new wheel and tire combination that was anywhere from 23.37 inches in total diameter to 24.81 inches in total diameter.
#### Tips and warnings
• Ensure the width and offset of your new wheels will fit your vehicle properly as well.
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• Most recent | 587 | 2,342 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.71875 | 4 | CC-MAIN-2017-13 | latest | en | 0.917249 |
https://brainly.in/question/337818 | 1,484,566,228,000,000,000 | text/html | crawl-data/CC-MAIN-2017-04/segments/1484560279169.4/warc/CC-MAIN-20170116095119-00056-ip-10-171-10-70.ec2.internal.warc.gz | 778,515,612 | 10,531 | # An 800 car parking lot is divided into 3 sections ,there are 270 spots in section 1,and 150 more in section 2 than in section 3.how many spots are in section 2?
2
2016-05-06T07:47:28+05:30
It is given that in section 1 there are 270 spots.
So the remaining is 800-270 = 530.
Let the spots in section 3 be x
section 2 be x+150 (in the question it is given that section 2 is 150 more than section 3)
x+x+150 = 530
2x+150 = 530
2x = 530-150
2x = 380
x =190
Therefore the number of spots in section 2 = x+150 = 340.
The answer is 340.
2016-05-06T08:50:01+05:30
Total number of spots in the car parking lot= 800
total spots in section 1= 270
let the number of spots in section 3 be x.
no. of spots in section 2= x+150
a/q x + (x+150) + 270= 800
= 2x+ 150+ 270= 800
= 2x = 800 - 420
= x = 380/2
∴ x = 190
so number of spots in section 3= 190
so number of spots in section 2 = 190+150 = 340
( 190 + 340+ 270= 800)
plzz mark as brainliest! | 376 | 1,022 | {"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} | 4.375 | 4 | CC-MAIN-2017-04 | latest | en | 0.727504 |
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# Re: st: Mata computes the second derivative of x^2 as 1.9.
From Patrick Roland To statalist@hsphsun2.harvard.edu Subject Re: st: Mata computes the second derivative of x^2 as 1.9. Date Sun, 5 Aug 2012 12:09:41 -0700
```Sure, but i) it is disturbing that the default parameters work so
poorly on such a simple problem and ii) in more problems where the
derivatives are not known, one does not have the option of choosing
the tuning parameters so that the procedure matches the known results
as closely as possible.
On Sat, Aug 4, 2012 at 11:15 PM, Tirthankar Chakravarty
<tirthankar.chakravarty@gmail.com> wrote:
> As with any numerical approximation procedure, there are tuning
> parameters. Resetting the bounds for the optimal parameter (scale)
> search works for your simple example
>
> *---------------------------------------
> mata:
> mata clear
> void f(p,v){
> v = p^2
> }
> D = deriv_init()
> deriv_init_bounds(D, (1e-5, 1e-4))
> deriv_init_evaluator(D, &f())
> deriv_init_params(D, 0.5)
> deriv(D,2)
> deriv_result_delta(D)
> end
> *---------------------------------------
>
> complex multiparameter nonlinear problems, you should look at a
> numerical analysis text.
>
> T
>
> On Sat, Aug 4, 2012 at 8:22 PM, Patrick Roland
> <patrick.rolande@gmail.com> wrote:
>> Perhaps someone might be able to explain this anomaly. The following
>> code finds the second derivative of x^2 evaluated at 0.5, which should
>> of course be 2.
>>
>>
>> mata:
>>
>> void f(p,v){
>> v = p^2
>> }
>> D = deriv_init()
>> deriv_init_evaluator(D, &f())
>> deriv_init_params(D, 0.5)
>> deriv(D,2)
>>
>> end
>>
>> Instead, in both Stata 11.2 and 12.1, the answer is 1.906569171.
>>
>> I'm trying to compute the Hessian of a complicated nonlinear function
>> of several hundred arguments. It appears that Mata cannot correctly
>> find the second derivative of x^2. This does not inspire confidence.
>>
>> Hopefully I'm overlooking something simple here, in which case I would
>> greatly appreciate a correction.
>> *
>> * For searches and help try:
>> * http://www.stata.com/help.cgi?search
>> * http://www.stata.com/support/statalist/faq
>> * http://www.ats.ucla.edu/stat/stata/
>
>
>
> --
> Tirthankar Chakravarty
> tchakravarty@ucsd.edu
> tirthankar.chakravarty@gmail.com
> *
> * For searches and help try:
> * http://www.stata.com/help.cgi?search
> * http://www.stata.com/support/statalist/faq
> * http://www.ats.ucla.edu/stat/stata/
*
* For searches and help try:
* http://www.stata.com/help.cgi?search
* http://www.stata.com/support/statalist/faq
* http://www.ats.ucla.edu/stat/stata/
``` | 806 | 2,822 | {"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-2015-48 | longest | en | 0.824802 |
http://www.in2013dollars.com/1950-AUD-in-1999 | 1,550,448,859,000,000,000 | text/html | crawl-data/CC-MAIN-2019-09/segments/1550247483873.51/warc/CC-MAIN-20190217233327-20190218015327-00315.warc.gz | 374,080,205 | 9,438 | # AU\$100 in 1950 → AU\$1,554.55 in 1999
## Australia Inflation Calculator
AU\$
### Australian Inflation Rate, AU\$100 in 1950 to 1999
According to the Bureau of Statistics consumer price index, prices in 1999 are 1,454.55% higher than average prices throughout 1950. The dollar experienced an average inflation rate of 5.76% per year during this period.
In other words, AU\$100 in 1950 is equivalent in purchasing power to AU\$1,554.55 in 1999, a difference of AU\$1,454.55 over 49 years.
The 1950 inflation rate was 10.00%. The inflation rate in 1999 was 1.48%. The 1999 inflation rate is lower compared to the average inflation rate of 2.62% per year between 1999 and 2019.
Cumulative price change 1,454.55% Average inflation rate 5.76% Converted amount (AU\$100 base) AU\$1,554.55 Price difference (AU\$100 base) AU\$1,454.55 CPI in 1950 4.400 CPI in 1999 68.400 Inflation in 1950 10.00% Inflation in 1999 1.48%
### Buying power of AU\$100 in 1950
This chart shows calculations of buying power equivalence over time for AU\$100 in 1950 (price index tracking began in 1922).
According to the Bureau of Statistics, each of these AUD amounts below is equal in terms of what it could buy at the time:
Year AUD Value Inflation Rate
1950 AU\$100.00 10.00%
1951 AU\$118.18 18.18%
1952 AU\$138.64 17.31%
1953 AU\$145.45 4.92%
1954 AU\$147.73 1.56%
1955 AU\$150.00 1.54%
1956 AU\$159.09 6.06%
1957 AU\$163.64 2.86%
1958 AU\$163.64 0.00%
1959 AU\$168.18 2.78%
1960 AU\$175.00 4.05%
1961 AU\$177.27 1.30%
1962 AU\$177.27 0.00%
1963 AU\$179.55 1.28%
1964 AU\$184.09 2.53%
1965 AU\$190.91 3.70%
1966 AU\$195.45 2.38%
1967 AU\$202.27 3.49%
1968 AU\$209.09 3.37%
1969 AU\$215.91 3.26%
1970 AU\$222.73 3.16%
1971 AU\$236.36 6.12%
1972 AU\$250.00 5.77%
1973 AU\$272.73 9.09%
1974 AU\$315.91 15.83%
1975 AU\$363.64 15.11%
1976 AU\$411.36 13.13%
1977 AU\$461.36 12.15%
1978 AU\$497.73 7.88%
1979 AU\$543.18 9.13%
1980 AU\$600.00 10.46%
1981 AU\$656.82 9.47%
1982 AU\$729.55 11.07%
1983 AU\$804.55 10.28%
1984 AU\$836.36 3.95%
1985 AU\$890.91 6.52%
1986 AU\$972.73 9.18%
1987 AU\$1,054.55 8.41%
1988 AU\$1,131.82 7.33%
1989 AU\$1,215.91 7.43%
1990 AU\$1,306.82 7.48%
1991 AU\$1,347.73 3.13%
1992 AU\$1,361.36 1.01%
1993 AU\$1,384.09 1.67%
1994 AU\$1,411.36 1.97%
1995 AU\$1,477.27 4.67%
1996 AU\$1,515.91 2.62%
1997 AU\$1,520.45 0.30%
1998 AU\$1,531.82 0.75%
1999 AU\$1,554.55 1.48%
2000 AU\$1,625.00 4.53%
2001 AU\$1,695.45 4.34%
2002 AU\$1,747.73 3.08%
2003 AU\$1,795.45 2.73%
2004 AU\$1,836.36 2.28%
2005 AU\$1,886.36 2.72%
2006 AU\$1,952.27 3.49%
2007 AU\$1,997.73 2.33%
2008 AU\$2,086.36 4.44%
2009 AU\$2,122.73 1.74%
2010 AU\$2,184.09 2.89%
2011 AU\$2,256.82 3.33%
2012 AU\$2,295.45 1.71%
2013 AU\$2,352.27 2.48%
2014 AU\$2,411.36 2.51%
2015 AU\$2,447.73 1.51%
2016 AU\$2,479.55 1.30%
2017 AU\$2,527.27 1.92%
2018 AU\$2,559.09 1.26%
2019 AU\$2,607.71 1.90%*
* Compared to previous annual rate. Not final. See inflation summary for latest 12-month trailing value.
### How to Calculate Inflation Rate for AU\$100, 1950 to 1999
This inflation calculator uses the following inflation rate formula:
CPI in 1999CPI in 1950
×
1950 AUD value
=
1999 AUD value
Then plug in historical CPI values. The Australian CPI was 4.4 in the year 1950 and 68.4 in 1999:
68.44.4
×
AU\$100
=
AU\$1,554.55
AU\$100 in 1950 has the same "purchasing power" or "buying power" as AU\$1,554.55 in 1999.
To get the total inflation rate for the 49 years between 1950 and 1999, we use the following formula:
CPI in 1999 - CPI in 1950CPI in 1950
×
100
=
Cumulative inflation rate (49 years)
Plugging in the values to this equation, we get:
68.4 - 4.44.4
×
100
=
1,455%
Politics and news often influence economic performance. Here's what was happening at the time:
• Jerusalem is proclaimed the capital of Israel by Knesset
• North Korea invades South Korea.
• Harry Truman announces that America will seek to develop a hydrogen bomb.
• Chinese forces occupy Tibet.
### Data Source & Citation
Raw data for these calculations comes from the government of Australia's annual (CPI) as provided by the Reserve Bank of Australia. The consumer price index was established in 1922 and is tracked by Australian Bureau of Statistics (ABS).
You may use the following MLA citation for this page: “1950 dollars in 1999 | Australia Inflation Calculator.” U.S. Official Inflation Data, Alioth Finance, 18 Feb. 2019, https://www.officialdata.org/1950-AUD-in-1999.
in2013dollars.com is a reference website maintained by the Official Data Foundation. | 1,713 | 4,516 | {"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-2019-09 | latest | en | 0.785744 |
https://www.spa020.nl/stable/4627/45yturv3zuqk.html | 1,638,271,142,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964358973.70/warc/CC-MAIN-20211130110936-20211130140936-00102.warc.gz | 1,123,681,315 | 6,374 | Crushing & Grinding Machines
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https://www.scribd.com/document/332953695/mathlessonplan-1 | 1,548,247,483,000,000,000 | text/html | crawl-data/CC-MAIN-2019-04/segments/1547584331733.89/warc/CC-MAIN-20190123105843-20190123131843-00247.warc.gz | 919,606,044 | 46,074 | You are on page 1of 2
# Money Unit
Duration: 5 days
Standard from Curriculum Framework
3.8
The student will determine, by counting, the value of a collection of bills and coins whose total
value is \$5.00 or less, compare the value of the bills and coins, and make change.
Essential Understandings
➔ Understand that a collection of coins and bills has a value that can be counted.
➔ Understand how to make change from \$5.00 or less.
Essential Knowledge and Skills
➔ Count the value of collections of coins and bills up to \$5.00.
➔ Compare the values of two sets of coins or bills, up to \$5.00, using the terms greater than, less than, and
equal to.
➔ Make change from \$5.00 or less.
Lessons
Target
1
I can identify and give the value of dollar bills, quarters,
dimes, nickels, and pennies.
2
I can accurately count a set of money.
3
I can compare two sets of money.
4
I can calculate the correct amount of change.
5
Review and Quiz
Lesson 1: Identify and Give Value
Learning Target: I can identify and give the value of dollar bills, quarters, dimes,
nickels, and pennies.
Whole Group
Small Group
1. Number Talks (5mins)
a. Practice decomposing and composing numbers to make ten.
2. Discuss
a. Tell students that we will be learning about money this week. Ask
students why they think it’s important to know how to count money and
know the value of it. Give students a variety of situations when knowing
how to count money will be useful.
3. Introduce
a. Read the learning target. Do a quick assessment of how well they know
the names of the coins and dollar bills.
b. BrainPop: https://jr.brainpop.com/math/money/dollarsandcents/
4. Practice
a. Go over each coins (penny, nickel, dime, quarter, half dollar, and golden
dollar) *briefly talk about the half-dollar and golden dollar. Go over one
dollar bill and five dollar bill.
i.
Have a picture (front and back), discuss the person on the coins
and bills, size, color, and worth.
b. Have just the picture of the coins and “quiz” students.
i.
Matching game
1. Money Flaps
a. Students will glue in their foldable in their math notebook.
b. Hand out one and five dollar bills.
c. Go over coins and dollar bills by presenting one at a time (real money)
and identifying the name and value. Go over how students can identify
each one (George Washington on a quarter).
d. Students will write the facts under the appropriate flap.
2. Review
a. Show one coin or bill to students and have them identify and give the
value of each.
Seat Work
1. Decomposing Quick Check
2. IXL: P.1 & P.6
Hands- On
1. Math Power Towers (addition and subtraction facts)
a. Use calculators to check.
Technology
1. Reflex Math/ Dreambox | 689 | 2,689 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.4375 | 4 | CC-MAIN-2019-04 | latest | en | 0.885096 |
https://la.mathworks.com/matlabcentral/cody/problems/44360-pentagonal-numbers/solutions/1981343 | 1,606,652,740,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141197593.33/warc/CC-MAIN-20201129093434-20201129123434-00116.warc.gz | 353,066,142 | 18,155 | Cody
# Problem 44360. Pentagonal Numbers
Solution 1981343
Submitted on 18 Oct 2019 by Renato SL
This solution is locked. To view this solution, you need to provide a solution of the same size or smaller.
### Test Suite
Test Status Code Input and Output
1 Pass
x1 = 1; x2 = 25; [p,d] = pentagonal_numbers(x1,x2) assert(isequal(p,[1,5,12,22])) assert(isequal(d,[0,1,0,0]))
p = 1 5 12 22 d = 1×4 logical array 0 1 0 0 p = 1 5 12 22 d = 1×4 logical array 0 1 0 0
2 Pass
x1 = 1; x2 = 4; [p,d] = pentagonal_numbers(x1,x2) assert(isequal(p,1)) assert(isequal(d,0))
p = 1 d = logical 0 p = 1 d = logical 0
3 Pass
x1 = 10; x2 = 40; [p,d] = pentagonal_numbers(x1,x2) assert(isequal(p,[12,22,35])) assert(isequal(d,[0,0,1]))
p = 12 22 35 d = 1×3 logical array 0 0 1 p = 12 22 35 d = 1×3 logical array 0 0 1
4 Pass
x1 = 10; x2 = 99; [p,d] = pentagonal_numbers(x1,x2) assert(isequal(p,[12,22,35,51,70,92])) assert(isequal(d,[0,0,1,0,1,0]))
p = 12 22 35 51 70 92 d = 1×6 logical array 0 0 1 0 1 0 p = 12 22 35 51 70 92 d = 1×6 logical array 0 0 1 0 1 0
5 Pass
x1 = 100; x2 = 999; [p,d] = pentagonal_numbers(x1,x2) assert(isequal(p,[117,145,176,210,247,287,330,376,425,477,532,590,651,715,782,852,925])) assert(isequal(d,[0,1,0,1,0,0,1,0,1,0,0,1,0,1,0,0,1]))
p = 117 145 176 210 247 287 330 376 425 477 532 590 651 715 782 852 925 d = 1×17 logical array 0 1 0 1 0 0 1 0 1 0 0 1 0 1 0 0 1 p = 117 145 176 210 247 287 330 376 425 477 532 590 651 715 782 852 925 d = 1×17 logical array 0 1 0 1 0 0 1 0 1 0 0 1 0 1 0 0 1
6 Pass
x1 = 40; x2 = 50; [p,d] = pentagonal_numbers(x1,x2) assert(isempty(p)) assert(isempty(d))
p = [] d = 0×0 empty logical array p = [] d = 0×0 empty logical array
7 Pass
x1 = 1000; x2 = 1500; [p,d] = pentagonal_numbers(x1,x2) assert(isequal(p,[1001,1080,1162,1247,1335,1426])) assert(isequal(d,[0,1,0,0,1,0]))
p = 1001 1080 1162 1247 1335 1426 d = 1×6 logical array 0 1 0 0 1 0 p = 1001 1080 1162 1247 1335 1426 d = 1×6 logical array 0 1 0 0 1 0
8 Pass
x1 = 1500; x2 = 3000; [p,d] = pentagonal_numbers(x1,x2) assert(isequal(p,[1520,1617,1717,1820,1926,2035,2147,2262,2380,2501,2625,2752,2882])) assert(isequal(d,[1,0,0,1,0,1,0,0,1,0,1,0,0]))
p = 1520 1617 1717 1820 1926 2035 2147 2262 2380 2501 2625 2752 2882 d = 1×13 logical array 1 0 0 1 0 1 0 0 1 0 1 0 0 p = 1520 1617 1717 1820 1926 2035 2147 2262 2380 2501 2625 2752 2882 d = 1×13 logical array 1 0 0 1 0 1 0 0 1 0 1 0 0
9 Pass
x1 = 1; x2 = 3000; [p,d] = pentagonal_numbers(x1,x2) assert(isequal(p,[1,5,12,22,35,51,70,92,117,145,176,210,247,287,330,376,425,477,532,590,651,715,782,852,925,1001,1080,1162,1247,1335,1426,1520,1617,1717,1820,1926,2035,2147,2262,2380,2501,2625,2752,2882])) assert(isequal(d,[0,1,0,0,1,0,1,0,0,1,0,1,0,0,1,0,1,0,0,1,0,1,0,0,1,0,1,0,0,1,0,1,0,0,1,0,1,0,0,1,0,1,0,0]))
p = Columns 1 through 15 1 5 12 22 35 51 70 92 117 145 176 210 247 287 330 Columns 16 through 30 376 425 477 532 590 651 715 782 852 925 1001 1080 1162 1247 1335 Columns 31 through 44 1426 1520 1617 1717 1820 1926 2035 2147 2262 2380 2501 2625 2752 2882 d = 1×44 logical array 0 1 0 0 1 0 1 0 0 1 0 1 0 0 1 0 1 0 0 1 0 1 0 0 1 0 1 0 0 1 0 1 0 0 1 0 1 0 0 1 0 1 0 0 p = Columns 1 through 15 1 5 12 22 35 51 70 92 117 145 176 210 247 287 330 Columns 16 through 30 376 425 477 532 590 651 715 782 852 925 1001 1080 1162 1247 1335 Columns 31 through 44 1426 1520 1617 1717 1820 1926 2035 2147 2262 2380 2501 2625 2752 2882 d = 1×44 logical array 0 1 0 0 1 0 1 0 0 1 0 1 0 0 1 0 1 0 0 1 0 1 0 0 1 0 1 0 0 1 0 1 0 0 1 0 1 0 0 1 0 1 0 0
10 Pass
x1 = 10000; x2 = 12000; [p,d] = pentagonal_numbers(x1,x2) assert(isequal(p,[10045,10292,10542,10795,11051,11310,11572,11837])) assert(isequal(d,[1,0,0,1,0,1,0,0]))
p = 10045 10292 10542 10795 11051 11310 11572 11837 d = 1×8 logical array 1 0 0 1 0 1 0 0 p = 10045 10292 10542 10795 11051 11310 11572 11837 d = 1×8 logical array 1 0 0 1 0 1 0 0
11 Pass
x1 = 100000; x2 = 110000; [p,d] = pentagonal_numbers(x1,x2) assert(isequal(p,[100492,101270,102051,102835,103622,104412,105205,106001,106800,107602,108407,109215])) assert(isequal(d,[0,1,0,1,0,0,1,0,1,0,0,1]))
p = 100492 101270 102051 102835 103622 104412 105205 106001 106800 107602 108407 109215 d = 1×12 logical array 0 1 0 1 0 0 1 0 1 0 0 1 p = 100492 101270 102051 102835 103622 104412 105205 106001 106800 107602 108407 109215 d = 1×12 logical array 0 1 0 1 0 0 1 0 1 0 0 1
12 Pass
x1 = 1000000; x2 = 1010101; [p,d] = pentagonal_numbers(x1,x2) assert(isequal(p,[1000825,1003277,1005732,1008190])) assert(isequal(d,[1,0,0,1]))
p = 1000825 1003277 1005732 1008190 d = 1×4 logical array 1 0 0 1 p = 1000825 1003277 1005732 1008190 d = 1×4 logical array 1 0 0 1
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## Ensemble Methods: Foundations and Algorithms
Ensemble learning is a kind of state-of-the-art machine learning method. I know it is very a very expensive book and it is illegal to share it openly. But, I guess nowadays, it is a common sense that every book has its pdf version online right? That is the reason why we love the internet. Data Scientist and Machine Learning Technologist: Translating modern machine learning and computer vision techniques into engineering and bringing ideas to life to built a better future.
## Popular Ensemble Methods: An Empirical Study
Quick jump to page content. Home Archives Vol. Published: Aug 1, Main Article Content D. Abstract An ensemble consists of a set of individually trained classifiers such as neural networks or decision trees whose predictions are combined when classifying novel instances. Previous research has shown that an ensemble is often more accurate than any of the single classifiers in the ensemble.
## [PDF] Ensemble Methods: Foundations and Algorithms (Chapman Hall/CRC Data Mining and Knowledge
Embed Size px x x x x Professor Zhous book is a comprehensive introduction to ensemble methods in machine learning. It reviews the latest research in this exciting area. I learned a lot reading it! Thomas G.
Skip to search form Skip to main content You are currently offline. Some features of the site may not work correctly. Lughofer and B. Lughofer , B.
In statistics and machine learning , ensemble methods use multiple learning algorithms to obtain better predictive performance than could be obtained from any of the constituent learning algorithms alone. Supervised learning algorithms perform the task of searching through a hypothesis space to find a suitable hypothesis that will make good predictions with a particular problem. Ensembles combine multiple hypotheses to form a hopefully better hypothesis. The term ensemble is usually reserved for methods that generate multiple hypotheses using the same base learner. Evaluating the prediction of an ensemble typically requires more computation than evaluating the prediction of a single model.
#### See a Problem?
- Наверное, увидел включенный монитор. - Черт возьми! - выругался коммандер. - Вчера вечером я специально позвонил дежурному лаборатории систем безопасности и попросил его сегодня не выходить на работу. Сьюзан это не удивило. Она не могла припомнить, чтобы когда-то отменялось дежурство, но Стратмор, очевидно, не хотел присутствия непосвященных. Он и мысли не допускал о том, что кто-то из сотрудников лаборатории узнает о Цифровой крепости. - Наверное, стоит выключить ТРАНСТЕКСТ, - предложила Сьюзан.
До поворота еще минуты две. Он знал, что этого времени у него. Сзади его нагоняло такси. Он смотрел на приближающиеся огни центра города и молил Бога, чтобы он дал ему добраться туда живым. Беккер проехал уже половину пути, когда услышал сзади металлический скрежет, прижался к рулю и до отказа открыл дроссель.
Тогда он дал бы нам ключ, чтобы мы могли уничтожить вирус. Сьюзан стало абсолютно очевидно, что план Танкадо ужасным образом рухнул. Он не собирался умирать. Он рассчитывал, сидя в испанском баре, услышать по Си-эн-эн пресс-конференцию об американском сверхсекретном компьютере, способном взломать любые шифры. После этого он позвонил бы Стратмору, считал пароль с кольца на своем пальце и в последнюю минуту спас главный банк данных АНБ. Вдоволь посмеявшись, он исчез бы насовсем, превратившись в легенду Фонда электронных границ. Сьюзан стукнула кулаком по столу: - Нам необходимо это кольцо.
Да.
Выслушай меня, Мидж. Направь мне официальный запрос. В понедельник я проверю твою машину. А пока сваливай-ка ты отсюда домой. Сегодня же суббота.
Вы можете заметить, - продолжал Смит, - что взгляд его устремлен. Он ни разу не посмотрел по сторонам. - Это так важно? - полувопросительно произнес Джабба.
Беккер удивленно посмотрел на. - Разве. Я думал, что он похоронен в Доминиканской Республике. - Да нет же, черт возьми. И кто только распустил этот слух.
Я не хотел, чтобы ты узнала об этом. Я был уверен, что он тебе все рассказал.
Еще один любитель молоденьких девочек, - подумал. - Ну. Сеньор?. - Буисан, - сказал Беккер. - Мигель Буисан.
Куда она могла уйти. Неужели уехала без меня в Стоун-Мэнор. - Эй! - услышал он за спиной сердитый женский голос и чуть не подпрыгнул от неожиданности.
Беккер, отступая к стене, вновь обрел способность мыслить четко и ясно. Он почувствовал жжение в боку, дотронулся до больного места и посмотрел на руку. Между пальцами и на кольце Танкадо была кровь. У него закружилась голова.
Наполнив тяжелый хрустальный стакан водой из фонтанчика, Беккер сделал несколько жадных глотков, потянулся и расправил плечи, стараясь сбросить алкогольное оцепенение, после чего поставил стакан на столик и направился к выходу. Когда он проходил мимо лифта, дверцы открылись. В кабине стоял какой-то мужчина.
Сьюзан замерла. Мгновение спустя, как в одном из самых страшных детских кошмаров, перед ней возникло чье-то лицо. Зеленоватое, оно было похоже на призрак. Это было лицо демона, черты которого деформировали черные тени. Сьюзан отпрянула и попыталась бежать, но призрак схватил ее за руку.
Конечно, нет! - возмущенно ответила девушка. Она смотрела на него невинными глазами, и Беккер почувствовал, что она держит его за дурака. - Да будет. На вид вы человек состоятельный. Дайте немножко денег, чтобы я могла вернуться домой.
Не успел Стратмор ее остановить, как она скользнула в образовавшийся проем. Он попытался что-то сказать, но Сьюзан была полна решимости. Ей хотелось поскорее оказаться в Третьем узле, и она достаточно хорошо изучила своего шефа, чтобы знать: Стратмор никуда не уйдет, пока она не разыщет ключ, спрятанный где-то в компьютере Хейла.
Трудно даже представить, что происходит там, внизу. - Я пробовал, - прошептал Стратмор еле слышно. Ей еще не приходилось слышать, чтобы он так .
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◆ clavsy_rook()
subroutine clavsy_rook ( character UPLO, character TRANS, character DIAG, integer N, integer NRHS, complex, dimension( lda, * ) A, integer LDA, integer, dimension( * ) IPIV, complex, dimension( ldb, * ) B, integer LDB, integer INFO )
CLAVSY_ROOK
Purpose:
``` CLAVSY_ROOK performs one of the matrix-vector operations
x := A*x or x := A'*x,
where x is an N element vector and A is one of the factors
from the block U*D*U' or L*D*L' factorization computed by CSYTRF_ROOK.
If TRANS = 'N', multiplies by U or U * D (or L or L * D)
If TRANS = 'T', multiplies by U' or D * U' (or L' or D * L')```
Parameters
[in] UPLO ``` UPLO is CHARACTER*1 Specifies whether the factor stored in A is upper or lower triangular. = 'U': Upper triangular = 'L': Lower triangular``` [in] TRANS ``` TRANS is CHARACTER*1 Specifies the operation to be performed: = 'N': x := A*x = 'T': x := A'*x``` [in] DIAG ``` DIAG is CHARACTER*1 Specifies whether or not the diagonal blocks are unit matrices. If the diagonal blocks are assumed to be unit, then A = U or A = L, otherwise A = U*D or A = L*D. = 'U': Diagonal blocks are assumed to be unit matrices. = 'N': Diagonal blocks are assumed to be non-unit matrices.``` [in] N ``` N is INTEGER The number of rows and columns of the matrix A. N >= 0.``` [in] NRHS ``` NRHS is INTEGER The number of right hand sides, i.e., the number of vectors x to be multiplied by A. NRHS >= 0.``` [in] A ``` A is COMPLEX array, dimension (LDA,N) The block diagonal matrix D and the multipliers used to obtain the factor U or L as computed by CSYTRF_ROOK. Stored as a 2-D triangular matrix.``` [in] LDA ``` LDA is INTEGER The leading dimension of the array A. LDA >= max(1,N).``` [in] IPIV ``` IPIV is INTEGER array, dimension (N) Details of the interchanges and the block structure of D, as determined by CSYTRF_ROOK. If UPLO = 'U': If IPIV(k) > 0, then rows and columns k and IPIV(k) were interchanged and D(k,k) is a 1-by-1 diagonal block. (If IPIV( k ) = k, no interchange was done). If IPIV(k) < 0 and IPIV(k-1) < 0, then rows and columns k and -IPIV(k) were interchanged and rows and columns k-1 and -IPIV(k-1) were inerchaged, D(k-1:k,k-1:k) is a 2-by-2 diagonal block. If UPLO = 'L': If IPIV(k) > 0, then rows and columns k and IPIV(k) were interchanged and D(k,k) is a 1-by-1 diagonal block. (If IPIV( k ) = k, no interchange was done). If IPIV(k) < 0 and IPIV(k+1) < 0, then rows and columns k and -IPIV(k) were interchanged and rows and columns k+1 and -IPIV(k+1) were inerchaged, D(k:k+1,k:k+1) is a 2-by-2 diagonal block.``` [in,out] B ``` B is COMPLEX array, dimension (LDB,NRHS) On entry, B contains NRHS vectors of length N. On exit, B is overwritten with the product A * B.``` [in] LDB ``` LDB is INTEGER The leading dimension of the array B. LDB >= max(1,N).``` [out] INFO ``` INFO is INTEGER = 0: successful exit < 0: if INFO = -k, the k-th argument had an illegal value```
Definition at line 153 of file clavsy_rook.f.
155 *
156 * -- LAPACK test routine --
157 * -- LAPACK is a software package provided by Univ. of Tennessee, --
158 * -- Univ. of California Berkeley, Univ. of Colorado Denver and NAG Ltd..--
159 *
160 * .. Scalar Arguments ..
161 CHARACTER DIAG, TRANS, UPLO
162 INTEGER INFO, LDA, LDB, N, NRHS
163 * ..
164 * .. Array Arguments ..
165 INTEGER IPIV( * )
166 COMPLEX A( LDA, * ), B( LDB, * )
167 * ..
168 *
169 * =====================================================================
170 *
171 * .. Parameters ..
172 COMPLEX CONE
173 parameter( cone = ( 1.0e+0, 0.0e+0 ) )
174 * ..
175 * .. Local Scalars ..
176 LOGICAL NOUNIT
177 INTEGER J, K, KP
178 COMPLEX D11, D12, D21, D22, T1, T2
179 * ..
180 * .. External Functions ..
181 LOGICAL LSAME
182 EXTERNAL lsame
183 * ..
184 * .. External Subroutines ..
185 EXTERNAL cgemv, cgeru, cscal, cswap, xerbla
186 * ..
187 * .. Intrinsic Functions ..
188 INTRINSIC abs, max
189 * ..
190 * .. Executable Statements ..
191 *
192 * Test the input parameters.
193 *
194 info = 0
195 IF( .NOT.lsame( uplo, 'U' ) .AND. .NOT.lsame( uplo, 'L' ) ) THEN
196 info = -1
197 ELSE IF( .NOT.lsame( trans, 'N' ) .AND. .NOT.lsame( trans, 'T' ) )
198 \$ THEN
199 info = -2
200 ELSE IF( .NOT.lsame( diag, 'U' ) .AND. .NOT.lsame( diag, 'N' ) )
201 \$ THEN
202 info = -3
203 ELSE IF( n.LT.0 ) THEN
204 info = -4
205 ELSE IF( lda.LT.max( 1, n ) ) THEN
206 info = -6
207 ELSE IF( ldb.LT.max( 1, n ) ) THEN
208 info = -9
209 END IF
210 IF( info.NE.0 ) THEN
211 CALL xerbla( 'CLAVSY_ROOK ', -info )
212 RETURN
213 END IF
214 *
215 * Quick return if possible.
216 *
217 IF( n.EQ.0 )
218 \$ RETURN
219 *
220 nounit = lsame( diag, 'N' )
221 *------------------------------------------
222 *
223 * Compute B := A * B (No transpose)
224 *
225 *------------------------------------------
226 IF( lsame( trans, 'N' ) ) THEN
227 *
228 * Compute B := U*B
229 * where U = P(m)*inv(U(m))* ... *P(1)*inv(U(1))
230 *
231 IF( lsame( uplo, 'U' ) ) THEN
232 *
233 * Loop forward applying the transformations.
234 *
235 k = 1
236 10 CONTINUE
237 IF( k.GT.n )
238 \$ GO TO 30
239 IF( ipiv( k ).GT.0 ) THEN
240 *
241 * 1 x 1 pivot block
242 *
243 * Multiply by the diagonal element if forming U * D.
244 *
245 IF( nounit )
246 \$ CALL cscal( nrhs, a( k, k ), b( k, 1 ), ldb )
247 *
248 * Multiply by P(K) * inv(U(K)) if K > 1.
249 *
250 IF( k.GT.1 ) THEN
251 *
252 * Apply the transformation.
253 *
254 CALL cgeru( k-1, nrhs, cone, a( 1, k ), 1, b( k, 1 ),
255 \$ ldb, b( 1, 1 ), ldb )
256 *
257 * Interchange if P(K) != I.
258 *
259 kp = ipiv( k )
260 IF( kp.NE.k )
261 \$ CALL cswap( nrhs, b( k, 1 ), ldb, b( kp, 1 ), ldb )
262 END IF
263 k = k + 1
264 ELSE
265 *
266 * 2 x 2 pivot block
267 *
268 * Multiply by the diagonal block if forming U * D.
269 *
270 IF( nounit ) THEN
271 d11 = a( k, k )
272 d22 = a( k+1, k+1 )
273 d12 = a( k, k+1 )
274 d21 = d12
275 DO 20 j = 1, nrhs
276 t1 = b( k, j )
277 t2 = b( k+1, j )
278 b( k, j ) = d11*t1 + d12*t2
279 b( k+1, j ) = d21*t1 + d22*t2
280 20 CONTINUE
281 END IF
282 *
283 * Multiply by P(K) * inv(U(K)) if K > 1.
284 *
285 IF( k.GT.1 ) THEN
286 *
287 * Apply the transformations.
288 *
289 CALL cgeru( k-1, nrhs, cone, a( 1, k ), 1, b( k, 1 ),
290 \$ ldb, b( 1, 1 ), ldb )
291 CALL cgeru( k-1, nrhs, cone, a( 1, k+1 ), 1,
292 \$ b( k+1, 1 ), ldb, b( 1, 1 ), ldb )
293 *
294 * Interchange if a permutation was applied at the
295 * K-th step of the factorization.
296 *
297 * Swap the first of pair with IMAXth
298 *
299 kp = abs( ipiv( k ) )
300 IF( kp.NE.k )
301 \$ CALL cswap( nrhs, b( k, 1 ), ldb, b( kp, 1 ), ldb )
302 *
303 * NOW swap the first of pair with Pth
304 *
305 kp = abs( ipiv( k+1 ) )
306 IF( kp.NE.k+1 )
307 \$ CALL cswap( nrhs, b( k+1, 1 ), ldb, b( kp, 1 ),
308 \$ ldb )
309 END IF
310 k = k + 2
311 END IF
312 GO TO 10
313 30 CONTINUE
314 *
315 * Compute B := L*B
316 * where L = P(1)*inv(L(1))* ... *P(m)*inv(L(m)) .
317 *
318 ELSE
319 *
320 * Loop backward applying the transformations to B.
321 *
322 k = n
323 40 CONTINUE
324 IF( k.LT.1 )
325 \$ GO TO 60
326 *
327 * Test the pivot index. If greater than zero, a 1 x 1
328 * pivot was used, otherwise a 2 x 2 pivot was used.
329 *
330 IF( ipiv( k ).GT.0 ) THEN
331 *
332 * 1 x 1 pivot block:
333 *
334 * Multiply by the diagonal element if forming L * D.
335 *
336 IF( nounit )
337 \$ CALL cscal( nrhs, a( k, k ), b( k, 1 ), ldb )
338 *
339 * Multiply by P(K) * inv(L(K)) if K < N.
340 *
341 IF( k.NE.n ) THEN
342 kp = ipiv( k )
343 *
344 * Apply the transformation.
345 *
346 CALL cgeru( n-k, nrhs, cone, a( k+1, k ), 1,
347 \$ b( k, 1 ), ldb, b( k+1, 1 ), ldb )
348 *
349 * Interchange if a permutation was applied at the
350 * K-th step of the factorization.
351 *
352 IF( kp.NE.k )
353 \$ CALL cswap( nrhs, b( k, 1 ), ldb, b( kp, 1 ), ldb )
354 END IF
355 k = k - 1
356 *
357 ELSE
358 *
359 * 2 x 2 pivot block:
360 *
361 * Multiply by the diagonal block if forming L * D.
362 *
363 IF( nounit ) THEN
364 d11 = a( k-1, k-1 )
365 d22 = a( k, k )
366 d21 = a( k, k-1 )
367 d12 = d21
368 DO 50 j = 1, nrhs
369 t1 = b( k-1, j )
370 t2 = b( k, j )
371 b( k-1, j ) = d11*t1 + d12*t2
372 b( k, j ) = d21*t1 + d22*t2
373 50 CONTINUE
374 END IF
375 *
376 * Multiply by P(K) * inv(L(K)) if K < N.
377 *
378 IF( k.NE.n ) THEN
379 *
380 * Apply the transformation.
381 *
382 CALL cgeru( n-k, nrhs, cone, a( k+1, k ), 1,
383 \$ b( k, 1 ), ldb, b( k+1, 1 ), ldb )
384 CALL cgeru( n-k, nrhs, cone, a( k+1, k-1 ), 1,
385 \$ b( k-1, 1 ), ldb, b( k+1, 1 ), ldb )
386 *
387 * Interchange if a permutation was applied at the
388 * K-th step of the factorization.
389 *
390 * Swap the second of pair with IMAXth
391 *
392 kp = abs( ipiv( k ) )
393 IF( kp.NE.k )
394 \$ CALL cswap( nrhs, b( k, 1 ), ldb, b( kp, 1 ), ldb )
395 *
396 * NOW swap the first of pair with Pth
397 *
398 kp = abs( ipiv( k-1 ) )
399 IF( kp.NE.k-1 )
400 \$ CALL cswap( nrhs, b( k-1, 1 ), ldb, b( kp, 1 ),
401 \$ ldb )
402 END IF
403 k = k - 2
404 END IF
405 GO TO 40
406 60 CONTINUE
407 END IF
408 *----------------------------------------
409 *
410 * Compute B := A' * B (transpose)
411 *
412 *----------------------------------------
413 ELSE IF( lsame( trans, 'T' ) ) THEN
414 *
415 * Form B := U'*B
416 * where U = P(m)*inv(U(m))* ... *P(1)*inv(U(1))
417 * and U' = inv(U'(1))*P(1)* ... *inv(U'(m))*P(m)
418 *
419 IF( lsame( uplo, 'U' ) ) THEN
420 *
421 * Loop backward applying the transformations.
422 *
423 k = n
424 70 IF( k.LT.1 )
425 \$ GO TO 90
426 *
427 * 1 x 1 pivot block.
428 *
429 IF( ipiv( k ).GT.0 ) THEN
430 IF( k.GT.1 ) THEN
431 *
432 * Interchange if P(K) != I.
433 *
434 kp = ipiv( k )
435 IF( kp.NE.k )
436 \$ CALL cswap( nrhs, b( k, 1 ), ldb, b( kp, 1 ), ldb )
437 *
438 * Apply the transformation
439 *
440 CALL cgemv( 'Transpose', k-1, nrhs, cone, b, ldb,
441 \$ a( 1, k ), 1, cone, b( k, 1 ), ldb )
442 END IF
443 IF( nounit )
444 \$ CALL cscal( nrhs, a( k, k ), b( k, 1 ), ldb )
445 k = k - 1
446 *
447 * 2 x 2 pivot block.
448 *
449 ELSE
450 IF( k.GT.2 ) THEN
451 *
452 * Swap the second of pair with Pth
453 *
454 kp = abs( ipiv( k ) )
455 IF( kp.NE.k )
456 \$ CALL cswap( nrhs, b( k, 1 ), ldb, b( kp, 1 ), ldb )
457 *
458 * Now swap the first of pair with IMAX(r)th
459 *
460 kp = abs( ipiv( k-1 ) )
461 IF( kp.NE.k-1 )
462 \$ CALL cswap( nrhs, b( k-1, 1 ), ldb, b( kp, 1 ),
463 \$ ldb )
464 *
465 * Apply the transformations
466 *
467 CALL cgemv( 'Transpose', k-2, nrhs, cone, b, ldb,
468 \$ a( 1, k ), 1, cone, b( k, 1 ), ldb )
469 CALL cgemv( 'Transpose', k-2, nrhs, cone, b, ldb,
470 \$ a( 1, k-1 ), 1, cone, b( k-1, 1 ), ldb )
471 END IF
472 *
473 * Multiply by the diagonal block if non-unit.
474 *
475 IF( nounit ) THEN
476 d11 = a( k-1, k-1 )
477 d22 = a( k, k )
478 d12 = a( k-1, k )
479 d21 = d12
480 DO 80 j = 1, nrhs
481 t1 = b( k-1, j )
482 t2 = b( k, j )
483 b( k-1, j ) = d11*t1 + d12*t2
484 b( k, j ) = d21*t1 + d22*t2
485 80 CONTINUE
486 END IF
487 k = k - 2
488 END IF
489 GO TO 70
490 90 CONTINUE
491 *
492 * Form B := L'*B
493 * where L = P(1)*inv(L(1))* ... *P(m)*inv(L(m))
494 * and L' = inv(L'(m))*P(m)* ... *inv(L'(1))*P(1)
495 *
496 ELSE
497 *
498 * Loop forward applying the L-transformations.
499 *
500 k = 1
501 100 CONTINUE
502 IF( k.GT.n )
503 \$ GO TO 120
504 *
505 * 1 x 1 pivot block
506 *
507 IF( ipiv( k ).GT.0 ) THEN
508 IF( k.LT.n ) THEN
509 *
510 * Interchange if P(K) != I.
511 *
512 kp = ipiv( k )
513 IF( kp.NE.k )
514 \$ CALL cswap( nrhs, b( k, 1 ), ldb, b( kp, 1 ), ldb )
515 *
516 * Apply the transformation
517 *
518 CALL cgemv( 'Transpose', n-k, nrhs, cone, b( k+1, 1 ),
519 \$ ldb, a( k+1, k ), 1, cone, b( k, 1 ), ldb )
520 END IF
521 IF( nounit )
522 \$ CALL cscal( nrhs, a( k, k ), b( k, 1 ), ldb )
523 k = k + 1
524 *
525 * 2 x 2 pivot block.
526 *
527 ELSE
528 IF( k.LT.n-1 ) THEN
529 *
530 * Swap the first of pair with Pth
531 *
532 kp = abs( ipiv( k ) )
533 IF( kp.NE.k )
534 \$ CALL cswap( nrhs, b( k, 1 ), ldb, b( kp, 1 ), ldb )
535 *
536 * Now swap the second of pair with IMAX(r)th
537 *
538 kp = abs( ipiv( k+1 ) )
539 IF( kp.NE.k+1 )
540 \$ CALL cswap( nrhs, b( k+1, 1 ), ldb, b( kp, 1 ),
541 \$ ldb )
542 *
543 * Apply the transformation
544 *
545 CALL cgemv( 'Transpose', n-k-1, nrhs, cone,
546 \$ b( k+2, 1 ), ldb, a( k+2, k+1 ), 1, cone,
547 \$ b( k+1, 1 ), ldb )
548 CALL cgemv( 'Transpose', n-k-1, nrhs, cone,
549 \$ b( k+2, 1 ), ldb, a( k+2, k ), 1, cone,
550 \$ b( k, 1 ), ldb )
551 END IF
552 *
553 * Multiply by the diagonal block if non-unit.
554 *
555 IF( nounit ) THEN
556 d11 = a( k, k )
557 d22 = a( k+1, k+1 )
558 d21 = a( k+1, k )
559 d12 = d21
560 DO 110 j = 1, nrhs
561 t1 = b( k, j )
562 t2 = b( k+1, j )
563 b( k, j ) = d11*t1 + d12*t2
564 b( k+1, j ) = d21*t1 + d22*t2
565 110 CONTINUE
566 END IF
567 k = k + 2
568 END IF
569 GO TO 100
570 120 CONTINUE
571 END IF
572 END IF
573 RETURN
574 *
575 * End of CLAVSY_ROOK
576 *
subroutine xerbla(SRNAME, INFO)
XERBLA
Definition: xerbla.f:60
logical function lsame(CA, CB)
LSAME
Definition: lsame.f:53
subroutine cswap(N, CX, INCX, CY, INCY)
CSWAP
Definition: cswap.f:81
subroutine cscal(N, CA, CX, INCX)
CSCAL
Definition: cscal.f:78
subroutine cgemv(TRANS, M, N, ALPHA, A, LDA, X, INCX, BETA, Y, INCY)
CGEMV
Definition: cgemv.f:158
subroutine cgeru(M, N, ALPHA, X, INCX, Y, INCY, A, LDA)
CGERU
Definition: cgeru.f:130
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https://develop.openfoam.com/Development/openfoam/-/commit/09f5d38fb451cf1060a37f1fa316f9aa3c00bfaa | 1,675,422,157,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764500044.66/warc/CC-MAIN-20230203091020-20230203121020-00153.warc.gz | 229,317,195 | 29,879 | Commit 09f5d38f by henry
### Minor consistency changes.
parent 9a7ca2b9
// Solve the Momentum equation // Solve the Momentum equation tmp UEqn ( fvm::ddt(rho, U) + fvm::div(phi, U) + turbulence->divDevRhoReff(U) ); tmp UEqn ( fvm::ddt(rho, U) + fvm::div(phi, U) + turbulence->divDevRhoReff(U) ); //if (oCorr != nOuterCorr-1) if (oCorr == nOuterCorr-1) { UEqn().relax(1); } else { UEqn().relax(); } volScalarField rUA = 1.0/UEqn().A(); if (momentumPredictor) { if (oCorr == nOuterCorr-1) { UEqn().relax(); solve(UEqn() == -fvc::grad(p), mesh.solver("UFinal")); } if (momentumPredictor) else { if (oCorr == nOuterCorr-1) { solve(UEqn() == -fvc::grad(p), mesh.solver("UFinal")); } else { solve(UEqn() == -fvc::grad(p)); } solve(UEqn() == -fvc::grad(p)); } } else { U = rUA*(UEqn().H() - fvc::grad(p)); U.correctBoundaryConditions(); }
... ... @@ -29,7 +29,12 @@ if (transonic) - fvm::laplacian(rho*rUA, p) ); if (oCorr == nOuterCorr-1 && corr == nCorr-1 && nonOrth == nNonOrthCorr) if ( oCorr == nOuterCorr-1 && corr == nCorr-1 && nonOrth == nNonOrthCorr ) { pEqn.solve(mesh.solver("pFinal")); } ... ... @@ -46,7 +51,7 @@ if (transonic) } else { phi = phi = fvc::interpolate(rho)* ( (fvc::interpolate(U) & mesh.Sf()) ... ... @@ -57,6 +62,7 @@ else for (int nonOrth=0; nonOrth<=nNonOrthCorr; nonOrth++) { // Pressure corrector fvScalarMatrix pEqn ( fvm::ddt(psi, p) ... ... @@ -64,7 +70,12 @@ else - fvm::laplacian(rho*rUA, p) ); if (oCorr == nOuterCorr-1 && corr == nCorr-1 && nonOrth == nNonOrthCorr) if ( oCorr == nOuterCorr-1 && corr == nCorr-1 && nonOrth == nNonOrthCorr ) { pEqn.solve(mesh.solver("pFinal")); } ... ...
... ... @@ -68,9 +68,8 @@ int main(int argc, char *argv[]) rho.storePrevIter(); } // --- PIMPLE loop int oCorr=0; do // --- Pressure-velocity PIMPLE corrector loop for (int oCorr=0; oCorrcorrect(); } while (++oCorr < nOuterCorr); } runTime.write(); ... ...
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Finish editing this message first! | 748 | 2,014 | {"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-2023-06 | latest | en | 0.255905 |
rainwater-harvesting-4-homes.com | 1,603,510,937,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107881640.29/warc/CC-MAIN-20201024022853-20201024052853-00063.warc.gz | 89,219,484 | 7,402 | # How Much Water Pressure Will I Get?
Method of Measurement
The height of the column of water contained in the rain barrel will determine how much pressure you will get. The weight of water will create .43 PSI (lbs per sq ft) per foot of height above the point of “draw” (i.e. the location of the spigot, where water will be removed).
Example – Pressure at Spigot
Suppose you have a hose attached to a rain barrel on a stand so that the surface of the water is 10 feet off the ground. And the rain barrel has its spigot 6 feet off the ground. So you have four feet between the top of the water column and the point of draw (i.e. the spigot). If you able to measure the water pressure at the spigot, you find you have 1.72 PSI (4 ft x .43 PSI) of water pressure at the spigot…plenty to fill buckets, push waterthrough a garden house, power a drip irrigation system, etc.
Example – Pressure at End of Hose
Now let’s expand on this example a bit and add a 50 foot garden hose to the spigot and run it to your garden. Let’s say the point of discharge on the end of the hose is now 12 feet below the surface of the water. The math says that you will have about 5.16 PSI of pressure at the end of the hose.If you raise the end of the hose in the air, the pressure drops… lower the end of the hose and the pressure will rise. The length of the hose is immaterial in this example (except for some minor friction complexities).
What’s Normal Water Pressure?
As a point of comparison, normal household water pressure ranges from 20 PSI (mostly RV’s) to about 80 or so PSI. You would need at least 80 feet of fall to run a garden sprinkler or hose sprayer… but the gravity option works well for drip irrigation, hand watering, etc. However, with the addition of a water pump, you can increase the PSI so that you can power your lawn sprinkler or garden hose.
Height vs. Volume
Think of it this way… the more water HEIGHT you stack atop the point of release, the more pressure you will get. The VOLUME of water above the point of release is immaterial. So, the higher you mount your rain barrel, the more water pressure you will generate. | 509 | 2,136 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.859375 | 4 | CC-MAIN-2020-45 | longest | en | 0.90045 |
http://www.java-forums.org/new-java/66240-number-prime-composite-between-two-given-limits-using-nested-while-loop-print.html | 1,477,264,108,000,000,000 | text/html | crawl-data/CC-MAIN-2016-44/segments/1476988719437.30/warc/CC-MAIN-20161020183839-00103-ip-10-171-6-4.ec2.internal.warc.gz | 517,816,254 | 2,724 | # A number is Prime or Composite between two given limits using Nested-while loop.
• 12-11-2012, 12:47 PM
UmairBaloch
A number is Prime or Composite between two given limits using Nested-while loop.
Hi guys,I'm trying to make a program that takes input from user and determines if a number is Prime or composite, I have to give two limits, upper and lower one, and I have to make the program using nested while loop
import java.util.Scanner;
class Pr{
public static void main(String a[]){
int L1, L2, count, check;
Scanner scan = new Scanner(System.in);
System.out.print("Enter the lower limit: ");
L1 = scan.nextInt();
System.out.print("Enter the upper limit: ");
L2 = scan.nextInt();
System.out.print("Enter the Number: ");
int num = scan.nextInt();
while (L1<=L2){
check = 0;
count = 2;
while(count<L1){
if (L1%count==0)
check++;
count++;
}
if (check>0)
System.out.print("Composite");
else
System.out.print("Prime");
}
}
}
it is sort of showing reverse result, eg 64 is prime and 79 is composite,
so any of you guys might wanna help, will be highly appreciated
• 12-11-2012, 05:19 PM
DarrylBurke
Re: A number is Prime or Composite between two given limits using Nested-while loop.
Do we need to repeat this advice every time you post code?
db
• 12-11-2012, 07:03 PM
JosAH
Re: A number is Prime or Composite between two given limits using Nested-while loop.
Why are you asking the user for a 'Number' (and store it in 'num')? You don't do anything with the bounds L1 and L2 either so your outer loop never ends (or it won't even start, depending on the values of L1 and L2)
kind regards,
Jos | 429 | 1,608 | {"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} | 2.96875 | 3 | CC-MAIN-2016-44 | latest | en | 0.812876 |
https://brainmass.com/business/weighted-average-cost-of-capital/weighted-average-contribution-margin-438801 | 1,545,006,778,000,000,000 | text/html | crawl-data/CC-MAIN-2018-51/segments/1544376828018.77/warc/CC-MAIN-20181216234902-20181217020902-00178.warc.gz | 544,847,458 | 18,927 | Explore BrainMass
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Weighted average contribution margin
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Jamal & Co. makes and sells two types of shoes, Plain and Fancy. Data concerning these products are as follows:
Plain Fancy
Unit selling price \$19.00 \$32.00
Variable cost per unit 12.00 23.00
Sixty percent of the unit sales are Plain, and annual fixed expenses are \$46,800.
The weighted-average unit contribution margin is:
Solution Preview
Jamal & Co. makes and sells two types of shoes, Plain and Fancy. Data concerning these products are as follows:
Plain Fancy
Unit selling ...
Solution Summary
Solution helps in computing weighted average contribution margin
\$2.19
Al Corp: sales mix and weighted average contribution margin
Al Corporation sells three products J, K, and L. The following information was taken from a recent budget:
J K L
Unit sales 40,000 130,000 30,000
Selling price \$60 \$80 \$75
Variable cost 40 65 50
Total fixed costs are anticipated to be \$2,450,000.
1. Determine Al's sales mix.
2. Determine the weighted average contribution margin.
3. Calculate the number of units of J, K, and L that must be sold to break even.
4. If Al's desires to increase company profitability, should it attempt to increase or decrease the sales of product K relative to those of J and L? Please explain.
View Full Posting Details | 331 | 1,427 | {"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} | 2.78125 | 3 | CC-MAIN-2018-51 | latest | en | 0.886915 |
http://projecteuler.net/problem=346 | 1,409,430,725,000,000,000 | text/html | crawl-data/CC-MAIN-2014-35/segments/1408500835699.86/warc/CC-MAIN-20140820021355-00460-ip-10-180-136-8.ec2.internal.warc.gz | 154,963,922 | 2,173 | ## Strong Repunits
### Problem 346
Published on Saturday, 3rd September 2011, 04:00 pm; Solved by 1311
The number 7 is special, because 7 is 111 written in base 2, and 11 written in base 6
(i.e. 710 = 116 = 1112). In other words, 7 is a repunit in at least two bases b > 1.
We shall call a positive integer with this property a strong repunit. It can be verified that there are 8 strong repunits below 50: {1,7,13,15,21,31,40,43}.
Furthermore, the sum of all strong repunits below 1000 equals 15864.
Find the sum of all strong repunits below 1012. | 175 | 552 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.609375 | 4 | CC-MAIN-2014-35 | latest | en | 0.875182 |
https://metacpan.org/pod/release/KRYDE/Math-PlanePath-115/lib/Math/PlanePath/HexSpiralSkewed.pm | 1,529,516,709,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267863830.1/warc/CC-MAIN-20180620163310-20180620183310-00462.warc.gz | 658,292,317 | 13,481 | ++ed by:
Kevin Ryde
and 1 contributors
# NAME
Math::PlanePath::HexSpiralSkewed -- integer points around a skewed hexagonal spiral
# SYNOPSIS
`````` use Math::PlanePath::HexSpiralSkewed;
my \$path = Math::PlanePath::HexSpiralSkewed->new;
my (\$x, \$y) = \$path->n_to_xy (123);``````
# DESCRIPTION
This path makes a hexagonal spiral with points skewed so as to fit a square grid and fully cover the plane.
`````` 13--12--11 ... 2
| \ \
14 4---3 10 23 1
| | \ \ \
15 5 1---2 9 22 <- Y=0
\ \ | |
16 6---7---8 21 -1
\ |
17--18--19--20 -2
^ ^ ^ ^ ^ ^
-2 -1 X=0 1 2 3 ...``````
The kinds of N=3*k^2 numbers which fall on straight lines in the plain `HexSpiral` also fall on straight lines when skewed. See Math::PlanePath::HexSpiral for notes on this.
## Skew
The skewed path is the same shape as the plain `HexSpiral`, but fits more points on a square grid. The skew pushes the top horizontal to the left, as shown by the following parts, and the bottom horizontal is similarly skewed but to the right.
`````` HexSpiralSkewed HexSpiral
13--12--11 13--12--11
| \ / \
14 10 14 10
| \ / \
15 9 15 9
-2 -1 X=0 1 2 -4 -3 -2 X=0 2 3 4``````
In general the coordinates can be converted each way by
`````` plain X,Y -> skewed (X-Y)/2, Y
skewed X,Y -> plain 2*X+Y, Y``````
# Corners
`HexSpiralSkewed` is similar to the `SquareSpiral` but cuts off the top-right and bottom-left corners so that each loop is 6 steps longer than the previous, whereas for the `SquareSpiral` it's 8. See "Corners" in Math::PlanePath::SquareSpiral for other corner cutting.
## Wider
An optional `wider` parameter makes the path wider, stretched along the top and bottom horizontals. For example
`` \$path = Math::PlanePath::HexSpiralSkewed->new (wider => 2);``
gives
`````` 21--20--19--18--17 2
| \
22 8---7---6---5 16 1
| | \ \
23 9 1---2---3---4 15 <- Y=0
\ \ |
24 10--11--12--13--14 ... -1
\ |
25--26--27--28--29--30 -2
^ ^ ^ ^ ^ ^ ^ ^
-4 -3 -2 -1 X=0 1 2 3 ...``````
The centre horizontal from N=1 is extended by `wider` many further places, then the path loops around that shape. The starting point 1 is shifted to the left by wider/2 places (rounded up to an integer) to keep the spiral centred on the origin X=0,Y=0.
Each loop is still 6 longer than the previous, since the widening is basically a constant amount added into each loop. The result is the same as the plain `HexSpiral` of the same widening too. The effect looks better in the plain `HexSpiral`.
## N Start
The default is to number points starting N=1 as shown above. An optional `n_start` can give a different start with the same shape etc. For example to start at 0,
`````` n_start => 0
27 26 25 24 3
28 12 11 10 23 2
29 13 3 2 9 22 1
30 14 4 0 1 8 21 ... <- Y=0
31 15 5 6 7 20 39 -1
32 16 17 18 19 38 -2
33 34 35 36 37 -3
-3 -2 -1 X=0 1 2 3 4``````
In this numbering the X axis N=0,1,8,21,etc is the octagonal numbers 3*X*(X+1).
# FUNCTIONS
See "FUNCTIONS" in Math::PlanePath for behaviour common to all path classes.
`\$path = Math::PlanePath::HexSpiralSkewed->new ()`
`\$path = Math::PlanePath::HexSpiralSkewed->new (wider => \$w)`
Create and return a new hexagon spiral object. An optional `wider` parameter widens the spiral path, it defaults to 0 which is no widening.
`\$n = \$path->xy_to_n (\$x,\$y)`
Return the point number for coordinates `\$x,\$y`. `\$x` and `\$y` are each rounded to the nearest integer, which has the effect of treating each point in the path as a square of side 1.
# OEIS
Entries in Sloane's Online Encyclopedia of Integer Sequences related to this path include
`````` A056105 N on X axis, 3n^2-2n+1
A056106 N on Y axis, 3n^2-n+1
A056107 N on North-West diagonal, 3n^2+1
A056108 N on X negative axis, 3n^2+n+1
A056109 N on Y negative axis, 3n^2+2n+1
A003215 N on South-East diagonal, centred hexagonals
n_start=0
A000567 N on X axis, octagonal numbers
A049450 N on Y axis
A049451 N on X negative axis
A045944 N on Y negative axis, octagonal numbers second kind
A062783 N on X=Y diagonal north-east
A033428 N on north-west diagonal, 3*k^2
A063436 N on south-west diagonal
A028896 N on south-east diagonal``````
# HOME PAGE
http://user42.tuxfamily.org/math-planepath/index.html
# LICENSE
Copyright 2010, 2011, 2012, 2013, 2014 Kevin Ryde
This file is part of Math-PlanePath.
Math-PlanePath is free software; you can redistribute it and/or modify it under the terms of the GNU General Public License as published by the Free Software Foundation; either version 3, or (at your option) any later version.
Math-PlanePath is distributed in the hope that it will be useful, but WITHOUT ANY WARRANTY; without even the implied warranty of MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE. See the GNU General Public License for more details.
You should have received a copy of the GNU General Public License along with Math-PlanePath. If not, see <http://www.gnu.org/licenses/>. | 1,761 | 5,567 | {"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} | 2.9375 | 3 | CC-MAIN-2018-26 | latest | en | 0.794008 |
https://techcommunity.microsoft.com/t5/excel/extract-multiple-matching-texts-in-separate-in-2-columns-with/td-p/3674157 | 1,670,244,573,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446711016.32/warc/CC-MAIN-20221205100449-20221205130449-00290.warc.gz | 577,715,111 | 59,946 | SOLVED
# Extract multiple matching texts in separate in 2 columns (with multiple texts separate by comma)
Occasional Contributor
# Extract multiple matching texts in separate in 2 columns (with multiple texts separate by comma)
Hi,
I have sample data as below, in the 3rd column would want to filter out all text that matching between 1st and 2nd column. Is there any formula function that can help to achieve this?
1st column 2nd column matching a,b,c a,c,z a,c c,z.d c,z c,z d,e,e e,d,k d,e b,a,c a,b a,b a,b,c a,c,z a,c c,z.d c,z c,z d,e,e e,d,k d,e b,a,c a,b a,b a,b,c a,c,z a,c c,z.d c,z c,z d,e,e e,d,k d,e b,a,c a,b a,b
6 Replies
# Re: Extract multiple matching texts in separate in 2 columns (with multiple texts separate by comma)
What output do you expect from your given sample? Can you make a 4th column showing your desired result?
best response confirmed by HZ2506 (Occasional Contributor)
Solution
# Re: Extract multiple matching texts in separate in 2 columns (with multiple texts separate by comma)
``=TEXTJOIN(",",,UNIQUE(FILTER(TRANSPOSE(TEXTSPLIT(B2,",")),TRANSPOSE(ISNUMBER(SEARCH(TEXTSPLIT(B2,","),A2))))))``
An alternative could be this formula.
# Re: Extract multiple matching texts in separate in 2 columns (with multiple texts separate by comma)
@Harun24HRHi, column no. 3 is the desire output that i expect....thank you.
# Re: Extract multiple matching texts in separate in 2 columns (with multiple texts separate by comma)
wow this formula work, thanks a lot, this really help.....
# Re: Extract multiple matching texts in separate in 2 columns (with multiple texts separate by comma)
@HZ2506 Oh! That's okay. You may try below formula. Check the attached file.
``=LET(x,TOCOL(TEXTSPLIT(A2,{",","."})),y,COUNTIFS(B2,"*"& x &"*"),z,FILTER(x,y),TEXTJOIN(",",TRUE,UNIQUE(z)))``
# Re: Extract multiple matching texts in separate in 2 columns (with multiple texts separate by comma)
wow!, this formula also work, thanks a lot, really help..... | 529 | 1,987 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.859375 | 3 | CC-MAIN-2022-49 | latest | en | 0.842439 |
https://linear-equation.com/of-a-linear-equation/side-side-side-similarity/9th-grade-math-worksheets.html | 1,722,645,630,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640353668.0/warc/CC-MAIN-20240802234508-20240803024508-00824.warc.gz | 299,190,531 | 11,150 | Try the Free Math Solver or Scroll down to Tutorials!
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