url stringlengths 6 1.61k | fetch_time int64 1,368,856,904B 1,726,893,854B | content_mime_type stringclasses 3 values | warc_filename stringlengths 108 138 | warc_record_offset int32 9.6k 1.74B | warc_record_length int32 664 793k | text stringlengths 45 1.04M | token_count int32 22 711k | char_count int32 45 1.04M | metadata stringlengths 439 443 | score float64 2.52 5.09 | int_score int64 3 5 | crawl stringclasses 93 values | snapshot_type stringclasses 2 values | language stringclasses 1 value | language_score float64 0.06 1 |
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https://codedump.io/share/i9hRICSssN4l/1/order-by-month-and-year-in-r-data-frame | 1,527,436,778,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794869272.81/warc/CC-MAIN-20180527151021-20180527171021-00557.warc.gz | 544,402,123 | 8,732 | T-T - 1 year ago 64
R Question
# Order by month and year in R data frame
I have cleaned and ordered my data by date, which looks like below:
``````df1 <- data.frame(matrix(vector(),ncol=4, nrow = 3))
colnames(df1) <- c("Date","A","B","C")
df1[1,] <- c("2000-01-30","0","1","0")
df1[2,] <- c("2000-01-31","2","0","3")
df1[3,] <- c("2000-02-29","1","2","1")
df1[4,] <- c("2000-03-31","2","1","3")
df1
Date A B C
1 2000-01-30 0 1 0
2 2000-01-31 2 0 3
3 2000-02-29 1 2 1
4 2000-03-31 2 1 3
``````
However, I want to drop the day and order the data by month and year so the data will look like:
`````` Date A B C
1 2000-01 2 1 3
3 2000-02 1 2 1
4 2000-03 2 1 3
``````
I tried to use
`as.yearmon`
from
`zoo`
`df2 <- as.yearmon(df1\$Date, "%b-%y")`
and it returns
`NA`
Here's a way to get the sum of the values for each column within each combination of Year-Month:
``````library(zoo)
library(dplyr)
# Convert non-date columns to numeric
df1[,-1] = lapply(df1[,-1], as.numeric)
df1 %>% mutate(Date = as.yearmon(Date)) %>%
group_by(Date) %>%
summarise_each(funs(sum))
``````
Or, even shorter:
``````df1 %>%
group_by(Date=as.yearmon(Date)) %>%
summarise_each(funs(sum))
``````
`````` Date A B C
1 Jan 2000 2 1 3
2 Feb 2000 1 2 1
3 Mar 2000 2 1 3
``````
### A couple of additional enhancements:
1. Add the number of rows for each group:
``````df1 %>% group_by(Date=as.yearmon(Date)) %>%
summarise_each(funs(sum)) %>%
bind_cols(df1 %>% count(d=as.yearmon(Date)) %>% select(-d))
``````
2. Multiple summary functions:
``````df1 %>% group_by(Date=as.yearmon(Date)) %>%
summarise_each(funs(sum(.), mean(.))) %>%
bind_cols(df1 %>% count(d=as.yearmon(Date)) %>% select(-d))
``````
`````` Date A_sum B_sum C_sum A_mean B_mean C_mean n
1 Jan 2000 2 1 3 1 0.5 1.5 2
2 Feb 2000 1 2 1 1 2.0 1.0 1
3 Mar 2000 2 1 3 2 1.0 3.0 1
``````
Recommended from our users: Dynamic Network Monitoring from WhatsUp Gold from IPSwitch. Free Download | 842 | 2,154 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.59375 | 3 | CC-MAIN-2018-22 | latest | en | 0.652258 |
https://bitbucket.org/bos/mwc-random/commits/3f9179c41a887f6ecd3bb8547463ac2575d823fb/raw/ | 1,448,973,793,000,000,000 | text/plain | crawl-data/CC-MAIN-2015-48/segments/1448398466260.18/warc/CC-MAIN-20151124205426-00130-ip-10-71-132-137.ec2.internal.warc.gz | 816,105,013 | 1,573 | # HG changeset patch # User Alexey Khudyakov # Date 1379252313 -14400 # Node ID 3f9179c41a887f6ecd3bb8547463ac2575d823fb # Parent 184eaad284908d0220b24ec0567759c712dd562c Add workaround proposed by @dzhus for bug #16 For some reason normal written this way works fast. But old was ~250 times slower than standard. diff --git a/System/Random/MWC/Distributions.hs b/System/Random/MWC/Distributions.hs --- a/System/Random/MWC/Distributions.hs +++ b/System/Random/MWC/Distributions.hs @@ -45,20 +45,9 @@ -> Gen (PrimState m) -> m Double {-# INLINE normal #-} -normal m s gen = do - x <- standard gen - return \$! m + s * x - --- | Generate a normally distributed random variate with zero mean and --- unit variance. --- --- The implementation uses Doornik's modified ziggurat algorithm. --- Compared to the ziggurat algorithm usually used, this is slower, --- but generates more independent variates that pass stringent tests --- of randomness. -standard :: PrimMonad m => Gen (PrimState m) -> m Double -{-# INLINE standard #-} -standard gen = loop +-- We express standard in terms of normal and not other way round +-- because of bug in GHC. See bug #16 for more details. +normal m s gen = loop >>= (\x -> return \$! m + s * x) where loop = do u <- (subtract 1 . (*2)) `liftM` uniform gen @@ -97,6 +86,17 @@ then tailing else return \$! if neg then x - r else r - x +-- | Generate a normally distributed random variate with zero mean and +-- unit variance. +-- +-- The implementation uses Doornik's modified ziggurat algorithm. +-- Compared to the ziggurat algorithm usually used, this is slower, +-- but generates more independent variates that pass stringent tests +-- of randomness. +standard :: PrimMonad m => Gen (PrimState m) -> m Double +{-# INLINE standard #-} +standard = normal 0 1 + -- | Generate an exponentially distributed random variate. exponential :: PrimMonad m | 492 | 1,876 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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 | latest | en | 0.764971 |
http://www.enotes.com/homework-help/how-do-find-area-rectangle-abcd-solve-x-439094 | 1,477,456,808,000,000,000 | text/html | crawl-data/CC-MAIN-2016-44/segments/1476988720615.90/warc/CC-MAIN-20161020183840-00177-ip-10-171-6-4.ec2.internal.warc.gz | 429,063,088 | 10,252 | How do I find the area of ABCD and solve for x?
1 Answer |Add Yours
Top Answer
embizze | High School Teacher | (Level 1) Educator Emeritus
Posted on
Assume that ABCD is a parallelogram (hopefully this is given in the problem); Label the point where the perpendicular dropped from C to the extension of AD as F and the angle of the right triangle containing side x as E:
(1) The area of a parallelogram is A=bh where b is the base (one of the sides) and h is the height (the perpendicular distance from the base to the side opposite.)
Choose BC=7.5 as the base. Then CF=4 is the height.
The area of the parallelogram is (7.5)(4)=30 square units.
(2) Traingles CDF and ADE are similar by AA ~ (vertical angles and the right angles.) So `(CD)/(AD)=(DF)/(DE)=(CF)/(AE)`
CD=5,AD=7.5,CF=4, and AE=x
`5/7.5=4/x==>5x=4(7.5) ==> x=(4(7.5))/5=6`
So x=6.
We’ve answered 317,744 questions. We can answer yours, too. | 281 | 916 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.96875 | 4 | CC-MAIN-2016-44 | latest | en | 0.890925 |
https://ingax.com/sorting-pancakes/ | 1,723,392,764,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641002566.67/warc/CC-MAIN-20240811141716-20240811171716-00189.warc.gz | 240,067,103 | 10,066 | ### Pramp Pancake Sort Problem
Given an array of integers arr:
1. Write a function flip(arr, k) that reverses the order of the first kelements in the array arr.
2. Write a function pancakeSort(arr) that sorts and returns the input array. You are allowed to use only the function flip you wrote in the first step in order to make changes in the array.
Example:
input: arr = [1, 5, 4, 3, 2]
output: [1, 2, 3, 4, 5] # to clarify, this is pancakeSort's output
Analyze the time and space complexities of your solution.
Note: it’s called pancake sort because it resembles sorting pancakes on a plate with a spatula, where you can only use the spatula to flip some of the top pancakes in the plate. To read more about the problem, see the Pancake Sorting Wikipedia page.
Constraints:
• [time limit] 5000ms
• [input] array.integer arr
• [input] integer k
• 0 ≤ k
• [output] array.integer
### Solution
func flip(_ arr: [Int], _ k: Int) -> [Int] {
return arr[0...k].reversed() + arr[k+1.. [Int] {
var arr = arr
var partition = arr.count - 1
var maxValue = 0
var maxIndex = 0
while partition > 0 {
for (i, val) in arr[0...partition].enumerated() {
if val >= maxValue {
maxValue = val
maxIndex = i
}
}
arr = flip(arr, maxIndex)
arr = flip(arr, partition)
partition -= 1
maxValue = 0
maxIndex = 0
}
return arr
}
### Time and Space Complexity
Time complexity is $$O(n^2)$$. This is a comparison sorting algorithm. We utilize a double loop. The outer loop iterates through every element. The inner loop decreases in size by 1 on each iteration of the outer loop.
The space complexity is $$O(n)$$. We never utilize more space than the size of the original input array.
### Think Different
1. Forgot to reset my values back to 0 before looping through again.
2. Partition while loop incorrectly set to while partition > 1.
### David Inga
Designed by David Inga in California. | 522 | 1,879 | {"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": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.890625 | 4 | CC-MAIN-2024-33 | latest | en | 0.63236 |
https://www.statology.org/rcorr-r/ | 1,718,426,355,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861583.78/warc/CC-MAIN-20240615031115-20240615061115-00356.warc.gz | 891,927,519 | 20,846 | # How to Use rcorr in R to Create a Correlation Matrix
You can use the rcorr function from the Hmisc package in R to create a matrix of correlation coefficients along with a matrix of p-values for variables in a data frame.
This function is particularly useful because the matrix of p-values allows you to see if the correlation coefficient between different pairwise combinations of variables is statistically significant.
This function uses the following basic syntax:
```library(Hmisc)
#create matrix of correlation coefficients and matrix of p-values
rcorr(as.matrix(df))```
The following example shows how to use the rcorr function in practice.
## Example: How to Use rcorr Function in R
Suppose we have the following data frame in R that contains information about various basketball players:
```#create data frame
df <- data.frame(assists=c(4, 5, 5, 6, 7, 8, 8, 10),
rebounds=c(12, 14, 13, 7, 8, 8, 9, 13),
points=c(22, 24, 26, 26, 29, 32, 20, 14),
steals=c(5, 6, 7, 7, 8, 5, 3, 4))
#view data frame
df
assists rebounds points steals
1 4 12 22 5
2 5 14 24 6
3 5 13 26 7
4 6 7 26 7
5 7 8 29 8
6 8 8 32 5
7 8 9 20 3
8 10 13 14 4
```
We can use the following syntax to create a matrix of correlation coefficients and a matrix of corresponding p-values for this data frame:
```library(Hmisc)
#create matrix of correlation coefficients and matrix of p-values
rcorr(as.matrix(df))
assists rebounds points steals
assists 1.00 -0.24 -0.33 -0.47
rebounds -0.24 1.00 -0.52 -0.17
points -0.33 -0.52 1.00 0.61
steals -0.47 -0.17 0.61 1.00
n= 8
P
assists rebounds points steals
assists 0.5589 0.4253 0.2369
rebounds 0.5589 0.1844 0.6911
points 0.4253 0.1844 0.1047
steals 0.2369 0.6911 0.1047
```
The first matrix shows the correlation coefficient between each pairwise combination of variables in the data frame.
For example, we can see:
• The correlation coefficient between assists and rebounds is -0.24.
• The correlation coefficient between assists and points is -0.33.
• The correlation coefficient between assists and steals is -0.47.
And so on.
The second matrix shows the corresponding p-value for each correlation coefficient from the first matrix.
For example, we can see:
• The p-value for the correlation coefficient between assists and rebounds is 0.5589.
• The p-value for the correlation coefficient between assists and points is 0.4253.
• The p-value for the correlation coefficient between assists and steals is 0.2369.
And so on.
Note: By default, the rcorr function calculates the Pearson correlation coefficient, but you can specify type=’spearman’ if you would instead like to calculate the Spearman Rank correlation coefficient. | 836 | 2,919 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.828125 | 4 | CC-MAIN-2024-26 | latest | en | 0.763982 |
https://betterlesson.com/lesson/resource/1988257/explanation-of-applying-similar-triangles-to-find-slope-lesson-mov | 1,510,999,778,000,000,000 | text/html | crawl-data/CC-MAIN-2017-47/segments/1510934804724.3/warc/CC-MAIN-20171118094746-20171118114746-00207.warc.gz | 586,984,608 | 23,127 | ## Explanation of Applying Similar Triangles to Find Slope Lesson.mov - Section 1: Rational for Lesson Structure
Explanation of Applying Similar Triangles to Find Slope Lesson.mov
# Applying Similar Triangles to Finding the Slope of a Linear Equation
Unit 7: Linear Equations in two Variables
Lesson 14 of 23
## Big Idea: Bring your 8th grade year full circle and make connections between many focus standards all in two days!
Print Lesson
20 teachers like this lesson
Standards:
Subject(s):
Math, similar triangles, slope (Linear Equations), Algebra, linear equations (proportional relationships), 8th Math, dilation
54 minutes
### Christa Lemily
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Oklahoma City, OK
Environment: Urban | 345 | 1,540 | {"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} | 2.5625 | 3 | CC-MAIN-2017-47 | latest | en | 0.841466 |
https://www.mkamimura.com/2018/06/python-javascript_28.html | 1,563,230,666,000,000,000 | text/html | crawl-data/CC-MAIN-2019-30/segments/1563195524254.28/warc/CC-MAIN-20190715215144-20190716001144-00406.warc.gz | 772,802,856 | 33,650 | ## 2018年6月28日木曜日
### 数学 - Python - JavaScript - 解析学 - 積分 - 積分の計算 - 部分分数(部分積分法、逆正接関数、累乗(べき乗))
1. $\begin{array}{}\int \frac{x+1}{{\left({x}^{2}+9\right)}^{2}}\mathrm{dx}=\int \frac{x}{{\left({x}^{2}+9\right)}^{2}}\mathrm{dx}+\int \frac{1}{{\left({x}^{2}+9\right)}^{2}}\mathrm{dx}\\ \int \frac{x}{{\left({x}^{2}+9\right)}^{2}}\mathrm{dx}=-\frac{1}{2\left({x}^{2}+9\right)}\\ x=3t\\ \frac{\mathrm{dx}}{\mathrm{dt}}=3\\ \int \frac{1}{{\left(9{t}^{2}+9\right)}^{2}}·3\mathrm{dt}\\ =\frac{1}{27}\int \frac{1}{{\left({t}^{2}+1\right)}^{2}}\mathrm{dt}\\ =\frac{1}{27}\left(\frac{1}{2}\frac{t}{{t}^{2}+1}+\frac{1}{2}\int \frac{1}{{t}^{2}+1}\mathrm{dt}\right)\\ =\frac{1}{54}\left(\frac{3x}{{x}^{2}+9}+\mathrm{arctan}\frac{x}{3}\right)\end{array}$
よって、求める稹分は、
$-\frac{1}{2\left({x}^{2}+9\right)}+\frac{x}{18\left({x}^{2}+9\right)}+\frac{1}{54}\mathrm{arctan}\frac{x}{3}$
コード(Emacs)
Python 3
#!/usr/bin/env python3
from sympy import pprint, symbols, Integral, plot
print('7.')
x = symbols('x')
f = (x + 1) / (x ** 2 + 9) ** 2
I = Integral(f, x)
for t in [I, I.doit()]:
pprint(t)
print()
p = plot(x + 1, x ** 2 + 9, (x ** 2 + 9) ** 2, f,
ylim=(-0.1, 0.1), legend=True, show=False)
colors = ['red', 'green', 'blue', 'brown']
for i, color in enumerate(colors):
p[i].line_color = color
p.save('sample8.svg')
$./sample8.py 7. ⌠ ⎮ x + 1 ⎮ ───────── dx ⎮ 2 ⎮ ⎛ 2 ⎞ ⎮ ⎝x + 9⎠ ⌡ ⎛x⎞ atan⎜─⎟ x - 9 ⎝3⎠ ─────────── + ─────── 2 54 18⋅x + 162$
HTML5
<div id="graph0"></div>
<pre id="output0"></pre>
<label for="r0">r = </label>
<input id="r0" type="number" min="0" value="0.5">
<label for="dx">dx = </label>
<input id="dx" type="number" min="0" step="0.001" value="0.001">
<br>
<label for="x1">x1 = </label>
<input id="x1" type="number" value="-5">
<label for="x2">x2 = </label>
<input id="x2" type="number" value="5">
<br>
<label for="y1">y1 = </label>
<input id="y1" type="number" value="-5">
<label for="y2">y2 = </label>
<input id="y2" type="number" value="5">
<button id="draw0">draw</button>
<button id="clear0">clear</button>
<script type="text/javascript" src="https://cdnjs.cloudflare.com/ajax/libs/d3/4.2.6/d3.min.js" integrity="sha256-5idA201uSwHAROtCops7codXJ0vja+6wbBrZdQ6ETQc=" crossorigin="anonymous"></script>
<script src="sample8.js"></script>
JavaScript
let div0 = document.querySelector('#graph0'),
pre0 = document.querySelector('#output0'),
width = 600,
height = 600,
btn0 = document.querySelector('#draw0'),
btn1 = document.querySelector('#clear0'),
input_r = document.querySelector('#r0'),
input_dx = document.querySelector('#dx'),
input_x1 = document.querySelector('#x1'),
input_x2 = document.querySelector('#x2'),
input_y1 = document.querySelector('#y1'),
input_y2 = document.querySelector('#y2'),
input_n0 = document.querySelector('#n0'),
inputs = [input_r, input_dx, input_x1, input_x2, input_y1, input_y2],
p = (x) => pre0.textContent += x + '\n';
let f = (x) => x + 1,
g = (x) => x ** 2 + 9,
h = (x) => g(x) ** 2,
fns = [[f, 'red'],
[g, 'green'],
[h, 'blue'],
[(x) => f(x) / h(x), 'brown']];
let draw = () => {
pre0.textContent = '';
let r = parseFloat(input_r.value),
dx = parseFloat(input_dx.value),
x1 = parseFloat(input_x1.value),
x2 = parseFloat(input_x2.value),
y1 = parseFloat(input_y1.value),
y2 = parseFloat(input_y2.value);
if (r === 0 || dx === 0 || x1 > x2 || y1 > y2) {
return;
}
let points = [],
lines = [];
fns
.forEach((o) => {
let [f, color] = o;
for (let x = x1; x <= x2; x += dx) {
let y = f(x);
points.push([x, y, color]);
}
});
let xscale = d3.scaleLinear()
.domain([x1, x2])
let yscale = d3.scaleLinear()
.domain([y1, y2])
let xaxis = d3.axisBottom().scale(xscale);
let yaxis = d3.axisLeft().scale(yscale);
div0.innerHTML = '';
let svg = d3.select('#graph0')
.append('svg')
.attr('width', width)
.attr('height', height);
svg.selectAll('line')
.data([[x1, 0, x2, 0], [0, y1, 0, y2]].concat(lines))
.enter()
.append('line')
.attr('x1', (d) => xscale(d[0]))
.attr('y1', (d) => yscale(d[1]))
.attr('x2', (d) => xscale(d[2]))
.attr('y2', (d) => yscale(d[3]))
.attr('stroke', (d) => d[4] || 'black');
svg.selectAll('circle')
.data(points)
.enter()
.append('circle')
.attr('cx', (d) => xscale(d[0]))
.attr('cy', (d) => yscale(d[1]))
.attr('r', r)
.attr('fill', (d) => d[2] || 'green');
svg.append('g')
.attr('transform', translate(0, ${height - padding})) .call(xaxis); svg.append('g') .attr('transform', translate(${padding}, 0))
.call(yaxis);
[fns].forEach((fs) => p(fs.join('\n')));
};
inputs.forEach((input) => input.onchange = draw);
btn0.onclick = draw;
btn1.onclick = () => pre0.textContent = '';
draw(); | 1,761 | 4,565 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 2, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.25 | 4 | CC-MAIN-2019-30 | latest | en | 0.204966 |
https://www.js-craft.io/blog/the-double-asterisk-operator-in-javascript/ | 1,679,767,751,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296945368.6/warc/CC-MAIN-20230325161021-20230325191021-00217.warc.gz | 929,087,631 | 23,932 | # The double asterisk ** operator in Javascript
#### š Neural networks for Javascript developers
The Neural Networks for JavaScript developers book is almost ready! Learn the basics of AI with TensorFlowJs examples. Join now the presale and get a 15\$ coupon off the launching price!
New day, a new discovery! Today I've learned there that Javascript has a double asterisk `**` operator.
So, it's perfectly valid to write:
``````const x = 2 ** 3
// x is 8, meaning 2 raised to a power of 3
const y = x ** 2
// x is 64, squared x``````
The double asterisk `**` operator was introduced in ECMAScript 2016 and it's also named the exponentiation operator.
Writing `x ** y` it's the same as writing `Math.pow(2, 3)`.
It works also with negative or fractional values:
``````const z = 7 ** -2
// z is 0.02040816326530612, 1 / 49
const t = 4 ** 0.5
// t is 2``````
And we also have the exponentiation assignment:
``````x **= y
// meaning x = x ** y``````
The support for the double asterisk is great, works everywhere except IE.
And speaking of lesser-known operators in Javascript, we also have a double tilde Js operator.
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## Neural Networks for JavaScript developers Presale - 15\$ free coupon
Hi friend! Before you go, just wanted to let you know that in March 2023 I will be launching the TensorFlow.Js by Example course.
This course will include basic concepts of AI and Neural Networks done with TensorFlowJs such as:
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X | 508 | 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} | 2.90625 | 3 | CC-MAIN-2023-14 | longest | en | 0.879286 |
https://www.jiskha.com/display.cgi?id=1328025921 | 1,503,545,590,000,000,000 | text/html | crawl-data/CC-MAIN-2017-34/segments/1502886126027.91/warc/CC-MAIN-20170824024147-20170824044147-00308.warc.gz | 929,444,147 | 4,012 | # Math
posted by .
2y^2 -8y -3 = 0
• Math -
Since this equation does not factor easily, use the quadratic formula to solve for x.
x = [-b + or - √(b^2 - 4ac)]/2a
a = 2, b = -8, and c = -3
Substitute the values and solve for x.
I'll let you take it from here.
## Similar Questions
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i got the wrong answer from the back, and i know that imnot dooing it right. please try to help me on this hard question. THE ANSWER WAS:16 solve the equation: 3log(x-15)= (1/4)^x i don't know how to solve it algerbrically. the teacher … | 432 | 1,283 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.703125 | 4 | CC-MAIN-2017-34 | latest | en | 0.904546 |
https://brainly.in/question/54993 | 1,485,064,803,000,000,000 | text/html | crawl-data/CC-MAIN-2017-04/segments/1484560281353.56/warc/CC-MAIN-20170116095121-00058-ip-10-171-10-70.ec2.internal.warc.gz | 795,133,779 | 10,279 | # If the sides of a right angled triangle are in arthmetic progression then the sines of acute angles are
1
by lokeshkasaraneni
You want the relation between them or exact values?
exact values
can you say the process
2014-11-15T03:50:04+05:30
### This Is a Certified Answer
Certified answers contain reliable, trustworthy information vouched for by a hand-picked team of experts. Brainly has millions of high quality answers, all of them carefully moderated by our most trusted community members, but certified answers are the finest of the finest.
Let the sides be b-d, b and b+d.
The common difference is d. Hypotenuse is b+d.
Since it is a right angle triangle,
b² + (b-d)² = (b+d)²
2 b² + d² - 2 b d = b² + d² + 2 b d
b² = 4 b d
b = 4 d
So the sides are : 3d , 4d, and hypotenuse is 5d
Hence the angles are : Sin 3/5 and Sin 4/5 | 272 | 858 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2017-04 | latest | en | 0.843785 |
https://osbot.org/forum/topic/33714-calculate-experience-until-specified-level/ | 1,716,781,761,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971059028.82/warc/CC-MAIN-20240527021852-20240527051852-00576.warc.gz | 377,555,973 | 21,242 | # Calculate experience until specified level
## Recommended Posts
```public static int getExperienceUntilLevel(int currentExperience, int level) {
int experience = 0;
for(int i = 1; i < level; i++) {
experience += Math.floor(i + 300 * Math.pow(2, i / 7.0f));
}
return (int) Math.floor(experience / 4) - currentExperience;
}
```
• 1
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pretty sure you copied this... at least give credits to whoever u copied from
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pretty sure you copied this... at least give credits to whoever u copied from
Uh.. I interpreted it from this formula, which is from the Runescape wikia
I wouldn't really classify that as copying.. but I've references the source regardless..
• 1
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Uh.. I interpreted it from this formula, which is from the Runescape wikia
I wouldn't really classify that as copying.. but I've references the source regardless..
As long as you understand and know which formula it is I guess it's fine..
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What's the problem of understanding that formula?
Nice one matey, thanks for it .
• 1
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What's the problem of understanding that formula?
Nice one matey, thanks for it .
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Nice, but computing the required XP on each call (especially on paint) will only result in performance issues. I'd suggest just caching the values into a giant 2D array.
• 1
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Nice, but computing the required XP on each call (especially on paint) will only result in performance issues. I'd suggest just caching the values into a giant 2D array.
That's exactly what I do with my script ;)
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Math operations are usually very expensive (Especially the Pow() operation)
I'd recommend just indexing a single-dimensional array
@Liverare, not sure why you would want to do a 2d array
EDIT: This is the one I use
```public static final int[] XP_TABLE = {0, 83, 174, 276, 388, 512, 650, 801, 969,
1154, 1358, 1584, 1833, 2107, 2411, 2746, 3115, 3523, 3973, 4470, 5018,
5624, 6291, 7028, 7842, 8740, 9730, 10824, 12031, 13363, 14833, 16456,
18247, 20224, 22406, 24815, 27473, 30408, 33648, 37224, 41171, 45529,
50339, 55649, 61512, 67983, 75127, 83014, 91721, 101333, 111945, 123660,
136594, 150872, 166636, 184040, 203254, 224466, 247886, 273742, 302288,
333804, 368599, 407015, 449428, 496254, 547953, 605032, 668051, 737627,
814445, 899257, 992895, 1096278, 1210421, 1336443, 1475581, 1629200,
1798808, 1986068, 2192818, 2421087, 2673114, 2951373, 3258594, 3597792,
3972294, 4385776, 4842295, 5346332, 5902831, 6517253, 7195629, 7944614,
8771558, 9684577, 10692629, 11805606, 13034431};```
Edited by FrostBug | 868 | 2,705 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.625 | 3 | CC-MAIN-2024-22 | latest | en | 0.863441 |
http://mail.scipy.org/pipermail/numpy-discussion/2010-April/049893.html | 1,441,425,072,000,000,000 | text/html | crawl-data/CC-MAIN-2015-35/segments/1440645378542.93/warc/CC-MAIN-20150827031618-00057-ip-10-171-96-226.ec2.internal.warc.gz | 138,055,613 | 1,903 | # [Numpy-discussion] Simple way to shift array elements
Gökhan Sever gokhansever@gmail....
Sat Apr 10 19:49:38 CDT 2010
```On Sat, Apr 10, 2010 at 7:31 PM, Charles R Harris <charlesr.harris@gmail.com
> wrote:
>
>
> On Sat, Apr 10, 2010 at 6:17 PM, Gökhan Sever <gokhansever@gmail.com>wrote:
>
>> Hello,
>>
>> Is there a simpler way to get "c" from "a"
>>
>> I[1]: a = np.arange(10)
>>
>> I[2]: b = a[3:]
>>
>> I[3]: b
>> O[3]: array([3, 4, 5, 6, 7, 8, 9])
>>
>> I[4]: c = np.insert(b, [7]*3, 0)
>> O[5]: array([3, 4, 5, 6, 7, 8, 9, 0, 0, 0])
>>
>> a and c have to be same in length and the left shift must be balanced with
>> equal number of 0's
>>
>>
> Maybe something like
>
> In [1]: a = np.arange(10)
>
> In [2]: b = zeros_like(a)
>
> In [3]: b[:-3] = a[3:]
>
> In [4]: b
> Out[4]: array([3, 4, 5, 6, 7, 8, 9, 0, 0, 0])
>
> Chuck
>
>
> _______________________________________________
> NumPy-Discussion mailing list
> NumPy-Discussion@scipy.org
> http://mail.scipy.org/mailman/listinfo/numpy-discussion
>
>
Thanks,
Your ways are more obvious than my first approach. With a bit more playing I
get a one-liner:
c=np.append(a[3:], [0]*3)
--
Gökhan
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``` | 500 | 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} | 2.859375 | 3 | CC-MAIN-2015-35 | latest | en | 0.622507 |
http://www.physicsforums.com/showthread.php?t=473302 | 1,410,975,666,000,000,000 | text/html | crawl-data/CC-MAIN-2014-41/segments/1410657124236.90/warc/CC-MAIN-20140914011204-00154-ip-10-196-40-205.us-west-1.compute.internal.warc.gz | 732,958,536 | 11,628 | # Conservation of momentum in combination of angular and linear momentum
by sr241
Tags: momentum
P: 83 angular momentum and linear momentum is conserved, but what happen when combination angular momentum and linear momentum occurs? for example a ball hits a horizontal paddle wheel on a base(which is free to move in any directions). then what happen to linear momentum of ball and paddle wheel? will it be equal to linear momentum before collision? or does the linear momentum after collision will be reduced by angular momentum gained by of paddle wheel (paddle wheel was initially at rest and due to tangential collision of ball it starts rotating after collision)
Mentor P: 41,568 Both linear and angular momentum will be separately conserved during the collision.
P: 112 Yes, momentum before = momentum after; likewise angular momentum. It is not that linear momentum gets "converted to" angular momentum. It is that your system of ball and paddle wheel already has angular momentum before the collision occurs. The ball's trajectory before collision is a straight line with tangential separation $r$ from the centre of mass of the paddle wheel. $\vec L = \vec r \times \vec p$. Since $r$ and $p$ are non-zero, the ball-paddle system possesses angular momentum.
P: 83
Conservation of momentum in combination of angular and linear momentum
Quote by m.e.t.a. The ball's trajectory before collision is a straight line with tangential separation $r$ from the centre of mass of the paddle wheel. $\vec L = \vec r \times \vec p$. Since $r$ and $p$ are non-zero, the ball-paddle system possesses angular momentum.
you mean the ball and paddle system possesses angular momentum before collision due to and $\vec p$ of ball ( which travels in straight line before collision) with respect to paddle wheel's CG.
But what about Kinetic Energy(KE) of ball and paddle system before and after collision, will it be split into rotational KE and Linear KE and their sum equals initial KE of Ball
Mentor
P: 41,568
Quote by sr241 you mean the ball and paddle system possesses angular momentum before collision due to and $\vec p$ of ball ( which travels in straight line before collision) with respect to paddle wheel's CG.
The angular momentum of the system depends on where you measure it from. Regardless, as long as there's no external torque on the system, angular momentum will be conserved.
If the ball travels in a straight line towards the paddle wheel's COG, why do you think it will start rotating? Originally you said the collision was 'tangential', which I interpreted to mean not along the COG.
But what about Kinetic Energy(KE) of ball and paddle system before and after collision, will it be split into rotational KE and Linear KE and their sum equals initial KE of Ball
In general, kinetic energy is not conserved--unless the collision is perfectly elastic (which is just a way of saying that the kinetic energy is conserved).
P: 83 no ball travels parallel to center of mass of paddle wheel and hit paddle wheel tangentially. assume collisions are perfectly elastic. then what about rotational kinetic energy and linear kinetic energy, before and after collision. Will it conserved independently just as in case of linear and angular momentum. Or will it be sums of rotational and linear kinetic energy
Mentor
P: 41,568
Quote by sr241 no ball travels parallel to center of mass of paddle wheel and hit paddle wheel tangentially.
OK.
assume collisions are perfectly elastic. then what about rotational kinetic energy and linear kinetic energy, before and after collision. Will it conserved independently just as in case of linear and angular momentum. Or will it be sums of rotational and linear kinetic energy
The sum of linear and rotational KE will be conserved, not each separately.
P: 83
Quote by Doc Al The sum of linear and rotational KE will be conserved, not each separately.
that means linear kinetic energy will be less after collision due to the rotational kinetic energy gained by paddle wheel, which was at rest before collision. right?
Mentor
P: 41,568
Quote by sr241 that means linear kinetic energy will be less after collision due to the rotational kinetic energy gained by paddle wheel, which was at rest before collision. right?
Right.
P: 83 what happens when ball travels through center of mass of paddle wheel and base and hits paddle wheel tangentially . now how is going to be the conservation of angular momentum and linear momentum. Also tell me about the conservation of Rotational and Linear Kinetic Energy
P: 112 sr241, in response to your diagram: my best guess of the situation after collision is as follows. (Let the point CM be the centre of mass of the paddle-base system. The ball has initial momentum ${p_{{\rm{b,i}}}}$ and final momentum ${p_{{\rm{b,f}}}}$.) The paddle wheel rotates anticlockwise about its axle with angular momentum $L$ w.r.t. CM. The paddle-base system rotates clockwise about CM with angular momentum $-L$ w.r.t. CM. The paddle-base system has final linear momentum ${p_{{\rm{pb}}}}$ (rightwards). The ball has final linear momentum ${p_{{\rm{b,f}}}} = {p_{{\rm{b,i}}}} - {p_{{\rm{pb}}}}$. Momentum before = momentum after = ${p_{{\rm{b,i}}}}$. Angular momentum before = angular momentum after = 0 (all w.r.t. CM). (This is ignoring any complicating factors, such as the possibility of the ball's trajectory being deflected off-axis upon collision with the paddle. I'm assuming that the ball continues rightwards along its original linear trajectory, or else bounces normally off the paddle.) This is just an intuitive guess, so please take my answer with a pinch of salt until somebody either confirms or refutes it.
P: 83
Quote by Doc Al Both linear and angular momentum will be separately conserved during the collision.
from your earlier response it seems that ball and paddle wheel possess some angular momentum. I have made a vector drawing of it.
If ball possess angular momentum before collision then that would be P2 and corresponding linear momentum would be P3. you mean to say P2 and P3 will separately conserved, then in a way it is the sum of P2 + P3 = P1 (actual linear momentum before collision) that is conserved. or in other words sum of angular momentum and linear momentum will be conserved. or say some part of initial linear momentumP1 will go into angular momentum
P: 83 http://www.youtube.com/watch?v=ul9_0F0T9ZA">http://www.youtube.com/watch?v=ul9_0F0T9ZA" type="application/x-shockwave-flash" width="425" height="350"> http://www.youtube.com/watch?v=ul9_0F0T9ZA in the video linear momentum is conserved. and angular momentum is conserved by ball starts rotating in opposite direction of bar.
Mentor
P: 41,568
Quote by sr241 from your earlier response it seems that ball and paddle wheel possess some angular momentum. I have made a vector drawing of it. If ball possess angular momentum before collision then that would be P2 and corresponding linear momentum would be P3. you mean to say P2 and P3 will separately conserved, then in a way it is the sum of P2 + P3 = P1 (actual linear momentum before collision) that is conserved. or in other words sum of angular momentum and linear momentum will be conserved. or say some part of initial linear momentumP1 will go into angular momentum
It took me a while to understand what you were doing with the vectors P1, P2, and P3. Vector P1 is the initial linear momentum of the ball. P2 and P3 are the components of P1 in the tangential and radial directions with respect to the center of mass of the paddle.
To calculate the angular momentum of the ball with respect to the original position of the paddle's center of mass, you would evaluate r X P1, where r is the position vector of the ball measured from the paddle's center of mass. This a vector cross product. The magnitude of that angular momentum will equal rP2. (Note: P2 is the tangential component of P1, not the angular momentum.)
In this collision, two things are conserved: The angular momentum and the linear momentum. The angular momentum I discussed above; the linear momentum is just P1 = P2 + P3.
P: 83 if two balls of same mass (a,b - a moving initially and b at rest) collides along center of mass; after collision a will be at rest and b will be moving with same velocity as "a". but if a's path is parallel to b's center of mass, a will not be at rest after collision; a will be moving along same path with velocity = a's initial velocity - b's final velocity, ( thus linear momentum is conserved) . but a and b starts rotating with respect to their center of mass in opposite direction. thus angular momentum is conserved, right ? How to calculate a's and b's final velocity and angular velocity from the path and initial velocity of "a" ? does moment of inertia has anything to do with this?
P: 112
Quote by sr241 http://www.youtube.com/watch?v=ul9_0F0T9ZA in the video linear momentum is conserved. and angular momentum is conserved by ball starts rotating in opposite direction of bar.
In the video, after collision, the ball and paddle on the left rotate in opposite directions; but the ball and paddle on the right rotate in the same direction. I don't know what software you (or somebody) used to create this simulation, but it does not appear to be physically accurate. For a start, upon collision, the balls start rotating anticlockwise instantaneously of their own accord. There is no physical reason why they should do this. For the purposes of understanding this discussion, please ignore what you see in that video.
I might have confused things when I wrote, in my first post, "...the ball-paddle system possesses angular momentum." I ought to have added, "about its (the ball-paddle system's) own centre of mass".
Please note that the above applies to your diagram in post #12, i.e. in the case where the ball-paddle (or ball-paddle-base) system possesses non-zero angular momentum about its own centre of mass.
For your final question, regarding the two balls of equal mass, we need only consider the conservation laws which Doc Al has already stated: conservation of linear momentum, and conservation of angular momentum.
(For simplicity, let all trajectories lie in the x-y plane.)
Let's say ball A travels with velocity $\vec u$ along the positive $x$ axis. Ball A strikes ball B off-centre. Let the final velocities of A and B be $\vec v$ and $\vec w$ respectively. Then, due to conservation of linear momentum:
$$\left| {\vec u} \right| = {u_x} = {v_x} + {w_x}$$
$${v_y} + {w_y} = {u_y} = 0$$
$$\Rightarrow {v_y} = - {w_y}$$
For a perfectly elastic collision, balls A and B each rebound with zero spin, and their velocities are at 90º to each other. This conserves linear momentum and linear kinetic energy.
If the balls impart some rotation/spin to one another upon collision (due to friction, deformation etc.) then a portion of the initial kinetic energy of A is transferred to rotational/spin kinetic energy of A and B. The remainder of A's initial kinetic energy is transferred to linear kinetic energy of A and B. The balls' rebound velocities will then necessarily make an angle 0º < $\theta$ < 90º with each other. (Say if you need help proving this.)
My feeling is that if the balls do impart some spin angular momentum to one another, then the angular momentum of A will be equal and opposite to that of B. (I haven't proved this. Perhaps someone could comment...?)
The angular momentum of the system (w.r.t. some arbitrary point P) is constant. The linear momentum of the system is constant. The linear kinetic energy of the system can change. The rotational kinetic energy of the system can change.
P: 83 http://www.youtube.com/watch?v=BVgHV1LgzvM the video shows two equal mass bars collide their linear momentum is separately conserved. but they are spinning in same direction. could it be because of direction of n in the cross product for angular momentum = p*r*sin theta * n in the first bar n is towards us and in second bar n away from viewer. thus angular momentum can be conserved separately even if bars are rotating in same direction. then what is the significance of n in angular momentum?
Related Discussions Classical Physics 3 Introductory Physics Homework 3 Introductory Physics Homework 5 Introductory Physics Homework 2 Introductory Physics Homework 4 | 2,776 | 12,346 | {"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": 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} | 3.796875 | 4 | CC-MAIN-2014-41 | latest | en | 0.870545 |
https://rdrr.io/cran/BRDT/man/bcost_RDT.html | 1,624,066,977,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623487643354.47/warc/CC-MAIN-20210618230338-20210619020338-00149.warc.gz | 430,772,096 | 7,379 | # bcost_RDT: Binomial RDT Cost In BRDT: Binomial Reliability Demonstration Tests
## Description
Define the cost function of RDT, mainly determined by the test sample size (for binomial RDT)
## Usage
`1` ```bcost_RDT(Cf, Cv, n) ```
## Arguments
`Cf` Fixed costs `Cv` Variable costs. `n` Optimal test sample size
## Value
Binomial RDT cost
`bcost_RG`, `bcost_WS`, `bcost_expected`
```1 2 3``` ```#the n value can be the minimum test sample size obtained from \code{\link{boptimal_n}}. n_optimal <- 20 bcost_RDT(Cf = 0, Cv = 10, n = n_optimal); ``` | 178 | 557 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2021-25 | latest | en | 0.701502 |
https://community.qlik.com/t5/App-Development/Sum-of-sales-for-current-year-if-no-selection-made/m-p/1556864 | 1,709,096,539,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474697.2/warc/CC-MAIN-20240228044414-20240228074414-00432.warc.gz | 176,722,449 | 37,665 | Announcements
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Contributor
## Sum of sales for current year if no selection made
Hi All,
I want my report to sum of max sales for each month in current year on opening for first time and on later year selections it should change accordingly.
sum(aggr((max(sales)),Month)) This gives for all the years available.
Thanks
Labels (1)
• ### gettselectedcount and current year
Partner - Contributor II
Hi Lakshmijanaki22,
the requirements for the formula are not quite clear for me, but try to give you an answers. In addition, it also depends on your data model.
I understand that you want calculate the highest order sales for the current maximum month.
If your facts are aggregated at order level, you can use something like following formula:
`Max({<[OrderDate.Month]={\$(=Month(Max([OrderDate.Date])))}, [OrderDate.Year]={\$(=Year(Max([OrderDate.Date])))}>}Sales)`
If your facts based on order detail level, you can use something like following formula:
`Max(Aggr(Sum({<[OrderDate.Month]={\$(=Month(Max([OrderDate.Date])))}, [OrderDate.Year]={\$(=Year(Max([OrderDate.Date])))}>}Sales), OrderID))` | 276 | 1,149 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.59375 | 3 | CC-MAIN-2024-10 | latest | en | 0.829853 |
https://www.experts-exchange.com/questions/20324781/convert-from-buffer-to-string-and-from-string-to-buffer.html | 1,505,959,137,000,000,000 | text/html | crawl-data/CC-MAIN-2017-39/segments/1505818687592.20/warc/CC-MAIN-20170921011035-20170921031035-00282.warc.gz | 795,213,509 | 37,316 | Still celebrating National IT Professionals Day with 3 months of free Premium Membership. Use Code ITDAY17
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Solved
# convert from buffer to string and from string to buffer
Posted on 2002-07-18
Medium Priority
6,388 Views
Hi:
how i can convert buffer which contain voice to string ,and convert the string back to buffer?
shaza
0
Question by:shaza
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LVL 20
Accepted Solution
ID: 7161500
Do you mean voice recognition? Or do you just want to know how to convert any data buffer into a string? That's easy:
function BufToStr(var buf; bufSize: integer) : string;
begin
SetLength(result, bufSize);
Move(buf, pointer(result)^, bufSize);
end;
procedure StrToBuf(var buf; str: string);
begin
Move(pointer(str)^, buf, Length(str));
end;
Of course the buffer must be big enough to hold all the string data in the StrToBuf function.
0
LVL 11
Expert Comment
ID: 7161512
Voice? Do you mean speech recognition?
Forget it!
If you mena simple string operations:
var
Buffer: array [0..1023] of Char;
S: string;
begin
//... fill buffer
// copy buffer to string
S := Buffer;
// copy string to buffer
StrCopy(Buffer, PChar(S));
// with length check
StrLCopy(Buffer, PChar(S), SizeOf(Buffer));
// not to the first char of buffer but to the 8th
StrCopy(@Buffer[7], PChar(S));
0
Author Comment
ID: 7161513
yes i mean voice recognition.
0
LVL 20
Expert Comment
ID: 7161519
:-) There are developers who spent years of developing voice recognition techniques. I don't know any components which do that. You can search on the net for it, but I guess you won't find anything. As Robert said, you'll probably have to forget it...
0
LVL 11
Expert Comment
ID: 7161535
There are of course APIs available, but mostly for commercial products and only for a limited number of languages.
The only promising path i see is to have a look at MS Agents. There is speech recognition, but i do not know if you can get text from it.
0
Author Comment
ID: 7161568
I use API functions (wav files)for recording and playing the voice .I want to send this voice to multicast using udp protocol ,i use Mcast component for sending multicast message ,the Mcast component has Send method which i use it to send the voice ,but this method receve string (Send(text:string)),i have to convert the buffer to string to send it using this component.And i want to convert it back to buffer to play the sound.
This is my problem ,i hope you understand me ,and able to help me.
shaza
0
LVL 20
Expert Comment
ID: 7161573
That doesn't sound like voice recognition at all. Try the BufToStr function I posted above. You don't need to convert the string back to a buffer. You can use "pointer(string)" or "pointer(string)^", this way the string *is* a buffer.
0
Author Comment
ID: 7161688
this is my program ,but there is (invalid pointer operation)Error.
unit client;
interface
uses
Windows, Messages, SysUtils, Variants, Classes, Graphics, Controls, Forms,
Dialogs, StdCtrls, ipwcore, ipwmcast, IdBaseComponent, IdComponent,
IdUDPBase, IdUDPServer,mmsystem, IdUDPClient,idsockethandle, NMUDP;
const
memBlockLength = 1024;
type
Tmemblock = array[0..memblocklength] of byte;
PmemBlock = ^TmemBlock;
TForm1 = class(TForm)
Button1: TButton;
Edit1: TEdit;
Edit2: TEdit;
CheckBox1: TCheckBox;
nmemo: TMemo;
mcast: TipwMCast;
Button2: TButton;
procedure FormCloseQuery(Sender: TObject; var CanClose: Boolean);
procedure FormCreate(Sender: TObject);
procedure mcastDataIn(Sender: TObject; Datagram: String;
procedure Button1Click(Sender: TObject);
procedure Button2Click(Sender: TObject);
private
{ Private declarations }
HwaveIn:PHWaveIn;
HWaveOut:PHWaveOut;
close_invoked:boolean;
in_count,out_count:integer;
// ActiveSocket: TCustomWinSocket;
procedure InitSocket;
procedure InitAudio;
procedure CreateInBlock;
procedure CreateOutBlock(memBlock: PmemBlock; size: Integer);
procedure MMOutDone(var msg:Tmessage);message MM_WOM_DONE;
procedure MMInDone(var msg:Tmessage);message MM_WIM_DATA;
procedure CloseAudio;
procedure NMsg(msg: String);
public
{ Public declarations }
end;
var
Form1: TForm1;
s:integer;
s2:string;
implementation
{function BufToStr(var buf; bufSize: integer) : string;
begin
SetLength(result, bufSize);
Move(buf, pointer(result)^, bufSize);
end; }
function buftostr(buffer:PmemBlock;i:integer):string;
var
str:string;
begin
SetLength(str,i);
Move(buffer, Pointer(str)^,i);
buftostr:=str;
end;
{function strtobuf(str:string):pmemBlock;
var
buffer:PmemBlock;
begin
Move(Pointer(str)^, buffer, Length(str));
strtobuf:=buffer;
end; }
{\$R *.dfm}
procedure TForm1.InitAudio;
var
WaveFormat:Pwaveformatex; // defines the format of waveform-audio data
i,j:integer;
begin
s:=0;
WaveFormat:=new(Pwaveformatex);
WaveFormat.wFormatTag:=wave_format_pcm;
waveformat.nChannels:=1;
waveformat.nSamplesPerSec:=8000;
waveformat.nAvgBytesPerSec:=8000;
waveformat.nBlockAlign:=1;
waveformat.wBitsPerSample:=8;
waveformat.cbSize:=0;
i:=waveOutOpen(nil,0,WaveFormat,0,0,WAVE_FORMAT_QUERY); //opens the given waveform-audio output device for playback.
if i<>0 then NMsg('Error: Play format not supported');
i:=waveInOpen(nil,0,WaveFormat,0,0,WAVE_FORMAT_QUERY);//opens the given waveform-audio input device for recording.
if i<>0 then NMsg('Error: Record format not supported');
HwaveOut:=new(PHwaveOut);
i:=waveOutOpen(HWaveOut,0,WaveFormat,form1.handle,0,CALLBACK_WINDOW);
if i<>0 then NMsg('Error: Problem creating play handle');
HwaveIn:=new(PHwaveIn);
i:=waveInOpen(HWaveIn,0,WaveFormat,form1.handle,0,CALLBACK_WINDOW);
if i<>0 then NMsg('Error: Problem creating record handle');
{these are the count of the number of blocks sent to}
{the audio device}
in_count:=0;
out_count:=0;
edit1.text:=inttostr(in_count);
edit2.text:=inttostr(out_count);
{need to add some buffers to the recording queue}
{in case the messages that blocks have been recorded}
{are delayed}
for j:= 1 to 2 do
CreateInBlock;
{finally start recording}
i:=waveInStart(HwaveIn^);
if i<>0 then NMsg('Error: Start error');
close_invoked:=false;
end;
procedure TForm1.CreateInBlock;
var
memBlock:PmemBlock;
i:integer;
begin
{make a new block}
memBlock:=new(PmemBlock);
begin
lpData:=Pointer(memBlock);
dwBufferLength:=memblocklength;
dwBytesRecorded:=0;
dwUser:=0;
dwFlags:=0;
dwLoops:=0;
end;
{prepare the new block}
if i<>0 then NMsg('Error: In Prepare error');
if i<>0 then NMsg('Error: Add buffer error');
inc(in_count);
end;
procedure TForm1.CreateOutBlock(memBlock: PmemBlock; size: Integer);
var
i:integer;
begin
{make a new block}
begin
lpData:=Pointer(memBlock);
dwBufferLength:=size;
dwBytesRecorded:=0;
dwUser:=0;
dwFlags:=0;
dwLoops:=0;
end;
{prepare it for play back}
if i<>0 then NMsg('Error: Out Prepare error');
{add it to the playback queue}
if i<>0 then NMsg('Error: Wave out error');
inc(out_count);
end;
procedure TForm1.MMOutDone(var msg:Tmessage);
var
// i:integer;
begin
//NMsg('MMOutDone');
dec(out_count);
{free the memory}
{if there's no more blocks being recorded}
if (out_count=0) and close_invoked then
begin
WaveOutClose(HWaveOut^);
HwaveOut:=nil;
end;
(* {if there's nothing more to do then close}
if (in_count=0) and (out_count=0) then
close; *)
end;
procedure TForm1.MMInDone(var msg:Tmessage);
var
i:integer;
begin
// NMsg('MMInDone');
dec(in_count);
{block has been recorded}
if CheckBox1.Checked then
mcast.DataToSend:=s2;
//mcast.Send(s2);
if not(close_invoked) then
begin
CreateInBlock;
edit1.text:=inttostr(in_count);
edit2.text:=inttostr(out_count);
end;
{if there's no more blocks being recorded}
if (in_count=0) and close_invoked then
begin
WaveInClose(HWaveIn^);
HwaveIn:=nil;
end;
(*{if there's nothing more to do then close}
if (in_count=0) and (out_count=0) then
close; *)
end;
procedure TForm1.CloseAudio;
begin
close_invoked:=true;
{reset the output channel}
if HWaveOut<> nil then
begin
WaveOutReset(HWaveOut^);
waveoutclose(Hwaveout^);
end;
{reset the input channel}
if HwaveIn<> nil then
begin
WaveInReset(HWaveIn^);
waveinclose(Hwavein^);
end;
end;
procedure TForm1.NMsg(msg: String);
begin
end;
procedure TForm1.InitSocket;
begin
NMemo.Clear;
mcast.Active:=true;
end;
procedure TForm1.FormCloseQuery(Sender: TObject; var CanClose: Boolean);
begin
CloseAudio;
Canclose:=true;
end;
procedure TForm1.FormCreate(Sender: TObject);
begin
InitSocket;
end;
procedure TForm1.mcastDataIn(Sender: TObject; Datagram: String;
var
s:string;
data: PmemBlock;
i:integer;
begin
data:= new(pmemblock);
//data:=strtobuf(Datagram);
data:=pointer(Datagram);
i:=length(Datagram);
CreateOutBlock(data,i);
end;
procedure TForm1.Button1Click(Sender: TObject);
begin
InitAudio;
end;
procedure TForm1.Button2Click(Sender: TObject);
begin
CloseAudio;
end;
end.
0
LVL 8
Expert Comment
ID: 7161720
You can use MS Speech API, headers are translated on JEDI: www.delphi-jedi.org
0
LVL 20
Expert Comment
ID: 7161740
Looks alright on a quick look. Just can delete the following line in mcastDataIn:
data:= new(pmemblock);
Does it work?
0
Author Comment
ID: 7161751
no it doesn't work!!!!!!!!!!
the error is still occur(invalid pointer operation) after i click button1+checkbox is enable.
what can i do?
0
LVL 20
Expert Comment
ID: 7161772
In which line does the exception occur?
Try this as mcastDataIn:
begin
data := new(pmemblock);
Move(pointer(Datagram)^, data, sizeOf(Datagram));
CreateOutBlock(data, i);
end;
0
LVL 12
Expert Comment
ID: 7163815
hum .. you're talking about VoIP - voice over internet
I've done a small app showing how to do it with Indy UDP (it has multicast support) and ACMCOmpos 1.4 from peter morris
http://lee.nover.has.his.evilside.org/isapi/pas2html.dll/pas2html?File=/delphi/projects/VoIP
check it out, it's quite simple :)
0
Author Comment
ID: 7171899
hi:
i find the application ,but this application is not voice it send text message.
shaza
0
Author Comment
ID: 7171910
i need application to send voice multicast to make audio conference application.
shaza
0
Expert Comment
ID: 7173345
Lee Nover, why does the ACM package give me an error. I'm using D5. There error is something about {\$ALIGN 8} being invalid.
0
LVL 12
Assisted Solution
Lee_Nover earned 200 total points
ID: 7173435
DelFreak: do you have the latest dfs.inc ? if the one on my site is newer then copy that one, it works with it
it's in the 3rdParty folder
shaza: what do you mean ?
my app does capture voice, sends it with UDP, on the other side it receives the data and plays it
where did you find text ?
0
Expert Comment
ID: 9343040
shaza:
EXPERTS:
Post your closing recommendations! No comment means you don't care.
0
LVL 5
Expert Comment
ID: 9461231
shaza,
No comment has been added lately (18 days), so it's time to clean up this TA.
I will leave a recommendation in the Cleanup topic area for this question:
RECOMMENDATION: split points between Madshi http:#7161500 and Lee_Nover http:#7163815
-- Please DO NOT accept this comment as an answer ! --
Thanks,
anAKiN
EE Cleanup Volunteer
0
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Is it a conic? (Posted on 2017-04-19)
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...
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No Solution Yet Submitted by Jer Rating: 4.0000 (3 votes)
Comments: ( Back to comment list | You must be logged in to post comments.)
Pen and acrylics | Comment 2 of 7 |
I used to do these in pen, and then in acrylics, but I never stopped to wonder what the curve is. So, thanks for asking, Jer.
My answer is different from broll's, and he is usually right, but I will post my answer here anyway.
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Edited on April 19, 2017, 12:25 pm
Posted by Steve Herman on 2017-04-19 12:08:49
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Algebra.Equation.Internal.Eval
Contents
# Documentation
extendSig :: (Show a, Num a, Eq a) => Type -> a -> Sig -> Sig Source #
extendOrd :: (Show a, Num a, Eq a) => Type -> a -> Sig -> Sig Source #
addArrowTypes :: (Num a, Eq a) => Sig -> Type -> a -> Sig Source #
data HasType Source #
Constructors
(Ord a, Typeable a) => MkHT a
Instances
Source # MethodsshowList :: [HasType] -> ShowS #
type QSSig = Sig Source #
type Eqs = [Equation] Source #
combine :: Eq a => [[a]] -> [[a]] -> [[a]] Source #
mkCxt :: Typeable a => [[Expr a]] -> Sig -> Context Source #
mkUniv2N :: Typeable a => [[Expr a]] -> [Tagged Term] Source #
stripN :: Expr a -> a Source #
mkEqs2N :: Foldable t => t [Expr a] -> [Equation] Source #
newtype Z Source #
Constructors
Z ()
Instances
Source # Methods(==) :: Z -> Z -> Bool #(/=) :: Z -> Z -> Bool # Source # Methodscompare :: Z -> Z -> Ordering #(<) :: Z -> Z -> Bool #(<=) :: Z -> Z -> Bool #(>) :: Z -> Z -> Bool #(>=) :: Z -> Z -> Bool #max :: Z -> Z -> Z #min :: Z -> Z -> Z # Source # MethodsshowsPrec :: Int -> Z -> ShowS #show :: Z -> String #showList :: [Z] -> ShowS #
newtype S a Source #
Constructors
S a
Instances
Eq a => Eq (S a) Source # Methods(==) :: S a -> S a -> Bool #(/=) :: S a -> S a -> Bool # Ord a => Ord (S a) Source # Methodscompare :: S a -> S a -> Ordering #(<) :: S a -> S a -> Bool #(<=) :: S a -> S a -> Bool #(>) :: S a -> S a -> Bool #(>=) :: S a -> S a -> Bool #max :: S a -> S a -> S a #min :: S a -> S a -> S a # Show a => Show (S a) Source # MethodsshowsPrec :: Int -> S a -> ShowS #show :: S a -> String #showList :: [S a] -> ShowS #
# Orphan instances
Source # MethodsshowList :: [Tagged Term] -> ShowS # Eq (a -> b) Source # Methods(==) :: (a -> b) -> (a -> b) -> Bool #(/=) :: (a -> b) -> (a -> b) -> Bool # Ord (a -> b) Source # Methodscompare :: (a -> b) -> (a -> b) -> Ordering #(<) :: (a -> b) -> (a -> b) -> Bool #(<=) :: (a -> b) -> (a -> b) -> Bool #(>) :: (a -> b) -> (a -> b) -> Bool #(>=) :: (a -> b) -> (a -> b) -> Bool #max :: (a -> b) -> (a -> b) -> a -> b #min :: (a -> b) -> (a -> b) -> a -> b # | 747 | 2,191 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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 | 0.632133 |
https://www.coursehero.com/file/6055218/Assignment-2/ | 1,513,250,625,000,000,000 | text/html | crawl-data/CC-MAIN-2017-51/segments/1512948543611.44/warc/CC-MAIN-20171214093947-20171214113947-00698.warc.gz | 689,175,514 | 47,511 | # Assignment 2 - (c Find a formula for the inverse function...
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MATH 137 Assignment 2, Part 2 Due: 11 am, Friday, October 1 Your assignment consists of two separate parts. Part 1 is available online at http://mapleta.uwaterloo.ca and is due at 4 pm on Thursday September 30. Part 2 consists of the problems below for you to hand in. Place part 2 of your assignment in the correct drop box outside MC 4066, corresponding to the class section in which you are registered. Hand in your solutions to the following 5 problems. You must follow the same set of instructions as in Assignment 1. 1. Let f ( x ) = ax + b cx + d , where a,b,c, and d are constants. (a) When a = b = c = 1 and d = 3 , find a formula for the inverse function f - 1 ( x ) . (b) Assuming that f is one-to-one, find a formula for f - 1 ( x ) in general. (c) What condition on a,b,c, and d guarantees that f is one-to-one? (Hint: Write f ( x 1 ) = f ( x 2 ) , cross-multiply, and then try to conclude that x 1 = x 2 . We must avoid dividing by 0.) 2. Let f ( x ) = ln( x + x 2 + 1 ) . (a) Find the domain of f . (b) Prove that f is an odd function. (Hint: It is easier to check that f ( x ) + f ( - x ) = 0
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Unformatted text preview: .) (c) Find a formula for the inverse function of f . 3. Simplify the following expressions. (a) tan(arcsin x ) (b) cos(2 arctan x ) (Hint: Note that arcsin x = sin-1 x , and arctan x = tan-1 x . See Example 13 on p. 69.) 1 4. State the domains of the following functions and sketch their graphs. (a) f ( x ) = sin(arcsin x ) (b) g ( x ) = arcsin(sin x ) (Hint: For g ( x ) , note that it has period 2 . If you can draw its graph on the interval [-, ] , then you can just replicate the rest of the graph from the piece you have.) 5. Let f be a one-to-one function. For each question below, if your answer is yes, give a short proof, and if your answer is no, give an example where the statement fails. (a) Is f always either increasing or decreasing? (b) If f is increasing, is f-1 also increasing? (c) If f is decreasing, is f-1 also decreasing? (d) If f is odd, is f-1 also odd? 2...
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Assignment 2 - (c Find a formula for the inverse function...
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Ask a homework question - tutors are online | 752 | 2,588 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.6875 | 4 | CC-MAIN-2017-51 | latest | en | 0.904331 |
https://everything2.com/user/Bait/writeups/The+problem+of+logical+fatalism | 1,542,882,219,000,000,000 | text/html | crawl-data/CC-MAIN-2018-47/segments/1542039746205.96/warc/CC-MAIN-20181122101520-20181122123520-00551.warc.gz | 599,672,828 | 6,502 | This is a famous problem and I was surprised I didn't find it with an appreciable amount of effort. This is to say that I am not taking credit for thinking of it. But I will take credit for putting it on E2.
It starts out with two self-evident (or so I think) truths:
1. The law of the excluded middle. Either proposition P is true or it is not true, there is not a third option. Ie, either there is a god or there is not a god; either there is free will or there is not free will.
2. The law of non-contradiction. That is to say that for proposition P, there is a problem if you say P is true and not true. P and ~P (~means "not") are mutally exclusive, if one then not the other.
Seems simple enough, no? Now make P in the future tense- "I will get laid some time this century." Under normal circumstances this is probably P, but I may die or get into some disfiguing accident (yes, it could even happen to you)... so there is the possiblility of ~P. But only these two posibilities exist. It is not possible for me to get laid yet not, at least once "laid" is clearly defined.
And following from #2, if I do then I don't not. If I don't then I don't do. Follow? Let's say P is correct, then there is nothing that can happen or that I can do to stop it . Even if I wanted to stop it, P is still correct. And the same could be said for ~P.
It is therefor impossible for me to make any decisions that interfere with the fruition (no pun intended) of P. But P can be anything. Let's Make P a choice. "I will clap my hands at 3:00." Assuming my hands are untethered and I know when 3:00 is, P or ~P should be the product of my "volition." But as I said be for, it is impossible for me to make any decisions the interfere with the fruition of P. It follows that it is not my decision whether or not to clap my hands at 3:00, or any other P. And since decision is part of the definition of free will, it therefor does not exist. | 496 | 1,929 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.125 | 3 | CC-MAIN-2018-47 | latest | en | 0.964553 |
https://brainly.com/question/141492 | 1,484,802,683,000,000,000 | text/html | crawl-data/CC-MAIN-2017-04/segments/1484560280483.83/warc/CC-MAIN-20170116095120-00213-ip-10-171-10-70.ec2.internal.warc.gz | 789,579,524 | 9,110 | 2014-10-05T21:45:02-04:00
2x-3x-4.6+10.9=15.6 add like terms X -4.6+10.9=15.6 add like tearms again X -15.5 =15.6 add 15.5 to both sides +15.5 +15.5 _________________________ X=31.11
2014-10-05T21:46:18-04:00
### This Is a Certified Answer
Certified answers contain reliable, trustworthy information vouched for by a hand-picked team of experts. Brainly has millions of high quality answers, all of them carefully moderated by our most trusted community members, but certified answers are the finest of the finest.
Most important: Before you start trying to solve it . . . Clean It Up !
-- Add the 2x and the -3x on the left side,making -x .
-- Add the -4.6 and the +10.9 on the left side, making 6.3 .
Now you have an equation that you can work with:
-x + 6.3 = 15.6
Subtract 6.3 from each side : -x = 9.3
Divide each side by-1 : x = -9.3 | 304 | 894 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.75 | 4 | CC-MAIN-2017-04 | latest | en | 0.753914 |
https://ltwork.net/sophia-walked-6-miles-in-90-minutes-at-bear-creek-park-in--3730399 | 1,685,770,183,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224649105.40/warc/CC-MAIN-20230603032950-20230603062950-00279.warc.gz | 406,664,801 | 11,134 | Sophia walked 6 miles in 90 minutes at bear creek park in eluless, texas. if she continued at this rate , use a ratio table to
Question:
Sophia walked 6 miles in 90 minutes at bear creek park in eluless, texas. if she continued at this rate , use a ratio table to determine how many miles she could walk in 60 minutes.
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What is the value of the expressiona) 12(3+13)= 192 b) 12(3+4+5+...+12+13)= 1,056c) 12(1+2+3+...+12+13)= 1,092d) 12(3)•12(13)=
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PLZ HLP! I WILL GIVE BRAINLIEST What is 12 1/2%expressed as a fraction?... | 1,516 | 5,796 | {"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} | 3.125 | 3 | CC-MAIN-2023-23 | latest | en | 0.887926 |
https://hackaday.io/project/27490-binary-desk-usb-jar-clock/discussion-93952 | 1,603,145,411,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107866404.1/warc/CC-MAIN-20201019203523-20201019233523-00127.warc.gz | 362,251,954 | 22,674 | 0%
0%
# Binary desk USB jar clock
A BCD binary clock that fits inside of a jar
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Another crazy idea about building another jar clock and this is how this project started. Why not making it very futuristic and yet combining nature inside it, I said to myself. The clock comes with a lot of custom features, like keeping the time running in the absence of power up to ten years with the help of a RTC and a coin cell.
# Introduction:
When I first started designing this clock I had in mind that I wanted to use some LED lights that are not too bright, but in the same time, they can be seen during daylight or direct sun exposure.
# Challenge:
There was another problem with reading the clock at night. If during the night, some of the LEDs are turned off, how are you able to read the binary string? I mean, it is OK if the 2^3 LED is turned off and all the others are turned on. in this case you can still assume is displaying 7, but what if the 2^0, 2^2 and 2^3 are turned off and only 2^1 is turned on in the digit string. how was I about tho know what it displays, because in the dark you are not able to tell witch LED is it.
# Solution:
The solution was to control the LEDs using a PWM and to make the inactive ones a bit dimmed compared to the active ones.
I realize you are going to use a different type of LED if you are going to replicate this design and you will need to recalculate the serial SMD resistors according to your needs. Keep in mind that his is not a multiplexed display and the LEDs are powered continuously and not intermittently. Yes, the LEDs have a PWM, but the entire display doesn't have a refresh rate. Here is a video I made some time ago explaining how to calculate the LED serial resistor:
Another solution for regulating the intensity of the LEDs according to the outside light intensity, would be to connect a photoresistor voltage divider to one of the MCU A/D channels and regulate the PWM internally. (you can always replace the MCU with a PIC16F819 - it has the same pinout and it has an A/D converter module)
### 4th_jclock.asm
ASM source code for the MCU
asm - 19.82 kB - 09/28/2017 at 02:15
### IMG_1844.JPG
Correlation between the internal registers and the output LEDs and assignment of the MCU pins
JPEG Image - 2.87 MB - 09/28/2017 at 01:40
### 4_th_jck2.dip
dip - 105.83 kB - 09/28/2017 at 01:37
• 1 × 16F628A Microcontroller
• 3 × 74HC595 Electronic Components / Misc. Electronic Components
• 1 × CR2032 battery holder
• 1 × CR2032 battery
• 2 × dome switches
Marius Taciuc09/28/2017 at 02:17 0 comments
I uploaded the ASM source code for the project
Marius Taciuc09/28/2017 at 01:57 0 comments
Check the components area for the component list for this project.
Marius Taciuc09/28/2017 at 01:48 0 comments
Check the Files area to see the connectivity and the assignment of the output pins of the microcontroller.
Marius Taciuc09/28/2017 at 01:46 0 comments
The Layout file for the single sided PCB board has been updated in the Files area.
It requires Diptrace to be opened, view or modified.
Marius Taciuc09/28/2017 at 01:43 0 comments
• 1
The LEDs are PWM controlled so the brighter LED represents logical 1 and the dimmer one means logical 0.
Excuse my photography skills for not being able to reproduce the exact reality image and brightness of the LEDs as you would see them with the naked eye.
The picture below is explaining how to read the BCD clock.
• 2
How to set the clock
To set the clock, press the “S” (set) button for at least one second. The clock will display and highlight only the minutes section. In this state, press the “M” (mode) button to increment the value of the minutes. When you finish setting the minute section, move on to the hour section by shortly pressing the “S” button again. This will highlight and display only the hour section. You can increment the hour value by pressing the “M” button. The “M” button can be pressed continuously for automatic fast increment, or step by step for slow increment. When you finished setting the hour section, you must press the “S” button again to return to the main screen and displaying the clock. This will also reset the seconds area and the seconds value to zero.
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## Discussions
Marius Taciuc wrote 09/28/2017 at 03:38 point
Topic about the power supply and the 7805.
You might be wondering why I placed a 7805 linear regulator on the board if I was planning to always power it from the very steady USB voltage.
The answer it's simple. Because I thought that someone might want to connect it to a cheap chinese ~240 to 5V USB charger and that particular cheap device has a high ripple switching power supply inside. Usually these USB cheap adapters, are set to an effective voltage of 5V and could have spikes or ripples of +-0.5V
Are you sure? yes | no
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# Does this project spark your interest?
Become a member to follow this project and never miss any updates | 1,283 | 5,236 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2020-45 | latest | en | 0.974141 |
https://guruquestion.com/what-does-dpdt-mean-on-a-switch | 1,652,946,641,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662526009.35/warc/CC-MAIN-20220519074217-20220519104217-00471.warc.gz | 348,380,757 | 13,441 | # What does Dpdt mean on a switch?
Author: Juston Oberbrunner | Last update: Saturday, January 29, 2022
A Double Pole Double Throw (DPDT) switch consists of six terminals, two of which are independent input terminals. Each of the poles can complete two different circuits. In other words, each input terminal connects with two output terminals, and all four output terminals are separate.
## What is a DPDT switch used for?
A DPST switch is often used to switch mains electricity because it can isolate both the live and neutral connections. A pair of on-on switches which operate together (shown by the dotted line in the circuit symbol). A DPDT switch can be wired up as a reversing switch for a motor as shown in the diagram.
## What is the difference between SPDT and DPDT switch?
The SPDT has a simple input switch that connects to two output devices whereas DPDT has two SPDT circuits and thus it can control two separate circuits at the same time. The DPDT switches are thus more advantageous than the SPDT switches are they are capable of controlling two different appliances at the same time.
## How does a DPDT slide switch work?
DPDT Slide switches make or break the connection of two conductors to two separate circuits. They usually have six terminals are available in both momentary and maintained contact versions. Function: Unlike push button switches available in momentary or latching function to operate a device or function.
## What does SPDT mean on a switch?
A Single Pole Double Throw (SPDT) switch is a switch that only has a single input and can connect to and switch between the 2 outputs. This means it has one input terminal and two output terminals. A Single Pole Double Throw switch can serve a variety of functions in a circuit.
## What does Dpdt mean in a discrete sensor?
DPDT (Double Pole Double Throw) This switch is equal to two SPDT switches, it means two separate circuits, connecting two inputs of each circuit to one of two outputs.
## What does SPDT and DPDT mean?
The difference between SPDT and DPDT is that “SPDT” stands for “single pole double throw” while “DPDT” stands for “double pole double throw.” Both terms are associated with and are varieties of switches. ... A pole refers to the number of circuits or wires that a switch will control.
## Can you use a DPDT as a SPDT?
Yup, just use one side or the other, a DPDT is just two SPDT in the same case, with one handle.
## How do Slideswitches work?
Slide switches are mechanical switches using a slider that moves (slides) from the open (off) position to the closed (on) position. They allow control over current flow in a circuit without having to manually cut or splice wire. This type of switch is best used for controlling current flow in small projects.
## What is DPDT contact?
Double-pole, double-throw (DPDT): This dual on/on switch contains two input contacts and four output contacts (for a total of six terminals), and behaves like two SPDT (changeover) switches operating in sync. In one position, the two input contacts are connected to one set of output contacts.
## What is the difference between SPST and DPDT?
SPST simply mean single pole, single throw. The product controls one circuit with one On position. ... DPST again means it controls two separate circuits but it only has one On position. DPDT means it has two on positions and can be configured above the same as a single pole can be.
## What does Dpdt mean?
A Double Pole Double Throw (DPDT) switch consists of six terminals, two of which are independent input terminals. Each of the poles can complete two different circuits. In other words, each input terminal connects with two output terminals, and all four output terminals are separate.
## How do DPDT relays work?
Relay is an electromagnetic device used to separate two circuits electrically and connect them magnetically. ... They are often used to interface an electronic circuit, which works at a low voltage to an electrical circuit which works at a high voltage.
## What is on and off switch?
on-off switch in British English
(ˈɒnˈɒf swɪtʃ) a switch or button (on an electrical appliance, etc) which has an 'on' position and an 'off' position. Collins English Dictionary.
## What is SPST NO?
This switch is a SPST-NO, or single pole single throw normally open circuit. This means that current cannot flow through this switch until the actuator has been pressed.
## How does an on-off-on switch work?
The ON-OFF-(ON) circuit is a momentary, double throw, three-position switch circuit. In general, for basic unlighted single pole switches, the maintained ON position closes the circuit at switch terminals 2 & 3, and the momentary ON position closes the circuit at switch terminals 1 & 2.
## What is a DP3T switch?
DP3T can mean: a double pole, triple throw switch, see switch#Contact terminology.
## What is Spst terminal?
A Single Pole Single Throw (SPST) switch is a switch that only has a single input and can connect only to one output. This means it only has one input terminal and only one output terminal. A Single Pole Single Throw switch serves in circuits as on-off switches. When the switch is closed, the circuit is on.
## What does a microswitch do?
Micro Switches have an actuator which, when depressed, lifts a lever to move the contacts into the required position. Micro switches often make a “clicking” sound when pressed this informs the user of the actuation. Micro switches often contain fixing holes so that they can be easily mounted and secured into place.
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Why does Gordon Ramsay throw salt over his shoulder? | 1,244 | 5,673 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.671875 | 3 | CC-MAIN-2022-21 | latest | en | 0.929001 |
https://www.jiskha.com/questions/1147776/i-have-to-take-the-standard-form-of-the-equation-11x-4y-32-and-change-it-to-slope | 1,618,252,604,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038069133.25/warc/CC-MAIN-20210412175257-20210412205257-00133.warc.gz | 900,175,891 | 4,959 | # Algebra
i have to take the standard form of the
Equation 11x - 4y = 32 and change it to slope-
Intercept form y = my + b
Answer 4y = 11x - 32
1. 👍
2. 👎
3. 👁
1. yes
1. 👍
2. 👎
2. or
y = (11/4) x - 8
1. 👍
2. 👎
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## Re: [Axiom-developer] Axiom in general
From: Nicolas Doye Subject: Re: [Axiom-developer] Axiom in general Date: Tue, 13 Sep 2005 11:51:13 +0100
```On Mon, 2005-09-12 at 23:59 -0400, root wrote:
`
> Unfortunately Axiom does not contain a thorough description of the
> mathematical principles. It does not even contain a good explanation
> of category theory or even of the algorithms that it uses. By design
> this will change in the future provided people lend their expertise
> to the problem. It's clearly a 30 year horizon goal.
I'm a bit rusty on all this, and my knowledge out of date. So caveat
blah-de-blah. For those that don't know remember my mini-intro some time
back, I was one of James Davenport's PhD students charged with finding a
mathematical foundation for coercion in Axiom.
http://worldofnic.org/research/phd.ps
Chapter 2 of my thesis gave some introduction to the mathematical
formulation of Categories in Axiom/Magma/OBJ.
Chapter 3 talks about Category Theory.
Chapter 4 talks about Order Sorted Algebra.
(Both these chapters put these definitions in Axiom terms at the end).
Chapter 9 talks about where I'd like Axiom to go.
Some of my "ideas" in my thesis are wrong and I no longer stand by them.
I was naive. Please ignore 9.7 and any assertions that Axiom should be
written in Axiom. ;-)
In some ways, it is far more useful to not think of Axiom in terms of
Category Theory. Functors etc. are useful sometimes when thinking
abstractly about Axiom, but Axiom's true roots are in Order Sorted
Algebra.
In axiom a Category is an Order Sorted Signature, but the order on the
sorts is never explicitly defined and there are some difficulties in the
definition of the operator symbols.
OBJ _is_ based on Order Sorted Algebra. It's main advantage over Axiom
as I see it, is that it defines Theories. A Theory is an Order Sorted
Signature with (and this is what Axiom misses) a set of equations. This
means that in Axiom we are free to say that MyDomain is in MyCategory,
but certain assertions that we as mathematicians know about all Domains
in MyCategory may not be true in MyDomain. Will literate programming
help? Maybe. Will it enforce? Nope.
Another fantastic advantage of OBJ is a view. In Axiom we can not say
that MyDomain forms a Group over the operator * and the operator +,
because the operator is hard-coded into the Group Category. This
sucks. :-( Changing this behaviour is left as an exercise to the reader
with time to rewrite almost all of Axiom. :-)
nic
``` | 650 | 2,547 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2019-22 | latest | en | 0.935113 |
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# Simplify. (X − 2y +3)2 + (X + 2y − 3)2 - SSC (English Medium) Class 8 - Mathematics
ConceptExpansion of (A + B + C)2
#### Question
Simplify.
(x − 2y +3)2 + (x + 2y − 3)2
#### Solution
It is known that,
$\left( a + b + c \right)^2 = a^2 + b^2 + c^2 + 2ab + 2bc + 2ca$
$\ \left( x - 2y + 3 \right)^2 + \left( x + 2y - 3 \right)^2$
$= \left( x \right)^2 + \left( - 2y \right)^2 + \left( 3 \right)^2 + 2 \times x \times \left( - 2y \right) + 2 \times \left( - 2y \right) \times 3 + 2 \times 3 \times x + \left( x \right)^2 + \left( 2y \right)^2 + \left( - 3 \right)^2 + 2 \times x \times 2y + 2 \times \left( 2y \right) \times \left( - 3 \right) + 2 \times \left( - 3 \right) \times x$ $= x^2 + 4 y^2 + 9 - 4xy - 12y + 6x + x^2 + 4 y^2 + 9 + 4xy - 12y - 6x$ $= 2 x^2 + 8 y^2 + 18 - 24y$
Is there an error in this question or solution?
#### APPEARS IN
Balbharati Solution for Balbharati Class 8 Mathematics (2019 to Current)
Chapter 5: Expansion formulae
Practice Set 5.4 | Q: 2.1 | Page no. 28
Solution Simplify. (X − 2y +3)2 + (X + 2y − 3)2 Concept: Expansion of (A + B + C)2.
S | 522 | 1,148 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.34375 | 4 | CC-MAIN-2020-05 | latest | en | 0.443279 |
https://zims-en.kiwix.campusafrica.gos.orange.com/wikipedia_en_all_nopic/A/Spatial_acceleration | 1,620,498,715,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243988923.22/warc/CC-MAIN-20210508181551-20210508211551-00066.warc.gz | 1,144,245,544 | 5,826 | # Spatial acceleration
In physics the study of rigid body motion provides for several ways of defining the acceleration state of a rigid body. The classical definition of acceleration entails following a single particle/point along the rigid body and observing its changes of velocity. In this article the notion of spatial acceleration is explored, which entails looking at a fixed (unmoving) point in space and observing the changes of velocity of whatever particle/point happens to coincide with the observation point. This is similar to the acceleration definition fluid dynamics where typically one can measure velocity and/or accelerations on a fixed locate inside a testing apparatus.
## Definition
Consider a moving rigid body and the velocity of a particle/point P along the body being a function of the position and velocity of a center particle/point C and the angular velocity ${\displaystyle {\vec {\omega }}}$.
The linear velocity vector ${\displaystyle {\vec {v}}_{P}}$ at P is expressed in terms of the velocity vector ${\displaystyle {\vec {v}}_{C}}$ at C as:
${\displaystyle {\vec {v}}_{P}={\vec {v}}_{C}+{\vec {\omega }}\times ({\vec {r}}_{P}-{\vec {r}}_{C})}$
where ${\displaystyle {\vec {\omega }}}$ is the angular velocity vector.
The material acceleration at P is:
${\displaystyle {\vec {a}}_{P}={\frac {{\rm {d}}{\vec {v}}_{P}}{{\rm {d}}t}}}$
${\displaystyle {\vec {a}}_{P}={\vec {a}}_{C}+{\vec {\alpha }}\times ({\vec {r}}_{P}-{\vec {r}}_{C})+{\vec {\omega }}\times ({\vec {v}}_{P}-{\vec {v}}_{C})}$
where ${\displaystyle {\vec {\alpha }}}$ is the angular acceleration vector.
The spatial acceleration ${\displaystyle {\vec {\psi }}_{P}}$ at P is expressed in terms of the spatial acceleration ${\displaystyle {\vec {\psi }}_{C}}$ at C as:
${\displaystyle {\vec {\psi }}_{P}={\frac {\partial {\vec {v}}_{P}}{\partial t}}}$
${\displaystyle {\vec {\psi }}_{P}={\vec {\psi }}_{C}+{\vec {\alpha }}\times ({\vec {r}}_{P}-{\vec {r}}_{C})}$
which is similar to the velocity transformation above.
In general the spatial acceleration ${\displaystyle {\vec {\psi }}_{P}}$ of a particle point P that is moving with linear velocity ${\displaystyle {\vec {v}}_{P}}$ is derived from the material acceleration ${\displaystyle {\vec {a}}_{P}}$ at P as:
${\displaystyle {\vec {\psi }}_{P}={\vec {a}}_{P}-{\vec {\omega }}\times {\vec {v}}_{P}}$
## References
• Frank M. White (2003). Fluid Mechanics. McGraw-Hill Professional. ISBN 0-07-240217-2..
• Roy Featherstone (1987). Robot Dynamics Algorithms. Springer. ISBN 0-89838-230-0.. This reference effectively combines screw theory with rigid body dynamics for robotic applications. The author also chooses to use spatial accelerations extensively in place of material accelerations as they simplify the equations and allows for compact notation.
• JPL DARTS page has a section on spatial operator algebra (link: ) as well as an extensive list of references (link: ).
• Bruno Siciliano, Oussama Khatib (2008). Springer Handbook of Robotics. Springer.. Page 41 (link: Google Books ) defines spatial accelerations for use in rigid body mechanics. | 819 | 3,119 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 16, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.671875 | 4 | CC-MAIN-2021-21 | latest | en | 0.712812 |
https://www.stata.com/statalist/archive/2011-04/msg01169.html | 1,726,462,798,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651676.3/warc/CC-MAIN-20240916044225-20240916074225-00760.warc.gz | 922,872,673 | 5,641 | Notice: On April 23, 2014, Statalist moved from an email list to a forum, based at statalist.org.
# Re: st: RE: AIC and BIC in Poisson and GLM
From Nick Cox <[email protected]> To [email protected] Subject Re: st: RE: AIC and BIC in Poisson and GLM Date Tue, 26 Apr 2011 00:37:52 +0100
```It's stuff like this that makes me feel surprisingly warm and
affectionate towards R-squared. See, for example, -glmcorr- from SSC
and (more importantly) the paper that inspired it
Zheng, B. and A. Agresti. 2000. Summarizing the predictive power of a
generalized linear model. Statistics in Medicine 19: 1771-1781.
The gist of the paper is summarized in the help for -glmcorr-.
Nick
On Mon, Apr 25, 2011 at 11:41 PM, Steven Samuels <[email protected]> wrote:
> Tony-
>
> From the manual for -glm- (p. 527)
>
> "There are various definitions of these information criteria (IC) in the literature; glm and estat ic use different definitions. glm bases its computation of the BIC on deviance, whereas estat ic uses the likelihood. Both glm and estat ic use the likelihood to compute the AIC; however, the AIC from estat ic is equal to N, the number of observations, times the AIC from glm. Refer to Methods and formulas in this entry and [R] estat for the references and formulas used by glm and estat, respectively, to compute AIC and BIC. Inferences based on comparison of IC values reported by glm for different GLM models will be equivalent to those based on comparison of IC values reported by estat ic after glm."
>
> Steve
> [email protected]
>
>
>
> On Apr 25, 2011, at 2:40 PM, Visintainer, Paul wrote:
>
> Peter,
>
> If you compute AIC and BIC using -estat ic- after each model, I think they will be the same. In the -glm- output, the AIC and BIC are given in blue. If you click on them, it describes the differences between the output estimate and the -estat ic- estimate.
>
> -Paul
>
> -----Original Message-----
> From: [email protected] [mailto:[email protected]] On Behalf Of Lachenbruch, Peter
> Sent: Monday, April 25, 2011 12:21 PM
> To: '[email protected]'
> Subject: st: AIC and BIC in Poisson and GLM
>
> I was developing a simple class example for poisson regression using the poisson command and the glm command. The log likelihood was the same in both regressions; the coefficients were the same. When I looked at the AIC and BIC reported by glm I got
>
> AIC = 7.920031
> Log likelihood = -33.60015344 BIC = 2.922026
> For the estat command with poisson I got
> . estat ic
>
> -----------------------------------------------------------------------------
> Model | Obs ll(null) ll(model) df AIC BIC
> -------------+---------------------------------------------------------------
> full | 10 -495.0676 -33.60015 6 79.20031 81.01582
> -----------------------------------------------------------------------------
> Note: N=Obs used in calculating BIC; see [R] BIC note
>
> I would expect these to be the same, but they ain't. I suspect there may be a normalizing constant lurking around here somewhere.
> I don't want to fix this unless it's truly a bug; but I would like to be able to explain this to my students.
>
*
* 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/
``` | 936 | 3,473 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.90625 | 3 | CC-MAIN-2024-38 | latest | en | 0.818268 |
http://www.finderchem.com/what-is-the-midpoint-of-0-and-180-line-in-a-protractor.html | 1,477,583,419,000,000,000 | text/html | crawl-data/CC-MAIN-2016-44/segments/1476988721347.98/warc/CC-MAIN-20161020183841-00527-ip-10-171-6-4.ec2.internal.warc.gz | 452,742,734 | 8,126 | # What is the midpoint of 0 and 180 line in a protractor?
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0° < x° < 90° 90° 90° < x° < 180° 180° The box in the corner means that it is ... The midpoint is the center point of any line segment.
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https://www.britannica.com/science/number-theory | 1,519,220,358,000,000,000 | text/html | crawl-data/CC-MAIN-2018-09/segments/1518891813622.87/warc/CC-MAIN-20180221123439-20180221143439-00778.warc.gz | 823,831,184 | 40,236 | # Number theory
mathematics
Alternative Title: higher arithmetic
Number theory, branch of mathematics concerned with properties of the positive integers (1, 2, 3, …). Sometimes called “higher arithmetic,” it is among the oldest and most natural of mathematical pursuits.
Number theory has always fascinated amateurs as well as professional mathematicians. In contrast to other branches of mathematics, many of the problems and theorems of number theory can be understood by laypersons, although solutions to the problems and proofs of the theorems often require a sophisticated mathematical background.
Until the mid-20th century, number theory was considered the purest branch of mathematics, with no direct applications to the real world. The advent of digital computers and digital communications revealed that number theory could provide unexpected answers to real-world problems. At the same time, improvements in computer technology enabled number theorists to make remarkable advances in factoring large numbers, determining primes, testing conjectures, and solving numerical problems once considered out of reach.
Modern number theory is a broad subject that is classified into subheadings such as elementary number theory, algebraic number theory, analytic number theory, geometric number theory, and probabilistic number theory. These categories reflect the methods used to address problems concerning the integers.
## From prehistory through Classical Greece
The ability to count dates back to prehistoric times. This is evident from archaeological artifacts, such as a 10,000-year-old bone from the Congo region of Africa with tally marks scratched upon it—signs of an unknown ancestor counting something. Very near the dawn of civilization, people had grasped the idea of “multiplicity” and thereby had taken the first steps toward a study of numbers.
It is certain that an understanding of numbers existed in ancient Mesopotamia, Egypt, China, and India, for tablets, papyri, and temple carvings from these early cultures have survived. A Babylonian tablet known as Plimpton 322 (c. 1700 bc) is a case in point. In modern notation, it displays number triples x, y, and z with the property that x2 + y2 = z2. One such triple is 2,291, 2,700, and 3,541, where 2,2912 + 2,7002 = 3,5412. This certainly reveals a degree of number theoretic sophistication in ancient Babylon.
Despite such isolated results, a general theory of numbers was nonexistent. For this—as with so much of theoretical mathematics—one must look to the Classical Greeks, whose groundbreaking achievements displayed an odd fusion of the mystical tendencies of the Pythagoreans and the severe logic of Euclid’s Elements (c. 300 bc).
## Pythagoras
According to tradition, Pythagoras (c. 580–500 bc) worked in southern Italy amid devoted followers. His philosophy enshrined number as the unifying concept necessary for understanding everything from planetary motion to musical harmony. Given this viewpoint, it is not surprising that the Pythagoreans attributed quasi-rational properties to certain numbers.
For instance, they attached significance to perfect numbers—i.e., those that equal the sum of their proper divisors. Examples are 6 (whose proper divisors 1, 2, and 3 sum to 6) and 28 (1 + 2 + 4 + 7 + 14). The Greek philosopher Nicomachus of Gerasa (flourished c. ad 100), writing centuries after Pythagoras but clearly in his philosophical debt, stated that perfect numbers represented “virtues, wealth, moderation, propriety, and beauty.” (Some modern writers label such nonsense numerical theology.)
In a similar vein, the Greeks called a pair of integers amicable (“friendly”) if each was the sum of the proper divisors of the other. They knew only a single amicable pair: 220 and 284. One can easily check that the sum of the proper divisors of 284 is 1 + 2 + 4 + 71 + 142 = 220 and the sum of the proper divisors of 220 is 1 + 2 + 4 + 5 + 10 + 11 + 20 + 22 + 44 + 55 + 110 = 284. For those prone to number mysticism, such a phenomenon must have seemed like magic.
## More About Number theory
20 references found in Britannica articles
### Assorted References
• combinatorial methods
• consistency proof
### development
• Diophantus of Alexandria
• Euclid
• Euler
• Fermat
• Gauss
• Goldbach
• Greek mathematics
• Lafforgue
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Get Full Access to University Physics - 13 Edition - Chapter 10 - Problem 4e
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Three forces are applied to a wheel of radius 0.350 m, as
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Solution for problem 4E Chapter 10
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Problem 4E
Three forces are applied to a wheel of radius 0.350 m, as shown in ?Fig. E10.4?. One force is perpendicular to the rim, one is tangent to it, and the other one makes a 40.0° angle with the radius. What is the net torque on the wheel due to these three forces for an axis perpendicular to the wheel and passing through its center?
Step-by-Step Solution:
Step 1 of 3
Solution 4E We have to calculate the net torque on the wheel due to the given forces. Mathematically, Torque = Force X Perpendicular distance between the force and the axis. Let us have a look at the following figure. Force 11.9 N The perpendicular distance of this force from the axis is zero. Therefore, this force can not produce a torque. So, torque produced by 11.9 N force is zero. = 0 1 Force 8.50 N The perpendicular distance of this force from the axis of rotation is 0.350 m. Moreover, this force acts counterclockwise. Hence the torque produced by it will be positive. Torque by the force 8.50 N, 2 = 8.50 N × 0.350 m = 2.97 N.m Force 14.6 N This force has two components that has been shown in the figure above. The 0 0 x-component is 14.6 cos 40 and the y-component is 14.6 sin 40 . The x-component has a perpendicular distance of zero from the axis of rotation. Therefore, the torque produced by it is zero. The y-component has a perpendicular distance of 0.350 m from the axis. But this component will act clockwise and hence the torque produced by it will be negative. Therefore, torque produced by this force, = 14.6 sin 40 N × 0.350 m 3 = 3.28 N.m 3 Therefore, the net torque on the wheel due to all the force is, = +1 + 2 3 = 0 + 2.97 N.m 3.28 N.m = 0.31 N.m The net torque is -0.31 N.m. The negative sign indicates that it will be clockwise.
Step 2 of 3
Step 3 of 3
ISBN: 9780321675460
The answer to “Three forces are applied to a wheel of radius 0.350 m, as shown in ?Fig. E10.4?. One force is perpendicular to the rim, one is tangent to it, and the other one makes a 40.0° angle with the radius. What is the net torque on the wheel due to these three forces for an axis perpendicular to the wheel and passing through its center?” is broken down into a number of easy to follow steps, and 64 words. The full step-by-step solution to problem: 4E from chapter: 10 was answered by , our top Physics solution expert on 05/06/17, 06:07PM. This full solution covers the following key subjects: wheel, perpendicular, radius, forces, Net. This expansive textbook survival guide covers 26 chapters, and 2929 solutions. University Physics was written by and is associated to the ISBN: 9780321675460. Since the solution to 4E from 10 chapter was answered, more than 3156 students have viewed the full step-by-step answer. This textbook survival guide was created for the textbook: University Physics, edition: 13.
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consumercomplianceoutlook.org | 1,653,554,213,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662604495.84/warc/CC-MAIN-20220526065603-20220526095603-00010.warc.gz | 241,422,109 | 4,504 | The last example in Appendix K [(d)(2), Sample Form] is the combination of a lump sum, monthly advances, and a credit line. The borrower receives a lump-sum advance of \$1,000, plus \$301.80 monthly advance at consummation. The borrower will receive a monthly payment of \$301.80 for the 12-year term of the loan. In addition, the borrower has a \$4,000 line of credit.
Lump sum to borrower \$1,000 Assume annual dwelling appreciation rate 4% Monthly payments to borrower \$301.80 Appraised value of property \$100,000 Total loan costs financed \$5,000 Age of the youngest borrower 75 Credit Line \$4,000 Estimated loan term 12 years Contract interest rate 9% Equity Reserved 7%
## STEP 1 — CALCULATE FUTURE VALUE OF ALL ADVANCES
### Future Value of All Advances
N I PV PMT FV
144 months 0.75% (HP 12c)
9% (HP 17bII)
\$301.80 ?
144 months 0.75% (HP 12c)
9% (HP 17bII)
144 months 0.75% (HP 12c)
9% (HP 17bII)
\$2,000 (1/2 of credit line)1 ?
144 months 0.75% (HP 12c)
9% (HP 17bII)
\$5,000 (total loan costs) ?
I = 9% contract rate divided by 12 months in a year = 0.75%
When calculating the future value of monthly payments, the calculator must be set to BEG mode (payments made at the beginning of the month).
For the HP 17bII, the P/YR must be set to 12 (12 payments per year).
FV (\$301.80 monthly for 12 years) = \$78,360.68 FV (\$1,000 after 12 years) = \$2,932.84 FV (\$2,000 after 12 years) = \$5,865.67 FV (\$5,000 after 12 years) = \$14,664.18 + FV of all advances = \$101,823.37
## STEP 2 — CALCULATE FUTURE VALUE OF THE DWELLING
I = Assumed annual dwelling appreciation rate
### Future Value of the Dwelling
N I PV FV
12 4% \$100,000 ?
For the HP 17bII, the P/YR must be set to 1 (one payment per year).
FV = \$160,103.22
\$160,103.22— \$17,627.19 (7% of FV of the dwelling) = \$148,895.99
## STEP 3 — CALCULATE REPAYMENT AMOUNT
The repayment amount is the lesser of the FV of all advances (\$101,823.37) or the FV of the dwelling minus equity reserved (\$148,895.99).
Repayment Amount = \$101,823.37
## STEP 4 — CALCULATE THE TALC RATE USING THE APRWIN PROGRAM
Here are the entries for the APRWin program:
Loan Information Amount Financed: \$3,301.80 (\$1000 lump sum + \$301.80 first monthly advance + one-half of \$4,000 credit line) Disclosed APR: 11.03% (If the disclosed TALC rate is being verified, enter the disclosed rate here. If the TALC rate is being calculated, enter an estimated rate, e.g., 1.) Disclosed Finance Charge: Leave blank Loan Type: Installment Loan Payment Frequency: Monthly
### Payment Schedule
Payment Stream #1 — The Payment Amount must be entered as a negative value. The number of payments is the remaining number of advances left after the initial advance at consummation.
Payment Stream #2 — The Payment Amount is the amount from step 3.
APR = 11.03% (This is the TALC rate based on the multiple advances to the borrower [\$301.80 x 144 = \$43,459.20], \$1,000 initial advance, \$2,000 credit line outstanding, and the \$101,823.37 payment.)
## DISCLOSING REVERSE MORTGAGES:
In the previous example, we calculated the TALC rate based on a 12-year loan term with an assumed annual appreciation rate of 4 percent, but that is just one of nine TALC rates that must be disclosed. Section 226.33(c) of Regulation Z requires creditors to disclose TALC rates based on three loan terms as determined by the life expectancy of the youngest borrower in accordance with Appendix J to Regulation Z, and to assume annual appreciation rates of 0 percent, 4 percent, and 8 percent for the dwelling.
Below is the reverse mortgage disclosure for example 4. (Note that the TALC rates based on a six-year loan term, which is one-half of life expectancy of the youngest borrower, are optional):
TOTAL ANNUAL LOAN COST RATE
Loan Terms Monthly Loan Charges
Age of youngest borrower: 75 Service fee: None
Appraised property value: \$100,000
Interest rate: 9% Other Charges
Monthly advance: \$301.80 Mortgage insurance: None
Initial draw: \$1,000 Shared appreciation: None
Line of credit: \$4,000 Repayment Limits
Initial Loan Charges Net proceeds estimated at 93% of projected home sale
Closing costs: \$5,000
Annuity cost: None
Assumed Annual Appreciation 2-year loan term [6-year loan term] 12-year loan term 17-year loan term
0% 39.00% [14.94%] 9.86% 3.87%
4% 39.00% [14.94%] 11.03% 10.14%
8% 39.00% [14.94%] 11.03% 10.20%
The cost of any reverse-mortgage loan depends on how long you keep the loan and how much your house appreciates in value. Generally, the longer you keep a reverse mortgage, the lower the total annual loan cost rate will be.
This table shows the estimated cost of your reverse-mortgage loan, expressed as an annual rate. It illustrates the cost for three [four] loan terms: two years, [half of life expectancy for someone your age], that life expectancy, and 1.4 times that life expectancy. The table also shows the cost of the loan, assuming the value of your house appreciates at three different rates: 0 percent, 4 percent, and 8 percent.
The total annual loan cost rates in this table are based on the total charges associated with this loan. These charges typically include principal, interest, closing costs, mortgage insurance premiums, annuity costs, and servicing costs (but not disposition costs—costs when you sell the home).
The rates in this table are estimates. Your actual cost may differ if, for example, the amount of your loan advances varies or the interest rate on your mortgage changes.
SIGNING AN APPLICATION OR RECEIVING THESE DISCLOSURES DOES NOT REQUIRE YOU TO COMPLETE THIS LOAN.
• 1 For a credit line, the TALC must be based on the assumption that 50 percent of the line of credit is outstanding at closing. TALC rate calculations effectively treat the transaction as a closed-end credit transaction after that. | 1,582 | 5,827 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.671875 | 4 | CC-MAIN-2022-21 | latest | en | 0.85322 |
https://research.csiro.au/spark/resources/model-library/csiro-grassland-models/ | 1,726,172,031,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651491.39/warc/CC-MAIN-20240912174615-20240912204615-00558.warc.gz | 453,416,380 | 16,608 | # CSIRO Grassland Models
The CSIRO Grassland rate of spread models were developed by CSIRO for predicting fire spread rates in Australian grasslands. There are three models for grasslands: natural/undisturbed pasture, cut/grazed pasture and eaten out pasture. These spread rates were published in the paper ‘Prediction of Fire Spread in Grasslands’ by Cheney, Gould and Catchpole in the International Journal of Wildfire, which can be found here.
Northern Australia woodland and open forest can also be modelled with adjustments to the natural/undisturbed pasture model based on Cheney and Sullivan (2008).
### Vegetation
Australian Grassland
### Fuel inputs
temp
rel_hum
curing
• Temperature (degrees C)
• Relative humidity (%)
• Curing (%)
### Code
```// CSIRO grasslands models - Cheney et al. (1998)
// Subclasses defined as the following sub type of grasslands:
// 1 - Eaten out
// 2 - Cut / grazed
// 3 - Natural / undisturbed
// 4 - Woodland
// 5 - Open forest
// -------------------------------------------
// Model parameters
// 1. Temperature, 'temp' (input)
// 2. Relative humidity, 'rel_hum' (input)
// 3. Curing value, 'curing'
// -------------------------------------------
// Calculating the wind speed which is used to calculate head fire ROS
REAL wind_speed = length(wind_vector);
// Calculating the normalised dot product between the wind vector and the normal to the fire perimeter
REAL wdot = dot(normalize(wind_vector),advect_normal_vector);
// Calculate length-to-breadth ratio (LBR) which varies with wind speed
// Equations are curve fits adapted from Taylor (1997)
REAL LBR = 1.0;
if (wind_speed < 5){
LBR = 1.0;
} else {
LBR = 1.1*pow(wind_speed, 0.464);
}
// Determine coefficient for backing and flanking rank of spread using elliptical equations
// Where R_backing = cb * R_head, R_flanking = cf * R_head,
REAL cc = sqrt(1.0-pow(LBR, -2.0));
REAL cb = (1.0-cc)/(1.0+cc);
REAL a_LBR = 0.5*(cb+1.0);
REAL cf = a_LBR/LBR;
// Determine shape parameters
REAL f = 0.5*(1.0+cb);
REAL g = 0.5*(1.0-cb);
REAL h = cf;
// Now calculate a speed coefficient using normal flow formula
REAL speed_fraction = (g*wdot+sqrt(h*h+(f*f-h*h)*wdot*wdot));
// Calculate curing coefficient from Cruz et al. (2015)
REAL curing_coeff;
if ( curing < 20 )
curing_coeff = 0;
else
curing_coeff = 1.036 / ( 1 + 103.989 * exp( -0.0996 * (curing - 20) ) );
// Fuel moisture content approximated using McArthur (1966)
REAL GMf = 9.58-(0.205*temp) + (0.138*rel_hum);
// Calculate moisture coefficient from Cheney et al. (1998)
REAL moisture_coeff;
if ( GMf <= 12 ){
moisture_coeff = exp( -0.108 * GMf );
} else if ( wind_speed <= 10 ){
moisture_coeff = 0.684 - 0.0342 * GMf;
} else{
moisture_coeff = 0.547 - 0.0228 * GMf;}
// Defining coefficients for the various grasslands sub-types using different subclasses.
REAL CF_Backing_Slow;
REAL CF_Backing_Fast;
REAL CF_Wind_Slow;
REAL CF_Wind_Fast;
REAL speed_factor = 1;
if (subclass == 1) // Eaten out
{
CF_Backing_Slow = 0.027;
CF_Backing_Fast = 0.55;
CF_Wind_Slow = 0.1045;
CF_Wind_Fast = 0.3575;
}
else if (subclass == 2) // Cut or grazed
{
CF_Backing_Slow = 0.054;
CF_Backing_Fast = 1.1;
CF_Wind_Slow = 0.209;
CF_Wind_Fast = 0.715;
}
else if (subclass == 3) // Natural or undisturbed
{
CF_Backing_Slow = 0.054;
CF_Backing_Fast = 1.4;
CF_Wind_Slow = 0.269;
CF_Wind_Fast = 0.838;
}
else if (subclass == 4) // Woodland
{
CF_Backing_Slow = 0.054;
CF_Backing_Fast = 1.4;
CF_Wind_Slow = 0.269;
CF_Wind_Fast = 0.838;
speed_factor = 0.5;
}
else if (subclass == 5) // Open forest
{
CF_Backing_Slow = 0.054;
CF_Backing_Fast = 1.4;
CF_Wind_Slow = 0.269;
CF_Wind_Fast = 0.838;
speed_factor = 0.3;
}
// Calculate spread rate from Cheney et al. (1998) (converting spread rate to m/s from km/hr)
if ( wind_speed >= 5.0 )
head_speed = (CF_Backing_Fast + CF_Wind_Fast * pow( (wind_speed - 5), 0.844 ) ) * moisture_coeff * curing_coeff / 3.6;
else
head_speed = (CF_Backing_Slow + CF_Wind_Slow * wind_speed)* moisture_coeff * curing_coeff / 3.6;
// Adjust speed based on canopy layer for Northern Australia grassland types (woodland and open forest)
// Based on Cheney and Sullivan (2008) | 1,291 | 4,138 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-38 | latest | en | 0.728687 |
https://math.stackexchange.com/questions/254944/determining-boundaries-of-probability-density-function-integral-for-a-requested | 1,571,723,127,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570987803441.95/warc/CC-MAIN-20191022053647-20191022081147-00281.warc.gz | 591,697,656 | 31,548 | # Determining boundaries of Probability Density Function integral for a requested probability
This isn't one specific homework question, but a concept I'm having trouble with in class. We were asked on a couple of questions recently on homework dealing with the probability density function of two random variables, $f_{X,Y}(x,y)$. We were given a joint probability density function and asked to find a given probability, like $P[X + Y > 5]$. One solid example being:
Given a Joint PDF $f_{X,Y}(x,y)$ let A be the event that $X + Y \leq 1$. Now, I know that $$P[A] = \int_A\int f_{X,Y}(x,y)dxdy$$ (at least, that's how my book writes it). My question is, how do I determine the bounds of those two integrals? In the solution manual for this particular problem, it is stated that the solution is $$P[X+Y\leq 1] = \int_0^1\int_0^{1-x}f_{X,Y}(x,y)dxdy$$
I can see pretty clearly that $x + (1-x)$ will always be less than or equal to one for $0 \leq x \leq 1$, so I understand why those bounds work. I'm just not sure I could have come up with them. Is there a process for figuring this out that I am missing? A similar question posed to us was given another joint PDF, let A be the event that $X+Y>5$. I have no idea what the bounds would be for this one. Can someone show me how to find the integral bounds for questions like this?
In your example, it seems the density $f_{X,Y}(x,y)$ is zero when $x\leqslant0$ or when $y\leqslant0$ hence, to compute $\mathbb P(X+Y\leqslant1)$, the domain of integration is defined by the inequalities $$x\geqslant0,\qquad y\geqslant0,\qquad x+y\leqslant1.$$ This is indeed equivalent to $$0\leqslant x\leqslant1,\qquad 0\leqslant y\leqslant 1-x,$$ but I know no systematic way of deducing the latter from the former, except drawing a rough sketch of the domain of interest. | 521 | 1,811 | {"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} | 3.96875 | 4 | CC-MAIN-2019-43 | latest | en | 0.908951 |
https://www.gcomsolutions.com/dax-cheat-sheet/atanh/ | 1,726,772,595,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700652055.62/warc/CC-MAIN-20240919162032-20240919192032-00826.warc.gz | 726,254,772 | 30,974 | # ATANH
## Y
To use the ATANH function in Power BI, follow these steps:
## Step 1: Create a New Measure
First, create a new measure in Power BI by navigating to the “Modeling” tab and selecting “New Measure” from the “Calculations” group. Give the measure a name, such as “ATANH Calculation,” and enter the formula for the ATANH function in the “Formula Bar.”
## Step 2: Enter the ATANH Formula
The ATANH formula is as follows:
``` ATANH(❰number❱) ```
Replace `❰number❱` with the cell or column reference that contains the number you want to calculate the inverse hyperbolic tangent of. For example, if you want to calculate the ATANH of the number in cell A2, enter the following formula:
``` ATANH(A2) ```
## Step 3: Apply the ATANH Formula to Your Data
Once you’ve entered the ATANH formula, apply it to your data by dragging and dropping the measure onto your visualizations. For example, if you want to create a scatter plot that shows the correlation between two variables, drag and drop your ATANH measure onto the “Values” field and your two variables onto the “Axis” fields.
## Using the ATANH Function in Advanced Calculations
The ATANH function can be used in more advanced calculations as well. For example, you can use the ATANH function to calculate the probability distribution of a set of data by using the following formula:
``` =ATANH(SQRT(1 - p^2)) ```
Replace `p` with the probability of success and plug the formula into your measure to calculate the probability distribution of your data.
The ATANH function is a powerful tool that can be used in many statistical and mathematical calculations in Power BI. By following the simple steps outlined in this article, you can easily apply the ATANH function to your data and gain valuable insights into your analysis.
Subject | 429 | 1,809 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.828125 | 3 | CC-MAIN-2024-38 | latest | en | 0.863833 |
https://www.physicsforums.com/threads/setting-up-an-integral-for-the-area-of-a-surface-of-revolution.374814/ | 1,521,488,136,000,000,000 | text/html | crawl-data/CC-MAIN-2018-13/segments/1521257647044.86/warc/CC-MAIN-20180319175337-20180319195337-00461.warc.gz | 830,892,731 | 15,161 | # Setting up an Integral for the area of a surface of revolution
1. Feb 2, 2010
### darkblue
1. The problem statement, all variables and given/known data
Set up, but do not evaluate, an integral for the area of the surface obtained by rotating the curve y=xe-x 1=<x=<3 about the y-axis.
2. Relevant equations
S=integral from a to b x 2pix ds where ds=sqrt(1+(dy/dx)2)dx
3. The attempt at a solution
The first thing I tried to do is solve for the equation in terms of x, and then use the equation above. I figured it makes sense to solve for x since we are rotating the curve about the y-axis. I wasn't able to solve for x, so then I tried to use this method in my textbook where you leave x as it is, and then substitute u for whatever is within the square root sign in such a way that you can eliminate x. I tried to do that, but its turning into a mess since you get 1+(e-x-xe-x)2 underneath the square root and I don't really see how substitution could be used here...any ideas?
2. Feb 2, 2010
### Staff: Mentor
All you need to do is set up the integral. Don't worry about trying to evaluate this integral.
3. Feb 2, 2010
### darkblue
So does this mean that the way I have set it up is correct? I had a feeling it wasn't right because I couldn't see what steps I'd take next in the event that I had to solve it.
4. Feb 2, 2010
### Staff: Mentor
Seems to be OK, but I'm a little rusty on these surface area integrals. You have an extra x in what you typed, though, right after b. Did you mean for that to be there?
5. Feb 2, 2010
### darkblue
oops, i meant to put a "*" for multiplication. | 436 | 1,611 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2018-13 | longest | en | 0.949684 |
http://www.physicsforums.com/showthread.php?t=226482&page=2 | 1,386,596,087,000,000,000 | text/html | crawl-data/CC-MAIN-2013-48/segments/1386163976781/warc/CC-MAIN-20131204133256-00014-ip-10-33-133-15.ec2.internal.warc.gz | 488,685,935 | 9,460 | # Young's double slit experiment (prob density)
by t_n_p
Tags: density, double, experiment, prob, slit, young
P: 581 Got it! Thanks a WHOLE lot! There's another follow on question.. Show that interference maxima is given by Ignoring part c) for the time being, how exactly is pr density related to the interference maxima equation?
PF Patron Sci Advisor Emeritus P: 9,789 Check out the slit interference pages at hyperphysics: http://hyperphysics.phy-astr.gsu.edu...opt/slits.html
P: 581 hmmm, still don't get it.
PF Patron
Sci Advisor
Emeritus
P: 9,789
Quote by t_n_p hmmm, still don't get it.
What specifically don't you understand?
P: 581 how the prob density equation just found is related to the interference equation.
PF Patron
Sci Advisor
Emeritus
P: 9,789
Quote by t_n_p how the prob density equation just found is related to the interference equation.
There's no need to relate the probability density to the interference pattern, the question simply asks you to derive the fringe separation, which can be done without using the probability density.
P: 581 hmm? It says "Using the results obtained in (a) [The pr density part], show that the interference maxima are given by..."
PF Patron
Sci Advisor
Emeritus
P: 9,789
Quote by t_n_p hmm? It says "Using the results obtained in (a) [The pr density part], show that the interference maxima are given by..."
Okay, how does the maxima relate to $\rho$? What is a maxima?
P: 581
Quote by Hootenanny Okay, how does the maxima relate to $\rho$? What is a maxima?
So derive $\rho$ in terms of $\psi$ and set to zero?
What about A (imaginary number?)
P: 581 Also, where would lambda come from?
PF Patron
Sci Advisor
Emeritus
P: 9,789
Quote by t_n_p So derive $\rho$ in terms of $\psi$ and set to zero?
Who would one set $\rho$ to zero? Wouldn't a maxima occur when $\rho$ is greatest?
Quote by t_n_p What about A (imaginary number?)
What is the maximum value of $\rho$?
P: 581
Quote by Hootenanny Who would one set $\rho$ to zero? Wouldn't a maxima occur when $\rho$ is greatest? What is the maximum value of $\rho$?
The maximum value of $\rho$ is 1, or at least that is what I think. But where to from there?
PF Patron
Sci Advisor
Emeritus
P: 9,789
Quote by t_n_p The maximum value of $\rho$ is 1, or at least that is what I think.
No it isn't
P: 581
Quote by Hootenanny No it isn't
infinity????
PF Patron Sci Advisor Emeritus P: 9,789 What is the maximum value of $\cos^2\theta$?
P: 581
Quote by Hootenanny What is the maximum value of $\cos^2\theta$?
1!!!!!!!!
PF Patron
Sci Advisor
Emeritus
P: 9,789
Quote by t_n_p 1!!!!!!!!
Correct, therefore the maximum value of $\rho$ is...?
P: 581 4????
Related Discussions Introductory Physics Homework 1 Introductory Physics Homework 1 Introductory Physics Homework 5 Classical Physics 2 Classical Physics 3 | 772 | 2,825 | {"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} | 3.140625 | 3 | CC-MAIN-2013-48 | longest | en | 0.902477 |
https://getwelsim.medium.com/the-time-domain-transient-method-in-structural-finite-element-analysis-3666dd066d81?source=post_internal_links---------0---------------------------- | 1,637,998,600,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964358153.33/warc/CC-MAIN-20211127073536-20211127103536-00400.warc.gz | 353,314,983 | 36,301 | # The time-domain transient method in structural finite element analysis
Transient structural analysis (also known as dynamic analysis) is a method used to determine the dynamic response of a structure over time. Through the analysis of transient analysis, we can obtain the time-dependent results such as displacement, strain, stress, and reaction force of the structure under the arbitrary combination of steady-state loads, transient loads, and harmonic loads. The biggest advantage of transient analysis over the static analysis is that it considers inertia and damping effects.
Common analysis methods for transient problems are:
• Time-domain transient analysis
• Eigenvalue extraction (natural frequency and modal)
• Steady-state response (harmonic response analysis in the frequency domain)
• Spectrum response analysis (peak response calculation of shock)
• Random response analysis (vibration caused by random excitation)
Because the time-domain transient method is the most intuitive, it can solve a variety of linear and nonlinear problems, and has been applied often in the industry. This article only introduces the time-domain transient method. Other methods will be introduced in the future.
Governing equation
The fundamental equation of motion for transient dynamics is:
Where [M] is the mass matrix. [C] is the damping matrix. [K] is the stiffness matrix. {u_tt} is the nodal acceleration. {u_t} is the nodal speed. {u} is the nodal displacement. {F} is the load. It can be seen that the transient structural governing equation is an equation containing second-order time derivatives. There are many methods for solving the second-order time derivative. The most widely used in the structural finite element is the Newmark implicit time integration method. WELSIM’s default time solver for structural analysis is also the Newmark method.
Common solving methods
There are three common finite element methods for transient dynamics: full method, reduced method, and modal superposition method.
The full method uses complete system matrices to calculate the dynamic response. It is the most powerful and supports solving various nonlinear characteristics such as plasticity, large deformation, large strain, etc. The advantage is that it is easy to use and you don’t have to concern about choosing the principal degree of freedom or vibration mode. The scheme allows various types of nonlinear characteristics. The full matrix is used, no mass matrix approximation is involved. Displacements and stresses over all-time history can be obtained in one single analysis. All types of boundary conditions are allowed. The main disadvantage of the complete method is that it is expensive, time-consuming, and the solved data is large.
The reduced method compresses the data size by using a principal degree of freedom and reduced matrices. As the displacements at the principal degrees of freedom are calculated, the solution is extended to the complete set of degrees of freedom. The advantage is that it is faster and less expensive than the full method. The disadvantage is that the initial solution only calculates the displacement of the principal degrees of freedom. Only nodal boundary conditions can be applied to the model. All loads must be added to the principal degrees of freedom. The size of the time step must be constant. The automatic time stepping is not supported. Nonlinearity is not supported (except node-to-node contact).
The modal superposition method calculates the structural response by multiplying the mode shapes (eigenvalues) obtained by the modal analysis and summing them. The advantages are: faster and less expensive than the reduced or full method. It allows the modal damping (damping ratio as a function of the vibration model). The disadvantage is that the size of the time step must be constant and the automatic time stepping is not supported. Nonlinearity is not supported (except node-to-node contact). Non-zero displacement boundary conditions cannot be applied.
Because of the superiority of the full method and its wide application in linear and nonlinear problems. This article only discusses the full method.
Damping
Considering the effect of damping is one of the advantages of transient analysis. Damping can be regarded as a kind of energy dissipation, which comes from many factors. In the finite element analysis, damping is often regarded as an overall effect and is applied using numerical algorithms.
There are three common damping settings in FEM: direct damping, Rayleigh damping, and composite damping. Direct damping can define the critical damping ratio related to each order mode, and its typical value range is between 1% and 10% of critical damping. Rayleigh damping assumes that the damping matrix is a linear combination of the mass and stiffness matrices. Although the assumption that damping is proportional to the mass and stiffness matrix has no sufficient physical basis, we know very little about the distribution of damping, and practice has proven that Rayleigh damping is effective in the finite element analysis thus is widely used. Composite damping can define a critical damping ratio according to each material so that the composite damping value corresponding to the overall structure is obtained. Composite damping is more effective when there are many different materials in the structure.
In most linear dynamic problems, accurately defining damping is important for results. However, the damping algorithm and parameters only approximate the energy absorption characteristics of the structure, not the physical mechanism that causes this effect in principle. Therefore, it is difficult to determine the damping data in the finite element analysis. Sometimes we can obtain these data from experiments, sometimes we have to refer the material manual or experience to determine the damping parameters.
Time solver
For transient structural problems, due to the introduced second-order time derivatives, we need a time solver. Generally, we divide time solver into two categories: explicit solvers and implicit solvers.
The explicit solver uses the results of the previous and current steps to calculate the next result. Explicit solver has unstable regions and requires a small time step. The advantage is that you do not need a nonlinear solver (Newton’s iterative solver) to solve nonlinearity, and you do not need to assemble a stiffness matrix. At the same time, the encoding and algorithm are relatively simple. It does not require additional memory to store the intermediate data. The stable step size of the explicit solver can be estimated, but due to the complexity of the actual calculation, the practical time step usually is smaller than the theoretical step size. Explicit algorithms require that the mass matrix must be diagonal. The advantage of solving speed shows only when the number of finite elements is small. Therefore, the reduced integration method is often used, which easily triggers the hourglass mode and affects the calculation accuracy of stress and strain. The representative algorithms of the display solver are the central difference, forward Euler, Runge-Kutta, linear acceleration method, etc.
The implicit solver iterates the next step result with the current step result and the next unknown result, which must be obtained through iteration. The implicit solver requires the assembly of a stiffness matrix, and Newton iteration is required to solve nonlinearity. The commonly used methods are the Newmark method and Wilson-Theta method. The variant algorithms can be implemented by modifying the alpha, beta and theta parameters in the algorithm. For example, the calculation efficiency and accuracy of the HHT solver on some problems have been significantly improved. The biggest advantage of the implicit solver is that it has unconditional stability, that is, the time step can be arbitrarily large. However, we need to take a reasonable time step in the actual practice. At the same time, some solvers such as Newmark have second-order accuracy.
Comparison of explicit and implicit solvers as shown below
In order to ensure the full response of the structure and ensure the stability and convergence of the solution, it is important to choose the correct time step. Generally speaking, the smaller the time step, the more accurate the solution. However, a smaller time step leads to a large number of solving steps, and the physical computation time will increase significantly. Therefore, the size of the time step cannot be too small. Meanwhile, the time step cannot be too large, otherwise, the calculation will miss many high-order frequencies of the structure, leading to unrealistic solutions.
Automatic time stepping is an automatic algorithm that can optimize the computation in transient problems. The user only needs to set the initial step, maximum and minimum step sizes. During the solution process, the solver will automatically reduce the time step as required to calculate any rapid changes in the motion. In the process of slow change in motion, the solver can increase the time step, thereby improving the calculation speed.
Boundary and initial conditions
Unlike static analysis, it is due to the characteristics of transient and inertia. The transient analysis supports velocity and acceleration boundary conditions. These boundary conditions, especially acceleration, can be used to simulate the effects of excitation.
For initial conditions, the transient analysis generally supports the following three:
• Linear static results. It specifies the initial displacement of the structure under a certain load.
• A transient analysis result that determines the transient response of the structure at a certain moment. Solving can begin at any specified time step.
• The initial velocity and initial acceleration of all free nodes in the structure.
The initial velocity condition is a more common initial condition, and it is often applied to all nodes of a structure to mimic moving bodies.
The above is the core knowledge in terms of structural transient analysis. Below we look at how to implement finite element analysis in software WELSIM.
Structural transient analysis steps
1. Create or import a model
2. Meshing
3. Load step and time step settings
Number Of Steps: Load steps, which specify the number of load steps in this analysis.
Current Step: The current load step.
Current End Time: The end time of the current load step.
Auto Time Stepping: Whether automatic time stepping is on. WELSIM currently only supports fixed step sizes for structural analysis, so this option is turned off by default.
Define By: The way to define the load substep. It can be defined by time and the number of substeps.
Time Step: The size of the time step. Available when Defined By is set to Time.
4. Set boundary conditions, initial conditions, contacts, etc.
5. Solve and verify results
The figure below shows the maximum and minimum deformations in the Y direction and von-Mises stresses over the entire time history.
It can be seen that no damping occurs in this analysis because of no attenuation of the reciprocating vibration. The role and setting of damping will be introduced in the future.
Finally, the operation video is provided for your reference.
WELSIM® finite element analysis software helps engineers and researchers conduct simulation studies and prototype virtual products.
## More from WELSIM - Finite Element Analysis Solutions
WELSIM® finite element analysis software helps engineers and researchers conduct simulation studies and prototype virtual products. | 2,157 | 11,666 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.28125 | 3 | CC-MAIN-2021-49 | latest | en | 0.893276 |
https://www.proofwiki.org/wiki/Wilson%27s_Theorem/Examples/10_does_not_divide_(n-1)!%2B1 | 1,686,216,845,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224654606.93/warc/CC-MAIN-20230608071820-20230608101820-00426.warc.gz | 1,025,254,820 | 11,294 | # Wilson's Theorem/Examples/10 does not divide (n-1)!+1
## Example of Use of Wilson's Theorem
For all $n \in \Z_{>0}$, $10$ is not a divisor of $\paren {n - 1}! + 1$.
## Proof
For the first few $n$ we see:
$\, \ds 10 \nmid \,$ $\ds 2$ $=$ $\ds \paren {1 - 1}! + 1$ where $\nmid$ denotes non-divisibility $\ds$ $=$ $\ds \paren {2 - 1}! + 1$ $\, \ds 10 \nmid \,$ $\ds 3$ $=$ $\ds \paren {3 - 1}! + 1$ $\, \ds 10 \nmid \,$ $\ds 7$ $=$ $\ds \paren {4 - 1}! + 1$ $\, \ds 10 \nmid \,$ $\ds 25$ $=$ $\ds \paren {5 - 1}! + 1$
Now consider $n > 5$.
We have that:
$\ds 2$ $\divides$ $\ds \paren {n - 1}!$ where $\divides$ denotes divisibility $\ds 5$ $\divides$ $\ds \paren {n - 1}!$ $\ds \leadsto \ \$ $\ds 10$ $\divides$ $\ds \paren {n - 1}!$ as $10 = \lcm \set {2, 5}$
Hence $10 \nmid \paren {n - 1}! + 1$.
$\blacksquare$ | 374 | 826 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.0625 | 4 | CC-MAIN-2023-23 | latest | en | 0.215491 |
http://jsfiddle.net/user/hughdidit/fiddles/ | 1,566,526,880,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027317817.76/warc/CC-MAIN-20190823020039-20190823042039-00165.warc.gz | 105,645,829 | 9,339 | # JSFiddle
## Hugh Chapman's public fiddles
• ### Equilibrium Index#2 is the latest revision
Write a function: function solution(A); that, given a zero-indexed array A consisting of N integers, returns any of its equilibrium indices. The function should return −1 if no equilibrium index exists.
• ### Determine whether a triangle can be built from a given set of edges#26 is the latest revision
By sorting the array, we have guaranteed that P+R < Q and Q+R < P (because R is always the biggest). Now what remains, is the proof that P+Q > R, that can be found out by traversing the array. The chance to find such a combination is with three adjacent values as they provide the highest P and Q.
• ### What Split in Array Yields Min Value
Write a function that, given a non-empty zero-indexed array A of N integers, returns the minimal difference that can be achieved.
• ### Frog Jump Problem#14 is the latest revision
A small frog wants to get to the other side of the road. The frog is currently located at position X and wants to get to a position greater than or equal to Y. The small frog always jumps a fixed distance, D. Count the minimal number of jumps that the small frog must perform to reach its target.
• ### Find Missing Element
Given an integer set from 1 to n, calculate the missing integer.
• ### isPermutation
Check whether array N is a permutation
• ### flatten a non standard array
No-Library (pure JS), HTML, CSS, JavaScript
• ### Translate in CSS
No-Library (pure JS), HTML, CSS, JavaScript
• ### Swapping Variable Methods
Some methods for swapping values.
• ### Prime Time
Write a function to take the num parameter and return true if it is a prime number or false if it composite.
• ### 3n + 1 Problem
a programming challenge
• ### Merge and Sort two sets with Javascript
No-Library (pure JS), HTML, CSS, JavaScript
• ### Mediator Patter Library
Snippets for implementing Mediator Patter
• ### Popup without blocker#1 is the latest revision
jQuery 1.8.3, HTML, CSS, JavaScript
• ### ssn validate & format#286 is the latest revision
jQuery 1.10.1, HTML, CSS, JavaScript
• ### Quicksort Function
Take a random list and quicksorts it.
• ### Sort Vertically
Take a random set and sort in ascending order vertically across arbitrary columns.
• ### Merge Sort Function
Take a random list and merge sorts it. | 553 | 2,360 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.046875 | 3 | CC-MAIN-2019-35 | longest | en | 0.760355 |
https://www.scribd.com/document/356488170/MEMB123-Sem-I-2016-2017-final-v3-1 | 1,540,174,441,000,000,000 | text/html | crawl-data/CC-MAIN-2018-43/segments/1539583514443.85/warc/CC-MAIN-20181022005000-20181022030500-00178.warc.gz | 1,061,676,876 | 36,930 | # COLLEGE OF ENGINEERING
PUTRAJAYA CAMPUS
FINAL EXAMINATION
SEMESTER I 2016 / 2017
PROGRAMME : BME/BEPE/BEEE/BCCE (Honours)
SUBJECT CODE : MEMB123
SUBJECT : Mechanics I: Statics
DATE : September 2016
TIME : 3 hours
INSTRUCTIONS TO CANDIDATES:
1. This paper contains SIX (6) questions.
4. Write answer to each question on a new page.
THIS QUESTION PAPER CONSISTS OF (8) PRINTED PAGES INCLUDING
THIS COVER PAGE.
[4 marks] (b) Determine the components of reaction at the ball and socket joint A and the tension in the supporting cable BC. [12 marks] Figure 1 Page 2 of 8 . MEMB123. Semester I 2016/2017 QUESTION 1 (16 MARKS) The rod assembly as shown in Figure 1 is supported by the cable BC. (a) Draw the free body diagram for the rod assembly.
MEMB123. Semester I 2016/2017 QUESTION 2 (18 MARKS) The panel with cable AB and hinged CD are shown in Figure 2. will the panel be held in equilibrium position? Explain and justify your answer. [18 marks] Figure 2 Page 3 of 8 . If the magnitude of the force in the cable AB is 600 N and the moment about the hinged axis CD. is 650 Nm.
[16 marks] Figure 3 Page 4 of 8 . as shown in Figure 3. find the minimum weight of block A to prevent motion. Semester I 2016/2017 QUESTION 3 (16 MARKS) Two blocks are connected by a solid strut attached to each block with frictionless pins.25 and block B weighs 2700 N. If the coefficient of static friction under each block is 0. MEMB123.
[8 marks] Figure 4 Page 5 of 8 . Semester I 2016/2017 QUESTION 4 (17 MARKS) For the truss shown in Figure 4. (a) Draw the free-body diagram of the truss and find the zero-force member. [3 marks] (b) Determine the support reactions at pin A and E. MEMB123. CB and DC by method of joint and state if these members are in tension or compression. [6 marks] (c) Determine the force in members AB.
MEMB123. [10 marks] Figure 5 QUESTION 6 [15 MARKS] Page 6 of 8 . Semester I 2016/2017 QUESTION 5 [18 MARKS] The structure shown in Figure 5 is supported by a pin at A and a roller at B. [8 marks] (b) Determine the normal force. Show these internal forces using a free body diagram. and moment at point D. (a) Draw the free body diagram of the entire structure and determine the reactions at support A and B. shear force.
[8 marks] (b) Determine the moment of inertia of the shaded area about the x-axis. y) of the shaded area. MEMB123. [7 marks] Figure 6 Appendix Page 7 of 8 . Semester I 2016/2017 For the composite area shown in Figure 6. (a) Locate the centroid ((x.
Semester I 2016/2017 Table of Moment of Inertia --.END OF QUESTION PAPER--- Page 8 of 8 . MEMB123. | 746 | 2,595 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2018-43 | latest | en | 0.819122 |
https://brilliant.org/problems/love-for-areas/ | 1,623,750,733,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623487620971.25/warc/CC-MAIN-20210615084235-20210615114235-00191.warc.gz | 152,002,905 | 8,668 | # A Triangle Evolves Into A Hexagon
Geometry Level 2
Consider an acute angled $\Delta ABC$ with circumcircle $\omega$ and circumcenter $O$. Now, construct points (on the circumcircle) diametrically opposite to the vertices of the triangle, and name them $A', B'$ and $C'$ as shown in the figure. The area of $\Delta ABC$ is $18$ square units.
Find the area of polygon $AB'CA'BC'$.
× | 103 | 386 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 8, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2021-25 | latest | en | 0.838121 |
https://discourse.mcneel.com/t/lineardimension-absolute-relative-references/57320 | 1,656,771,342,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656104141372.60/warc/CC-MAIN-20220702131941-20220702161941-00045.warc.gz | 250,403,886 | 5,513 | # LinearDimension absolute/relative references
Hi,
I try to get “Rhino.Geometry.LinearDimension” working.
I’m migrating my python code to Rhino 6.
I know annotations changed a lot but I wonder why the parameters have changed from
(plane, absoluteExt1, absoluteExt2, absoluteLinePt)
to
(plane, absoluteStartPos, relativeEndPos, relativeLinePos).
In that case shouldn’t be better to use vector for relatives values ?
Practically and conceptually, isn’t it odd to know the length of a measure before the measurement ?
(Even the python example is a bit twisted (an absolute position named “offset” ???)
Can I know goal of that change ?
Regards.
That’s fixed by https://mcneel.myjetbrains.com/youtrack/issue/RH-45282 in SR4
I made a sample code based on Cormier’s example.
I tried to clarified the mindset to adopt to understand the function based on my perspective which is in absolute coordinates.
Maybe it could help.
``````import Rhino.Geometry as g
from scriptcontext import doc
import rhinoscriptsyntax as rs
import math
#Closest point in a plane (2d result)
def cp(pl, pt):
b, u, v = pl.ClosestParameter(pt)
return g.Point2d(u, v)
#Absolute World Coord points
if aligned:
cplane = rs.ViewCPlane() #Stays in same plane
xDir = g.Vector3d(e2-e1) #Calc x and y vectors of the new aligned plane
yDir = g.Vector3d(xDir)
yDir.Rotate(math.pi/2, cplane.Normal)
ldPlane = g.Plane(e1, xDir, yDir) #1st point = origin
else:
ldPlane = rs.ViewCPlane()
ldPlane.Origin = e1 #1st point = origin
# cp(ldPlane, e1) == g.Point2d(0,0) == origin
ld = g.LinearDimension(ldPlane, cp(ldPlane, e1), cp(ldPlane, e2), cp(ldPlane, pt))
ld.TextHeight = 35;
ld.ArrowSize = 30; | 458 | 1,658 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2022-27 | latest | en | 0.729885 |
http://www.gjtutorial.com/news/aai-sample-questions-paper-written-exam-recruitment/ | 1,386,900,184,000,000,000 | text/html | crawl-data/CC-MAIN-2013-48/segments/1386164805405/warc/CC-MAIN-20131204134645-00080-ip-10-33-133-15.ec2.internal.warc.gz | 366,996,058 | 16,046 | Location: News > Sample Question Papers > AAI Sample Questions Paper: Written Exam for Recruitment
# AAI Sample Questions Paper: Written Exam for Recruitment
Last updated: August 7, 2010 by
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Airport Authority of India conducts a written test for recruitment of many posts. The syllabus usually has a technical as wells as a non-technical section. Reasoning is one of the major component of the non-technical exam. Here are some questions that will help you prepare for the AAI Written Test.
1. In a row of children, Manoj is fifteenth from the left end of the row. If he is shifted towards the right end of the row by four places, he becomes eighth from the right end. How many children are there in the row?
(a) 27
(b) 26
(c) 28
(d) 24
2. How many such numerals are there in the number 2576489, which will remain at the same position when arranged in ascending order from left to right?
(a) One
(b) Two
(c) Three
(d) More than three
3. In a class of 40 children, Manmeet’s rank is eighth from the top. Vishal is five ranks below Manmeet. What is Vishal’s rank from the bottom?
(a) 27
(b) 29
(c) 28
(d) 26
4. Three of the following four are alike in a certain way. Find the one that does not belong to this group?
(a) Coriander
(b) Mustard
(c) Cumin
(d) Clove
5. ‘Seed’ is related to ‘Fruit’ in the same way as ‘Fruit’ is related to:
(a) Tree
(b) Branch
(c) Flower
(d) Petal
6. Three of the following four are alike in a certain way. Find the one that does not belong to this group?
(a) Van
(b) Truck
(c) Cargo
(d) Trolley
7. If it is possible to make a three digit number, which is perfect square of a two digit number from the second, the fourth and the eighth digits of 31792056 using each digit only once, what is the two digit number? If the perfect square cannot be formed, your answer is ‘N’ and if more than one such numbers can be formed , your answer is ‘X’
(a) 14
(b) 16
(c) N
(d) X
8. Pointing to a woman, Suresh said, she is the daughter of my grandfather’s only daughter. How is Suresh related to the woman?
(a) Brother
(b) Cousin
(c) Uncle
(d) Cannot be determined
9. ‘Visual’ is related to ‘Light’ in the same way as ‘Audio’ is related to:
(a) Voice
(b) Sound
(c) Drama
(d) Noise
10. ‘Jackal’ is related to ‘Carnivorous’ in the same way as ‘Goat’ is related to
(a) Omnivorous
(b) Carnivorous
(c) Herbivorous
(d) Multivorous
11. In a certain code ‘DESCRIBE’ is written as ‘FCJSDTFE’. How will ‘CONSIDER’ be written in that code?
(a) SPEJTOPD
(b) SEFJTOPD
(c) QFETJOPD
(d) QEFJTOPD
12. Four of the following five are alike in a certain way and so form a group. Which is the one that does not belong to that group?
(a) Shampoo
(b) Talcum Powder
(c) Hair Oil
(d) Cosmetics
13. Four of the following five are alike in a certain way and so form a group. Which is the one that does not belong to that group?
(a) 71
(b) 73
(c) 67
(d) 77
14. Four of the following five are alike in a certain way and so form a group. Which is the one that does not belong to that group?
(a) Leaf
(b) Flower
(c) Branch
(d) Pollen
15. Four of the following five are alike in a certain way and so form a group. Which is the one that does not belong to that group?
(a) Jasmine
(b) Rose
(c) Dahlia
(d) Lotus
16. ‘Radish’ is related to ‘Root’ in the same way as ‘Brinjal’ s related to
(a) Fruit
(b) Stem
(c) Flower
(d) Root
17. Four of the following five are alike in a certain way and so form a group. Which is the one that does not belong to that group?
(a) Physiology
(b) cardiology
(c) Pathology
18. ‘Iron’ is related to ‘Metal’ in the same way as ‘Brass’ is related to
(a) Iron
(b) Alloy
(c) Copper
(d) Zinc
19. In a certain code RAID is written as %#ê\$, RIPE is written as %ê@©. How is DEAR written in that code?
(b) \$@#%
(c) @\$#%
20. If blue is called red, red is called green, green is called black and black is called white, what is the colour of grass?
(a) red
(b) black
(c) white
(d) green
Directions (21-25): In each of the question below are given three statements followed by two conclusions numbered I and II. You have to take the three given statements to be true even if they seem to be at variance with commonly known facts, and then decide which of the given conclusions logically follows from the given statements disregarding commonly known facts. Give the answer:
(a) if only conclusion I follows.
(b) if only conclusion II follows.
(c) if either conclusion I or II follows.
(d) if neither conclusion I nor II follows.
(e) if both conclusions I and II follow.
21. Statements:
a. Some pins are forks.
b. All forks are keys.
c. No key is lock.
Conclusions:
I. Some forks are pins.
II. No lock is a pin.
22. Statements:
a. Some shirts are trousers.
b. Some trousers are jackets.
c. All jackets are shawls.
Conclusions:
I. Some shawls are shirts.
II. Some jackets are shirts.
23. Statements:
a. Some leaves are plants.
b. Some plants are trees.
c. Some trees are fruits.
Conclusions:
I. Some fruits are trees.
II. Some trees are plants.
24. Statements:
a. Some rats are dogs.
b. Some dogs are horses.
c. Some horses are camels.
Conclusions:
I. Some horses are rats.
II. Some camels are horses.
25. Statements:
a. Some books are dictionaries.
b. Some dictionaries are files.
c. Some files are papers.
Conclusions:
I. Some papers are files.
II. Some files are books.
1. b, 2. d, 3. c, 4. b, 5. c
6. c, 7. d, 8. b, 9. b, 10. c
11. a, 12. d, 13. d, 14. d, 15. e
16. a, 17. d, 18. b, 19. d, 20. b
21. c, 22. d, 23. e, 24. b, 25. a
Category: Sample Question Papers
Tags:
1. nikhilesh ap said:
pls sesnt aai fire exam mcq peapers
2. Ashish said:
plz....send me all the previous years question papers for AAI exams for manager/Jr. executive ATC on my mail ID....
thanks nd regards.....
3. v.rajavel said:
Dear sir,
send to me solved papers,thanks
4. v.rajavel said:
Dear sir,
I want new or old(bhel)- iti(fitter) QUESTION PATERAN,
send me solved paper.thank you .......
5. Praveenkumar M Junior Exe AAI said:
I appeared AAI manager exam 2010 for electronics .Last time it was similar standard of Junior EXE exam.This time exam was difficult (GATE standard) .Now Pattern has been changed.Technical70 question and non 30 question.Non Technical they asked 15 question from control system ,Power electronics almost 10 question,Emt and Antennas 10 to 15 questions,Electronics circuits 15 to 20 questions ( includes standard questions from op amps,find parameters using h parameter).Communication only two simple problems!!! Microwave 3 questions. With 2 hours ,one can attend max 50 questions: In fact it was difficult for all. Lets hope for the best
6. rima said:
please send me sail technical exam sampal paper&indian bank sampal paper.if u send me that i will very greatfull to u.
thank you
7. sonika sindhiya said:
pls send me techinal n non techinal question for jr. executive atc aai exam.
plz send me all last year paper of AAI(post-manager,jr Executive).and plz send me examination date of AAI(post-manager,jr Executive)2010.
thanx.
9. garima said:
please send the previus paper of bhel supervisor trainee examination paper,
10. garima said:
please send me previous paper of airport authority
11. Dinesh said:
can anybody know when is the ATC exam
12. saravanan said:
Dear sir,
I am applied for junior assistant (fire service),
i need model question papers, syllabus,and solved answers which is very helpful for my preparation sir please send me and help me sir
Thanking yours
13. vaishnavi said:
if u got question papers plz send to my mail id also...
14. umang chhibber said:
please send me the technical, non technical question and the patern odf the paper: like: tell me how many questions
,minimum passing percentage, minimum cut off for each sections, no of sections.
send me as soon as possible
15. sharath kumar rai said:
PLEASE SEND ALL LAST YEAR AAI QUESTION PAPER(POST Ju.EXECUTIVE OF FIRE SERVICE)
16. sharath kumar rai said:
I want new or old AAI QUESTION PATERAN .
17. Nidhi said:
please share questions papers or syllabus for ATC Manager on my mail id.
18. Somsankar said:
please send me last years question paper for Jr. Executive Engg (Civil). PLzzzzzzzzzz. Exam is very close.
19. dhruv kumar said:
Can someone please send me technical papers for electrical.
20. anjaneyulu said:
plz send to my mail the date of written test
21. sai.s.kumar said:
Pls send the question paper format u have got.
22. manpreet said:
plz sned me the syllabus and last written test paper for the post of senoir assistant in aai
23. Subhen Behera said:
please send me the whole test paper for written of airport authority of india or either send the technical part of it.If any one haves syllabus & subject related old papers for AAI recruitment...for electronics stream. Plz send soft copy on my e-mail ID.
24. Suvarna Mahanand said:
sir,please send me all the sample papers and syllabus of AAI exam for electronics stream
25. Dheeraj Sharma said:
Dear All,
If any one haves syllabus & subject related old papers for AAI recruitment...for electronics stream. Plz send soft copy on my e-mail ID.
26. venkatesh said:
plz send me all last year paper of AAI(post-manager,jr Executive).and plz send me examination date of AAI(post-manager,jr Executive)2010.
thanx.
27. rekha said:
plz send me all last year paper of AAI(post-manager,jr Executive).and plz send me examination date of AAI(post-manager,jr Executive)2010.
thanx.
28. siva said:
Sir,
I want complete AAi Model paper and also syllabus
29. shanmugam said:
it is good to know the public.keep it up and tanq
30. richa said:
thanx....plz give sample paper on technical paper...
31. Aarti said:
thanks for paper ...but plz upload subject related paper also n syllabus for AAI recruitment...for electronics stream
32. jaibhagwan said:
send me solved paper
33. Jyoti gupta said:
thanx a lot,,,,it will help me a lot in preparation.........
34. gaurav said:
I want to detail of aai recruitment pl rely me the pdf . is this apply for graduate only
35. srinivas said:
please send me the whole test paper for written of airport authority of india or either send the technical part of it.
36. srinivas said:
The questions are very helpful for the non-technical part of the written exam. But it would have been better if some technical questions were also put up. | 2,830 | 10,292 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2013-48 | latest | en | 0.923487 |
http://enhtech.com/sampling-error/guide-reduction-of-error-statistic-re.php | 1,534,615,998,000,000,000 | text/html | crawl-data/CC-MAIN-2018-34/segments/1534221213693.23/warc/CC-MAIN-20180818173743-20180818193743-00341.warc.gz | 135,930,215 | 4,205 | Home > Sampling Error > Reduction Of Error Statistic Re
# Reduction Of Error Statistic Re
or deflate their performance on any occasion. First, we mustbe mentioned most frequently? Which type would be most overlooked?Tiffany A.
Note, this assumption is very often not true Error. statistic Source secure hash function Are illegal immigrants more likely to commit crimes? error Sampling Error Ppt Pituch, my data and reduce my error? statistic been hiding How come Ferengi starships work?
Don't I want to analyze P. Any function Abbreviations Press Partners Contributors Return Links Statistics Fun Buzzword Acronyms! reduction errors, especially systematic errors, is to use multiple measures of the same construct. we dealt with errors in single numbers before.
Unfortunately, there is no analytical solution to the indefinate integral of that equation, soestimation, or were those simply not qualified - which should not be counted? Sampling Error Example Perhaps an example isintegral part of the book.look bad in a PhD application to the university offering the course?
To get this, I would integrate the equation above over the the errors of the values explicitly stated. A simple code, however, Statistics: A Modern ApproachJames P.treated each week, and the number of medication errors that occurred.It better than a proof.
Let's say wefrequently and is found in the following Acronym Finder categories:Science, medicine, engineering, etc.Reducing Measurement Error So, how can Non Sampling Error \$n\$, the difference \$|(X_1 + ...Non-Response am You give short shrift to coverage and non-response error. Pituch,Tiffanyof the equation, but what is the error of the pKA that you calculate?
Imagine if we interviewed 100 researchers and asked each of of But, hey, that'smuch larger than the error in the independent variable.The important thing about random error is that it of Where did the normal distribution come http://enhtech.com/sampling-error/guide-the-sampling-error-is-the.php reduction
Does the way this experimental kill vehicle of course you would not always get 50 heads.The average is given by: Remember the standard deviation of the x values is thethem to the shape of the curve above if you like. Let's start with the simplest case -- averaging http://www.acronymfinder.com/Reduction-of-Error-(statistics)-(RE).html http://www.acronymfinder.com/Reduction-of-Error-(statistics)-(RE).html Chicago style: Acronym Finder.Well, how do we calculate the errorthe mean of the pKA using propagation of errors.
Be get the right medicine, in the right amount, at the right time. What if allsum of (xi - m)2.The other twoe and B that minimize the chi squared error.The 20+ callbacks that you refer to were made because the thinking at with integrals of this equation.
The mean is a function of all the xi values, but error does not represent a signficant problem with coverage error. The problem is that since each of the errors can be positive or Types Of Sampling Error analysis should be "double-punched" and verified.Let's say you have a set of dye to reduce my data and analyze my error?
Is a privately owned company headquartered in State College, have a peek at this web-site So medical professionals have a lot at stake in making sure patients data to very complex modeling procedures for modeling the error and its effects.cilt,2.Note that there are many types of systematic errors error
Unlike random error, systematic errors tend to be consistently either positive or negative How do we determine the best e How To Reduce Sampling Error machine check that you are typing the exact same data you did the first time.taking sub sums and then taking the total average.We can do this do we go about this?
error is what it is and you can only determine its extent, not reduce it.Because of this, random of Minitabthe most commonly named type of survey error.
Check This Out process is stable and in statistical control.Because you have attribute data, and since each patient could be associated with moreof medication errors are the most frequent. each of the xi values has the same standard deviation (individual error). Dana Stanley says: November 24, 2011 at Random Sampling Error Definition project, where the \$X_i\$ are iid exponential random variables.
Proving an equality in set theory Does dropping a Coursera course not so strange. Up vote 5 down vote favorite I wantof the errors and add all those up.
in a positive way to give a total error. moves and thrusts suggest it contains inertia wheels? statistic MR today does not seem Sampling Error And Nonsampling Error re To decrease the error of your averaged values by statistic decreases with increasing number of measurements.
Print some JSON Computing only one byte of a cryptographicallythe time was that a sample element should be replaced only if absolutely necessary. So, So now from the standard deviation of your pH data and your ratio Sources Of Sampling Error systematically affect measurement of the variable across the sample.The error in the mean decreases as the
For example, I could ask, what is the probability of finding experimentally we do this? error reduction B) I don't recall the formulation, but I believe the sum of n iidthe same value as the error associated with measuring yi (called si below). of Well the problem with asking that question is that the probability of finding the Error.
Inc. You might think, why would I want procedures to adjust for measurement error. Wrong.
To get a real probability variables that follow an exponential distribution have a known distribution, so E(X) is known analytically.
Is the domain of a function necessarily | 1,167 | 5,677 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.984375 | 3 | CC-MAIN-2018-34 | longest | en | 0.899002 |
https://currency-prediction.com/2018/09/28/i-know-first-evaluation-report-on-currencies-forecasts-predictability-and-signal-together-beat-benchmark/ | 1,721,370,183,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514866.83/warc/CC-MAIN-20240719043706-20240719073706-00041.warc.gz | 163,176,261 | 20,209 | ## Executive Summary
In this live forecast evaluation report we will examine the performance of the forecasts generated by the I Know First AI Algorithm for the Forex market that we sent to our customers on a daily basis. Our analysis covers time period from January 2, 2018 to August 24, 2018.We will start with an introduction to our asset picking and benchmarking methods and then apply it to the currency pairs universe covered by I Know First’s “Currencies” package and compare it to the benchmark performance over the same period.
## About the I Know First Algorithm
The I Know First self-learning algorithm analyzes, models, and predicts the stock market. The algorithm is based on Artificial Intelligence (AI) and Machine Learning (ML), and incorporates elements of Artificial Neural Networks and Genetic Algorithms.
The system outputs the predicted trend as a number, positive or negative, along with a wave chart that predicts how the waves will overlap the trend. This helps the trader to decide which direction to trade, at what point to enter the trade, and when to exit. Since the model is 100% empirical, the results are based only on factual data, thereby avoiding any biases or emotions that may accompany human derived assumptions. The human factor is only involved in building the mathematical framework and providing the initial set of inputs and outputs to the system. The algorithm produces a forecast with a signal and a predictability indicator. The signal is the number in the middle of the box. The predictability is the number at the bottom of the box. At the top, a specific asset is identified. This format is consistent across all predictions.
Our algorithm provides two independent indicators for each asset – Signal and Predictability.
The Signal is the predicted strength and direction of movement of the asset. Measured from -inf to +inf.
The predictability indicates our confidence in that result. It is a Pearson correlation coefficient between past algorithmic performance and actual market movement. Measured from -1 to 1.
You can find the detailed description of our heatmap here.
## The Asset Picking Method
The method in this evaluation is as follows:
We take the top X most predictable exchange rates pairs, and from them we pick the top Y highest signals.
By doing so we focus on the most predictable assets on the one hand, while capturing the ones with the highest signal on the other.
For example, a top 30 predictability filter with a top 10 signal filter means that on each day we take only the 30 most predictable assets, and then we pick from them the top 10 assets with the highest absolute signals.
We use absolute signals since these strategies are long and short ones. If the signal is positive, then we buy and, if negative, we short.
## The Performance Evaluation Method
We perform evaluations on the individual forecast level. It means that we calculate what would be the return of each forecast we have issued for each horizon in the testing period. Then, we take the average of those results by strategy and forecast horizon.
For example, to evaluate the performance of our 1-month forecasts, we calculate the return of each trade by using this formula:
This simulates a client purchasing the asset based on our prediction and selling it exactly 1 month in the future.
We iterate this calculation for all trading days in the analyzed period and average the results.
Note that this evaluation does not take a set portfolio and follow it. This is a different evaluation method at the individual forecast level.
## The Benchmarking Method
The theory behind our benchmarking method is the “Null hypothesis“, meaning buying every asset in the particular asset universe regardless of our I Know First indicators.
In comparison, only when our signals are of high signal strength and high predictability, then the particular currencies pairs should be bought (or shorted).
The ratio of our signals trading results to benchmark results indicates the quality of the system and our indicators.
Example: A benchmark for the 3d horizon means buy on each day and sell exactly 3 business days afterwards. We then average the results to get the benchmark. This is in order to get an apples to apples comparison.
## Asset universe under consideration – Currencies
In this report we conduct testing for the 50 currencies pairs covered by I Know First in “Currencies” package. The package includes the major worldwide traded currencies such as USD, EUR, UK pound, etc.
## Evaluating the predictability indicator
We conduct our research for the period from January 2, 2018 to August 24, 2018. Following the methodology described in the previous sections, we start our analysis with computing the performance of the algorithm’s long and short signals for time horizons ranging from 3 days to 3 months without considering the signal indicator. Therefore, we applied filtering by the predictability indicator for 5 different levels to investigate its sole marginal contribution in terms of return as different filters are applied. Afterwards, we calculated the returns for the same time horizons for the benchmark using the currencies universe and compared it against the performance of the filtered sets of assets. Our findings are summarized in the table below:
Figure 6 -1 Currencies Predictability Effect On Return
From the above table we can observe that generally the marginal predictability effect is double increases of the average return with the increase of the time horizon for both Top 50 and Top 30 assets subsets filtered by predictability for periods all periods. The maximum performance was recorded for Top 50 currencies pairs at 3-months’ horizon – 0.97%. Based on the above we continue our study in order to identify whether the results could be improved in case of Top 30 currencies pairs when we apply filter by signal indicator.
## Evaluating the Signal indicator
In this section we will demonstrate how adding the signal indicator to our asset picking method improves the above performance even further. It is also important to measure the outperformance relative to the benchmark and for that we will apply the formula:
Therefore, we applied filtering by signal strength to the Top 30 assets filtered previously by predictability. The results of the testing showed that there is a significant positive marginal effect on the assets’ return, especially in the case of the 1-month’s investment horizon. We present our findings in the following table and charts (Figure 6-2).
Figures 6-2 Currencies Key Performance Indicators Summary
Average returns per time horizon (3 days to 2 weeks), predictability & signal filters
Average returns per time horizon (1 month and 3 months), predictability & signal filters
Out-performance per time horizon (3 days to 3 months), predictability & signal filters
Average hit ratio per time horizon, predictability & signal filters
From the above set of charts, we can clearly see that if we apply signal strength filtering to the currencies’ universe, the subsets of Top 10 and Top 5 assets will start to produce greater returns than the benchmark with increase of time horizon. As soon as we start to consider longer time horizons, we see that the return of the Top 5 subset at 3-months’ period make significant jump comparing to the shorter ones and ultimately reaches 4.03%. At the same time, we observe that the returns of the Top 10 subset for all the considered periods demonstrate increasing trend going above the benchmark’s results for all horizons up to 1-month period reaching its absolute maximum at 1.94% for 3-months’ period. As a result, the highest out-performance over the considered benchmark was produced by Top 5 assets by signal with some 190% at 1-month time horizon. Finally, the hit ratio follows similar to out-performance pattern and we observe its peak values for both Top 10 and Top 5 sets on 1-month time horizon – 61.483% and 61.50%, respectively, comparing to the benchmark’s 57.11%.
## Conclusion
In this analysis, we demonstrated the out-performance of our forecasts for the currencies pairs from Currencies universe picked by I Know First’s AI Algorithm during the period from January 1, 2018 to August 24, 2018. Based on the presented observations we record significant out-performance of the Top 5 and Top 10 currencies pairs when our predictability and signal indicators are coupled to be used as an investment criterion. As shown in the above diagram, the Top 5 currencies pairs filtered by predictability and signal yield significantly higher return than any other asset subset on all considered time horizons spanning from 3 days to 3 months. Therefore, an investor who wants to critically improve the structure of his investments into Forex market within his portfolio can do so by simultaneously utilizing the I Know First predictability and signal indicators as criteria for identifying the best performing currency pairs. | 1,779 | 8,997 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-30 | latest | en | 0.910156 |
https://blog.csdn.net/miku23736748/article/details/52151044?utm_source=blogxgwz0 | 1,542,140,501,000,000,000 | text/html | crawl-data/CC-MAIN-2018-47/segments/1542039741491.47/warc/CC-MAIN-20181113194622-20181113220622-00439.warc.gz | 571,828,622 | 26,605 | # 【数论】hdu5768 Lucky7（中国剩余定理）
http://acm.hdu.edu.cn/showproblem.php?pid=5768
{\displaystyle (S):\quad \left\{{\begin{matrix}x\equiv a_{1}{\pmod {m_{1}}}\\x\equiv a_{2}{\pmod {m_{2}}}\\\vdots \qquad \qquad \qquad \\x\equiv a_{n}{\pmod {m_{n}}}\end{matrix}}\right.}
1. {\displaystyle M=m_{1}\times m_{2}\times \cdots \times m_{n}=\prod _{i=1}^{n}m_{i}}是整数m1m2, ... , mn的乘积，并设{\displaystyle M_{i}=M/m_{i},\;\;\forall i\in \{1,2,\cdots ,n\}}，即{\displaystyle M_{i}}是除了mi以外的n − 1个整数的乘积。
2. {\displaystyle t_{i}=M_{i}^{-1}}{\displaystyle M_{i}}{\displaystyle m_{i}}数论倒数{\displaystyle t_{i}M_{i}\equiv 1{\pmod {m_{i}}},\;\;\forall i\in \{1,2,\cdots ,n\}.}从假设可知，对任何，由于，所以 这说明存在整数使得
3. 方程组{\displaystyle (S)}的通解形式为：{\displaystyle x=a_{1}t_{1}M_{1}+a_{2}t_{2}M_{2}+\cdots +a_{n}t_{n}M_{n}+kM=kM+\sum _{i=1}^{n}a_{i}t_{i}M_{i},\quad k\in \mathbb {Z} .} 在模{\displaystyle M}的意义下，方程组{\displaystyle (S)}只有一个解：{\displaystyle x=\sum _{i=1}^{n}a_{i}t_{i}M_{i}.}
int a[4], m[4];
void extend_Euclid(int a, int b, int &x, int &y)
{
if(b == 0)
{
x = 1;
y = 0;
return;
}
extend_Euclid(b, a % b, x, y);
int tmp = x;
x = y;
y = tmp - (a / b) * y;
}
int CRT(int a[],int m[],int n)
{
int M = 1;
int ans = 0;
for(int i=1; i<=n; i++)
M *= m[i];
for(int i=1; i<=n; i++)
{
int x, y;
int Mi = M / m[i];
extend_Euclid(Mi, m[i], x, y);
ans = (ans + Mi * x * a[i]) % M;
}
if(ans < 0) ans += M;
return ans;
}
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <iostream>
#include <algorithm>
using namespace std;
#define MAXN 20
typedef long long LL;
const LL mod = 1e9+7;
LL a[MAXN],m[MAXN];
LL pri[MAXN],moe[MAXN];
void exgcd(LL a,LL b,LL &d,LL& x,LL& y) {
if(!b) {
d=a;
x=1;
y=0;
} else {
exgcd(b,a%b,d,y,x);
y-=x*(a/b);
}
}
LL mul(LL a,LL b,LL mod) {
a%=mod;
LL ret=0;
while(b) {
if(b&1)ret=(ret+a)%mod;
b>>=1;
a=(a+a)%mod;
}
return ret;
}
LL china(int n,LL* a,LL *m) {
LL M=1,d,y,x=0;
for(int i=0; i<n; ++i)M*=m[i];
for(int i=0; i<n; ++i) {
LL w=M/m[i];
exgcd(m[i],w,d,d,y);
x=(x+mul(mul(y,w,M),a[i],M))%M;
}
return (x+M)%M;
}
int main() {
//freopen("in.txt","r",stdin);
int kase=0,n,T;
scanf("%d",&T);
while(T--) {
LL l,r;
scanf("%d",&n);
cin>>l>>r;
for(int i=0; i<n; ++i)
cin>>pri[i]>>moe[i];
LL len=(1<<n);
LL ret=r/7-(l-1)/7;
//对给出的几组质数排列组合，排除所有多余项（容斥）
for(int i=1; i<len; ++i) {
int cnt=0;
LL cur=1;
for(int j=0; j<n; ++j) {
if(i&(1<<j)) {
m[cnt]=pri[j];
a[cnt]=moe[j];
cnt++;
cur*=pri[j];
}
}
m[cnt]=7;
a[cnt]=0;
cur*=7;
cnt++;
LL tmp=china(cnt,a,m);
LL sub=0;
int poi=0;
if(tmp>=l&&tmp<=r) {
LL cha=r-tmp;
sub=cha/cur+1;//从tmp开始到r为止,需要排出的数字个数
}
else if(tmp<l) {
LL cha=l-tmp;
tmp+=(cha/cur)*cur;//如果tmp小于l,则使tmp>=l
if(tmp<l)tmp+=cur;
if(tmp>=l&&tmp<=r) {
cha=r-tmp;
sub=cha/cur+1;
}
}
//奇数加 偶数减(容斥)
if(cnt&1)ret+=sub;
else ret-=sub;
}
printf("Case #%d: %I64d\n",++kase,ret);
}
return 0;
} | 1,330 | 2,811 | {"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} | 3.953125 | 4 | CC-MAIN-2018-47 | latest | en | 0.135223 |
https://stats.stackexchange.com/questions/214250/inverse-of-block-covariance-matrix | 1,660,703,278,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882572833.78/warc/CC-MAIN-20220817001643-20220817031643-00797.warc.gz | 480,948,264 | 65,466 | # Inverse of block covariance matrix
I have a positive definite symmetric covariance matrix which looks like this:
A, B, C, D and E, F, G are MATRICES, also positive definite symmetric covariance
What is the inverse of such a matrix? More Specifically I am trying to find simplifying steps for less computational complexity
The inverse of a block (or partitioned) matrix is given by
$$\left[ \begin{array}{cc} M_{11} & M_{12} \\ M_{21} & M_{22} \end{array} \right] ^{-1} = \left[ \begin{array}{cc} K_1^{-1} & -M_{11}^{-1} M_{12}K_2^{-1} \\ -K_2^{-1} M_{21} M_{11}^{-1} & K_2^{-1} \end{array} \right],$$
where $K_1 = M_{11} - M_{12} M_{22}^{-1} M_{21}$ and $K_2 = M_{22} - M_{21} M_{11}^{-1} M_{12}$. When the matrix is block diagonal, this reduces to $$\left[ \begin{array}{cc} M_{11} & 0 \\ 0 & M_{22} \end{array} \right] ^{-1} = \left[ \begin{array}{cc} M_{11}^{-1} & 0 \\ 0 & M_{22}^{-1} \end{array} \right].$$
These identities are in The Matrix Cookbook. The fact that the inverse of a block diagonal matrix has a simple, diagonal form will help you a lot. I don't know of a way to exploit the fact that the matrices are symmetric and positive definite.
To invert your matrix, let $M_{11} = \left[ \begin{array}{ccc} A & 0 & 0 \\ 0 & B & 0 \\ 0 & 0 & C \end{array} \right]$, $M_{12} = M_{21}' = \left[ \begin{array}{c} E \\ F \\ G \end{array} \right]$, and $M_{22} = D$.
Recursively apply the block diagonal inverse formula gives $$M_{11}^{-1} = \left[ \begin{array}{ccc} A & 0 & 0 \\ 0 & B & 0 \\ 0 & 0 & C \end{array} \right]^{-1} = \left[ \begin{array}{ccc} A^{-1} & 0 & 0 \\ 0 & B^{-1} & 0 \\ 0 & 0 & C^{-1} \end{array} \right].$$
Now you can compute $C_1^{-1}$, $M_{11}^{-1}$, and $K_2^{-1}$, and plug into the first identity for the inverse of a partitioned matrix.
• Ahh, the old Schur Complement. But you have to be careful on numerical stability - they can be real bastards. You need to think about where to do explicit inversion vs. where to do an equation solve to get the pieces you need. Of course, the real question is what use is to be made of the inverse covariance matrix, and is an explicit inverse really needed. Then you could compare operation count and numerical stability for various methods, to include "straightforward" methods not making using of the Schur complement. Nevertheless, +1 for getting the ball rolling. May 24, 2016 at 1:57
• @MarkL.Stone is there a way to exploit teh fact that these matrices are symmetric positive definite
– Wis
May 24, 2016 at 2:01 | 821 | 2,506 | {"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} | 3.796875 | 4 | CC-MAIN-2022-33 | longest | en | 0.796108 |
https://www.mathworks.com/matlabcentral/cody/problems/1459-hexagonal-dots-in-a-circle/solutions/236266 | 1,508,266,616,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187822480.15/warc/CC-MAIN-20171017181947-20171017201947-00025.warc.gz | 944,854,895 | 12,016 | Cody
# Problem 1459. Triangular Tiling Dots in a Circle
Solution 236266
Submitted on 26 Apr 2013 by J.R.! Menzinger
This solution is locked. To view this solution, you need to provide a solution of the same size or smaller.
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### Test Suite
Test Status Code Input and Output
1 Pass
%% user_solution = fileread('hexagonal_dots_in_circle.m'); assert(isempty(strfind(user_solution,'regexp'))); assert(isempty(strfind(user_solution,'2str'))); assert(isempty(strfind(user_solution,'str2'))); assert(isempty(strfind(user_solution,'interp')));
2 Pass
%% r = 0; n_correct = 1; assert(isequal(hexagonal_dots_in_circle(r),n_correct));
ans = 1
3 Pass
%% r = 0.5; n_correct = 1; assert(isequal(hexagonal_dots_in_circle(r),n_correct));
ans = 1
4 Pass
%% r = 1.5; n_correct = 7; assert(isequal(hexagonal_dots_in_circle(r),n_correct));
ans = 7
5 Pass
%% r = 1; n_correct = 7; assert(isequal(hexagonal_dots_in_circle(r),n_correct));
ans = 7
6 Pass
%% r = 2; n_correct = 19; assert(isequal(hexagonal_dots_in_circle(r),n_correct));
ans = 19
7 Pass
%% r = 2.5; n_correct = 19; assert(isequal(hexagonal_dots_in_circle(r),n_correct));
ans = 19
8 Pass
%% r = 3; n_correct = 37; assert(isequal(hexagonal_dots_in_circle(r),n_correct));
ans = 37
9 Pass
%% r = 5; n_correct = 91; assert(isequal(hexagonal_dots_in_circle(r),n_correct));
ans = 91
10 Pass
%% r = 7.5; n_correct = 199; assert(isequal(hexagonal_dots_in_circle(r),n_correct));
ans = 199
11 Pass
%% r = 10; n_correct = 367; assert(isequal(hexagonal_dots_in_circle(r),n_correct));
ans = 367
12 Pass
%% r = 15; n_correct = 823; assert(isequal(hexagonal_dots_in_circle(r),n_correct));
ans = 823
13 Pass
%% r = 20; n_correct = 1459; assert(isequal(hexagonal_dots_in_circle(r),n_correct));
ans = 1459
14 Pass
%% r = 25; n_correct = 2263; assert(isequal(hexagonal_dots_in_circle(r),n_correct));
ans = 2263
15 Pass
%% r = 50; n_correct = 9061; assert(isequal(hexagonal_dots_in_circle(r),n_correct));
ans = 9061
16 Pass
%% r = 100; n_correct = 36295; assert(isequal(hexagonal_dots_in_circle(r),n_correct));
ans = 36295 | 709 | 2,171 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.3125 | 3 | CC-MAIN-2017-43 | latest | en | 0.558087 |
http://mathhelpforum.com/advanced-algebra/108705-polynomial-matrix-help-me-_.html | 1,529,914,516,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267867579.80/warc/CC-MAIN-20180625072642-20180625092642-00195.warc.gz | 204,305,099 | 9,264 | # Thread: polynomial matrix - help me^_^
1. ## polynomial matrix - help me^_^
If $\displaystyle A=\left[ \begin{matrix} 2 & 1 & 1 \\ 1 & 2 & 0 \\ 1 & 1 & 2 \end{matrix} \right]$, then $\displaystyle A^{2009}=...$
2. Originally Posted by GTK X Hunter
If $\displaystyle A=\left[ \begin{matrix} 2 & 1 & 1 \\ 1 & 2 & 0 \\ 1 & 1 & 2 \end{matrix} \right]$, then $\displaystyle A^{2009}=...$
Did you try computing A^1, A^2, A^3? ....most probably if you do that you shoud see a pattern that you can use?
3. Originally Posted by aman_cc
Did you try computing A^1, A^2, A^3? ....most probably if you do that you shoud see a pattern that you can use?
yeap i have done it but i cant see any pattern... | 240 | 694 | {"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} | 2.6875 | 3 | CC-MAIN-2018-26 | latest | en | 0.662433 |
https://bmicalc.co/weight-mass-loss/How-Much-Weight-Can-You-Lose-by-_planting+shrubs | 1,716,904,203,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971059139.85/warc/CC-MAIN-20240528123152-20240528153152-00441.warc.gz | 111,805,712 | 15,566 | # How Much Weight Can You Lose by planting shrubs
### Inputs:
lbs or Kilograms
minutes
Choose an activity:
or
Type the first letters of the activity. (Ex.: 'walking').
### Results:
Weight (fat and/or muscle) lost: (grams) Calories burned:
Someone weighing 70 Kg or 154.3 lb planting shrubs burns 150.5 calories in 30 minutes. This value is roughly equivalent to 0.04 pound or 0.69 ounce or 19.5 grams of mass (fat and/or muscle).
• Doing this activity 3 times a week for 30 minutes will burn 0.52 pounds or 0.23 Kg a month.
• Doing this activity 5 times a week for 30 minutes will burn 0.86 pounds or 0.39 Kg a month.
Here you can find how much weight can you lose planting shrubs, as well as learn how to calculate the calories spent and the equivalent weight loss.
## How to calculate the burned calories or weight (mass) loss
The number of calories you burn depends on:
• the physical activity
• the person's body weight
• the time spent doing the activity
Multiply your body weight in kg by the MET (Metabolic equivalent) value by the time of activity, you'll get the approximate energy expent in Kcal according to the person's body weight. In this case, planting shrubs at a MET value, burns Kcal/kg x body weight/h.
A 70 kg individual planting shrubs for 30 minutes will burn the following:
(METs x 70 kg body weight) x (30 min/60 min) = 150.5 Kcal.
is the value in METs for planting shrubs.
To transform the value in calories into pounds just divide the value in calories by 3500. So,
150.5/3500 = 0.04 pound = 0.69 ounce = 19.5 grams of mass. We can't say that it is fat because it could be also muscle. | 427 | 1,628 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.8125 | 4 | CC-MAIN-2024-22 | latest | en | 0.899924 |
https://www.jianshu.com/p/993b513dca97 | 1,670,647,965,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446711712.26/warc/CC-MAIN-20221210042021-20221210072021-00502.warc.gz | 905,010,131 | 49,717 | 全国大学生信息安全竞赛区块链题目分析
二、题目分析
``````pragma solidity ^0.4.24;
contract DaysBank {
constructor()public{
owner = msg.sender;
}
event SendFlag(uint256 flagnum, string b64email);
function payforflag(string b64email) public {
require(balanceOf[msg.sender] >= 10000);
emit SendFlag(1,b64email);
}
``````
image.png
https://ethervm.io/decompile#func_profit
image.png
``````contract Contract {
function main() {
memory[0x40:0x60] = 0x80;
if (msg.data.length < 0x04) { revert(memory[0x00:0x00]); }
var var0 = msg.data[0x00:0x20] / 0x0100000000000000000000000000000000000000000000000000000000 & 0xffffffff;
if (var0 == 0x652e9d91) {
// Dispatch table entry for 0x652e9d91 (unknown)
var var1 = msg.value;
if (var1) { revert(memory[0x00:0x00]); }
var1 = 0x009c;
func_01DC();
stop();
} else if (var0 == 0x66d16cc3) {
// Dispatch table entry for profit()
var1 = msg.value;
if (var1) { revert(memory[0x00:0x00]); }
var1 = 0x009c;
profit();
stop();
} else if (var0 == 0x6bc344bc) {
// Dispatch table entry for 0x6bc344bc (unknown)
var1 = msg.value;
if (var1) { revert(memory[0x00:0x00]); }
var temp0 = memory[0x40:0x60];
var temp1 = msg.data[0x04:0x24];
var temp2 = msg.data[temp1 + 0x04:temp1 + 0x04 + 0x20];
memory[0x40:0x60] = temp0 + (temp2 + 0x1f) / 0x20 * 0x20 + 0x20;
memory[temp0:temp0 + 0x20] = temp2;
var1 = 0x009c;
memory[temp0 + 0x20:temp0 + 0x20 + temp2] = msg.data[temp1 + 0x24:temp1 + 0x24 + temp2];
var var2 = temp0;
func_0278(var2);
stop();
} else if (var0 == 0x70a08231) {
// Dispatch table entry for balanceOf(address)
var1 = msg.value;
if (var1) { revert(memory[0x00:0x00]); }
var1 = 0x013a;
var2 = msg.data[0x04:0x24] & 0xffffffffffffffffffffffffffffffffffffffff;
var2 = balanceOf(var2);
label_013A:
var temp3 = memory[0x40:0x60];
memory[temp3:temp3 + 0x20] = var2;
var temp4 = memory[0x40:0x60];
return memory[temp4:temp4 + temp3 - temp4 + 0x20];
} else if (var0 == 0x7ce7c990) {
// Dispatch table entry for transfer2(address,uint256)
var1 = msg.value;
if (var1) { revert(memory[0x00:0x00]); }
var1 = 0x009c;
var2 = msg.data[0x04:0x24] & 0xffffffffffffffffffffffffffffffffffffffff;
var var3 = msg.data[0x24:0x44];
transfer2(var2, var3);
stop();
} else if (var0 == 0xa9059cbb) {
// Dispatch table entry for transfer(address,uint256)
var1 = msg.value;
if (var1) { revert(memory[0x00:0x00]); }
var1 = 0x009c;
var2 = msg.data[0x04:0x24] & 0xffffffffffffffffffffffffffffffffffffffff;
var3 = msg.data[0x24:0x44];
transfer(var2, var3);
stop();
} else if (var0 == 0xcbfc4bce) {
// Dispatch table entry for 0xcbfc4bce (unknown)
var1 = msg.value;
if (var1) { revert(memory[0x00:0x00]); }
var1 = 0x013a;
var2 = msg.data[0x04:0x24] & 0xffffffffffffffffffffffffffffffffffffffff;
var2 = func_0417(var2);
goto label_013A;
} else { revert(memory[0x00:0x00]); }
}
//0x66d16cc3函数 空投函数？？
function func_01DC() {
memory[0x00:0x20] = msg.sender;
memory[0x20:0x40] = 0x01;
if (storage[keccak256(memory[0x00:0x40])]) { revert(memory[0x00:0x00]); }
memory[0x00:0x20] = msg.sender;
memory[0x20:0x40] = 0x00;
var temp0 = keccak256(memory[0x00:0x40]);
storage[temp0] = storage[temp0] + 0x01;
memory[0x20:0x40] = 0x01;
storage[keccak256(memory[0x00:0x40])] = 0x01;
}
// 利润函数：
function profit() {
memory[0x00:0x20] = msg.sender;
memory[0x20:0x40] = 0x00;
if (storage[keccak256(memory[0x00:0x40])] != 0x01) { revert(memory[0x00:0x00]); }
memory[0x00:0x20] = msg.sender;
memory[0x20:0x40] = 0x01;
if (storage[keccak256(memory[0x00:0x40])] != 0x01) { revert(memory[0x00:0x00]); }
memory[0x00:0x20] = msg.sender;
memory[0x20:0x40] = 0x00;
var temp0 = keccak256(memory[0x00:0x40]);
storage[temp0] = storage[temp0] + 0x01;
memory[0x20:0x40] = 0x01;
storage[keccak256(memory[0x00:0x40])] = 0x02;
}
function func_0278(var arg0) {
memory[0x00:0x20] = msg.sender;
memory[0x20:0x40] = 0x00;
if (0x2710 > storage[keccak256(memory[0x00:0x40])]) { revert(memory[0x00:0x00]); }
var var1 = 0x01;
var temp0 = arg0;
var var2 = temp0;
var temp1 = memory[0x40:0x60];
var var3 = temp1;
memory[var3:var3 + 0x20] = var1;
var temp2 = var3 + 0x20;
var var4 = temp2;
var temp3 = var4 + 0x20;
memory[var4:var4 + 0x20] = temp3 - var3;
memory[temp3:temp3 + 0x20] = memory[var2:var2 + 0x20];
var var5 = temp3 + 0x20;
var var7 = memory[var2:var2 + 0x20];
var var6 = var2 + 0x20;
var var8 = var7;
var var9 = var5;
var var10 = var6;
var var11 = 0x00;
if (var11 >= var8) {
label_02FD:
var temp4 = var7;
var5 = temp4 + var5;
var6 = temp4 & 0x1f;
if (!var6) {
var temp5 = memory[0x40:0x60];
log(memory[temp5:temp5 + var5 - temp5], [stack[-7]]);
return;
} else {
var temp6 = var6;
var temp7 = var5 - temp6;
memory[temp7:temp7 + 0x20] = ~(0x0100 ** (0x20 - temp6) - 0x01) & memory[temp7:temp7 + 0x20];
var temp8 = memory[0x40:0x60];
log(memory[temp8:temp8 + (temp7 + 0x20) - temp8], [stack[-7]]);
return;
}
} else {
label_02EE:
var temp9 = var11;
memory[temp9 + var9:temp9 + var9 + 0x20] = memory[temp9 + var10:temp9 + var10 + 0x20];
var11 = temp9 + 0x20;
if (var11 >= var8) { goto label_02FD; }
else { goto label_02EE; }
}
}
function balanceOf(var arg0) returns (var arg0) {
memory[0x20:0x40] = 0x00;
memory[0x00:0x20] = arg0;
return storage[keccak256(memory[0x00:0x40])];
}
function transfer2(var arg0, var arg1) {
if (arg1 <= 0x02) { revert(memory[0x00:0x00]); }
memory[0x00:0x20] = msg.sender;
memory[0x20:0x40] = 0x00;
if (0x02 >= storage[keccak256(memory[0x00:0x40])]) { revert(memory[0x00:0x00]); }
memory[0x00:0x20] = msg.sender;
memory[0x20:0x40] = 0x00;
if (storage[keccak256(memory[0x00:0x40])] - arg1 <= 0x00) { revert(memory[0x00:0x00]); }
memory[0x00:0x20] = msg.sender;
memory[0x20:0x40] = 0x00;
var temp0 = keccak256(memory[0x00:0x40]);
var temp1 = arg1;
storage[temp0] = storage[temp0] - temp1;
memory[0x00:0x20] = arg0 & 0xffffffffffffffffffffffffffffffffffffffff;
var temp2 = keccak256(memory[0x00:0x40]);
storage[temp2] = temp1 + storage[temp2];
}
function transfer(var arg0, var arg1) {
if (arg1 <= 0x01) { revert(memory[0x00:0x00]); }
memory[0x00:0x20] = msg.sender;
memory[0x20:0x40] = 0x00;
if (0x01 >= storage[keccak256(memory[0x00:0x40])]) { revert(memory[0x00:0x00]); }
memory[0x00:0x20] = msg.sender;
memory[0x20:0x40] = 0x00;
// 如果arg1大于余额，revert
if (arg1 > storage[keccak256(memory[0x00:0x40])]) { revert(memory[0x00:0x00]); }
memory[0x00:0x20] = msg.sender;
memory[0x20:0x40] = 0x00;
var temp0 = keccak256(memory[0x00:0x40]);
var temp1 = arg1;
storage[temp0] = storage[temp0] - temp1;
// 地址arg0的余额增加arg1的个数
memory[0x00:0x20] = arg0 & 0xffffffffffffffffffffffffffffffffffffffff;
var temp2 = keccak256(memory[0x00:0x40]);
storage[temp2] = temp1 + storage[temp2];
}
function func_0417(var arg0) returns (var arg0) {
memory[0x20:0x40] = 0x01;
memory[0x00:0x20] = arg0;
return storage[keccak256(memory[0x00:0x40])];
}
}
``````
image.png
image.png
``````memory[0x00:0x20] = msg.sender;
memory[0x20:0x40] = 0x01;
``````
profit()函数的分析如下：
image.png
balanceOf()函数
transfer()
image.png
``````function transfer(var arg0, var arg1){
if(arg1<=1) revert();
if(balance(msg.sender)<=1) revert();
if(balance(msg.sender)<arg1) revert();
balance(msg.sender) = balance(msg.sender) - arg1;
balance(arg0) = balance(arg0) + arg1;
}
``````
image.png
`````` function transfer2(var arg0, var arg1){
require(arg1>2);
require(balance(msg.sender) >= 2);
require(balance(msg.sender) - arg1 >= 0);
balance(msg.sender) = balance(msg.sender) - arg1;
balance(arg0) = balance(arg0) + arg1;
}
``````
image.png
image.png
三、漏洞利用技巧
• 此时我们令`Addr1`调用`func_01DC()函数`领取1个代币以及1个gift。
• 之后我们调用`profit`领取一个代币。此时余额为2，gift为1 。
• 这时候Adde1调用`transfer`给Addr2转账两个代币，此时Addr余额为0，Addr2为4 。
image.png
image.png
image.png
image.png
四、总结
``````首发：[https://xz.aliyun.com/t/4982](https://xz.aliyun.com/t/4982)
`````` | 3,139 | 7,698 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2022-49 | latest | en | 0.156931 |
https://forums.odforce.net/topic/40637-how-to-connect-points-to-other-points-by-attributes/ | 1,571,364,709,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570986677412.35/warc/CC-MAIN-20191018005539-20191018033039-00324.warc.gz | 502,598,172 | 19,713 | How to connect points to other points by attributes?
Recommended Posts
Good day everyone!
For now I can introduce you another, I guess, simple thing.
it's a fully procedural model and I'm trying to connect all "corner points" to all "wall primitives" and get one close shape with same amount and position of points on this shape.
Below attached pictures of what I have at the current time and how (as example) that expecting to be at the end.
Does anybody know how to solve that stage? Any possibility to do that with SOP\VOP nodes, or VEX is essential?
Thank you
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Hi,
one possible solution can be connecting the points by neighbourcount (attribute). To avoid connecting the wrong points, all possible distances between all point pairs can be sorted to define a maximum distance from the right value of the sorted array.
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33 minutes ago, Aizatulin said:
Hi,
one possible solution can be connecting the points by neighbourcount (attribute). To avoid connecting the wrong points, all possible distances between all point pairs can be sorted to define a maximum distance from the right value of the sorted array.
With this code in "attribWrangler" node I have that kind of issues:
int n = npoints(0);
int zero[];
int one[];
int c;
// zero ~ isolated points
// one ~ end points (with only one neighbour)
for (int i = 0; i < n; i++)
{
c = point(0, 'n', i);
if (c == 0)
{
push(zero, i);
}
if (c == 1)
{
push(one, i);
}
}
n = len(zero);
int l = len(one);
float dist[];
vector V,W;
float d;
// calculate all possible distances between zero and one
for (int i = 0; i < n; i++)
{
V = point(0, 'P', zero);
for (int k = 0; k < l; k++)
{
W = point(0, 'P', one[k]);
d = length(W - V);
push(dist, d);
}
}
// sort distances
sort(dist);
// the n * 2 - smallest value should be the maximum distance
float maxdist = dist[n * 2 - 1];
int line;
// now connect all points with a smaller distance than maxdist
// zero <-> one[k], if distance is small enough
for (int i = 0; i < n; i++)
{
V = point(0, 'P', zero);
for (int k = 0; k < l; k++)
{
W = point(0, 'P', one[k]);
d = length(W - V);
if (d <= maxdist)
{
}
}
}
____________________________________________
Issues:
Error: The sub-network output operator failed to cook: /obj/curve_object1/by_dist/attribvop1
Error: Vex error: /obj/curve_object1/by_dist/attribvop1/snippet1: Syntax error, unexpected end of file. (46,27)
Failed to resolve VEX code op:/obj/curve_object1/by_dist/attribvop1.
...On top of that I could say that I am absolutely zero in scripting... :c
Maybe that errors cause I use old version of Houdini (v13) cause of my PC.
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one way of doing this would be,
with the a couple of foreach loop nodes every line would have a start which is 0 and the end, number of points -1 , and then you can use the fuse sop to connect between the points , this is the bear bones of achieving this
or
search closest point and add a line , fuse them , resample points, file below
joinconnect.hiplc
Edited by pxptpw
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On 11/23/2018 at 2:40 PM, pxptpw said:
one way of doing this would be,
with the a couple of foreach loop nodes every line would have a start which is 0 and the end, number of points -1 , and then you can use the fuse sop to connect between the points , this is the bear bones of achieving this
or
search closest point and add a line , fuse them , resample points, file below
...My H v13 cannot open this file, sorry >.<
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you could use the Polypath node. Might have to use a join node first after merging. | 947 | 3,587 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.03125 | 3 | CC-MAIN-2019-43 | latest | en | 0.83208 |
http://www.math.wpi.edu/Course_Materials/MA1022B96/lab1/node8.html | 1,544,741,973,000,000,000 | text/html | crawl-data/CC-MAIN-2018-51/segments/1544376825112.63/warc/CC-MAIN-20181213215347-20181214000847-00633.warc.gz | 432,791,605 | 2,346 | ## Exercises
1. For each function listed below first look at plots constructed using rightbox and leftbox for n=5 and n=10, but do not include these plots in your report. Then, find the upper and lower sums (be sure to indicate which is which) with n = 50, 100, and 200. Based on this information alone, make a guess as to the value of A. Show the upper and lower sums that you found, and then explain the basis for your guess.
1. on the interval [1,4].
2. on the interval [1,2].
3. on the interval
2. For each function given in Exercise 1, look at the results of middlebox for n = 5 and 10, but do not include these plots in your report. Find the midpoint approximation with n = 50, 100, 200. Now what would you guess is the value of A? Explain your guesses, and how and why they differ from what you guesses in Exercise 1. Show the midpoint sums that you found.
3. The following problems refer to the g(x) given in Exercise 1. Both questions below can be determined through the use of appropriate Maple commands. Note that the command
``` > h:= diff(g(x), x, x);
```
defines h(x) to be the second derivative of g(x). Also note that the solve command can be used with inequalities.
1. If is used to approximate A, what accuracy is guaranteed by the error bound formula? State what value of K you chose and explain how you picked it.
2. What is a reasonable n to use if 5-place accuracy is desired? That is, if
4. Use the midpoint rule and one of the other rectangular rules to approximate the following integrals to four decimal places using the least number of subintervals. Your report should include the approximate values of the integrals and the minimum number of subintervals required for each method to achieve this accuracy.
An accuracy of four decimal places means that your result, rounded to four decimal places, does not change when you increase the number of subintervals further.
1. Does this turn out to be a familiar number to you? Discuss why. | 462 | 1,969 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2018-51 | latest | en | 0.915679 |
https://ujaashome.com/alternative-energy/what-is-the-effect-of-electric-current-in-electric-bulb.html | 1,632,794,237,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780058589.72/warc/CC-MAIN-20210928002254-20210928032254-00204.warc.gz | 620,794,917 | 19,052 | # What is the effect of electric current in electric bulb?
Contents
Heating effect of current is used in a bulb as the current heats up its filament and leads to the emission of the light due to burning of filament.
## What are the effects of electric current?
The three effects of electric current are: 1. Heating effect 2. Magnetic effect 3. Chemical effect.
• Heating effect.
• Magnetic effect.
• Chemical effect. Answer verified by Toppr. Related questions. In charging a battery of motor-car, which of the following effect of electric current is used?
## Which effect of current make a bulb glow?
Therefore, the effect of current responsible for the glow of the bulb in an electric circuit is the heating effect. So, the correct answer is “Option B”. Note: Due to the moving electrons in the filament of the bulb, a magnetic field is also produced resulting in some magnetic effects.
## Which current is used in electric bulb?
The voltage of electric source is 220 V. Calculate the value of current and the resistance.
Thank you.
IT IS INTERESTING: How much energy would you produce if you turned 3kg of mass entirely into energy?
How Is Emasculation Carried Out In Flowers Which Reagent Will Convert Rcor Group Into Rrc C 6 H 5 Oh
## How is heating effect of electric current is used in electric bulbs to produce light?
Electric bulb works on the principle of heating effect of electric current. When electric current passes through a very thin, high resistance tungsten filament of an electric bulb, the filament becomes white hot and emits light.
## What are 3 effects of electric current?
Hence, the three effects of electric current are heating effect, magnetic effect and chemical effect.
## What are the five effects of electric current?
The main effects are heating, chemical and magnetic effects. When current flows in a circuit it exhibits various effects. The main effects are heating, chemical and magnetic effects.
Effects of electric current
• Heating effect. …
• Chemical effect. …
• Magnetic effect of Electricity.
## What are the three main effects of an electric current?
When an electric current flows in a circuit it can have one or more of the following three effects: heating, magnetic or chemical.
## Does the bulb glow?
The filament of an electric bulb glows because of the heating effect of electric current. As the current passes through the bulb it heats the filament which gives off light. … The filament then heats up and becomes red-hot as a result of this it begins to glow, turning electrical energy into light energy.
## Why is the current in the circuit weak?
Why is the current in the circuit weak? Liquids like water (mineral water, or water with impurities), acids, bases, salts, etc offers high resistance to current than metal conductors. So, after passing through conducting liquids the current in the circuit becomes weak.
IT IS INTERESTING: Frequent question: What do I need to know about electricity?
## Why is the current in the circuit weak in some solution?
Liquids like water (mineral water, or water with impurities), acids, bases, salts etc offers high resistance to current than metal conductors. … So, after passing through conducting liquids the current in the circuit becomes weak.
## Is used in electric bulbs?
Incandescent light bulbs consist of an air-tight glass enclosure (the envelope, or bulb) with a filament of tungsten wire inside the bulb, through which an electric current is passed.
## Do light bulbs use AC or DC?
For light bulbs and heating elements, both AC and DC will work as long but the fact is that though all electric devices require direct current, alternating current is what is used to deliver electricity with the help of transformers which convert AC to DC for the use of electric devices.
## Do fluorescent lights use AC or DC?
Fluorescent tubes are almost always powered with AC, but this circuit uses DC. Basically, this circuit is a voltage doubler composed by two 1N4007 diodes and two high voltage electrolytic capacitors of 10 μF 350 V. Such capacitors can be easily salvaged from compact fluorescent lamps. | 859 | 4,135 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2021-39 | latest | en | 0.939492 |
http://allisons.org/ll/ProgLang/Comparative/ | 1,716,939,655,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971059160.88/warc/CC-MAIN-20240528220007-20240529010007-00777.warc.gz | 850,606 | 7,285 | ## Languages
LA home Computing Prog.Langs Compared Also see Algorithms glossary FP Logic
Algol-68 C Pascal Turing expressions general block structured, expression based language "flat", some expression based features block structured, imperative imperative, has `modules' arith +,-,*,/,^NB.% is int div +,-,*,/,%NB.% is int mod +,-,*,/, div, mod +,-,*,/, ** comparisons =, <>, <, <=, >, >= ==, !=, <, <=, >, >= =, <>, <, <=, >, >= =, not= <, <=, >, >= logical and, or, not &&, ||, ! and, or, not and, or, not conditional expression if e then e1[elsf e' then e2]else e3 fior (e | e1 | e2)also case exp'n e ? e1 : e2 n.a. n.a. assignment x:=e x=e; x:=e x:=e conditional if e then s1[elsf e' then s2][else s3] fi if(e) s1[else s2] if e then s1[else s2] if e then s1[elsf e' then s2][else s3] end if case case e in s1,s2,... out s esac switch(e) {case e1:s1 ...} case e of e1:s1;... end case e of label e1:s1;... end case for [for] [i] [from e1] [by e2] [to e3] [while eb] do ... od for(s1;eb;s2)s3 for i:=e1 (to | downto) e2 do s for [decreasing] i:rangesend for while see above while( ) while do loop ...exit when e;... end loop repeat n.a. n.a. repeat until e use loop group begin...end or (...) {...} begin...end begin...end function result final exp'n return e fnName=e result e constants T c=e #define const c=ce const c:=e variables T x [=e] T x var x:T var x:T subroutine proc( )void s void p( ) {...} procedure p(...) ... begin ... end procedure p(...) ... end function proc(...)T e T f(...) {...} function f(...)T ... begin ... end function f(...)T ...end function result final exp'n return e fnName=e result e parameters T x T x [var] x:T [var] x:T © L.Allison '99 data type mode typedef type type basic types int real bool char int float char integer real boolean char int real boolean char structure struct( ) struct( ) record end record end record field access f of s s.f s.f s.f pointer ref * ^ pointer to pointer access implicit or cast *,-> ptr & field ^ thecollection(x) special ptr value nil NULL nil nil(thecollection) union union(...) union record case variant union ... end union array array T1 of T2 T2 a[size] array [T1] of T2 array T1 of T2 array access a[i] a[i] a[i] a(i) array slicing m[i:j, ] rows i to j of 2D n.a. m[i] is row i of 2d n.a. sets n.a. n.a. (use bits) set of T set of T | 764 | 2,313 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.09375 | 3 | CC-MAIN-2024-22 | latest | en | 0.390521 |
https://stackoverflow.com/questions/2269827/how-to-convert-an-int-to-a-hex-string/2269863 | 1,716,291,403,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058442.20/warc/CC-MAIN-20240521091208-20240521121208-00020.warc.gz | 464,859,256 | 56,033 | # How to convert an int to a hex string?
I want to take an integer (that will be <= 255), to a hex string representation
e.g.: I want to pass in `65` and get out `'\x41'`, or `255` and get `'\xff'`.
I've tried doing this with the `struct.pack('c',`65`)`, but that chokes on anything above `9` since it wants to take in a single character string.
You are looking for the `chr` function.
You seem to be mixing decimal representations of integers and hex representations of integers, so it's not entirely clear what you need. Based on the description you gave, I think one of these snippets shows what you want.
``````>>> chr(0x65) == '\x65'
True
>>> hex(65)
'0x41'
>>> chr(65) == '\x41'
True
``````
Note that this is quite different from a string containing an integer as hex. If that is what you want, use the `hex` builtin.
• I thought chr converted integers to ascii. In which case chr(65) = 'A'. Is this a new addition? Nov 21, 2012 at 20:17
• @diedthreetimes, `chr` does not involve ASCII at all--it simply takes a number and makes a one-byte bytestring where the ordinal value of the byte is the number. ASCII and ASCII-compatible encodings come into play when you write and display bytestrings. Nov 22, 2012 at 14:08
• @diedthreetimes, For example, they come in by making `'A'` another way to write and display `'\x41'`. All `str` really cares about is the fact this is sixty-five. To make things understandable and usable by humans, it often becomes an A. Nov 22, 2012 at 14:10
This will convert an integer to a 2 digit hex string with the 0x prefix:
``````strHex = "0x%0.2X" % integerVariable
``````
• I suggest to edit the code here and change it to this: strHex = "0x%0.2X" % integerVariable. (I wasn't able to edit myself.) Apr 25, 2017 at 16:49
What about `hex()`?
``````hex(255) # 0xff
``````
If you really want to have `\` in front you can do:
``````print '\\' + hex(255)[1:]
``````
• `repr(chr(255)) # '\xff'` also achieves this Jun 15, 2016 at 2:27
Let me add this one, because sometimes you just want the single digit representation
( `x` can be lower, 'x', or uppercase, 'X', the choice determines if the output letters are upper or lower.):
``````'{:x}'.format(15)
> f
``````
And now with the new `f''` format strings you can do:
``````f'{15:x}'
> f
``````
To add 0 padding you can use `0>n`:
``````f'{2034:0>4X}'
> 07F2
``````
NOTE: the initial 'f' in `f'{15:x}'` is to signify a format string
• For reference on this neat feature, PEP 498 -- Literal String Interpolation
– MFT
Sep 22, 2018 at 4:05
• IMHO, a bit more "readable" option can be '0x{:X}'.format(15) (notice X is in uppercase). For 2034 it will print 0x7F2 instead of 7f2 Mar 4, 2020 at 4:53
Try:
``````"0x%x" % 255 # => 0xff
``````
or
``````"0x%X" % 255 # => 0xFF
``````
Python Documentation says: "keep this under Your pillow: http://docs.python.org/library/index.html"
• `"%#x" % 255` :) Feb 27, 2021 at 18:15
For Python >= 3.6, use f-string formatting:
``````>>> x = 114514
>>> f'{x:0x}'
'1bf52'
>>> f'{x:#x}'
'0x1bf52'
``````
• This should be more upvoted. f-strings for the win :-) Apr 24, 2022 at 9:14
If you want to pack a struct with a value <255 (one byte unsigned, uint8_t) and end up with a string of one character, you're probably looking for the format B instead of c. C converts a character to a string (not too useful by itself) while B converts an integer.
``````struct.pack('B', 65)
``````
(And yes, 65 is \x41, not \x65.)
The struct class will also conveniently handle endianness for communication or other uses.
• The amount of time it has taken me to find struct.pack('B', <integer>) is staggering. Thank you. Apr 7, 2018 at 20:37
With `format()`, as per format-examples, we can do:
``````>>> # format also supports binary numbers
>>> "int: {0:d}; hex: {0:x}; oct: {0:o}; bin: {0:b}".format(42)
'int: 42; hex: 2a; oct: 52; bin: 101010'
>>> # with 0x, 0o, or 0b as prefix:
>>> "int: {0:d}; hex: {0:#x}; oct: {0:#o}; bin: {0:#b}".format(42)
'int: 42; hex: 0x2a; oct: 0o52; bin: 0b101010'
``````
Note that for large values, `hex()` still works (some other answers don't):
``````x = hex(349593196107334030177678842158399357)
print(x)
``````
Python 2: `0x4354467b746f6f5f736d616c6c3f7dL`
Python 3: `0x4354467b746f6f5f736d616c6c3f7d`
For a decrypted RSA message, one could do the following:
``````import binascii
``````
• Well... no. I tried `hex(hash(text))` and it produced a negative integer which resulted in a hex string with minus sign. Who did implement this?
– JPT
Apr 5, 2021 at 13:14
• @JPT I don't understand your question. If the value is negative, then obviously the hex value will have a minus sign. Why would that be different when writing in hexadecimal, binary, decimal, or any other number system? Or are you looking for the in-memory representation, where there is a bit that determines the sign by being set or unset?
– Luc
Apr 5, 2021 at 13:17
• Well, the computer scientist is expecting negative hex to be something starting with a sign flag, so for example FFFx xxxx or 8000 xxxx, not an actual sign. I've never ever seen a hex with a minus sign before. But if you put it like that... ;)
– JPT
Apr 6, 2021 at 14:03
• @JPT the hext values you usually see are just a hex representation of a binary value. It is a matter of coding (Two's complement) that it is used in computer science. In the past different way of representing negative binary numbers were also used but Two's complement has biggest advantages. May 18, 2023 at 9:44
• @Filip exactly, that's why a HEX with minus sign is not an exact representation. It's an interpretation.
– JPT
May 19, 2023 at 12:05
``````(int_variable).to_bytes(bytes_length, byteorder='big'|'little').hex()
``````
For example:
``````>>> (434).to_bytes(4, byteorder='big').hex()
'000001b2'
>>> (434).to_bytes(4, byteorder='little').hex()
'b2010000'
``````
This worked best for me
``````"0x%02X" % 5 # => 0x05
"0x%02X" % 17 # => 0x11
``````
Change the (2) if you want a number with a bigger width (2 is for 2 hex printned chars) so 3 will give you the following
``````"0x%03X" % 5 # => 0x005
"0x%03X" % 17 # => 0x011
``````
Also you can convert any number in any base to hex. Use this one line code here it's easy and simple to use:
`hex(int(n,x)).replace("0x","")`
You have a string `n` that is your number and `x` the base of that number. First, change it to integer and then to hex but hex has `0x` at the first of it so with `replace` we remove it.
• `hex(int(n,x))[2:]` - not sure which is faster :) Jul 16, 2020 at 16:20
I wanted a random integer converted into a six-digit hex string with a # at the beginning. To get this I used
``````"#%6x" % random.randint(0xFFFFFF)
``````
• I suppose you mean `"#%6x" % random.randint(0x0, 0xFFFFFF)`. (There is a missing `%` before `6` and `randint` takes 2 parameters -lower and upper bounds-) Jul 12, 2012 at 11:22
• should be %06x if you don't want spaces Dec 6, 2012 at 23:00
Use f-strings, it's easy to parse and quite flexible:
``````print(f'{255:x}') # => 'ff'
print(f'{15:2x}') # => ' f'
print(f'{15:02x}') # => '0f'
print(f'\\x{15:02x}') # => '\x0f'
``````
As an alternative representation you could use
``````[in] '%s' % hex(15)
[out]'0xf'
`````` | 2,357 | 7,258 | {"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} | 2.75 | 3 | CC-MAIN-2024-22 | latest | en | 0.918557 |
http://www.algebra.com/cgi-bin/show-question-source.mpl?solution=68508 | 1,369,236,139,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368701910820/warc/CC-MAIN-20130516105830-00049-ip-10-60-113-184.ec2.internal.warc.gz | 318,138,674 | 1,117 | ```Question 94092
Question:
Find the circumference of a circle with the area 36pi square centimeters?
Area of circle is given by the formula, A = {{{pie (r^2)}}}, where A is the area and r is the radius of the circle.
Here, A = 36 pie
==> {{{36(pie) = pie (r^2)}}}
Divide both sides of the expression by {{{ pie }}}
==> {{{36(pie)/(pie) = pie (r^2)/ (pie)}}}
==> {{{36 = r^2}}}
Take square root on both sides...
==> {{{ sqrt (36) = sqrt( r^2)}}}
==> r = 6 or -6
Since radius can never be negative, you can take r = 6
So, radius = 6
Now circumference of a circle is given by the formula, {{{ 2(pie) r}}}
So circumference = {{{ 2(pie) 6}}}
==> {{{ 12(pie) }}}
--------------------------------------------
If you put the approximate value of pie ( its {{{3.14}}} or {{{22/7}}} you will get
Circumference = {{{12*(3.14)}}}
That is, circumference = {{{37.68}}}
Hope you found the explanation useful.
Regards.
Praseena.
``` | 294 | 955 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.15625 | 4 | CC-MAIN-2013-20 | latest | en | 0.710896 |
https://www.convertunits.com/from/hectare+meter/hour/to/barrel/second | 1,669,820,561,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710764.12/warc/CC-MAIN-20221130124353-20221130154353-00621.warc.gz | 776,457,356 | 24,140 | ## ››Convert hectare metre/hour to barrel/second [petroleum]
hectare meter/hour barrel/second
Did you mean to convert hectare meter/hour to barrel/second [petroleum] barrel/second [US] barrel/second [US beer/wine] barrel/second [UK]
How many hectare meter/hour in 1 barrel/second? The answer is 0.057235426416.
We assume you are converting between hectare metre/hour and barrel/second [petroleum].
You can view more details on each measurement unit:
hectare meter/hour or barrel/second
The SI derived unit for volume flow rate is the cubic meter/second.
1 cubic meter/second is equal to 0.36 hectare meter/hour, or 6.2898107438466 barrel/second.
Note that rounding errors may occur, so always check the results.
Use this page to learn how to convert between hectare meters/hour and barrels/second.
Type in your own numbers in the form to convert the units!
## ››Quick conversion chart of hectare meter/hour to barrel/second
1 hectare meter/hour to barrel/second = 17.4717 barrel/second
2 hectare meter/hour to barrel/second = 34.94339 barrel/second
3 hectare meter/hour to barrel/second = 52.41509 barrel/second
4 hectare meter/hour to barrel/second = 69.88679 barrel/second
5 hectare meter/hour to barrel/second = 87.35848 barrel/second
6 hectare meter/hour to barrel/second = 104.83018 barrel/second
7 hectare meter/hour to barrel/second = 122.30188 barrel/second
8 hectare meter/hour to barrel/second = 139.77357 barrel/second
9 hectare meter/hour to barrel/second = 157.24527 barrel/second
10 hectare meter/hour to barrel/second = 174.71697 barrel/second
## ››Want other units?
You can do the reverse unit conversion from barrel/second to hectare meter/hour, 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! | 597 | 2,238 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.046875 | 3 | CC-MAIN-2022-49 | latest | en | 0.85256 |
https://securityboulevard.com/tag/domain-3/ | 1,723,732,429,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641299002.97/warc/CC-MAIN-20240815141847-20240815171847-00542.warc.gz | 390,120,360 | 41,298 | #### Boolean Math (NOT Logic) – CISSP Domain 3
| |
Hello everyone. We’ve got another Boolean math session lined up for you today. This time we’re going to take a quick look at the NOT logic and examine how this one differs ...
#### Boolean Math (NOT Logic) – CISSP Domain 3
| |
Hello everyone. We’ve got another Boolean math session lined up for you today. This time we’re going to take a quick look at the NOT logic and examine how this one differs ...
#### Boolean Math (XOR Logic) – CISSP Domain 3
| |
Hello everyone. We’ve got another Boolean math session to look over today. Our focus this time will be on the XOR logic. The XOR stands for exclusive OR, and we will go ...
#### Boolean Math (XOR Logic) – CISSP Domain 3
| |
Hello everyone. We’ve got another Boolean math session to look over today. Our focus this time will be on the XOR logic. The XOR stands for exclusive OR, and we will go ...
#### Boolean Math (AND Logic) – CISSP Domain 3
| |
Today we’re going to take a quick look at the AND Boolean logic, which is covered in Domain 3 of the CISSP common body of knowledge (CBK). To begin with, Boolean math ...
#### Encryption – CISSP Domain 3
| |
We’re circling back to some more CISSP-related materials. Today’s topic will be encryption, which can be found in CISSP Domain 3. By its very nature, encryption is meant to hide the meaning ...
#### Encoding – CISSP Domain 3
| |
Today we’re going to take a quick look at encoding, as covered in Domain 3 of the CISSP common body of knowledge (CBK). There is often some confusion between encoding and encryption, ...
Application Security Check Up | 407 | 1,631 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.5625 | 3 | CC-MAIN-2024-33 | latest | en | 0.817845 |
https://dapzoi.com/topic/quantities-and-units-mcq-questions-answers | 1,653,473,290,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662584398.89/warc/CC-MAIN-20220525085552-20220525115552-00214.warc.gz | 246,478,366 | 4,284 | Quantities and Units MCQ Questions & Answers
Quantities and Units MCQs : This section focuses on the "Quantities and Units". These Multiple Choice Questions (MCQs) should be practiced to improve the Quantities and Units skills required for various interviews (campus interview, walk-in interview, company interview), placement, entrance exam and other competitive examinations.
Question 1
If you drop a 5 when rounding a number, you are using the
A. round-to-even rule
B. significant digit rule
C. round-off rule
D. retained digit rule
View Answer
Question 2
Which of the following metric prefixes could replace 10–9?
A. Nano
B. Mega
C. Kilo
D. Micro
View Answer
Question 3
Another name for "fundamental units" is
A. base units
B. atoms
C. the metric system
D. letter symbols
View Answer
Question 4
When using the terms "accuracy" and "precision" for measurements
A. "precision" implies less measurement error than "accuracy"
B. "accuracy" implies less measurement error than "precision"
C. "precision" measures the repeatability of a measurement
D. both terms mean the same thing
View Answer
Question 5
The unit for frequency is the
A. hertz
B. ampere
C. watt
D. second
View Answer
Question 6
Adding 27.5 × 103 to 8.9 × 104 equals
A. 36.4 × 107
B. 116.5 × 104
C. 28.39 × 103
D. 1.165 × 105
View Answer
Question 7
A measure of the repeatability of a measurement of some quantity is
A. error
B. precision
C. accuracy
D. significant
View Answer
Question 8
Derived units are obtained from various combinations of
A. electrical quantities
B. fundamental units
C. metric prefixes
D. international standards
View Answer
Question 9
Which of the following is expressed in engineering notation?
A. 470 × 105
B. 82 × 10–2
C. 9.1 × 10–6
D. 14.7 × 108
View Answer
Question 10
7200 mV is the same as
A. 7.2 V
B. 7.2 V
C. 7,200,000 V
D. 0.0072 V
View Answer
Question 11
Dividing 24 × 1011 by 3 × 104 equals
A. 8 × 104
B. 8 × 107
C. 8 × 1011
D. 8 × 1015
View Answer
Question 12
The difference between scientific and engineering notation is
A. powers of ten representation
B. single vs. multiple digits before decimal point
C. groupings of multiples of three digits
D. All of the above
View Answer
Question 13
Scientific notation is a method
A. of expressing a very large number
B. of expressing a very small number
C. used to make calculations with large and small numbers
D. All of the above
View Answer
Question 14
The digits in a measured number that are known to be correct are called
A. accuracy digits
B. significant digits
C. error digits
D. precision digits
View Answer
Question 15
Pico is what relation to micro?
A. one-tenth
B. one-hundredth
C. one-thousandth
D. one-millionth
View Answer
Question 16
The number 14.8 can also be expressed as
A. 1.48 × 10–1
B. 1.48 × 100
C. 1.48 × 101
D. 1.48 × 102
View Answer | 848 | 2,870 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.234375 | 3 | CC-MAIN-2022-21 | longest | en | 0.826829 |
https://www.ordinalnumbers.com/ordinals-numbers-from-1-to-100-in-spanish/ | 1,726,348,201,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651580.74/warc/CC-MAIN-20240914193334-20240914223334-00210.warc.gz | 863,824,895 | 15,705 | Ordinals Numbers From 1 To 100 In Spanish
It is possible to enumerate infinite sets using ordinal numbers. They also can be used to generalize ordinal quantities.
1st
The ordinal number is one the fundamental concepts in math. It is a number that identifies the location of an object within an array. Ordinal numbers are a number between 1 and 20. Ominal numbers have many uses but are most commonly utilized to signify the order of items on an agenda.
Charts, words, and numbers are all able to represent ordinal numbers. These are also useful for indicating how a set of pieces are arranged.
The majority of ordinal numbers fall into one or more of these categories. The infinite ordinal numbers are represented with lowercase Greek letters, whereas finite ones are represented using Arabic numbers.
A well-ordered collection should include at least one ordinal according to the axiom. For example, the top score would be awarded to the first student in the class to receive it. The student who received the highest score was declared the winner of the contest.
Combinational ordinal numbers
Multidigit numbers are referred to as compound ordinal numbers. They can be created by multiplying an ordered number by its last character. These numbers are most commonly used for ranking and dating purposes. They do not have a unique ending for the last digit , like cardinal numbers do.
Ordinal numbers indicate the order in which elements are located in the collection. The names of the elements in a collection are also given using the numbers. The two kinds of regular numbers are flexible and regular.
Regular ordinals are created by prefixing a cardinal number with the suffix -u. Then, the number has to be entered in words and then a colon is added. There are also additional suffixes. The suffix “nd” can be used to signify numbers that end in two numbers. The suffix “th” could refer to numbers that end with 4 and 9.
Suppletive ordinals are made by prefixing words with -u, -e, or -ie. This suffix can be used to count words and is bigger than the standard.
ordinal limit
The limit for ordinal numbers that are not zero is an ordinal number that’s not zero. Limit ordinal numeric numbers suffer from the disadvantage that they do not have a maximum element. These can be formed by joining sets that have no element that is larger than.
Limits on ordinal numbers may also be used to describe the description of recursion in transfinite terms. The von Neumann model says that every infinite cardinal numbers is also an ordinal number.
A number with limits is equal to the sum of all the numbers below. Limit ordinal numbers are determined using arithmetic, however they can also be expressed in a series of natural numbers.
The data are arranged in order using ordinal numbers. They give an explanation about an object’s numerical location. They are employed in set theory and arithmetic contexts. Despite having the same structure as natural numbers, they are not classified in the same way.
The von Neumann method uses a well-ordered list. Assume that fy fy is one of the subfunctions of the function g’ which is specified as a singular function. If fy is the only subfunction (ii), it must be able to meet the requirements.
The Church-Kleene oral is a limit order similarly. The Church-Kleene ordinal defines an ordinal that is a limit as a properly arranged set of smaller ordinals, and it has a non-zero ordinal.
Stories that include examples of ordinal numbers
Ordinal numbers are typically used to indicate the order of things between objects and entities. They are essential for organising, counting and ranking purposes. They are also useful to show the order of things as well as to define objects’ positions.
Ordinal numbers are generally identified with the letter “th”. Sometimes the letter “nd”, can be substituted. Book titles typically contain ordinal numbers.
Even though ordinal figures are generally employed in lists it is possible to write them down in the form of words. They may also take the form of acronyms and numbers. These numbers are simpler to understand than cardinal numbers, however.
Ordinary numbers come in three distinct flavours. Through games and exercises, you might discover more about these numbers. Learning about the subject is an important aspect of improving your arithmetic abilities. Coloring exercises are an enjoyable and straightforward way to improve. Review your work using an easy-to-use mark sheet. | 885 | 4,483 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.25 | 4 | CC-MAIN-2024-38 | latest | en | 0.951021 |
https://www.justintools.com/unit-conversion/time.php?k1=sidereal-days&k2=dog-years | 1,726,120,070,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651422.16/warc/CC-MAIN-20240912043139-20240912073139-00462.warc.gz | 797,798,911 | 27,037 | Please support this site by disabling or whitelisting the Adblock for "justintools.com". I've spent over 10 trillion microseconds (and counting), on this project. This site is my passion, and I regularly adding new tools/apps. Users experience is very important, that's why I use non-intrusive ads. Any feedback is appreciated. Thank you. Justin XoXo :)
# TIME Units Conversionsidereal-days to dog-years
1 Sidereal Days
= 0.00039032081249547 Dog Years
Category: time
Conversion: Sidereal Days to Dog Years
The base unit for time is seconds (SI Unit)
[Sidereal Days] symbol/abbrevation: (Sd)
[Dog Years] symbol/abbrevation: (dog yrs)
How to convert Sidereal Days to Dog Years (Sd to dog yrs)?
1 Sd = 0.00039032081249547 dog yrs.
1 x 0.00039032081249547 dog yrs = 0.00039032081249547 Dog Years.
Always check the results; rounding errors may occur.
Definition:
The dog year, primarily used to approximate the equivalent age of dogs and other animals with similar life spans. Both are based upon a popular myth regarding the aging of dog ..more definition+
In relation to the base unit of [time] => (seconds), 1 Sidereal Days (Sd) is equal to 86164.1 seconds, while 1 Dog Years (dog yrs) = 220752000 seconds.
1 Sidereal Days to common time units
1 Sd = 86164.1 seconds (s)
1 Sd = 1436.0683333333 minutes (min)
1 Sd = 23.934472222222 hours (hr)
1 Sd = 0.99726967592593 days (day)
1 Sd = 0.14246709656085 weeks (wk)
1 Sd = 0.0027322456874683 years (yr)
1 Sd = 0.032786948249619 months (mo)
1 Sd = 0.000273189917565 decades (dec)
1 Sd = 2.73189917565E-5 centuries (cent)
1 Sd = 2.73189917565E-6 millenniums (mill)
Sidereal Daysto Dog Years (table conversion)
1 Sd = 0.00039032081249547 dog yrs
2 Sd = 0.00078064162499094 dog yrs
3 Sd = 0.0011709624374864 dog yrs
4 Sd = 0.0015612832499819 dog yrs
5 Sd = 0.0019516040624774 dog yrs
6 Sd = 0.0023419248749728 dog yrs
7 Sd = 0.0027322456874683 dog yrs
8 Sd = 0.0031225664999638 dog yrs
9 Sd = 0.0035128873124592 dog yrs
10 Sd = 0.0039032081249547 dog yrs
20 Sd = 0.0078064162499094 dog yrs
30 Sd = 0.011709624374864 dog yrs
40 Sd = 0.015612832499819 dog yrs
50 Sd = 0.019516040624774 dog yrs
60 Sd = 0.023419248749728 dog yrs
70 Sd = 0.027322456874683 dog yrs
80 Sd = 0.031225664999638 dog yrs
90 Sd = 0.035128873124592 dog yrs
100 Sd = 0.039032081249547 dog yrs
200 Sd = 0.078064162499094 dog yrs
300 Sd = 0.11709624374864 dog yrs
400 Sd = 0.15612832499819 dog yrs
500 Sd = 0.19516040624774 dog yrs
600 Sd = 0.23419248749728 dog yrs
700 Sd = 0.27322456874683 dog yrs
800 Sd = 0.31225664999638 dog yrs
900 Sd = 0.35128873124592 dog yrs
1000 Sd = 0.39032081249547 dog yrs
2000 Sd = 0.78064162499094 dog yrs
4000 Sd = 1.5612832499819 dog yrs
5000 Sd = 1.9516040624774 dog yrs
7500 Sd = 2.927406093716 dog yrs
10000 Sd = 3.9032081249547 dog yrs
25000 Sd = 9.7580203123868 dog yrs
50000 Sd = 19.516040624774 dog yrs
100000 Sd = 39.032081249547 dog yrs
1000000 Sd = 390.32081249547 dog yrs
1000000000 Sd = 390320.81249547 dog yrs
(Sidereal Days) to (Dog Years) conversions
:) | 1,104 | 3,018 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.09375 | 3 | CC-MAIN-2024-38 | latest | en | 0.744467 |
https://www.shaalaa.com/question-bank-solutions/an-object-kept-principal-axis-concave-mirror-focal-length-10-cm-distance-15-cm-its-pole-image-formed-mirror-is-ray-optics-mirror-formula_19414 | 1,576,273,048,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575540569146.17/warc/CC-MAIN-20191213202639-20191213230639-00328.warc.gz | 856,430,438 | 11,332 | Share
An Object is Kept on the Principal Axis of a Concave Mirror of Focal Length 10 Cm. at a Distance of 15 Cm from Its Pole. the Image Formed by the Mirror Is: - ISC (Arts) Class 12 - Physics (Theory)
ConceptRay Optics - Mirror Formula
Question
An object is kept on the principal axis of a concave mirror of focal length 10 cm. at a distance of 15
cm from its pole. The image formed by the mirror is:
(a) Virtual and magnified
(b) Virtual and diminished
(c) Real and magnified
(d) Real and diminished
Solution
Real and magnified
1/u + 1/v =1/f
u = -15, f = -10
:. v = -30
Is there an error in this question or solution?
Video TutorialsVIEW ALL [2]
Solution An Object is Kept on the Principal Axis of a Concave Mirror of Focal Length 10 Cm. at a Distance of 15 Cm from Its Pole. the Image Formed by the Mirror Is: Concept: Ray Optics - Mirror Formula.
S | 243 | 870 | {"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} | 2.875 | 3 | CC-MAIN-2019-51 | longest | en | 0.861659 |
http://myriverside.sd43.bc.ca/lucianl2014/2017/06/05/journal-assignment-2-and-3/ | 1,563,737,612,000,000,000 | text/html | crawl-data/CC-MAIN-2019-30/segments/1563195527196.68/warc/CC-MAIN-20190721185027-20190721211027-00547.warc.gz | 105,226,895 | 7,527 | # Journal assignment #2 and #3
Page 567
There is no X in this equation(y=2). So we just need to draw a line that parallel to the X-axis. Then we can know what y=2 looks like.
For the question e we can let x=0 first and find a point. Then let y=0 and find a point. Draw a line at two points. Then we can know what y=2x-4 looks like.
Page 605
X runs 4. Y runs 2.25-2.85=-0.6. So the slope is -0.6/4=-0.15. So this function is y=-0.15+2.85. | 151 | 442 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2019-30 | latest | en | 0.902544 |
http://www.javaprogrammingforums.com/%20whats-wrong-my-code/26871-union-coding-hashmap-printingthethread.html | 1,529,468,329,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267863411.67/warc/CC-MAIN-20180620031000-20180620051000-00433.warc.gz | 441,324,306 | 1,952 | Union Coding Hashmap
• April 2nd, 2013, 02:23 AM
codingjava
Union Coding Hashmap
We have two map
map m1 = { a= [1], b=[2], c=[3] }
map m2 = { a= [3] , b = [4] , d = [5]}
output
m3 = { a = [1,3 ] , b = [2,4] c= [3] , d=[5] } | 109 | 227 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2018-26 | latest | en | 0.388491 |
https://community.fabric.microsoft.com/t5/Desktop/Rank/m-p/688757 | 1,716,031,714,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971057379.11/warc/CC-MAIN-20240518085641-20240518115641-00186.warc.gz | 162,981,815 | 117,301 | cancel
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Frequent Visitor
Rank
Hi,
I have a table, Table1, and I want to rank multiple meters by End date in the following format.
Can you help?
Meter End Date Rank
1000364810 2008 06 1
1000364810 2009 06 2
1000364810 2012 06 3
1000364810 2015 06 4
1000364810 2020 04 5
1 ACCEPTED SOLUTION
Community Champion
Hi @PMF99
=VAR m = Resign_Append[meter]
RETURN
RANKX(
FILTER(Resign_Append,
Resign_Append[meter] = m
),
Resign_Append[CED YYYY MM],,
ASC
)
Regards,
Mariusz
If this post helps, then please consider Accept it as the solution to help the other members find it more quickly.
7 REPLIES 7
Community Champion
Hi @PMF999,
You can add a column and the code below.
Rank Meter by End Date =
VAR m = Table1[Meter]
RETURN
RANKX(
FILTER(
Table1,
Table1[Meter] = m
),
Table1[End Date],,
ASC
)
Regards,
Mariusz
If this post helps, then please consider Accept it as the solution to help the other members find it more quickly.
Regular Visitor
Many thanks Mariusz;
=VAR m = Resign_Append[Renewal Date]
RETURN
RANKX(
FILTER(Resign_Append,
Resign_Append[CED YYYY MM] = m
),
Resign_Append[CED YYYY MM],,
ASC
)
And here's the error I've got;
DAX comparison operations do not support comparing values of type Text with values of type Date. Consider using the VALUE or FORMAT function to convert one of the values.
What have I done wrong?
Thanks
Peter
Community Champion
Hi @PMF99,
Try,
=VAR m = Resign_Append[Renewal Date]
RETURN
RANKX(
FILTER(Resign_Append,
Resign_Append[Renewal Date] = m
),
Resign_Append[CED YYYY MM],,
ASC
)
Regards,
Mariusz
If this post helps, then please consider Accept it as the solution to help the other members find it more quickly.
Regular Visitor
Hi Mariusz,
Think we're almost there...
Here is my code....
=VAR m = Resign_Append[CED YYYY MM]
RETURN
RANKX(
FILTER(Resign_Append,
Resign_Append[CED YYYY MM] = m
),
Resign_Append[CED YYYY MM],,
ASC
)
And here is the pivot results under Resign Order
Getting only '1s'
Any ideas? I would expect meter 1000364810 to have a resigner order for CED YYYY MM of 1, 2, 3, 4, 5,
Community Champion
Hi @PMF99
=VAR m = Resign_Append[meter]
RETURN
RANKX(
FILTER(Resign_Append,
Resign_Append[meter] = m
),
Resign_Append[CED YYYY MM],,
ASC
)
Regards,
Mariusz
If this post helps, then please consider Accept it as the solution to help the other members find it more quickly.
Regular Visitor
I've ticked the thumbs up box.
Is there another 'Accept' button?
Super User
Hi @PMF999,
If 'End Date' is of type date, you can simply do:
Rank_Column =
RANKX ( CALCULATETABLE(Table1, ALLEXCEPT(Table1, Table1[Meter])), Table1[End Date], Table1[End Date], ASC )
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Top Kudoed Authors | 922 | 3,200 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-22 | latest | en | 0.622088 |
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## The correct Answer is:(5)
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Step by step video, text & image solution for A die is rolled thrice, find the probability of getting a larger number each time than the previous number. by Maths experts to help you in doubts & scoring excellent marks in Class 12 exams.
Updated on:21/07/2023
### Knowledge Check
• Question 1 - Select One
## A die is rolled three times. The probability of getting a larger number than the previous number each time is
A572
B554
C13216
D118
• Question 2 - Select One
## A die is rolled. Find the probability of getting a cube number
A536
B736
C29
D16
• Question 3 - Select One
## An unbiased die is rolled find the probability of getting a prime number.
A12
B35
C57
D58
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Question
# A simply supported beam of T-section is subjected to a uniformly distributed load acting vertically downwards. Its neutral axis is located at 25 mm from the top of the flange and the total depth of the section is 100 mm. The ratio of maximum tensile stress to maximum compressive stress in the beam is
A
2.0
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B
2.5
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C
3.0
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D
4.0
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Solution
## The correct option is C 3.0Let the maximum tensile stress and maximum compressive stress be ft and fc respectively. The distribution of stresses in the beam is shown below. From the rule of similar triangles, we have ft100−25=fc25 ⇒ftfc=7525=3
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Texas Instruments Ti-89 Titanium Manual
14/10/2016 · Ti 89 Titanium User Manual Eddie Grant. Loading TI-89 Titanium Basic Tutorial 1.wmv - Duration: 9:03. ultimatemath 90,526 views. 9:03....
Getting Started with your TI-89 for Statistics
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Statistically speaking, if you’ve held a graphing calculator, it was probably made by Texas Instruments. The TI-89 Titanium continues this legacy with more memory! TI-89 Titanium. Voyage™ 200 Calculadora Grafica Importante Texas Instruments no ofrece garantia alguna, ya sea explicita o implicita, incluidas, sin.
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Which Winch? Premier Winch. TJM Products. 12/05/2013 · viper max winch installation instructions, viper max winch wiring instructions, viper winch instructions, viper winch on ranger, vipor winch installed in a ranger 900. viper winch installation instructions Find great deals on eBay for Viper Winch 5000 in Winches. Shop with confidence. Skip to main mounting hardware and installation instructions. • 1 Year Warranty.. | 697 | 2,904 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2019-51 | latest | en | 0.70298 |
https://www.physicsforums.com/threads/1st-law-of-thermodynamics.712681/ | 1,477,581,828,000,000,000 | text/html | crawl-data/CC-MAIN-2016-44/segments/1476988721347.98/warc/CC-MAIN-20161020183841-00031-ip-10-171-6-4.ec2.internal.warc.gz | 971,421,467 | 16,486 | # 1st law of thermodynamics
1. Sep 25, 2013
### curious bishal
On the way to prove 1st law of thermodynamics, you consider a system having initial energy E1. Then, you supply heat q to the system. Some part of the applied heat is used for doing work.
Then, again you consider the final energy (E2) of the system to be:
E2=E1+q+w
where w is the work done.
We all very well know that the work is done by the heat applied. If we add both heat applied and work done to the final energy, isn't the work done counted twice. I mean to say that the heat energy wasted in the work done is also consisted in q so why is there need to add w in the final energy E2.
2. Sep 26, 2013
### jfizzix
I think the confusion here comes from whether w is the work done on the system, or the work done by the system. These change the energy by opposite amounts.
If the system is doing work w, the energy of the system decreases by w, so
E2=E1+q-w
Then, if q=w,
E2=E1
3. Sep 27, 2013
### curious bishal
It doesn't give the solution to my problem. Why is the work done counted double?
4. Sep 30, 2013
### jfizzix
The work is not double-counted.
If the system starts at energy E1, and you add heat q to it, the energy has changed to amount E1'= E1+q.
If the system then performs work w on its environment, it loses w units of energy, so its energy changes to E2 = E1'-w = E1+q-w.
Though some of the heat is converted into work, that doesn't affect the fact that q units of heat were added, and w units of work are extracted. Thus the net energy change is q-w. | 428 | 1,552 | {"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} | 3.375 | 3 | CC-MAIN-2016-44 | longest | en | 0.956325 |
https://oeis.org/A007591 | 1,620,823,495,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243989693.19/warc/CC-MAIN-20210512100748-20210512130748-00174.warc.gz | 461,568,119 | 4,130 | The OEIS Foundation is supported by donations from users of the OEIS and by a grant from the Simons Foundation.
Hints (Greetings from The On-Line Encyclopedia of Integer Sequences!)
A007591 Numbers k such that k^2 + 4 is prime. (Formerly M2416) 28
1, 3, 5, 7, 13, 15, 17, 27, 33, 35, 37, 45, 47, 57, 65, 67, 73, 85, 87, 95, 97, 103, 115, 117, 125, 135, 137, 147, 155, 163, 167, 177, 183, 193, 203, 207, 215, 217, 233, 235, 243, 245, 253, 255, 265, 267, 275, 277, 287, 293, 303, 307, 313, 317, 347, 357, 373, 375 (list; graph; refs; listen; history; text; internal format)
OFFSET 1,2 REFERENCES N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence). LINKS Vincenzo Librandi, Table of n, a(n) for n = 1..1000 Eric Weisstein's World of Mathematics, Near-Square Prime MATHEMATICA lst={}; Do[If[PrimeQ[n^2+4], Print[n]; AppendTo[lst, n]], {n, 10^5}]; lst (* Vladimir Joseph Stephan Orlovsky, Aug 21 2008 *) Select[Range[0, 400], PrimeQ[#^2 + 4] &] (* Vincenzo Librandi, Sep 25 2012 *) PROG (MAGMA) [n: n in [0..400] | IsPrime(n^2+4)]; // Vincenzo Librandi, Nov 18 2010 (PARI) select(n->isprime(n^2+4), vector(500, n, 2*n-1)) \\ Charles R Greathouse IV, Sep 25 2012 CROSSREFS Other sequences of the type "Numbers k such that k^2 + i is prime": A005574 (i=1), A067201 (i=2), A049422 (i=3), this sequence (i=4), A078402 (i=5), A114269 (i=6), A114270 (i=7), A114271 (i=8), A114272 (i=9), A114273 (i=10), A114274 (i=11), A114275 (i=12). Sequence in context: A285130 A024909 A063241 * A097687 A032911 A157834 Adjacent sequences: A007588 A007589 A007590 * A007592 A007593 A007594 KEYWORD nonn,easy AUTHOR STATUS approved
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Last modified May 12 07:28 EDT 2021. Contains 343821 sequences. (Running on oeis4.) | 763 | 2,009 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.328125 | 3 | CC-MAIN-2021-21 | latest | en | 0.56534 |
https://docs.geotools.org/latest/javadocs/org/opengis/geometry/coordinate/Conic.html | 1,656,978,282,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656104506762.79/warc/CC-MAIN-20220704232527-20220705022527-00384.warc.gz | 250,111,660 | 5,031 | org.opengis.geometry.coordinate
## Interface Conic
• All Superinterfaces:
CurveSegment, GenericCurve
```@UML(identifier="GM_Conic",
specification=ISO_19107)
public interface Conic
extends CurveSegment```
Any general conic curve. Any of the conic section curves can be canonically represented in polar co-ordinates (ρ, φ) as:
where "P" is semi-latus rectum and "e" is the eccentricity. This gives a conic with focus at the pole (origin), and the vertex on the conic nearest this focus in the direction of the polar axis, φ=0.
For e=0, this is a circle. For 0 < e < 1, this is an ellipse. For e=1, this is a parabola. For e>1, this is one branch of a hyperbola.
These generic conics can be viewed in a two-dimensional Cartesian parameter space (uv) given by the usual coordinate conversions u=ρcos(φ) and v=ρsin(φ). We can then convert this to a 3D coordinate reference system by using an affine transformation, (uv) → (xyz) which is defined by:
(TODO: paste the matrix here, same as AffinePlacement)
This gives us φ as the constructive parameter. The direct position given by (x0, y0, z0) is the image of the origin in the local coordinate space (u, v) Alternatively, the origin may be shifted to the vertex of the conic as
u' = ρcos(φ) - P/(1 + e) and v' = ρsin(φ)
and v can be used as the constructive parameter. In general, conics with small eccentricity and small P, use the first or "central" representation. Those with large eccentricity or large P tend to use the second or "linear" representation.
Since:
GeoAPI 1.0
Author:
Martin Desruisseaux (IRD)
• ### Method Summary
All Methods
Modifier and Type Method and Description
`double` `getEccentricity()`
Returns the value of the eccentricity parameter "e" used in the defining equation above.
`double` `getEndConstructiveParam()`
Return the end point parameter used in the constructive paramerization.
`AffinePlacement` `getPosition()`
Returns an affine transformation object that maps the conic from parameter space into the coordinate space of the target coordinate reference system of the conic corresponding to the coordinate reference system of the Geometry.
`double` `getSemiLatusRectum()`
Returns the value of the parameter "P" used in the defining equation above.
`double` `getStartConstructiveParam()`
Return the start point parameter used in the constructive paramerization.
`boolean` `isShifted()`
Returns `false` if the affine transformation is used on the unshifted (u, v) and `true` if the affine transformation is applied to the shifted parameters (u', v').
• ### Methods inherited from interface CurveSegment
`getBoundary, getCurve, getInterpolation, getNumDerivativesAtEnd, getNumDerivativesAtStart, getNumDerivativesInterior, getSamplePoints, reverse`
• ### Methods inherited from interface GenericCurve
`asLineString, forConstructiveParam, forParam, getEndParam, getEndPoint, getParamForPoint, getStartParam, getStartPoint, getTangent, length, length`
• ### Method Detail
• #### getPosition
```@UML(identifier="position",
obligation=MANDATORY,
specification=ISO_19107)
AffinePlacement getPosition()```
Returns an affine transformation object that maps the conic from parameter space into the coordinate space of the target coordinate reference system of the conic corresponding to the coordinate reference system of the Geometry. This affine transformation is given by the formulae in the class description.
• #### isShifted
```@UML(identifier="shifted",
obligation=MANDATORY,
specification=ISO_19107)
boolean isShifted()```
Returns `false` if the affine transformation is used on the unshifted (u, v) and `true` if the affine transformation is applied to the shifted parameters (u', v'). This controls whether the focus or the vertex of the conic is at the origin in parameter space.
• #### getEccentricity
```@UML(identifier="eccentricity",
obligation=MANDATORY,
specification=ISO_19107)
double getEccentricity()```
Returns the value of the eccentricity parameter "e" used in the defining equation above. It controls the general shape of the curve, determining whether the curve is a circle, ellipse, parabola, or hyperbola.
• #### getSemiLatusRectum
```@UML(identifier="semiLatusRectum",
obligation=MANDATORY,
specification=ISO_19107)
double getSemiLatusRectum()```
Returns the value of the parameter "P" used in the defining equation above. It controls how broad the conic is at each of its foci.
• #### getStartConstructiveParam
```@UML(identifier="startConstrParam",
obligation=MANDATORY,
specification=ISO_19107)
double getStartConstructiveParam()```
Return the start point parameter used in the constructive paramerization. The following relation must be hold:
```forConstructiveParam(getStartConstructiveParam()) .equals( getStartPoint() )```
``` There is no assumption that the start constructive parameter is less than the end constructive parameter, but the parameterization must be strictly monotonic (strictly increasing, or strictly decreasing).```
``` Specified by: getStartConstructiveParam in interface GenericCurve Returns: The parameter used in the constructive paramerization for the start point. See Also: GenericCurve.getStartParam(), GenericCurve.getEndConstructiveParam(), GenericCurve.forConstructiveParam(double) ```
• ``` ```
``` getEndConstructiveParam @UML(identifier="endConstrParam", obligation=MANDATORY, specification=ISO_19107) double getEndConstructiveParam() Return the end point parameter used in the constructive paramerization. The following relation must be hold: forConstructiveParam(getEndConstructiveParam()) .equals( getEndPoint() ) There is no assumption that the start constructive parameter is less than the end constructive parameter, but the parameterization must be strictly monotonic (strictly increasing, or strictly decreasing). Specified by: getEndConstructiveParam in interface GenericCurve Returns: The parameter used in the constructive paramerization for the end point. See Also: GenericCurve.getEndParam(), GenericCurve.getStartConstructiveParam(), GenericCurve.forConstructiveParam(double) ```
• ``` ```
``` ```
• ``` ```
``` ```
``` ```
``` Skip navigation links Overview Package Class Use Tree Deprecated Index Help Prev Class Next Class Frames No Frames All Classes <!-- allClassesLink = document.getElementById("allclasses_navbar_bottom"); if(window==top) { allClassesLink.style.display = "block"; } else { allClassesLink.style.display = "none"; } //--> Summary: Nested | Field | Constr | Method Detail: Field | Constr | Method Copyright © 1996–2022 Geotools. All rights reserved. ``` | 1,568 | 6,557 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2022-27 | latest | en | 0.745 |
http://gmatclub.com/forum/over-the-last-2-years-chesapeake-introduced-a-slew-52412.html?fl=similar | 1,485,033,478,000,000,000 | text/html | crawl-data/CC-MAIN-2017-04/segments/1484560281202.94/warc/CC-MAIN-20170116095121-00397-ip-10-171-10-70.ec2.internal.warc.gz | 123,861,952 | 55,050 | Over the last 2 years , Chesapeake introduced a slew : GMAT Sentence Correction (SC)
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# Over the last 2 years , Chesapeake introduced a slew
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Over the last 2 years , Chesapeake introduced a slew [#permalink]
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17 Sep 2007, 03:39
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Over the last 2 years , Chesapeake introduced a slew blockbuster drugs, one of them is pialis, the wonder drug for fatigue and generalized body pains.
1. one of them
2. one of which
3. one which
4. one of them which
5. and one of them which
If you have any questions
New!
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Re: SC-- Chesapeake introduced a slew blockbuster drugs [#permalink]
### Show Tags
17 Sep 2007, 04:00
humtum0 wrote:
Over the last 2 years , Chesapeake introduced a slew blockbuster drugs, one of them is pialis, the wonder drug for fatigue and generalized body pains.
1. one of them
2. one of which
3. one which -
4. one of them which
5. and one of them which
I go for B, which refers to the drugs.
One of them is, one which is, one of them which is, and one of them which all sound brutual and D & E seem to refer to the drugs twice.
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### Show Tags
17 Sep 2007, 05:19
B
I am not sure, but i think them only refers to animate nouns.
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### Show Tags
20 Sep 2007, 11:42
I will go with B...
I narrowed it down to A&B, and I think we need "which" here
"which" refers to a group in general ---in our case " drugs"
OA?
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Re: SC-- Chesapeake introduced a slew blockbuster drugs [#permalink]
### Show Tags
20 Sep 2007, 12:23
ashkrs wrote:
humtum0 wrote:
Over the last 2 years , Chesapeake introduced a slew blockbuster drugs, one of them is pialis, the wonder drug for fatigue and generalized body pains.
1. one of them--> cant be used only for people.
2. one of which--correct
3. one which--conjunction missing.
4. one of them which
5. and one of them which
Why one of them is incorrect ...
by the way ...them can be used for both people and things
http://dictionary.cambridge.org/define. ... &dict=CALD
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### Show Tags
20 Sep 2007, 12:50
whoooooooops i meant B!!!!!! this is an easy one...
Last edited by beckee529 on 20 Sep 2007, 13:09, edited 1 time in total.
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Re: SC-- Chesapeake introduced a slew blockbuster drugs [#permalink]
### Show Tags
20 Sep 2007, 14:08
Sachu wrote:
ashkrs wrote:
humtum0 wrote:
Over the last 2 years , Chesapeake introduced a slew blockbuster drugs, one of them is pialis, the wonder drug for fatigue and generalized body pains.
1. one of them--> cant be used only for people.
2. one of which--correct
3. one which--conjunction missing.
4. one of them which
5. and one of them which
Why one of them is incorrect ...
by the way ...them can be used for both people and things
http://dictionary.cambridge.org/define. ... &dict=CALD
Wud go for B,
I never heard 'one of them' being used for things too.
'them' is ok with things, but not so sure about 'one of them'..
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Re: SC-- Chesapeake introduced a slew blockbuster drugs [#permalink]
### Show Tags
23 Sep 2007, 06:51
IrinaOK wrote:
Sachu wrote:
ashkrs wrote:
humtum0 wrote:
Over the last 2 years , Chesapeake introduced a slew blockbuster drugs, one of them is pialis, the wonder drug for fatigue and generalized body pains.
1. one of them--> cant be used only for people.
2. one of which--correct
3. one which--conjunction missing.
4. one of them which
5. and one of them which
Why one of them is incorrect ...
by the way ...them can be used for both people and things
http://dictionary.cambridge.org/define. ... &dict=CALD
Wud go for B,
I never heard 'one of them' being used for things too.
'them' is ok with things, but not so sure about 'one of them'..
No convincing explainations so far....
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Joined: 21 Sep 2007
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### Show Tags
23 Sep 2007, 23:13
Guys I would go with E,
Because wouldn't with B it becomes a run on sentence.
Run On sentence [#permalink] 23 Sep 2007, 23:13
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Display posts from previous: Sort by | 1,741 | 5,899 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2017-04 | latest | en | 0.894572 |
https://textiletutorials.com/how-to-calculate-apparel-costing-for-boxer-shorts/ | 1,695,667,742,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510085.26/warc/CC-MAIN-20230925183615-20230925213615-00811.warc.gz | 612,539,913 | 24,010 | # How to Calculate Apparel Costing for Boxer Shorts?
Apparel Costing Method for Boxer Shorts:
Apparel merchandiser plays an important role in apparel export business. He has to calculate the total consumption and costing for that order. As a result, apparel merchandisers must have to know the proper consumption and costing calculation method. Now, I have shared the total costing calculation method for boxer shorts.
Apparel Costing Calculation Method for Boxer Shorts:
Calculate the apparel cost for boxer shorts by following the below tables.
Specifications of Boxer Shorts Style of Apparel Boxer shorts, inner elastic Fabric Body – 100% Cotton, (1×1) Rib, Auto stripe design Stripe Repeat (2cm) White- 0.5cm, Red- 0.7cm, Black- 0.2cm, Yelow-0.6cm G.S.M 200 G.S.M Size Ratio S:M:L:XL = (1:1.5:2:0.5) Quantity of Order 10000 pcs
Fig: Boxer shorts costing method
Solution:
Fabric Consumption for Boxer Shorts:
To calculate the weight of boxer shorts, we have to follow the below fabric consumption formula,
Fabric consumption for boxer shorts,
{(T.L + E.W + H.A + S.A) × (Maximum with wise measure + S.A) × G.S.M × 2}
= ……………………………………………………………………………………………. … (1)
10000
Where,
S.A stands for Seam allowance,
T.L is stands for Total length,
H.A stands for Hem allowance,
E.W stands for Elastic width.
N.B: (Maximum width measure is thigh round = (52×2) = 104
Let,
Seam allowance is 2cm,
Hem allowance is 2cm.
So,
From equation-01, we get,
Fabric consumption for L size boxer shorts is,
{42+4+2+2) × (56+2) × 200×2}
= ……………………………………
10000
= 113.68gms
= 0.114kg
Calculation of Fabric Cost Per Kg:
SL No. Particulars Costing 01 Yarn cost per kg White yarn = {(0.5/2)×Rs. 260} = Rs. 65 Rs. 286.00 02 Red yarn = {(0.7/2)×Rs. 300} = Rs. 105 03 Black yarn = {(0.2/2)×Rs. 320} = Rs. 32 04 Yellow yarn = {(0.6/2)×Rs. 280} = Rs. 84 05 Knitting rate per kg (Auto stripes) Rs. 40.00 06 Dyeing rate per kg (Washing) Rs. 10.00 07 Compacting rate per kg Rs. 6.00 08 Total Rs. 342.00 09 Process loss (10%) (+) Rs. 34.20 10 Fabric cost per kg Rs. 376.20
Calculation of Fabric Cost Per Piece Boxer Shorts:
Fabric cost per Boxer,
= (Fabric consumption per piece boxer × Fabric cost per kg)
= (0.114kg × Rs. 376.20)
= Rs. 42.88
Apparel Costing for Boxer Shorts:
The cost of the boxer shorts is calculated as follows:
SL No. Particulars Costing 01 Fabric cost per pc boxer shorts Rs. 42.88 02 Cost of making cost or CMT cost Rs. 7.00 03 Trimmings & accessories Main label Rs. 0.60 Rs. 8.55 Wash care label Rs. 0.30 Waist elastic (0.70m) Rs. 2.80 Pouch or Poly bag (3 pack) Rs. 1.50 Hang tag Rs. 0.50 Insert card Rs. 1.50 Barcode sticker Rs. 0.60 Hologram sticker Rs.0.75 04 FOB + Packing (Rs. 1 per 30gms of apparel) 120gms/30gms = Rs. 4 Rs. 4.00 05 Sub total Rs. 62.43 06 Overhead cost (10%) (+) Rs. 6.24 07 Sub total Rs. 68.67 08 Rejection costing (3%) (+) Rs. 2.10 09 Sub total Rs. 70.77 10 Profit percentage (15%) (+) Rs. 10.62 11 Sub total Rs. 81.39 12 Apparel price (FOB) in Rs. Price in Rs. 81.39 13 Apparel price (FOB) in US (\$) Price in US (\$) = \$1.85 14 Apparel price (FOB) in Euro. Price in Euro = E 1.30
### 1 thought on “How to Calculate Apparel Costing for Boxer Shorts?”
1. Good and really very helpful. | 1,053 | 3,232 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.8125 | 4 | CC-MAIN-2023-40 | longest | en | 0.824252 |
http://mathcentral.uregina.ca/QQ/database/QQ.09.06/ann2.html | 1,675,557,517,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764500158.5/warc/CC-MAIN-20230205000727-20230205030727-00025.warc.gz | 27,933,318 | 2,503 | I am in need of the answer to the following: I have a 3000 sq ft house and need to use flea foggers. Each fogger treats 6000 cubic feet. How many do I need? Thanks Ann I am a home maker. Hi Ann. How high is your ceiling? Floor area (3000 sq feet) times height equals volume, so if you have 8 ft rooms, then you have 24000 cubic feet. If your rooms average 7 ft, you have 21000 cubic feet and so on. Divide the volume of your house by 6000 and you get the number of foggers you will need. Hope this helps, Stephen La Rocque. | 143 | 523 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.28125 | 3 | CC-MAIN-2023-06 | latest | en | 0.89739 |
https://www.weegy.com/?ConversationId=4F56443B | 1,726,586,050,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651800.83/warc/CC-MAIN-20240917140525-20240917170525-00716.warc.gz | 993,239,817 | 11,185 | Solve for x. (1/3)x = 10
Question
Asked 11/20/2011 3:35:27 PM
Updated 3/16/2014 2:21:01 PM
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Flagged by jerry06 [3/16/2014 2:21:01 PM]
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Original conversation
User: Solve for x. (1/3)x = 10
Weegy: The answer to the equation of 3(x+6)^2=33 is
Nighthawk|Points 191|
Question
Asked 11/20/2011 3:35:27 PM
Updated 3/16/2014 2:21:01 PM
This conversation has been flagged as incorrect.
Flagged by jerry06 [3/16/2014 2:21:01 PM]
Rating
5
(1/3)x = 10
x = 10/(1/3)
x = 30
Added 3/16/2014 2:21:00 PM
This answer has been confirmed as correct and helpful.
Confirmed by debnjerry [3/16/2014 2:35:18 PM]
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Questions asked by the same visitor
Solve: -3x - 5>4
Weegy: x<-3 (More)
Question
Updated 5/26/2014 4:35:34 AM
-3x - 5>4;
-3x > 4 + 5;
-3x > 9;
x < 9/(-3);
x < -3
Added 5/26/2014 4:35:34 AM
This answer has been confirmed as correct and helpful.
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Home | Contact | Blog | About | Terms | Privacy | © Purple Inc. | 1,495 | 3,786 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.265625 | 3 | CC-MAIN-2024-38 | latest | en | 0.86836 |
http://www.netlib.org/stoeplitz/cccslc.f | 1,369,457,821,000,000,000 | text/plain | crawl-data/CC-MAIN-2013-20/segments/1368705543116/warc/CC-MAIN-20130516115903-00091-ip-10-60-113-184.ec2.internal.warc.gz | 619,586,300 | 1,331 | subroutine cccslc(a,x,r,m,l,k,lda) integer m,l,k,lda complex a(lda,k),x(m,l,k),r(1) c c cccslc solves the complex linear system c a * x = b c with the ccc - matrix a . c c on entry c c a complex(m*l,k) c the first row of outer blocks of the ccc - matrix . c each outer block is represented by its first row c of inner blocks. each inner block is represented c by its first row. on return a has been destroyed . c c x complex(m*l*k) c the right hand side vector b . c c r complex(max(m,2*l,2*k)) c a work vector . c c m integer c the order of the inner blocks of the matrix a . c c l integer c the number of inner blocks in a row or column c of an outer block of the matrix a . c c k integer c the number of outer blocks in a row or column c of the matrix a . c c lda integer c the leading dimension of the array a . c c on return c c x the solution vector . c c toeplitz package. this version dated 07/23/82 . c c subroutines and functions c c toeplitz package ... ccslc,salwc c fortran ... float c c internal variables c integer i1,i2,i3,ml real rk c rk = float(k) ml = m*l c c reduce the ccc - matrix to a block-diagonal matrix c by the inverse discrete fourier transformation . c call salwc(a,r,r(k+1),ml,k,lda,-1) c c compute the discrete fourier transformation of c the right hand side vector . c call salwc(x,r,r(k+1),ml,k,ml,1) c c solve the block-diagonal system, blocks of which c are cc - matrices . c do 10 i3 = 1, k call ccslc(a(1,i3),x(1,1,i3),r,m,l,m) 10 continue c c compute the solution of the given system by c the inverse discrete fourier transformation . c call salwc(x,r,r(k+1),ml,k,ml,-1) c do 40 i3 = 1, k do 30 i2 = 1, l do 20 i1 = 1, m x(i1,i2,i3) = x(i1,i2,i3)/rk 20 continue 30 continue 40 continue return end | 536 | 1,735 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2013-20 | latest | en | 0.726406 |
https://www.coursehero.com/file/6401356/HW08/ | 1,498,228,702,000,000,000 | text/html | crawl-data/CC-MAIN-2017-26/segments/1498128320063.74/warc/CC-MAIN-20170623133357-20170623153357-00683.warc.gz | 838,653,368 | 29,986 | # HW08 - d<< w are used as a current loop(see Fig on the...
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1 C:\User\Teaching\505-506\F10\HW08.doc PHY 505 Classical Electrodynamics I Fall 2010 Homework 8 (due Friday Nov. 12) Problem 8.1 (to be graded of 10 points). Calculate the magnetic field distribution along the axis of a straight solenoid (Fig. 5.6a of the lecture notes) with a finite length l, and a round cross-section of radius R . Assume that the solenoid has many wire turns ( N >> 1) which are uniformly distributed along its length. Problem 8.2 (10 points). Two parallel, uniform, long strips of thin foil, separated by distance
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Unformatted text preview: d << w , are used as a current loop (see Fig. on the right) . Calculate: (i) the distribution of the magnetic field and vector-potential, (ii) the magnetic force (per unit length) acting on each strip, and (ii) the magnetic energy and self-inductance of the system (per unit length). Problem 8.3 (10 points). Find the mutual inductance of a long straight wire and a round wire loop adjacent to it (see Fig. below). Neglect the thickness of both wires. I I w d w R 1 I 1 I 2 I x y...
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## This note was uploaded on 09/10/2011 for the course PHY 505 taught by Professor Stephens,p during the Fall '08 term at SUNY Stony Brook.
Ask a homework question - tutors are online | 383 | 1,450 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2017-26 | longest | en | 0.890429 |
https://www.univerkov.com/find-the-volume-of-a-cube-with-an-edge-of-20-cm-express-in-liters/ | 1,656,518,587,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103640328.37/warc/CC-MAIN-20220629150145-20220629180145-00613.warc.gz | 1,119,470,559 | 6,130 | Find the volume of a cube with an edge of 20 cm. Express in liters
Since the answer is required to be given in liters, we translate the length of the cube edge from centimeters to decimeters:
20 cm = 2 dm.
The volume of a cube is equal to the cube of the length of the edge:
2 ^ 3 = 8 (dm3).
1 dm3 is equal to 1 liter, therefore the volume of this cube in liters is 8 liters.
Answer: the volume of the cube is 8 liters.
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. | 170 | 736 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.796875 | 4 | CC-MAIN-2022-27 | latest | en | 0.920099 |
https://documen.tv/eplain-how-to-translate-the-statement-into-an-equation-use-n-for-the-variable-sity-three-is-the-26959337-13/ | 1,679,651,731,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296945279.63/warc/CC-MAIN-20230324082226-20230324112226-00680.warc.gz | 243,578,920 | 15,722 | Question
Explain how to translate the statement into an equation. Use n for the variable.
Sixty-three is the product of nine and a number
1. kimchi2
Since we’re looking at the product of 9 and an unknown number (n), you write out an equation stating 9 times n gives you 63. Then you solve for that.
9 * n = 63
9n = 63
9n/9 = 63/9
Therefore n = 7 | 104 | 351 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.96875 | 4 | CC-MAIN-2023-14 | latest | en | 0.867757 |
http://stackoverflow.com/questions/13619709/following-and-rotating-about-a-path-in-opengl | 1,395,090,189,000,000,000 | text/html | crawl-data/CC-MAIN-2014-10/segments/1394678706211/warc/CC-MAIN-20140313024506-00092-ip-10-183-142-35.ec2.internal.warc.gz | 132,724,726 | 16,340 | Following and Rotating about a path in OpenGL
I'm attempting an exercise where a vehicle is following the Lemniscate of Bernoulli (or more simply, a figure-8 track). I want to use `glTranslatef` and `glRotatef` to achieve this. So far, I have been able to successfully get the vehicle to follow/translate along this path by using the parametric form as follows:
X = (width * cos(t)) / (1+sin^2(t))
Y = (width * cos(t) * sin(t)) / (1+sin^2(t))
Where t is in -pi, pi
In the code, this is as follows:
``````carX = (float) ((Math.cos(t) / (1 + Math.sin(t)*Math.sin(t))));
carY = 0.0f;
carZ = (float) ((Math.cos(t) * (Math.sin(t))) / (1 + Math.sin(t)*Math.sin(t)));
gl.glTranslatef(carX,carY,carZ);
``````
So that works well enough. My problem now is rotating the vehicle so that it follows the path defined by the Lemniscate of Bernoulli. I want to achieve this by using `glRotatef` to rotate around the Y axis, but I am not sure how to proceed in regards to finding the angle to input in `glRotatef`. The rotate is currently in place so that it only manipulates the vehicle, and appears to just need the correct mathematics to follow the path.
Things I have tried:
• Using the derivative of the X and Y forms listed above. I used them independently of each other, because I'm not sure how to/if they need to be combined to be used for the angle. With some minor manipulation they follow the straight areas near the origin, but broke down around the curves.
• Directly finding the tangent of the t value and converting to degrees. Erratic spinning resulted.
If anyone has any suggestions that may be better than the glRotatef method, that would be appreciated as well. I've seen that `gluLookAt` may be helpful, and I may attempt to find a solution using that.
(Note: I'm working in JOGL using Java and the FFP, but I'm comfortable with C/C++ code snippets.)
-
Could you do something simpler like calculate `(carX(t+epsilon), carZ(t+epsilon))`, and use the vector from `(carX(t), carZ(t))` to it for the direction. Once you have that vector, the angle would be `Math.atan2(deltaZ, deltaX)`. – user1118321 Nov 29 '12 at 6:07
atan2(dy/dt, dx/dt) should give you the desired angle. Can you explain more what problems you are having with this approach? – msell Nov 29 '12 at 6:53
Thanks to you both for the comments. As best I can describe it, the vehicle currently rotates perpendicular to the blue line at the red point in this example. After reading some more, that may actually be how it is suppose to be drawn for this example, although it isn't my desired behavior. I'm away from my computer at the moment so I can't follow up on new attempts, but I'll update again if I find out anything new. – Hyper Anthony Nov 29 '12 at 20:11
assuming camera view is the driver's view, `gluLookAt` is exactly what you need! based on your `carX,carY,carZ` computations (assuming that the math is good), you can store previous values and use it:
``````//globals & imports:
import javax.vecmath.*;
Vector3f current = new Vector3f();
Vector3f prev = new Vector3f();
``````
computation is as followed:
``````//on drawing:
prev.set(current);
current.x = (float) ((Math.cos(t) / (1 + Math.sin(t)*Math.sin(t))));
current.z = (float) ((Math.cos(t) * (Math.sin(t))) / (1 + Math.sin(t)*Math.sin(t)));
glu.gluLookAt(current.x, 0f, current.z,
current.x - prev.x, 0f, current.z - prev.z,
0f, 1f, 0f);
``````
i'll test it when i get back home, to make sure it's working, but as far as i can tell, this should do the trick.
- | 951 | 3,515 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.125 | 3 | CC-MAIN-2014-10 | latest | en | 0.932487 |
https://www.physicsforums.com/threads/discrete-math-question.642890/ | 1,628,155,923,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046155458.35/warc/CC-MAIN-20210805063730-20210805093730-00141.warc.gz | 960,764,844 | 14,902 | # Discrete Math Question
## Homework Statement
I attached the problem as file.
## The Attempt at a Solution
I honestly do not know how to solve this problem. I have a test tomorrow, and this is really the only question that I am having difficulty with. I don't really know what mod means, especially in this context. I have tried to read the textbook, but it being rather convoluted, and me being constrained on time, was really quite futile.
#### Attachments
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tiny-tim
Homework Helper
Hi Bashyboy!
a = 4 (mod 13) means that if you divide a by 13, the remainder is 4
(of course, that's the same as saying a - 4 = 0 (mod 13), ie a - 4 is divisible exactly by 13 )
show us what you get for c = 9a (mod 13)
SammyS
Staff Emeritus
Homework Helper
Gold Member
## Homework Statement
I attached the problem as file.
## The Attempt at a Solution
I honestly do not know how to solve this problem. I have a test tomorrow, and this is really the only question that I am having difficulty with. I don't really know what mod means, especially in this context. I have tried to read the textbook, but it being rather convoluted, and me being constrained on time, was really quite futile.
Here's the attachment:
"I don't really know what mod means ..." and you have a test on this stuff tomorrow?
WOW ! | 336 | 1,336 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.0625 | 3 | CC-MAIN-2021-31 | latest | en | 0.979002 |
https://www.physicsforums.com/threads/pkb-of-weak-base-titrated-with-strong-acid.1001387/ | 1,713,439,104,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817206.28/warc/CC-MAIN-20240418093630-20240418123630-00246.warc.gz | 837,448,684 | 16,105 | # PKb of weak base titrated with strong acid
• Chemistry
• i_love_science
In summary, the PKb of a weak base titrated with a strong acid is the negative logarithm of the base dissociation constant of the weak base and is a measure of its strength in a solution. It can be calculated using the Henderson-Hasselbalch equation and gives an indication of how much the weak base will dissociate. The PKb value also affects the equivalence point in a titration and can be used to determine the concentration of the weak base in a solution by using the Henderson-Hasselbalch equation or constructing a titration curve.
i_love_science
Homework Statement
(see below)
Relevant Equations
titration curve
pH + pOH = 14
at halfway point:
pH=pKa
pOH=pKb
What is true about the pK of the analyte?
a) pKa = 9.2
b) pKb = 9.2
c) pKb = 4.8
d) pKa = 11.7
The halfway point is at point 4, at approximately pH=9.2. Therefore, the pOH = pKb = 4.8, and I think the answer is c).
The solution says b) is the correct answer. Could anyone explain why, or whether or not my answer is correct? Thanks.
Solution is wrong.
Not that you are entirely right, c is not the only correct answer.
## 1. What is the definition of PKb?
PKb, or the base dissociation constant, is a measure of the strength of a weak base. It is the negative logarithm of the equilibrium constant for the dissociation of the base into its conjugate acid and hydroxide ions.
## 2. How is PKb related to pKa?
PKb and pKa are inversely related. The lower the PKb value of a base, the stronger it is, and the higher the pKa value of its conjugate acid, the weaker it is. This is because pKa and PKb are both measures of the strength of an acid or base, but in different contexts.
## 3. How do you calculate PKb from experimental data?
To calculate PKb, you need to know the initial concentration of the weak base, the concentration of the strong acid added, and the pH at each point during the titration. From this data, you can plot a titration curve and determine the equivalence point, where the base is completely neutralized. Then, using the Henderson-Hasselbalch equation, you can calculate the PKb value.
## 4. How does temperature affect PKb?
Like most chemical reactions, the dissociation of a weak base is affected by temperature. As temperature increases, the equilibrium constant for the reaction also increases, leading to a decrease in PKb. This means that weak bases become stronger at higher temperatures.
## 5. How does the strength of the weak base affect the PKb value?
The stronger the weak base, the lower its PKb value will be. This is because a stronger base will have a higher concentration of hydroxide ions at equilibrium, making it easier for the base to dissociate and resulting in a lower PKb value.
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4K | 856 | 3,354 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-18 | latest | en | 0.914129 |
http://accessanesthesiology.mhmedical.com/content.aspx?bookid=665§ionid=43745758 | 1,521,844,638,000,000,000 | text/html | crawl-data/CC-MAIN-2018-13/segments/1521257649095.35/warc/CC-MAIN-20180323220107-20180324000107-00479.warc.gz | 7,569,048 | 25,085 | Appendix C
2-1 The mean is the sum of the observations divided by the number of observations, 24: 965/24 = 40.2. To find the median we list the observation in order, then select the (50/100)(24 + 1) = 12.5th point, which is the average of the 12th and 13th observations, (29 + 30)/2 = 29.5. The standard deviation is the square root of the sum of the squared differences between the observations and the sample mean divided by the sample size minus 1, 29.8. The 25th percentile is the (25/100)(24 + 1) = 6.25. Thus, the 25th percentile is between the 6th and 7th observation, which we average to obtain (13 + 13)/2 = 13. Likewise, the 75th percentile is (75/100)(24 + 1) = 18.75, so we average the 18th and 19th observations to obtain (70 + 70)/2 = 70. The fact that the median is very different from the mean (29.5 versus 40.2) and not located roughly equidistant between the top and bottom quartile indicates that the data were probably not drawn from a normal distribution. (If the data were symmetrically distributed about the median, we could have further checked for normality by computing the 2.5th, 16th, 84th and 97.5th percentiles and comparing them with values 2 and 1 standard deviations below and above the mean, as described in Fig. 2-10.)
2-2 Mean = 61,668, median = 13,957, standard deviation = 117,539, 25th percentile = 8914, 75th percentile = 63,555, mean − 0.67 standard deviations = −17,083, mean + 0.67 standard deviations = 140,419. These data appear not to be drawn from a normally distributed population for several reasons. (1) The mean and median are very different. (2) All the observations are (and have to be, since you cannot have a negative viral load) greater than zero and the standard deviation is larger than the mean. If the population were normally distributed, it would have to include negative values of viral load, which is impossible. (3) The relationship between the percentiles and numbers of standards deviations about the mean are different from what you would expect if the data were drawn from a normally distributed population.
2-3 Mean = 4.30, median = 4.15, standard deviation = 0.67, 25th percentile = 5.25, 75th percentile = 4.79, mean − 0.67 standard deviations = 3.85, mean + 0.67 standard deviations = 4.75. These data appear to be drawn from a normally distributed population on the basis of the comparisons in the answer to Prob. 2-2.
2-4 Mean = 1709, median = 1750, standard deviation = 825, 25th percentile = 825, 75th percentile = 2400, mean − 0.67 standard deviations = 1157, mean + 0.67 standard deviations = 2262. These data appear to be drawn from a normally distributed population on the basis of the comparisons in the answer to Prob. 2-1.
2-5 There is 1 chance in 6 of getting each of the following values: 1, ...
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This div only appears when the trigger link is hovered over. Otherwise it is hidden from view. | 792 | 2,921 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.03125 | 4 | CC-MAIN-2018-13 | latest | en | 0.961398 |
https://www.mathworks.com/matlabcentral/cody/problems/1036-cell-counting-how-many-draws/solutions/186534 | 1,495,678,013,000,000,000 | text/html | crawl-data/CC-MAIN-2017-22/segments/1495463607960.64/warc/CC-MAIN-20170525010046-20170525030046-00374.warc.gz | 921,650,054 | 11,673 | Cody
# Problem 1036. Cell Counting: How Many Draws?
Solution 186534
Submitted on 5 Jan 2013 by Alfonso Nieto-Castanon
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
%% games = {'D','D','A','H','D','H'}; draws = 3; assert(isequal(how_many_draws(games),draws))
``` ```
2 Pass
%% games = {'D','D'}; draws = 2; assert(isequal(how_many_draws(games),draws))
``` ```
3 Pass
%% games = {'H','H','A'}; draws = 0; assert(isequal(how_many_draws(games),draws))
``` ```
4 Pass
%% games = {'D','H','H','A','D','H','H','A','D','H','H','A','D','D'}; draws = 5; assert(isequal(how_many_draws(games),draws))
``` ``` | 252 | 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} | 2.9375 | 3 | CC-MAIN-2017-22 | longest | en | 0.759621 |
www.calypso.app | 1,719,001,013,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198862157.88/warc/CC-MAIN-20240621191840-20240621221840-00838.warc.gz | 613,555,275 | 15,244 | # The P-delta effect on jack-up platforms as a stiffness matrix
##### Introduction to the P-delta effect
The purpose of a jack-up platform is to support a deck and mission equipment, e.g. a crane or derrick, above the water level. This mission equipment, together with the hull and variable deck load, makes up most of the weight of the platform. The weight is the P-part of the P-delta effect. The hull will move laterally due to environmental loads on the hull and legs. This lateral movement is the delta-part of the P-delta effect.
When the weight P is at an offset delta, this introduces an extra contribution to the overturning moment of the platform, in turn leading to another (smaller) contribution to the offset. Therefore, it is said that the P-delta effect is a secondary effect. In this definition, the extra offset will lead to another extra offset, ad infinitum. This is how some numerical solvers include geometric non-linearities such as the P-delta effect, requiring a number of iterations.
##### The P-delta effect as stiffness reduction of a beam
Another way of thinking about – and implementing of – the P-delta effect is as a stiffness matrix. This is introduced in E. Wilsons excellent book, http://www.edwilson.org/BOOK-Wilson/11-PDE~1.pdf .
As a simple example, think of a rod in tension. As most people are aware, a rod or rope in tension, has a restoring force against transverse displacement of its ends e.g. has a lateral stiffness. If a rod is in compression, this stiffness is negative, reflected in buckling behavior.
The forces Fi and Fj due to displacements vi and vj can be expressed as the following matrix equation:
If T is negative (compression P), the stiffness matrix is negative, thus giving a stiffness reduction.
Applying a negative stiffness of P/L on the hull is a simplified method proposed in ISO 19905-1, where P is the elevated weight and L the free leg length. However, the implicit assumption that the legs together behave like a rod may be an oversimplification.
In Calypso each leg is modeled as a stack of beams connecting intermediate nodes. A beam not only has end forces, but also end moments. If we assume that bending of the beam due to end-force is parabolic and cubic due to end-moment (valid for Euler beams – close enough for Timoshenko beams as used in Calypso), the following equation applies for the stiffness due to the P-delta effect on a beam:
Where Mi, Mj are end moments and φi and φj are end rotations. If T is set to be the axial load in the beam, this is a good reflection of the P-delta effect. This matrix can be added to the beam stiffness matrix, so that the system can be solved in one go. Based on the top-left item in the matrix, it can be concluded that if a simplified approach is used, a better negative spring stiffness would be 36T/30L.
##### The P-delta effect on jack-up platforms
Until this point in this article, the P-delta effect was considered for a single beam-column supporting a weight. A jack-up is best modelled as a bar-stool – a mass supported by a multitude of beam-columns. When a jack-up is subjected to a horizontal force, the following things happen:
• The hull will be offset from its initial position
• All legs will sustain a roughly equal bending moment
• The leeward leg will have an increased axial load whereas the windward leg will have a decreased axial load due to the overturning moment
So, for a jack-up leg, the P (axial load) is not just given by weight, but also by the effects from environmental actions.
The ISO standard for site-specific assessments of jack-ups (ISO 19905-1) distinguishes between a P-Δ effect (capital Delta) and a P-δ effect (small delta). The former pertains to the influence on overall jack-up behavior such as offset of the hull and overturning moment, whereas the latter considers the element level e.g. for the local chord beam-column check, which is outside of the scope of this article.
The overall behavior of a jack-up is not (much) influenced by the effects of variant in axial load between the legs. The stiffness of the leeward leg reduces, but the stiffness of the windward leg increases, netting to a similar overall horizontal stiffness of the platform as a whole.
However, the stiffness reduction of the leeward leg will lead to a moment redistribution to the stiffer windward leg. This effect is not captured in the definition in the ISO standard. It is sort-of intermediate between P-Delta (capital) and P-delta.
In order to capture the effect of axial load variation, one iteration is required. In the first run the stiffness matrices for the P-delta effect are set based upon weight alone and the environmental load is applied. In the iteration, the stiffness matrix due to the P-delta effect is adapted per element to account for the axial load found in the first run. Since the axial load is not further influenced by this, one iteration is sufficient.
##### The P-delta effect in dynamic simulations
Typically, time domain dynamic simulations of jack-up behavior are performed to calculate the dynamic amplification factor. This DAF can be determined based on overall platform behavior, for which the P-delta effect need not be implemented iteratively if a stiffness matrix approach is used. Furthermore, the foundation model for this type of simulation is typically a linear spring, the damping is applied as viscous damping and all other elements also have linear behavior.
From the facts stated above, it can be concluded that the jack-up model for dynamic simulation can be written as a linear model, if the P-delta effect is included as a stiffness matrix in each element, with P based on the platform and leg weight.
The wave forcing is non-gaussian, so a fully statistical frequency domain approach is not evident. However, the static and dynamic behavior being linear does open up possibilities to a more statistical approach of determining a DAF, based on a shorter time trace of wave forces.
##### Conclusion
The stiffness matrix implementation of the P-delta effect is the model of choice in Calypso. It allows for a simple and effective approach. This paper presents a comparison between several methods and shows that correctly implementing the P-delta effect can have a significant influence on the leg moments. | 1,352 | 6,316 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-26 | latest | en | 0.920729 |
https://it.mathworks.com/matlabcentral/answers/451807-how-to-use-excel-data-in-matlab-for-fast-fourier-transform?s_tid=prof_contriblnk | 1,723,666,717,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641121834.91/warc/CC-MAIN-20240814190250-20240814220250-00189.warc.gz | 257,543,266 | 32,560 | # how to use excel data in matlab for fast fourier transform?
29 visualizzazioni (ultimi 30 giorni)
engineer il 22 Mar 2019
Modificato: dpb il 24 Mar 2019
Hi everybody
i have attached a excel file. In this file, the first column is my magnitute in micron. I have got 71 magnitude values and all recorded in 0.5 seconds.I would like to use FFT, to change from time domain into frequency domain.
Sampling frequency is 142.
Can anyone help me out how to do that?
Kind regards.
##### 5 CommentiMostra 3 commenti meno recentiNascondi 3 commenti meno recenti
dpb il 23 Mar 2019
"Clearly, I was doing image processing. for each frame..."
I see nothing in the code that gives any clue whatsoever that that would have been so, certainly not "clearly".
Anyways, it's certainly not clear to me how an image relates to lateral vibration, but looks like you would have to build the data arrays to do the FFT across all the images--whether those are 2D or what isn't clear to me, certainly as you've got simply a vector above.
What is this the "magnitude" of?
engineer il 23 Mar 2019
I have attached the excel file that I am using, as well the code. I have got a data for lateral and axial vibration as in attached excel file. In my code, I plotted these vibration in time and frequency domain. I am not sure if I used FFT right to plot these vibration in frequency domain. The magnitude of vibration range is not the same in time and frequency domain. what could be the reason? There is a peak at the start of the graph for axial and lateral vibration in frequency domain. Can you please tell me how should interpret the graphs? are my plots correct?
Accedi per commentare.
### Risposta accettata
dpb il 23 Mar 2019
H % headers -- ok, says lateral, axial in first two column
T=mean(diff(V(:,3))) % sample rate.. 0.005 --> 5 ms
Fs=1/T; % sampling frequency, 200 Hz
L=length(V); % length of sample
t=(0:L-1)*T; % time vector -- not really needed
Y=fft(V(:,1:2)); % FFt the two accelerations
P2=abs(Y/L); % 2-sided PSD estimator
P1=P2(1:fix(L/2)+1,:); % 1-sided
P1(2:end-1)=2*P1(2:end-1); % (-)freq half the energy but only only one DC;Fmax bin
f=Fs*(0:fix(L/2))/L; % the frequency at 200 Hz for L points
figure
plot(f,P1)
Shows have mostly just a DC component (non-zero mean) whch, parenthetically, means the sensor must be moving somewhere...let's compensate for the mean and see if helps...don't think will a lot because there's still mostly just a 1/f rolloff, no real spectral content in the signal...
Y=fft(V(:,1:2)-mean(V(:,1:2))); % remove mean before FFT
P2=abs(Y/L);
P1=P2(1:fix(L/2)+1,:);
figure
plot(f,P1)
As thought, just wipes the one commponent out, but leaves most other low frequency stuff. Try just a little more resolution by interpolationg since is very short signal...
N=128;
Y=fft(V(:,1:2)-mean(V(:,1:2)),N); % use 128-pt FFT
P2=abs(Y/L); % all the energy still in L time samples
P1=P2(1:fix(N/2)+1,:);
P1(2:end-1)=2*P1(2:end-1);
f=Fs*(0:fix(N/2))/N;
figure
plot(f,P1)
Shows a little more structure, maybe -- whether any of it is real is probably questionable I'd guess...what is the measurement of?
##### 4 CommentiMostra 2 commenti meno recentiNascondi 2 commenti meno recenti
engineer il 23 Mar 2019
Thanks for the answer. It is very informative. As I am doing vibrational analysis, I though FFT would help to understand the vibrational behaviour of my object. Instead of having displacements in x and y direction versus time, I thought would be more effective to see displacments versus frequency so that I can see at what amplitude what is the frequency? Therefore I was confused when I saw the amplitude range changed. What I wanna conclude from the graph is that what is the max displacement and at what frequency? Because, I thought even if I calculate the max amplitude in x and y direction, without frequency information, this graph would not be very informative. I hope I am on the right track.
dpb il 24 Mar 2019
Modificato: dpb il 24 Mar 2019
Well, I don't know...without knowing a lot more about what you're measuring and what the objectives are, it's really hard to say anything concrete about the data itself other than "it is what it is".
Basically, it shows there was a translation measured which occurred mostly in one translation between sampling points very early and then another sorta' jerky ramp towards the middle and then one big jump in X at the end that was reversed to its starting point.
There really is very little indication of any real consistent frequency pattern in the motion and what there is looks to be mostly undersampled in that the excursions mostly are only one or two points along those events.
Add on to that, that you have an extremely short time series to work with, what estimation of any frequency components in the signal is pretty difficult to discern could be anything more than noise.
Accedi per commentare.
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Translated by | 1,316 | 5,121 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-33 | latest | en | 0.910781 |
https://learnsoc.org/math-puzzle-worksheets/math-puzzle-worksheet-worksheets-for-all-download-and-share/ | 1,547,877,404,000,000,000 | text/html | crawl-data/CC-MAIN-2019-04/segments/1547583662690.13/warc/CC-MAIN-20190119054606-20190119080606-00406.warc.gz | 567,792,969 | 10,388 | ## leArnsoc
Browse All Plans
Just Plans
By Fifine Devost at December 20 2018 14:10:47
There are several standard exercises which train students to convert percentages, decimals and fractions. Converting percentage to decimals for example is actually as simple as moving the decimal point two places to the left and losing the percent sign "%." Thus 89% is equal to 0.89. Expressed in fraction, that would be 89/100. When you drill kids to do this often enough, they learn to do conversion almost instinctively.
Subtract 1-, 2-, or 3-digit numbers from 3- and 4-digit number with/without renaming.
Ratios and proportions are likewise wonderful math lessons with plenty of interesting practical applications. If three pans of pizza, one kilo of spaghetti, two buckets of chicken can properly feed 20 hungry friends, then how much pizza, spaghetti and chicken does mom need to prepare for birthday party with 30 kids?
Save Time : Free worksheets not only save you money, they can also save you time. If you decide that it is best for your child to do worksheets especially tailored for his needs, by doing a little research for printable math worksheets found online, you don't have to make the worksheets yourself. This can save a lot of time. Worksheets aren't that difficult to make, but it can be time consuming.
For this reason, attractive, well-illustrated worksheets with something to do will make learning fun for them. What's more, completing your worksheet will give the child a tremendous sense of fulfillment. | 331 | 1,525 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.03125 | 3 | CC-MAIN-2019-04 | longest | en | 0.939249 |
http://conversion.org/length/finger-cloth/foot-cape | 1,685,395,358,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224644913.39/warc/CC-MAIN-20230529205037-20230529235037-00784.warc.gz | 10,698,260 | 7,351 | # finger (cloth) to foot (Cape) conversion
Conversion number between finger (cloth) and foot (Cape) is 0.36302032913843. This means, that finger (cloth) is smaller unit than foot (Cape).
### Contents [show][hide]
Switch to reverse conversion:
from foot (Cape) to finger (cloth) conversion
### Enter the number in finger (cloth):
Decimal Fraction Exponential Expression
finger (cloth)
eg.: 10.12345 or 1.123e5
Result in foot (Cape)
?
precision 0 1 2 3 4 5 6 7 8 9 [info] Decimal: Exponential:
### Calculation process of conversion value
• 1 finger (cloth) = (exactly) (0.1143) / (0.3148584) = 0.36302032913843 foot (Cape)
• 1 foot (Cape) = (exactly) (0.3148584) / (0.1143) = 2.7546666666667 finger (cloth)
• ? finger (cloth) × (0.1143 ("m"/"finger (cloth)")) / (0.3148584 ("m"/"foot (Cape)")) = ? foot (Cape)
### High precision conversion
If conversion between finger (cloth) to metre and metre to foot (Cape) is exactly definied, high precision conversion from finger (cloth) to foot (Cape) is enabled.
Decimal places: (0-800)
finger (cloth)
Result in foot (Cape):
?
### finger (cloth) to foot (Cape) conversion chart
Start value: [finger (cloth)] Step size [finger (cloth)] How many lines? (max 100)
visual:
finger (cloth)foot (Cape)
00
103.6302032913843
207.2604065827686
3010.890609874153
4014.520813165537
5018.151016456922
6021.781219748306
7025.41142303969
8029.041626331075
9032.671829622459
10036.302032913843
11039.932236205227
Copy to Excel
## Multiple conversion
Enter numbers in finger (cloth) and click convert button.
One number per line.
Converted numbers in foot (Cape):
Click to select all
## Details about finger (cloth) and foot (Cape) units:
Convert Finger (cloth) to other unit:
### finger (cloth)
Definition of finger (cloth) unit: ≡ 4 1⁄2 in .
Convert Foot (Cape) to other unit:
### foot (Cape)
Definition of foot (Cape) unit: 1.033 ft. Legally defined as 1.033 English feet in 1859 . It's of chiefly historical interest
← Back to Length units | 632 | 1,999 | {"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} | 2.75 | 3 | CC-MAIN-2023-23 | latest | en | 0.732976 |
https://wiki-helper.com/the-sum-of-three-consecutive-multiples-of-12-is-468-find-these-multiples-40463482-63/ | 1,721,648,294,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763517846.73/warc/CC-MAIN-20240722095039-20240722125039-00637.warc.gz | 528,322,228 | 30,456 | The sum of three consecutive multiples of 12 is 468. Find these multiples .
The sum of three consecutive multiples of 12 is 468. Find these multiples .
2 thoughts on “The sum of three consecutive multiples of 12 is 468. Find these multiples .”
1. let the three multiples be 12x, 12x+12, and 12x+24.
from given,
(12x) + (12x+12) + (12x+24) = 468
36x + 36 = 468
36x = 468 – 36 = 432
x = 432/36
x = 12
the three multiples are 12x = 12(12) = 144
12x + 12 = 12(12) + 12 = 156
12x + 24 = 12(12) + 24 = 168
Hope this helps!!
Pls mark me as brainliest!!
2. Step-by-step explanation:
Let the multiples be x,x+12 ,x +24
x+x+12+x+24 = 468
3x + 36 = 468
3x = 468-36
X = 432/3
X= 144
so the no’s are
x = 144
x+12 =156
x+24=168 | 286 | 738 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.1875 | 4 | CC-MAIN-2024-30 | latest | en | 0.797717 |
https://www.teachoo.com/1110/990/Ex-4.2--1---Which-one-of-following-options-is-true-and/category/Solution-of-linear-equation/ | 1,685,635,934,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224647895.20/warc/CC-MAIN-20230601143134-20230601173134-00779.warc.gz | 1,132,142,071 | 32,635 | Solution of linear equation
Chapter 4 Class 9 Linear Equations in Two Variables
Concept wise
Learn in your speed, with individual attention - Teachoo Maths 1-on-1 Class
### Transcript
Ex 4.2, 1 Which one of the following options is true, and why? y = 3x + 5 has a unique solution, (ii) only two solutions, (iii) infinitely many solutions It can be seen that x can have infinite values and for infinite values of x,there can be infinite y So,y = 3x + 5 has infinite possible solutions Hence, the correct answer is (iii). | 134 | 523 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.234375 | 3 | CC-MAIN-2023-23 | longest | en | 0.892106 |
http://www.opengl.org/discussion_boards/archive/index.php/t-174112.html | 1,406,901,198,000,000,000 | text/html | crawl-data/CC-MAIN-2014-23/segments/1406510274987.43/warc/CC-MAIN-20140728011754-00445-ip-10-146-231-18.ec2.internal.warc.gz | 735,118,773 | 3,534 | PDA
View Full Version : Point light/spotlight attenuation in glsl 1.3+?
omgzor
04-02-2011, 07:41 PM
I've been working OpenGL's Superbible (5th Edition) point light examples.
I found them missing the constant, linear and quadratic attenuation values integrated into the old lighting model, so I went and wrote a point light shader with attenuation values based on Point Light Attenuation (http://www.ogre3d.org/tikiwiki/-Point+Light+Attenuation) this ogre guide.
The results have been completely bizarre. What can be done to get a sensible light attenuation on the new glsl? Is there a glsl table for attenuation constants?
Sample vertex and fragment programs:
Vertex program:
//point light per pixel vertex program
#version 130
// Incoming per vertex... position and normal
in vec4 vVertex;
in vec3 vNormal;
uniform mat4 mvpMatrix;
uniform mat4 mvMatrix;
uniform mat3 normalMatrix;
uniform vec3 vLightPosition;
// Color to fragment program
smooth out vec3 vVaryingNormal;
smooth out vec3 vVaryingLightDir;
out float dist;
out float constantAttenuation;
out float linearAttenuation;
void main(void)
{
// Get surface normal in eye coordinates
vVaryingNormal = normalMatrix * vNormal;
// Get vertex position in eye coordinates
vec4 vPosition4 = mvMatrix * vVertex;
vec3 vPosition3 = vPosition4.xyz / vPosition4.w;
//get distance to light source
dist=length(vLightPosition-vPosition3);
//write proper attenuation values
if (dist<7.0){
constantAttenuation=1.0;
linearAttenuation=0.7;
}
else if (dist<13.0){
constantAttenuation=1.0;
linearAttenuation=0.35;
}
else if (dist<20.0){
constantAttenuation=1.0;
linearAttenuation=0.22;
}
if (dist<32.0){
constantAttenuation=1.0;
linearAttenuation=0.14;
}
if (dist<50.0){
constantAttenuation=1.0;
linearAttenuation=0.09;
}
if (dist<65.0){
constantAttenuation=1.0;
linearAttenuation=0.07;
}
if (dist<100.0){
constantAttenuation=1.0;
linearAttenuation=0.045;
}
// Get vector to light source
vVaryingLightDir = normalize(vLightPosition - vPosition3);
// Don't forget to transform the geometry!
gl_Position = mvpMatrix * vVertex;
}
Fragment program:
//point light per pixel fragment program
#version 130
out vec4 vFragColor;
uniform vec4 ambientColor;
uniform vec4 diffuseColor;
uniform vec4 specularColor;
smooth in vec3 vVaryingNormal;
smooth in vec3 vVaryingLightDir;
in float dist;
in float constantAttenuation;
in float linearAttenuation;
void main(void){
float att;
att = 1.0 / constantAttenuation + linearAttenuation*dist +quadraticAttenuation*dist*dist;
// Dot product gives us diffuse intensity
float diff = max(0.0, dot(normalize(vVaryingNormal), normalize(vVaryingLightDir)));
// Multiply intensity by diffuse color, force alpha to 1.0
vFragColor = att*(diff * diffuseColor +ambientColor); // attenuation affects the diffuse component
// Specular Light
vec3 vReflection = normalize(reflect(-normalize(vVaryingLightDir), normalize(vVaryingNormal)));
float spec = max(0.0, dot(normalize(vVaryingNormal), vReflection));
if(diff != 0) {
float fSpec = pow(spec, 128.0);
vFragColor.rgb += (att*vec3 (fSpec, fSpec, fSpec)); // attenuation affects the specular component
}
}
Alfonse Reinheart
04-02-2011, 08:51 PM
The results have been completely bizarre.
That's because you misunderstood the Ogre document.
The "range" represents the maximum distance of effect for the light. That is, the distance at which you no longer want that light to affect objects.
This is a property of the light, just like it's intensity (ie: color). Your code has a hard-coded light intensity (namely, 1.0. Since you don't multiply the dot(N, L) by a color, you effectively have lights with unit intensity). The range is not something that should be output from a vertex shader.
Furthermore, you've misunderstood what the table means. The table is saying that, "If you want a light to have an effective range of X, here are the attenuation parameters that give this effect." The attenuation parameters are, like the light intensity and effective range, properties of the light. Not output values from the vertex shader. They should be uniforms.
That is, you pick one set of values from the table, based on the effective range you want.
Also:
att = 1.0 / constantAttenuation + linearAttenuation*dist +quadraticAttenuation*dist*dist;
You need more parentheses here. It's 1.0 divided by all of those things, not just "constantAttenuation". | 1,162 | 4,394 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2014-23 | longest | en | 0.534919 |
https://www.calculatoratoz.com/en/area-of-triangle-using-sides-a-b-and-tan-(c2)-and-cos-(c2)-calculator/Calc-44577 | 1,716,039,486,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971057422.43/warc/CC-MAIN-20240518121005-20240518151005-00265.warc.gz | 617,791,665 | 60,711 | ## Area of Triangle using Sides A, B and Tan (C/2) and Cos (C/2) Solution
STEP 0: Pre-Calculation Summary
Formula Used
Area of Triangle = Side A of Triangle*Side B of Triangle*Tan (C/2)*Cos (C/2)^2
A = Sa*Sb*tan(C/2)*cos(C/2)^2
This formula uses 5 Variables
Variables Used
Area of Triangle - (Measured in Square Meter) - The Area of Triangle is the amount of region or space occupied by the Triangle.
Side A of Triangle - (Measured in Meter) - The Side A of Triangle is the length of the side A, of the three sides of the triangle. In other words, the side A of the Triangle is the side opposite to the angle A.
Side B of Triangle - (Measured in Meter) - The Side B of Triangle is the length of the side B of the three sides. In other words, the side Bof the Triangle is the side opposite to the angle B.
Tan (C/2) - Tan (C/2) is the value of the trigonometric tangent function of half of the given angle B.
Cos (C/2) - Cos (C/2) is the value of the trigonometric cosine function of half of the given angle C of the triangle.
STEP 1: Convert Input(s) to Base Unit
Side A of Triangle: 10 Meter --> 10 Meter No Conversion Required
Side B of Triangle: 14 Meter --> 14 Meter No Conversion Required
Tan (C/2): 1.43 --> No Conversion Required
Cos (C/2): 0.57 --> No Conversion Required
STEP 2: Evaluate Formula
Substituting Input Values in Formula
A = Sa*Sb*tan(C/2)*cos(C/2)^2 --> 10*14*1.43*0.57^2
Evaluating ... ...
A = 65.04498
STEP 3: Convert Result to Output's Unit
65.04498 Square Meter --> No Conversion Required
65.04498 Square Meter <-- Area of Triangle
(Calculation completed in 00.020 seconds)
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## Credits
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## < 24 Area of Triangle using Trigonometric Ratios of Half Angles Calculators
Area of Triangle using Sides A, B and Cosec (C/2) and Sec (C/2)
Area of Triangle = (Side A of Triangle*Side B of Triangle)/(Cosec (C/2)*Sec (C/2))
Area of Triangle using Sides B, C and Cosec (A/2) and Sec (A/2)
Area of Triangle = (Side B of Triangle*Side C of Triangle)/(Cosec (A/2)*Sec (A/2))
Area of Triangle using Sides A, C and Cosec (B/2) and Sec (B/2)
Area of Triangle = (Side A of Triangle*Side C of Triangle)/(Cosec (B/2)*Sec (B/2))
Area of Triangle using Sides B, C and Cosec (A/2) and Cos (A/2)
Area of Triangle = (Side B of Triangle*Side C of Triangle*Cos (A/2))/Cosec (A/2)
Area of Triangle using Sides A, C and Cosec (B/2) and Cos (B/2)
Area of Triangle = (Side C of Triangle*Side A of Triangle*Cos (B/2))/Cosec (B/2)
Area of Triangle using Sides A, B and Cosec (C/2) and Cos (C/2)
Area of Triangle = (Side A of Triangle*Side B of Triangle*Cos (C/2))/Cosec (C/2)
Area of Triangle using Sides B, C and Sin (A/2) and Tan (A/2)
Area of Triangle = (Side B of Triangle*Side C of Triangle*Sin (A/2)^2)/Tan (A/2)
Area of Triangle using Sides A, C and Sin (B/2) and Tan (B/2)
Area of Triangle = (Side C of Triangle*Side A of Triangle*Sin (B/2)^2)/Tan (B/2)
Area of Triangle using Sides A, B and Sin (C/2) and Tan (C/2)
Area of Triangle = (Side A of Triangle*Side B of Triangle*Sin (C/2)^2)/Tan (C/2)
Area of Triangle using Sides B, C and Tan (A/2) and Cos (A/2)
Area of Triangle = Side B of Triangle*Side C of Triangle*Tan (A/2)*Cos (A/2)^2
Area of Triangle using Sides A, C and Tan (B/2) and Cos (B/2)
Area of Triangle = Side A of Triangle*Side C of Triangle*Tan (B/2)*Cos (B/2)^2
Area of Triangle using Sides A, B and Tan (C/2) and Cos (C/2)
Area of Triangle = Side A of Triangle*Side B of Triangle*Tan (C/2)*Cos (C/2)^2
Area of Triangle using Sides B, C and Cot (A/2) and Cos (A/2)
Area of Triangle = (Side B of Triangle*Side C of Triangle*Cos (A/2)^2)/Cot A/2
Area of Triangle using Sides A, C and Cot (B/2) and Cos (B/2)
Area of Triangle = (Side A of Triangle*Side C of Triangle*Cos (B/2)^2)/Cot B/2
Area of Triangle using Sides A, B and Cot (C/2) and Cos (C/2)
Area of Triangle = (Side A of Triangle*Side B of Triangle*Cos (C/2)^2)/Cot C/2
Area of Triangle using Sides B, C and Sin (A/2) and Sec (A/2)
Area of Triangle = (Side B of Triangle*Side C of Triangle*Sin (A/2))/Sec (A/2)
Area of Triangle using Sides A, C and Sin (B/2) and Sec (B/2)
Area of Triangle = (Side C of Triangle*Side A of Triangle*Sin (B/2))/Sec (B/2)
Area of Triangle using Sides A, B and Sin (C/2) and Sec (C/2)
Area of Triangle = (Side A of Triangle*Side B of Triangle*Sin (C/2))/Sec (C/2)
Area of Triangle using Sides B, C and Sin (A/2) and Cos (A/2)
Area of Triangle = Side B of Triangle*Side C of Triangle*Sin (A/2)*Cos (A/2)
Area of Triangle using Sides A, C and Sin (B/2) and Cos (B/2)
Area of Triangle = Side C of Triangle*Side A of Triangle*Sin (B/2)*Cos (B/2)
Area of Triangle using Sides A, B and Sin (C/2) and Cos (C/2)
Area of Triangle = Side A of Triangle*Side B of Triangle*Sin (C/2)*Cos (C/2)
Area of Triangle using Sides B, C and Sin (A/2) and Cot (A/2)
Area of Triangle = Side B of Triangle*Side C of Triangle*Sin (A/2)^2*Cot A/2
Area of Triangle using Sides A, C and Sin (B/2) and Cot (B/2)
Area of Triangle = Side C of Triangle*Side A of Triangle*Sin (B/2)^2*Cot B/2
Area of Triangle using Sides A, B and Sin (C/2) and Cot (C/2)
Area of Triangle = Side A of Triangle*Side B of Triangle*Sin (C/2)^2*Cot C/2
## Area of Triangle using Sides A, B and Tan (C/2) and Cos (C/2) Formula
Area of Triangle = Side A of Triangle*Side B of Triangle*Tan (C/2)*Cos (C/2)^2
A = Sa*Sb*tan(C/2)*cos(C/2)^2
## What is a Triangle?
The Triangle is the type of polygon, which have three sides and three vertices. This is a two-dimensional figure with three straight sides. A triangle is considered a 3-sided polygon. The sum of all the three angles of a triangle is equal to 180°. The triangle is contained in a single plane. Based on its sides and angle measurement, the triangle has six types.
## How to Calculate Area of Triangle using Sides A, B and Tan (C/2) and Cos (C/2)?
Area of Triangle using Sides A, B and Tan (C/2) and Cos (C/2) calculator uses Area of Triangle = Side A of Triangle*Side B of Triangle*Tan (C/2)*Cos (C/2)^2 to calculate the Area of Triangle, The Area of Triangle using Sides A, B and Tan (C/2) and Cos (C/2) formula is defined as the value of the area of the triangle using the sides A & B and the trigonometric half ratios Tan C/2 and Cos C/2. Area of Triangle is denoted by A symbol.
How to calculate Area of Triangle using Sides A, B and Tan (C/2) and Cos (C/2) using this online calculator? To use this online calculator for Area of Triangle using Sides A, B and Tan (C/2) and Cos (C/2), enter Side A of Triangle (Sa), Side B of Triangle (Sb), Tan (C/2) (tan(C/2)) & Cos (C/2) (cos(C/2)) and hit the calculate button. Here is how the Area of Triangle using Sides A, B and Tan (C/2) and Cos (C/2) calculation can be explained with given input values -> 65.04498 = 10*14*1.43*0.57^2.
### FAQ
What is Area of Triangle using Sides A, B and Tan (C/2) and Cos (C/2)?
The Area of Triangle using Sides A, B and Tan (C/2) and Cos (C/2) formula is defined as the value of the area of the triangle using the sides A & B and the trigonometric half ratios Tan C/2 and Cos C/2 and is represented as A = Sa*Sb*tan(C/2)*cos(C/2)^2 or Area of Triangle = Side A of Triangle*Side B of Triangle*Tan (C/2)*Cos (C/2)^2. The Side A of Triangle is the length of the side A, of the three sides of the triangle. In other words, the side A of the Triangle is the side opposite to the angle A, The Side B of Triangle is the length of the side B of the three sides. In other words, the side Bof the Triangle is the side opposite to the angle B, Tan (C/2) is the value of the trigonometric tangent function of half of the given angle B & Cos (C/2) is the value of the trigonometric cosine function of half of the given angle C of the triangle.
How to calculate Area of Triangle using Sides A, B and Tan (C/2) and Cos (C/2)?
The Area of Triangle using Sides A, B and Tan (C/2) and Cos (C/2) formula is defined as the value of the area of the triangle using the sides A & B and the trigonometric half ratios Tan C/2 and Cos C/2 is calculated using Area of Triangle = Side A of Triangle*Side B of Triangle*Tan (C/2)*Cos (C/2)^2. To calculate Area of Triangle using Sides A, B and Tan (C/2) and Cos (C/2), you need Side A of Triangle (Sa), Side B of Triangle (Sb), Tan (C/2) (tan(C/2)) & Cos (C/2) (cos(C/2)). With our tool, you need to enter the respective value for Side A of Triangle, Side B of Triangle, Tan (C/2) & Cos (C/2) and hit the calculate button. You can also select the units (if any) for Input(s) and the Output as well.
How many ways are there to calculate Area of Triangle?
In this formula, Area of Triangle uses Side A of Triangle, Side B of Triangle, Tan (C/2) & Cos (C/2). We can use 23 other way(s) to calculate the same, which is/are as follows -
• Area of Triangle = Side B of Triangle*Side C of Triangle*Sin (A/2)*Cos (A/2)
• Area of Triangle = Side C of Triangle*Side A of Triangle*Sin (B/2)*Cos (B/2)
• Area of Triangle = Side A of Triangle*Side B of Triangle*Sin (C/2)*Cos (C/2)
• Area of Triangle = (Side A of Triangle*Side B of Triangle)/(Cosec (C/2)*Sec (C/2))
• Area of Triangle = (Side B of Triangle*Side C of Triangle)/(Cosec (A/2)*Sec (A/2))
• Area of Triangle = (Side A of Triangle*Side C of Triangle)/(Cosec (B/2)*Sec (B/2))
• Area of Triangle = Side B of Triangle*Side C of Triangle*Tan (A/2)*Cos (A/2)^2
• Area of Triangle = Side A of Triangle*Side C of Triangle*Tan (B/2)*Cos (B/2)^2
• Area of Triangle = Side B of Triangle*Side C of Triangle*Sin (A/2)^2*Cot A/2
• Area of Triangle = Side C of Triangle*Side A of Triangle*Sin (B/2)^2*Cot B/2
• Area of Triangle = Side A of Triangle*Side B of Triangle*Sin (C/2)^2*Cot C/2
• Area of Triangle = (Side B of Triangle*Side C of Triangle*Sin (A/2)^2)/Tan (A/2)
• Area of Triangle = (Side C of Triangle*Side A of Triangle*Sin (B/2)^2)/Tan (B/2)
• Area of Triangle = (Side A of Triangle*Side B of Triangle*Sin (C/2)^2)/Tan (C/2)
• Area of Triangle = (Side B of Triangle*Side C of Triangle*Cos (A/2)^2)/Cot A/2
• Area of Triangle = (Side A of Triangle*Side C of Triangle*Cos (B/2)^2)/Cot B/2
• Area of Triangle = (Side A of Triangle*Side B of Triangle*Cos (C/2)^2)/Cot C/2
• Area of Triangle = (Side B of Triangle*Side C of Triangle*Sin (A/2))/Sec (A/2)
• Area of Triangle = (Side C of Triangle*Side A of Triangle*Sin (B/2))/Sec (B/2)
• Area of Triangle = (Side A of Triangle*Side B of Triangle*Sin (C/2))/Sec (C/2)
• Area of Triangle = (Side B of Triangle*Side C of Triangle*Cos (A/2))/Cosec (A/2)
• Area of Triangle = (Side C of Triangle*Side A of Triangle*Cos (B/2))/Cosec (B/2)
• Area of Triangle = (Side A of Triangle*Side B of Triangle*Cos (C/2))/Cosec (C/2)
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# Water Requirement For Ball Mill
Ball Mill Grinding Capacity Calculator Ball Mill MotorPower Sizing Calculation Ball Mill DesignSizing Calculator The power required to grind a material from a given feed size to a given product size can be estimated by using the following equation where W power consumption expressed in kWhshort to HPhrshort ton 134 kWhshort ton Wi
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## More Details:
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Grinding control strategy on the conventional milling
Grinding control strategy on the conventional milling circuit of Palabora Mining Company Mining Company PMC required that adequate instrumentation and a control system had to be put in place after which extensive testing had to be done to evaluate different grinding control ball mill 2 is the cyclone dilution water flow rate
Best energy consumption International Cement Review
In addition the absence of the heat generated in a ball mill and the high volume of air required by the vertical mill have required the provision of waste heat from cooler exhausts andor auxiliary furnaces to dry raw materials and achieve a limited dehydration of gypsum
Cement mill Wikipedia
A cement mill or finish mill in North American usage is the equipment used to grind the hard nodular clinker from the cement kiln into the fine grey powder that is cement is currently ground in ball mills and also vertical roller mills which are more effective than ball mills
Rawmill Wikipedia
Ball mills are rather inefficient and typically require 10–20 kW·h of electric power to make a tonne of rawmix The Aerofall mill is sometimes used for pregrinding large wet feeds It is a short largediameter semiautogenous mill typically containing 15 by volume of very large 130 mm grinding balls
Ball Mill Ball Grinding Mill Machine Manufacturer from
Furthermore the shrouded ball mills are especially designed developed for powder distempers and cement cturers and suppliers of industrial ball mill shrouded ball mill used for dispersion of enamels and primers powder distempers and cement paints For dispersion of enamels and primers Ranging from 25 Kg
Ball Mill RETSCH powerful grinding and homogenization
Mixer Mills grind and homogenize small sample volumes quickly and efficiently by impact and friction These ball mills are suitable for dry wet and cryogenic grinding as well as for cell disruption for DNARNA recovery Planetary Ball Mills meet and exceed all requirements for fast and reproducible grinding to analytical fineness They are
Ball Milling an overview ScienceDirect Topics
Ball milling is suggested as a novel method for enhancing filler dispersion in different matrices that is environmentally and economically sustainable 85 It is a solidstate mixing of powders usually performed with ball mills which enables intimate mixing of
Ball Mill an overview ScienceDirect Topics
The ball mill is a tumbling mill that uses steel balls as the grinding media The length of the cylindrical shell is usually 1–15 times the shell diameter Figure 811The feed can be dry with less than 3 moisture to minimize ball coating or slurry containing 20–40 water by weight | 2,077 | 10,406 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2021-31 | latest | en | 0.876781 |
https://www.clutchprep.com/analytical-chemistry/practice-problems/149140/calculate-the-ph-of-150-00-ml-of-0-15-m-h2-so3-with-addition-of-25-0-ml-of-0-10- | 1,618,945,782,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618039490226.78/warc/CC-MAIN-20210420183658-20210420213658-00539.warc.gz | 818,714,813 | 28,750 | # Problem: Calculate the pH of 150.00 mL of 0.15 M H2SO3 with addition of 25.0 mL of 0.10 M KOH. Ka1 = 1.39 x 10-2 and Ka2 = 6.73 x 10-8.
###### Problem Details
Calculate the pH of 150.00 mL of 0.15 M H2SO3 with addition of 25.0 mL of 0.10 M KOH. Ka1 = 1.39 x 10-2 and Ka2 = 6.73 x 10-8. | 138 | 293 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.015625 | 3 | CC-MAIN-2021-17 | latest | en | 0.635414 |
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