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https://www.convertunits.com/from/metro/to/hvat | 1,618,757,107,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038492417.61/warc/CC-MAIN-20210418133614-20210418163614-00371.warc.gz | 801,197,398 | 16,585 | ## ››Convert metro to hvat [Croatia]
metro hvat
How many metro in 1 hvat? The answer is 1.8965.
We assume you are converting between metro and hvat [Croatia].
You can view more details on each measurement unit:
metro or hvat
The SI base unit for length is the metre.
1 metre is equal to 1 metro, or 0.52728710783021 hvat.
Note that rounding errors may occur, so always check the results.
Use this page to learn how to convert between metros and hvat [Croatia].
Type in your own numbers in the form to convert the units!
## ››Quick conversion chart of metro to hvat
1 metro to hvat = 0.52729 hvat
5 metro to hvat = 2.63644 hvat
10 metro to hvat = 5.27287 hvat
20 metro to hvat = 10.54574 hvat
30 metro to hvat = 15.81861 hvat
40 metro to hvat = 21.09148 hvat
50 metro to hvat = 26.36436 hvat
75 metro to hvat = 39.54653 hvat
100 metro to hvat = 52.72871 hvat
## ››Want other units?
You can do the reverse unit conversion from hvat to metro, or enter any two units below:
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## ››Definition: Metro
El metro es la unidad de longitud del Sistema Internacional de Unidades. Se define como la longitud del trayecto recorrido en el vacĂo por la luz durante un tiempo de 1/299 792 458 de segundo (unidad de tiempo) (aproximadamente 3,34 ns).
## ››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! | 504 | 1,776 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.734375 | 3 | CC-MAIN-2021-17 | latest | en | 0.49179 |
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https://www.mrexcel.com/board/threads/return-a-blank-if-error-or-0.1111109/?mode=threaded&s=09f4338705743689e836fea832db8aa2 | 1,579,532,775,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579250598800.30/warc/CC-MAIN-20200120135447-20200120164447-00417.warc.gz | 1,003,098,349 | 15,399 | # Return a Blank if Error OR 0
#### The_T
##### New Member
Hi guys,
Wondering if anybody a help with the following:
I am currently using the following formula to return blanks when an error results from 'E25' being blank:
=IFERROR(VLOOKUP(E25,'WORKSHEET'!\$A\$1:\$S\$1000,19,FALSE,"")
However, when E25 is not blank but row '19' in '\$A\$1:\$S\$1000' has no results, the formula returns a '0'.
In the case of there being no result to return, I'd like there also to be a blank (and for there to only not be a blank when there is a result to return - 'YES or 'NO' being the only 2 returnable options in this case).
I have tried OR and AND formulae but with no success.
I'm tired so maybe I have missed something.
Any help would be much appreciated.
#### Fluff
##### MrExcel MVP, Moderator
=IF(VLOOKUP(E25,WORKSHEET!\$A\$1:\$S\$1000,19,FALSE)=0,"",IFERROR(VLOOKUP(E25,WORKSHEET!\$A\$1:\$S\$1000,19,FALSE),""))
#### steve the fish
##### Well-known Member
Hi Fluff. That would error before the iferror ever had its chance. Maybe wrap the if rather than the vlookup.
=IFERROR(IF(VLOOKUP(E25,WORKSHEET!\$A\$1:\$S\$1000,19,FALSE)=0,"",VLOOKUP(E25,WORKSHEET!\$A\$1:\$S\$1000,19,FALSE)),"")
#### Fluff
##### MrExcel MVP, Moderator
Excellent point, I obviously wasn't thinking clearly :banghead:
#### steve the fish
##### Well-known Member
I forget to think all the time
You could also do this if the lookup is producing text results:
=IFERROR(""&VLOOKUP(E25,WORKSHEET!\$A\$1:\$S\$1000,19,FALSE),"")
#### The_T
##### New Member
Hi guys, thanks a lot for the replies.
Will give both options a go and get back to you!
#### Fluff
##### MrExcel MVP, Moderator
Don't bother trying my suggestion from post#2, as Steve pointed out it won't always work.
#### The_T
##### New Member
Steve the Fish - Your solution works perfectly. You are clearly either a genius or some sort of divine entity.
Cheers!
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I calculated the derivative of $$\frac{x}{7}*e^{-2x^2}$$ and got $$\frac{1}{7}e^{-2x^2}(1-4x^2)$$ (I included it cause if I got that wrong calculating the rest is pointless)
I don't know how to find the limit of this function: $$\frac{1}{7}e^{-2x^2}(1-4x^2)$$ I tried splitting it into two but I still don't know how to handle this$$\lim_{x \to \infty} \frac{1}{7}e^{-2x^2}+\lim_{x \to \infty} \frac{1}{7}xe^{-2x^2}*(-4x)$$
• Exponential growth is more powerful than polynomial growth. $e^{-2x^2}$ goes to $0$ more powerfully than $(1-4x^2)$ goes to $-\infty$. Their product goes to $0.$ – Doug M Jan 3 at 15:45
• Is $*$ multiplication or convolution in your question? – Asaf Karagila Jan 4 at 7:50
Using L'Hôpital's rule , we get \begin{align} \lim_{x\to \infty}\frac{1-4x^2}{7e^{2x^2}}&=\lim_{x\to \infty}\frac{-8x}{28\cdot xe^{2x^2}}\\ &=\lim_{x\to \infty}\frac{-8}{28e^{2x^2}}=0 \end{align}.
• L'Hôpital feels overkill for this problem (and, indeed, for most similar problems). KM101's solution is much simpler. – Andreas Rejbrand Jan 4 at 7:01
Hint: Rewrite the expression as
$$\frac{1}{7}e^{-2x^2}\left(1-4x^2\right) = \frac{1-4x^2}{7e^{2x^2}}$$
Now, notice the growth of the numerator and denominator. Which grows more quickly?
Then arrange it as:
$$\frac{1-4x^2}{7e^{2x^2}}$$
Use that, due to exponentiation having a much greater effect than indices, $$e^{2x^2}>>4x^2$$ for sufficiently large $$x$$, to show this limit is very clearly $$0$$
Note that for $$x\ge0$$, \begin{align} e^x &=1+x+\frac{x^2}2+\dots\\ &\ge\frac{x^2}2\tag1 \end{align} Thus, $$e^{-2x^2}\le\frac1{2x^4}\tag2$$ Applying $$(2)$$ to the expression for the derivative gives \begin{align} \left|\frac17e^{-2x^2}\!\!\left(1-4x^2\right)\right| &\le\frac{4x^2+1}7\frac1{2x^4}\\ &=\frac2{7x^2}+\frac1{14x^4}\tag3 \end{align}
You pretty much have the solution. The following limit: $$\frac{1}{7}\lim_{x\to \infty}\frac{1}{e^{2x^2}}=0.$$
I do not believe that was your problem. The last bit is what you were probably having trouble with. I would try u substitution. Let $$u=x^2$$ then we have the following: $$-\frac{4}{7} \lim_{u\to\infty}\frac{u}{e^{2u}}=0$$by L'Hopitals Rule.
• Thank you I completely forgot the L'Hopitals Rule – B. Czostek Jan 3 at 16:04
• @B.Czostek I think you'd be better off rarely remembering iL'Hopital. Most limit problems have easier and more informative solutions - as the other answers show for this one. – Ethan Bolker Jan 4 at 1:34 | 949 | 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": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 18, "wp-katex-eq": 0, "align": 3, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.09375 | 4 | CC-MAIN-2019-35 | latest | en | 0.753508 |
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Geometry End-of-Course Assessment
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What does it take to really win the game?
## Components of the Final Score
The final score of the team will be a simple summation of the scores of individual Cryptonites in the team. The score that each Cryptonite will get in the game will have the following parts:
• Base Score
• Rarity Bonus
• Level Bonus
• Power Ups
Let’s go over each one of them using an example. Suppose you create a team with the following composition:
1. Sandbox (SAND) — Legendary — Level 1 — Knight
2. Cardano (ADA) — Epic — Level 0
3. Basic Attention Token (BAT) — Epic — Level 2 — Defender
4. Aave (AAVE) — Rare — Level 3
5. Binance (BNB) — Common — Level 4
### Base Score
For computing the base score of a Cryptonite, we find the price movement of the underlying token in every 15 min interval.
Here is an example:
• Price Movement — Let us suppose ADA moved up by 0.1% between 10–10:15 pm, then moved up by 0.2% from 10:15 pm to 10:30 pm, then moved down -0.1% between 10:30 pm — 10:45 pm.
• Scaling the score — Each 15 min interval score is multiplied by 10 to scale it up. So, the 15 min interval scores of ADA would look like, 0.1*10 = 1, 0.2 * 10 = 2 and -0.1 * 10 = -1.
• Cumulative score — The base score of the Cryptonite is simply a summation of these 15 min interval scores over the duration of the game (essentially a summation of these 15 min interval scores for each token during the 15 hours of the game time). From these 15 mins blocks the data is refreshed on the product every 30 secs.
Here are the base scores for each Cryptonite:
### Rarity & Level Bonus
Owning Rarer NFTs provides advantages in gameplay. The rarer the NFT, the more powerful it is. Based on the rarity of the Cryptonite, the following rarity bonuses are provided:
Each level provides an incremental bonus of 0.5% in the gameplay.
The formula will start looking like this:
And the rarity & level bonus values will start looking like this:
### Power Ups
And the final piece is the power ups. Each player gets to use 2 power ups on the team — a Knight and a Defender. To know more, read here.
• Knight — For our team, SAND is the Knight, so, it simply gets a 100% bonus on the base score.
• Defender — For our team, BAT is the Defender. A defender revises the base score. It changes the -ve scores in the 15 mins block to 0 but keeps adding the +ve score. So something like shields you against the (-ve)s while giving the advantage to the (+ve)s. Let us suppose the 15 min interval scores of BAT looked something like this: -1.1, -2.1, -1.45, -1.87, 1.09, 0, 30, 9.69… the sum of which was 29.04. Now using defender, all negative values will become zero and the score series would look like 0, 0, 0, 0, 1.09, 0, 30, 9.69… and the base score changes to 40.78. So, here is what the revised score would look like:
### Final Score
The final score for each Cryptonite is simply a summation of Base Score, Rarity & Level Bonus and Power Up. The final score for the team is the summation of scores for each Cryptonite.
Here is what it would look like:
Yay! Now that you understand how the scoring works, it is time to find out the rewards of the game.
Last updated | 902 | 3,171 | {"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} | 3.359375 | 3 | CC-MAIN-2024-38 | latest | en | 0.85724 |
http://queenrocks.tk/tefip/what-is-a-normal-z-score-vacu.php | 1,544,627,550,000,000,000 | text/html | crawl-data/CC-MAIN-2018-51/segments/1544376823895.25/warc/CC-MAIN-20181212134123-20181212155623-00227.warc.gz | 247,864,927 | 4,748 | # What is a normal z score
The standard normal distribution is a normal distribution represented in z scores.The table in the frame below shows the probabilities for the standard normal distribution.Each Z score represents a unit of standard deviation away from the mean.This is a PowerPoint presentation that can be used to introduce normal distributions.
The test is most accurate for women of average stature, between 5 feet and 5 feet 10 inches tall.Because the normal distribution approximates many natural phenomena so well, it has developed into a.
### Introduction to Normal Distribution and z-score by
STANDARD NORMAL DISTRIBUTION: Table Values Represent AREA to the LEFT of the Z score. Z.00.01.02.03.04.05.06.07.08.09. -3.9.00005.00005.00004.00004.00004.00004.00004.00004.00003.00003.
### WORKSHEET: (Normal Distribution and Z scores) Conversion
So a z-score of 2.0 means the measurement is 2 standard deviations away from the mean.
If IQ tests have a If IQ tests have a score of a mean of 100 and SD of 16 points.
### Descriptive Statistics: Measures of Central Tendency
Calculating Z-Score This video covers how to calculate z-score in a few different situations.Then have the students calculate their own z scores. III. The Central Limit Theorem 1.
### Z Score Table - Definition, Steps & Percentile Table
This standard is a mean of zero and a standard deviation of 1.Example finding the minimum z-score and numerical threshold to be in a given percentile in a normal distribution.This applet will allow you to explore the normal distribution by changing values of the mean, the standard deviation, the observation, or z itself, and examine the areas under the curve.
### Z-Score: Definition, Formula and Calculation - Statistics
Z scores are especially informative when the distribution to which they refer is normal.
Note that the probabilities given in this table represent the area to the LEFT of the z-score.The diagram also illustrates how the Z scores fall on the normal distribution.A z-score of 1.5 is 1.5 standard deviations above and below the mean.Normal Distribution - Real-World Problems Using z Values Now that you can calculate probabilities using a z table, it is time to use this knowledge to solve real-world problems.The first step is to standardize the target variable value into a standard normal random variable (Z Score) using the known standard deviation and mean.We again use IQ scores, with a mean of 100 and a standard deviation of 15, to calculate some probabilities.
Normal distribution The normal distribution is the most widely known and used of all distributions.
### Normal distribution
The required inputs of this calculator are a random or individual sample of the population, population mean and the population standard deviation.A standard score (aka, a z-score) indicates how many standard deviations an element is from the mean.
A standard score can be calculated from the following formula.You can average together the Z-scores, then divide by sqrt(N) to get the Z-score then normal p-value of the sample mean that you just drew.
### Difference Between Z Score and T Score
The bone density test results have two different reports, the T-score and the Z-score.For data points that are below the mean, the Z-score is negative.
A standard normal variable has zero mean and variance of one (consequently its standard deviation is also one).CHAPTER 6 THE NORMAL CURVE, STANDARDIZATION, AND z SCORES 51 Assume that your class is the population and provide the students with the population mean.You might prefer to work with Z-scores over actual drawn values of the distribution for a variety of reasons, including computational.In your textbook on page. 398-399 you will find the z-score table.Z Scores and the Normal Distribution Z score is a common standard score used most often in research.A z-score of 1 means that an observation is 1 standard deviation away from the mean.The lesson covers what a normal distribution is and then talks through how to convert to a standardised normal distribution (find the z-score). | 855 | 4,091 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2018-51 | latest | en | 0.832724 |
https://www.solumaths.com/en/math-apps/calc-online/tan | 1,632,493,692,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780057558.23/warc/CC-MAIN-20210924140738-20210924170738-00348.warc.gz | 1,012,176,703 | 12,190 | # tangent
## Calculus tan
The tan trigonometric function to calculate the tangent of an angle in radians, degrees or gradians.
tan(0), returns 0
### Function : tan
#### Summary :
The tan trigonometric function to calculate the tangent of an angle in radians, degrees or gradians.
tan online
# Tangent function
The calculator allows to use most of the trigonometric functions, it is possible to calculate the tan, the sine and the cosine of an angle through the functions of the same name..
The trigonometric function tangent noted tan, allows to calculate the tangent of an angle online , it is possible to use different angular units :
• radians wich is the angular unit by default,
• degrees or
## Calculation of the tangent
### Tangent calculating an angle in radians
The tangent calculator allows through the tan function to calculate online the tangent of an angle in radians, you must first select the desired unit by clicking on the options button calculation module. After that, you can start your calculations.
To calculate tangent online of pi/6, enter tan(pi/6), after calculation, the result sqrt(3)/3 is returned.
Note that the tangent function is able to recognize some special angles and make the calculations with special associated values in exact form.
### Calculate the tangent of an angle in degrees
To calculate the tangent of an angle in degrees, you must first select the desired unit by clicking on the options button calculation module. After that, you can start your calculus.
To calculate tangent of 60, enter tan(60), after calculation, the restults sqrt(3) is returned.
### Calculate the tangent of an angle in gradians
To calculate the tangent of an angle in gradians, you must first select the desired unit by clicking on the options button calculation module. After that, you can start your calculus.
To calculate tangent of 50, enter tan(50), after computation, the result 1 is returned.
Note that the tangent function is able to recognize some special angles and do the calculus with special associated exact values.
## Special tangent values
The tangent admits some special values which the calculator is able to determine in exact forms. Here is the list of the special tangent values:
## Derivative of tangent
The derivative of the tangent is equal to 1/cos(x)^2.
## Antiderivative of tangent
The antiderivative of the tangent is equal to -ln(cos(x)).
## Properties of the tangent function
The tangent function is an odd function, for every real x, tan(-x)=-tan(x). The consequence for the curve representative of the tangent function is that it admits the origin of the reference point as point of symmetry.
The tan trigonometric function to calculate the tangent of an angle in radians, degrees or gradians.
#### Syntax :
tan(x), where x is the measure of an angle in degrees, radians, or gradians.
#### Examples :
tan(0), returns 0
#### Derivative tangent :
To differentiate function tangent online, it is possible to use the derivative calculator which allows the calculation of the derivative of the tangent function
The derivative of tan(x) is derivative_calculator(tan(x))=1/cos(x)^2
#### Antiderivative tangent :
Antiderivative calculator allows to calculate an antiderivative of tangent function.
An antiderivative of tan(x) is antiderivative_calculator(tan(x))=-ln(cos(x))
#### Limit tangent :
The limit calculator allows the calculation of limits of the tangent function.
The limit of tan(x) is limit_calculator(tan(x))
#### Inverse function tangent :
The inverse function of tangent is the arctangent function noted arctan.
#### Graphic tangent :
The graphing calculator is able to plot tangent function in its definition interval.
#### Property of the function tangent :
The tangent function is an odd function.
Calculate online with tan (tangent)
• Arccosine : arccos. The arccos function allows the calculation of the arc cosine of a number. The arccosine function is the inverse functions of the cosine function.
• Arcsine : arcsin. The arcsin function allows the calculation of the arc sine of a number. The arcsine function is the inverse function of the sine function.
• Arctangent : arctan. The arctan function allows the calculation of the arctangent of a number. The arctangent function is the inverse functions of the tangent function.
• Cosine : cos. The cos trigonometric function calculates the cosine of an angle in radians, degrees or gradians.
• Cosecant : cosec. The trigonometric function sec allows to calculate the secant of an angle expressed in radians, degrees, or grades.
• Cotangent : cotan. The cotan trigonometric function to calculate the cotangent of an angle in radians, degrees or gradians.
• Trigonometric expansion : expand_trigo. The calculator makes it possible to obtain the trigonometric expansion of an expression.
• Trigonometric linearization : linearization_trigo. Calculator that allows to linearize a trigonometric expression.
• Secant : sec. The trigonometric function sec allows to calculate the secant of an angle expressed in radians, degrees, or grades.
• Simplify Calculator : simplify. Calculator wich can simplify an algebraic expression online.
• Sine : sin. The sin trigonometric function to calculate the sine of an angle in radians, degrees or gradians.
• Tangent : tan. The tan trigonometric function to calculate the tangent of an angle in radians, degrees or gradians.
• Trigonometric Calculator : trig_calculator. Calculator wich uses trigonometric formula to simplify trigonometric expression. | 1,183 | 5,553 | {"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} | 3.40625 | 3 | CC-MAIN-2021-39 | latest | en | 0.761944 |
https://how-what-do.com/13170590-how-to-build-a-bocce-court-10-steps | 1,675,854,386,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764500758.20/warc/CC-MAIN-20230208092053-20230208122053-00748.warc.gz | 319,861,784 | 10,818 | # How to Build a Bocce Court: 10 Steps
Bocce ball is a game of Italian origin played traditionally on a flat surface covered with sand or trimmed grass and bordered by a wooden border. The player must throw several balls of different sizes around the court (called a court) and calculate the points according to the positions of the balls. To build a court, measure the dimensions of the court. Then insert the wooden edge and fill it with a base layer of stones and a top layer of fine sand or oyster shells.
## Steps
### Part 1 of 3: Plotting the court dimensions
#### Step 1. Decide the size of the court
An official bocce court is around 28 x 4 meters. However, you can build the court any size you want. As per the general rule of thumb, the ratio of court width to length should be between 1:5 and 1:7.
### Build a court that fits the dimensions of your garden (or the place you are going to use). If your space is limited, build the court no more than 1.5 x 6 meters
#### Step 2. Measure the length and width of the court
Use a tape measure to calculate the length and total width of the pitch on the ground where you want to build. Remember that you will need to add the dimensions of the auxiliary material (for example, the court structure) in the total width and length of the measurements. If using 10 x 10 cm boards in the frame, add 8 inches to the length and width of the measurements.
### For example, if you are going to build a court of the regulated size, the measurements of the outside dimensions of the court should be 28 x 4.2 meters
#### Step 3. Place stakes in the corners of the court
After measuring the length and width of the court, use a hammer to drive a stake into the ground in each corner. These markings will indicate the size of the court and the area you need to dig.
### Part 2 of 3: Creating the court
#### Step 1. Dig the court area
The best way to flatten and level the court space is to dig about two to four inches from the ground. Thus, it will be possible to remove stones or thickets from the ground, which can crack the surface of the court and create a hole in the place when placing the subsequent layers.
### Choose a flat area to build the court. Otherwise, the playing surface will be crooked. Dig less than 10 centimeters if you think your court is likely to get wet during the rainy season
#### Step 2. Install and connect the wooden edges
It is possible to use other types of wood, however, pressure treated boards are the best choice. Lay the 4 x 10 cm boards along the edges of the excavated bocce hole and arrange them end to end. Use a carpenter's level to ensure all planks are level with the floor. Secure each plank to adjacent edges using 10-inch nails.
### Go to a building supply store to buy wood. Before going to the site, calculate how many boards you will be using. If you're building a standard court and you're buying 10-by-10-foot planks, you'll need about 14 sections of wood
#### Step 3. Place a washed gravel base
The base layer of the court must be built with relatively large and resistant stones. Use a shovel to make a 2.5-inch layer of stone across the entire court. If you have dug the area of the court 10 centimeters deep, place approximately 5 centimeters of large stone layer.
### At stores specializing in construction and gardening materials, it is possible to buy 2.5 centimeters of washed gravel. If you cannot find the material, contact a quarry or a place that sells gravel
#### Step 4. Place a second layer of gravel or gravel
When the 2.5 cm layer of stone has covered the entire dug hole, add a second layer that will contain smaller stones. There is no standard size required; you can use gravel or gravel of 1.2 centimeters. Use the shovel to cover the court with gravel. The layer must be at least 2 centimeters thick.
### Part 3 of 3: Placing the surface
#### Step 1. Place a layer of oyster shells if you are building an official sized court
To build a standard court (if you are interested in becoming a professional bocce player) cover the court with processed oyster shells. The layer should be about 0.6 centimeters thick. This material has a very fine grain and is resistant to storms.
• Unless you live in an international coastal region, you will need to ship these processed shells to your address.
• Contact companies in the coastal region that work with this product to purchase the material.
#### Step 2. Throw in the sand as you cover the course layer
If you're not interested in making a block with standard dimensions - and want to save money - skip the oyster shells step. A layer of ordinary sand can be used to make a playing surface. Pour sand at least 1.2 centimeters thick over the underlying layer of rocks or gravel. Use the shovel to spread the sand and press it into all corners of the court.
### The downside to using sand is that a storm can make the terrain soggy and compact. Sweep the sand so that it dries and reduces its compactness
#### Step 3. Finish the court with synthetic grass
If you don't want to play in processed oyster shells or sand, apply synthetic grass to the surface. This material is very affordable and easy to maintain; if branches or debris fall on the site, it will be very simple to clean.
• Buy artificial grass at garden stores or plant nurseries.
• Depending on the size of your court and the size of the sections of grass you plan to purchase, it may be best to purchase several sections of grass.
## Tips
• If you decide that you are not skilled enough to perform this task, call a professional and have him set up the court.
• It is possible - and quite fun - to play bocce without a court. You can play on any field or grass (which is trimmed). | 1,307 | 5,740 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2023-06 | latest | en | 0.90745 |
https://www.twoboysandadad.com/math | 1,553,640,200,000,000,000 | text/html | crawl-data/CC-MAIN-2019-13/segments/1552912206677.94/warc/CC-MAIN-20190326220507-20190327002507-00067.warc.gz | 944,961,670 | 122,923 | ## No Tricks – Just How to Round on a Number Line
It wasn’t until a few years after the Common Core was adopted in California did I start teaching students how to round on a number line. Before that, what did I use? Rounding rhymes, tricks,…
## Unitizing? Yes! It’s an Important Big Math Idea
Unitizing is not only a mathematical term, but it’s also a business term. Have you ever wondered when businesses ship hundreds of boxes, how do they keep track? By unitizing! The shipping company will unitize…
## 7 Helpful Posts for Teaching Multiplication
Teaching multiplication can take many forms. Teachers are always finding new and engaging ways of teaching multiplication. Whether it be using manipulatives, arrays, songs, tips or strategies, there is no one way of teaching multiplication.…
## 3 Tried and True Ways on How to Teach Multiplication
The gateway to learning algebra and higher forms of mathematics is multiplication. It is critically important that elementary school children learn the concept of multiplication as well as, just learning the multiplication facts. I am…
## How to Use the Compensation Strategy for Addition
What are some of the strategies you teach your students or children to add two and three digit numbers? Do you use compensation? Or do you begin teaching with the standard algorithm? Teaching students to…
## Making a Ten is an Important MUST HAVE Mental Math Strategy
Just the other day, I taught a lesson about making a TEN to add sums greater than ten. For example, with 8 + 6, you could increase the 8 to a 10 and reduce the 6…
Most teachers report that students struggle with multiplication fluency. In fact, the problem is pervasive enough that even my son’s eighth-grade math teacher said the same thing at his Back To School Night presentation to the…
## Place Value Doesn’t Have to be Boring!
Teaching second grade has given me a much better understanding of how second graders are prepared for third grade. If you examine the Common Core State Standards for Math, you will see that place value…
## Save Time, Money and Your Sanity with Teachers Pay Teachers
What is it that teachers need more of? Time and money! There’s never enough time in a day to teach everything we need to teach nor is there sufficient time in the day to prepare…
## How to Teach the Area of Irregular Shapes
My class has been learning to find the area of shapes for a few weeks now. But now it was time to move on to find the area of irregular shapes. So far, my students…
## Comparing Fractions Using Google Slides and Virtual Manipulatives
Now that my students were familiar with comparing and ordering fractions, the next step was to have them comparing fractions using Google Slides. It is one thing to be able to show how to compare…
## Comparing Fractions with Manipulatives and Math Talk
In Part 1 of How to Compare Fractions, I explained how I used fraction strips to have students explore fraction sizes. Comparing fractions and their sizes visually helped my students establish a benchmark fraction of…
## Compare Fractions in Simple Ways with Manipulatives
Before I started teaching my third graders to compare fractions, I wanted to assess what they already knew about fractions. True story. I asked my students to write in their journal and complete this sentence:…
## Distributive Property of Multiplication – How To Break It Down
Did you ever think that as a third-grade teacher or even as an elementary teacher you would be teaching the Distributive Property of Multiplication? When I started teaching over 30 years ago, there weren’t even…
## Learning Multiplication is More than Memorization!
I have a vivid memory of learning multiplication in third grade. My teacher, Mrs. Bowman, drew three circles on the chalkboard. Then she put five milk bottles in each one. She said this is 3… | 837 | 3,879 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2019-13 | longest | en | 0.918337 |
https://vietnoodlebar.net/slide/decision-theory-for-conservation-biology-drmon8 | 1,619,174,242,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618039617701.99/warc/CC-MAIN-20210423101141-20210423131141-00595.warc.gz | 692,881,209 | 50,151 | # Decision Theory for Conservation Biology
What is a metapopulation? And why should I care? Hugh Possingham and friends How to manage a metapopulation Problem 1 Michael Westphal (UC Berkeley), Drew Tyre (U Nebraska), Scott Field (UQ) Can we make metapopulation theory useful? Specifically: how to reconstruct
habitat for a small metapopulation Part of general problem of optimal landscape design the dynamics of how to reconstruct landscapes Minimising the extinction probability of one part of the Mount Lofty Ranges Emu-wren population. Metapopulation dynamics based on Stochastoc Patch Occupancy Model (SPOM) of Day and Possingham (1995) Optimisation using Stochastic Dynamic Programming (SDP) see Possingham (1996)
The Mount Lofty Ranges, South Australia Hughs birthplace MLR Southern Emu Wren
Small passerine (Australian malurid) Very weak flyer Restricted to swamps/fens Listed as Critically Endangered subspecies About 450 left; hard to see or hear Has a recovery team (flagship) The Cleland Gully Metapopulation; basically isolated
Figure shows options Where should we revegetate now, and in the future? Does it depend on the state of the metapopulation? Stochastic Patch Occupancy Model (Day and Possingham, 1995) State at time, t,
(0,1,0,0,1,0) Intermediate states Extinction process (0,1,0,0,1,0) (0,1,0,0,0,0) (0,0,0,0,1,0)
Plus fire Colonization process State at time, t+1, (0,1,1,0,1,0) The SPOM A lot of population states, 2n, where n is the number of patches. The transition matrix is 2n by 2n in size (128 by 128 in this case). A chain binomial model; SPOM has recolonisation
and local extinction where functional forms and parameterization follow Moilanen and Hanski Overall transition matrix, a combination of extinction and recolonization, depends on the landscape state, a consequence of past restoration activities Decision theory steps Set objective (minimize extinction prob) Define state variables (population and landscape states) and control variables (options for restoration)
Describe state dynamics the SPOM Set constraints (one action per 5 years) Solve: in this case SDP Control options (one per 5 years, about 1ha reveg) E5: largest patch bigger, can do 6 times E2: most connected patch bigger, 6 times C5: connect largest patch
C2: connect patches1,2,3 E7: make new patch DN: do nothing Management trajectories: 1 only largest patch occupied C5 E5 E5 E5 E5
E5 E5 E7 DN Management trajectories: 2 all patches occupied E2 DN
C2 E5 E5 E2 C5 E7 E5
E5 E5 E5 E5 E2 Take home message Metapopulation state matters Actions justifiable but no clear sweeping generalisation, no simple rule of thumb!
Previous work has assumed that landscape and population dynamics are uncoupled. This paper represents the first spatially explicit optimal landscape design for a threatened species. Other issues Computational problems Problems, models and algorithms what are they? Optimal translocation strategies
Problem 2 Brigitte Tenhumberg, Drew Tyre (U Nebraska), Katriona Shea (Penn State) Consider the Arabian Oryx Oryx leucoryx if we know how many are in the wild, and in a zoo, and we know birth and death rates in the zoo and the wild, how many should we translocate to or from the wild to maximise persistence of the wild population Oryx problem
Growth rate R = 0.85 Capacity = 50 Growth rate R = 1.3 Capacity = 20 ?? Zoo Population Wild Population
Wild Population Result base parameters R C Captive Population R = release, mainly when
population in zoo is near capacity C = capture, mainly when zoo population small, capture entire wild population when this would roughly fill the zoo If zoo growth rate changes, results change but for a new species we wont know R in the zoo Enter active adaptive management,
Management with a plan for learning Metapopulaton dynamics in a dynamic landscape What do mussels, Leadbeaters possum and annual herbs have in common? Empirical conversations over a long time Eradicate, Exploit, Conserve Pure Ecological Theory
+ = Applied Theoretical Ecology Decision Theory
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• Indoor air quality certificate. The GREENGUARD Indoor Air Quality Certification Program gives assurance that products designed for use in office environments and other indoor spaces meet strict chemical emissions limits, which contribute to the creation of healthier interiors. | 1,413 | 6,069 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.640625 | 3 | CC-MAIN-2021-17 | latest | en | 0.814713 |
https://www.ece.uic.edu/bin/view/ECE/ECE310?cover=print | 1,601,131,459,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400244231.61/warc/CC-MAIN-20200926134026-20200926164026-00023.warc.gz | 814,101,334 | 3,858 | ECE 310 - Discrete and Continuous Signals and Systems
Description: Credit 3. Signals; systems; convolution; discrete and continuous Fourier series and transforms; Z-transforms; Laplace transforms; sampling; frequency response; applications; computer simulations.
Prerequisite: MATH 220; and Credit or concurrent registration in ECE 225; or Credit or concurrent registration in ECE 210 for non-ECE students.
Topics:
1. Signals:
unit step; unit impulse; exponential;
sinusoidal; complex exponential;
linear combination; time-shifting
scaling, and reversal
2. Differential Equations
initial conditions; complete response;
zero-state and zero-input response
3. Systems
linearity; time-invariance; casuality;
stability; unit-step and unit-impulse
response; convolution sum and convolution
integral
4. Discrete and Continuous Fourier Series
periodic signals; convergence; properties
5. Discrete and Continuous Fourier Transform
aperiodic signals; periodic signals;
properties; convolution; multiplication
6. Z-Transform and Laplace Transform
region of convergence; inverse transform;
relationship to the Fourier transform;
properties; convolution; multiplication;
unilateral transform
7. Sampling
sampling theorem; decimation and
interpolation; aliasing
8. Filtering
ideal filters; magnitude-phase
representation; low-pass filters; band-pass
filters; high-pass filters
9. Modulation
amplitude modulation; frequency modulation;
pulse-amplitude modulation; pulse-code
modulation; frequency-division multiplexing;
time-division multiplexing; time-division
multiplexing
10. Feedback
stability
11. Circuits
sinusoidal and periodic signal response;
frequency response; transfer function
12. Computer Simulations
MATLAB program simulations throughout course
This topic: ECE > WebHome > Courses > ECE310
Topic revision: r2 - 2011-03-08 - 18:18:29 - Main.ala
Copyright 2016 The Board of Trustees of the University of Illinois.webmaster@cs.uic.edu WISESTHelping Women Faculty AdvanceFunded by NSF | 448 | 1,998 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2020-40 | latest | en | 0.718633 |
https://docs.mosek.com/latest/toolbox/tutorial-pow-shared.html | 1,686,029,840,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224652235.2/warc/CC-MAIN-20230606045924-20230606075924-00172.warc.gz | 252,013,438 | 6,446 | 6.4 Power Cone Optimization¶
The structure of a typical conic optimization problem is
$\begin{split}\begin{array}{lccccl} \mbox{minimize} & & & c^T x+c^f & & \\ \mbox{subject to} & l^c & \leq & A x & \leq & u^c, \\ & l^x & \leq & x & \leq & u^x, \\ & & & Fx+g & \in & \D, \end{array}\end{split}$
(see Sec. 12 (Problem Formulation and Solutions) for detailed formulations). We recommend Sec. 6.2 (From Linear to Conic Optimization) for a tutorial on how problems of that form are represented in MOSEK and what data structures are relevant. Here we discuss how to set-up problems with the primal/dual power cones.
MOSEK supports the primal and dual power cones, defined as below:
• Primal power cone:
$\POW_n^{\alpha_k} = \left\{ x\in \real^n ~:~ \prod_{i=0}^{n_\ell-1} x_i^{\beta_i} \geq \sqrt{\sum_{j=n_\ell}^{n-1}x_j^2},\ x_0\ldots,x_{n_\ell-1}\geq 0 \right\}$
where $$s = \sum_i \alpha_i$$ and $$\beta_i = \alpha_i / s$$, so that $$\sum_i \beta_i=1$$.
• Dual power cone:
$(\POW_n^{\alpha_k}) = \left\{ x\in \real^n ~:~ \prod_{i=0}^{n_\ell-1} \left(\frac{x_i}{\beta_i}\right)^{\beta_i} \geq \sqrt{\sum_{j=n_\ell}^{n-1}x_j^2},\ x_0\ldots,x_{n_\ell-1}\geq 0 \right\}$
where $$s = \sum_i \alpha_i$$ and $$\beta_i = \alpha_i / s$$, so that $$\sum_i \beta_i=1$$.
Perhaps the most important special case is the three-dimensional power cone family:
$\POW_3^{\alpha,1-\alpha} = \left\lbrace x \in \real^3: x_0^\alpha x_1^{1-\alpha}\geq |x_2|,\ x_0,x_1\geq 0 \right\rbrace.$
which has the corresponding dual cone:
For example, the conic constraint $$(x,y,z)\in\POW_3^{0.25,0.75}$$ is equivalent to $$x^{0.25}y^{0.75}\geq |z|$$, or simply $$xy^3\geq z^4$$ with $$x,y\geq 0$$.
For other types of cones supported by MOSEK, see Sec. 15.8 (Supported domains) and the other tutorials in this chapter. Different cone types can appear together in one optimization problem.
6.4.1 Example POW1¶
Consider the following optimization problem which involves powers of variables:
(6.9)$\begin{split}\begin{array} {lrcl} \mbox{maximize} & x_0^{0.2}x_1^{0.8} + x_2^{0.4} - x_0 & & \\ \mbox{subject to} & x_0+x_1+\frac12 x_2 & = & 2, \\ & x_0,x_1,x_2 & \geq & 0. \end{array}\end{split}$
We convert (6.9) into affine conic form using auxiliary variables as bounds for the power expressions:
(6.10)$\begin{split}\begin{array} {lrcl} \mbox{maximize} & x_3 + x_4 - x_0 & & \\ \mbox{subject to} & x_0+x_1+\frac12 x_2 & = & 2, \\ & (x_0,x_1,x_3) & \in & \POW_3^{0.2,0.8}, \\ & (x_2,1.0,x_4) & \in & \POW_3^{0.4,0.6}. \end{array}\end{split}$
The two conic constraints shown in (6.10) can be expressed in the ACC form as shown in (6.11):
(6.11)$\begin{split}\left[\begin{array}{ccccc}1&0&0&0&0\\0&1&0&0&0\\0&0&0&1&0\\0&0&1&0&0\\0&0&0&0&0\\0&0&0&0&1\end{array}\right] \left[\begin{array}{c}x_0\\x_1\\x_2\\x_3\\x_4\end{array}\right] + \left[\begin{array}{c}0\\0\\0\\0\\1\\0\end{array}\right] \in \POW^{0.2,0.8}_3 \times \POW^{0.4,0.6}_3.\end{split}$
Setting up the linear part
The linear parts (constraints, variables, objective) are set up exactly the same way as for linear problems, and we refer to Sec. 6.1 (Linear Optimization) for all the details. The same applies to technical aspects such as defining an optimization problem, retrieving the solution and so on.
Setting up the conic constraints
To define the conic constraints, we set the prob.f and prob.g in such a way that $$Fx+g$$ is the vector consisting of the six affine expressions appearing in the conic constraints of (6.10) in the same order. The domains and dimensions of affine conic constraints are specified using the structure accs. Each power cone is specified using its dimension (3), followed by the number of additional parameters (2) and finally those parameters (the $$\alpha$$ exponents for each cone).
Listing 6.7 demonstrates how to solve the example (6.9) using MOSEK.
Listing 6.7 Script implementing problem (6.9). Click here to download.
function pow1()
clear prob;
[r, res] = mosekopt('symbcon');
% Specify the non-conic part of the problem.
% Variables number 1,2,3 correspond to x,y,z, variables 4,5 are auxiliary
prob.c = [-1 0 0 1 1];
prob.a = [1 1 0.5 0 0];
prob.blc = [2.0];
prob.buc = [2.0];
prob.blx = [-inf -inf -inf -inf -inf];
prob.bux = [ inf inf inf inf inf];
% Specify the cones as affine conic constraints.
% Two conic constrains with the power cone, both of dimension 3
prob.accs = [res.symbcon.MSK_DOMAIN_PRIMAL_POWER_CONE 3 2 0.2 0.8 res.symbcon.MSK_DOMAIN_PRIMAL_POWER_CONE 3 2 0.4 0.6];
% The matrices such that f * x + g = [x(1), x(2), x(4), x(3), 1, x(5)]
prob.f = sparse( [1, 2, 3, 4, 6], [1, 2, 4, 3, 5], ones(1, 5) );
prob.g = [0 0 0 0 1 0];
% That means
%
% (x(1), x(2), x(4)) \ in PPOW_3(0.2, 0.8)
% (x(3), 1, x(5)) \ in PPOW_3(0.4, 0.6)
%
% which is equivalent to
%
% |x(4)| <= x(1)^0.2 * x(2)^0.8
% |x(5)| <= x(3)^0.4
% Optimize the problem.
[r,res]=mosekopt('maximize',prob);
% Display the primal solution.
res.sol.itr.xx'
For a step by step introduction to formulating problems with affine conic constraints (ACC) see also Sec. 6.2 (From Linear to Conic Optimization). | 1,904 | 5,105 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.671875 | 4 | CC-MAIN-2023-23 | latest | en | 0.692078 |
https://mathfilefoldergames.com/2016/11/16/madeline-hunters-8-step-lesson-plan-format/ | 1,696,292,425,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233511023.76/warc/CC-MAIN-20231002232712-20231003022712-00461.warc.gz | 404,517,269 | 21,670 | Madeline Hunter’s 8-Step Lesson Plan Format
Over the past twenty years, teachers have been following the teachings of Dr. Madeline Hunter based on her belief that in order to be effective a teacher must plan a lesson according to a certain methodology. This methodology required a lesson contain eight elements that would maximize and enhance learning. She labeled and described each of the eight elements and launched a teacher training career. Her eight elements of lesson design has certainly stood the test of time and is still used in colleges’ teacher training courses and by school districts’ judging of the effectiveness of teachers.
It was never Hunter’s intention that every lesson must include all of the eight elements to be effective. Hunter actually never labeled her suggested lesson plan elements as the “Eight-Step Lesson Plan”. The misunderstanding began in 1976 when Hunter and Doug Russel wrote an article entitled “Planning for Effective Instruction Design”. In this article they described the elements that should be considered in the design of a lesson. They stated that the inclusion of the following elements would increase the probability of the students’ success of reaching the objective of the lesson: objective, anticipatory set, input, modeling, checking for understanding, guided practice, independent practice and closure.
Ever since then, teachers have been trying to fit all the elements into all their lesson plans, and administrators looked for the including of all seven steps in teachers’ lessons. It wasn’t long before teachers complained that all the steps didn’t seem to be appropriate and should not always be followed step by step. Once a teacher fully understands each of the elements, he or she can better select the ones that would lead to the students’ reaching the lesson objective. For the purpose of your understanding each element, they are defined below with an example applied to the element. For the sake of demonstration, examples are based on a classroom lesson on fractions and percentages.
Note: Materials needed for this 45 minute lesson include: a jar of M&Ms or Skittles, markers or crayons in various colors, calculator and worksheet for recording calculations.
1. Anticipatory Set:
The purpose of the anticipatory set is to both get the students focused on what they will learn with today’s lesson and to get them interested in the topic as well.
Before the lesson begins, the teacher initiates a short activity designed to get the students’ attention. It can be a hand-out given to the students as they enter the classroom, one or two problems presented on an overhead transparency, review questions on yesterday’s lesson on the front board or today’s lesson agenda written on the front board.
For example: The teacher will be teaching a lesson on percentages and fractions. In preparing for the lesson before class, the teacher places 4 or 5 transparent jars filled with a mixture of various colored M&Ms or Skittles on a desk or table at the front of the class. Four questions are written on the front board:
• “What fraction would represent the number of red M&Ms (or Skittles) in the first jar?“
• “How could we determine the fraction to represent the number of red M&Ms (or Skittles) in the first jar?”
• “What is the percentage of blue M&Ms (or Skittles) in your group’s jar?”
• “How could we determine the percentage of blue M&Ms (or Skittles) in your group’s jar?”
2. Objective and Purpose:
When students know what they will be learning and why, they will learn more effectively. The students need to know what they will learn, how they will learn it, and why they are learning it.
The objectives of the M&Ms lesson are:
• The students will know what a fraction is.
• The students will be able to calculate a fraction.
• The students will know what a percent is.
• The students will be able to calculate a percent.
• The students will know how percents and fractions are related.
• The students will be able to use fractions to calculate percents.
3. Input:
The teacher presents and/or develops the new knowledge, skill or process to the students in an effective manner. The effective manner could be through discussion, questions and answers, discovery, discussion, listening, reading or observing. The input should include vocabulary, concepts and skills the students need to understand in order to be successful at accomplishing the objectives.
Through questions and answers for the lesson on fractions and percents, the teacher will lead students in a discussion of fractions and percents, how percents are calculated and how percents and fractions are related.
4. Modeling:
Modeling makes it possible for students to actually “see” what they will be doing and learning. The students’ understanding of the concepts is increased by the teacher demonstrating them.
Using a transparency of the worksheet, the first jar of M&Ms (or Skittles) and questions and answers, the teacher demonstrates how to get answers to the first two questions written on the front board.
• “What fraction would represent the number of red M&Ms (or Skittles) in the first jar?“
• “How could we determine the fraction to represent the number of red M&Ms (or Skittles) in the first jar?”
5. Checking for Understanding:
It is essential that the teacher makes sure the students understand what has been presented. She or he applies questioning strategies in order to determine the lesson pace.
6. Guided Practice:
The new learning is practiced by the students as the teacher supervises them directly. Using a hear and see approach, the teacher leads the students through steps to perform the skill introduced.
At this time, the teacher can arrange students into cooperative groups for the fraction and percent lesson. A jar of M&Ms (or Skittles), worksheets and markers or crayons would be distributed to each group. As the teacher demonstrates figuring out the answers to the second two questions written on the front board, the students also follow in order to determine the answers to the second two questions:
• “What is the percentage of blue M&Ms (or Skittles) in your group’s jar?”
• “How could we determine the percentage of blue M&Ms (or Skittles) of your group’s jar?”
7. Independent Practice & Application of Principles:
The teacher assigns independent practice when convinced the students fully understand the new material. The students now practice the skills taught by the previous steps. The purpose of this section is to reinforce learning through practice. It can be homework, individual work or group work done in class. It can be a means for students to apply the principles of the lesson they have learned.
Each group is directed by the teacher to select another color of the M&Ms (or Skittles) and answer all four questions written on the front board. The students record their calculation results on the worksheet. The worksheets can be used by the teacher to evaluate the success and progress of the students.
8. Closure:
Most lessons end with closure which basically wraps up the lesson. It can entail a question such as “What have we learned today?” It can also be a wrap up statement given orally by the teacher and shown on a transparency on the overhead at the same time. It can be a kind of final “check for understanding” since it reviews and clarifies the lesson’s key points. It also provides a cue to the students that the lesson is ending.
Note: Every lesson does not necessarily include all eight lesson elements. There are also times when one of these elements occurs more than once in a lesson. Since the eight elements are only meant to guide the teacher’s thinking when planning a specific lesson, the list shouldn’t be used as a rigid formula. There can also be times when it can take more than one class to include all the elements.
Using Math Games in Madeline Hunter’s 8-Step Lesson Plan Format
When it comes to lesson plans that work, Madeline Hunter’s 8-step format for lesson plans has been known to withstand the test of time. Many school districts and individual teachers consider this the format for a successful lesson. The question now is, “How do Math Games fit into Hunter’s lesson format?”
Step 1: It is essential that you, the teacher, know the objective of your lesson. Only then can you select the file folder game to match the objective of your lesson. The purpose of using the file folder games is not to fill time. If they are to have any value to the lesson, they must be an integral part of the lesson plan.
Step 2: Customize the game level. Since you know the level of your students, you can revise the game’s rules sufficiently to be sure your students are challenged. Adaptation suggestions for all of the 42 games are included in the book.
Step 3: You are now ready to begin the lesson plan. Hunter’s lesson plan format begins with the “Anticipatory Set”. The purpose of the anticipatory set is to grab the attention of the students. This is where your creativity comes into the plan. You can have a problem on the front board for students to attempt to solve, or you can simply present the game.
Step 4: Once you have the students’ attention, it is time to let them know exactly what your objective is for the lesson. After introducing the vocabulary, concepts and skills to be taught by playing the game, the students should be ready to play.
Step 5: If necessary, you can demonstrate or model how the game is played. This can be the “input” segment of the lesson.
Step 6: To play the game, arrange students in cooperative groups. Allow the students to play independently while learning. The teacher can at that time circulate around the room in order to offer help to students in the “guided practice” segment of the lesson.
Step 7: The closure of the lesson can be to have students report to you what they learned by playing the game.
Step 8: One of the simpler games can be sent home for homework and independent practice with an application of principles. | 2,050 | 10,017 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2023-40 | latest | en | 0.957011 |
https://www.numbersaplenty.com/320080 | 1,719,108,120,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198862425.28/warc/CC-MAIN-20240623001858-20240623031858-00179.warc.gz | 820,896,485 | 3,257 | Search a number
320080 = 2454001
BaseRepresentation
bin1001110001001010000
3121021001211
41032021100
540220310
610505504
72502115
oct1161120
9537054
10320080
111a9532
12135294
13b28c7
148490c
1564c8a
hex4e250
320080 has 20 divisors (see below), whose sum is σ = 744372. Its totient is φ = 128000.
The previous prime is 320063. The next prime is 320081. The reversal of 320080 is 80023.
It can be written as a sum of positive squares in 2 ways, for example, as 53824 + 266256 = 232^2 + 516^2 .
It is a tau number, because it is divible by the number of its divisors (20).
It is a self number, because there is not a number n which added to its sum of digits gives 320080.
It is a congruent number.
It is not an unprimeable number, because it can be changed into a prime (320081) by changing a digit.
320080 is an untouchable number, because it is not equal to the sum of proper divisors of any number.
It is a pernicious number, because its binary representation contains a prime number (7) of ones.
It is a polite number, since it can be written in 3 ways as a sum of consecutive naturals, for example, 1921 + ... + 2080.
2320080 is an apocalyptic number.
It is an amenable number.
320080 is an abundant number, since it is smaller than the sum of its proper divisors (424292).
It is a pseudoperfect number, because it is the sum of a subset of its proper divisors.
320080 is a wasteful number, since it uses less digits than its factorization.
320080 is an odious number, because the sum of its binary digits is odd.
The sum of its prime factors is 4014 (or 4008 counting only the distinct ones).
The product of its (nonzero) digits is 48, while the sum is 13.
The square root of 320080 is about 565.7561312085. The cubic root of 320080 is about 68.4047373120.
The spelling of 320080 in words is "three hundred twenty thousand, eighty". | 536 | 1,857 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.484375 | 3 | CC-MAIN-2024-26 | latest | en | 0.90511 |
https://mph-to-kmh.com/500-km-to-miles/ | 1,719,267,030,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198865490.6/warc/CC-MAIN-20240624214047-20240625004047-00464.warc.gz | 354,760,545 | 15,377 | # Converting 500 km to Miles: The Ultimate Guide to Simple and Accurate Conversion Rates
## Converting 500 km to miles: A comprehensive guide for accurate calculations
### Understanding the conversion factor
Converting kilometers to miles may seem complex at first, but with the right knowledge, it can be a straightforward process. To convert 500 kilometers to miles, we must understand the conversion factor. The conversion factor between kilometers and miles is approximately 0.6214. This means that for every kilometer, there are 0.6214 miles.
### Step-by-step calculation
To convert 500 kilometers to miles, simply multiply the distance in kilometers by the conversion factor. In this case, we will multiply 500 by 0.6214. The calculation looks like this: 500 km x 0.6214 = 310.7 miles. Therefore, 500 kilometers is equivalent to approximately 310.7 miles.
### Importance of accurate calculations
Accurate conversions are crucial when dealing with international distances, travel planning, or any situation requiring precise measurements. Whether you’re trying to understand road trip distances or need to convert distances for business purposes, it’s essential to obtain accurate results. By following the correct conversion methods, such as the one explained above, you can ensure accurate calculations.
## The simplicity of converting 500 km to miles: A step-by-step tutorial
### Step 1: Understand the Conversion Factor
To convert kilometers to miles, you need to know the conversion factor between the two units. The conversion factor for kilometers to miles is 0.621371. This means that for every kilometer, there are approximately 0.621371 miles. Understanding this conversion factor is crucial for accurately converting 500 km to miles.
### Step 2: Multiply by the Conversion Factor
Now that you know the conversion factor, you can easily convert 500 km to miles. Simply multiply 500 by the conversion factor, 0.621371. Using a calculator or doing the math manually, you’ll find that 500 km is equivalent to approximately 310.6855 miles.
Note: When converting kilometers to miles, it’s important to remember that the result will be an approximation since the conversion factor is not an exact value.
### Step 3: Double-Check Your Conversion
To ensure the accuracy of your conversion, it’s always a good idea to double-check your work. Here’s a simple way to do that for converting 500 km to miles:
– Divide the result (310.6855 miles) by the conversion factor (0.621371).
– You should get approximately 500 km. If the result is close to 500 km, then your conversion is correct.
Pro Tip: It’s helpful to memorize or bookmark the conversion factor for kilometers to miles for quick and easy calculations in the future.
By following these steps, you can easily convert 500 km to miles. This knowledge is valuable when working with different measurement systems, planning road trips, or understanding distances in different countries.
## Understanding the conversion process: 500 km to miles and its significance
When it comes to understanding the conversion process from kilometers to miles, one specific measurement that often comes up is 500 km. This distance is significant for various reasons and being able to convert it accurately to miles is important in many industries and everyday life.
Converting 500 km to miles allows us to easily compare and understand distances. For example, let’s say you’re planning a road trip in the United States and your destination is 500 km away. The conversion to miles would give you a better idea of the distance you’ll be traveling, making it easier to plan for rest stops, fuel stops, and estimating travel times.
To convert 500 km to miles, you can use the following formula: miles = kilometers * 0.621371192. This conversion factor ensures accurate results, allowing you to quickly and confidently convert 500 km to miles in any situation.
You may also be interested in: Converting 500 kg to lbs: Quick and Easy Guide for Accurate Weight Conversion
Understanding the conversion process and knowing how to convert 500 km to miles is not only helpful for road trips, but it also has significant applications in various industries, such as aviation and logistics. Properly converting distances ensures safety and efficiency, especially when it comes to flight planning, route optimization, and fuel consumption calculations. By having a clear and accurate understanding of the conversion process, professionals can make informed decisions that have a direct impact on their operations and the overall success of their projects.
You may also be interested in: Mastering the Conversion: Unlocking the Secrets of 400 Meters to Miles
## Convert 500 km to miles effortlessly with these helpful tips and tricks
### Understanding the Conversion Factor
When it comes to converting kilometers to miles, understanding the conversion factor is key. The conversion factor for kilometers to miles is 0.62137119, which means that for every kilometer, there are approximately 0.62 miles. This conversion factor remains constant and can be used to convert any distance from kilometers to miles. To convert 500 km to miles, simply multiply 500 by the conversion factor, which will give you the equivalent distance in miles.
Pro Tip: Remembering the conversion factor can save you time and effort when converting kilometers to miles. You can also use online conversion calculators or mobile apps to quickly get the converted value.
### The Calculation Method
To convert 500 km to miles manually, you can multiply the distance in kilometers by the conversion factor. In the case of 500 km, the calculation would be 500 km * 0.62137119, which equals approximately 310.686 miles. Keep in mind that the result might have decimals, so you might need to round it off depending on your requirements.
Pro Tip: If you require a more precise result, you can use the rounded value of the conversion factor or use more decimal places in the calculation.
### Common Mistakes to Avoid
While converting kilometers to miles is a straightforward process, there are a few common mistakes to watch out for. One common mistake is using the incorrect conversion factor. Make sure you are using the correct value, as using an outdated or incorrect conversion factor can lead to inaccurate results. Another mistake to avoid is forgetting to include the unit of measurement. Always double-check that you include the appropriate unit when stating the converted distance, whether it’s in kilometers or miles.
Pro Tip: It’s always a good idea to double-check your calculations using online conversion tools or comparing your results with known conversion values to ensure accuracy.
By following these helpful tips and tricks, converting 500 km to miles can be a breeze. Remember the conversion factor, apply the calculation method correctly, and be mindful of common mistakes. Whether you need to convert distances for travel planning or other purposes, having a solid understanding of this conversion can come in handy.
## Unlocking the mystery: Converting 500 km to miles made easy
### The conversion formula
Converting kilometers to miles may seem like a daunting task, but it is actually quite simple once you know the formula. The basic conversion formula is as follows:
Miles = Kilometers * 0.621371
With this formula, you can easily convert any number of kilometers to miles, including 500 km. All you need to do is multiply the given number of kilometers by the conversion factor of 0.621371.
### Why convert?
You might be wondering why anyone would need to convert kilometers to miles in the first place. Well, the answer is simple – it’s all about understanding and convenience. While most countries around the world use kilometers as their primary unit of distance, the United States still predominantly uses miles. Therefore, if you’re traveling or trying to understand distances in the US, it’s essential to be able to convert between the two units.
You may also be interested in: Discover How to Convert 37.6 °C to °F - The Ultimate Guide for Accurate Temperature Conversions
### A practical example: converting 500 km to miles
Let’s take a practical example of converting 500 kilometers to miles. Using the formula mentioned earlier, we can easily calculate the result:
Miles = 500 km * 0.621371
Simplifying the equation, we find that 500 kilometers is approximately equal to 310.7 miles. So, if you ever come across a distance of 500 km and need to know the equivalent in miles, you now have the knowledge to make the conversion effortlessly.
By understanding the simple conversion formula and the reasons behind the need for conversion, you can unlock the mystery of converting kilometers to miles. Whether you’re a traveler, a student, or simply curious about distances, having this knowledge will make your life easier when dealing with different units of measurement. | 1,796 | 8,947 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.15625 | 4 | CC-MAIN-2024-26 | latest | en | 0.876224 |
http://photo.stackexchange.com/questions/30730/what-magnification-would-i-get-with-extension-tubes/30733 | 1,455,248,734,000,000,000 | text/html | crawl-data/CC-MAIN-2016-07/segments/1454701163421.31/warc/CC-MAIN-20160205193923-00088-ip-10-236-182-209.ec2.internal.warc.gz | 175,400,025 | 17,386 | # What magnification would I get with extension tubes?
I have a Nikon 60mm f/2.8 Micro lens (DX sensor) and I haven't been able to exactly determine what magnification I can get with Kenko 12mm, 20mm, 36mm extension tubes. Suppose I would use all of them together.
Also how many stops of light would I lose using them?
Depending on the answer, is it worth it to obtain the tubes or the macro effect wouldn't be so different and I would lost a lot of light instead? I would buy them only if I could create images with magnification similar to this one
-
possible duplicate of How can I calculate what the effect of an extension tube will be? – mattdm Dec 22 '12 at 14:29
@mattdm I still don't know if it helps on macro lenses (which can focus at close distance without help) or I have to use some zoom (70-300) with the tubes to get good magnification. – TTT Dec 22 '12 at 14:44
A "macro lens" is just a (well-corrected) ordinary lens with a built-in, adjustable extension tube. Adding additional tubes will get you closer than 1:1 (by how much depends on the actual focal length of the lens at its 1:1 focusing distance), as with any other lens. And, as with any other lens, you may hit a point where the required subject distance from the optical centre of the lens for a given magnification lies inside of the lens, at which point the lens would need to be reversed. – user2719 Dec 22 '12 at 19:18
## 1 Answer
The previous post link talks about magnification.
As far as light loss, I would first suggest you just use the meter in your camera. It should show the relative setting no matter if you have neutral density filters or extension tubes attached.
If you really want to calculate this stuff manually, adding extension tubes simply alters the focal length of a lens. And the aperture of a lens is a ratio of its focal length.
Using the formula N = f/D (F-stop = focal length divided by diameter of lens opening), your 60mm f/2.8 lens has an pupil diamter of 21.43mm.
Now, you add 12mm of extension and your focal length becomes 72mm, but the physical aperture and diameter sizes of the lens did not change.
This means that your new f-stop value, with 12mm extension, becomes f/3.5 which is 72mm / 21.43 opening size, or, 2/3 of a stop less light.
- | 565 | 2,267 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.65625 | 3 | CC-MAIN-2016-07 | longest | en | 0.95119 |
https://8thshore.com/betting/how-many-ways-can-you-roll-a-9-with-3-dice.html | 1,638,071,084,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964358443.87/warc/CC-MAIN-20211128013650-20211128043650-00355.warc.gz | 161,330,802 | 18,286 | # How many ways can you roll a 9 with 3 dice?
Contents
There are 216 possible ways to roll three dice in order. The possible ways to roll a 9 are: 1–2–6 (6 combinations)
## How many ways can you roll a 9?
The probability of throwing any given total is the number of ways to throw that total divided by the total number of combinations (36).
Probabilities for the two dice.
Total Number of combinations Probability
7 6 16.67%
8 5 13.89%
9 4 11.11%
10 3 8.33%
## How many ways can you get 10 with 3 dice?
There are 135 successes and 216 possible outcomes, so the probability we seek is 135/216 = 62.5%. When the first die shows 1, there are 1+2+3+4 successes, corresponding to the values on the second die: 3,4,5 or 6.
## How many ways can you roll an 8 with 3 dice?
The total number of ways to roll an 8 with 3 dice is therefore 21, and the probability of rolling an 8 is 21/216, which is less than 5/36.
## Why is 7 the most common dice roll?
So why is 7 the most common dice roll for two dice? Seven it the most common dice roll with two dice because it has the most number of different combinations that add up to seven. For example, a player can roll 1 and 6; 2 and 5; 3 and 4; 4 and 3; 5 and 2; and 6 and 1. … No other dice total has that many combinations.
IT IS INTERESTING: Quick Answer: Can I apply for DV lottery every year?
## What is the probability that you will roll a 9?
Two (6-sided) dice roll probability table
Roll a… Probability
6 15/36 (41.667%)
7 21/36 (58.333%)
8 26/36 (72.222%)
9 30/36 (83.333%)
## How many ways can you roll a 7?
There are six ways to make the seven. By knowing how the numbers are made, you can calculate the odds of making any numberbefore the seven is rolled. Since the number 7 can be rolled six ways, you dividethe number six by the number of ways a number is rolled.
## How many ways can you roll a 13 with 3 dice?
Three dice are rolled. What is the probability of getting a sum of 13? – Quora. There are a total of 216 combinations in throwing of 3 dice. Therefore there are a total of 3 + 6 + 6 + 3 + 3 = 21 such combinations.
## How many ways can you roll a 7 with 3 dice?
Originally Answered: When 3 dice are rolled, what is the probability of getting a sum of 7? Four combinations of numbers from three normal dice give a total of 7. 115, 124, 133 and 223. This gives 15 distinct ways of obtaining a total of 7.
## How do you find the probability of 3 dice?
We divide the total number of ways to obtain each sum by the total number of outcomes in the sample space, or 216. The results are: Probability of a sum of 3: 1/216 = 0.5% Probability of a sum of 4: 3/216 = 1.4% | 764 | 2,646 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.40625 | 4 | CC-MAIN-2021-49 | latest | en | 0.914016 |
https://www.codeblogbt.com/archives/520211 | 1,550,286,464,000,000,000 | text/html | crawl-data/CC-MAIN-2019-09/segments/1550247479838.37/warc/CC-MAIN-20190216024809-20190216050809-00457.warc.gz | 792,933,385 | 8,832 | Article From:https://www.cnblogs.com/sunrisepeak/p/9652968.html
• An example is given to illustrate the process of algorithm design and the techniques and ideas used in design and analysis (c / C + +)
Example: finding the greatest common divisor of two positive integers.
1. Mathematical model: an integer of a, B & gt; 0, finding that c, C can divide a, B and a / C and B / C are mutually prime (no common number).
2. Problem analysis:
• Decomposition factor method: the greatest divisor of a and B.
• Decomposition quality factor method: multiplication of a and B divisibility
• Short division: all common divisors are multiplied.
• Rotary phase division:
3. Algorithm description:
Decomposition factor method:
1. Decomposition factor method: ☛source code
int a, b, flag;
input(a,b)
if(a>b)
a，bExchange valuefor(Decreasing from a cycle to searching the greatest common factor.if(Can divide a at the same time, B is the common factor.Jump out of circulation;Print (flag);
2. Algorithm description:
• Define flag for marking common factor (or a loop), easy to understand and define flag variables.
• Compare the size of a and B, ensure that a is the smallest number, and facilitate the following loop (s-code provides one.Two variable exchange valueMethods)
• The greatest common factor is required.Reverse searchReduce search times (from big to small).
• If you want to type A, b value (that is, multiple sets of data) and a or B equal to 0, end the input.while(cin >> a >> b && a && b)Input mode: any conditional condition in while is false stop cycle (0 is false, non 0 is true).
3. Algorithmic analysis: It is best to cycle once, the worst cycle a, then T (n) = 1/n (1 +…. + n) = (1/n) * n * (n + 1) / 2 = 1/2 * n + 1 / 2 time complexity is O (n).
Decomposition mass factor method:
1. Decomposition mass factor method:
#include<iostream>
#include<cmath>
using namespace std;
bool pNumber(int n) //Quality of judgment
{
for(int i = 2; i <= sqrt(n); i++)
if(n%i == 0)
return false;
return true;
}
void exchange(int &x, int &y) //Swap the value of X, y
{
int temp = x;
x = y;
y = temp;
}
int main()
{
int a, b;
while(cin >> a >> b && a != 0 && b != 0){
int flag = 1;
if(a > b)
exchange(a, b);
for(int i = 2; i <= a; i++)
if(a%i == 0 && b%i == 0 && pNumber(i)){
flag *= i;
// cout << "this is debug!----" <<i << endl;
}
cout << flag << endl;
}
return 0;
}
2. Algorithm description:
• The modularization idea is used to modularize some general functions of main function (judging prime number and exchanging variable value), and subdivide them step by step from top to bottom.
• Reduce the number of cycles by virtue of the advantages of short circuit and (& &).:pNumber()The number of cycles caused by different functions is different.if(a%i == 0 && b%i == 0 && pNumber(i)< if(pNumber(i) &&a%i == 0 && b%i == 0 )
3. Algorithm analysis: suppose that I can divide a at the same time, and the probability of B is p.:O(n) = n^(5/2).
Short division:
1. Short division:
#include<iostream>
#include<cmath>
using namespace std;
int main()
{
int a, b, t, i;
cin >> a >> b;
t = 1;
for(i = 2; i <= a && i <= b; i++)
while(a%i == 0 && b%i == 0)
{
t = t*i;
a = a/i;
b = b/i;
}
cout << t << endl;
return 0;
}
Rotary phase division:
1. Rotary phase division:
#include<iostream>
#include<cmath>
using namespace std;
int main()
{
int a, b, c;
cin >> a >> b;
if(b == 0 || a == 0)
{
cout << "data error!" << endl;
return 0;
}
else
{
c = a % b;
while(c != 0)
{
a = b;
b = c;
c = a %b;
}
}
cout << b;
return 0;
}
☛source code——For reference only | 1,018 | 3,559 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2019-09 | latest | en | 0.732271 |
https://physics.stackexchange.com/questions/287311/most-general-form-for-the-wave-equation | 1,721,211,923,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514759.39/warc/CC-MAIN-20240717090242-20240717120242-00775.warc.gz | 398,700,405 | 41,103 | # Most general form for the wave equation
What is the most general form for the wave equation ? Is it $\frac{\partial^2 \Psi}{\partial t^2}-v^2\nabla^2\Psi=0$?
For instance, can $\frac{\partial^2 \Psi}{\partial t^2}-v^2\nabla^2\Psi=cte$ be a wave equation? If yes, what is the solution in that case.
Mason handled the distinction between inhomogeneous and homogeneous differential equations, but if one is speaking of the most general possible form of the wave equation, it is,
$$\square\phi^{i_1\dots i_m}_{j_1 \dots j_n}(x) = f^{i_1 \dots i_m}_{j_1 \dots j_n}(x)$$
where both fields are rank $(m,n)$ tensors, acted upon by the Laplace-Beltrami operator $\square = \nabla^a \nabla_a$ whose action on the tensors depends on both the metric and their rank. For a scalar field with metric $\eta_{\mu \nu}$, it reduces to the most familiar form of the wave equation, $(\partial^2_t - \nabla^2)\phi = f$. (The above can also be recast in the language of differential forms.)
However, in a way this does not cover all the possibilities. For example, in general relativity, for a perturbation $h_{ab}$ of the metric, the first order change in the curvature is,
$$\delta R_{ab} \propto \Delta_L h_{ab} = \square h_{ab} -2 \nabla_{(a} \nabla^c \bar{h}_{b)c} -2 R_{d(a} h^d_{b)} +2 R_{acbd}h^{cd}$$
which is understood as the curved space 'wave operator' in the literature because it certainly admits wave solutions but is clearly not equivalent to the wave equation above as it contains other terms involving curvature tensors. Thus, the 'most general form' of the wave equation isn't something we can really write down, unless your idea of it is strictly $(\partial^2_t - \nabla^2)\phi = f$.
I'm not sure what you mean by $cte$, but I'm assuming it's some constant but I may be misinterpreting
We often talk about two classes of differential equation, homogeneous and inhomogeneous. This distinction is the root of your question, $$\frac{1}{v^2}(\partial_t)^2 f(\vec{r},t) - \nabla ^2 f(\vec{r},t) = 0$$ is the homogeneous form of the wave equation, whereas $$\frac{1}{v^2}(\partial_t)^2 f(\vec{r},t) - \nabla ^2 f(\vec{r},t) = u(\vec{r},t)$$ is the inhomogeneous wave equation ($u(\vec{r},t)$ can also be constant if we like). This arises all over the place. One example is that electromagnetic radiation in the presence of charges and currents is governed by the inhomogeneous wave equation, the homogeneous form is only valid when $\rho=0$ and $\vec{J}=0$. Depending on who you ask, I think most people would still say the the inhomogeneous wave equation is a wave equation, but that's up to taste as it's solutions can end up having a very different character to the homogeneous ones.
In general there's not much I can say about these solutions since they'll depend heavily on the form of $u$, though I'm sure some googling will give you plenty of examples.
• Perfect. And what about damped wave equation ? What is its form ? Commented Oct 19, 2016 at 7:18
......................... | 841 | 2,992 | {"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": 2, "x-ck12": 0, "texerror": 0} | 3.140625 | 3 | CC-MAIN-2024-30 | latest | en | 0.887399 |
https://www.numberempire.com/8412075 | 1,623,498,014,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623487582767.0/warc/CC-MAIN-20210612103920-20210612133920-00313.warc.gz | 828,020,476 | 7,715 | Home | Menu | Get Involved | Contact webmaster
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# Number 8412075
eight million four hundred twelve thousand seventy five
### Properties of the number 8412075
Factorization 3 * 3 * 5 * 5 * 7 * 7 * 7 * 109 Divisors 1, 3, 5, 7, 9, 15, 21, 25, 35, 45, 49, 63, 75, 105, 109, 147, 175, 225, 245, 315, 327, 343, 441, 525, 545, 735, 763, 981, 1029, 1225, 1575, 1635, 1715, 2205, 2289, 2725, 3087, 3675, 3815, 4905, 5145, 5341, 6867, 8175, 8575, 11025, 11445, 15435, 16023, 19075, 24525, 25725, 26705, 34335, 37387, 48069, 57225, 77175, 80115, 112161, 133525, 171675, 186935, 240345, 336483, 400575, 560805, 934675, 1201725, 1682415, 2804025, 8412075 Count of divisors 72 Sum of divisors 17732000 Previous integer 8412074 Next integer 8412076 Is prime? NO Previous prime 8412049 Next prime 8412077 8412075th prime 149390837 Is a Fibonacci number? NO Is a Bell number? NO Is a Catalan number? NO Is a factorial? NO Is a regular number? NO Is a perfect number? NO Polygonal number (s < 11)? NO Binary 100000000101101110101011 Octal 40055653 Duodecimal 2998123 Hexadecimal 805bab Square 70763005805625 Square root 2900.357736556 Natural logarithm 15.9451787316 Decimal logarithm 6.9249031360948 Sine 0.91034078492925 Cosine -0.41385946321716 Tangent -2.1996374756123
Number 8412075 is pronounced eight million four hundred twelve thousand seventy five. Number 8412075 is a composite number. Factors of 8412075 are 3 * 3 * 5 * 5 * 7 * 7 * 7 * 109. Number 8412075 has 72 divisors: 1, 3, 5, 7, 9, 15, 21, 25, 35, 45, 49, 63, 75, 105, 109, 147, 175, 225, 245, 315, 327, 343, 441, 525, 545, 735, 763, 981, 1029, 1225, 1575, 1635, 1715, 2205, 2289, 2725, 3087, 3675, 3815, 4905, 5145, 5341, 6867, 8175, 8575, 11025, 11445, 15435, 16023, 19075, 24525, 25725, 26705, 34335, 37387, 48069, 57225, 77175, 80115, 112161, 133525, 171675, 186935, 240345, 336483, 400575, 560805, 934675, 1201725, 1682415, 2804025, 8412075. Sum of the divisors is 17732000. Number 8412075 is not a Fibonacci number. It is not a Bell number. Number 8412075 is not a Catalan number. Number 8412075 is not a regular number (Hamming number). It is a not factorial of any number. Number 8412075 is an abundant number and therefore is not a perfect number. Binary numeral for number 8412075 is 100000000101101110101011. Octal numeral is 40055653. Duodecimal value is 2998123. Hexadecimal representation is 805bab. Square of the number 8412075 is 70763005805625. Square root of the number 8412075 is 2900.357736556. Natural logarithm of 8412075 is 15.9451787316 Decimal logarithm of the number 8412075 is 6.9249031360948 Sine of 8412075 is 0.91034078492925. Cosine of the number 8412075 is -0.41385946321716. Tangent of the number 8412075 is -2.1996374756123
### Number properties
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Examples: 3628800, 9876543211, 12586269025 | 1,138 | 2,793 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2021-25 | latest | en | 0.478627 |
https://custom-essay-writing-service.net/business-and-finance-basics-mathematics-homework-help/ | 1,628,137,502,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046155322.12/warc/CC-MAIN-20210805032134-20210805062134-00146.warc.gz | 197,367,910 | 18,087 | # Business and finance basics | Mathematics homework help
1. If a car is depreciated in four years, what is the rate of depreciation using twice the straight-line rate?
A. 100%
B. 50%
C. 75%
D. 25%
2. In using horizontal analysis, comparative reports are
A. often used.
B. infrequently used.
C. always used.
D. never used.
3. Depreciation expense is located on the
A. the accounts payable documentation.
B. income statement.
C. balance sheet.
D. the accounts receivable documention.
4. The acid test ratio does not include
A. cash.
B. inventory.
C. supplies.
D. accounts receivable.
5. When are annuity due payments made?
A. At the beginning of the period
B. Monthly
C. At the end of the period
D. Yearly
6. Use the following information to answer the question:
Cost of car: \$26,000
Residual value: \$6,000
Life: 5 years
Using the given information, determine the depreciation expense for the first year straight-line method?
A. \$5,200
B. \$6,000
C. \$4,400
D. \$4,000
7. Ben Brown bought a home for \$225,000. He put down 20%. The mortgage is at 6 ½% for 30 years.
Using the tables in the Business Math Handbook that accompanies the course textbook, determine his
monthly payment.
A. \$1,216.80
B. \$1,319.40
C. \$1,139.40
D. \$1,319.04
8. At the beginning of each year for 14 years, Sherry Kardell invested \$400 that earns 10% annually. What
is the future value of Sherry’s account in 14 years?
A. \$12,709
B. \$13,100
C. \$14,000
D. \$12,309
9. Megan Mei is charged 2 points on a \$120,000 loan at the time of closing. The original price of the home
before the down payment was \$140,000. How much do the points in dollars cost Megan?
A. \$4,200
B. \$8,200
C. \$2,400
D. \$2,800
10. Dan Miller bought a new Toyota truck for \$28,000. Dan made a down payment of \$6,000 and paid
\$390 monthly for 70 months. What is the total finance charge?
A. \$13,300
B. \$11,300
C. \$27,300
D. \$5,300
11. Open credit in a revolving charge plan results in
A. as many charged purchases till credit limit is reached.
B. as many cash purchases till credit limit is reached.
C. one purchase per month.
D. the U.S. Rule being applied to each purchase.
12. Which one of the following methods is not based on the passage of time?
A. Units-of-production method
B. Declining-balance method
C. Straight-line method
D. None of these
13. Dick Hercher bought a home in Homewood, Illinois, for \$230,000. He put down 20% and obtained a
mortgage for 25 years at 8%. What is the total interest cost of the loan?
A. \$184,000.00
B. \$327,372.80
C. \$242,411.00
D. \$242,144.00
14. Jay Corporation has earned \$175,900 after tax. The accountant calculated the return on equity as
12.5%. Jay Corporation’s stockholders’ equity to the nearest dollar is
A. \$140,720,000.
B. \$140,720.
C. \$14,720.
D. \$1,407,200.
15. Lee Company has a current ratio of 2.65. The acid test ratio is 2.01. The current liabilities of Lee are
\$45,000. Assuming there are no prepaid expenses, the dollar amount of merchandise inventory is
A. \$28,800.
B. \$90,450.
C. \$90,540.
D. \$28,008.
16. Depreciation expense in the declining-balance method is calculated by the depreciation rate
A. times accumulated depreciation at year end.
B. times book value at beginning of year.
C. divided by book value at beginning of year.
D. plus book value at end of year.
17. Abe Aster bought a new split level for \$200,000. Abe put down 30%. Assuming a rate of 111∕2% on a
30-year mortgage, use the tables in the Business Math Handbook that accompanies the course textbook to
determine Abe’s monthly payment.
A. \$1,387.40
B. \$1,367.80
C. \$1,982.00
D. \$1,423.80
18. Jen purchased a condo in Naples, Florida, for \$699,000. She put 20% down and financed the rest at
5% for 35 years. What are Jen’s total finance charges?
A. \$606,823.20
B. \$600,000.00
C. \$457,425.60
D. \$626,863.20
19. At the beginning of each year, Bill Ross invests \$1,400 semiannually at 8% for nine years. Using the
tables in the Business Math Handbook that accompanies the course textbook, determine the cash value of
the annuity due at the end of the ninth year.
A. \$38,739.68
B. \$37,399.68
C. \$37,939.86
D. \$37,339.68
20. Using the tables in the Business Math Handbook that accompanies the course textbook, determine the
difference between the monthly payments on a \$120,000 home at 61∕2% and at 8% for 25 years.
A. \$115.20
B. \$91.12
C. \$81.12
D. \$151.02
21. Cost of merchandise sold equals beginning inventory
A. minus net purchases plus ending inventory.
B. minus net purchases minus ending inventory.
C. plus net purchases plus ending inventory.
D. plus net purchases minus ending inventory.
22. An annuity due can use the ordinary annuity table if one extra period is added and
A. three payments are subtracted from total value.
B. one payment is subtracted from total value.
C. two payments are added to total value.
D. one payment is added to total value.
23. Use the following information and the tables in the Business Math Handbook that accompanies the
course textbook to answer the question.
\$140.10 per month
Cash price: \$5,600
Down payment: \$0
Cash or trade months with bank-approved credit; amount financed: \$5,600
Finance charge: \$2,806
Total payments: \$8,406
What is the APR by table lookup?
A. 16.75%–17.00%
B. 17.25%–17.50%
C. 17.00%–17.25%
D. 16.50%–16.75%
24. What is a sinking fund?
A. It requires one lump sum payment at the beginning.
B. It doesn’t compound its money.
C. It’s not really an annuity.
D. It aids in meeting a future obligation.
25. A truck costs \$16,000 with a residual value of \$1,000. It has an estimated useful life of five years. If
the truck was bought on July 3, what would be the book value at the end of year 1 using straight-line rate?
A. \$12,500
B. \$16,000
C. \$14,500
D. \$1,500
Business and Finance Basics III and Statistics
1. Jim opened a new pizza shop. He insures his store for \$90,000.00 for fire. What is his premium if the
rate per \$100.00 is \$0.83?
A. \$74.70
B. \$700.00
C. \$74,700.00
D. \$747.00
2. Clay’s Fishing Shop’s beginning inventory is \$70,000 and ending inventory is \$36,500. What was Clay’s
average inventory?
A. \$48,000
B. \$53,250
C. \$18,250
D. \$35,000
3. Jay Miller insured his pizza shop for \$200,000 for fire insurance at an annual rate per \$100 of \$.49. At
the end of 10 months, Jay canceled the policy since his pizza shop went out of business. Using the tables in
the Business Math Handbook that accompanies the course textbook, determine the refund to Jay.
A. \$186.20
B. \$852.60
C. \$980
D. \$127.40
4. Personal property items do not include
A. furniture.
B. land.
C. jewelry.
D. autos.
5. Crestwood Paint Supply had a beginning inventory of 10 cans of paint at \$25.00 per can. They
purchased 20 cans during the month at \$30.00 per can. They had an ending inventory valued at \$500. How
much paint in dollars was used for the month?
A. \$850
B. \$1,350
C. \$350
D. \$250
6. Which one of the following statements is true about reduced paid-up insurance?
B. It results in a face amount less than the original amount .
C. It means the original face amount is continued for a certain number of years.
D. It continues for 20 years.
7. Bee Sting bought 400 shares of Google at \$399.75 per share. Assume a commission of 2% of the
purchase price. What is the total to Bee?
A. \$156,702
B. \$163,098
C. \$159,900
D. \$163,980
A. They contribute directly to the running of a business.
B. They’re directly related to a specific department.
C. They contribute indirectly to the running of a business.
D. They’re directly related to a specific product.
9. The cost ratio in the retail method is found by the cost of goods available for sale at cost divided by the
A. cost of goods available for sale at retail.
B. net sales.
C. ending inventory at retail.
D. net purchases at cost.
10. What is the retail method?
A. It doesn’t require a cost ratio.
B. It aids a company in not having to calculate an inventory cost for each individual item.
C. It’s not an estimate.
D. It eliminates the need to take a physical inventory.
11. Bauer Supply had total cost of goods sold of \$1,400 with 140 units available for sales. What was the
average cost per unit?
A. \$14
B. \$10
C. 14.10
D. \$140
12. Calculate the median from the following numbers: 16 + 9 + 10 + 5 + 4.
A. 5
B. 10
C. 9
D. 4
13. To avoid distortion of extreme values, a good indicator would be the
A. mode.
B. median.
C. weighted-mean.
D. mean.
14. Which one of the following items is subject to sales tax in the District of Columbia?
A. Roast beef
B. Milk
C. Shampoo
D. Tomatoes
15. Which one of the following statements is true of specific identification?
A. Ending inventory isn’t associated with specific purchase prices.
B. Low-cost items aren’t used in this method.
C. The specific purchase invoice prices aren’t used.
D. Flow of goods and flow of cost are the same.
16. The tax rate of \$.6943 in decimal can be expressed per \$100 as
A. \$69.43.
B. \$6.943.
C. \$690.3.
D. \$69.43 mills.
17. In terms of premium cost, the most expensive type of insurance is _______ insurance.
A. 20-payment life
B. 20-year endowment
C. straight-life
D. term
18. With net sales of \$40,000, beginning inventory at retail of \$14,000, ending inventory at retail of
\$20,000, and cost of goods sold of \$19,500, what is the inventory turnover at retail rounded to the nearest
hundredth?
A. 5.15
B. 5.23
C. 3.25
D. 2.35
19. The range of 35, 22, 43, 18, 22, 27, 48, 39, 31, and 16 is
A. 32.
B. 29.
C. 30.
D. 22.
20. Which one of the following statements is true of preferred stock?
A. It never receives dividends in arrears.
B. It never has a preference to dividends over common stockholders.
C. It has equal rights to common stock.
D. It can be cumulative.
21. Find the mean for the following numbers and do not round your answer to the nearest whole number:
38 + 18.05 + 25 + 26 + 46
A. 23.74
B. 32.14
C. 30.61
D. 21.45
22. Total sales of \$400,000 that included a 6% sales tax yields actual sales of
A. \$42,800.
B. \$37,537.58.
C. \$377,358.49.
D. \$48,200.
23. A bond quote of 82.25 in dollars is equal to
A. \$822.50.
B. \$82.25.
End of exam
C. \$8.25.
D. \$8,025.50.
24. Suppose Department A is 8,000 square feet, Department B is 5,000 square feet, and Department C is
6,000 square feet. What is the percent of overhead expense applied to Department C? (Round your answer
to the nearest whole percent.)
A. 68%
B. 26%
C. 32%
D. 42%
25. Mike’s condo has a market value of \$310,000. The property in Mike’s area is assessed at 40% of the
market value. The tax rate is \$145.10 per \$1,000 of assessed valuation. The tax for Mike is
A. \$16,992.40.
B. \$7,999.30.
C. \$7,999.40.
D. \$17,992.40. | 3,193 | 10,663 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4 | 4 | CC-MAIN-2021-31 | latest | en | 0.90808 |
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# simple q..
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28 Dec 2008, 20:28
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is there a typo in the attached q.
both A and B seem to be correct?
Thanks
Jack
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### Show Tags
28 Dec 2008, 21:52
jackychamp wrote:
Attachment:
q.GIF
is there a typo in the attached q.
both A and B seem to be correct?
Thanks
Jack
yup. both A and B cannot be the values of x.
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28 Dec 2008, 21:54
The OA is "A", so i guess its typo in the q.
Thanks
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28 Dec 2008, 23:31
A can be the value of x. But, B cannot be. as from B x = 1/4 and sqrt(x) = 1/2 and x < 1/2. Hence, to me, there is no typo in the question.
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28 Dec 2008, 23:42
yes but 1/2 < 1/2 is false...
hence according to q it cannot represent x.
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29 Dec 2008, 10:15
jackychamp wrote:
Attachment:
q.GIF
is there a typo in the attached q.
both A and B seem to be correct?
Thanks
Jack
A. if x = 1/4, sqrtx = 1/2
B. if x = 1/2, sqrtx = 1/sqrt2, which is greater than 1/2.
so x = 1/4 is definitely not possible but x = 1/2 is even not possible.
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29 Dec 2008, 22:34
I think, I need a break. I am making such silly mistakes.
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11 Jan 2010, 18:20
jackychamp wrote:
Attachment:
q.GIF
is there a typo in the attached q.
both A and B seem to be correct?
Thanks
Jack
Correct me if I'm wrong but if you square everything you end up with
0<x<1/4
so unless it said, 0<x (less than or equal to)1/4, you will have two answers 1/2 or 1/4.....
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### Show Tags
14 Jan 2010, 02:55
Both A & B CANNOT be correct since:
0 < sqrt of x < 1/2
A.- 1/2 => sqrt of 1/2 = 0,707 therefore sqrt of 1/2 is more than 1/2 (0,5).
B.- 1/4 => sqrt of 1/4 = 0,5 therefore sqrt of 1/4 is equal to 1/2 (0,5).
Could you tell us the source of this question?
Re: simple q.. [#permalink] 14 Jan 2010, 02:55
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Display posts from previous: Sort by | 1,389 | 3,991 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.640625 | 4 | CC-MAIN-2017-04 | latest | en | 0.855489 |
https://www.nv5geospatialsoftware.com/docs/IMSL_ANOVA1.html | 1,726,720,187,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651981.99/warc/CC-MAIN-20240919025412-20240919055412-00867.warc.gz | 845,122,680 | 126,636 | The IMSL_ANOVA1 function analyzes a one-way classification model.
The IMSL_ANOVA1 function performs an analysis of variance of responses from a one-way classification design. The model is:
yij = μi + eiji = 1, 2, ..., k; j = 1, 2, ..., ni
where the observed value yij constitutes the j-th response in the i-th group, μi denotes the population mean for the i-th group, and the eij arguments are errors that are identically and independently distributed normal with mean 0 and variance σ2. The IMSL_ANOVA1 function requires the yij observed responses as input into a single vector y with responses in each group occupying contiguous locations. The analysis of variance table is computed along with the group sample means and standard deviations. A discussion of formulas and interpretations for the one-way analysis of variance problem appears in most statistics texts, e.g., Snedecor and Cochran (1967, Chapter 10).
The IMSL_ANOVA1 function computes simultaneous confidence intervals on all:
pairwise comparisons of k means μ1, μ2, ..., μk in the one-way analysis of variance model. Any of several methods can be chosen. A good review of these methods is given by Stoline (1981). The methods also are discussed in many statistics texts, e.g., Kirk (1982, pp. 114–127).
Let s2 be the estimated variance of a single observation. Let n be the degrees of freedom associated with s2. Let:
The methods are summarized as follows:
Tukey Method: The Tukey method gives the narrowest simultaneous confidence intervals for all pairwise differences of means mi – mj in balanced
(n1 = n2 = ... nk = n) one-way designs. The method is exact and uses the Studentized range distribution. The formula for the difference μ i – μj is given by the following:
where is the (1 – a ) 100 percentage point of the Studentized range distribution with parameters k and n.
Tukey-Kramer Method: The Tukey-Kramer method is an approximate extension of the Tukey method for the unbalanced case. (The method simplifies to the Tukey method for the balanced case.) The method always produces confidence intervals narrower than the Dunn-Sidak and Bonferroni methods. Hayter (1984) proved that the method is conservative, i.e., the method guarantees a confidence (1 – a) 100. Hayter's proof gave further support to earlier recommendations for its use (Stoline 1981). (Methods that are currently better are restricted to special cases and only offer improvement in severely unbalanced cases; see, for example, Spurrier and Isham 1985.) The formula for the difference μi – μj is given by the following:
Dunn-Sidák Method: The Dunn-Sidak method is a conservative method. The method gives wider intervals than the Tukey-Kramer method. (For large n and small a and k, the difference is only slight.) The method is slightly better than the Bonferroni method and is based on an improved Bonferroni (multiplicative) inequality (Miller 1980, pp. 101, 254–255). The method uses the t distribution (see IMSL_TCDF. The formula for the difference μi – μj is given by the following:
where tf;v is the 100f percentage point of the t distribution with n degrees of freedom.
Bonferroni Method: The Bonferroni method is a conservative method based on the Bonferroni (additive) inequality (Miller, p. 8). The method uses the t distribution. The formula for the difference μi – μj is given by the following:
Scheffé Method: The Scheffé method is an overly conservative method for simultaneous confidence intervals on pairwise difference of means. The method is applicable for simultaneous confidence intervals on all contrasts, i.e., all linear combinations:
where the following is true:
This method can be recommended here only if a large number of confidence intervals on contrasts, in addition to the pairwise differences of means, are to be constructed. The method uses the F distribution (see IMSL_FCDF. The formula for the difference μi – μj is given by the following:
where:
is the (1 – a) 100 percentage point of the F distribution with k – 1 and n degrees of freedom.
One-at-a-Time t Method (Fisher's LSD): The One-at-a-Time t method is appropriate for constructing a single confidence interval. The confidence percentage input is appropriate for one interval at a time. The method has been used widely in conjunction with the overall test of the null hypothesis μ1 = μ2 = ... = μk by the use of the F statistic. Fisher's LSD (least significant difference) test is a two-stage test that proceeds to make pairwise comparisons of means only if the overall F test is significant. Milliken and Johnson (1984, p. 31) recommend LSD comparisons after a significant F only if the number of comparisons is small and the comparisons were planned prior to the analysis. If many unplanned comparisons are made, they recommend Scheffé's method. If the F test is insignificant, a few planned comparisons for differences in means can still be performed by using either Tukey, Tukey-Kramer, Dunn-Sidak or Bonferroni methods. Because the F test is insignificant, Scheffé's method does not yield any significant differences. The formula for the difference μi – μj is given by the following:
## Examples
### Example 1
This example computes a one-way analysis of variance for data discussed by Searle (1971, Table 5.1, pp. 165–179). The responses are plant weights for six plants of three different types shown in the following table: three normal, two off-types, and one aberrant.
Normal Off-Type Aberrant 101 84 32 105 88 94
`n = [3,2,1]`
`y = [101.0, 105.0, 94.0, 84.0, 88.0, 32.0]`
`PRINT,';p-value = ', IMSL_ANOVA1(n, y)`
IDL prints:
`p-value = 0.00276887`
### Example 2: Multiple Comparisons
Simultaneous confidence intervals are generated for the measurements of cold-cranking power for five models of automobile batteries shown in the table below. Nelson (1989, pp. 232–241) provided the data and approach.
Model 1 Model 2 Model 3 Model 4 Model 5 41 42 27 48 28 43 43 26 45 32 42 46 28 51 37 46 38 27 46 25
The Tukey method is chosen for the analysis of pairwise comparisons, with a confidence level of 99 percent. The means and their confidence limits are output. First, a procedure to print out the results is defined.
`.RUN`
`PRO print_results, anova_table, diff_means`
` anova_labels = [';df for among groups', \$`
` ';df for within groups', 'total (corrected) df', \$`
` ';ss for among groups', 'ss for within groups', \$`
` ';total (corrected) ss', 'mean square among groups', \$`
` ';mean square within groups', 'F-statistic', \$`
` ';P-value', 'R-squared (in percent)', \$`
` ';adjusted R-squared (in percent)', \$`
` ';est. std of within group error', 'overall mean of y', \$`
` ';coef. of variation (in percent)']`
` PRINT, '; * *Analysis of Variance * *'`
` FOR i = 0, 14 DO PM, anova_labels(i), \$`
` anova_table(i), FORMAT = ';(a40,f20.2)'`
` PRINT`
` ; Print the analysis of variance table.`
` PRINT, '; * *Differences of Means * *'`
` PRINT, ';groups', 'difference', 'lower limit', 'upper limit'`
` PM, diff_means, FORMAT = ';(2i3, x, f9.2, 4x, f9.2, 5x, f9.2)'`
` ; Print the differences of means.`
`END`
` `
`n = [4, 4, 4, 4, 4]`
`y = [41, 43, 42, 46, 42, 43, 46, 38, 27, 26, 28, 27, \$`
` 48, 45, 51, 46, 28, 32, 37, 25]`
`p_value = IMSL_ANOVA1(n, y, Confidence = 99.0, \$`
` Anova_Table = anova_table, Tukey = diff_means)`
`; Call IMSL_ANOVA1.`
`print_results, anova_table, diff_means`
` `
`; Output the results.`
` `
`* *Analysis of Variance * *`
` df for among groups 4.00`
` df for within groups 15.00`
` total (corrected) df 19.00`
` ss for among groups 1242.20`
` ss for within groups 150.75`
` total (corrected) ss 1392.95`
` mean square among groups 310.55`
` mean square within groups 10.05`
` F-statistic 30.90`
` P-value 0.00`
` R-squared (in percent) 89.18`
` adjusted R-squared (in percent) 86.29`
` est. std of within group error 3.17`
` overall mean of y 38.05`
` coef. of variation (in percent) 8.33`
`* *Differences of Means * *`
`groups difference lower limit upper limit`
` 1 2 0.75 -8.05 9.55`
` 1 3 16.00 7.20 24.80`
` 1 4 -4.50 -13.30 4.30`
` 1 5 12.50 3.70 21.30`
` 2 3 15.25 6.45 24.05`
` 2 4 -5.25 -14.05 3.55`
` 2 5 11.75 2.95 20.55`
` 3 4 -20.50 -29.30 -11.70`
` 3 5 -3.50 -12.30 5.30`
` 4 5 17.00 8.20 25.80`
## Syntax
Result = IMSL_ANOVA1(N, Y[, ANOVA_TABLE=variable] [, BONFERRONI=variable] [, CONFIDENCE=value] [, /DOUBLE] [, DUNN_SIDAK=variable] [, GROUP_COUNTS=variable] [, GROUP_MEANS=variable] [, GROUP_STD_DEV=variable] [, ONE_AT_A_TIME=variable] [, SCHEFFE=variable] [, TUKEY=variable])
## Return Value
The p-value for the F-statistic.
## Arguments
### N
One-dimensional array containing the number of responses for each group.
### Y
One-dimensional array of length:
n(0) + n(1) + ...+ n(N_ELEMENTS(n) – 1)
containing the responses for each group.
## Keywords
### ANOVA_TABLE (optional)
Named variable into which the analysis of variance table is stored. The analysis of variance statistics are as follows:
• 0: Degrees of freedom for the model
• 1: Degrees of freedom for error
• 2: Total (corrected) degrees of freedom
• 3: Sum of squares for the model
• 4: Sum of squares for error
• 5: Total (corrected) sum of squares
• 6: Model mean square
• 7: Error mean square
• 8: Overall F-statistic
• 9: p-value
• 10: R2 (in percent)
• 11: Adjusted R2 (in percent)
• 12: Estimate of the standard deviation
• 13: Overall mean of y
• 14: Coefficient of variation (in percent)
### BONFERRONI (optional)
Named variable into which the array containing the statistics relating to the difference of means is stored. On return, the named variable contains an array of size:
where ngroups = N_ELEMENTS (n).
• 0: Group number for the i-th mean
• 1: Group number for the j-th mean
• 2: Difference of means (i-th mean) - (j-th mean)
• 3: Lower confidence limit for the difference
• 4: Upper confidence limit for the difference
The IMSL_ANOVA1 function computes confidence intervals on all pairwise differences of means using one of six methods: Tukey, Tukey-Kramer, Dunn-Sidák, Bonferroni, Scheffé, or Fisher's LSD (One-at-a-Time). If Tukey is specified, Tukey confidence intervals are calculated if the group sizes are equal; otherwise, the Tukey-Kramer confidence intervals are calculated.
### CONFIDENCE (optional)
Confidence level for the simultaneous interval estimation. If Tukey is specified, Confidence must be in the range [90.0, 99.0); otherwise, Confidence is in the range [0.0, 100.0). Default: 95.0
### DOUBLE (optional)
If present and nonzero, then double precision is used.
### DUNN_SIDAK (optional)
Named variable into which the array containing the statistics relating to the difference of means is stored. On return, the named variable contains an array of size:
where ngroups = N_ELEMENTS (n).
• 0: group number for the i-th mean
• 1: group number for the j-th mean
• 2: difference of means (i-th mean) - (j-th mean)
• 3: lower confidence limit for the difference
• 4: upper confidence limit for the difference
The IMSL_ANOVA1 function computes confidence intervals on all pairwise differences of means using one of six methods: Tukey, Tukey-Kramer, Dunn-Sidák, Bonferroni, Scheffé, or Fisher's LSD (One-at-a-Time). If Tukey is specified, Tukey confidence intervals are calculated if the group sizes are equal; otherwise, the Tukey-Kramer confidence intervals are calculated.
### GROUP_COUNTS (optional)
Named variable into which the array containing the number of nonmissing observations for the groups is stored.
### GROUP_MEANS (optional)
Named variable into which the array containing the group means is stored.
### GROUP_STD_DEV (optional)
Named variable into which the array containing the group standard deviations is stored.
### ONE_AT_A_TIME (optional)
Named variable into which the array containing the statistics relating to the difference of means is stored. On return, the named variable contains an array of size:
where ngroups = N_ELEMENTS (n).
• 0: group number for the i-th mean
• 1: group number for the j-th mean
• 2: difference of means (i-th mean) - (j-th mean)
• 3: lower confidence limit for the difference
• 4: upper confidence limit for the difference
The IMSL_ANOVA1 function computes confidence intervals on all pairwise differences of means using one of six methods: Tukey, Tukey-Kramer, Dunn-Sidák, Bonferroni, Scheffé, or Fisher's LSD (One-at-a-Time). If Tukey is specified, Tukey confidence intervals are calculated if the group sizes are equal; otherwise, the Tukey-Kramer confidence intervals are calculated.
### SCHEFFE (optional)
Named variable into which the array containing the statistics relating to the difference of means is stored. On return, the named variable contains an array of size:
where ngroups = N_ELEMENTS (n).
• 0: group number for the i-th mean
• 1: group number for the j-th mean
• 2: difference of means (i-th mean) - (j-th mean)
• 3: lower confidence limit for the difference
• 4: upper confidence limit for the difference
The IMSL_ANOVA1 function computes confidence intervals on all pairwise differences of means using one of six methods: Tukey, Tukey-Kramer, Dunn-Sidák, Bonferroni, Scheffé, or Fisher's LSD (One-at-a-Time). If Tukey is specified, Tukey confidence intervals are calculated if the group sizes are equal; otherwise, the Tukey-Kramer confidence intervals are calculated.
### TUKEY (optional)
Named variable into which the array containing the statistics relating to the difference of means is stored. On return, the named variable contains an array of size:
where ngroups = N_ELEMENTS (n).
• 0: Group number for the i-th mean
• 1: Group number for the j-th mean
• 2: Difference of means (i-th mean) - (j-th mean)
• 3: Lower confidence limit for the difference
• 4: Upper confidence limit for the difference
The IMSL_ANOVA1 function computes confidence intervals on all pairwise differences of means using one of six methods: Tukey, Tukey-Kramer, Dunn-Sidák, Bonferroni, Scheffé, or Fisher's LSD (One-at-a-Time). If Tukey is specified, Tukey confidence intervals are calculated if the group sizes are equal; otherwise, the Tukey-Kramer confidence intervals are calculated.
## Version History
6.4 Introduced | 3,912 | 14,622 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.6875 | 3 | CC-MAIN-2024-38 | latest | en | 0.870103 |
https://math.answers.com/Q/What_are_the_factors_of_65_and_195 | 1,701,174,482,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679099514.72/warc/CC-MAIN-20231128115347-20231128145347-00225.warc.gz | 453,058,165 | 44,539 | 0
# What are the factors of 65 and 195?
Wiki User
8y ago
The factors of 65 are:
1, 5, 13, 65
The factors of 195 are:
1, 3, 5, 13, 15, 39, 65, 195
Wiki User
8y ago
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Q: What are the factors of 65 and 195?
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Related questions
### Factors of 195?
The factors of 195 are 1, 3, 5, 13, 15, 39, 65, and 195.
### What are the factors and multiples of the number 65?
Factors: 1, 5, 13, 65 Multiples: 65, 130, 195 and so on.
### What are the factors and prime factors of 195?
The factors of 195 are 1, 3, 5, 13, 15, 39, 65, and 195. The factor pairs of 195 are 1 x 195, 3 x 65, 5 x 39, and 13 x 15.The prime factors of 195 are 3, 5, and 13. The prime factorization of 195 is 3 x 5 x 13.The prime factors of 195 are 3, 5 and 13.1,3,5,13,15,39,65,195
### What are the factors of 195?
1, 3, 5, 13, 15, 39, 65, and 195
### What are the factors of 195 that equal 1?
The factors of 195 are 1, 3, 5, 13, 15, 39, 65, 195. No two of them equal one unless you multiply one by itself.
3
### How many factors are in 195what?
8 factors 1, 3, 5, 13, 15, 39, 65, 195
260
### What times what equals to 195?
Pairs of positive integer factors of 195 are: 1 x 195 = 195 3 x 65 = 195 5 x 39 = 195 13 x 15 = 195
### What are the least common multiples of 15 and 65?
65, 130, 195 The LCM is 195.
### What is the greatest common factor of 65 and 195?
The greatest common factor of 65 and 195 is 65.
### Factors of 195 and 330?
The largest companies to sell products on the Internet What are you waiting for: Share - Shop - Buy The companies are global and 100% guaranteed | 596 | 1,608 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.984375 | 3 | CC-MAIN-2023-50 | latest | en | 0.900104 |
https://www.urbandictionary.com/define.php?term=BOJANGULATION | 1,596,932,991,000,000,000 | text/html | crawl-data/CC-MAIN-2020-34/segments/1596439738366.27/warc/CC-MAIN-20200808224308-20200809014308-00519.warc.gz | 883,328,826 | 21,980 | Top definition
The way of life that a Bojangler lives: worthless, slow, dimwitted, and often mistaken for retardation
"As Tito walked into the dimly lit bar he encountered some sort of BOJANGULATION by what appeared to be a retard, but it was just Ole ManNigga Bojangles, hunched over at the bar , foaming at the mouth, and just wating to die.
by DaCapt'mn & Brova' Man Tito March 12, 2008
Get a BOJANGULATION mug for your coworker Callisto.
Aug 8 Word of the Day
A method of guessing on a multiple choice test that involves looking at the position of the second hand. If the hand is between 12 and 3 the guess is A. If the hand is between 3 and 6 the guess is B. Between 6 and 9 guess C. Between 9 and 12 guess D.
I passed my test! Thanks to the Clock Method
by ET4444 November 12, 2007
Get a Clock Method mug for your mom Julia. | 233 | 830 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2020-34 | latest | en | 0.943342 |
https://programmersheaven.com/discussion/424957/help-wanted-for-my-little-program | 1,540,059,398,000,000,000 | text/html | crawl-data/CC-MAIN-2018-43/segments/1539583513009.81/warc/CC-MAIN-20181020163619-20181020185119-00257.warc.gz | 794,322,389 | 10,728 | # Help wanted for my little program
Hi everyone!
I am a beginner in c programming, and I would be very very grateful if someone coule take a look at my little program which I wrote and which unfortunately does not really work.
The program calculates the running time and the annuity of a credit, after receiving the necessary values.
However, the formula for the running time does not work. I tried to rewrite this formula, maybe I have made a mistake:
in my programm n is the running time, R is the annuity, S is the amount of the credit, m is the fixed fee and i is the lending rate in per cent.
[code]
#include
#include
double R,m,S,i;
double annuity()
{
double R = 12*m;
return R;
}
int term()
{
int n = -(log(1-(i*S/R))/log(1+i));
return n;
}
main()
{
printf("
Fill in a value for S: ");
scanf("%lf",&S);
printf("
Fill in a value for m: ");
scanf("%lf",&m);
printf("
Fill in a value for i: ");
scanf("%lf",&i);
printf("
The annuity is %.2lf",annuity());
printf("
The running time is %i
", term());
}
[/code] | 267 | 1,023 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.578125 | 3 | CC-MAIN-2018-43 | latest | en | 0.925707 |
https://math.stackexchange.com/questions/382356/logical-explanation-of-eulers-formula/1803399 | 1,713,875,809,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296818474.95/warc/CC-MAIN-20240423095619-20240423125619-00676.warc.gz | 355,094,893 | 36,943 | # Logical explanation of Euler's formula
This question is a about (if not proving) at least guessing the Euler's formula. I don't want the proof using the infinite sums.
We can guess by logic that for example that the equation $x^2+1=\sqrt{x}$ has no real solutions because $x^2=\sqrt{x}$ has 2 solutions $x=0, x=1$ but by adding 1 on the left side, we cancel these 2 solutions, so there are no solutions.
I want to know if there is a way to guess by logic that $e^{ix}=\cos(x)+i\sin(x)$. I guess that the most important here here will be $\frac{d}{dx}e^x=e^x$. And suggestions?
• I'm not sure what you mean by "logical" here. Does it have to do with mathematical logic (i.e. formulating the argument in a certain logic and language, and some particular theory), or are you just looking for an explanation appealing intuitive logic about what's correct? May 5, 2013 at 17:28
• You need to choose a definition for the exponential. You don't want to us the series definition, so which would you like? May 5, 2013 at 17:29
• @AsafKaragila It's actually your second statement explanation appealing intuitive logic about what's correct. May 5, 2013 at 17:32
• The equation $x^2+1 = \sqrt{x}$ does have solutions. Squaring both sides and rearranging gives $x^4 + 2x^2 -x + 1 = 0$. By the fundamental theorem of algebra, this has - when counted with multiplicity - four solutions over the complex plane. The solutions to $x^2 + 1 = \sqrt{x}$ will be solutions of $x^4 + 2x^2 -x + 1 = 0$, although the converse need not be true. May 5, 2013 at 17:41
• I think a nice (and rigorous) way is showing that $f(\theta)=\cos\left(\theta\right)+i\sin\left(\theta\right)$ satifies both $f'\left(\theta\right)=i\cdot f\left(\theta\right)$ and $f\left(0\right)=1$. Showing that $\left(f\left(\theta\right)\right)^n=f\left(n\theta\right)$ also feeds intuitions. May 5, 2013 at 18:19
The power series argument, while simple, is indeed unenlightening. You can easily show that $f(\theta)=\cos\left(\theta\right)+i\sin\left(\theta\right)$ satifies both $f'\left(\theta\right)=i\cdot f\left(\theta\right)$ and $f\left(0\right)=1$, and it looks remarkably similar to one definition of $\gamma(t)=e^{\alpha t}$: the function $\gamma\,\colon\mathbb{C}\rightarrow\mathbb{C}$ that both $\left(e^{\alpha t}\right)'=\alpha e^{\alpha t}$ and $\gamma(0)=1$.
This isn't a proof, but an illustration as why the formula is defined in that way:
Consider $h(z) = A^z$, A > 0. This is well-defined for z real, and you want to extend it to when z is complex. It's enough to define it whenever z is pure imaginary.
Define $f(x) := Re(h(ix))$, $g(x) := Im(h(ix))$ So $h(ix) == f(x) + i*g(x)$
1) $h(0) == 1$, so $f(0) == 1$, $g(0) == 0$.
2) The distinction between i and -i is arbitrary, so when extending real-valued functions to the complex numbers, you usually want $h(\bar{z}) == \bar{h(z)}$, so $f(-x) == f(x)$, $g(-x) == -g(x)$
3) $A^m * A^n == A^{m+n}$, so $h(i(x+y)) == h(ix) * h(iy)$ $f(x+y) + i*g(x+y) == (f(x) + i*g(x)) (f(y) + i*g(y))$ $f(x+y) + i*g(x+y) == f(x)*f(y) + i*g(x)*f(y) + i*f(x)*g(y) - g(x)*g(y)$ $f(x+y) + i*g(x+y) == (f(x)*f(y) - g(x)*g(y)) + i*(g(x)*f(y) + f(x)*g(y))$
Equating real and imaginary parts: $f(x+y) == f(x)*f(y) - g(x)*g(y)$ $g(x+y) == f(x)*g(h) + f(y)*g(x)$
Those formulas should be very familiar, and they suggest that $A^{ix}$ behaves very much like $cos(x) + i*sin(x)$.
($cos(kx) + i*sin(kx)$) satisfies all the formulas for any $k$. Determining which value of $k$ corresponds to a particular value of $A$ requires calculus or the equivalent.) | 1,164 | 3,557 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.890625 | 4 | CC-MAIN-2024-18 | latest | en | 0.896507 |
http://icpc.njust.edu.cn/Problem/CF/107A/ | 1,569,074,547,000,000,000 | text/html | crawl-data/CC-MAIN-2019-39/segments/1568514574501.78/warc/CC-MAIN-20190921125334-20190921151334-00434.warc.gz | 92,755,155 | 10,402 | # Dorm Water Supply
Time Limit: 2 seconds
Memory Limit: 256 megabytes
## Description
The German University in Cairo (GUC) dorm houses are numbered from 1 to n. Underground water pipes connect these houses together. Each pipe has certain direction (water can flow only in this direction and not vice versa), and diameter (which characterizes the maximal amount of water it can handle).
For each house, there is at most one pipe going into it and at most one pipe going out of it. With the new semester starting, GUC student and dorm resident, Lulu, wants to install tanks and taps at the dorms. For every house with an outgoing water pipe and without an incoming water pipe, Lulu should install a water tank at that house. For every house with an incoming water pipe and without an outgoing water pipe, Lulu should install a water tap at that house. Each tank house will convey water to all houses that have a sequence of pipes from the tank to it. Accordingly, each tap house will receive water originating from some tank house.
In order to avoid pipes from bursting one week later (like what happened last semester), Lulu also has to consider the diameter of the pipes. The amount of water each tank conveys should not exceed the diameter of the pipes connecting a tank to its corresponding tap. Lulu wants to find the maximal amount of water that can be safely conveyed from each tank to its corresponding tap.
## Input
The first line contains two space-separated integers n and p (1 ≤ n ≤ 1000, 0 ≤ p ≤ n) — the number of houses and the number of pipes correspondingly.
Then p lines follow — the description of p pipes. The i-th line contains three integers ai bi di, indicating a pipe of diameter di going from house ai to house bi (1 ≤ ai, bi ≤ n, ai ≠ bi, 1 ≤ di ≤ 106).
It is guaranteed that for each house there is at most one pipe going into it and at most one pipe going out of it.
## Output
Print integer t in the first line — the number of tank-tap pairs of houses.
For the next t lines, print 3 integers per line, separated by spaces: tanki, tapi, and diameteri, where tanki ≠ tapi (1 ≤ i ≤ t). Here tanki and tapi are indexes of tank and tap houses respectively, and diameteri is the maximum amount of water that can be conveyed. All the t lines should be ordered (increasingly) by tanki.
## Sample Input
Input3 21 2 102 3 20Output11 3 10Input3 31 2 202 3 103 1 5Output0Input4 21 2 603 4 50Output21 2 603 4 50
## Sample Output
None
None
None | 622 | 2,474 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 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} | 3.0625 | 3 | CC-MAIN-2019-39 | longest | en | 0.923592 |
https://byjusexamprep.com/gate-ee-2022-electromagnetic-field-quiz-9-i-0f346650-af32-11eb-b203-0540030d3801 | 1,632,167,982,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780057091.31/warc/CC-MAIN-20210920191528-20210920221528-00145.warc.gz | 204,058,726 | 49,428 | Time Left - 15:00 mins
# GATE EE 2022: Electromagnetic field Quiz 9 (App update required to attempt this test)
Attempt now to get your rank among 119 students!
Question 1
The static electric field inside a dielectric medium with relative permittivity, εr = 2.25, expressed in cylindrical coordinate system is given by the following expression
Where ar, aϕ, az are unit vectors along r, ϕ and z directions, respectively. If the above expression represents a valid electrostatic field inside the medium, then the volume charge density associated with this field in terms of free space permittivity, ε0, in SI unit is given by :
Question 2
Four electric charges +q, –q, +q, –q are placed at the corner of a square of side 2L as shown below
The electric potential at point A, midway between 2 charges +q and –q is
Question 3
For spherical surface of radius (r) 4 cm, uniform spherical surface charge density is 20 × 10–6 C/m2. If 1 × 10–3 Joule of energy is found in the region 0.06 < r < a, then ‘a’ is _________ m.
Question 4
The insulation of a concentric cable of core diameter 'd' is made up of 2 layers of different insulating material each of thickness t. In what may should the permittivitiesand of the materials differ in order that the maximum dielectric stress shall be the same in each?
Question 5
Two identical uniformly distributed line charges are situated along X and Y axis, The charge density being ρ = 30 μc/m. Obtain the resultant electric flux density at point (3, 4, 5) due to line charges__
Question 6
In a non-magnetic medium V/m. Then the value of in intrinsic impedance is _________.
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GradeStack Learning Pvt. Ltd.Windsor IT Park, Tower - A, 2nd Floor, Sector 125, Noida, Uttar Pradesh 201303 [email protected] | 506 | 1,901 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2021-39 | latest | en | 0.864689 |
https://whatisconvert.com/27-kilograms-in-long-tons | 1,719,117,773,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198862430.93/warc/CC-MAIN-20240623033236-20240623063236-00360.warc.gz | 551,880,145 | 7,301 | ## Convert 27 Kilograms to Long Tons
To calculate 27 Kilograms to the corresponding value in Long Tons, multiply the quantity in Kilograms by 0.00098420652761106 (conversion factor). In this case we should multiply 27 Kilograms by 0.00098420652761106 to get the equivalent result in Long Tons:
27 Kilograms x 0.00098420652761106 = 0.026573576245499 Long Tons
27 Kilograms is equivalent to 0.026573576245499 Long Tons.
## How to convert from Kilograms to Long Tons
The conversion factor from Kilograms to Long Tons is 0.00098420652761106. To find out how many Kilograms in Long Tons, multiply by the conversion factor or use the Mass converter above. Twenty-seven Kilograms is equivalent to zero point zero two six six Long Tons.
## Definition of Kilogram
The kilogram (or kilogramme, SI symbol: kg), also known as the kilo, is the fundamental unit of mass in the International System of Units. Defined as being equal to the mass of the International Prototype Kilogram (IPK), that is almost exactly equal to the mass of one liter of water. The kilogram is the only SI base unit using an SI prefix ("kilo", symbol "k") as part of its name. The stability of kilogram is really important, for four of the seven fundamental units in the SI system are defined relative to it.
## Definition of Long Ton
A long ton is defined as exactly 2,240 pounds. The long ton arises from the traditional British measurement system: A long ton is 20 cwt, each of which is 8 stone (1 stone = 14 pounds). Thus a long ton is 20 × 8 × 14 lb = 2,240 lb. Long ton, also known as the imperial ton or displacement ton is the name for the unit called the "ton" in the avoirdupois or Imperial system of measurements standardised in the thirteenth century that is used in the United Kingdom
## Using the Kilograms to Long Tons converter you can get answers to questions like the following:
• How many Long Tons are in 27 Kilograms?
• 27 Kilograms is equal to how many Long Tons?
• How to convert 27 Kilograms to Long Tons?
• How many is 27 Kilograms in Long Tons?
• What is 27 Kilograms in Long Tons?
• How much is 27 Kilograms in Long Tons?
• How many uk ton are in 27 kg?
• 27 kg is equal to how many uk ton?
• How to convert 27 kg to uk ton?
• How many is 27 kg in uk ton?
• What is 27 kg in uk ton?
• How much is 27 kg in uk ton? | 609 | 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} | 3.640625 | 4 | CC-MAIN-2024-26 | latest | en | 0.863829 |
https://kerodon.net/tag/01US | 1,709,263,621,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474948.91/warc/CC-MAIN-20240301030138-20240301060138-00612.warc.gz | 334,486,325 | 11,413 | # Kerodon
$\Newextarrow{\xRightarrow}{5,5}{0x21D2}$ $\newcommand\empty{}$
### 5.3.7 Application: Path Fibrations
Recall that every morphism of Kan complexes $f: X \rightarrow Y$ admits a canonical factorization
$X \xrightarrow {\delta } P(f) \xrightarrow {\pi } Y,$
where $\delta$ is a homotopy equivalence and $\pi$ is the path fibration
$P(f) = X \times _{ \operatorname{Fun}( \{ 0\} , Y)} \operatorname{Fun}( \Delta ^1, Y) \rightarrow \operatorname{Fun}( \{ 1\} , Y) \simeq Y$
of Example 3.1.7.10. Note that the simplicial set $P(f) = X \operatorname{\vec{\times }}_{Y} Y$ is an example of an oriented fiber product (Definition 4.6.4.1), which is defined for any morphism of simplicial sets $f: X \rightarrow Y$. Beware that if $X$ and $Y$ are not Kan complexes, then $\delta$ need not be a homotopy equivalence and $\pi$ need not be a Kan fibration. However, if $X = \operatorname{\mathcal{C}}$ and $Y = \operatorname{\mathcal{D}}$ are $\infty$-categories, then we have the following weaker statements:
$(a)$
The functor $\delta : \operatorname{\mathcal{C}}\rightarrow \operatorname{\mathcal{C}}\operatorname{\vec{\times }}_{\operatorname{\mathcal{D}}} \operatorname{\mathcal{D}}$ is fully faithful, and its essential image is the homotopy fiber product $\operatorname{\mathcal{C}}\times ^{\mathrm{h}}_{\operatorname{\mathcal{D}}} \operatorname{\mathcal{D}}$ of Construction 4.5.2.1 (Corollary 4.5.2.22).
$(b)$
The functor $\pi : \operatorname{\mathcal{C}}\operatorname{\vec{\times }}_{\operatorname{\mathcal{D}}} \operatorname{\mathcal{D}}\rightarrow \operatorname{\mathcal{D}}$ is a cocartesian fibration of $\infty$-categories (Corollary 5.3.7.3).
Moreover, the oriented fiber product $\operatorname{\mathcal{C}}\operatorname{\vec{\times }}_{\operatorname{\mathcal{D}}} \operatorname{\mathcal{D}}$ can be characterized by a universal mapping property: roughly speaking, the diagonal map $\delta$ exhibits the cocartesian fibration $\pi$ as freely generated by the functor $f$ (Theorem 5.3.7.6).
Our starting point is the following observation:
Lemma 5.3.7.1. Let $U: \operatorname{\mathcal{C}}\rightarrow \operatorname{\mathcal{D}}$ be an inner fibration of simplicial sets, let $e$ be an edge of $\operatorname{Fun}( \Delta ^1, \operatorname{\mathcal{C}})$, and let
$V: \operatorname{Fun}( \Delta ^1, \operatorname{\mathcal{C}}) = \operatorname{\mathcal{C}}\operatorname{\vec{\times }}_{\operatorname{\mathcal{C}}} \operatorname{\mathcal{C}}\rightarrow \operatorname{\mathcal{C}}\operatorname{\vec{\times }}_{\operatorname{\mathcal{D}}} \operatorname{\mathcal{C}}$
denote the morphism induced by $U$. Let $\operatorname{ev}_0, \operatorname{ev}_1: \operatorname{Fun}(\Delta ^1, \operatorname{\mathcal{C}}) \rightarrow \operatorname{\mathcal{C}}$ be the evaluation maps. If $\operatorname{ev}_0(e)$ is $U$-cocartesian, then $e$ is $V$-cocartesian. If $\operatorname{ev}_1(e)$ is $U$-cartesian, then $e$ is $V$-cartesian.
Proof. Assume that $\operatorname{ev}_0(e)$ is $U$-cocartesian; we will show that $e$ is $V$-cocartesian (the second assertion follows by a similar argument). Let $n \geq 2$; we wish to show that every lifting problem
5.34
$$\begin{gathered}\label{equation:cocartesian-in-oriented-fiber-product-ultimate} \xymatrix@R =50pt@C=50pt{ \Lambda ^{n}_{0} \ar [r]^-{ \sigma _0 } \ar [r] \ar [d] & \operatorname{Fun}( \Delta ^1, \operatorname{\mathcal{C}}) \ar [d]^-{ V } \\ \Delta ^{n} \ar [r]^-{ \overline{\sigma } } \ar@ {-->}[ur]^{\sigma } & \operatorname{\mathcal{C}}\operatorname{\vec{\times }}_{\operatorname{\mathcal{D}}} \operatorname{\mathcal{C}}} \end{gathered}$$
admits a solution, provided that the composite map
$\Delta ^1 \simeq \operatorname{N}_{\bullet }( \{ 0 < 1 \} ) \hookrightarrow \Lambda ^{n}_0 \xrightarrow { \tau _0 } \operatorname{Fun}( \Delta ^1, \operatorname{\mathcal{C}})$
coincides with $e$. Let $X(0)$ denote the simplicial subset of $\Delta ^1 \times \Delta ^ n$ given by the union of $\operatorname{\partial \Delta }^1 \times \Delta ^ n$ with $\Delta ^1 \times \Lambda ^{n}_0$. Unwinding the definitions, we can rewrite (5.34) as a lifting problem
$\xymatrix@R =50pt@C=50pt{ X(0) \ar [r]^-{ \tau _0 } \ar [r] \ar [d] & \operatorname{\mathcal{C}}\ar [d]^-{ U } \\ \Delta ^1 \times \Delta ^{n} \ar [r]^-{ \overline{\tau } } \ar@ {-->}[ur]^{\tau } & \operatorname{\mathcal{D}}. }$
Choose a filtration
$X(0) \subset X(1) \subset X(2) \subset \cdots \subset X(t) = \Delta ^{1} \times \Delta ^{n}$
satisfying the requirements of Lemma 4.4.4.7. We will complete the proof by showing that, for each $s \leq t$, the morphism $\tau _0$ admits an extension $\tau _ s: X(s) \rightarrow \operatorname{\mathcal{C}}$ satisfying $U \circ \tau _{s} = \overline{\tau }|_{ X(s) }$. The proof proceeds by induction on $s$, the case $s = 0$ being vacuous. Let us therefore assume that $0 < s \leq t$ and that $\tau _0$ has already been extended to a morphism of simplicial sets $\tau _{s-1}: X(s-1) \rightarrow \operatorname{\mathcal{C}}$. By construction, we have a pushout diagram of simplicial sets
$\xymatrix@R =50pt@C=50pt{ \Lambda ^{q}_{p} \ar [r]^-{\varphi _0} \ar [d] & X(s-1) \ar [d] \\ \Delta ^{q} \ar [r]^{\varphi } & X(s) }$
for some $q \geq 2$ and $0 \leq p < q$. Consequently, to prove the existence of $\tau _ s$, it will suffice to show that $\tau _{s-1} \circ \varphi _0$ can be extended to a $q$-simplex of $\operatorname{\mathcal{C}}$ lying over the simplex $\overline{\tau } \circ \varphi : \Delta ^ q \rightarrow \operatorname{\mathcal{D}}$. For $p \neq 0$, the existence of this extension follows from our assumption that $U$ is an inner fibration. To handle the case $p= 0$, we observe that the morphism $\varphi$ carries the initial edge of $\Delta ^{q}$ to the edge $(0,0) \rightarrow (0,1)$ of $\Delta ^{1} \times \Delta ^ n$, so that $\tau _{s-1} \circ \varphi _0$ carries the initial edge of $\Delta ^ q$ to the edge $\operatorname{ev}_0(e)$ of $\operatorname{\mathcal{C}}$, which is $U$-cocartesian by assumption. $\square$
Proposition 5.3.7.2. Suppose we are given a commutative diagram of simplicial sets
$\xymatrix { \operatorname{\mathcal{C}}_0 \ar [d]^{U_0} \ar [r]^{F_0} & \operatorname{\mathcal{C}}\ar [d]^{U} & \operatorname{\mathcal{C}}_1 \ar [l]_{F_1} \ar [d]_{U_1} \\ \operatorname{\mathcal{D}}_0 \ar [r]^{G_0} & \operatorname{\mathcal{D}}& \operatorname{\mathcal{D}}_1 \ar [l]_{G_1} }$
where $U_0$, $U_1$, and $U$ are cocartesian fibrations, and $F_0$ carries $U_0$-cocartesian edges of $\operatorname{\mathcal{C}}_0$ to $U$-cocartesian edges of $\operatorname{\mathcal{C}}$. Then the induced map
$V: \operatorname{\mathcal{C}}_0 \operatorname{\vec{\times }}_{\operatorname{\mathcal{C}}} \operatorname{\mathcal{C}}_1 \rightarrow \operatorname{\mathcal{D}}_0 \operatorname{\vec{\times }}_{\operatorname{\mathcal{D}}} \operatorname{\mathcal{D}}_1$
is a cocartesian fibration of simplicial sets. Moreover, an edge $e$ of $\operatorname{\mathcal{C}}_0 \operatorname{\vec{\times }}_{\operatorname{\mathcal{C}}} \operatorname{\mathcal{C}}_1$ is $V$-cocartesian if and only if it satisfies the following condition:
$(\ast )$
Let $e_0$ and $e_1$ denote the images of $e$ in $\operatorname{\mathcal{C}}_0$ and $\operatorname{\mathcal{C}}_1$, respectively. Then $e_0$ is $U_0$-cocartesian and $e_1$ is $U_1$-cocartesian.
Proof. Let us say that an edge $e$ of $\operatorname{\mathcal{C}}_0 \operatorname{\vec{\times }}_{\operatorname{\mathcal{C}}} \operatorname{\mathcal{C}}_1$ is special if it satisfies condition $(\ast )$. We first show that if $e$ is a special edge of $\operatorname{\mathcal{C}}_0 \operatorname{\vec{\times }}_{\operatorname{\mathcal{C}}} \operatorname{\mathcal{C}}_1$, then $e$ is $V$-cocartesian. Let $e_0$ and $e_1$ denote the images of $e$ in $\operatorname{\mathcal{C}}_0$ and $\operatorname{\mathcal{C}}_1$, respectively. Note that $V$ factors as a composition
$\operatorname{\mathcal{C}}_0 \operatorname{\vec{\times }}_{ \operatorname{\mathcal{C}}} \operatorname{\mathcal{C}}_1 \xrightarrow {V'} \operatorname{\mathcal{C}}_0 \operatorname{\vec{\times }}_{\operatorname{\mathcal{D}}} \operatorname{\mathcal{C}}_1 \xrightarrow {V''} \operatorname{\mathcal{D}}_0 \operatorname{\vec{\times }}_{\operatorname{\mathcal{D}}} \operatorname{\mathcal{D}}_1.$
Here $V'$ is a pullback of the projection map $\overline{V}': \operatorname{Fun}( \Delta ^1, \operatorname{\mathcal{C}}) = \operatorname{\mathcal{C}}\operatorname{\vec{\times }}_{\operatorname{\mathcal{C}}} \operatorname{\mathcal{C}}\rightarrow \operatorname{\mathcal{C}}\operatorname{\vec{\times }}_{\operatorname{\mathcal{D}}} \operatorname{\mathcal{C}}$. Since $F( e_0 )$ is $U$-cocartesian, Lemma 5.3.7.1 implies that $e$ is $V'$-cocartesian. Moreover, $V''$ is a pullback of the product map $(U_0 \times U_1): \operatorname{\mathcal{C}}_0 \times \operatorname{\mathcal{C}}_1 \rightarrow \operatorname{\mathcal{D}}_0 \times \operatorname{\mathcal{D}}_1$. By assumption, $e_0$ is $U_0$-cocartesian and $e_1$ is $U_1$-cocartesian. It follows that $V'(e)$ is $V''$-cocartesian, so that $e$ is $V$-cocartesian by virtue of Remark 5.1.1.6.
Since $U_0$, $U_1$, and $U$ are inner fibrations, the morphisms $\overline{V}'$ and $(U_0 \times U_1)$ are also inner fibrations (see Proposition 4.1.4.1). It follows that $V'$ and $V''$ are inner fibrations (Remark 4.1.1.5), so that $V$ is an inner fibration (Remark 4.1.1.8). To show that $V$ is a cocartesian fibration, it will suffice to show that if $C$ is an object of $\operatorname{\mathcal{C}}_0 \operatorname{\vec{\times }}_{ \operatorname{\mathcal{C}}} \operatorname{\mathcal{C}}_1$ and $\overline{e}: V(C) \rightarrow \overline{C}'$ is an edge of $\operatorname{\mathcal{D}}_0 \operatorname{\vec{\times }}_{\operatorname{\mathcal{D}}} \operatorname{\mathcal{D}}_1$, then there exists a special edge $e: C \rightarrow C'$ satisfying $V(e) = \overline{e}$. Let us identify $C$ with a triple $(C_0, C_1, u)$ where $C_0$ is a vertex of $\operatorname{\mathcal{C}}_0$, $C_1$ is a vertex of $\operatorname{\mathcal{C}}_1$, and $u: F_0(C_0) \rightarrow F_1(C_1)$ is an edge of $\operatorname{\mathcal{C}}$. Similarly, we can identify $\overline{C}'$ with a triple $(\overline{C}'_0, \overline{C}'_1, \overline{u}' )$ where $\overline{C}_0$ is a vertex of $\operatorname{\mathcal{D}}_0$, $\overline{C}'_1$ is a vertex of $\operatorname{\mathcal{D}}_1$, and $\overline{u}': G_0( \overline{C}'_0 ) \rightarrow G_1( \overline{C}'_1 )$ is an edge of $\operatorname{\mathcal{D}}$. The edge $\overline{e}$ has images $\overline{e}_{0}: U_0(C_0) \rightarrow \overline{C}'_0$ an $\overline{e}_{1}: U_1(C_1) \rightarrow \overline{C}'_{1}$ in $\operatorname{\mathcal{D}}_0$ and $\operatorname{\mathcal{D}}_1$-respectively. Since $U_0$ is a cocartesian fibration, we can lift $\overline{e}_0$ to a $U_0$-cocartesian edge $e_0: C_0 \rightarrow C'_0$ of $\operatorname{\mathcal{C}}_0$. Similarly, we can lift $\overline{e}_1$ to a $U_1$-cocartesian edge $e_1: C_1 \rightarrow C'_1$ of $\operatorname{\mathcal{C}}_1$. The edge $\overline{e}$ also determines a map $\Delta ^1 \times \Delta ^1 \rightarrow \operatorname{\mathcal{D}}$, which we depict informally in the diagram
$\xymatrix { (G_0 \circ U_0)( C_0 ) \ar [r]^-{U(u)} \ar [d]^{ G_0(\overline{e}_0) } & (G_1 \circ U_1)(C_1) \ar [d]^{ G_1(\overline{e}_1) } \\ G_0(\overline{C}'_0) \ar [r]^{\overline{u}'} & G_1( \overline{C}'_{1} ). }$
Using our assumption that $U$ is an inner fibration, we can lift the upper right triangle to a $2$-simplex $\sigma$:
$\xymatrix { F_0(C_0) \ar [r]^{u} \ar@ {-->}[dr]^{v} & F_1(C_1) \ar [d]^{ F_1( e_1 ) } \\ & F_1( C'_1 ) }$
of the simplicial set $\operatorname{\mathcal{C}}$. Using the fact that $F_0( e_0 )$ is $U$-cocartesian, we can lift the lower triangle to a $2$-simplex $\tau$
$\xymatrix { F_0(C_0) \ar [d]^{ F_0(e_0 ) } \ar [dr]^{v} & \\ F_0( C'_0 ) \ar@ {-->}[r]^{w} & F_1( C'_1 ) }$
of $\operatorname{\mathcal{C}}$. Setting $C' = (C'_0, C'_1, w) \in \operatorname{\mathcal{C}}_0 \operatorname{\vec{\times }}_{\operatorname{\mathcal{C}}} \operatorname{\mathcal{C}}_1$, we observe that the tuple $( e_0, e_1, \sigma , \tau )$ determines a special edge $e: C \rightarrow C'$ satisfying $V(e) = \overline{e}$.
We now complete the proof by showing that every $V$-cocartesian edge $f: C \rightarrow C''$ in $\operatorname{\mathcal{C}}_0 \operatorname{\vec{\times }}_{\operatorname{\mathcal{C}}} \operatorname{\mathcal{C}}_1$ is special. Using the preceding argument, we can choose a special edge $e: C \rightarrow C'$ satisfying $V(e) = V(f)$. Set $\overline{C}' = V( C' ) = V(C'' )$. Applying Remark 5.1.3.8, we deduce that there is a $2$-simplex $\rho :$
$\xymatrix { & C' \ar [dr]^{s} & \\ C \ar [ur]^{ e } \ar [rr]^{f} & & C'' }$
of the simplicial set $\operatorname{\mathcal{C}}_0 \operatorname{\vec{\times }}_{\operatorname{\mathcal{C}}} \operatorname{\mathcal{C}}_1$, where $s$ is an isomorphism in the $\infty$-category $V^{-1}( \{ \overline{C}' \} )$. Applying Example 5.1.3.6, we deduce that the images of $s$ in $\operatorname{\mathcal{C}}_0$ is $U_0$-cocartesian, and the image of $s$ in $\operatorname{\mathcal{C}}_1$ is $U_1$-cocartesian. Since the collections of $U_0$-cocartesian and $U_1$-cocartesian edges are closed under composition (Corollary 5.1.2.4), we conclude that $f$ is also special. $\square$
Corollary 5.3.7.3. Let $F_0: \operatorname{\mathcal{C}}_0 \rightarrow \operatorname{\mathcal{C}}$ and $F_1: \operatorname{\mathcal{C}}_1 \rightarrow \operatorname{\mathcal{C}}$ be morphisms of simplicial sets and let
$\operatorname{\mathcal{C}}_0 \xleftarrow { \pi } \operatorname{\mathcal{C}}_0 \operatorname{\vec{\times }}_{\operatorname{\mathcal{C}}} \operatorname{\mathcal{C}}_1 \xrightarrow {\pi '} \operatorname{\mathcal{C}}_1$
denote the projection maps. Then:
$(1)$
If $\operatorname{\mathcal{C}}_0$ and $\operatorname{\mathcal{C}}$ are $\infty$-categories, then $\pi '$ is a cocartesian fibration of simplicial sets. Moreover, an edge $e$ of $\operatorname{\mathcal{C}}_0 \operatorname{\vec{\times }}_{\operatorname{\mathcal{C}}} \operatorname{\mathcal{C}}_1$ is $\pi '$-cocartesian if and only if $\pi (e)$ is an isomorphism in the $\infty$-category $\operatorname{\mathcal{C}}_0$.
$(2)$
If $\operatorname{\mathcal{C}}_1$ and $\operatorname{\mathcal{C}}$ are $\infty$-categories, then the evaluation map $\pi : \operatorname{\mathcal{C}}_0 \operatorname{\vec{\times }}_{\operatorname{\mathcal{C}}} \operatorname{\mathcal{C}}_1$ is a cartesian fibration of simplicial sets. Moreover, an edge $e$ of $\operatorname{\mathcal{C}}_0 \operatorname{\vec{\times }}_{\operatorname{\mathcal{C}}} \operatorname{\mathcal{C}}_1$ is $\pi$-cartesian if and only if $\pi '(e)$ is an isomorphism in the $\infty$-category $\operatorname{\mathcal{C}}_1$.
Proof. Assertion $(1)$ follows by applying Proposition 5.3.7.2 in the special case $\operatorname{\mathcal{D}}_0 = \operatorname{\mathcal{D}}= \Delta ^0$ and $\operatorname{\mathcal{D}}_1 = \operatorname{\mathcal{C}}_1$. Assertion $(2)$ follows by a similar argument. $\square$
Example 5.3.7.4. Let $\operatorname{\mathcal{C}}$ be an $\infty$-category. Applying Corollary 5.3.7.3 in the case where both $F$ and $G$ are the identity functor $\operatorname{id}: \operatorname{\mathcal{C}}\rightarrow \operatorname{\mathcal{C}}$, we deduce that the evaluation functor
$\operatorname{Fun}( \Delta ^1, \operatorname{\mathcal{C}}) \rightarrow \operatorname{Fun}( \{ 0\} , \operatorname{\mathcal{C}}) \simeq \operatorname{\mathcal{C}}$
is a cartesian fibration of $\infty$-categories, and the evaluation functor
$\operatorname{Fun}( \Delta ^1, \operatorname{\mathcal{C}}) \rightarrow \operatorname{Fun}( \{ 1\} , \operatorname{\mathcal{C}}) \simeq \operatorname{\mathcal{C}}$
is a cocartesian fibration of $\infty$-categories.
Corollary 5.3.7.5. Let $\operatorname{\mathcal{C}}$ be an $\infty$-category and let $K$ be a simplicial set. Then:
$(1)$
The restriction map $U: \operatorname{Fun}(K^{\triangleleft }, \operatorname{\mathcal{C}}) \rightarrow \operatorname{Fun}(K, \operatorname{\mathcal{C}})$ is a cocartesian fibration. Moreover, a morphism $e$ of $\operatorname{Fun}(K^{\triangleleft }, \operatorname{\mathcal{C}})$ is $U$-cocartesian if and only if it carries the cone point ${\bf 0} \in K^{\triangleleft }$ to an isomorphism in $\operatorname{\mathcal{C}}$.
$(2)$
The restriction map $V: \operatorname{Fun}(K^{\triangleright }, \operatorname{\mathcal{C}}) \rightarrow \operatorname{Fun}(K, \operatorname{\mathcal{C}})$ is a cartesian fibration. Moreover, a morphism $e$ of $\operatorname{Fun}(K^{\triangleright }, \operatorname{\mathcal{C}})$ is $U$-cartesian if and only if it carries the cone point ${\bf 1} \in K^{\triangleright }$ to an isomorphism in $\operatorname{\mathcal{C}}$.
Proof. We will prove $(1)$; the proof of $(2)$ is similar. Let $\Delta ^0 \diamond K$ denote the blunt join of Notation 4.5.8.3, and let $c: \Delta ^0 \diamond K \rightarrow \Delta ^{0} \star K = K^{\triangleleft }$ be the categorical equivalence of Theorem 4.5.8.8. We have a commutative diagram of $\infty$-categories
$\xymatrix@R =50pt@C=50pt{ \operatorname{Fun}( K^{\triangleleft }, \operatorname{\mathcal{C}}) \ar [dr]^{U} \ar [rr]^-{ \circ c} & & \operatorname{Fun}( \Delta ^0 \diamond K, \operatorname{\mathcal{C}}) \ar [dl]^{U'} \\ & \operatorname{Fun}(K, \operatorname{\mathcal{C}}) & }$
where the horizontal map is an equivalence of $\infty$-categories (Proposition 4.5.3.8) and the vertical maps are isofibrations (Corollary 4.4.5.3). Unwinding the definitions, we can identify $\operatorname{Fun}( \Delta ^0 \diamond K, \operatorname{\mathcal{C}})$ with the oriented fiber product $\operatorname{\mathcal{C}}\operatorname{\vec{\times }}_{ \operatorname{Fun}(K, \operatorname{\mathcal{C}}) } \operatorname{Fun}(K, \operatorname{\mathcal{C}})$. Under this identification, the functor $U'$ is given by projection onto the second factor, and is therefore a cocartesian fibration (Corollaryt 5.3.7.3). Applying Corollary 5.1.6.2, we deduce that $U$ is also a cocartesian fibration. Moreover, a morphism $e$ of $\operatorname{Fun}(K^{\triangleleft }, \operatorname{\mathcal{C}})$ is $U$-cocartesian if and only if its image in $\operatorname{Fun}( \Delta ^0 \diamond K, \operatorname{\mathcal{C}})$ is $U'$-cocartesian (Proposition 5.1.6.6). Using the criterion of Corollary 5.3.7.3, we see that this is equivalent to the requirement that $e$ carries the cone point ${\bf 0} \in K^{\triangleleft }$ to an isomorphism in $\operatorname{\mathcal{C}}$. $\square$
Theorem 5.3.7.6. Let $F: \operatorname{\mathcal{C}}\rightarrow \operatorname{\mathcal{D}}$ be a functor of $\infty$-categories, let $\pi : \operatorname{\mathcal{C}}\operatorname{\vec{\times }}_{\operatorname{\mathcal{D}}} \operatorname{\mathcal{D}}\rightarrow \operatorname{\mathcal{D}}$ be given by projection onto the second factor, let $\delta : \operatorname{\mathcal{C}}\hookrightarrow \operatorname{\mathcal{C}}\operatorname{\vec{\times }}_{\operatorname{\mathcal{D}}} \operatorname{\mathcal{D}}$ be the diagonal map. For every cocartesian fibration $U: \operatorname{\mathcal{E}}\rightarrow \operatorname{\mathcal{D}}$, precomposition with $\delta$ induces a trivial Kan fibration of $\infty$-categories
$\operatorname{Fun}^{\operatorname{CCart}}_{ / \operatorname{\mathcal{D}}}( \operatorname{\mathcal{C}}\operatorname{\vec{\times }}_{\operatorname{\mathcal{D}}} \operatorname{\mathcal{D}}, \operatorname{\mathcal{E}}) \rightarrow \operatorname{Fun}_{ / \operatorname{\mathcal{D}}}( \operatorname{\mathcal{C}}, \operatorname{\mathcal{E}}).$
Our proof of Theorem 5.3.7.6 will make use of an auxiliary construction.
Notation 5.3.7.7 (Cocartesian Direct Images). Let $U: \operatorname{\mathcal{D}}\rightarrow \operatorname{\mathcal{C}}$ be a morphism of simplicial sets $\operatorname{id}_{\operatorname{\mathcal{D}}}: \operatorname{\mathcal{D}}\rightarrow \operatorname{\mathcal{D}}$ determines a section of the projection map $\pi : \operatorname{Fun}( \operatorname{\mathcal{D}}/\operatorname{\mathcal{C}}, \operatorname{\mathcal{D}}) \rightarrow \operatorname{\mathcal{C}}$. For every morphism of simplicial sets $V: \operatorname{\mathcal{E}}\rightarrow \operatorname{\mathcal{D}}$, we let $\operatorname{Res}_{\operatorname{\mathcal{D}}/\operatorname{\mathcal{C}}}(\operatorname{\mathcal{E}})$ denote the fiber product $\operatorname{\mathcal{C}}\times _{ \operatorname{Fun}(\operatorname{\mathcal{D}}/\operatorname{\mathcal{C}}, \operatorname{\mathcal{D}}) } \operatorname{Fun}(\operatorname{\mathcal{D}}/\operatorname{\mathcal{C}}, \operatorname{\mathcal{E}})$. Unwinding the definitions, we see that vertices of $\operatorname{Res}_{\operatorname{\mathcal{D}}/\operatorname{\mathcal{C}}}(\operatorname{\mathcal{E}})$ can be identified with pairs $(C,F)$, where $C$ is a vertex of $\operatorname{\mathcal{C}}$ and
$F: \operatorname{\mathcal{D}}_{C} = \{ C\} \times _{\operatorname{\mathcal{C}}} \operatorname{\mathcal{D}}\rightarrow \{ C\} \times _{\operatorname{\mathcal{C}}} \operatorname{\mathcal{E}}= \operatorname{\mathcal{E}}_{C}$
is a section of the map $V|_{\operatorname{\mathcal{E}}_ C}: \operatorname{\mathcal{E}}_ C \rightarrow \operatorname{\mathcal{D}}_{C}$. If $V$ is a cocartesian fibration, we let $\operatorname{Res}^{\operatorname{CCart}}_{\operatorname{\mathcal{D}}/\operatorname{\mathcal{C}}}( \operatorname{\mathcal{E}})$ denote the full simplicial subset of $\operatorname{Res}_{\operatorname{\mathcal{D}}/\operatorname{\mathcal{C}}}(\operatorname{\mathcal{E}})$ spanned by those vertices $(C,F)$ where $F$ carries each edge of $\operatorname{\mathcal{D}}_{C}$ to $V_{C}$-cocartesian edge of $\operatorname{\mathcal{E}}_{C}$. We will refer to $\operatorname{Res}_{\operatorname{\mathcal{D}}/\operatorname{\mathcal{C}}}^{\operatorname{CCart}}( \operatorname{\mathcal{E}})$ as the cocartesian direct image of $\operatorname{\mathcal{E}}$ along $U$.
Remark 5.3.7.8. Let $U: \operatorname{\mathcal{D}}\rightarrow \operatorname{\mathcal{C}}$ be a morphism of simplicial sets and let $V: \operatorname{\mathcal{E}}\rightarrow \operatorname{\mathcal{D}}$ be a cocartesian fibration of simplicial sets. Then the projection map $\pi : \operatorname{Res}_{\operatorname{\mathcal{D}}/\operatorname{\mathcal{C}}}( \operatorname{\mathcal{E}}) \rightarrow \operatorname{\mathcal{C}}$ restricts to a projection map $\pi ^{\operatorname{CCart}}: \operatorname{Res}_{\operatorname{\mathcal{D}}/\operatorname{\mathcal{C}}}^{\operatorname{CCart}}( \operatorname{\mathcal{E}}) \rightarrow \operatorname{\mathcal{C}}$. Moreover, for each vertex $C \in \operatorname{\mathcal{C}}$, the canonical isomorphism $\{ C\} \times _{\operatorname{\mathcal{C}}} \operatorname{Res}_{\operatorname{\mathcal{D}}/\operatorname{\mathcal{C}}}( \operatorname{\mathcal{E}}) \simeq \operatorname{Fun}_{ / \operatorname{\mathcal{D}}_ C }( \operatorname{\mathcal{D}}_ C, \operatorname{\mathcal{E}}_ C )$ restricts to an isomorphism of full subcategories $\{ C\} \times _{\operatorname{\mathcal{C}}} \operatorname{Res}_{\operatorname{\mathcal{D}}/\operatorname{\mathcal{C}}}^{\operatorname{CCart}}(\operatorname{\mathcal{E}}) \simeq \operatorname{Fun}^{\operatorname{CCart}}_{/\operatorname{\mathcal{D}}_ C }( \operatorname{\mathcal{D}}_ C, \operatorname{\mathcal{E}}_ C )$.
Proposition 5.3.7.9. Let $V: \operatorname{\mathcal{E}}\rightarrow \operatorname{\mathcal{D}}$ be a cocartesian fibration of simplicial sets, let $U: \operatorname{\mathcal{D}}\rightarrow \operatorname{\mathcal{C}}$ be a cartesian fibration of simplicial sets. Then:
$(1)$
The projection map $\pi : \operatorname{Res}_{\operatorname{\mathcal{D}}/\operatorname{\mathcal{C}}}( \operatorname{\mathcal{E}}) \rightarrow \operatorname{\mathcal{C}}$ is a cocartesian fibration of simplicial sets.
$(2)$
Let $e: X \rightarrow Y$ be a $\pi$-cocartesian edge of the simplicial set $\operatorname{Res}_{\operatorname{\mathcal{D}}/\operatorname{\mathcal{C}}}(\operatorname{\mathcal{E}})$. If $X$ belongs to the simplicial subset $\operatorname{Res}_{\operatorname{\mathcal{D}}/\operatorname{\mathcal{C}}}^{\operatorname{CCart}}( \operatorname{\mathcal{E}})$, then $Y$ also belongs to the simplicial subset $\operatorname{Res}_{\operatorname{\mathcal{D}}/\operatorname{\mathcal{C}}}^{\operatorname{CCart}}( \operatorname{\mathcal{E}})$.
$(3)$
The morphism $\pi$ restricts to a cocartesian fibration $\pi ^{\operatorname{CCart}}: \operatorname{Res}_{\operatorname{\mathcal{D}}/\operatorname{\mathcal{C}}}^{\operatorname{CCart}}( \operatorname{\mathcal{E}}) \rightarrow \operatorname{\mathcal{C}}$.
$(4)$
An edge of the simplicial set $\operatorname{Res}_{\operatorname{\mathcal{D}}/\operatorname{\mathcal{C}}}^{\operatorname{CCart}}( \operatorname{\mathcal{E}})$ is $\pi ^{\operatorname{CCart}}$-cocartesian if and only if it is $\pi$-cocartesian.
Proof. Assertion $(1)$ follows from Proposition 5.3.6.6 (after passing to opposite simplicial sets). To prove $(2)$, we may assume without loss of generality that $\operatorname{\mathcal{C}}= \Delta ^1$ and $\pi (e)$ is the nondegenerate edge of $\operatorname{\mathcal{C}}$. In this case, the simplicial sets $\operatorname{\mathcal{D}}$ and $\operatorname{\mathcal{E}}$ are $\infty$-categories, and we can identify the edge $e$ with a morphism of simplicial sets $E: \operatorname{\mathcal{D}}\rightarrow \operatorname{\mathcal{E}}$ satisfying $V \circ E = \operatorname{id}_{\operatorname{\mathcal{D}}}$. Let $u: D \rightarrow D'$ be a morphism in the $\infty$-category $\operatorname{\mathcal{D}}_1 = \{ 1\} \times _{\operatorname{\mathcal{C}}} \operatorname{\mathcal{D}}$; we wish to show that $E(u)$ is a $V$-cocartesian morphism of $\operatorname{\mathcal{E}}$. To prove this, let $G: \operatorname{\mathcal{D}}_{1} \rightarrow \operatorname{\mathcal{D}}_0 = \{ 0\} \times _{\operatorname{\mathcal{C}}} \operatorname{\mathcal{D}}$ be given by contravariant transport along the nondegenerate edge of $\operatorname{\mathcal{C}}$, so that we have a commutative diagram
$\xymatrix@R =50pt@C=50pt{ G(D) \ar [d]^{ G(u) } \ar [r] & D \ar [d]^{u} \\ G(D') \ar [r] & D', }$
in the $\infty$-category where the horizontal maps are $U$-cartesian. Our assumption that $e$ is $\pi$-cocartesian guarantees that the functor $E$ carries $U$-cartesian morphisms of $\operatorname{\mathcal{D}}$ to $V$-cocartesian morphisms of $\operatorname{\mathcal{E}}$ (Proposition 5.3.6.6). We therefore obtain a commutative diagram
$\xymatrix@R =50pt@C=50pt{ (E \circ G)(D) \ar [d]^{ (E \circ G)(u) } \ar [r] & E(D) \ar [d]^{ E(u) } \\ (E \circ G)(D') \ar [r] & E(D' ), }$
where the horizontal maps are $V$-cocartesian. By virtue of Corollary 5.1.2.4, it will suffice to show that the morphism $(E \circ G)(u)$ is $V$-cocartesian, which follows from our assumption that $X$ belongs to $\operatorname{Res}_{\operatorname{\mathcal{D}}/\operatorname{\mathcal{C}}}^{\operatorname{CCart}}( \operatorname{\mathcal{E}})$. This completes the proof of $(2)$; assertions $(3)$ and $(4)$ then follow by applying Proposition 5.1.4.16. $\square$
In the situation of Proposition 5.3.7.9, the cocartesian direct image $\operatorname{Res}_{\operatorname{\mathcal{D}}/\operatorname{\mathcal{C}}}^{\operatorname{CCart}}( \operatorname{\mathcal{E}})$ can be characterized by a universal property:
Proposition 5.3.7.10. Let $V: \operatorname{\mathcal{E}}\rightarrow \operatorname{\mathcal{D}}$ be a cocartesian fibration of simplicial sets and let $U: \operatorname{\mathcal{D}}\rightarrow \operatorname{\mathcal{C}}$ be a cartesian fibration of simplicial sets. For every cocartesian fibration of simplicial sets $W: \operatorname{\mathcal{C}}' \rightarrow \operatorname{\mathcal{C}}$, the canonical isomorphism
$\operatorname{Fun}_{ / \operatorname{\mathcal{C}}}( \operatorname{\mathcal{C}}', \operatorname{Res}_{\operatorname{\mathcal{D}}/\operatorname{\mathcal{C}}}( \operatorname{\mathcal{E}}) ) \xrightarrow {\sim } \operatorname{Fun}_{/\operatorname{\mathcal{D}}}( \operatorname{\mathcal{C}}' \times _{\operatorname{\mathcal{C}}} \operatorname{\mathcal{D}}, \operatorname{\mathcal{E}})$
restricts to an isomorphism of full simplicial subsets
$\operatorname{Fun}^{\operatorname{CCart}}_{/\operatorname{\mathcal{C}}}( \operatorname{\mathcal{C}}', \operatorname{Res}_{\operatorname{\mathcal{D}}/\operatorname{\mathcal{C}}}^{\operatorname{CCart}}( \operatorname{\mathcal{E}}) ) \xrightarrow {\sim } \operatorname{Fun}^{\operatorname{CCart}}_{/\operatorname{\mathcal{D}}}( \operatorname{\mathcal{C}}' \times _{\operatorname{\mathcal{C}}} \operatorname{\mathcal{D}}, \operatorname{\mathcal{E}}).$
Proof. Let $\pi : \operatorname{Res}_{\operatorname{\mathcal{D}}/\operatorname{\mathcal{C}}}(\operatorname{\mathcal{E}}) \rightarrow \operatorname{\mathcal{C}}$ denote the projection map and let $f: \operatorname{\mathcal{C}}' \rightarrow \operatorname{Res}_{\operatorname{\mathcal{D}}/\operatorname{\mathcal{C}}}(\operatorname{\mathcal{E}})$ be a morphism satisfying $\pi \circ f = W$, corresponding to a morphism of simplicial sets $F: \operatorname{\mathcal{C}}' \times _{\operatorname{\mathcal{C}}} \operatorname{\mathcal{D}}\rightarrow \operatorname{\mathcal{E}}$ for which $V \circ F$ is given by projection to the second factor. Note that we can regard $F$ as a vertex of the simplicial subset $\operatorname{Fun}^{\operatorname{CCart}}_{/\operatorname{\mathcal{D}}}( \operatorname{\mathcal{C}}' \times _{\operatorname{\mathcal{C}}} \operatorname{\mathcal{D}}, \operatorname{\mathcal{E}})$ if and only if it satisfies the following condition:
$(a)$
For every edge $(e',e)$ of the fiber product $\operatorname{\mathcal{C}}' \times _{\operatorname{\mathcal{C}}} \operatorname{\mathcal{D}}$ for which $e'$ is a $W$-cocartesian edge of $\operatorname{\mathcal{C}}'$, the image $F(e',e)$ is a $V$-cocartesian edge of $\operatorname{\mathcal{E}}$.
We wish to show that $(a)$ is equivalent to the following pair of conditions:
$(b)$
The morphism $f$ factors through the full simplicial subset $\operatorname{Res}^{\operatorname{CCart}}_{\operatorname{\mathcal{D}}/\operatorname{\mathcal{C}}}(\operatorname{\mathcal{E}}) \subseteq \operatorname{Res}_{\operatorname{\mathcal{D}}/\operatorname{\mathcal{C}}}(\operatorname{\mathcal{E}})$. In other words, for every edge $(e',e)$ of the fiber product $\operatorname{\mathcal{C}}' \times _{\operatorname{\mathcal{C}}} \operatorname{\mathcal{D}}$ for which $e'$ is a degenerate edge of $\operatorname{\mathcal{C}}'$, the image $F(e',e)$ is a $V$-cocartesian edge of $\operatorname{\mathcal{E}}$.
$(c)$
For every $W$-cocartesian edge $e'$ of $\operatorname{\mathcal{C}}'$, the image $f(e')$ is a $\pi |_{ \operatorname{Res}_{\operatorname{\mathcal{D}}/\operatorname{\mathcal{C}}}^{\operatorname{CCart}}( \operatorname{\mathcal{E}}) }$-cocartesian edge of $\operatorname{Res}^{\operatorname{CCart}}_{\operatorname{\mathcal{D}}/\operatorname{\mathcal{C}}}(\operatorname{\mathcal{E}})$. By virtue of Propositions 5.3.7.9 and 5.3.6.6, this is equivalent to the assertion that for every edge $(e',e)$ of the fiber product $\operatorname{\mathcal{C}}' \times _{\operatorname{\mathcal{C}}} \operatorname{\mathcal{D}}$ where $e'$ is $W$-cocartesian and $e$ is $U$-cartesian, the image $F(e',e)$ is a $V$-cocartesian edge of $\operatorname{\mathcal{E}}$.
The implications $(a) \Rightarrow (b)$ and $(a) \Rightarrow (c)$ are clear. For the converse, suppose that $(b)$ and $(c)$ are satisfied; we wish to prove $(a)$. Let $(e',e): (X',X) \rightarrow (Z',Z)$ be an edge of the fiber product $\operatorname{\mathcal{C}}' \times _{\operatorname{\mathcal{C}}} \operatorname{\mathcal{D}}$, where $e': X' \rightarrow Z'$ is $W$-cocartesian. Let $\overline{e} = U(e) = W(e')$ denote the corresponding edge of $\operatorname{\mathcal{C}}$. Since $U$ is a cartesian fibration, there exists a $U$-cartesian morphism $f: Y \rightarrow Z$ satisfying $U(f) = \overline{e}$. Let $\overline{\sigma }$ denote the left-degenerate $2$-simplex $s^{1}_0(\overline{e})$. Since $f$ is $U$-cartesian, we can lift $\overline{\sigma }$ to a $2$-simplex of $\operatorname{\mathcal{D}}$ as indicated in the diagram
$\xymatrix@R =50pt@C=50pt{ & Y \ar [dr]^{f} & \\ X \ar [ur] \ar [rr]^-{e} & & Z. }$
Writing $\sigma '$ for the left-degenerate $2$-simplex $s^{1}_0(e')$ of $\operatorname{\mathcal{C}}'$, we obtain a $2$-simplex $\tau = F( \sigma ', \sigma )$ of $\operatorname{\mathcal{E}}$. It follows from assumption $(b)$ that the restriction $\tau |_{ \operatorname{N}_{\bullet }( \{ 0 < 1\} ) }$ is a $V$-cocartesian edge of $\operatorname{\mathcal{E}}$, and from assumption $(c)$ that the restriction $\tau |_{ \operatorname{N}_{\bullet }( \{ 1 < 2 \} ) }$ is a $V$-cocartesian edge of $\operatorname{\mathcal{E}}$. Applying Proposition 5.1.4.12, we conclude that $F(e',e) = \tau |_{ \operatorname{N}_{\bullet }( \{ 0 < 2 \} ) }$ is also a $V$-cocartesian edge of $\operatorname{\mathcal{E}}$. $\square$
Proof of Theorem 5.3.7.6. Let $F: \operatorname{\mathcal{C}}\rightarrow \operatorname{\mathcal{D}}$ be a functor of $\infty$-categories, let $U: \operatorname{\mathcal{E}}\rightarrow \operatorname{\mathcal{D}}$ be a cocartesian fibration of $\infty$-categories, and let $\delta : \operatorname{\mathcal{C}}\hookrightarrow \operatorname{\mathcal{C}}\operatorname{\vec{\times }}_{\operatorname{\mathcal{D}}} \operatorname{\mathcal{D}}$ be the diagonal embedding. Since $U$ is an isofibration (Proposition 5.1.4.8), the restriction map $\overline{\theta }: \operatorname{Fun}_{ / \operatorname{\mathcal{D}}}( \operatorname{\mathcal{C}}\operatorname{\vec{\times }}_{\operatorname{\mathcal{D}}} \operatorname{\mathcal{D}}, \operatorname{\mathcal{E}}) \rightarrow \operatorname{Fun}_{ / \operatorname{\mathcal{D}}}( \operatorname{\mathcal{C}}, \operatorname{\mathcal{E}})$ is also an isofibration (Corollary 4.5.5.16). Because $\operatorname{Fun}^{\operatorname{CCart}}_{ / \operatorname{\mathcal{D}}}( \operatorname{\mathcal{C}}\operatorname{\vec{\times }}_{\operatorname{\mathcal{D}}} \operatorname{\mathcal{D}}, \operatorname{\mathcal{E}})$ is a replete full subcategory of $\operatorname{Fun}_{ / \operatorname{\mathcal{D}}}( \operatorname{\mathcal{C}}\operatorname{\vec{\times }}_{\operatorname{\mathcal{D}}} \operatorname{\mathcal{D}}, \operatorname{\mathcal{E}})$, it follows that $\overline{\theta }$ restricts to an isofibration $\theta : \operatorname{Fun}^{\operatorname{CCart}}_{ / \operatorname{\mathcal{D}}}( \operatorname{\mathcal{C}}\operatorname{\vec{\times }}_{\operatorname{\mathcal{D}}} \operatorname{\mathcal{D}}, \operatorname{\mathcal{E}}) \rightarrow \operatorname{Fun}_{ / \operatorname{\mathcal{D}}}( \operatorname{\mathcal{C}}, \operatorname{\mathcal{E}})$. To prove Theorem 5.3.7.6, we will show that $\theta$ is an equivalence of $\infty$-categories (it is then automatically a trivial Kan fibration of simplicial sets: see Proposition 4.5.5.20).
Note that the functor $U: \operatorname{\mathcal{E}}\rightarrow \operatorname{\mathcal{D}}$ induces cocartesian fibrations $U': \operatorname{\mathcal{C}}\operatorname{\vec{\times }}_{\operatorname{\mathcal{D}}} \operatorname{\mathcal{E}}\rightarrow \operatorname{\mathcal{C}}\operatorname{\vec{\times }}_{\operatorname{\mathcal{D}}} \operatorname{\mathcal{D}}$ and $U'': \operatorname{\mathcal{C}}\times _{\operatorname{\mathcal{D}}} \operatorname{\mathcal{E}}\rightarrow \operatorname{\mathcal{C}}$. Let $\pi ': \operatorname{\mathcal{C}}\operatorname{\vec{\times }}_{\operatorname{\mathcal{D}}} \operatorname{\mathcal{D}}\rightarrow \operatorname{\mathcal{C}}$ be given by projection onto the first factor, so that $\pi '$ is a cartesian fibration (Corollary 5.3.7.3). Let $\operatorname{\mathcal{M}}$ denote the cocartesian direct image $\operatorname{Res}^{\operatorname{CCart}}_{ \operatorname{\mathcal{C}}\operatorname{\vec{\times }}_{\operatorname{\mathcal{D}}} \operatorname{\mathcal{D}}/ \operatorname{\mathcal{C}}}( \operatorname{\mathcal{C}}\operatorname{\vec{\times }}_{\operatorname{\mathcal{D}}} \operatorname{\mathcal{E}})$ and let $T: \operatorname{\mathcal{M}}\rightarrow \operatorname{\mathcal{C}}$ be the projection map. Precomposition with the diagonal embedding $\delta : \operatorname{\mathcal{C}}\hookrightarrow \operatorname{\mathcal{C}}\operatorname{\vec{\times }}_{\operatorname{\mathcal{D}}} \operatorname{\mathcal{D}}$ induces a restriction functor
$\delta ^{\ast }: \operatorname{\mathcal{M}}\rightarrow \operatorname{Res}_{ \operatorname{\mathcal{C}}/\operatorname{\mathcal{C}}}( \operatorname{\mathcal{C}}\times _{\operatorname{\mathcal{D}}} \operatorname{\mathcal{E}}) = \operatorname{\mathcal{C}}\times _{\operatorname{\mathcal{D}}} \operatorname{\mathcal{E}}$
which fits into a commutative diagram
$\xymatrix@R =50pt@C=50pt{ \operatorname{\mathcal{M}}\ar [rr]^{ \delta ^{\ast } } \ar [dr]_{T} & & \operatorname{\mathcal{C}}\times _{\operatorname{\mathcal{D}}} \operatorname{\mathcal{E}}\ar [dl]^{U''} \\ & \operatorname{\mathcal{C}}& }$
It follows from Proposition 5.3.7.9 that $T$ is a cocartesian fibration and that $\delta ^{\ast }$ carries $T$-cocartesian morphisms of $\operatorname{\mathcal{M}}$ to $U''$-cocartesian morphisms of $\operatorname{\mathcal{C}}\times _{\operatorname{\mathcal{D}}} \operatorname{\mathcal{E}}$. Using Proposition 5.3.7.10, we can identify $\theta$ with the map
$\operatorname{Fun}_{ / \operatorname{\mathcal{C}}}( \operatorname{\mathcal{C}}, \operatorname{\mathcal{M}}) \rightarrow \operatorname{Fun}_{ / \operatorname{\mathcal{C}}}( \operatorname{\mathcal{C}}, \operatorname{\mathcal{C}}\times _{\operatorname{\mathcal{D}}}, \operatorname{\mathcal{E}}) \simeq \operatorname{Fun}_{ / \operatorname{\mathcal{D}}}( \operatorname{\mathcal{C}}, \operatorname{\mathcal{E}})$
given by postcomposition with $\delta ^{\ast }$. Consequently, to show that $\theta$ is an equivalence of $\infty$-categories, it will suffice to show that $\delta ^{\ast }$ is an equivalence of cocartesian fibrations over $\operatorname{\mathcal{C}}$. By virtue of Proposition 5.1.7.14), this can be checked fiberwise: that is, it suffices to show that for each object $C \in \operatorname{\mathcal{C}}$, the induced map of fibers
$\delta ^{\ast }_{C}: \{ C\} \times _{\operatorname{\mathcal{C}}} \operatorname{\mathcal{M}}\simeq \operatorname{Fun}^{\operatorname{CCart}}_{ / \operatorname{\mathcal{D}}}( \{ C\} \operatorname{\vec{\times }}_{\operatorname{\mathcal{D}}} \operatorname{\mathcal{D}}, \operatorname{\mathcal{E}}) \rightarrow \{ C\} \times _{\operatorname{\mathcal{D}}} \operatorname{\mathcal{E}}$
is an equivalence of $\infty$-categories. This is a special case of Corollary 5.3.1.22, since $\delta (C)$ is an initial object of the $\infty$-category $\{ C\} \operatorname{\vec{\times }}_{\operatorname{\mathcal{D}}} \operatorname{\mathcal{D}}$ (Proposition 4.6.7.22). $\square$ | 13,139 | 38,879 | {"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": 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": 1, "x-ck12": 0, "texerror": 0} | 2.796875 | 3 | CC-MAIN-2024-10 | longest | en | 0.557271 |
loqeculisyhe.swisseurasier.com | 1,568,686,165,000,000,000 | text/html | crawl-data/CC-MAIN-2019-39/segments/1568514573011.59/warc/CC-MAIN-20190917020816-20190917042816-00269.warc.gz | 570,613,125 | 4,387 | # Write an equation of the line from a graph
We know the slope and a point x,y. That's our y-intercept, right there at the origin. So if delta x is equal to 3. In this form, the y-intercept is b, which is the constant. Y is the dependent variable that changes in response to X.
If we run one, two, three. If we go over to the right by one, two, three, four. The next point is 2, 4. Looking at the graph, you can see that this graph never crosses the y-axis, therefore there is no y-intercept either.
The graph would look like this: Well where does this intersect the y-axis? And then the slope-- once again you see a negative sign. Where is the b? And let me draw a line. You want to get close. It's going to look something like that. Although you have the slope, you need the y-intercept.
The answer is the slope is 2 and the y-intercept is On the Cartesian plane[ edit ] Lines in a Cartesian plane or, more generally, in affine coordinatescan be described algebraically by linear equations. So delta y over delta x, When we go to the right, our change in x is 1.
In fact, Euclid did not use these definitions in this work and probably included them just to make it clear to the reader what was being discussed.
They go in opposite directions. When we go over by 3, we're going to go down by 2. Well you know that having a 0 in the denominator is a big no, no. Let's start at some reasonable point.
Now let's go the other way. I've already used orange, let me use this green color. The equation for a linear relationship should look like this: Let's look at some equations of lines knowing that this is the slope and this is the y-intercept-- that's the m, that's the b-- and actually graph them.
Add 2 to each y making them -2,5-4,8and -6, On the Cartesian plane[ edit ] Lines in a Cartesian plane or, more generally, in affine coordinatescan be described algebraically by linear equations.
This example is written in function notation, but is still linear. Your slope is the coefficient of your x term. Therefore, you need only two points. I just have to connect those dots. I keep doing that. So that's our first line.
Well, when x is equal to 0, y is equal to 1. What is the equation of this line in slope-intercept form? The numerator tells us that the y value for the next coordinate increase by 5, the denominator tells us that the x value for the next coordinate changes by 1, so we can add this values to our starting coordinate of 0, Let's do this second line.
So b is equal to 1. So then y is going to be equal to b.swisseurasier.com Write expressions that record operations with numbers and with letters standing for numbers. For example, express the calculation "Subtract y from 5" as 5 - y.
To establish a rule for the equation of a straight line, consider the previous example. An increase in distance by 1 km results in an increase in cost of \$3. swisseurasier.com Write expressions that record operations with numbers and with letters standing for numbers.
For example, express the calculation "Subtract y from 5" as 5 - y. Equation of a Line from 2 Points. First, let's see it in action.
Here are two points (you can drag them) and the equation of the line through them.
Graphing a linear equation written in slope-intercept form, y= mx+b is easy! Remember the structure of y=mx+b and that graphing it will always give you a straight line. Line graphs provide a visual representation of the relationship between variables and how that relationship changes.
For example, you might make a line graph to show how an animal's growth rate varies over time, or how a city's average high temperature varies from month to month. You can also graph.
Write an equation of the line from a graph
Rated 0/5 based on 52 review | 887 | 3,749 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.59375 | 5 | CC-MAIN-2019-39 | latest | en | 0.961104 |
http://www.emathematics.net/parabola1.php?a=2&tipo=exploring | 1,369,162,916,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368700438490/warc/CC-MAIN-20130516103358-00060-ip-10-60-113-184.ec2.internal.warc.gz | 445,190,929 | 13,549 | User:
Documento sin título
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Exploring Quadratic Graphs
The general quadratic function is y=ax2+bx+c
It has one of two basic graphs shapes, as shown below:
It is a symmetrical "U"-shape or "hump"-shape, depending on the sign of a. Another way of saying the same thing is that the y-values drop to a minimum and then rise again if a is poitive, whereas they rise to a maximum and then fall if a is negative.
The roots or solutions of the equation y=ax2+bx+c are given by the intercepts on the x-axis, and by symmetry the maximum or minimum is always halfway between them. Some quadratics equations do not have real roots, and in these cases the graph simply does not cut the x-axis at all. | 595 | 3,013 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.453125 | 3 | CC-MAIN-2013-20 | latest | en | 0.77314 |
https://www.traditionaloven.com/metal/precious-metals/platinum/convert-cubic-millimetre-mm3-platinum-to-pound-lb-of-platinum.html | 1,643,347,269,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320305420.54/warc/CC-MAIN-20220128043801-20220128073801-00377.warc.gz | 1,062,445,269 | 13,824 | Platinum cubic millimeter to pounds of platinum converter
# platinum conversion
## Amount: cubic millimeter (mm3) of platinum volume Equals: 0.000047 pounds (lb) in platinum mass
Calculate pounds of platinum per cubic millimeter unit. The platinum converter.
TOGGLE : from pounds into cubic millimeters in the other way around.
### Enter a New cubic millimeter Amount of platinum to Convert From
* Enter whole numbers, decimals or fractions (ie: 6, 5.33, 17 3/8)
## platinum from cubic millimeter to pound Conversion Results :
Amount : cubic millimeter (mm3) of platinum
Equals: 0.000047 pounds (lb) in platinum
Fractions: 5/106383 pounds (lb) in platinum
CONVERT : between other platinum measuring units - complete list.
## Platinum Amounts (solid platinum)
Here the calculator is for platinum amounts (solid platinum volume; dense, precious, gray to white metal rare in abundance on the planet earth. Its annual production is only a very few hundred tons. It is a very highly valuable metal. Platinum performs real well in resisting corrosion. Not only beautiful jewellery is made out of platinum, this metal enjoys quite a wide variety of uses. For instance in electronics, chemical industries and also in chemotherapy applications against certain cancers. Traders invest money in platinum on commodity markets, in commodity future trading as this material is also one of the major precious commodity metals. Thinking of going into investing in stocks? It would be a wise idea to start learning at least basics at a commodity trading school first, to get used to the markets, then start with small investments. Only after sell and buy platinum.)
Is it possible to manage numerous units calculations, in relation to how heavy other volumes of platinum are, all on one page? The all in one Pt multiunit calculation tool makes it possible to manage just that.
Convert platinum measuring units between cubic millimeter (mm3) and pounds (lb) of platinum but in the other direction from pounds into cubic millimeters.
conversion result for platinum: From Symbol Equals Result To Symbol 1 cubic millimeter mm3 = 0.000047 pounds lb
# Precious metals: platinum conversion
This online platinum from mm3 into lb (precious metal) converter is a handy tool not just for certified or experienced professionals. It can help when selling scrap metals for recycling.
## Other applications of this platinum calculator are ...
With the above mentioned units calculating service it provides, this platinum converter proved to be useful also as a teaching tool:
1. in practicing cubic millimeters and pounds ( mm3 vs. lb ) exchange.
2. for conversion factors training exercises with converting mass/weights units vs. liquid/fluid volume units measures.
3. work with platinum's density values including other physical properties this metal has.
International unit symbols for these two platinum measurements are:
Abbreviation or prefix ( abbr. short brevis ), unit symbol, for cubic millimeter is: mm3
Abbreviation or prefix ( abbr. ) brevis - short unit symbol for pound is: lb
### One cubic millimeter of platinum converted to pound equals to 0.000047 lb
How many pounds of platinum are in 1 cubic millimeter? The answer is: The change of 1 mm3 ( cubic millimeter ) unit of a platinum amount equals = to 0.000047 lb ( pound ) as the equivalent measure for the same platinum type.
In principle with any measuring task, switched on professional people always ensure, and their success depends on, they get the most precise conversion results everywhere and every-time. Not only whenever possible, it's always so. Often having only a good idea ( or more ideas ) might not be perfect nor good enough solutions. Subjects of high economic value such as stocks, foreign exchange market and various units in precious metals trading, money, financing ( to list just several of all kinds of investments ), are way too important. Different matters seek an accurate financial advice first, with a plan. Especially precise prices-versus-sizes of platinum can have a crucial/pivotal role in investments. If there is an exact known measure in mm3 - cubic millimeters for platinum amount, the rule is that the cubic millimeter number gets converted into lb - pounds or any other unit of platinum absolutely exactly. It's like an insurance for a trader or investor who is buying. And a saving calculator for having a peace of mind by knowing more about the quantity of e.g. how much industrial commodities is being bought well before it is payed for. It is also a part of savings to my superannuation funds. "Super funds" as we call them in this country.
Conversion for how many pounds ( lb ) of platinum are contained in a cubic millimeter ( 1 mm3 ). Or, how much in pounds of platinum is in 1 cubic millimeter? To link to this platinum - cubic millimeter to pounds online precious metal converter for the answer, simply cut and paste the following.
The link to this tool will appear as: platinum from cubic millimeter (mm3) to pounds (lb) metal conversion.
I've done my best to build this site for you- Please send feedback to let me know how you enjoyed visiting. | 1,082 | 5,164 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.890625 | 3 | CC-MAIN-2022-05 | latest | en | 0.848895 |
http://searchengineland.com/google-conversion-rates-about-the-same-no-matter-ad-position-24172 | 1,369,183,804,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368700984410/warc/CC-MAIN-20130516104304-00080-ip-10-60-113-184.ec2.internal.warc.gz | 234,122,915 | 23,387 | Hal Varian, Google’s Chief Economist, posted some pretty interesting details on how the conversion rate of an ad differs based on the position of the ad in Google. Google’s data shows that overall the conversion rate does not vary all that much based on the ad position. Hal said, if “an ad that had a 1.0% conversion rate in the best position, would have about a 0.95% conversion rate in the worst position, on average.”
It is important to understand how Google came up with this data, so here is the full post:
Advertisers often ask us how conversion rates vary with position. Everyone is aware that higher positions tend to get more clicks and therefore more conversions in total. The question of interest is how does the conversion rate (conversions/clicks) vary with position?
This is a tricky question for several reasons. Since Google ranks ads by bid times ad quality, ads in higher positions tend to have higher quality and higher quality ads tend to have higher conversion rates. Thus you may see a correlation between auction position and conversion rates just due to this ad quality effect. However, the real question is how the conversion rate for the same ad would change if it were displayed in a different position.
Another difficulty is that the average position number reported by Google is that it is an average over all auctions in which you participate. If you increase your bid, it is quite possible to see your average position move lower on the page! The reason is that when you increase your bid, your ad will appear in new auctions, and it will tend to come in at the bottom of those new auctions. This effect can be large enough to push your overall average position down. See this FAQ for more on this issue.
We have used a statistical model to account for these effects and found that, on average, there is very little variation in conversion rates by position for the same ad. For example, for pages where 11 ads are shown the conversion rate varies by less than 5% across positions. In other words, an ad that had a 1.0% conversion rate in the best position, would have about a 0.95% conversion rate in the worst position, on average. Ads above the search results have a conversion rate within ±2% of right-hand side positions.
The bottom line: conversion rates don’t vary much by position.
I would be incredibly interested in seeing from the larger agencies who manage thousands of campaigns and track conversions for each of those clicks, if they agree. I will see if we can have one of our columnists in this area do a follow up with their own data.
About The Author: is Search Engine Land's News Editor and owns RustyBrick, a NY based web consulting firm. He also runs Search Engine Roundtable, a popular search blog on very advanced SEM topics. Barry's personal blog is named Cartoon Barry and he can be followed on Twitter here. For more background information on Barry, see his full bio over here.
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• http://www.cpcsearch.com Terry Whalen
I’d be more interested in what Hal has to say versus other agencies. You need an absolute TON of data to look at this – I don’t know of many agencies outside of maybe EF that have enough data. Part of the reason you need a ton of data for this is that you have to negate all the other reasons for changes in conversion rates, including ad rotation, changes on the landing page or to the funnel, conversion tracking changes, etc. With enough data from enough different accounts all the other variables can kind of cancel themselves out. But with smaller datasets, you’re likely to come to erroneous conclusions.
In addition, I think the only data that really matters is the data of the specific advertiser. It may be true that in aggregate, conversion rates don’t vary a whole lot by ad position. But for specific advertisers, or possibly for specific categories, this may not be true. The problem is that for specific advertisers, there usually isn’t enough data from the account to form any conclusions about conversion rate by ad position.
• Astru
Why is this so surprising? When people click on a search ad, why would they act differently on the advertiser’s site based on the initial ad placement? Either the site satisfies their need (and the visit results in a conversion) or not.
• mothner
Interesting, though somewhat not surprising in that one could argue that lower positioned ads would have actually a better conversion rate, because once the users has made it down to lower positioned ads, he either did not see what he wanted or may have clicked on higher ads and returned to the SERP, so when he finally does click on the lower ads he has exhausted other options and is more likely to convert for this ad.
Mike
• http://www.christophtrautmann.de Chris T.
Conversion Rate
Transactions
Bounce-Rate
etc.
- Chris
• http://www.periscopix.co.uk/ Periscopix
Alistair in our team did a bit of thinking about this after Google published this blog. He thinks that actually Google may have used *too much* data and ended up aggregating several distinct populations, effectively erasing the differences between them.
• http://www.summitwebconsultants.com SummitWebConsultants
I check the keyword position report for all of my ecommerce customers and have yet to see anything significant enough to warrant using position preference.
But I agree with Terry, every site is different and it’s something you should look at for yourself. I never assume my ad performance will mirror data that has been aggregated over different websites, industries, seasons ect. ect.
- Dan
• http://www.rimmkaufman.com George Michie
We reached this same conclusion years ago. Terry’s right, it takes a ton of data, and fancy stats to get to this conclusion.
Indeed, this is the reason we’ve advocated against position bidding since we started the business. See eg: http://www.rimmkaufman.com/rkgblog/2008/09/30/position-bidding/
What Hal didn’t mention, for obvious reasons, is that those “worst” positions are at the top of the page, and the bottom of the page is slightly better. Google doesn’t want folks chasing each other DOWN the page.
However, people shouldn’t avoid position 1 on these grounds. The differences between 1 and 10 are terrifically slight, and well within the statistical noise of any coherent bidding system.
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https://schoeller-up.at/liquid-volume-lcs/archive.php?tag=population-mean-calculator-bf2418 | 1,624,617,259,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623487630081.36/warc/CC-MAIN-20210625085140-20210625115140-00154.warc.gz | 458,051,768 | 10,780 | # population mean calculator
This video shows how to calculate the sample size n required to estimate the population mean µ. X ± Z: s √(n) Where: X is the … Because a population is usually very large or unknown, the population mean is usually an unknown constant. sum of all observations divided by the number of observations, but there … Let's suppose you have number of values, randomly drawn from some source population (these values are usually referred to as a sample). Calculating a confidence interval involves determining the sample mean, X̄, and the population standard deviation, σ, if possible. Instructions: This Normal Probability Calculator for Sampling Distributions will compute normal distribution probabilities for sample means $$\bar X$$, using the form below. Depending on which … For given sample you can calculate the mean and the standard deviation of the sample. Statistics - Statistics - Estimation of a population mean: The most fundamental point and interval estimation process involves the estimation of a population mean. Confidence Interval for Mean Calculator. Knowing the confidence interval for the mean of a normal population can be very useful for assessing the true nature of a population variable in analytics studies that use normally distributed sample data. If the population standard deviation cannot be used, then the sample standard deviation, s, can be used when the sample size is greater than 30. Standard Deviation and Mean. This is just a few minutes of a complete course. If the blood pressure of a … What is Sample Mean? A sample mean “x” is a point estimate of a population mean “μ.” A sample variance “s2” is a point estimate of a population variance “σ2.” When you look at this in a more formal perspective, the occurrence of the estimate is a result of the application of the point estimate to a sample data set. But you can use the standard deviation of your sample if you have enough observations: at least n=30, the more the better. getcalc.com's Mean (μ) Calculator is an online statistics & probability tool to measure the average of sample or population data or to summarize the common behavior of sample or finite population data in statistical experiments. An estimate of the population mean is the sample mean. Find the Population Mean of 13,23,12,44,55, μ = ( ∑ X ) / N So, the best estimate (population proportion) is 85. z-score is fixed for the confidence level (CL). There are two types of arithmetic mean: population mean (μ) and sample mean (x̄). So in this example, the ∑X is 224% and the number of observed values for the population is 12 as it comprises the return for the sto… Let us try to analyze the return of a stock XYZ for the last twelve years. In practice we may not necessarily know for certain what the population standard deviation really is. And the returns for the stock in the last twelve years are 12%, 25%, 16%, 14%, 40%, 15%, 13%, 17%, 23%, 13%, 17%, and 19%. Suppose it is of interest to estimate the population mean, μ, for a quantitative variable. Confidence interval calculator find out population mean of a given sample. Population Mean Calculator The population Mean formula is really useful in estimating an average of the group characteristic. point estimate of the population mean calculator: point estimate of population proportion: best point estimate: find the best point estimate of the population proportion p: find point estimate: point estimate of mean: point estimate calculator lower and upper bound: point estimate of population proportion calculator: estimate the slope of the tangent line: point estimate excel: point estimate confidence … Use this mean calculator to find the common behavior or the average of group elements in a statistical survey … The accuracy of a population mean is comparatively higher than the sample mean. This can be of numbers, people, objects, etc. Hence, N=5.µ=(50+55+45+60+40)/5 =250/5 =50So, the Calculation of population variance σ2 can be done as follows-σ2 = 250/5Population Variance σ2 will be-Population Variance (σ2 ) = 50The population variance is 50. For sample size greater than 30, the population standard deviation and the sample standard deviation will be similar. Using this information the 95% confidence interval is calculated as between 68.43 and 71.57mmHg. Population Mean is the average of a set of group characteristics. To calculate the mean, enter the numerical values in the box above. The sample mean is the average of all the items in a sample (a group of observations). The formula for population variance can be calculated by using the following five simple steps: Step 1: Calculate the mean (µ) of the given data. If you would like to estimate the mean systolic blood pressure of British adults with 95% confidence and a margin of error no larger than 2mmHg, how many samples are required? To force another distribution, choose it explicitly. Then fill in the standard deviation, the sample mean, $$\bar{x}$$, the sample size, $$n$$, the hypothesized population mean $$\mu_0$$, and indicate if the test is left tailed, <, right tailed, >, or two tailed, $$\neq$$. You can also see the work peformed for the calculation. Population Mean Confidence Interval Calculator. 3.3 Hypothesis Tests Concerning the Population Mean. A confidence interval corresponds to a region in which we are fairly confident that a population parameter is contained by. The Population Standard Deviation Calculator is used to calculate the population standard deviation of a set of numbers. How It Is Calculated. The population mean is denoted by μ The formula to calculate population mean is given by: where, Σx = Sum of all data N = Number of items in population In the below online population … In using this formula we are assuming that we know what this standard deviation is. In order to calculate the mean for the whole population, we need to find out the summation of all the observed values first. The formula for the confidence interval for one population mean, using the t-distribution, is. Using this information the 95% confidence interval is calculated as between 68.43 and 71.57mmHg. The sample mean is the average of all the items in a sample (a group of observations). (If you need to calculate mean and standard deviation from a set of raw scores, you can do so using our descriptive statistics tools.) But you can use the standard deviation of your sample if you have enough observations: at least n=30, the more the better. In this case, the sample mean, is 4.8; the sample standard deviation, s, is 0.4; the sample size, n, is 30; and the degrees of freedom, n – 1, is 29. A sample mean refers to the average of a set of data. Both population mean and sample mean are calculated in the same way except the symbols which define the formula varies. Select if the population standard deviation, $$\sigma$$, is known or unknown. You do not need to specify whether the data is from a population or a sample because this does not affect the calculation of the mean. | 1,521 | 7,027 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.546875 | 4 | CC-MAIN-2021-25 | latest | en | 0.853068 |
https://numbermatics.com/n/347412/ | 1,656,885,057,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656104249664.70/warc/CC-MAIN-20220703195118-20220703225118-00261.warc.gz | 480,910,516 | 6,559 | # 347412
## 347,412 is an even composite number composed of five prime numbers multiplied together.
What does the number 347412 look like?
This visualization shows the relationship between its 5 prime factors (large circles) and 48 divisors.
347412 is an even composite number. It is composed of five distinct prime numbers multiplied together. It has a total of forty-eight divisors.
## Prime factorization of 347412:
### 22 × 3 × 13 × 17 × 131
(2 × 2 × 3 × 13 × 17 × 131)
See below for interesting mathematical facts about the number 347412 from the Numbermatics database.
### Names of 347412
• Cardinal: 347412 can be written as Three hundred forty-seven thousand, four hundred twelve.
### Scientific notation
• Scientific notation: 3.47412 × 105
### Factors of 347412
• Number of distinct prime factors ω(n): 5
• Total number of prime factors Ω(n): 6
• Sum of prime factors: 166
### Divisors of 347412
• Number of divisors d(n): 48
• Complete list of divisors:
• Sum of all divisors σ(n): 931392
• Sum of proper divisors (its aliquot sum) s(n): 583980
• 347412 is an abundant number, because the sum of its proper divisors (583980) is greater than itself. Its abundance is 236568
### Bases of 347412
• Binary: 10101001101000101002
• Base-36: 7G2C
### Squares and roots of 347412
• 347412 squared (3474122) is 120695097744
• 347412 cubed (3474123) is 41930925297438528
• The square root of 347412 is 589.4166607757
• The cube root of 347412 is 70.2988582917
### Scales and comparisons
How big is 347412?
• 347,412 seconds is equal to 4 days, 30 minutes, 12 seconds.
• To count from 1 to 347,412 would take you about four days.
This is a very rough estimate, based on a speaking rate of half a second every third order of magnitude. If you speak quickly, you could probably say any randomly-chosen number between one and a thousand in around half a second. Very big numbers obviously take longer to say, so we add half a second for every extra x1000. (We do not count involuntary pauses, bathroom breaks or the necessity of sleep in our calculation!)
• A cube with a volume of 347412 cubic inches would be around 5.9 feet tall.
### Recreational maths with 347412
• 347412 backwards is 214743
• The number of decimal digits it has is: 6
• The sum of 347412's digits is 21
• More coming soon!
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The information we have on file for 347412 includes mathematical data and numerical statistics calculated using standard algorithms and methods. We are adding more all the time. If there are any features you would like to see, please contact us. Information provided for educational use, intellectual curiosity and fun!
Keywords: Divisors of 347412, math, Factors of 347412, curriculum, school, college, exams, university, Prime factorization of 347412, STEM, science, technology, engineering, physics, economics, calculator, three hundred forty-seven thousand, four hundred twelve.
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Some bits of this website may not work unless you switch it on. | 943 | 3,570 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2022-27 | latest | en | 0.860523 |
https://www.fmaths.com/interesting-about-math/what-are-the-different-functions-in-math.html | 1,638,205,850,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964358786.67/warc/CC-MAIN-20211129164711-20211129194711-00427.warc.gz | 873,453,554 | 11,034 | # What Are The Different Functions In Math?
## What are the 4 types of functions?
The various types of functions are as follows:
• Many to one function.
• One to one function.
• Onto function.
• One and onto function.
• Constant function.
• Identity function.
• Polynomial function.
## What are the 3 types of functions?
There are 3 types of functions:
• Linear.
• Exponential.
## What are the 8 types of functions?
The eight types are linear, power, quadratic, polynomial, rational, exponential, logarithmic, and sinusoidal.
## How many types of functions are there?
There can be 4 different types of user-defined functions, they are: Function with no arguments and no return value. Function with no arguments and a return value. Function with arguments and no return value.
## What are the two main types of functions?
What are the two main types of functions? Explanation: Built-in functions and user defined ones. The built-in functions are part of the Python language.
## What is the most basic function?
The parent function is the most basic function in the family of functions, the function from which all the other functions in the family can be derived. A family of functions is a group of functions that can all be derived from transforming a single function called the parent function.
You might be interested: Quick Answer: How To Type Math Symbols In Word?
## What are the 12 basic functions?
Precalculus: The Twelve Basic Functions Identity Function Squaring Function Cubing Function Inverse Function Square Root Functio.
## What are the 7 parent functions?
The following figures show the graphs of parent functions: linear, quadratic, cubic, absolute, reciprocal, exponential, logarithmic, square root, sine, cosine, tangent. Scroll down the page for more examples and solutions.
## What is a common function?
Some relations that classify as functions map several x-values to the same y-value. If x = -2 is substituted into the same function and also results in y=8, that is acceptable. Several domain values can map to the same range value.
## WHAT IS function and its type?
1. Injective (One-to-One) Functions: A function in which one element of Domain Set is connected to one element of Co-Domain Set. 2. Surjective (Onto) Functions: A function in which every element of Co-Domain Set has one pre-image.
## What are four examples of functions?
we could define a function where the domain X is again the set of people but the codomain is a set of number. For example, let the codomain Y be the set of whole numbers and define the function c so that for any person x, the function output c(x) is the number of children of the person x.
## What is a general mathematical function?
In mathematics, a function is a binary relation between two sets that associates to each element of the first set exactly one element of the second set. Typical examples are functions from integers to integers, or from the real numbers to real numbers. For example, the position of a planet is a function of time.
## What is function type of function?
In computer science and mathematical logic, a function type (or arrow type or exponential) is the type of a variable or parameter to which a function has or can be assigned, or an argument or result type of a higher-order function taking or returning a function.
## What is a function list?
Function List Panel is a zone to display all the functions (or method) found in the current file. Function list contains a search engine (by using regular expression) and a panel to display the search result ( function list ).
## What are the special types of functions?
Special function
• Function.
• Gamma function.
• Bessel function.
• Hypergeometric function.
• Elliptic function. | 792 | 3,758 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.46875 | 4 | CC-MAIN-2021-49 | latest | en | 0.875709 |
https://quant.stackexchange.com/questions/15719/girsanovs-theorem-change-of-measure | 1,620,979,239,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243991648.40/warc/CC-MAIN-20210514060536-20210514090536-00321.warc.gz | 481,461,458 | 38,952 | # Girsanov's Theorem - Change of Measure
I have trouble understanding Girsanov's theorem. The Radon Nikodym process $Z$ is defined by:
$$Z(t)=\exp\left(-\int_0^t\phi(u) \, \mathrm dW(u) - \int_0^t\frac{\phi^2(u)}{2} \, \mathrm du\right)$$
Now $\hat P$ is a new probability measure. The trouble is I am not understanding how to go from old $P$ to the new one. The old $P$ is normally distributed with mean $0$ and variance $t$. Now say I want to know the new probability for an infinitesimally small interval around $0.2$. For that I need to know the value of $Z$ at this interval (event you may say). And then I can multiply (integrate) the value of $Z$ with old $P$, and get new $\hat P$.
Assume $t$ is fixed.
I have no idea how to calculate the value of $Z$ for this interval/event. Help would be appreciated.
Let's distinguish: measure $P$ gives probability over the paths, hence you can't really say that it has certain mean and variance: that would apply to a measure $P_t$ which restricts $P$ to the time instance $t$. Now, if you know that $P_t$ has density $f$ (with respect to Lebesgue measure) and $\frac{\mathrm d\hat P_t}{\mathrm d P_t} = g$ then $\hat P_t$ has density $$\frac{\mathrm d\hat P_t}{\mathrm d \lambda} = \frac{\mathrm d\hat P_t}{\mathrm d P_t}\cdot \frac{\mathrm d P_t}{\mathrm d \lambda} = g\cdot f.$$ Unfortunately, I do not know whether you can get $g$ directly out of $Z$ - in fact, it does not seem to be that you can always do this in some nice analytical way since it involves computing rather peculiar conditional expectations. Another sanity check: if there would be an easy way to find $g$ out of $Z$, then just by knowing the density $f$ for the Geometric Brownian motion $\mathrm dX_t = \sigma X_t\,\mathrm dW_t$ would allow you to know densities for any process of the form $\mathrm dX_t = \mu_t\,\mathrm dt + \sigma X_t\,\mathrm dW_t$ for pretty much any adapted process $\mu_t$. I'm pretty sure that even if $\mu_t = \mu(X_t)$ there are a lot of cases where the densities are still not known.
Edit: to support the intuition above, I've computed $g$ in terms of $Z$, and indeed it looks pretty simple $g(x) = \Bbb E[Z_t|X_t = x]$ but its computation would be rather hard in general.
• Then what is the function of $Z$. In discrete time, it connects two measures. Here the function should also be the same. – user3001408 Dec 2 '14 at 11:28
• @user3001408: I assume that by function you mean purpose. RN derivatives are always used to construct one measure from another, e.g. here $\mathrm d\hat P := Z\,\mathrm dP$. so in your terms $Z$ connects $P$ and $\hat P$. Of course, in particular it also connects their marginal (or finite-dimensional) distributions $P_t$ and $\hat P_t$ in a sense that $\mathrm d\hat P_t = h(Z)\,\mathrm dP_t$ for some function $h$. My point is that it's quite hard to deduce the shape of function $h$. – Ulysses Dec 2 '14 at 11:40
• @user3001408: how would I know the distribution of the new probaibilty measure - why do you think it is possible? As I told you, we know distribution of GBM at any point of time, and we can obtain $\mathrm dX_t = (X_t-\sqrt{|X_t|})\mathrm dt + X_t\mathrm dW_t$ through the GBM using Girsanov's theorem, but that does really mean we have analytical formula for distribution of $X_t$, right? – Ulysses Dec 2 '14 at 11:52 | 957 | 3,324 | {"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} | 3.5625 | 4 | CC-MAIN-2021-21 | latest | en | 0.896199 |
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# Past Papers Examples Discussion & Preparation
Let's Discuss & solve here past papers examples:
Leasing
A Limited Company acquired an asset on lease; the fair value of the asset is Rs. 59,738. The Company has to pay Rs. 10,000 as down payment and Rs. 20,000 annually for three years. The interest rate implicit in the lease is 10% p.a. You are required to calculate the present value of Minimum Lease Payment….5 marks
Please have a look & correct me if I am wrong or missing any part.
Soluion:
Fair Value / Cost 59,738 No of Installments 3 Installment Amount 20,000 Down Payment 10,000 IRR p.a 10.00 Principal Outstanding (68,000 - 10,000) 49,738
Lease Rent Financial Charge Principal Principal Outstanding 49,738 20,000 4973.8 15,026 34,712 20,000 3471 16,529 18,183 20,000 1818 18,182 1 60,000 10,263 49,737
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### Replies to This Discussion
Please post your solved examples here to prepare for the final term
Principal Outstanding (68,000 - 10,000) 49,738
ye 68000 kahan se ayi, 49,738 is ri8 ans. but ap ke mutabiq to 58000 ata he....?
sorry woh amount dosray example ki thi, it's 59,738 -10,000 = 49,738
1
2
3 | 558 | 1,979 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.828125 | 3 | CC-MAIN-2020-05 | longest | en | 0.831086 |
https://tnboardsolutions.com/samacheer-kalvi-11th-maths-guide-chapter-3-ex-3-8/ | 1,716,229,141,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058293.53/warc/CC-MAIN-20240520173148-20240520203148-00665.warc.gz | 518,365,956 | 12,725 | Tamilnadu State Board New Syllabus Samacheer Kalvi 11th Maths Guide Pdf Chapter 3 Trigonometry Ex 3.8 Text Book Back Questions and Answers, Notes.
## Tamilnadu Samacheer Kalvi 11th Maths Solutions Chapter 3 Trigonometry Ex 3.8
Question 1.
Find the principal solution and general solutions of the following
(i) sin θ = – $$\frac{1}{\sqrt{2}}$$
(ii) cot θ = √3
(iii) tan θ = –$$\frac{1}{\sqrt{3}}$$
(i) sin θ = – $$\frac{1}{\sqrt{2}}$$
We know that principal of sin θ lies in $$\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$$
sin θ = – $$\frac{1}{\sqrt{2}}$$ < 0
∴ The principal value of sin θ lies in the IV quadrant.
sin θ = – $$\frac{1}{\sqrt{2}}$$
= – sin $$\left(\frac{\pi}{4}\right)$$
sin o = sin $$\left(-\frac{\pi}{4}\right)$$
Hence θ = $$-\frac{\pi}{4}$$ is the principal solution.
The general solution is
θ = nπ + (- 1)n . $$\left(-\frac{\pi}{4}\right)$$ , n ∈ Z
θ = nπ + (- 1)n + 1 . $$\frac{\pi}{4}$$ , n ∈ Z
(ii) cot θ = √3
The principal value of tan θ lies in $$\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$$
Since tan θ = $$\frac{1}{\sqrt{3}}$$ > 0
The principal value of tan θ lies in the I quadrant.
The general solution of tan θ is
θ = nπ + $$\frac{\pi}{6}$$ , n ∈ Z
(iii) tan θ = –$$\frac{1}{\sqrt{3}}$$
The principal value of tan θ lies in $$\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$$
Since tan θ = – $$\frac{1}{\sqrt{3}}$$ > 0
The principal value of tan θ lies in the IV quadrant.
The general solution of tan θ is
θ = nπ – $$\frac{\pi}{6}$$ , n ∈ Z
Question 2.
Solve the following equations for which solutions lies in the interval 0° ≤ 9 < 360°
(i) sin4x = sin2x
sin4x – sin2x = 0
sin2 x (sin2 x – 1) = 0
sin2 x [ – (1 – sin2 x)] = 0
sin2x × – cos2x = 0
– sin2x cos2x = 0
(sin x cos x)2 = 0
($$\frac { 1 }{ 2 }$$ × 2 sin cos x)2 = 0
$$\frac { 1 }{ 4 }$$ sin 2x = 0
sin 2x = 0
The general solution is
2x = nπ, n ∈ Z
x = $$\frac{\mathrm{n} \pi}{2}$$, n ∈ Z
(ii) 2 cos2x + 1 = – 3 cos x
2 cos2x + 1 = – 3 cos x
2 cos2x + 3 cos x + 1 = 0
2 cos2x + 2 cos x + cos x + 1 = 0
2 cos x (cos x + 1) + 1 (cos x + 1) = 0
(2 cos x + 1) (cos x + 1) = 0
2 cos x + 1 = 0 or cos x + 1 = 0
cos x = $$-\frac{1}{2}$$ or cos x = – 1
To find the solution of cos x = $$-\frac{1}{2}$$
cos x = $$-\frac{1}{2}$$
To find the solution of cos x = – 1
cos x = – 1
cos x = cos π
The general solution is
x = 2nπ ± π, n ∈ Z
x = 2nπ + π or x = 2nπ – π, n ∈ Z
Consider x = 2nπ + π
when n = 0 , x = 0 + π = π ∈ (0°, 360°)
when n = 1 , x = 2π + π = 3π ∉ (0°, 360°)
Consider x = 2nπ – π
when n = 0, x = 0 – π ∉ (0°, 360°)
when n = 1, x = 2π – π = π ∈ (0°, 360°)
when n = 2, x = 4π – π = 3π ∉ (0°, 360°)
∴ The required solution are x = $$\frac{2 \pi}{3}$$, $$\frac{4 \pi}{3}$$, π
(iii) 2 sin2x + 1 = 3 sin x
2 sin2x – 3 sin x + 1 = 0
2 sin2x – 2 sin x – sin x + 1 = 0
2 sin x (sin x – 1) – 1 (sin x – 1) = 0
(2 sin x – 1)(sin x – 1) = 0
2 sin x – 1 = 0 or sin x – 1 = 0
sin x = $$\frac { 1 }{ 2 }$$ or sin x = 1
To find the solution of sin x = $$\frac { 1 }{ 2 }$$
sin x = $$\frac { 1 }{ 2 }$$
sin x = sin $$\left(\frac{\pi}{6}\right)$$
The general solution is x = nπ + (-1)n$$\frac{\pi}{6}$$, n ∈ z
(iv) cos2x = 1 – 3 sin x
1 – 2 sin2x = 1 – 3 sinx
2 sin2 x – 3 sin x = 0
sin x(2 sin x – 3) = 0smx =
sin x = 0 or 2 sin x – 3 = 0
sin x = 0 or sin x = $$\frac{3}{2}$$
sin x = $$\frac{3}{2}$$ is not possible since sin x ≤ 1
∴ sin x = 0 = sin 0
The general solution is x = nit ,
When n = 0, x = 0 ∉ (0°, 360°)
When n = 1, x = π ∈ (0°, 360°)
When n = 2, x = 2π ∉ (0°, 360°)
∴ The required solutions is x = π
Question 3.
Solve the following equations:
(i) sin 5x – sin x = cos 3x
2 cos 3 x . sin 2x = cos 3 x
2 cos 3x . sin 2x – cos3x = 0
cos 3x (2 sin 2x – 1) = 0
cos 3x = 0 or 2 sin 2x – 1 = 0
cos 3x = 0 or sin 2x = $$\frac { 1 }{ 2 }$$
To find the general solution of cos 3x = 0
The general solution of cos 3x = 0 is
3x = (2n + 1)$$\frac{\pi}{2}$$, n ∈ Z
x = (2n + 1)$$\frac{\pi}{6}$$, n ∈ Z
To find the general solution of sin 2x = $$\frac{1}{2}$$
sin 2x = $$\frac{1}{2}$$
sin 2x = sin $$\left(\frac{\pi}{6}\right)$$
The general solution is
∴ The required solutions are
(ii) 2 cos2θ + 3 sin θ – 3 = θ
2 cos2θ + 3 sin θ – 3 = θ
2(1 – sin2θ)+ 3 sin θ – 3 = θ
2 – 2 sin2θ + 3 sin θ – 3 = θ
– 2 sin2θ + 3 sin θ – 1 = θ
2 sin2 θ – 3 sin θ + 1 = θ
2 sin2θ – 2 sin θ – sin θ + 1 = θ
2 sin θ (sin θ – 1) – (sin θ – 1) = θ
(2 sin θ – 1) (sin θ – 1) = 0
2 sin θ – 1 = 0 or sin θ – 1 = θ
sin θ = $$\frac { 1 }{ 2 }$$ or sin θ = 1
To find the general solution of’ sin θ = $$\frac { 1 }{ 2 }$$
sin θ = $$\frac { 1 }{ 2 }$$
sin θ = sin $$\frac{\pi}{6}$$
The general solution is θ = nπ + (-1)n$$\frac{\pi}{6}$$, n ∈ Z
To find the general solution of sin θ = 1
sin θ = 1
sin θ = $$\frac{\pi}{2}$$
The general solution is θ = nπ + (-1)n$$\frac{\pi}{2}$$, n ∈ Z
∴ The required solutions are
θ = nπ + (-1)n$$\frac{\pi}{6}$$, n ∈ Z (or)
θ = nπ + (-1)n$$\frac{\pi}{6}$$, n ∈ Z
(iii) cos θ + cos 3θ = 2 cos 2θ
cos 3θ + cos θ = 2 cos 2θ
2 cos 2θ . cos θ = 2 cos 2θ
cos 2θ . cos θ – cos 2θ = θ
cos 2θ (cos θ – 1) = θ
cos 2θ = θ or cos θ – 1 = θ
cos 2θ = θ or cos θ = 1
To find the general solution of cos 2θ = θ
The general solution is
2θ = (2n + 1)$$\frac{\pi}{2}$$, n ∈ Z
θ = (2n + 1)$$\frac{\pi}{4}$$, n ∈ Z
To find the general solution of cos θ = 1
cos θ = 1
cos θ = cos 0
The general solution is θ = 2nπ , n ∈ Z
∴ The required solutions are
θ = (2n + 1)$$\frac{\pi}{4}$$, n ∈ Z (or)
θ = 2nπ, n ∈ Z
(iv) sin θ + sin 3θ + sin 5θ = 0
sin 5θ + sin 3θ + sin θ = 0
2 sin 3θ . cos 2θ + sin 3θ = 0
sin 3θ (2 cos 2θ + 1) = θ
sin 3θ = 0 or 2 cos 2θ + 1 = θ
sin 3θ = 0 or cos 2θ = –$$\frac { 1 }{ 2 }$$
To find the general solution of sin 3θ = 0
The general solution is
3θ = nπ, n ∈ Z
θ = $$\frac{\mathbf{n} \pi}{3}$$, n ∈ Z
To find the general solution of cos 2θ = –$$\frac { 1 }{ 2 }$$
The general solution is
∴ The required solutions are
(v) sin 2θ – cos 2θ – sin θ + cos θ = θ
(vi) sin θ + cos θ = √2
The general solution is
(vii) sin θ + √3 cos θ = 1
Divide each term by 2
(viii) cot θ + cosec θ = √3
(ix) tan θ + tan $$\left(\theta+\frac{\pi}{3}\right)$$ + tan $$\left(\theta+\frac{2 \pi}{3}\right)$$ = √3
(x) cos 2θ = $$\frac{\sqrt{5}+1}{4}$$
we know cos 36° = $$\frac{\sqrt{5}+1}{4}$$, 36° = $$\frac{\pi}{5}$$
cos 2θ = cos 36° = cos $$\left(\frac{\pi}{5}\right)$$
The general solution is
2θ = 2nπ ± $$\frac{\pi}{5}$$, n ∈ Z
θ = nπ ± $$\frac{\pi}{10}$$, n ∈ Z
(xi) 2cos 2x – 7 cos x + 3 = 0
cos x = $$\frac { 1 }{ 2 }$$ or cos x = 3
∴ cos x = $$\frac { 1 }{ 2 }$$
cos x = cos $$\frac{\pi}{3}$$
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http://math.maine121.org/welcome/chapter4/ | 1,597,416,470,000,000,000 | text/html | crawl-data/CC-MAIN-2020-34/segments/1596439739328.66/warc/CC-MAIN-20200814130401-20200814160401-00523.warc.gz | 70,708,831 | 13,378 | # Roots of Polynomial Equations
Baby Mandelbrot found by calculating where the orbit of the critical point (the root of the polynomial equation z2 + c = c) does not tend to infinity
In Chapter 3, we discovered when we vertically translated, reflected or scaled linear (degree 1) equations, the transformed equations were equivalent, in that they had the same solutions. This allowed us to solve easier equations instead, which gave us the same solutions as the original equations . These transformations also explained the steps behind solving the linear equations algebraically.
Given the linear equation 2x + 3 = 9, we would translate the polynomial and the output value down 3 (subtract 3 from both sides) to get 2x = 6, and then scale the transformed polynomial and output value by 1/2 (divide both sides by 2) to get x = 3.
In other words, we would geometrically transform the polynomial into x, a 45º diagonal line through the origin where the input values and output values are always the same (get the x alone).
In this chapter, the question is, will the same approach work for quadratic (degree 2) and/or higher degree equations?
Given a quadratic equation, such as 6×2 – 5x + 3 = 7 , will translating and scaling, transform the polynomial to x?
Go on to begin answering this question.
To solve \$latex 6x^{2} – 5x + 3 = 7 \$ using the same approach as in Chapter 3, we would first vertically translate the polynomial and the output value down 3 to get \$latex 6x^{2} – 5x = 4\$ . The next step would be to vertically scale the translated polynomial and the translated output value.
Below is a snapshot of an applet that will allow you to vertically translate and then vertically scale the above example.
Run the applet “quadexample.html” , to see if vertically translating and scaling the polynomial and output value, works for solving quadratic equations, as it does for solving linear equations.
After recording and defending your answer, go on to the next page.
As we discovered, scaling makes the parabola (the graph of a quadratic) wider, but no matter how much we scale, it will never become a line. Even if scaling could turn a parabola into a line, there are two solutions, not one. The best this approach can do, is give an estimate of the values of the solutions.
From looking at the graph of \$latex 6x^{2} – 5x + 3 = 7 \$ on the right, it’s clear there are 2 solutions (there are two input values that have output values of 7). Whatever the solutions are, we would algebraically write them both as, x = value 1 and x = value 2.
However, x doesn’t just represent the two missing input values, x also represents two linear polynomials. This means there are two equivalent linear equations with the same solution values. In other words, the reason the above approach didn’t work, is because we need a way to transform the polynomial into two lines, not one.
Below are snapshots of an applet, which will randomly generate a quadratic equation, two equivalent linear equations, and show all three graphs.
In the applet “quadlines.html” , follow the directions after the 1st bullet, and answer the question after the 2nd bullet.
Use a copy of the pdf appletnotes to show your work and to record your answer to the given question.
After neatly showing your work, and recording your answer to the given question, go on to the next page.
In Chapter 2, we found out that all quadratic (degree 2) polynomials can be factored into two linear (degree 1) polynomials, which is how a quadratic equation can be written as two linear polynomials. Knowing that two linear factors exist is one thing though, being able to find them is another. So for now, as in Chapter 2, we’ll only consider a quadratic factorable, if the two linear polynomial factors have integer coefficients and constants. Even if we’re able to factor the quadratic polynomial in an equation into two linear polynomials, then what?
Solve the following factored quadratic equation algebraically (without graphing).
\$latex 6x^{2} + 7x – 20 = -17 \$ (2x + 5)(3x – 4) = -17 What two input values for x, will result in an output value of -17?
Use a copy of the pdf booknotes, to record your work and the steps which led you to the two solutions.
After finding both values for x which will make the above equations true, go on to the next page.
Chances are, it took a lot of guessing and checking to find the two solutions to the problem on the previous page.
Below is a snapshot of an applet, which will give us a hint for finding an easier way to solve factored quadratic equations.
Run the applet “quadtranslate2.html” , until you have an answer to the given question.
Use a copy of the pdf appletnotes1g, to record and defend your answer.
After recording and defending your answer to the given question, go on to the next page.
Below is a snapshot of an applet, which will either confirm or allow you to modify your answer from the last applet.
Run the applet “zpp.html” , and modify your answer as needed, to the question from the last applet (shown below).
“Since the plan is to factor the quadratic on the left side of the equation, into two linear polynomials, which equivalent vertically translated quadratic equation would you choose to solve, and why?”
After confirming or modifying your answer to the above question from the last applet, go on to the next page.
By vertically translating the polynomial and the output value in a quadratic equation, until the output value is a zero, turns a “guess and check” problem into a “go right to the solutions” problem. This is because the only way we can have a zero for a product, is if at least one of the factors is a zero (this is called the zero product property). Below is the example from earlier in the chapter.
\$latex 6x^{2} + 7x – 20 = -17 \$ (2x + 5)(3x – 4) = -17 What two input values for x will result in an output value of -17?
Instead of factoring and trying to solve the equation as it was given, we now know we can vertically translate, and then solve an equivalent equation which has an output value of zero.
\$latex 6x^{2} + 7x – 20 = -17 \$ \$latex 6x^{2} + 7x – 3 = 0 \$ \$latex (3x – 1)(2x + 3) = 0 \$
What’s nice about this is, we can use the zero product property to go right to the solutions, because the only way we can get an output value of zero, is if 3x – 1 = 0 or 2x + 3 = 0 .
3x – 1 = 0 2x + 3 = 0 3x = 1 2x = -3 x = 1/3 x = -3/2
Below is a snapshot of an applet which gives examples of using the zero product property to solve quadratic equations.
Run the applet “quadtranslate3.html” until you can solve a quadratic equation, without moving the slider.
Use a copy of the pdf appletnotes1g to copy down an example and record your work.
After successfully solving an example and recording your work neatly, go on to the next page.
Solutions to polynomials, set equal to zero, are called roots or zeros. When solving polynomial equations though, we can use the term solutions, roots, or zeros, to represent the values that make the equations true (for every polynomial equation with an output value not equal to zero, there’s an equivalent polynomial equation with an output value that is ).
To the right are snapshots of a program, which will randomly generate 7 quadratic equations, in factored form.
Run program 401 until you can successfully solve at least 5 of the 7 factored quadratic equations.
Use a copy of the pdf programnotes to record the equations, your work, and any discoveries.
After successfully solving at least 5 of the 7 problems, and neatly recording your work, go on to the next page.
Below are snapshots of a program, which will randomly generate 10 quadratic equations, with integer roots.
Run program 402 until you can successfully solve at least 8 of the 10 quadratic equations by factroring.
Use a copy of the pdf programnotes to record the equations, your work, and any discoveries.
After successfully solving at least 8 of the 10 equations, and neatly recording your work, go on to the next page.
Below are snapshots of a program, which will randomly generate 10 quadratic equations, with rational roots.
Run program 403 until you can successfully solve at least 8 of the 10 quadratic equations.
Use a copy of the pdf programnotes to record the equations, your work, and any discoveries.
After successfully solving at least 8 of the 10 equations, and neatly recording your work, go on to the next page.
As we discovered in chapter 2, sometimes the best we can do, is factor out the greatest common factor.
To the right is a snapshot of a program, which will randomly generate 10 quadratic equations, with rational roots.
Run program 404 until you can successfully solve at least 8 of the 10 quadratic equations.
Use a copy of the pdf programnotes to record the equations, your work, and any discoveries.
After successfully solving at least 8 of the 10 equations, and neatly recording your work, go on to the next page.
Below are examples of a program, which will randomly generate 10 sets of integers. The challenge is to find the simplest polynomial, whose roots, are those sets of integers.
What is the simplest polynomial with a root of 5? x = 5 x – 5 = 0 x – 5What is the simplest polynomial with roots of 1 and -3? x = 1 or x = -3 x-1=0 or x+3 = 0 (x – 1) (x + 3) = 0 \$latex x^{2} + 2x – 3 = 0 \$ \$latex x^{2} + 2x – 3\$
Run program 405 until you can successfully determine at least 8 of the 10 polynomials.
Use a copy of the pdf programnotes to neatly record your work and any discoveries.
After successfully determining at least 8 of the polynomials, and neatly recording your work, go on to the next page.
Below are examples of a program, which will randomly generate 10 sets of rational numbers. The challenge is to find the simplest polynomial, whose roots, are those sets of rational numbers.
What’s the simplest polynomial which has a root of -8/9 ? x = -8/9 9x = -8 9x + 8 = 0 9x + 8What is the simplest polynomial which has roots of 7/8 and -3/4 ? x = 7/8 or x = -3/4 8x = 7 or 4x = -3 8x – 7 = 0 or 4x + 3 = 0 (8x – 7)(4x + 3) = 0 \$latex 32x^{2} – 4x – 21 = 0 \$ \$latex 32x^{2} – 4x – 21 \$
Run program 406 until you can successfully determine at least 8 of the 10 polynomials.
Use a copy of the pdf programnotes to neatly record your work and any discoveries.
After successfully determining at least 8 of the polynomials, and neatly recording your work, go on to the next page.
So far we can solve any factorable quadratic equation, by using the zero product property.
In other words, we only know how to factor quadratics, which have linear factors with integer coefficients and constants.
In 8 guesses or less, we can factor the quadratic 2×2 – 11x – 21( 2x – 1 )( 1x + 21 ) = 2×2 + 41x – 21 ( 2x + 1 )( 1x – 21 ) = 2×2 – 41x – 21 ( 2x – 21 )( 1x + 1 ) = 2×2 – 19x – 21 ( 2x + 21 )( 1x – 1 ) = 2×2 + 19x – 21 ( 2x – 7 )( 1x + 3 ) = 2×2 – 1x – 21 ( 2x + 7 )( 1x – 3 ) = 2×2 + 1x – 21 ( 2x – 3 )( 1x + 7 ) = 2×2 + 11x – 21 ( 2x + 3 )( 1x – 7 ) = 2×2 – 11x – 21
Good luck though, trying to factor the quadratic, 4×2 – 7x – 3 , by guessing!
( 8x – 7 + √ 97 )( 8x – 7 – √ 97 )
We can solve any factorable quadratic equation, by vertically translating until the output value is zero, and then use the zero product property.
If the quadratic isn’t factorable though, having an output value of zero, isn’t good enough.
To the right are snapshots of an applet, which will allow us to see what happens to the quadratic equation, when we continue to use other transformations as well.
Run the applet “quadtransform.html” , until you have successfully completed all 3 steps.
Use a copy of the pdf appletnotes, to record any information which may be revealed, along with an example.
After you have neatly recorded any revealed information, along with an example, go on to the next page.
Now we know we can transform any quadratic equation, into a monic quadratic equation (where the lead coefficient is a positive one), with an output value of zero.
Below are snapshots of an applet, which will give examples of a special type of monic quadratic equation.
Run the applet “rorschach.html” , and use the slider until you can answer the 2 questions.
Use a copy of the pdf appletnotes1g to record both of your answers, an example, and any discoveries.
After recording an example, and defending your answers to the 2 questions, go on to the next page.
A monic quadratic equation where the coefficient of the x term is zero, and the constant is negative, is referred to as a difference of two squares, which can always be factored into x +/- the square root of the constant.
It even works if the constant isn’t a perfect square.
x2 – 25 = 0 x2 – 7 = 0 (x + 5)(x – 5) = 0 (x + √7)(x – √7) = 0 x = -5 or x = 5 x = -√7 or x = √7
We can also solve the equations by translating up and solving the equivalent linear equation (which also works for higher degree polynomial equations).
Below are snapshots of an applet which will give examples of monic polynomial equations of degree 1 through 5.
Run the applet “polyroots.html” to explore more examples, and then go on to the next page.
So now we know we can always transform a quadratic equation into a monic quadratic equation, and a monic quadratic equation with no x term, can always be factored.
If we could transform any monic quadratic equation, into a monic quadratic equation with no x term, we’d be all set.
Below are snapshots of an applet which will allow you to discover if this is possible, or if we’ll have to look for another way.
Run the applet “xminusn.html” , and use the slider until you can answer the 2 questions.
Use a copy of the pdf appletnotes1g, to record both answers, and to algebraically simplify an example.
After recording both answers, and defending them by simplifying an example algebraically, go on to the next page.
The good news is, it’s possible to transform a monic quadratic equation into a monic quadratic equation with no x term. The bad news is, without using an applet with a slider, how are we supposed to know how far to translate horizontally?
We know we want the vertex (the lowest or highest point) of the parabola (the graph of a quadratic) to be on the y-axis, so if we knew the x-value of the vertex of the original quadratic, we would know how far to translate.
In the example below, the x-value of the vertex of the original quadratic (in black) is 3, so we know we’ll have to translate horizontally to the left 3.
The problem is, without looking at the above graph, how would we know the x-value of the vertex was 3?
Below are snapshots of an applet, which will allow you to discover a connection (if there is one), between the roots of a monic quadratic, and the x-value of the vertex.
Run the applet “vertex.html” until you discover a connection between the roots and the vertex of a monic quadratic.
Use a copy of the pdf appletnotes1g to record your discovery, along with at least one example.
After recording your discovery, and defending it with at least one example, go on to the next page.
We now know, for any monic quadratic, there is a connection between the two roots and the x-value of the vertex.
Knowing the roots, we can just average them to find the x-value of the vertex, and know how far to horizontally translate.
The problem is, that’s the thing, we’re looking for a way to find the roots of unfactorable quadratic equations.
Hopefully, there’s another way of determining the x-value of the vertex, without having to know what the roots are.
To the right, are snapshots of an applet, which will allow you to search for a different way of determining the x-value of the vertex, without knowing the values of the two roots of a monic quadratic.
Run the applet “monicsumprod.html” to see if you can find another way to determine the x-value of the vertex, without knowing the values of the two roots.
Use a copy of the pdf appletnotes1g, to record your conclusion, and defend it with an example.
After searching for another way to determine the x-value of the vertex, without knowing the values of the two roots, and defending your conclusion with an example, go on to the next page.
We know the x-value of the vertex of a monic quadratic, is the average of the two roots, due to the symmetry of parabolas.
We also know, the average of two values, is just their sum divided by two. As we just discovered, this means all we have to do, is take half of the opposite of the x coefficient, to know how far to horizontally translate. Below is an example.
Below are snapshots of an applet which will show examples of solving monic quadratic equations, by translating horizontally. After getting used to this approach, we will be able to solve any quadratic equation, with any type of roots.
Run the applet “quadroots.html” , until you can successfully find the roots of a monic quadratic equation, by algebraically translating horizontally, without using the slider.
Use a copy of the pdf appletnotes, to neatly record your work for at least one example.
Challenge: Try the horizontal translation method on a problem from program 403 , which won’t be monic to start.
After finding the roots of a monic quadratic equation, by algebraically translating horizontally, and neatly recording your work before using the slider, go on to the next page.
Since translating horizontally can turn any monic quadratic equation into a difference of two squares equation, which can always be factored (even if the roots end up being irrational), we can now solve any quadratic equation. Of course, if the roots are rational numbers (integers and/or fractions), we can still factor and use the zero product property if we wish.
In other words, we can use the horizontal translation method on a general quadratic equation (one that doesn’t use specific values for the coefficients and constant), to create a formula for solving any quadratic equation.
Below are the first few steps for using the horizontal translation method on a general quadratic equation.
Challenge: continue the steps, until you have a formula for both roots of any quadratic equation.
Use a copy of the pdf booknotes to neatly show all of your steps.
To see how you did, open the pdf quadraticformula, before going on to the next page.
For objects in free fall, near the surface of the earth, there is a quadratic relationship between time and distance.
Ignoring friction, the quadratic h – 16t2 gives the height in feet of an object, after falling t seconds, from a height of h feet.
Below are snapshots of an applet, which will allow you to model a blue sphere being dropped from a given height.
Run the applet “gravity.html” to explore the quadratic relationship between time and distance of an object in free fall.
Choose a height (other than 25 ft), and determine how many seconds (rounded to the hundredths place) it would take for the blue sphere to hit the ground, and then use the “free fall time” slider to check your prediction.
Use a copy of the pdf appletnotes to make a sketch, to show all of your work, and to record any discoveries.
Determine how many seconds it would take an object to fall from 142 feet, and then open the pdf “Bowdoin.pdf” .
Determine how many seconds it would take an object to fall from 420 feet, and then open the pdf “tower.pdf” .
To learn how to make a ruler which measures time, open the pdf gravityruler
After determining the correct free fall time of the blue sphere, and neatly showing your work,go on to the next page.
Near the surface of the earth, the trajectory (path) of an object when propelled at an angle, can be described as parabolic.
Below are snapshots of an applet, which will challenge you to describe the trajectory of the water, coming out of the bubbler in front of the school’s 8th grade science room.
Run the applet “fountain.html” , until you make the blue parabola match the stream of water.
After finding the quadratic polynomial which describes the path of the water, go on to the next page.
The vertical and horizontal parts of the paths of projectiles are independent of each other, and can be studied separately.
Below are snapshots of an applet, which will let you further explore trajectories near the earth’s surface (ignoring friction).
Run the applet “trajectories.html” to see what you can discover.
Use a copy of the pdf “appletnotes(2g)” to record your discoveries.
After recording your discoveries, go on to the next page.
Earlier in the chapter, we discovered it’s possible to transform any quadratic equation into a monic one (where the lead coefficient is a positive 1). We also discovered it’s possible to solve (find the 2 roots for) any monic quadratic equation by translating horizontally. Not only that, you discovered the same approach can produce a formula for finding the two roots.
Formulas for the roots of cubic (degree 3) and quartic (degree 4) equations exist as well. It was proven by Niels Abel in the 1800’s though, that it’s impossible to create algebraic formulas based on coefficients, for finding all of the roots of polynomial equations of degree 5 or higher. Thanks to the Fundamental Theorem of Algebra, wich was first proved by Carl Gauss in 1799, we know for any polynomial equation, a root will always exist. Actually, a degree n polynomial equation will always have n roots (they could be complex numbers and/or repeats though). Many methods for finding estimates of roots to higher degree polynomial equations exist. For now though we’ll rely on the fact, if the input value for a polynomial results in an output value of zero, then that input value is a root. Two examples are shown below.
\$latex x^{3} – x^{2} – 4x + 4 \$ \$latex x^{3} – x^{2} – 4x + 4 \$ \$latex (2)^{3} – (2)^{2} – 4(2) + 4 \$ \$latex (5)^{3} – (5)^{2} – 4(5) + 4 \$ 8 – 4 – 8 + 4 125 – 25 – 20 + 4 0 ( so 2 is a root ) 84 ( so 5 is not a root )
It turns out, there’s a quicker way to determine if an input value is a root to a polynomial. It’s called synthetic division.
Below are snapshots of a program, which will randomly generate 10 synthetic division problems for you.
Run program 412 , until you discover how to use synthetic division, and get at least 8 of the 10 problems correct.
Use a copy of the pdf programnotes2g to record your work and discoveries.
Challenge: show why synthetic division works. Use x as the input value, and letters for the coefficients/constant.
Challenge: If an input of 5 results in a -3, and an input of 6 results in a 2, what do you know about the polynomial?
After using synthetic division correctly for at least 8 of the problems, and recording your work, go on to the next page.
To the right is a picture of the Mandelbrot set (the bug shaped object shown in black). Points in the c-plane are used for c in the quadratic z2 + c , and iterated using a seed of 0. If the orbit does not go off to infinity, the c-value is part of the Mandelbrot set.
The seed of 0 is called the critical point. It’s the root of the unique quadratic equation z2 + c = c .
If you thought the picture at the beginning of the chapter was interesting, open the spreadsheet “bigMset.numbers” , to explore a 26 ft by 13 ft picture of the Mandelbrot set.
Below is a snapshot of the chapter 4 programs menu program.
Run the program “4menu.bas”, or its alias to access every program used in chapter 4.
“If I feel unhappy, I do mathematics to become happy. If I am happy, I do mathematics to keep happy” — Alfred Renyi —
“The tantalizing and compelling pursuit of mathematical problems offers mental obsorption, peace of mind amid endless challenges, repose in activity, battle without conflict…” — Morris Kline —
“The Mandelbrot set is the most complex mathematical object known to mankind.” — Benoit Mandelbrot —
“There can be very little of present-day science and technology that is not dependent on complex numbers in one way or another.” — Keith Devlin —
“Indeed, nowadays no electical engineer could get along without complex numbers, and neither could anyone working in aerodynamics or fluid dynamics.” — Keith Devlin — | 6,133 | 24,591 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.46875 | 4 | CC-MAIN-2020-34 | latest | en | 0.888308 |
https://www.geeksforgeeks.org/python-add-list-elements-with-a-multi-list-based-on-index/ | 1,603,655,267,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107889651.52/warc/CC-MAIN-20201025183844-20201025213844-00640.warc.gz | 732,406,177 | 20,444 | # Python | Add list elements with a multi-list based on index
Given two lists, one is a simple list and second is a multi-list, the task is to add both lists based on index.
Example:
```Input:
List = [1, 2, 3, 4, 5, 6]
List of list = [[0], [0, 1, 2], [0, 1], [0, 1], [0, 1, 2], [0]]
Output:
[[1], [2, 3, 4], [3, 4], [4, 5], [5, 6, 7], [6]]
Explanation:
[1] = [1+0]
[2, 3, 4] = [0+2, 1+2, 2+2]
[3, 4] = [3+0, 3+1]
[4, 5] = [4+0, 4+1]
[5, 6, 7] = [5+0, 5+1, 5+2]
[6] = [6+0]
```
Let’s discuss some methods to do this task.
Method #1: Using iteration
`# Python code to add list elements ` `# with a multi-list based on index ` ` ` `# List initialization ` `List` `=` `[``1``, ``2``, ``3``, ``4``, ``5``, ``6``] ` `List_of_List ``=` `[[``0``], [``0``, ``1``, ``2``], [``0``, ``1``], ` ` ``[``0``, ``1``], [``0``, ``1``, ``2``], [``0``]] ` `Output ``=` `[] ` ` ` `# Iteration ` `for` `x ``in` `range``(``len``(``List``)): ` ` ``temp ``=` `[] ` ` ``for` `y ``in` `List_of_List[x]: ` ` ``temp.append(y ``+` `List``[x]) ` ` ``Output.append(temp) ` ` ` `# Printing ` `print``(``"Initial list is:"``, ``List``) ` `print``(``"Initial list of list is :"``, List_of_List) ` `print``(``"Output is"``, Output) `
Output:
```Initial list is: [1, 2, 3, 4, 5, 6]
Initial list of list is : [[0], [0, 1, 2], [0, 1], [0, 1], [0, 1, 2], [0]]
Output is [[1], [2, 3, 4], [3, 4], [4, 5], [5, 6, 7], [6]]
```
Method #2: Using enumerate()
`# Python code to add list elements ` `# with a multi-list based on index ` ` ` `# List initialization ` `List` `=` `[``1``, ``2``, ``3``, ``4``, ``5``, ``6``] ` `List_of_List ``=` `[[``0``], [``0``, ``1``, ``2``], [``0``, ``1``], [``0``, ``1``], [``0``, ``1``, ``2``], [``0``]] ` `Output ``=` `[] ` ` ` `# using enumerate ` `Output ``=` `[[elem ``+` `List``[x] ``for` `elem ``in` `y] ` ` ``for` `x, y ``in` `enumerate``(List_of_List)] ` ` ` ` ` `# Printing ` `print``(``"Initial list is:"``, ``List``) ` `print``(``"Initial list of list is :"``, List_of_List) ` `print``(``"Output is"``, Output) `
Output:
```Initial list is: [1, 2, 3, 4, 5, 6]
Initial list of list is : [[0], [0, 1, 2], [0, 1], [0, 1], [0, 1, 2], [0]]
Output is [[1], [2, 3, 4], [3, 4], [4, 5], [5, 6, 7], [6]]
```
Method #3: Using `Zip()`
`# Python code to add list elements ` `# with a multi-list based on index ` ` ` `# List initialization ` `List` `=` `[``1``, ``2``, ``3``, ``4``, ``5``, ``6``] ` `List_of_List ``=` `[[``0``], [``0``, ``1``, ``2``], [``0``, ``1``], [``0``, ``1``], [``0``, ``1``, ``2``], [``0``]] ` `Output ``=` `[] ` ` ` `# using zip ` `Output ``=` `[[z ``+` `x ``for` `z ``in` `y ]``for` `x, y ``in` ` ``zip``(``List``, List_of_List)] ` ` ` `# Printing ` `print``(``"Initial list is:"``, ``List``) ` `print``(``"Initial list of list is :"``, List_of_List) ` `print``(``"Output is"``, Output) `
Output:
```Initial list is: [1, 2, 3, 4, 5, 6]
Initial list of list is : [[0], [0, 1, 2], [0, 1], [0, 1], [0, 1, 2], [0]]
Output is [[1], [2, 3, 4], [3, 4], [4, 5], [5, 6, 7], [6]]
```
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Please write to us at contribute@geeksforgeeks.org to report any issue with the above content. | 1,460 | 3,614 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2020-45 | latest | en | 0.549993 |
https://justaaa.com/chemistry/287506-a-2050102m-solution-of-nacl-in-water-is-at-200c | 1,695,523,122,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233506559.11/warc/CC-MAIN-20230924023050-20230924053050-00308.warc.gz | 378,412,196 | 9,665 | Question
# A 2.050×10−2M solution of NaCl in water is at 20.0∘C. The sample was created by dissolving...
A 2.050×10−2M solution of NaCl in water is at 20.0∘C. The sample was created by dissolving a sample of NaCl in water and then bringing the volume up to 1.000 L. It was determined that the volume of water needed to do this was 999.4 mL . The density of water at 20.0∘C is 0.9982 g/mL.
Part A
Calculate the molality of the salt solution.
Express your answer to four significant figures and include the appropriate units.
Part B
Calculate the mole fraction of salt in this solution.
Express the mole fraction to four significant figures.
Part C
Calculate the concentration of the salt solution in percent by mass.
Express your answer to four significant figures and include the appropriate units.
Part D
Calculate the concentration of the salt solution in parts per million.
Express your answer as an integer to four significant figures and include the appropriate units.
A) Molality = mol of solute / mass of solvent.
Mol of solute = Molarity of solute x Volume of solution = (0.02050 M) x (1.000 L) = 0.02050 mol of solute
Mass of solvent = 999.4 mL x (0.9983 g / 1 mL) = 997.7 g = 0.9977 kg
Molality = 0.02050 mol / 0.9977 kg = 0.02055 mol / kg
B) Mol fraction of salt = mol of salt / mol total
Mol of water = 997.7 g x (1 mol H2O / 18.0154 g H2) = 55.38 mol H2O
Mol fraction of salt = 0.02050 mol / (0.02050 mol + 55.38 mol) = 0.0003700
C) Concentration by percent mass = (mass of salt / mass of solution) x 100%
Mass of salt = 0.02050 mol NaCl x (58.44 g NaCl / 1 mol NaCl) = 1.198 g NaCl
Mass of water = 997.7 g H2O
Mass of solution = 1.198 g + 997.7 g = 998.90 g
Therefore, (1.198 g / 998.90 g) x 100% = 0.1199 %
D)Parts per million = (g of solute / g of solution)*1000000 = (1.198 g / 998.90 g)*1000000 = 1199 ppm
#### Earn Coins
Coins can be redeemed for fabulous gifts. | 609 | 1,901 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.703125 | 4 | CC-MAIN-2023-40 | latest | en | 0.882597 |
http://ask.csdn.net/questions/242451 | 1,521,323,579,000,000,000 | text/html | crawl-data/CC-MAIN-2018-13/segments/1521257645362.1/warc/CC-MAIN-20180317214032-20180317234032-00514.warc.gz | 20,527,422 | 13,822 | void_cpp1 于 2016.03.13 16:17 提问
ACM的小伙伴进，uva-1584有些小问题求大神解惑
Some DNA sequences exist in circular forms as in the following figure, which shows a circular sequence `CGAGTCAGCT", that is, the last symbol`T" in `CGAGTCAGCT" is connected to the first symbol`C". We always read a circular sequence in the clockwise direction.
\epsfbox{p3225.eps}
Since it is not easy to store a circular sequence in a computer as it is, we decided to store it as a linear sequence. However, there can be many linear sequences that are obtained from a circular sequence by cutting any place of the circular sequence. Hence, we also decided to store the linear sequence that is lexicographically smallest among all linear sequences that can be obtained from a circular sequence.
Your task is to find the lexicographically smallest sequence from a given circular sequence. For the example in the figure, the lexicographically smallest sequence is ``AGCTCGAGTC". If there are two or more linear sequences that are lexicographically smallest, you are to find any one of them (in fact, they are the same).
Input
The input consists of T test cases. The number of test cases T is given on the first line of the input file. Each test case takes one line containing a circular sequence that is written as an arbitrary linear sequence. Since the circular sequences are DNA sequences, only four symbols, A, C, G and T, are allowed. Each sequence has length at least 2 and at most 100.
Output
Print exactly one line for each test case. The line is to contain the lexicographically smallest sequence for the test case.
The following shows sample input and output for two test cases.
Sample Input
2
CGAGTCAGCT
CTCC
Sample Output
AGCTCGAGTC
CCCT
``````#include <iostream>
#include <cstring>
using namespace std;
int small(const char* s,int p,int q);
#define maxn 105
int main()
{
int n;//测试数
cin>>n;
while(n--){
//为何不能用char *str={0};？
char str[maxn];
cin>>str;
int len=strlen(str);
int ans=0;
for(int i=1;i<len;i++){
if(small(str,i,ans))
ans=i;
}
for(int i=0;i<len;i++){
putchar(str[(ans+i)%len]);
}
putchar('\n');
}
return 0;
}
int small(const char* s,int p,int q){ //比较表示法p和表示法q字典序大小
int n=strlen(s);
for(int i=0;i<n;i++){
if(s[(p+i)%n]!=s[(q+i)%n])
return s[(p+i)%n]<s[(q+i)%n];
}
return 0;
}
``````
1个回答
leilba 2016.03.13 17:37
`````` char *str={0};
cin>>str;
``````
`````` char *str = new char[maxn];
cin>>str;
``````
void_cpp1 谢谢，我明白了 | 688 | 2,428 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.171875 | 3 | CC-MAIN-2018-13 | latest | en | 0.864012 |
https://gitlab.ift.uam-csic.es/scw2017/python/-/commit/2820c217acf504eb9612b2d4d6ee3214af1a4461.diff | 1,638,145,344,000,000,000 | text/plain | crawl-data/CC-MAIN-2021-49/segments/1637964358673.74/warc/CC-MAIN-20211128224316-20211129014316-00124.warc.gz | 360,623,505 | 10,049 | \n", + "
\n", + "
### Exercise - Types
\n", + "
\n", + "
\n", + " What type of value is 3.4?
\n", + " What type of value is 3.25 + 4?
\n", + " What type of value is 4/2?
\n", + "
\n", + "
" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "
\n", + "
\n", + "
### Exercise - Division Types
\n", + "
\n", + "
\n", + "In Python 3, the // operator performs integer (whole-number) floor division, the / operator performs floating-point division, and the ‘%’ (or modulo) operator calculates and returns the remainder from integer division.
\n", + "\n", + "However in Python2 (and other languages), the / operator between two integer types perform a floor (//) division. To perform a float division, we have to convert one of the integers to float\n", + "
\n", + "
\n" + ] + }, + { + "cell_type": "raw", + "metadata": {}, + "source": [ + "print('5 // 3 =', 5//3)\n", + "print('5 / 3 =', 5/3)\n", + "print('5 % 3 =', 5%3)\n", + "print('float(5)/3 =', float(5) / 3 )\n", + "\n", + "\n" + ] + }, + { + "cell_type": "markdown", + "metadata": { + "collapsed": true + }, + "source": [ + "\n", + "
\n", + "
\n", + "
### Exercise - Arithmetic with Different Types
\n", + "
\n", + "
\n", + " Which of the following will print 2.0? Note: there may be more than one right answer.
\n", + " first = 1.0
\n", + "second = \"1\"
\n", + "third = \"1.1\"
\n", + "
\n", + "
\n", + "1. first + float(second)
\n", + "2. float(second) + float(third)
\n", + "3. first + int(third)
\n", + "4. first + int(float(third))
\n", + "5. int(first) + int(float(third))
\n", + "6. 2.0 * second\n", + "
\n", + "
\n", + "\n", + "\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "
\n", + "
\n", + "
### Key Points
\n", + "
\n", + "
\n", + "
\n", + "
• Every value has a type.
• \n", + "
• Use the built-in function type() to find the type of a value.
• \n", + "
• Types control what operations can be done on values.
• \n", + "
• Strings can be added and multiplied.
• \n", + "
• Strings have a length (but numbers don’t).
• \n", + "
• Must convert numbers to strings or vice versa when operating on them.
• \n", + "
• Must convert numbers to strings or vice versa when operating on them.
• \n", + "
• Can mix integers and floats freely in operations.
• \n", + "
\n", + "
\n", + "
\n", + "
\n", + "
### Key Points
\n", + "
\n", + "
\n", + "
\n", + "
• \n", + "
• A function may take zero or more arguments.
• \n", + "
• Commonly-used built-in functions include max, min, and round.
• \n", + "
• Functions may have default values for some arguments.
• \n", + "
• Use the built-in function help to get help for a function.
• \n", + "
• The Jupyter Notebook has two ways more to get help.
• \n", + "
• Every function returns something.
• \n", + "
\n", + "
\n", + "
\n", + " Lists and character strings are both *collections*.\n", + "
" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Indexing beyond the end of the collection is an error.\n", + "\n", + "* Python reports an `IndexError` if we attempt to access a value that doesn't exist.\n", + " " + ] + }, + { + "cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true, + "scrolled": false + }, + "outputs": [], + "source": [] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Strides" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "* If we write a slice as [low:high:stride], what does stride do?" + ] + }, + { + "cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], + "source": [] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "
\n", + "
\n", + "
### Exercise - Fill in the Blanks
\n", + "
\n", + "
\n", + "\n", + "
\n", + "
" + ] + }, + { + "cell_type": "raw", + "metadata": {}, + "source": [ + "values = ____\n", + "values.____(1)\n", + "values.____(3)\n", + "values.____(5)\n", + "print('first time:', values)\n", + "values = values[____]\n", + "print('second time:', values)\n", + "\n", + "---RESULT---\n", + "first time: [1, 3, 5]\n", + "second time: [3, 5]" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "
\n", + "
\n", + "
### Exercise - How Large is a Slice?
\n", + "
\n", + "
\n", + "If ‘low’ and ‘high’ are both non-negative integers, how long is the list values[low:high]?\n", + "
\n", + "
" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "
\n", + "
\n", + "
### Exercise - Nested Lists
\n", + "
\n", + "
\n", + "\n", + "
\n", + "
" + ] + }, + { + "cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], + "source": [ + "x = [['pepper', 'zucchini', 'onion'],\n", + " ['cabbage', 'lettuce', 'garlic'],\n", + " ['apple', 'pear', 'banana']]" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "" + ] + }, + { + "cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], + "source": [ + "print(x)" + ] + }, + { + "cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], + "source": [ + "print([x[0]])" + ] + }, + { + "cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], + "source": [ + "print(x[0])" + ] + }, + { + "cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], + "source": [ + "print(x[0][0])" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "
\n", + "
\n", + "
### Key Points
\n", + "
\n", + "
\n", + "
\n", + "
• A list stores many values in a single structure..
• \n", + "
• Use an item’s index to fetch it from a list.
• \n", + "
• Lists’ values can be replaced by assigning to them.
• \n", + "
• Appending items to a list lengthens it.
• \n", + "
• Use del to remove items from a list entirely.
• \n", + "
• The empty list contains no values.
• \n", + "
• Lists may contain values of different types.
• \n", + "
• Character strings can be indexed like lists.
• \n", + "
• Character strings are immutable.
• \n", + "
• Indexing beyond the end of the collection is an error.
• \n", + "
\n", + "
\n", + "
\n", + "
\n", + "
### Exercise - Squared and cubed prime numbers
\n", + "
\n", + "
\n", + " We have a list of prime numbers, and we want to calculate the squared and cubed numbers for every prime number.\n", + "
\n", + "\n", + "
\n", + "
\n", + " \n", + " " + ] + }, + { + "cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], + "source": [] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "
\n", + "
\n", + "
### Exercise - Reverse a string
\n", + "
\n", + "
\n", + " Fill in the blanks in the program below so that it prints “nit” (the reverse of the original character string “tin”).
\n", + "
\n", + "
\n", + "\n" + ] + }, + { + "cell_type": "raw", + "metadata": {}, + "source": [ + "original = \"theoretical physics\"\n", + "result = ____\n", + "for char in original:\n", + " result = ____\n", + "print(result)" + ] + }, + { + "cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], + "source": [] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "
\n", + "
\n", + "
### Exercise - Practice Accumulating
\n", + "
\n", + "
\n", + " Fill in the blanks in each of the programs below to produce the indicated result.
\n", + "
\n", + "
" + ] + }, + { + "cell_type": "raw", + "metadata": {}, + "source": [ + "# Total length of the strings in the list: [\"red\", \"green\", \"blue\"] => 12\n", + "total = 0\n", + "for word in [\"red\", \"green\", \"blue\"]:\n", + " ____ = ____ + len(word)\n", + "print(total)" + ] + }, + { + "cell_type": "raw", + "metadata": {}, + "source": [ + "# List of word lengths: [\"red\", \"green\", \"blue\"] => [3, 5, 4]\n", + "lengths = ____\n", + "for word in [\"red\", \"green\", \"blue\"]:\n", + " lengths.____(____)\n", + "print(lengths)" + ] + }, + { + "cell_type": "raw", + "metadata": {}, + "source": [ + "# Concatenate all words: [\"red\", \"green\", \"blue\"] => \"redgreenblue\"\n", + "words = [\"red\", \"green\", \"blue\"]\n", + "result = ____\n", + "for ____ in ____:\n", + " ____\n", + "print(result)" + ] + }, + { + "cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], + "source": [ + "##Solution 1\n" + ] + }, + { + "cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], + "source": [ + "##Solution 2\n" + ] + }, + { + "cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], + "source": [ + "#solution 3\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "
\n", + "
\n", + "
### Key Points
\n", + "
\n", + "
\n", + "
\n", + "
• A for loop runs commands once for each value in a collection.
• \n", + "
• The first line of the for loop must end with a colon, and the body must be indented.
• \n", + "
• Indentation is always meaningful in Python.
• \n", + "
• A for loop is made up of a collection, a loop variable, and a body.
• \n", + "
• Loop variables can be called anything (but it is strongly advised to have a meaningful name to the looping variable).
• \n", + "
• The body of a loop can contain many statements.
• \n", + "
• Use range to iterate over a sequence of numbers.
• \n", + "
• The Accumulator pattern turns many values into one.
• \n", + "
\n", + "
\n", + "
\n", + "
\n", + "
### Exercise - Identifying Syntax Errors
\n", + "
\n", + "
\n", + "1. Read the code below and try to identify what the errors are without running it.
\n", + "2. Run the code and read the error message. Is it a SyntaxError or an IndentationError?
\n", + "3. Fix the error.
\n", + "4. Repeat steps 2 and 3 until you have fixed all the errors.
\n", + "def another_function
\n", + "
\n", + "
" + ] + }, + { + "cell_type": "raw", + "metadata": {}, + "source": [ + "def another_function\n", + " print(\"Syntax errors are annoying.\")\n", + " print(\"But at least python tells us about them!\")\n", + " print(\"So they are usually not too hard to fix.\")" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "
\n", + "
\n", + "
### Exercise - Simulating a dynamical system
\n", + "
\n", + "
\n", + "In mathematics, a dynamical system is a system in which a function describes the time dependence of a point in a geometrical space. Canonical example of a dynamical system is a system called the logistic map.
\n", + "\n", + "1. Define a function called **logistic_map** that takes two inputs: **X**, representing the state of the system at time t, and a parameter **r**. This function should return a value representing the state of the system at time **t+1**.
\n", + "2. Using a for loop, iterate the logistic_map function defined in part 1 starting from an **initial condition** of 0.5 for **T=10**, **100**, and **1000** periods. Store the intermediate results in a list so that after the for loop terminates you have accumulated a sequence of values representing the state of the logistic map at time t=0,1,…,T.
\n", + "3. Encapsulate the logic of your for loop into a function called **iterate** that takes the **initial condition** as its first input, the parameter **T** as its second input and the parameter **r** as its third input. The function should return the list of values representing the state of the logistic map at time t=0,1,…,T.
\n", + "
\n", + "
\n" + ] + }, + { + "cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], + "source": [ + "#Solucion" + ] + }, + { + "cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], + "source": [] + }, + { + "cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], + "source": [ + "\n", + " \n", + " " + ] + }, + { + "cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], + "source": [] + }, + { + "cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], + "source": [] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "
\n", + "
\n", + "
### Key Points
\n", + "
\n", + "
\n", + "
\n", + "
• Break programs down into functions to make them easier to understand.
• \n", + "
• Functions are reusable
• \n", + "
• Define a function using def with a name, parameters, and a block of code.
• \n", + "
• Every function returns something.
• \n", + "
• Arguments may have default values.
• \n", + "
• Any argument without default values is a required argument and MUST come before the argument with default values.
• \n", + "
\n", + "
\n", + "
\n", + "
\n", + "
### Exercise - What Is Truth?
\n", + "
\n", + "
\n", + "True and False are special words in Python called booleans which represent true and false statements. However, they aren’t the only values in Python that are true and false. In fact, any value can be used in an if or elif. After reading and running the code below, explain what the rule is for which values are considered true and which are considered false.
\n", + "\n", + "
\n", + "
\n" + ] + }, + { + "cell_type": "raw", + "metadata": { + "scrolled": true + }, + "source": [ + "if '':\n", + " print('empty string is true')\n", + "if 'word':\n", + " print('word is true')\n", + "if []:\n", + " print('empty list is true')\n", + "if [1, 2, 3]:\n", + " print('non-empty list is true')\n", + "if 0:\n", + " print('zero is true')\n", + "if 1:\n", + " print('one is true')" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "
\n", + "
\n", + "
### Exercise - Counting Vowels
\n", + "
\n", + "
\n", + "1. Write a loop that counts the number of vowels in a character string.
\n", + "2. Test it on a few individual words and full sentences.
\n", + "3. Once you are done, compare your solution to your neighbor’s.
\n", + "\n", + "
\n", + "
\n", + "\n", + "\n", + "\n" + ] + }, + { + "cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], + "source": [ + "#Solution\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "
\n", + "
\n", + "
### Key Points
\n", + "
\n", + "
\n", + "
\n", + "
• Use if condition to start a conditional statement, elif condition to provide additional tests, and else to provide a default.
• \n", + "
• The bodies of the branches of conditional statements must be indented.
• \n", + "
• Conditionals are often used inside loops.
• \n", + "
• Use == to test for equality.
• \n", + "
• X and Y is only true if both X and Y are true.
• \n", + "
• X or Y is true if either X or Y, or both, are true.
• \n", + "
• Zero, the empty string, and the empty list are considered false; all other numbers, strings, and lists are considered true.
• \n", + "
\n", + "
\n", + "
\n", + "\n" + ] + } + ], + "metadata": { + "kernelspec": { + "display_name": "Python 3", + "language": "python", + "name": "python3" + }, + "language_info": { + "codemirror_mode": { + "name": "ipython", + "version": 3 + }, + "file_extension": ".py", + "mimetype": "text/x-python", + "name": "python", + "nbconvert_exporter": "python", + "pygments_lexer": "ipython3", + "version": "3.6.1" + } + }, + "nbformat": 4, + "nbformat_minor": 2 +} diff --git a/09_scripts_acdf.ipynb b/09_scripts_acdf.ipynb new file mode 100644 index 0000000000000000000000000000000000000000..eb3e703f7bf14b53ab07681283b6117906cf0c76 --- /dev/null +++ b/09_scripts_acdf.ipynb @@ -0,0 +1,332 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Scripts in Python" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "* The Jupyter Notebook and other interactive tools are great for prototyping code and exploring data, but sooner or later we will want to use our program in a pipeline or run it in a shell script to process thousands of data and these programms won't need our interation." + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "
In this lesson we are switching from typing commands in a Python interpreter to typing commands in a shell terminal window (such as bash). When you see a \$ in front of a command that tells you to run that command in the shell rather than the Python interpreter.
" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Command-Line Programs\n", + "* Using the text editor of your choice, in our case nano editor, save the following in a text file called sys_version.py:" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Libraries\n", + "* Most of the power of a programming language is in its libraries.\n", + "* A library is a collection of files (called modules) that contains functions for use by other programs.\n", + "* A library is a collection of modules, but the terms are often used interchangeably, especially since many libraries only consist of a single module, so don’t worry if you mix them.\n", + "* Use import to load a library module into a program’s memory.\n", + "* Then refer to things from the module as module_name.thing_name.\n", + "* Python uses . to mean “part of”.\n", + "* Use help to learn about the contents of a library module." + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Print Python version " + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "
\n", + "\$ nano sys_version.py
" + ] + }, + { + "cell_type": "markdown", + "metadata": { + "collapsed": true + }, + "source": [ + "
\n", + "import sys
\n", + "print('version is', sys.version)
\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "
\n", + "\$ python sys_version.py
" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Print argument list" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "
\n", + "\$ nano argv_list.py
" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "
\n", + "import sys
\n", + "print('sys.argv is', sys.argv)
\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "
\n", + "\$ python argv_list.py
" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "
\n", + "\$ python argv_list.py this is a list of arguments
" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "
\n", + "\$ nano temperature.py
" + ] + }, + { + "cell_type": "raw", + "metadata": {}, + "source": [ + "\n", + "import sys\n", + "\n", + "def main():\n", + " #script = sys.argv[0]\n", + " temp = sys.argv[1]\n", + " print(temp)\n", + " \n" + ] + }, + { + "cell_type": "raw", + "metadata": {}, + "source": [ + "\n", + "import sys\n", + "\n", + "def main():\n", + " #script = sys.argv[0]\n", + " temp = sys.argv[1]\n", + " print(temp)\n", + "\n", + "main()\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "
\n", + "\$ conversion.py
" + ] + }, + { + "cell_type": "code", + "execution_count": null, + "metadata": { + "scrolled": true + }, + "outputs": [], + "source": [ + "import sys\n", + "\n", + "def fahr_to_kelvin(temp):\n", + " return ((temp - 32)* (5/9) + 273.15)\n", + "\n", + "print('Fahrenheit to Kelvin Conversion ',sys.argv[1],'-->',fahr_to_kelvin(56))\n", + "\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Counting the number of arguments" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "
\n", + "\$ conversion.py
" + ] + }, + { + "cell_type": "code", + "execution_count": null, + "metadata": { + "scrolled": true + }, + "outputs": [], + "source": [ + "#Count the number of arguments\n", + "import sys\n", + "def fahr_to_kelvin(temp):\n", + " return ((temp - 32) * (5/9)) + 273.15\n", + "\n", + "if ((len(sys.argv[1:])<1) or (len(sys.argv[1:])>=2)):\n", + " print(\"Error: 1 argument expected\")\n", + "else:\n", + " print('Fahrenheit to kelvin conversion',sys.argv[1], '-->',fahr_to_kelvin(int(sys.argv[1])))" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### List of arguments" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "
\n", + "\$ prime_numbers.py
" + ] + }, + { + "cell_type": "code", + "execution_count": null, + "metadata": { + "scrolled": true + }, + "outputs": [], + "source": [ + "import sys\n", + "\n", + "def prime_numbers(numbers):\n", + " for n in numbers:\n", + " n=int(n)\n", + " \n", + " if n<0: # meets first condition?\n", + "\n", + " # if a IS less than zero\n", + " print(n, 'is a negative number')\n", + "\n", + " elif n>0: # did not meet first condition. meets second condition?\n", + "\n", + " # if a ISN'T less than zero and IS more than zero\n", + " print(n, ' is a positive number')\n", + "\n", + " else: # met neither condition\n", + "\n", + " # if a ISN'T less than zero and ISN'T more than zero\n", + " print(n, ' must be zero!')\n", + " \n", + "\n", + "prime_numbers(sys.argv[1:])" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Dynamic system" + ] + }, + { + "cell_type": "code", + "execution_count": null, + "metadata": {}, + "outputs": [], + "source": [ + "def logistic_map(X, r):\n", + " return r * X * (1 - X)\n", + "\n", + "\n", + "\n", + " \n", + "def iterate(initial_condition, T, r):\n", + " trajectory = [initial_condition]\n", + " for t in range(1, T):\n", + " trajectory.append(logistic_map(trajectory[t-1], r))\n", + " return trajectory\n", + "\n", + "\n", + "initial_condition = 0.5\n", + "T = 10\n", + "r = 1.0 # r=4, sistema caotico\n", + "print(iterate(initial_condition,T,r))" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "
\n", + "
\n", + "
### RECAP
\n", + "
\n", + "
\n", + " \n", + "
\n", + "
\n" + ] + } + ], + "metadata": { + "kernelspec": { + "display_name": "Python 3", + "language": "python", + "name": "python3" + }, + "language_info": { + "codemirror_mode": { + "name": "ipython", + "version": 3 + }, + "file_extension": ".py", + "mimetype": "text/x-python", + "name": "python", + "nbconvert_exporter": "python", + "pygments_lexer": "ipython3", + "version": "3.6.1" + } + }, + "nbformat": 4, + "nbformat_minor": 2 +} | 7,132 | 22,466 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2021-49 | latest | en | 0.654371 |
https://www.physicsforums.com/threads/electric-potential-of-a-spherical-shell.747789/ | 1,508,458,734,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187823482.25/warc/CC-MAIN-20171019231858-20171020011858-00072.warc.gz | 959,396,087 | 18,604 | # Electric Potential of a Spherical Shell
1. Apr 8, 2014
### asap9993
1. The problem statement, all variables and given/known data
A conducting spherical shell has inner radius a, outer radius b, and has a +Q point charge at the center. A charge of -Q is put on the conductor.
a) What is the charge on the inner and outer surface of the shell?
b) What is the electric field everywhere?
c) What is the electric potential everywhere?
2. Relevant equations
Gauss's Law
V = -∫E$\bullet$dl
3. The attempt at a solution
a) From conservation of charge and the fact that it's a conductor, the charge on the inner surface of the shell is -Q and the charge on the outer surface is 0. In addition, there is no charge inside the shell.
b) For 0 < r < a, Gauss's Law gives E = kQ/(r^2)
For a < r < b, E = 0
For r > b, E = 0
c) This is where I'm stuck. E = 0 for r > b, so the potential difference is 0. But the question asks for the actual potential function. Normally, I would use V = kQ/r. But Q = 0 in this case, so V = 0. So that would mean V = 0 for a < r < b Lastly, for 0 < r < a,
V(r) = V(a) - kQ((1/a)-(1/r)) = 0 - kQ((1/a)-(1/r)) = kQ((1/r)-(1/a)).
I believe this is what the answer should be. Please, can anyone tell me if they agree?
Last edited: Apr 9, 2014
2. Apr 9, 2014
### Simon Bridge
First you say there is a charge of -Q inside the shell, then you say there is no charge inside the shell ... which is it?
The question is a tad vaguely worded - it is unclear what is meant by "inside" and "outside" in each case. Taken literally, "inside" would commonly be considered r<a and "outside" would be r>b ... so that would exclude the surface charges. a≤r≤b would be "within the shell itself". It is possible to answer all the sections using that definition ... it's an exercise in Gauss' Law.
You'll have to use the context provided by your lessons to correctly interpret the question.
... that is only for r<a, at r=a, isn't there an additional surface charge?
Note: what is the relationship between electric potential and electric feild?
3. Apr 9, 2014
### asap9993
Yes,sorry about that. I rephrased the question.
4. Apr 9, 2014
### Simon Bridge
No it doesn't. Check. Hint: E is a vector.
Also:
What you have written means that at r=a, E=kQ/a^2 (1st relation) and E=0 (2nd relation), at the same time: how can this be?
5. Apr 9, 2014
### asap9993
Right. E is indeed a vector that points radially outward in this case. I'm focused on its magnitude here. In vector form, there is an r roof vector included.
Ok I see that you are saying that E(r) is discontinuous at r = a. I'm guessing that E must be 0 at r = a since the enclosed net charge is 0, so the first relation should be E(0 < r < a) = kQ((1/r^2) - (1/a^2)). That would make it continuous. However, that contradicts Gauss's Law. That doesn't make sense. Also, why does E have to be continuous anyway? There are several examples in my textbook in which it is not. Are the charges correct?
Lastly, do you agree with my reasoning in the part on finding the potential? Suppose my electric fields are correct, then did I get the correct potentials?
6. Apr 9, 2014
### Simon Bridge
You have the idea, it is just you are not being careful about the boundaries.
The trick is to be careful about dividing the volume up.
For a gaussian surface that is a sphere centered on the origin of radius $0<r<a$,
What is the total charge enclosed?
What is the electric field due to that charge?
Again for $a\leq r < b$ ?
Again for $r\geq b$ ?
It may help to think of the surface charge as being infinitesimally inside the surface.
For the potentials, you seem to have chosen your reference potential at r=a ... I'm not really sure I've understood your reasoning: was there a specific reason for this?
Note: there is a relationship between the electric field and the potential.
Why not use it?
7. Apr 9, 2014
### ehild
You should state with respect to what is the potential difference zero. But you are right, if the electric field is zero which is the negative gradient of the potential, the potential does not change between infinity and r=b.
The potential of a point charge Q is kQ/r if the zero of the potential is at infinity.
Yes, the potential is also zero inside the conductor, as the electric field is zero, the potential is constant and the potential is a continuous function. If it is zero just outside the conductive shell, it is zero also inside. For the same reason, the potential is also zero inside the void at r=a. | 1,196 | 4,514 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.78125 | 4 | CC-MAIN-2017-43 | longest | en | 0.95086 |
https://nobeltextiles.com/qa/quick-answer-what-is-the-lcm-of-18-and-27.html | 1,600,905,061,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400212959.12/warc/CC-MAIN-20200923211300-20200924001300-00011.warc.gz | 538,895,605 | 7,294 | # Quick Answer: What Is The LCM Of 18 And 27?
## What is the LCM of 18 and 27 using prime factorization?
Alternatively, the lcm of 18 and 27 can be found using the prime factorization of 18 and 27: The prime factorization of 18 is: 2 x 3 x 3.
The prime factorization of 27 is: 3 x 3 x 3.
Eliminate the duplicate factors of the two lists, then multiply them once with the remaining factors of the lists to get lcm(18,18) = 54..
## What is the LCM of 18 and 15?
Multiply the factors to get the LCM. The LCM of 15 and 18 is 90.
## What is the LCM of 27 and 36?
108Least common multiple (LCM) of 27 and 36 is 108.
## What is the LCM of 9 18 and 27?
The least common multiple 54 is a product of common & odd prime factors between the integers which is divisible by each one an integer of this same group. The step by step work for LCM of 9, 18 and 27 may useful to understand how to find LCM for two or three numbers.
## What is the LCM of 7 18 and 21?
The least common multiple 126 is a product of common & odd prime factors between the integers which is divisible by each one an integer of this same group. The step by step work for LCM of 7, 18 and 21 may useful to understand how to find LCM for two or three numbers.
## What is the HCF of 18 and 24?
We found the factors and prime factorization of 18 and 24. The biggest common factor number is the GCF number. So the greatest common factor 18 and 24 is 6.
## What is the HCF of 27 and 36?
We found the factors and prime factorization of 27 and 36. The biggest common factor number is the GCF number. So the greatest common factor 27 and 36 is 9.
## What is the LCM of 24 and 36?
Explanation: The LCM of 24 and 36 is the smallest positive integer that divides the numbers 24 and 36 without a remainder. If you just want to know what is the least common multiple of 24 and 36, it is 72. Usually, this is written as.
## What is the LCM of 24 and 30?
Least common multiple (LCM) of 24 and 30 is 120.
## What is the LCM of 24 36 and 42?
Problem & Workout :LCM of 24, 36 and 42624362463237212 more rows
## What is the LCM of 12 and 18?
Least common multiple (LCM) of 12 and 18 is 36.
## What is the LCM for 18?
Least common multiple (LCM) of 18 and 24 is 72.
## What is the LCM for 18 and 30?
Least common multiple (LCM) of 18 and 30 is 90.
## What is the LCM of 20 and 30?
Least common multiple (LCM) of 20 and 30 is 60.
## What is the LCM of 16 and 18?
Least common multiple (LCM) of 18 and 16 is 144. | 744 | 2,480 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.59375 | 5 | CC-MAIN-2020-40 | latest | en | 0.944249 |
https://www.adda247.com/engineering-jobs/rrb-sse-syllabus/ | 1,725,836,533,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651035.2/warc/CC-MAIN-20240908213138-20240909003138-00514.warc.gz | 602,495,557 | 116,671 | Engineering Jobs » RRB SSE Syllabus 2024
# RRB SSE Syllabus 2024, Check Exam Pattern of RRB SSE
The Railway Recruitment Board will soon release the official notification for Senior Sectional Engineer on its official website @indianrailways.gov.in. The aspirants can get all the details related to RRB SSE Syllabus 2024 through this article.
## RRB SSE Syllabus 2024
The official notification for RRB SSE Recruitment 2024 will be out soon for the posts of Senior Sectional Engineer on its official website @indianrailways.gov.in. Those candidates are willing to make a career in Govt. Jobs apply for RRB SSE Recruitment 2024 for those having an engineering degree in various disciplines.
## RRB SSE 2024 Syllabus
The candidates interested in RRB SSE Recruitment 2024, must understand the RRB SSE Syllabus 2024 to plan their preparation accordingly. The syllabus gives an overview of the level of the question paper and section-wise weightage to prepare for the exam effectively. The article covers the RRB SSE Syllabus 2024 in detail to help the aspirants. Read the full article to check the topic-wise RRB SSE Syllabus 2024 and bookmark this website for further Engineering Job Updates.
## RRB SSE Syllabus 2024 Detailed Syllabus
This section covered the RRB SSE Syllabus 2024, candidate must read the section below to know the topic-wise syllabus for RRB SSE Exam 2024:
RRB SSE Syllabus 2024 Subjects Syllabus General Intelligence & Reasoning Analogy Puzzle Matrix Order & Ranking Verbal Reasoning Coding-Decoding Word Formation Blood Relation Syllogism Alphabet Series Venn Diagram Problem-Solving Direction & Distance Non-verbal Reasoning Arithmetic S.I & C.I Simplification Profit & Loss Number System Surds & Indices Sequence & Series Ratio & Proportion Percentage Number System Work & time Continuity Time & Distance Trigonometry Straight Lines Conic sections Probability 3 Dimensions Limits Permutation & Combination Probability Problem with the age Equation Set Theory Pipes & Cistern General Awareness Current affairs History Current events Geography Sports Banking Static GK Budget Science & Technology Important Days Books & Author General Science Basic concepts of science for class 12th Standard Questions will be asked from Physics, Chemistry, Biology &Zoology Mechanical Engineering Strength of Material Fluid Mechanics Thermodynamics Cycle Engineering Mechanics Theory of machines Heat transfer Steam tables Machine tool operations Casting Machining, Inspection, Metrology Electrical Engineering Electrical Machine Electronic Measurement Measurement System Signal System Control System Basics of circuit power electronic Electrical Digital Electronics Circuits Analog Electrical machines Power Civil Engineering Construction/Structure Engineering & Drawing Estimation Costing and valuation & building material Steel Structure Structure analysis Mechanics Solid mechanics Concrete structures Surveying Coordinate system Traffic engineering Water requirements Errors & Adjustment Measurement of Distance & Directions Hydraulics Hydrology Construction Highway Planning Projection Curves Environmental Engineering
## RRB SSE Exam Pattern 2024
Candidates who want to apply for RRB SSE 2024 Recruitment must know the important details regarding the RRB SSE Exam Pattern 2024. We have provided the detailed RRB SSE Exam Pattern 2024 below.
• The Nature of the Examination will be objective.
• The total number of questions asked in the RRB SSE 2024 will be 150.
• The time duration for the RRB SSE Exam is 120 minutes.
• There is a provision for negative marking in the RRB SSE Exam. 1/3rd of marks will be deducted for every wrong answer.
• There are 4 sections in the RRB SSE 2024 Exam. It will consist of Technical subjects, General Awareness, General Intelligence, and Reasoning and Arithmetic. The detailed pattern is tabulated below. Interested candidates can refer to the official notification for a more detailed syllabus.
RRB SSE Exam Pattern 2024 Subject name Number of Questions Maximum marks Technical Subjects 90 90 General Awareness 60 60 General Intelligence and Reasoning Arithmetic Total 150 150
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The RRB SSE 2024 notification will be published soon on its official website @indianrailways.gov.in. | 952 | 4,625 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.75 | 3 | CC-MAIN-2024-38 | latest | en | 0.807674 |
https://community.easymorph.com/t/check-a-column-for-unique-and-or-duplicate-values-and-group-by-non-unique-values/2505 | 1,685,730,383,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224648850.88/warc/CC-MAIN-20230602172755-20230602202755-00638.warc.gz | 213,365,718 | 5,366 | Check a column for unique and/or duplicate values and group by non-unique values
Hi,
I have a table like so,
Account Number Col1 Col2 Col3 Amount
0052364 Cash xx xx 120.00
0052364 Cash xx xx 138.50
0052364 Cash xx xx 141.21
I use this file to calculate a sum of Amount based on Account number so instead of 3 different values for the same account number I use group by ‘Account number’ to calculate the total amount. So the final output is:
Account Number Col1 Col2 Col3 Amount
0052364 xx xx xx 399.71
It works fine for now as there is just one account number in the file. But I need to prepare for a scenario where there might be a different account number in the file so I need to check if the account number is unique and if yes then group by account number, if not then group by similar account numbers.
Like:
Account Number Col1 Col2 Col3 Amount
0052364 Cash xx xx 120.00
0052364 Cash xx xx 138.50
0072930 Cash xx xx 141.21
0072930 Cash xx xx 100.00
If the file comes in with a different account number, how do I identify that and group similar account numbers together?
Thank you
R
As I understand it, you need to aggregate and group by the account in any case. The question is whether the aggregation returns 1 group (i.e. when the account is unique) or more than 1 group (i.e. when the account is not unique).
In this case, you can calculate a column with the total number of accounts, and then do conditional branching depending on whether you have 1 or more accounts in your dataset.
See below an example that does that:
non-unique-branching.morph (4.4 KB)
2 Likes
Ah ok, I understand. Thank you so much for your prompt response
R
You’re welcome | 420 | 1,672 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.578125 | 3 | CC-MAIN-2023-23 | latest | en | 0.773925 |
https://archived.hpcalc.org/museumforum/archive/index.php?thread-248903.html | 1,632,860,678,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780060882.17/warc/CC-MAIN-20210928184203-20210928214203-00176.warc.gz | 156,883,731 | 8,797 | # HP Forums
Full Version: HP PRIME CAS QUICK GUIDE [SPN]
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[Eng] A math teacher is writing a quick guide in Castilian of the commands and functions of the HP-Prime CAS
first version HP PRIME CAS QUICK GUIDE
[Spn] Un docente de matemáticas esta escribiendo una guía rápida en castellano de los comandos y funciones de la HP-Prime CAS
primera versión
GUÍA RÁPIDA DEL CAS DE LA HP PRIME (FUNCIONES BÁSICAS)
[Request for HP-Prime CAS Development Group]
0: Name all commands to lowercase, if mixed words change the first letter to uppercase. write in ALL CAPS is an old custom of HP48SX/GX/50G
for example:
DESOLVE renamed within the catalog to DEqSolve
DOT => DotProduct ~ MAPLE
CROOSS => CrossProduct
...
1: Do not use special symbols (->) in some cases , best use acronyms, FP, IP abbreviations was used for small screens, we now have an editor on a large screen
B->R => Bin2real
R->B => Real2bin
EVALC => EvalCplx
CSOLVE => CplxSolve
CZEROS => CplxZeros
NSOLVE => NumSolve
FP => FracPart
IP => IntPart
LEFT => LeftPart
RIGHT => RightPart
MID => MiddlePart
EXPR => ToExpr
CHAR => ToChar
2: differentiate two commands by extension (postfix #)
type
TYPE => Type2 or TypeNum
3: initiate important commands in UPPERCASE and lowercase secondary
IF THEN => If Then
EDITMAT => editMat
4: Differentiate Ends
IF THEN END => If Then EndIf
thanks
Quote:
GUÍA RÁPIDA DEL CAS DE LA HP PRIME (FUNCIONES BÁSICAS)
Por: Elkin Arbelaez Gaviria
abcuv(a(x),b(x),c(x)):ecuación polinómica de la forma a(x)*u+b(x)*v=c(x)
ABS(z):valor absoluto del argumento (número,expresión,lista,vector,matriz)
ACOSH(z):arco coseno hiperbólico del argumento
ACOT(z):arco cotangente del argumento
ACSC(z):arco cosecante del argumento
acos2asin(xpr):transforma una expresión con acos( ) en función de asin( )
acos2atan(xpr):transforma una expresión con acos( ) en función de atan( )
ADDCOL(mat,vec,num):inserta en una matriz un vector como columna en la
posición dada por un numero natural
ADDROW(mat,vec,num): inserta en una matriz un vector como fila en la pos.
ALOG(z):antilogaritmo decimal del argumento
AND:conjunción de 2 argumentos (números,condiciones lógicas,ecuaciones)
append(lst1,lst2) :anexa una lista después de otra (queda aumentada)
apply(fvar,lst):aplica una función predefinida a los elementos de una lista
approx(xpr) o evalf(xpr):aproximan una expresión a un valor numérico
arcLen(fun,var,a,b):longitud de arco de una función f(var) en intervalo [a;b]
ARG(z) o angle(z):ángulo polar del argumento (número o lista de complejos)
ASEC(z):arco secante del argumento
ASIN(z):arco seno del argumento
asin2acos(xpr):transforma una expresión con asin( ) en función de acos( )
asin2atan(xpr):transforma una expresión con asin( ) en función de atan( )
ASINH(z):arco seno hiperbólico del argumento
ATAN(z):arco tangente del argumento
atan2acos(xpr):transforma una expresión con atan( ) en función de acos( )
atan2asin(xpr):transforma una expresión con atan( ) en función de asin( )
ATANH(z):arco tangente hiperbólica del argumento
atrig2ln(xpr):transforma una expresión con funciones trig.inversas a logaritm.
B?R(num):convierte un número entero binario a real (punto flotante)
Beta(arg1,arg2):función beta de 2 argumentos reales
Canonical_form(p(x),x):forma canónica de un polinomio cuadrático
CEILING(num):retorna el menor entero = el argumento
cFactor(pol):trata de factorizar una expresión algebraica sobre el campo C
CHAR(vec):retorna la cadena correspondiente al vector de códigos ASCII
CHOOSE(var,titulo,item1,…,itemn):crea una caja de selección de datos
chinrem(mat):trata de resolver un sistema de congruencias polinómicas
coeff(pol,var):retorna lista de coeficientes de un polinomio de una variable
col(mat,num):extrae una columna de una matriz dada su posición
colDim(mat):número de columnas de una matriz
collect(xpr):agrupa una expresión algebraica
COLNORM(mat):norma por columnas de una matriz
comDenom(xpr):reduce una expresión a común denominador
companion(pol,var):matriz compañera de un polinomio de una variable p(var)
COMB(num):combinaciones de un número finito de elementos (sin repetición) complexroot(pol):aproxima raíces complejas de polinomio (multiples)
CONCAT(mat1,mat2):aumentada por filas de 2 arreglos (vectores o matrices)
content(pol):máximo común divisor (MCD) de los coeficientes de un polinomio
COS(z):coseno del argumento
COSH(z):coseno hiperbólico del argumento
cos2sintan(xpr):transforma una expresión con cos( ) en función de sin( ) y tan()
COT(z):cotangente del argumento
cpartfrac(frat):fracciones parciales complejas de una función racional
crationalroot(frat):aproxima raíces complejas de una función racional (simples)
CROSS(vec1,vec2):producto cruz de 2 vectores 3D (o de 2 listas 3D)
curl(fvec,lvar):rotacional de una función vectorial 3D en coord. cartesianas
CSC(z):cosecante del argumento
cSolve(eqn[lst],var[lvar]):slnes complejas de una ecuación (o de un sistema)
cZeros(xpr[lst],var[lvar]):ceros complejos de expresión (o lista de expresiones)
degree(pol):retorna el grado de un polinomio
DELCOL(mat,num):borra columna de una matriz ,dada su posición
DELROW(mat,num):borra fila de una matriz, dada su posición
denom(xpr):extrae el denominador de una expresión
deSolve(F(y,y’,y”,…)=0,x[xo],y[yo]):trata de resolver EDO (con o sin CI)
diag(vec):crea una matriz diagonal dado un vector (o retorna la diagonal
diff(xpr,var,n) o ?(xpr,var,n):derivada n-sima de expr. con respecto a var.
DIM(mat):lista de dimensiones de una matriz (numero de filas y de columnas)
divergence(fvec,lvar):divergencia de una función vectorial 3D en coord.cart
divis(pol):divisores (factores) de un polinomio
DOT(vec1,vec2):producto punto de 2 arreglos (listas,vectores o matrices)
egcd(pol1,pol2):MCD extendido de 2 polinomios
erf(arg):función error de argumento real
erfc(arg):función error complementaria de argumento real
EVAL(xpr):evalúa una expresión algebraica
evalc(xpr):evalúa una expresión numérica compleja
euler(num):función Totient de Euler (# de primos menores o iguales que arg)
exact(arg):retorna una fracción (racional) equivalente al argumento dado
expand(xpr):expande el argumento (expresión)
exp2pow(xpr):transforma una expresión con exp( ) en función de potencias
exp2trig(xpr):transforma una expresión con exp( ) en terminos de fun.trig.
factor(pol):trata de factorizar una expresión algebraica sobre el campo R
factorial(num):factorial del argumento (número complejo,lista compleja)
fcoeff(vec): retorna una función racional dado el vector de raíces fft(mat):transformada rápida de Fourier de un arreglo (vector o matriz) cuyas
dimensiones deben ser potencias enteras de 2
FLOOR(num):retorna el mayor entero = el argumento
fMax(fun,var):máximo local de una función f(var) sobre los reales
fMin(fun,var):mínimo local de una función f(var) sobre los reales
FNROOT(xpr,var,ini):aproxima una raíz de una expresión dado un val.inicial
FP(num):parte fraccionaria del argumento (número,lista,vector,matriz)
froot(fun):aproxima un vector de raíces y polos de una función racional
fsolve(xpr,var,ini,met):aproxima un cero de una expresión, dada una
estimación inicial (o intervalo) y el método (Newton,bisección,etc)
Gamma(z):función Gamma de argumento complejo
gauss(xpr,lvar):representación de una forma cuadrática como suma o dif.
gcd(pol1,pol2):MCD de 2 polinomios
GETKEY:retorna el código de la tecla presionada
gramschmidt(mat):retorna una base ortonormal de las columnas de una matriz
halftan(xpr):reemplaza en una expresión sin( ),cos( ) y tan( ) por tan(x/2)
halftan_hyp2exp(xpr):reemplaza en un expresión tanh(x/2) por exp( )
harmonic_conjugate(cfun):armónica conjugada de una función compleja
head(lst):retorna el primer elemento de una lista
hermite(n):n-simo polinomio ortogonal de Hermite en terminos de x
hessian(fun,lvar):matriz Hessiana de una función escalar multivariable
hyp2exp(xpr):transforma una expresión con funciones hiperbólicas a expon.
ibp ( ):integración por partes de una función f(x) de la forma UdV
ichinrem(vec1,vec2 ):trata de resolver un sistema de 2 congruencias en Z
idivis(num):retorna los divisores de un número entero positivo (natural)
ifactor(num):retorna la factorización de un número natural (como producto)
ifactors(num):retorna el vector de factores y exponentes de un numero natural
MCD(lst):MCD de una lista numérica real
ifft(mat):inversa de la transformada rápida de Fourier de un vector o matriz
ihermite(mat):forma de Hermite para una matriz
IM(arg):parte imaginaria del argumento (número,lista,vector,matriz)
INPUT(var,titulo,etiq):crea una caja de entrada de datos
INSTRING(str):posición de la primera aparición de un caractér en una cadena
Int(xpr,var,a,b) o
(xpr,var,a,b):integral de expresión con respecto a var.
inv(arg):recíproco (inverso) del argumento (número,lista,matriz)
ilaplace(F(s),s,t) o invlaplace(F(s),s,t):transform. inversa de Laplace de F(s)
invztrans(F(z),z,n):transformada z inversa de función F(z) en términos de n
IP(arg ) o iPart(arg ):parte entera del argumento (número,lista,vector,matriz)
iquo(num1,num2):cociente de la división de 2 números enteros
irem(num1,num2):residuo de la división de 2 números enteros
ismith(mat):forma normal de Smith de una matriz
isPrime(num):chequea si un numero natural es primo (test probabilístico)
ITERATE(xpr,num):evalúa iterativamente una expresión varias veces
ithprime(n):retorna el enésimo primo
lagrange(mat):polinomio interpolador de Lagrange para una matriz de puntos laguerre(n):n-simo polinomio ortogonal de Laguerre en términos de x
laplace(f(t),t,s):transformada de Laplace de una función f(t) en términos de s
laplacian(fun,lvar):laplaciano de función escalar multivariable en coord.cart
lcm(pol1,pol2):mínimo común múltiplo (MCM) de 2 polinomios
lcoeff(pol):coeficiente principal (el de mayor grado) de un polinomio
legendre(n):n-simo polinomio ortogonal de Legendre en términos de x
lgcd(lst):MCD de una lista de números reales
limit(fun,var,val):límite de una función cuando una variable tiende a un valor
list2mat(lst,nc):convierte una lista a matriz (con nc columnas)
LN(z):logaritmo natural del argumento
lncollect(xpr):agrupa una expresión que contiene funciones logarítmicas
lnexpand(xpr):expande una expresión que contiene funciones logarítmicas
LOG(z):logaritmo decimal del argumento
logb(z):logaritmo en base b del argumento
?LIST(lst) o sum(lst):suma de los elementos de una lista
LIST(lst) o deltalist(lst): primeras diferencias de una lista
?LIST(lst) o product(lst):producto de los elementos de una lista
LQ(A):descomposición P*A=L*Q de una matriz de mxn (P es de permutación)
lvar(xpr) o lname(xpr):retorna la lista de variables de una expresión algebraica
MAKELIST(fvar,ini,fin,[inc]):crea una lista a partir de una función predefinida
MAKEMAT(xpr,nf,nc):crea una matriz a partir de una expresión (dim=nf x nc)
MANT(arg):mantisa del argumento (número,lista,vector,matriz)
mat2list(mat ):convierte una matriz a lista (vector)
MAX(arg1,arg2):máximo de 2 argumentos (números,listas,matrices)
member(lst,elem):dada una lista y un elemento,retorna su posición en la lista
MIN(arg1,arg2):mínimo de 2 argumentos (números,listas,matrices)
MOD:residuo de la división de 2 números enteros ( num1 MOD num2)
MSGBOX(str o xpr):crea una caja de mensaje con la cadena o expresión dada
mult_conjugate(xpr):multiplica una expresión por el conjugado del denom.
nextprime(n):retorna el primer primo siguiente a un numero número natural n
norm(mat):norma (de Frobenius) de una matriz (o de un vector)
normalize(vec):normaliza un vector (retorna un vector unitario)
nSolve(eqn[lst],var[lvar]):solución numérica de ecuación (o de sistema)
NTHROOT:raiz enésima del argumento (n v arg)
numer(xpr):extrae el numerador de una expresión
odesolve(F(t,y),[t,y],[to,yo],tf ):trata de resolver numéricamente EDO y
sistemas de EDO de primer orden con CI (condiciones iniciales) dadas
OR:disyunción de 2 argumentos (números enteros,condiciones lógicas)
partfrac(frat):descomp. en fracciones parciales reales de una función racional
pa2b2(n):retorna el entero gaussiano correspondiente al numero primo n
pcoeff(vec):vector de coeficientes de un polinomio (dado el vector de raíces)
PERM(num):permutaciones de un número de elementos (sin repetición)
PIECEWISE(tst1,caso1,… ):chequea varios casos y retorna el verdadero pívot(mat,i,j):pivotea una matriz usando el elemento ij como pivote
pole(frat):aproxima los polos de una función racional
poly2symb(vec):convierte un vector de coeficientes a polinomio simbólico
POLYCOEF(pol):retorna la lista de coeficientes de un polinomio simbólico
POLYEVAL(pol,arg):evalua polinomio en un argumento (numero,lista,matriz)
POLYROOT(vec):retorna vector de raíces de un polinomio (dados los coef)
potential(fvec):potencial escalar de una función vectorial 3D en coord.cart
powexpand(xpr):expande expresiones que contienen potencias
powmod(b,e,m):función potencia del argumento modulo un número natural
pow2exp(xpr):transforma una expresión con potencias a exponenciales
prepend(lst1,lst2):anexa una lista antes de otra (queda aumentada)
preval(xpr,a,b):evalúa primitiva (antiderivada) en los extremos de un intervalo
prevprim(n):retorna el primer número primo anterior al numero natural n
PRINT(xpr):visualiza en la pantalla una expresión (o una cadena)
proot(vec):retorna el vector de raíces de un polinomio (dado el vector de coef)
propfrac(xpr):fracción propia de una fracción impropia (numérica o algebraica)
QR(A):descomposición A*P=Q*R de una matriz de mxn
quo(pol1,pol2):cociente de la división de 2 polinomios
RAND(n):genera un número pseudoaleatorio a partir de un valor n (semilla)
randMat(nf,nc):genera matriz aleatoria con coeficientes enteros (dim= nf x nc)
randPoly(var,n,a,b):crea vector aleatorio de dim n con coef.enteros ? [a;b]
RANK(mat):retorna el rango de una matriz (# de filas linealmente independ)
RE(arg):parte real del argumento (número,lista,vector,matriz)
REDIM(mat,nf,nc):redimensiona una matriz a las dimensiones dadas
rem(pol1,pol2):residuo de la división de 2 polinomios
remove(lst,elem):elimina (borra) un elemento de una lista, dada su posición
reorder(xpr):reordena una expresión según las variables
REPLACE(mat,pos,submat):reemplaza parte de una lista o de un arreglo
residue(frat):aproxima los residuos de una función racional
REVERSE(lst) o Revlist(lst):invierte el orden de los elementos de una lista
ROUND(n,m):redondea un argumento al número de cifras significativas dado.
romberg(fun,var,a,b ):integral definida de f(x) en [a;b] por met.de Romberg
rowNorm(mat):norma por filas de una matriz
RREF(mat):forma escalonada reducida por filas de una matriz
rSolve(eqn[sys],icond):trata de resolver ecuaciones (sistemas) recursivas
R?B(num):convierte un número real a binario
SEC(z):secante del argumento
seq(xpr[vec],var,ini,fin):crea un vector (matriz) a partir de una expr (vector)
series(fun,var):serie de Taylor y expansión asintótica de una función f(var)
SIGN(arg):signo del argumento (número,lista numérica)
simplify(xpr):simplifica expresiones
simult(A,b):solucionador de sistemas lineales en formato matricial Ax=b(TI-89)
SIN(z):seno del argumento sincos(xpr):transforma exponenciales y logaritmos complejos a sin( ) y cos( )
SINH(z):seno hiperbólico del argumento
sin2costan(xpr):transforma una expresión con sin( ) a cos( ) y tan( )
SIZE(str) o Length(str):#de caracteres de cadena o dimensiones de un arreglo
SORT(lst):ordena una lista en forma ascendente
spline(lstx,lsty,var,deg ):trazador cúbico natural para 2 listas de valores (x,y)
STRING(arg):convierte el argumento a cadena de caracteres
sturmseq(fun):secuencia de Sturm (cambios de signo) de una función f(var)
subMat(mat,p1,p2) :extrae una submatriz de otra mayor dadas pos1 y pos2
subst(xpr,var,val):permite substituir variables por valores en una expresión
SVD(mat):descomposición con valores singulares de una matriz (A=U*S*V)
SVL(mat):retorna los valores singulares de una matriz rectangular
SWAPCOL(mat,col1,col2):intercambiar 2 columnas de una matriz
SWAPROW(mat,fil1,fil2):intercambiar 2 filas de una matriz
sylvester(pol1,pol2):matriz de Sylvester de 2 polinomios
symb2poly(pol):convierte un polinomio simbólico a vector de coeficientes
TAIL(lst):retorna todos los elementos de una lista excepto el primero
TAN(z):tangente del argumento
TANH(z):tangente hiperbólica del argumento
tan2cossin2(xpr):transforma una expresión con tan( ) a cos( ) y sin²( )
tan2sincos(xpr):transforma una expresión con tan( ) a sin( ) y cos( )
tan2sincos2(xpr):transforma una expresión con tan( ) a sin( ) y cos²( )
taylor(fun,var=val,n):serie de Taylor de f(var) centrada en var=val y orden n
tchebyshev1(n):n-simo polinomio de Chebyshev de clase 1
tchebyshev2(n):n-simo polinomio de Chebyshev de clase 2
tCollect(xpr):agrupa expresiones trigonométricas
tExpand(xpr):expande expresiones trigonométricas
transpose(mat):transpuesta de una matriz (sin conjugar)
trigcos(xpr):simplifica una expresión trigonométrica a términos con cos( )
trigexpand(xpr):expande expresiones que contienen funciones trigonométricas
trigsin(xpr):simplifica una expresión trigonométrica a términos con sin( )
trigtan(xpr):simplifica una expresión trigonométrica a términos con tan( )
trig2exp(xpr):transforma una expresión con funciones trig. a exponenciales
trunc(arg,num):trunca el argumento a un número dado de cifras decimales
TYPE(arg):retorna el número de tipo de un argumento
tsimplify(xpr):simplifica expresiones con funciones trascendentes (log y exp)
vandermonde(lst):crea una matriz de Vandermonde (dada una lista numérica)
when(cond,xpr1,xpr2):permite definir funciones por tramos
vpotential(fvec):potencial vectorial de una función vectorial 3D en coord.cart
XOR:disyunción lógica exclusiva de 2 argumentos (números,condiciones)
zeta(z):funcion zeta de Riemann (para argumento real)
ztrans(f(n),n,z):transformada z de una función f(n) en términos de z
?(xpr,var,ini,fin):suma finita de una expresión con respecto a una variable | 5,515 | 18,261 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2021-39 | latest | en | 0.30614 |
http://blendernpr.org/store/?page_id=2318&share=google-plus-1 | 1,550,856,481,000,000,000 | text/html | crawl-data/CC-MAIN-2019-09/segments/1550247518497.90/warc/CC-MAIN-20190222155556-20190222181556-00003.warc.gz | 34,231,326 | 11,973 | # Pixel Math [ebook]
### The Problems:
Have you been changing blend types in mix node and hoping to get the correct one?
Yes. Even the pros do it in tutorial videos. 😳
Have you wondered how factor differ between the blend modes?
There are differences? Never knew anything about them. 😮
Have you read books or websites explaining about blend modes and wonder about the actual inner working, the algorithms and their limitations?
I have read Wikipedia and other websites, wasting hours. Still learn nothing. I think of the inner working all the time. Blend modes are math, and computers don’t do magic. 😈
Have you ever wanted to make your own blend mode?
Yes. If I’m equipped with the knowledge, I’ll make one and name it with my name. 😆
### The Solution:
Pixel Math (noun): Mathematical equations & algorithms which determine color behavior when two or more colors are layered on the same pixel location. Commonly known as blend modes.
#### Features:
• ###### Quick navigation provided.
Pixel Math Ebook Preview
Extremely beginner friendly.
Learn how each blend mode works.
Learn how factor works in different algorithms.
Discuss which blend modes are useful for which situation.
Outside reference for specific blend modes.
Learn pixel math in the process, it is easy and fun.
Pick up bits and pieces to make your own blend mode.
Node setup working in material and compositor.
Bonus chapter included.
Do it in code, or use the default blend modes, or make your own pixel math node group.
NO MORE wildly scrolling blend modes & hoping that one will come out right. 😉
### You can become a Pixel Math Wizard. 😎
Side reading: Background stories of the conception of this e-book
This ebook is not limited to non-photorealistic rendering (NPR). Pixel Math is used everywhere in Blender. For example: render passes are based on pixel math, texture blending, clamping/boosting of high dynamic range image, color grading, tone mapping etc.
Soft Light algorithm in node form
### Reviews:
Pixel math is an excellent guide to Blender’s powerful node based shading system. I highly recommend it to everyone interested in real time rendering or compositing. You are not going to find a more complete reference anywhere else. —Todor Imreorov
As both an artist and compositor I would have to recommend “Pixel Math”. Blend nodes have been covered much but never before was I able to read about them in such depth. Clever writing thats down to the point accompanied with plenty of visual guides made this a very easy to read and understand book. Blend nodes are obvious in their behavior when it comes to screen and multiply however when the other nodes are mentioned it opens up a world of possibilities that have never been covered so in depth before. I cannot recommend this book enough since color control and color math are two of the most important things when it comes to understanding pixel math. Blend modes offer power to generate isolations of values that can change the impact of an entire piece. I am honored to have been allowed the opportunity to get to review it also learn from this. Just the reviewing of this book expanded my pixel math to another level and its all thanks to the genius’ over at BNPR. —Jerry Perkins a.k.a MasterXeon1001
A handy reference of color blend modes in Blender, this small ebook explains in simple words all the math equations that govern different ways of blending colors. The book will be a good source of information for anyone who has experience of getting mad with color blending options in the compositor, material nodes, and Freestyle line color modifiers. — Tamito Kajiyama a.k.a T.K.
### Blend On & Do Pixel Math
Specs: Delivery: Online | Size: ~4MB | Format: True PDF
Page Count: 102 | Difficulty: Beginner to Advance
Note: Please open the “Buy Now” link in a new tab for secure purchase. | 824 | 3,848 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2019-09 | latest | en | 0.849325 |
https://www.vetrina-eventi.com/how-to-calculate-roi/ | 1,713,084,050,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296816875.61/warc/CC-MAIN-20240414064633-20240414094633-00463.warc.gz | 967,902,888 | 14,011 | # How to Calculate ROI: A Comprehensive Guide to Measuring Business Success
## I. Introduction
Return on Investment (ROI) is a fundamental financial concept that allows businesses and investors to determine the profitability of their investments. ROI is an essential concept for entrepreneurs, managers, and business owners as it helps them evaluate the effectiveness of their investments and make informed decisions. This article is a comprehensive guide to calculating ROI, featuring its definition, a step-by-step guide on how to calculate it, common mistakes to avoid, and real-world examples of how ROI can be used to make better business decisions.
## II. The Beginner’s Guide to Calculating ROI
Defining ROI: ROI is a financial metric that measures the return on an investment relative to its cost. It helps businesses and investors determine the profitability of their investments and evaluate their performance over time. ROI is expressed as a percentage, with higher values indicating better returns on investment.
The formula for calculating ROI: ROI = (Gain from Investment – Cost of Investment)/Cost of Investment x 100.
For example, suppose a business invests \$1000 in a marketing campaign that generates \$2500 in profits. The ROI would be (2500-1000)/1000 x 100 = 150%. This means that for every dollar invested, the business earned \$1.50 in profits.
It’s important to note that ROI can be calculated for any investment, whether it’s a marketing campaign, equipment purchase, or business expansion.
## III. 5 Common Mistakes to Avoid When Calculating ROI
Some common mistakes in calculating ROI include:
1. Using inaccurate or incomplete data
2. Not accounting for all costs and expenses
3. Misinterpreting cause and effect
4. Using a short-term focus instead of evaluating long-term outcomes
5. Comparing ROI values across industries with different cost structures
To avoid these mistakes, businesses should ensure that they have all the necessary data and use accurate calculations. Additionally, they should consider the long-term effects of their investments and compare their ROI values against other metrics to get a holistic view of their investments.
## IV. How to Use ROI to Make Better Business Decisions
ROI should be a fundamental component of financial analysis in business decision-making. It can be used in various ways to improve business profitability, including:
• Evaluating investment effectiveness and opportunities
• Determining which business units or products are the most profitable and deserve more investment
• Comparing investment options to determine which have a higher ROI
• Providing a framework for evaluating the effectiveness of existing business processes and strategies
Businesses can effectively interpret ROI data by comparing the ROI values to industry benchmarks and using multiple metrics to determine the overall performance of their investments.
Real-world examples of companies that have used ROI calculations to make better business decisions include Amazon, which used ROI to evaluate the profitability of its Prime membership program, and PwC, which leveraged ROI to evaluate the effectiveness of its marketing campaigns.
## V. ROI Calculators: Are They Worth Using?
ROI calculators are online tools that help businesses calculate ROI without manually inputting data. They can be an effective way to save time and effort, but they come with pros and cons.
Pros of using ROI calculators include:
• Efficiency: Calculations are completed quickly and accurately
• Accessibility: Calculations can be done from anywhere with an internet connection
• Standardization: Calculations are consistent and comparable
Cons of using ROI calculators include:
• Dependence: The accuracy of ROI values is dependent on the accuracy of the input data
• Limitations: Calculators may not take into account certain nuances, such as differences in industry cost structures
• Cost: Some calculators may require payment for access to premium features or accurate data
Businesses can find the best ROI calculators for their needs by considering factors such as accuracy, complexity, and cost. Mainstream financial software like QuickBooks or Sage50 will have ROI calculator functions built in.
## VI. Real-World Examples of ROI in Action
Several companies have used ROI calculations in their business decisions and seen success, including:
• Coca-Cola: Increased its sales by more than \$2.5 million by investing in a mobile app campaign with a 10:1 ROI
• Zillow: Increased its online visibility and revenue by over \$150 million through various SEO campaigns with a 586% ROI
• Nordstrom: Used ROI to evaluate the profitability of its digital marketing campaigns and reallocated marketing funds to optimize its campaigns better. It achieves a 23:1 ROI on its Facebook Ads, generating \$22.54 in revenue for every dollar spent on advertisements.
A few key takeaways from these examples are that businesses should focus on strategic investments that provide long-term value, regularly measure ROI, and adjust their investment strategies as needed.
## VII. The Role of ROI in Marketing Campaigns
ROI plays a crucial role in marketing campaigns, allowing businesses to evaluate the effectiveness of different campaigns and channels. Some strategies for tracking and measuring marketing ROI include:
• Determining upfront costs
• Setting goals and objectives
• Identifying the metrics that matter
• Measuring and analyzing campaign success
• Optimizing campaigns
Businesses can optimize their marketing strategies by using ROI data to determine which channels or campaigns are the most effective and reallocate resources accordingly.
## VIII. ROI vs. Other Metrics: What You Need to Know
ROI is just one of several metrics that businesses can use to measure their financial and operational performance. Other popular metrics include:
• Net present value (NPV)
• Internal rate of return (IRR)
• Payback period
• Customer lifetime value (CLV)
Each metric has its strengths and weaknesses, and businesses should choose the most appropriate metrics based on their specific needs and circumstances.
## IX. Conclusion
ROI is a critical metric for measuring business success and profitability. By calculating ROI correctly, businesses can make better-informed decisions, allocate resources more efficiently, and identify areas for improvement. In summary, businesses should:
• Understand the importance of ROI in evaluating investments and making data-driven decisions
• Know how to calculate ROI using a standard formula
• Avoid common mistakes in ROI calculations
• Use ROI in business strategy and decision-making
• Consider using ROI calculators for efficiency, with caution
By following these guidelines and leveraging real-world examples, businesses can effectively use ROI to improve their bottom line and drive long-term success.
Proudly powered by WordPress | Theme: Courier Blog by Crimson Themes. | 1,310 | 6,975 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-18 | latest | en | 0.910569 |
https://math.stackexchange.com/questions/2369184/how-jacobian-transformation-have-been-used-in-this-problem | 1,582,633,985,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875146066.89/warc/CC-MAIN-20200225110721-20200225140721-00094.warc.gz | 456,856,221 | 30,579 | # How jacobian transformation have been used in this problem
Let $t$ and $a$ be positive real numbers. Define $B_a = \{x=(x_1,x_2,x_3,\dots,x_n) \in\mathbb R^n\mid x_1^2 + x_2^2+\dots+x_n^2 \leq a^2\}$. Then for any compactly supported continous function $f$ on $\mathbb R^n$ Then the following is correct $\int_{ B_a} {f(xt) dx} =\int_{B_{ta}} {f(x) t^{-n} dx}$
But I can't understand.. I have seen its solution it has written directly Jacobian transformation $= t^{-n}$ How can we get $t^{-n}$ in RHS. Plz tell me how the Jacobian has been used
The determinant of the (Jacobian of the) linear map $T(x)=tx$ ($x\in\Bbb R^n$) is $t^n$. If you rewrote your equation as $$\int_{B_{ta}} f(x)\,dx = \int_{B_a} f(tx)t^n\,dx$$ it would fit the change-of-variables formula precisely (assuming $t>0$, of course). | 277 | 807 | {"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} | 3.34375 | 3 | CC-MAIN-2020-10 | latest | en | 0.805656 |
http://gogeometry.blogspot.com/2012/01/problem-716-intersecting-circles-center.html | 1,527,266,945,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794867140.87/warc/CC-MAIN-20180525160652-20180525180652-00482.warc.gz | 122,278,488 | 14,718 | Saturday, January 14, 2012
Problem 716. Intersecting Circles, Center, Radius, Perpendicular, 90 Degrees
Geometry Problem
Level: Mathematics Education, High School, Honors Geometry, College.
Click the figure below to see the problem 716 details.
1. Some clarifications and minor corrections of previous solution:
http://img849.imageshack.us/img849/782/problem716.png
Draw diameter FK of circle O’ . Connect KA and KB
KA perpen to FA , KB perpen to FB
From G &H draw GA’//KA and HB’//KB ( see picture) . GA’ cut FK at N
Note that GA’ perpen to FA , HB’ perpen to FB
1. Triangle FAG congruence to FBH (case AA)
Let FG/FA=FH/FB= k and m(HFB)=m(GFA)=x
So FA’/FA= FG/FA . cos(x)= k.cos(x)= FN/FK
And FB’/FB=FH/FB. cos(x)= k.cos(x) =FA’/FA=FN/FK
So GA’ and HB’ will intersect at point N on FK (properties of // lines and congruence triangles)
2. In triangle GHN, HA’ and GB’ are altitudes and F will be the orthocenter of triangle GHN so O’F perpen To GH
With the same reason as part 1, OC will perpen to ED and L will be the orthocenter of triangle C’MF’
And ML will perpen to C’F’
Peter Tran
2. Join BA and extend it.
Draw tangent at F to circle (O’) to meet BA extended at X.
Let tangent XF be extended to meet C’F’ at Y.
∠YFB = ∠FAB (angle in the alternate segment).
= ∠HAB = ∠HGB (angles in the same segment).
∴ XY ∥ MH.
In turn this implies F’O’F ⊥ MH.
By a similar argument C’OC ⊥ MD.
∴ L is the orthocenter of ∆ MC’F’ and
Hence ML ⊥ C’F’
3. Very good solution Pravin !
Peter Tran
4. Let CC’ meet circle O at P and DE at Q. Let F’F meet HG at R.
< EDA = < CBA = < CPA, hence APDQ is concyclic and so < PQF’ = < PAC = 90.
Similarly we can show that < C’RF = 90
Hence L is the orthocenter of Triangle MC’F’ and so ML is perpendicular to C’F’
Sumith Peiris
Moratuwa
Sri Lanka | 582 | 1,786 | {"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} | 4.125 | 4 | CC-MAIN-2018-22 | latest | en | 0.768298 |
https://forum.openframeworks.cc/t/drawing-a-circle-of-vertices-in-ofmesh/19214 | 1,656,782,590,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656104189587.61/warc/CC-MAIN-20220702162147-20220702192147-00642.warc.gz | 296,169,281 | 4,490 | # Drawing a circle of vertices in ofMesh
Hi everyone,
I am trying to add a circle of vertices to ofMesh. I have no idea why I am having such a hard time hahah. I want to draw the vertices with ofVec3f as I will draw more circles on zindex once I get this circle drawn. I suspect I am doing some super noob mistake… any help appreciated!
``````int radius = 40;
int circlepts = 40;
ofPoint startpt;
ofPoint circx, circy, circz;
void ofApp::setup(){
mesh.setMode(OF_PRIMITIVE_POINTS);
startpt = ofVec3f(0,0,0);
//draw first circle of vertices
for(int i = 0; i < circlepts; i++){
float angle = i * TWO_PI / circlepts;
cout << angle << endl;
circx = startpt + radius * cos(angle);
circy = startpt + radius * sin(angle);
circz = startpt;
ofVec3f p = (circx, circy, circz); | 227 | 778 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2022-27 | latest | en | 0.748835 |
https://www.nctm.org/Publications/Teaching-Children-Mathematics/Blog/Counting_-Why-is-it-Important-and-How-Do-We-Support-Children_-Part-1/ | 1,600,627,978,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400198287.23/warc/CC-MAIN-20200920161009-20200920191009-00331.warc.gz | 988,004,767 | 56,676 | Counting: Why is it Important and How Do We Support Children? Part 1
• # Counting: Why is it Important and How Do We Support Children? Part 1
By Lynsey Gibbons and Kendra Lomax, posted November 23, 2015 –
Counting is an essential building block of mathematics. For the next four entries to the Math Tasks to Talk About blog, we will discuss different aspects of counting. In this first post, we introduce the importance of counting. In subsequent posts, we dig deeper into the important concepts related to counting quantities (part 2) and suggest activities you can do with children in primary grades (part 3) and intermediate grades (part 4) to work on these concepts.
How long are these ropes?
Each autumn I (Lynsey) try to visit my sister and her family in central Pennsylvania. We always have a lovely time attending a Penn State football game, autumn festivals, and Halloween events with my niece and nephew. One evening while we were visiting this past year, Olivia (age 5) decided that she wanted to figure out the length of three different ropes—two jump ropes and a long shoelace. She laid each rope taut on the ground. She then disappeared to find some tape. When she returned, Olivia had a plan. She was going to wrap pieces of tape around each rope, side by side, from one end to the other.
Next, she planned to write a number on each piece to understand its length. After using about five pieces of tape, Olivia realized how laborious this would be and decided to abandon that plan. Instead, she wondered if she could use her feet to count how long each rope was. With my assistance to keep her balance, Olivia put each of her feet end to end to count how many of her foot lengths each rope was. She had some interesting noticings about the three ropes, including which rope was the longest.
Children’s Curiosity and Counting
Children are curious about the world. They want to make sense of what’s going on around them. Like Olivia, children often wonder about questions like what the length of something is, how tall someone is, who has more, how many geese there are, how many armholes are in different clothing items, and how long until we get somewhere. Quantifying stuff is an important part of our everyday lives. Adults have similar wonderings. Those who live in cities with a lot of traffic like we do, are often curious how many cars will make it through each cycle at this stop light, whether it’s worth driving across town to save ten cents per gallon on gasoline, and how much time is needed to drive across town to get that gas before the soccer game starts.
To quantify items, cultures have created different counting systems over time. (To learn more about different counting systems, see this example about the Central Alaskan Yupik.) In many cultures, families help children count their fingers, toys, people at the table, and other sets of objects. Questions concerning who has more and whether we have enough are part of the daily lives of children as young as two or three years old (Van de Walle, Karp, and Bay-Williams 2012). These experiences help children make sense of the important question, “How many?”
By the time children reach kindergarten, they begin to put their counting skills to work in solving simple problems that call for adding, subtracting, multiplying, or dividing amounts (Kilpatrick, Swaford, and Findell 2001). Throughout elementary school, children build on their early ideas of counting and quantity to understand the base-ten structure of our number system and use these understandings to engage in multidigit computation. It is not surprising that many mathematics education policy documents discuss the importance of counting (Kilpatrick, Swaford, and Findell 2001), and that counting standards show up in the early grades in mathematics content standards documents (e.g., Common Core State Standards and NCTM Focal Points). However, in our experience, children in intermediate grades need support with counting as well.
In the next three posts, we will explore important aspects of counting and consider instructional activities that teachers can do in classrooms with children to support their understanding of number. In the meantime, try getting curious about the world around you like Olivia did. Where are there opportunities to count, compare, measure, and wonder? How can these everyday experiences help us learn about counting and quantity?
We want to hear from you. Post your comments below or share your thoughts on Twitter @TCM_at_NCTM using #TCMcounting.
References
Kilpatrick, Jeremy, Jane Swafford, and Bradford Findell, eds. 2001. Adding It Up. Mathematics Learning Study Committee, Center for Education, Washington, DC: National Academies Press.
Van de Walle, John A., Karen S. Karp, and Jennifer M. Bay-Williams. 2012. Elementary and Middle School Mathematics: Teaching Developmentally. Boston, MA: Pearson.
Lynsey Gibbons, @lynseymathed, is an assistant professor in mathematics education at Boston University in Massachusetts. She is a former elementary school teacher and mathematics coach. Her current scholarly work seeks to understand how we can reorganize schools to support the learning of children and adults. Kendra Lomax, @kendralomax, is a math educator at the University of Washington in Seattle. She designs and facilitates professional learning opportunities around elementary mathematics through projects like TEDD.org. Curiosity about children’s mathematical thinking is at the heart of her work. The authors would like to note that they are continually learning about children and counting. They have learned a great deal from their colleagues, reading the mathematics education literature, and interacting with children about counting. The following colleagues have greatly informed their thinking about how to support children in finding the joy in mathematics and in counting in particular: Ruth Balf, Adrian Cunard, Megan Franke, Allison Hintz, Elham Kazemi, Becca Lewis, Teresa Lind, Angela Chan Turrou, and many teachers in the Seattle, Washington, area. | 1,263 | 6,076 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.28125 | 4 | CC-MAIN-2020-40 | longest | en | 0.968588 |
http://www.actuarialoutpost.com/actuarial_discussion_forum/showthread.php?s=8cc6259a3afbf748f18441b8d5881e9e&p=9502749 | 1,585,517,922,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585370496227.25/warc/CC-MAIN-20200329201741-20200329231741-00278.warc.gz | 202,049,360 | 9,156 | Actuarial Outpost SOA #87
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#1
12-21-2018, 01:10 PM
joelcheung SOA Join Date: Mar 2018 Location: Singapore Studying for Exam STAM College: Singapore Management University, sophomore Posts: 9
SOA #87
Looking at the 6th line in the solution, i.e. E(X-20)+ = E(X) - ....
I understand that this line is using the formula E[(X-20)+] = E(X) - E[(X^20)],
but how does that second part of the 6th line equate to E[(X^20)]??
And for this particular question, can I use the survival function (derived by finding the distribution function F(x) from the f(x) in the solution and then S(x)= 1 - F(x)), and then integrating the survival function from 20 to 80 for the flat f(x) portion + integrating the survival function from 82 to 120 for the declining f(x) portion?
Thank you!
#2
12-21-2018, 01:15 PM
daaaave David Revelle Join Date: Feb 2006 Posts: 3,103
Whenever X is a continuous loss variable,
$E[X \wedge 20] = \int_0^{20} x \cdot f(x) \, dx + 20 \Pr[X>20]$
In their expression, they are finding Pr[X>20] by doing 1 - Pr[X<=20] as the integral for P[X<=20] is easier than the integral for Pr[X>20].
Yes, you could use the survival function. I think it will take longer.
__________________ | 427 | 1,506 | {"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": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 1, "/images/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-2020-16 | latest | en | 0.818157 |
https://www.flexiprep.com/NIOS-Notes/Computer-Science/Ch-9-Control-Statements-Part-7.html | 1,582,108,216,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875144111.17/warc/CC-MAIN-20200219092153-20200219122153-00264.warc.gz | 753,054,878 | 8,366 | # NIOS Computer Science: Chapter 9 – Control Statements Part 7
Doorsteptutor material for UGC Computer-Science is prepared by world's top subject experts: Get detailed illustrated notes covering entire syllabus: point-by-point for high retention, fully solved questions with step-by-step explanation- practice your way to success, Get full length tests using official NTA interface: all topics with exact weightage, real exam experience, detailed analytics, comparison and rankings, & questions with full solutions.
#### While Loop
Syntax of while loop
while (condition)
{
statement(s);
}
The flow diagram indicates that a condition is first evaluated. If the condition is true, the loop body is executed and the condition is re-evaluated.
Hence, the loop body is executed repeatedly as long as the condition remains true. As soon as the condition becomes false, it comes out of the loop and goes to the statement next to the ‘while’ loop.
To make it more clearly, we take the following example.
### Example
To find the sum of first ten natural numbers i.e.
# include <iostream.h>
void main ( )
{
int n, total = 0 ;
n = 1 ;
while (n < = 10)
{
total + = n ;
n + + ;
}
cout << “sum of first ten natural number” << total :
}
The variable n is called a loop control variable since its value is used to control loop repetition. Normally, the three operations listed below must be performed on the loop control variable.
(1) Initialize the loop control variable
(2) Test the loop control variable
(3) Update the loop control variable
Operation:
(i) Must be performed before the loop is entered. Operation.
(ii) Must be performed before each execution of the loop body; depending on the result of this test, the loop will either be repeated or make an exit. Operation
(iii) Must be included as part of the loop body. Unless the loop control variable is updated in the loop body, its value cannot change and loop exit will never occur.
Developed by: | 430 | 1,974 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2020-10 | latest | en | 0.901038 |
https://www.blossomandsoar.co.uk/how-to-help-your-child-with-place-value-part-1/ | 1,600,839,766,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400209999.57/warc/CC-MAIN-20200923050545-20200923080545-00262.warc.gz | 745,556,757 | 8,652 | Place value is a really important aspect of Numeracy which helps to give children a solid foundation in their understanding of number. It also really helps with their written methods and mental maths methods.
What does place value, with whole numbers, look like in each year group?
• In Year 3, children are working on 3 digit numbers.
• In Year 4, children are working on 4 digit numbers. However, they may also start to look at 5 digit numbers.
• In Year 5, children will work with numbers up to 6 digits.
• In Year 6, children will work with numbers up to 7 digits.
Recognising the value of each digit in a number
Children will work to identify the value of each digit in a number, often through partitioning the number as shown below or by identifying the value of the digit underlined in a number. For example: 386 = the underlined digit is worth 80.
e.g. Year 3: 386 – using the chart below, this shows the 3 is worth 300, the 8 is worth 80 and the 6 is worth 6.
Another way of showing this would be partitioning 386 = 300 + 80 + 6
(H) Hundreds (T) Tens (O) Ones 3 8 6
Year 4: 2784 – using the chart below, this shows the 2 is worth 2000, the 7 is worth 700, the 8 is worth 80 and the 4 is worth 4.
Another way of showing this would be partitioning 2784 = 2000 + 700 + 80 + 4
Th (Thousands) H (Hundreds) T (Tens) O (Ones) 2 7 8 4
Year 5: 78 265 = 70 000 + 8000 + 200 + 60 + 5
TTh (Ten Thousands) Th (Thousands) H (Hundreds) T (Tens) O (Ones) 7 8 2 6 5
824 361 = 800 000 + 20 000 + 4000 + 300 + 60 + 1
HTh (Hundred Thousands) TTh (Ten Thousands) Th (Thousands) H (Hundreds) T (Tens) O (Ones) 8 2 4 3 6 1
Year 6: 5 672 817 = 5 000 000 + 600 000 + 70 000 + 2000 + 800 + 10 + 7
M (Millions) HTh (Hundred Thousands) TTh (Ten Thousands) Th (Thousands) H (Hundreds) T (Tens) O (Ones) 5 6 7 2 8 1 7
Children will also work on being able to read aloud a number that is written.
e.g Year 3: 456 = children would say four hundred and fifty six.
Year 4: 5628 = children would say five thousand, six hundred and twenty eight.
Year 5: 46 789 = children would say forty six thousand, seven hundred and eighty nine.
345 271 = children would say three hundred and forty five thousand, two hundred and seventy one.
Year 6: 6 892 654 = children would say six million, eight hundred and ninety two thousand, six hundred and fifty four.
Writing numbers
Children will then look at writing numbers, using words and numbers.
e.g. Year 3: 768 = children would write seven hundred and sixty eight.
Two hundred and thirty four = children would write 234.
Year 4: 8263 = children would write eight thousand, two hundred and sixty three.
Six thousand, seven hundred and twenty one = children would write 6721.
Year 5: 78 215 = children would write seventy eight thousand, two hundred and fifteen.
Fifty two thousand, six hundred and eighty nine = children would write 52 689.
234 156 = children would write two hundred and thirty four thousand, one hundred and fifty six.
Nine hundred and twenty two thousand, four hundred and twelve = children would write 922 412.
Year 6: 8 915 427 = children would write eight million, nine hundred and fifteen thousand, four hundred and twenty seven.
Seven million, one hundred and seventy six thousand, two hundred and fifty five = 7 176 255.
Important notes to remember: It is also great to try numbers with your child which include zeros within the number e.g. 608, 2305, 63 808, 204 506, 1 203 450 as this helps to ensure they understand zero is a place holder.
For example the value of the digits in 63 808 = 60 000 + 3000 + 800 + 8 and to say the number would be sixty three thousand, eight hundred and eight. Often children miss the zero out when reading and writing numbers, which results in a very different number so it is an important skill to practise. Some schools and teachers refer to zero as Zero the Hero who flies in as a place holder!
If your child is struggling with the numbers in their own year group, try the numbers in the previous year group and gradually build up from there (for year three children, try some 2 digit numbers). | 1,140 | 4,109 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.65625 | 5 | CC-MAIN-2020-40 | latest | en | 0.870899 |
https://blisswrite.org/2021/08/25/expert-answeracct122-spring-2014_national-brands-corporatio/ | 1,679,660,801,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296945282.33/warc/CC-MAIN-20230324113500-20230324143500-00771.warc.gz | 175,432,994 | 26,767 | Select Page
Solved by verified expert :ACCT122 – Spring 2014Due March 8, 2014 Project #2Preparation of a Statement of Cash Flows – Indirect MethodUsing the provided information for National Brands Corporation, answer the following questions:a. What is the book value of the equipment sold?b. What is the book value of the building sold?c. Prepare a statement of stockholders’ equityd. Prepare a statement of cash flows using the indirect method.Your answer may be submitted using Excel or Word; please make sure it is in the proper format.Plant assets:? Acquisitions for the year were \$47,000 cash.? Depreciation expense – \$4,100Common stock was issued for \$88,000.Equipment:? Acquisitions for the year were \$20,000 by executing a note payable.? Sold equipment for \$35,000 cash. Calculate the gain/loss (analyze the t-account).? Depreciation expense – \$4,000Buildings:? Sold a building for \$31,700 cash. Calculate the gain/loss (analyze the t-account).? Depreciation expense – \$6,000Treasury stock was sold for \$45,000 cash.Cash dividends of \$15,000 were paid.Net income/ (loss) for the year is calculated from the activity in Retained Earnings.National Brands CorporationComparative Balance SheetDecember 31, 2013December 31, 2013 December 31, 2012Cash \$180,200 \$15,000Accounts Receivable 15,000 28,600Inventory 23,800 17,000Prepaid expenses 3,000 5,700Equipment 16,000 36,400Plant assets 52,900 10,000Building 84,000 114,500\$374,900 \$227,200Accounts payable (merchandise) \$37,400 \$16,000Salary payable 5,500 4,000Accrued liabilities 15,000 19,300Income tax payable 3,800 3,000Notes payable 117,000 97,000Common stock 138,000 50,000Treasury stock 0 (45,000)Retained earnings 58,200 82,900\$374,900 \$227,200
Order your essay today and save 10% with the discount code ESSAYHELP | 498 | 1,795 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2023-14 | latest | en | 0.881991 |
https://www.physicsforums.com/threads/circuits-impedance-undefined-j8ohms-j8ohms.151879/ | 1,576,024,171,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575540529516.84/warc/CC-MAIN-20191210233444-20191211021444-00062.warc.gz | 804,524,737 | 17,024 | # Circuits: impedance = undefined? (j8ohms-j8ohms)
• Engineering
1. Homework Statement
I'm working a problem where I have a j8ohm and a -j8ohm in parallel, and I need to reduce the circuit to get the current in one of these branches. Being these numbers, I get a fraction with zero on the bottom. How should I treat this?
3. The Attempt at a Solution
I'm thinking it shorts it, but that's a guess
Related Engineering and Comp Sci Homework Help News on Phys.org
well I change my answer on the basis that I think the impedance is infinity...
treat as an open?
Okay, i'm convinced it's the open circuit, but I'm still curious: if you actually did this in a lab, and set the frequency to make these impedances, would it actually not let current get through? or is that more of the "ideal" vs. "real"
berkeman
Mentor
The +j and -j matching impedances represent a parallel LC circuit, correct? What is the impedance of that kind of circuit at resonance (where the +j and -j reactances match)? And what is the thing in a real circuit that keeps the input Z of a parallel LC circuit from being infinite?
Well...it's either the fact that you can't reach infinite impedance, so it will appear as very large, but finite.
Or is the fact that practical capacitors will have a small amount of leakage current
berkeman
Mentor
Well...it's either the fact that you can't reach infinite impedance, so it will appear as very large, but finite.
Or is the fact that practical capacitors will have a small amount of leakage current
I suppose the leakage current might be a factor, but it's a different property of both the inductor and capacitor that keeps it from being infinite. The thing that makes the Zin of a parallel LC look infinite is that you can get a current going back and forth, with the energy going back and forth between the charge stored on the cap and the magnetic field stored by the inductor, all with very little energy being input from outside.
Once you get the oscillation going, you can put almost no energy in to keep it going. But that's when you have an ideal cap and ideal inductor. Now what happens if you add in the DCR of the inductor and ESR of the cap? What would happen to the LC oscillation if you didn't put in any energy?
hmm...
I do believe energy associated with a resistor is turned into heat/friction, so since the practical models of C and L has resistance, and it's oscillating back and forth...
W = IVt, So to keep the current and voltage oscillating, you have to add power (w) because it's being converted (lost as far as circuits go)
am I in the ballpark here?
berkeman
Mentor
Yep, exactly. Now to finish figuring out your OP question about the parallel combination of the reactive phasors and what happens. Go ahead and draw the parallel LC circuit now with a resistor Rdcr ("DC Resistance") in series with the L, and a resistor Resr ("Equivalent Series Resistance) in series with the capacitor. Solve for the input impedance now, with those real-world resistances accounted for.
And then put in some real-world numbers, either by hand, or run them as SPICE simulations, to see what the input impedance is at resonance. Do you have any guesses about how the input Z at resonance is affected by Resr and Rdcr?
BTW, you can use these values for practical starting values:
C = 10uF, Resr = 1 Ohm (and step it up by 0.5 Ohms at a step)
L = 1mH, Rdcr = 1 Ohm (and step it up by 0.5 Ohms at a step) | 824 | 3,438 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.390625 | 3 | CC-MAIN-2019-51 | latest | en | 0.954312 |
https://tutorstips.com/question-17-chapter-3-of-class-12-part-1-vk-publication/ | 1,723,066,517,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640713269.38/warc/CC-MAIN-20240807205613-20240807235613-00099.warc.gz | 457,425,114 | 31,255 | # Question 17 Chapter 3 of Class 12 Part – 1 VK Publication
Question 17 Chapter 3 of Class 12 Part - 1 VK Publication
Question 17 Chapter 3 of Class 12 Part – 1
17. Amit, Vijay and Kuldeep are partners sharing profits and losses equally. From 1st April 2018 they decide to share future profits and losses in the ratio 3:2:1. Calculate Each partners gain/sacrifice due to change in ratio.
## The solution of Question 17 Chapter 3 of Class 12 Part – 1: –
Old Ratio of Amit, Vijay and Kuldeep= 1:1:1
New Ratio of Amit, Vijay and Kuldeep= 3:2:1
Sacrificing Share = Old Share – New share
Amit = 1 – 3 3 6
= 2 – 3 6
= -1 (Gaining) 10
Vijay = 1 – 2 3 6
= 2 – 2 6
= 0 (Nil) 10
Kuldeep = 1 – 1 3 6
= 2 – 1 6
= 1 (Sacrifice) 10
Amit has gained 1/6 th share, Kuldeep has sacrificed 1/6 th share
Comment if you have any questions.
Also, Check out the solved question of previous Chapters: – | 311 | 898 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-33 | latest | en | 0.904444 |
https://ged-testprep.com/question/for-the-following-numbers-which-is-the-correct-order-of-the-variables-6060866906619904/ | 1,717,003,518,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971059384.83/warc/CC-MAIN-20240529165728-20240529195728-00136.warc.gz | 238,176,481 | 21,161 | Scan QR code or get instant email to install app
Question:
# For the following numbers, which is the CORRECT order of the variables from smallest to largest? P = 0.0002 Q = -0.02 R = 0.2 S = -200 T = 0.5
A SQPRT
explanation
To solve these questions, we need to put the numbers in increasing order. This will allow you to easily compare which numbers are larger or smaller.
Then place them in order from least to greatest (like you would see on a number line):
-200
-0.02
0.0002
0.2
0.5
S
Q
P
R
T | 150 | 499 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.453125 | 3 | CC-MAIN-2024-22 | latest | en | 0.916443 |
https://se.mathworks.com/matlabcentral/cody/problems?action=index&controller=problems&sort=&term=tag%3A%22odd%22+difficulty_rating_bin%3Aeasy | 1,638,526,275,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964362619.23/warc/CC-MAIN-20211203091120-20211203121120-00101.warc.gz | 562,691,744 | 17,962 | Cody
# Problems
1 – 9 of 9
Problem Title Likes Solvers Difficulty
#### Problem 43140. Odd times even numbers in a matrix
Created by: Bert
Tags count, even, odd
0
41
#### Problem 44721. Seperate array to small section according to its index position
Created by: Binbin Qi
Tags even, odd
4
12
#### Problem 44158. Even or Odd
Created by: Said BOUREZG
0
59
#### Problem 43006. Odd row
Created by: Tomas Macha
Tags row, odd, elementary
0
47
Created by: Bert
0
64
Created by: Bert
0
53
#### Problem 44397. Determine whether the input is odd, even, or neither.
Created by: Harvey Eldridge
3
38
Created by: goc3
Tags even, odd
1
81
#### Problem 42083. Ring Matrix
Created by: Mayuri Patil
Tags matrix, odd, ring
1
25
1 – 9 of 9 | 234 | 761 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2021-49 | longest | en | 0.726605 |
https://www.teacherspayteachers.com/Product/4th-Grade-Rounding-Numbers-Digital-Practice-4NBT3-Google-Classroom-3284812 | 1,548,058,377,000,000,000 | text/html | crawl-data/CC-MAIN-2019-04/segments/1547583763839.28/warc/CC-MAIN-20190121070334-20190121092334-00525.warc.gz | 924,967,247 | 22,170 | Subject
Resource Type
Common Core Standards
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PDF (Acrobat) Document File
24 MB|21 practice slides, answer key
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1. Google Paperless Practice - 4th Grade Place Value Bundle {4.NBT.1, 4.NBT.2, 4.NBT.3} Engage your students in practicing key place value skills with these interactive DIGITAL resources that work with Google Slides™. No more copies to be made, no more printer ink, and no more lost papers! Plus, stu
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Engage your students with this interactive DIGITAL resource that works with Google Slides™. No more copies to be made, no more printer ink, and no more lost papers!
With this 21-slide digital resource, your students will practice rounding numbers within one million to any place value (with emphasis on rounding to the nearest thousand, ten thousand, and hundred thousand). Students will love interacting with the movable pieces and typing their responses on these slides!
This resource was created to support 4th grade Common Core standard 4.NBT.3: Use place value understanding to round multi-digit whole numbers to any place.
• Instructions for opening, sharing, and using this Google Slides™ file
• 21 interactive slides for your students to complete
If you have a Google classroom, this activity is sure to make practicing these key skills way more fun than if they were done with paper and pencil.
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4th Grade Understanding Place Value {4.NBT.1} Digital Practice
4th Grade Comparing & Ordering Larger Numbers {4.NBT.2} Digital Practice
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You can also check out our 3rd Grade Digital Resources and 5th Grade Digital Resources!
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Your students may also enjoy these 4th grade place value and operations activities from Games 4 Gains:
4th Grade Place Value Games Pack - BEST SELLER!
4th Grade Operations Games Pack - BEST SELLER!
Comparing Numbers Within 1,000,000 'Clip and Flip' Cards
Rounding Numbers Within 1,000,000 'Clip and Flip' Cards
"School Days" Place Value and Operations Color-by-Number Activities (Review of 3rd Grade Standards)
"School Days" Place Value and Operations Color-by-Number Activities (Review of 4th Grade Standards)
Adding Multi-Digit Whole Numbers Bump Games
Subtracting Multi-Digit Whole Numbers Bump Games
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\$4.00 | 796 | 3,522 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2019-04 | latest | en | 0.877998 |
https://hoho.vn/riddles/the-contestant-riddle | 1,532,354,032,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676596463.91/warc/CC-MAIN-20180723125929-20180723145929-00123.warc.gz | 663,563,813 | 4,782 | Câu hỏi:
In a reality show, there is a challenge for choosing teams. There are ten contestants: Hannah, Ryan, Jenny, Debstep, Markus, Ernesto, Georgia, Florida, California, and Jackson. The last two contestants remaining will be team captains. Team A will have the people who failed according to odd numbers (e.g. first faller, third, fifth). Team B will be the evens. Here are the clues: Hannah fails after Debstep || Ryan fails before Markus || Markus fails before Ernesto || Debstep fails after Ernesto || Florida fails after Georgia || California fails before Jackson || Jackson fails before Georgia || Hannah fails before California || Jenny fails after Florida. Who are the teams and who are the captains?
Đáp án:
* = captain Team A: Florida*, Jackson, Hannah, Ernesto, Ryan Team B: Jenny*, Georgia, California, Debstep, Markus
Chia sẻ với bạn bè | 193 | 852 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2018-30 | longest | en | 0.86658 |
https://www.quizzes.cc/metric/percentof.php?percent=4070&of=360000 | 1,591,296,560,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347445880.79/warc/CC-MAIN-20200604161214-20200604191214-00251.warc.gz | 834,589,838 | 4,279 | #### What is 4070 percent of 360,000?
How much is 4070 percent of 360000? Use the calculator below to calculate a percentage, either as a percentage of a number, such as 4070% of 360000 or the percentage of 2 numbers. Change the numbers to calculate different amounts. Simply type into the input boxes and the answer will update.
## 4070% of 360,000 = 14652000
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Calculating four thousand and seventy of three hundred and sixty thousand How to calculate 4070% of 360000? Simply divide the percent by 100 and multiply by the number. For example, 4070 /100 x 360000 = 14652000 or 40.7 x 360000 = 14652000
#### How much is 4070 percent of the following numbers?
4070% of 360000.01 = 1465200040.7 4070% of 360000.02 = 1465200081.4 4070% of 360000.03 = 1465200122.1 4070% of 360000.04 = 1465200162.8 4070% of 360000.05 = 1465200203.5 4070% of 360000.06 = 1465200244.2 4070% of 360000.07 = 1465200284.9 4070% of 360000.08 = 1465200325.6 4070% of 360000.09 = 1465200366.3 4070% of 360000.1 = 1465200407 4070% of 360000.11 = 1465200447.7 4070% of 360000.12 = 1465200488.4 4070% of 360000.13 = 1465200529.1 4070% of 360000.14 = 1465200569.8 4070% of 360000.15 = 1465200610.5 4070% of 360000.16 = 1465200651.2 4070% of 360000.17 = 1465200691.9 4070% of 360000.18 = 1465200732.6 4070% of 360000.19 = 1465200773.3 4070% of 360000.2 = 1465200814 4070% of 360000.21 = 1465200854.7 4070% of 360000.22 = 1465200895.4 4070% of 360000.23 = 1465200936.1 4070% of 360000.24 = 1465200976.8 4070% of 360000.25 = 1465201017.5
4070% of 360000.26 = 1465201058.2 4070% of 360000.27 = 1465201098.9 4070% of 360000.28 = 1465201139.6 4070% of 360000.29 = 1465201180.3 4070% of 360000.3 = 1465201221 4070% of 360000.31 = 1465201261.7 4070% of 360000.32 = 1465201302.4 4070% of 360000.33 = 1465201343.1 4070% of 360000.34 = 1465201383.8 4070% of 360000.35 = 1465201424.5 4070% of 360000.36 = 1465201465.2 4070% of 360000.37 = 1465201505.9 4070% of 360000.38 = 1465201546.6 4070% of 360000.39 = 1465201587.3 4070% of 360000.4 = 1465201628 4070% of 360000.41 = 1465201668.7 4070% of 360000.42 = 1465201709.4 4070% of 360000.43 = 1465201750.1 4070% of 360000.44 = 1465201790.8 4070% of 360000.45 = 1465201831.5 4070% of 360000.46 = 1465201872.2 4070% of 360000.47 = 1465201912.9 4070% of 360000.48 = 1465201953.6 4070% of 360000.49 = 1465201994.3 4070% of 360000.5 = 1465202035
4070% of 360000.51 = 1465202075.7 4070% of 360000.52 = 1465202116.4 4070% of 360000.53 = 1465202157.1 4070% of 360000.54 = 1465202197.8 4070% of 360000.55 = 1465202238.5 4070% of 360000.56 = 1465202279.2 4070% of 360000.57 = 1465202319.9 4070% of 360000.58 = 1465202360.6 4070% of 360000.59 = 1465202401.3 4070% of 360000.6 = 1465202442 4070% of 360000.61 = 1465202482.7 4070% of 360000.62 = 1465202523.4 4070% of 360000.63 = 1465202564.1 4070% of 360000.64 = 1465202604.8 4070% of 360000.65 = 1465202645.5 4070% of 360000.66 = 1465202686.2 4070% of 360000.67 = 1465202726.9 4070% of 360000.68 = 1465202767.6 4070% of 360000.69 = 1465202808.3 4070% of 360000.7 = 1465202849 4070% of 360000.71 = 1465202889.7 4070% of 360000.72 = 1465202930.4 4070% of 360000.73 = 1465202971.1 4070% of 360000.74 = 1465203011.8 4070% of 360000.75 = 1465203052.5
4070% of 360000.76 = 1465203093.2 4070% of 360000.77 = 1465203133.9 4070% of 360000.78 = 1465203174.6 4070% of 360000.79 = 1465203215.3 4070% of 360000.8 = 1465203256 4070% of 360000.81 = 1465203296.7 4070% of 360000.82 = 1465203337.4 4070% of 360000.83 = 1465203378.1 4070% of 360000.84 = 1465203418.8 4070% of 360000.85 = 1465203459.5 4070% of 360000.86 = 1465203500.2 4070% of 360000.87 = 1465203540.9 4070% of 360000.88 = 1465203581.6 4070% of 360000.89 = 1465203622.3 4070% of 360000.9 = 1465203663 4070% of 360000.91 = 1465203703.7 4070% of 360000.92 = 1465203744.4 4070% of 360000.93 = 1465203785.1 4070% of 360000.94 = 1465203825.8 4070% of 360000.95 = 1465203866.5 4070% of 360000.96 = 1465203907.2 4070% of 360000.97 = 1465203947.9 4070% of 360000.98 = 1465203988.6 4070% of 360000.99 = 1465204029.3 4070% of 360001 = 1465204070
1% of 360000 = 3600 2% of 360000 = 7200 3% of 360000 = 10800 4% of 360000 = 14400 5% of 360000 = 18000 6% of 360000 = 21600 7% of 360000 = 25200 8% of 360000 = 28800 9% of 360000 = 32400 10% of 360000 = 36000 11% of 360000 = 39600 12% of 360000 = 43200 13% of 360000 = 46800 14% of 360000 = 50400 15% of 360000 = 54000 16% of 360000 = 57600 17% of 360000 = 61200 18% of 360000 = 64800 19% of 360000 = 68400 20% of 360000 = 72000 21% of 360000 = 75600 22% of 360000 = 79200 23% of 360000 = 82800 24% of 360000 = 86400 25% of 360000 = 90000
26% of 360000 = 93600 27% of 360000 = 97200 28% of 360000 = 100800 29% of 360000 = 104400 30% of 360000 = 108000 31% of 360000 = 111600 32% of 360000 = 115200 33% of 360000 = 118800 34% of 360000 = 122400 35% of 360000 = 126000 36% of 360000 = 129600 37% of 360000 = 133200 38% of 360000 = 136800 39% of 360000 = 140400 40% of 360000 = 144000 41% of 360000 = 147600 42% of 360000 = 151200 43% of 360000 = 154800 44% of 360000 = 158400 45% of 360000 = 162000 46% of 360000 = 165600 47% of 360000 = 169200 48% of 360000 = 172800 49% of 360000 = 176400 50% of 360000 = 180000
51% of 360000 = 183600 52% of 360000 = 187200 53% of 360000 = 190800 54% of 360000 = 194400 55% of 360000 = 198000 56% of 360000 = 201600 57% of 360000 = 205200 58% of 360000 = 208800 59% of 360000 = 212400 60% of 360000 = 216000 61% of 360000 = 219600 62% of 360000 = 223200 63% of 360000 = 226800 64% of 360000 = 230400 65% of 360000 = 234000 66% of 360000 = 237600 67% of 360000 = 241200 68% of 360000 = 244800 69% of 360000 = 248400 70% of 360000 = 252000 71% of 360000 = 255600 72% of 360000 = 259200 73% of 360000 = 262800 74% of 360000 = 266400 75% of 360000 = 270000
76% of 360000 = 273600 77% of 360000 = 277200 78% of 360000 = 280800 79% of 360000 = 284400 80% of 360000 = 288000 81% of 360000 = 291600 82% of 360000 = 295200 83% of 360000 = 298800 84% of 360000 = 302400 85% of 360000 = 306000 86% of 360000 = 309600 87% of 360000 = 313200 88% of 360000 = 316800 89% of 360000 = 320400 90% of 360000 = 324000 91% of 360000 = 327600 92% of 360000 = 331200 93% of 360000 = 334800 94% of 360000 = 338400 95% of 360000 = 342000 96% of 360000 = 345600 97% of 360000 = 349200 98% of 360000 = 352800 99% of 360000 = 356400 100% of 360000 = 360000 | 3,094 | 6,442 | {"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} | 3.65625 | 4 | CC-MAIN-2020-24 | latest | en | 0.739713 |
https://www.wyzant.com/resources/answers/functions_and_domain?userid=8550839 | 1,529,275,497,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267859817.15/warc/CC-MAIN-20180617213237-20180617233237-00084.warc.gz | 964,652,784 | 13,503 | example: Calculate the limit of a function, while x approaches -1 f(x)=x+arccos(x/1-x^2)-1/(x^2)*e^(1/2*x^2) //the function arccosX (X is a real number) is defined for X∈[-1,1], but...
Let X : [0,1] → R. Find inverse images of all intervals [a,b] for X(x) = 2 *1[0,½) (x) + 3 * 1[½,1] (x) Can someone please provide a solution to this
If f(x) function with domain [0.infinty) and g(x) is (1-x)/floor(x) what is the domain of fx/gx
Make an input output table for the function y=-9.1x+7.22 using the domain -2, -1, 0, 1, 2 then state the range
Their are 2 answers. A. The independent variable is V; the dependent variable is h. The domain is D=? B. The Independent variable Is h; The dependent variable is V. The domain...
Please explain in detail how to solve this problem. thank you so much! Identifying functions from their equations Decide whether each relation defines y as a function of x. Give...
f(x)=(1-x)/7x, g(x)=1/(1+7x) Find (f*g)(x) and its domain Find (g*f)(x) and its domain | 327 | 999 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.28125 | 3 | CC-MAIN-2018-26 | latest | en | 0.748813 |
https://www.ademcetinkaya.com/2023/12/dbrgh-stock-strong-financial-position.html | 1,713,838,607,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296818452.78/warc/CC-MAIN-20240423002028-20240423032028-00616.warc.gz | 556,081,063 | 61,355 | Outlook: DigitalBridge Group Inc. 7.125% Series H is assigned short-term Caa2 & long-term B2 estimated rating.
AUC Score : What is AUC Score?
Short-Term Revised1 :
Dominant Strategy : Buy
Time series to forecast n: for Weeks2
Methodology : Ensemble Learning (ML)
Hypothesis Testing : Chi-Square
Surveillance : Major exchange and OTC
1The accuracy of the model is being monitored on a regular basis.(15-minute period)
2Time series is updated based on short-term trends.
Summary
DigitalBridge Group Inc. 7.125% Series H prediction model is evaluated with Ensemble Learning (ML) and Chi-Square1,2,3,4 and it is concluded that the DBRG^H stock is predictable in the short/long term. Ensemble learning is a machine learning (ML) technique that combines multiple models to create a single model that is more accurate than any of the individual models. This is done by combining the predictions of the individual models, typically using a voting scheme or a weighted average.5 According to price forecasts for 16 Weeks period, the dominant strategy among neural network is: Buy
Key Points
1. Ensemble Learning (ML) for DBRG^H stock price prediction process.
2. Chi-Square
3. Fundemental Analysis with Algorithmic Trading
4. What are the most successful trading algorithms?
5. Prediction Modeling
DBRG^H Stock Price Forecast
We consider DigitalBridge Group Inc. 7.125% Series H Decision Process with Ensemble Learning (ML) where A is the set of discrete actions of DBRG^H stock holders, F is the set of discrete states, P : S × F × S → R is the transition probability distribution, R : S × F → R is the reaction function, and γ ∈ [0, 1] is a move factor for expectation.1,2,3,4
Sample Set: Neural Network
Stock/Index: DBRG^H DigitalBridge Group Inc. 7.125% Series H
Time series to forecast: 16 Weeks
According to price forecasts, the dominant strategy among neural network is: Buy
F(Chi-Square)6,7= $\begin{array}{cccc}{p}_{a1}& {p}_{a2}& \dots & {p}_{1n}\\ & ⋮\\ {p}_{j1}& {p}_{j2}& \dots & {p}_{jn}\\ & ⋮\\ {p}_{k1}& {p}_{k2}& \dots & {p}_{kn}\\ & ⋮\\ {p}_{n1}& {p}_{n2}& \dots & {p}_{nn}\end{array}$ X R(Ensemble Learning (ML)) X S(n):→ 16 Weeks $∑ i = 1 n s i$
n:Time series to forecast
p:Price signals of DBRG^H stock
j:Nash equilibria (Neural Network)
k:Dominated move of DBRG^H stock holders
a:Best response for DBRG^H target price
Ensemble learning is a machine learning (ML) technique that combines multiple models to create a single model that is more accurate than any of the individual models. This is done by combining the predictions of the individual models, typically using a voting scheme or a weighted average.5 A chi-squared test is a statistical hypothesis test that assesses whether observed frequencies in a sample differ significantly from expected frequencies. It is one of the most widely used statistical tests in the social sciences and in many areas of observational research. The chi-squared test is a non-parametric test, meaning that it does not assume that the data is normally distributed. This makes it a versatile tool that can be used to analyze a wide variety of data. There are two main types of chi-squared tests: the chi-squared goodness of fit test and the chi-squared test of independence.6,7
For further technical information as per how our model work we invite you to visit the article below:
How do PredictiveAI algorithms actually work?
DBRG^H Stock Forecast (Buy or Sell) Strategic Interaction Table
Strategic Interaction Table Legend:
X axis: *Likelihood% (The higher the percentage value, the more likely the event will occur.)
Y axis: *Potential Impact% (The higher the percentage value, the more likely the price will deviate.)
Z axis (Grey to Black): *Technical Analysis%
Financial Data Adjustments for Ensemble Learning (ML) based DBRG^H Stock Prediction Model
1. If an entity originates a loan that bears an off-market interest rate (eg 5 per cent when the market rate for similar loans is 8 per cent), and receives an upfront fee as compensation, the entity recognises the loan at its fair value, ie net of the fee it receives.
2. Subject to the conditions in paragraphs 4.1.5 and 4.2.2, this Standard allows an entity to designate a financial asset, a financial liability, or a group of financial instruments (financial assets, financial liabilities or both) as at fair value through profit or loss provided that doing so results in more relevant information.
3. IFRS 17, issued in May 2017, amended paragraphs 2.1, B2.1, B2.4, B2.5 and B4.1.30, and added paragraph 3.3.5. Amendments to IFRS 17, issued in June 2020, further amended paragraph 2.1 and added paragraphs 7.2.36‒7.2.42. An entity shall apply those amendments when it applies IFRS 17.
4. For the purposes of applying the requirements in paragraphs 5.7.7 and 5.7.8, an accounting mismatch is not caused solely by the measurement method that an entity uses to determine the effects of changes in a liability's credit risk. An accounting mismatch in profit or loss would arise only when the effects of changes in the liability's credit risk (as defined in IFRS 7) are expected to be offset by changes in the fair value of another financial instrument. A mismatch that arises solely as a result of the measurement method (ie because an entity does not isolate changes in a liability's credit risk from some other changes in its fair value) does not affect the determination required by paragraphs 5.7.7 and 5.7.8. For example, an entity may not isolate changes in a liability's credit risk from changes in liquidity risk. If the entity presents the combined effect of both factors in other comprehensive income, a mismatch may occur because changes in liquidity risk may be included in the fair value measurement of the entity's financial assets and the entire fair value change of those assets is presented in profit or loss. However, such a mismatch is caused by measurement imprecision, not the offsetting relationship described in paragraph B5.7.6 and, therefore, does not affect the determination required by paragraphs 5.7.7 and 5.7.8.
*International Financial Reporting Standards (IFRS) adjustment process involves reviewing the company's financial statements and identifying any differences between the company's current accounting practices and the requirements of the IFRS. If there are any such differences, neural network makes adjustments to financial statements to bring them into compliance with the IFRS.
DBRG^H DigitalBridge Group Inc. 7.125% Series H Financial Analysis*
Rating Short-Term Long-Term Senior
Outlook*Caa2B2
Income StatementCBa2
Balance SheetCaa2B2
Leverage RatiosCCaa2
Cash FlowB3Caa2
Rates of Return and ProfitabilityCaa2Caa2
*Financial analysis is the process of evaluating a company's financial performance and position by neural network. It involves reviewing the company's financial statements, including the balance sheet, income statement, and cash flow statement, as well as other financial reports and documents.
How does neural network examine financial reports and understand financial state of the company?
References
1. D. Bertsekas. Dynamic programming and optimal control. Athena Scientific, 1995.
2. M. L. Littman. Friend-or-foe q-learning in general-sum games. In Proceedings of the Eighteenth International Conference on Machine Learning (ICML 2001), Williams College, Williamstown, MA, USA, June 28 - July 1, 2001, pages 322–328, 2001
3. Wager S, Athey S. 2017. Estimation and inference of heterogeneous treatment effects using random forests. J. Am. Stat. Assoc. 113:1228–42
4. J. Spall. Multivariate stochastic approximation using a simultaneous perturbation gradient approximation. IEEE Transactions on Automatic Control, 37(3):332–341, 1992.
5. Banerjee, A., J. J. Dolado, J. W. Galbraith, D. F. Hendry (1993), Co-integration, Error-correction, and the Econometric Analysis of Non-stationary Data. Oxford: Oxford University Press.
6. K. Tumer and D. Wolpert. A survey of collectives. In K. Tumer and D. Wolpert, editors, Collectives and the Design of Complex Systems, pages 1–42. Springer, 2004.
7. A. K. Agogino and K. Tumer. Analyzing and visualizing multiagent rewards in dynamic and stochastic environments. Journal of Autonomous Agents and Multi-Agent Systems, 17(2):320–338, 2008
Frequently Asked QuestionsQ: Is DBRG^H stock expected to rise?
A: DBRG^H stock prediction model is evaluated with Ensemble Learning (ML) and Chi-Square and it is concluded that dominant strategy for DBRG^H stock is Buy
Q: Is DBRG^H stock a buy or sell?
A: The dominant strategy among neural network is to Buy DBRG^H Stock.
Q: Is DigitalBridge Group Inc. 7.125% Series H stock a good investment?
A: The consensus rating for DigitalBridge Group Inc. 7.125% Series H is Buy and is assigned short-term Caa2 & long-term B2 estimated rating.
Q: What is the consensus rating of DBRG^H stock?
A: The consensus rating for DBRG^H is Buy.
Q: What is the forecast for DBRG^H stock?
A: DBRG^H target price forecast: Buy | 2,209 | 8,985 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 2, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.578125 | 3 | CC-MAIN-2024-18 | latest | en | 0.87852 |
http://www.emis.de/journals/JIS/keith.html | 1,529,384,950,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267861899.65/warc/CC-MAIN-20180619041206-20180619061206-00554.warc.gz | 408,908,910 | 11,045 | Journal of Integer Sequences, Vol. 1 (1998), Article 98.1.6
## On Repdigit Polygonal Numbers
### Mike Keith 4100 Vitae Springs Road Salem, OR 97306 Email address: domnei@aol.com
Abstract: We consider the problem of determining which polygonal numbers are repdigits (numbers consisting of a single repeated digit). An efficient algorithm for finding repdigit polygonal numbers is presented and used to provide a complete characterization of all 1526 such numbers with 50 or fewer digits. Several other new and intriguing integer sequences (such as the sequence of so-called primitive solutions) are also introduced.
## 1. Introduction
The polygonal numbers are illustrated in the figure below. The nth k-sided polygonal number, P(k,n), is the number of counters that can be arranged into a k-sided polygon with n counters along each side.
n=2 3 4
k=2 oo ooo oooo
2 3 4 ...
o
o o o
k=3 o o o o o o
o o o o o o o o o
3 6 10 ...
o o o o
o o o o o o o
k=4 o o o o o o o o o
o o o o o o o o o
4 9 16 ...
o
o o o
o o o o o o o
k=5 o o o o o o o o o o
o o o o o o o o o
o o o o o
o o o o
5 12 22 ...
Figure 1. The first few polygonal numbers
A formula for P(k,n) is:
P(k,n) = n((k-2)(n-1) + 2)/2 (n>=2 and k>=2) (1)
A repdigit is a number, like 3 or 55 or 999999, consisting of repetitions of a single digit. Some polygonal numbers are also repdigits, such as P(5,4)=22 shown in Figure 1, or more dramatically
P(8925662618878671, 387) = 666666666666666666666.
In this paper we consider some properties of these repdigit polygonal (RP) numbers as well as some new integer sequences which result from their study.
## 2. Finding RP Numbers, and the Combination Sequence
The primary problem is to determine which polygonal numbers are also repdigits. The converse is easy - every repdigit number is trivially a polygonal number, because every integer r is equal to both P(2,r) and P(r,2), as can be seen by the fact that the first row and column of Figure 1 are just the integers in order. However, nontrivial representations are also possible, as shown by (see Figure 1)
P(2,22) = P(5,4) = P(22,2) = 22
or
P(2,666) = P(3,36) = P(46,6) = P(223,3) = P(666,2) = 666.
We define the combination number of a given repdigit r, c(r), to be the number of pairs n, k such that P(n,k) = r. The above examples show that c(22)=3 and c(666)=5. The combination sequence is the sequence of combination numbers pertaining to the successive repdigits (which are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11, 22, 33, 44, 55, 66, 77, 88, 99, 111, 222, 333, etc., sequence A010785 in the On-Line Encyclopedia of Integer Sequences).
Here is an efficient procedure for computing the combination number of a given repdigit. Every repdigit is just some decimal digit, d, times a repunit number (a number of the form 111...1). We denote the repunit with m decimal digits by Rm. Then equation (1) becomes
dRm = n((k-2)(n-1) + 2)/2
Solving for k, we get
k = ( (2dRm)/n + 2n - 4 )/(n-1)
Denote the quotient (2dRm)/n by the symbol q. For a given Rm to be a polygonal number, we see that it is necessary and sufficient that the two following conditions hold:
2dRm = 0 (mod n), (a) q + 2n - 4 = 0 (mod n-1). (b)
Condition (a) is required for q to be an integer, and (b) is required for k to be an integer. Note that (b) can be rewritten as
q = 2 (mod n-1). (b')
Condition (a) means that the only possible values of n are the divisors of 2dRm. We must therefore factorize Rm. Fortunately, the prime factorizations of the repunits can be readily obtained from a number of sources, or calculated easily (at least for the first hundred or so repunits) using modern factorization algorithms. We assume that a table of repunit factorizations is provided. We can now find all (k,n) pairs such that P(k,n) equals a given repdigit number dRm by the following simple procedure:
Put all the prime factors of R(m) in a list.
Adjoin 2 and the prime factors of d.
Form all possible divisors n of 2*d*R(m) by taking all
combinations of primes from this list.
For each trial divisor n, compute q.
If q = 2 mod (n-1), this is a success: (k,n) is an RP number.
This algorithm must be programmed in a language that supports arbitrary-precision arithmetic. We used UBASIC, and were able to find all repdigit polygonal numbers with 50 or fewer decimal digits in a couple of hours on a home PC. From this we can tabulate the first 500 values of the combination sequence (A033618), which are as follows:
d
1 2 3 4 5 6 7 8 9
m
1 0 1 2 2 2 3 2 2 3
2 2 3 3 2 4 5 2 3 3
3 4 3 4 3 4 5 3 3 3
4 3 2 3 3 3 6 3 2 3
5 2 3 3 3 3 4 2 3 3
6 7 3 4 3 4 5 4 3 4
7 2 2 3 3 3 4 3 3 3
8 3 4 3 2 3 6 2 3 3
9 6 3 4 3 4 5 4 3 3
10 3 2 3 3 3 5 3 2 4
11 2 3 3 3 4 4 2 3 3
12 6 3 5 3 6 5 4 3 3
13 5 2 3 3 3 4 3 3 3
14 3 4 3 2 3 6 2 3 3
15 5 3 4 3 4 5 3 4 3
16 5 3 5 3 3 5 4 2 3
17 2 3 3 3 3 4 2 3 3
18 6 3 4 4 4 5 5 3 3
19 2 2 3 3 3 4 3 3 3
20 3 4 3 2 4 6 3 3 3
21 5 3 4 3 4 7 3 3 3
22 3 2 3 3 3 5 3 2 3
23 2 3 3 3 3 4 2 3 3
24 6 3 4 3 4 5 4 4 3
25 3 2 3 3 3 4 3 3 3
26 3 4 3 2 3 6 2 3 5
27 6 4 4 3 4 5 4 3 3
28 4 2 4 3 3 5 3 2 3
29 3 3 3 3 4 4 2 3 3
30 6 3 4 3 4 5 4 3 5
31 2 2 3 3 3 4 3 3 3
32 3 4 3 2 3 6 3 3 3
33 6 3 4 3 4 5 4 3 3
34 3 2 3 3 3 5 3 2 3
35 2 3 3 3 4 4 2 3 3
36 9 3 5 3 4 5 5 3 4
37 2 2 3 3 3 4 3 3 3
38 3 4 3 2 4 6 3 3 3
39 5 3 4 3 4 6 3 3 3
40 4 2 3 3 3 5 4 2 3
41 3 3 3 3 3 4 2 3 3
42 6 3 6 3 4 5 4 3 4
43 3 2 3 3 3 4 3 3 3
44 3 4 3 3 3 6 2 3 4
45 7 3 4 3 4 6 7 3 3
46 4 2 3 3 3 6 3 2 3
47 2 3 3 3 4 4 2 3 3
48 6 3 4 3 4 6 5 3 3
49 4 2 3 3 3 4 3 3 3
50 3 4 3 2 3 6 2 3 3
Table 1. c(r), the number of ways of expressing each repdigit, r = dRm ,
as a polygonal number.
Every term in this sequence (after the first two) is at least 2, since every repdigit number r equals both P(r,2) and P(2,r). The largest term so far in this sequence is the (unique) 9 at m=36, d=1. This says that there are 9 different ways of representing the number 11111 11111 11111 11111 11111 11111 11111 1 as a polygonal number. The value 8 does not yet occur, although presumably it will eventually. The average of all the terms in the sequence so far is close to 3 (about 3.06).
Here are a few related sequences. Summing up each row, we obtain the number of polygonal numbers which are m-digit repdigits (for m=1, 2, 3, ..., A033702):
17 27 32 28 26 37 26 29 35 28 27 38 29 29 34 33 26 37 26 31 35 27 26 36 27
31 36 29 28 37 26 30 35 27 27 41 26 31 34 29 27 38 27 31 40 29 27 37 28 29
The partial sums of this sequence give the number of polygonal numbers which are repdigits with m or fewer digits (A033703):
RPN(m) = 17 44 76 104 130 167 193 222 257 285
312 350 379 408 442 475 501 538 564 595
630 657 683 719 746 777 813 842 870 907
933 963 998 1025 1052 1093 1119 1150 1184 1213
1240 1278 1305 1336 1376 1405 1432 1469 1497 1526
In particular, there are precisely 1526 distinct repdigit polygonal numbers with 50 or fewer digits.
## 3. Primitive Repdigit Polygonal Numbers
So far we have not actually displayed a listing of the 1526 repdigit polygonal numbers with 50 or less digits. The reason for this is that we can describe these numbers using a much smaller list, by recognizing that many of these RP numbers are related.
For example, consider:
P(2,3) = 33
P(12,3) = 333
P(112,3) = 3333
etc.
and
P(5,4) = 22
P(3705,4) = 22222
P(3703705,4) = 22222222
etc.
In both cases, as we shall see below, there is an infinite sequence of RP numbers, where n remains constant, k steadily increases and the repdigit number has the same base digit d but gets p digits longer at each step. We refer to p as the period of the infinite RP number sequence.
In turns out that almost every repdigit polygonal number is a member of one of these infinite sequences. We call the first term in such a sequence the primitive RP number. We distinguish between two types: a simple primitive RP number, in which k=2 or n=2, and a fancy primitive number (the rest). The reason for this distinction is that all the simple primitives are obvious (since the k=2 and n=2 polygonal numbers are just the integers in order), and hence the fancy primitives are the only ones that really need to be enumerated.
Before enumerating all the primitive RP numbers less than 1050, we first explain why these infinite sequences of solutions occurs.
Suppose we have any RP number, which as we have seen must satisfy conditions (a) and (b'). Also, n is a product of certain prime divisors of 2dRm. In general n will consist of zero or more factors of 2d combined with zero or more factors of Rm.
Lemma 1: If Rm is the smallest repunit divisible by a prime f, then Rcm is also divisible by f, for all c >= 1.
Proof: By induction on c. If Rcm is divisible by f, then so is R(c+1)m , because
R(c+1)m = 10mRcm + Rm
and both terms on the right side are divisible by f (by the induction hypothesis).
For example, the 3rd repunit, 111, factors as 3 x 37. Both of these factors are present in the 6th repunit, 111111 (which = 3 x 7 x 11 x 13 x 37), the 9th repunit 111111111 (which = 3 x 3 x 37 x 333667), and so on.
Consider now our RP number, for which n contains a subset of the prime factors of the repunit Rm. Although m may not be the smallest index in which each of these prime factors occurs, we know from the lemma that each prime factor in this subset will occur among the higher repunits with some period. Define Prep to be the LCM of all these periods. Then it must be the case that every Prep-th repunit after this one is divisible by n, since each of these contains the prime factors needed for n to divide it evenly.
In summary: if we start from a given RP number and form larger ones in which d and n remain the same and the length of the repunit increases in steps of Prep, every one of these will satisfy condition (a).
Example: consider P(41139,74)=111111111. Since n = 74 = 2 x 37, and the 2 can be obtained from 2d, the only repunit factor n contains is 37. We know from the previous example that 37 occurs as a prime factor of every 3rd repunit (because the lowest repunit in which it appears is R3); therefore Prep = 3. This means that 111111111111 and every 3rd repunit thereafter will be divisible by 37, and will therefore satisfy condition (a). For instance, 2 x 111111111111 is exactly divisible by 74.
What about condition (b')?
We have a series of repunits which satisfy condition (a): r0 = Rm, r1 = Rm+Prep, r2 = Rm+2Prep, etc., and at the first step condition (b') is satisfied: q = 2dr0/n = 2 mod n-1. We ask: what is the sequence of q values (q0, q1, q2, ...) that correspond to the repdigits r0, r1, r2, ...? To answer this, note that each repunit in the sequence is related to the previous one by
ri = 10Prepri-1 + RPrep ,
i.e.
(2dri)/n = (2d10Prepri-1)/n + (2dRPrep)/n ,
or
qi = qi-110Prep + q0 mod 10Prep
For (b') to be satisifed we need qi = 2 mod n-1 for some larger i. This will be the case (and there will be an infinite sequences of cases for which it is true) if the following condition holds:
Ring Period Condition: Working in the ring of integers mod n-1, start with the value 2 and successively apply the linear recurrence qi = aqi-1 + b, where a = 10Prep mod n-1 and b = (q0 mod 10Prep) mod n-1. If the sequence of values q0 (= 2) -> q1 -> q2 -> ... eventually returns to the value 2 (which means it satisfies condition (b')) then the primitive repdigit polygonal number under consideration generates an infinite sequence of RP numbers.
We refer to the period with which 2 repeats in the sequence of q's as the ring period Pring. Because the sequence of q's is itself spaced with a period of Prep within the repdigits, the full period with which additional RP numbers appear beyond a primitive one is the product of these two periods: p = PrepPring.
Continuing the previous example, we may now determine the sequence of RP numbers generated by the primitive solution P(41139,74)=111111111, for which Prep = 3. We compute a = 103 mod 73 = 51 and b = ((2 x 111111111)/74 mod 103 ) mod 73 = 2. Starting with 2 and applying the function 51x + 2 iteratively, we produce:
starting value: 2 51*2+2 = 104 = 31 mod 73 51*31+2 = 1583 = 50 mod 73 51*50+2 = 2552 = 70 mod 73 51*70+2 = 3572 = 68 mod 73 51*68+2 = 3470 = 39 mod 73 51*39+2 = 1991 = 20 mod 73 51*20+2 = 1022 = 0 mod 73 51*0 +2 = 2 = 2 mod 73
so Pring = 8. The full period for this primitive solution is therefore 3 x 8 = 24. We obtain the following sequence of RP numbers:
P(41139,74) = 111111111
P(41137027438397301411000041139,74) = 111111111111111111111111111111111
etc.
where each repdigit in the sequence (and, consequently, each value of k) is 24 digits longer than the previous one.
We can now concisely describe all RP numbers of 50 digits or less. First, take all those which are generated by the simple primitive solutions with k=2 and n equal to some repdigit number. Here are the periods of the small simple primitive solutions:
Value of d
m 1 2 3 4 5 6 7 8 9
1 1 3 1 1 3 6 **
2 2 6 ** 42 54 6 18 84 42
3 6 48 123 663 69 18 96 2658 498
4 12 2220 336 2220 2776 420 972 17772 1428
5 20 36990 480 31710 33810 495 4860 49380 3205
Table 2. Periods of the simple primitive RP numbers 3,4,5,...99999 with k=2.
All of the larger simple primitive solution have a period of at least 498, and so are not relevant for RP numbers with 50 digits or less. There is one exception: the d=1 primitives, all of which (as can be seen from the table) have relatively small periods. In fact, it is easy to show that those solutions have a period of exactly m(m-1).
In addition to these simple primitives, we also have to consider the remaining simple primitives, which are simply those of the form k=d, n=2, repdigit=d, and all the RP numbers generated by these (which are of the form k=ddd...d, n=2, repdigit=ddd...d).
Finally, take all those generated by the fancy primitive solutions. Table 3 below gives the complete list of fancy primitive RP numbers of 50 digits or less (A033704 and A033705) with their periods.
k n RP number Period
3 3 6 1
4 3 9 1
5 4 22 3
3 10 55 9
3 11 66 2
16 4 88 3
38 3 111 3
9 6 111 3
75 3 222 3
11 9 333 3
149 3 444 3
186 3 555 3
3 36 666 6
260 3 777 3
297 3 888 3
9 44 6666 42
1589 8 44444 6
531 21 111111 6
131 42 111111 30
1475 33 777777 6
514 63 999999 30
41139 74 111111111 24
21604940 9 777777777 9
65359479 18 9999999999 16
170677592 63 333333333333 30
933706818 35 555555555555 48
5378862 455 555555555555 678
806321563 53 1111111111111 78
360633274 79 1111111111111 78
199660579 106 1111111111111 78
3220611916266 24 888888888888888 66
63890006966 187 1111111111111111 240
975514583945 68 2222222222222222 528
8944083 27302 3333333333333333 104368
34829977467 438 3333333333333333 792
57189542483662 17 7777777777777777 16
1610305958132047 24 444444444444444444 66
8925662618878671 387 666666666666666666666 1344
3561667376774099913 707 888888888888888888888888 96
880855486848827581349 477 99999999999999999999999999 624
77645779951859616429849 54 111111111111111111111111111 351
633111744222855333966447 27 222222222222222222222222222 54
8210180623973727422003286 29 3333333333333333333333333333 84
2183081749534812567698 3191 11111111111111111111111111111 812
233754090696587190275829829 93 999999999999999999999999999999 330
56287290329843521332883037 189 999999999999999999999999999999 138
9649728635845433403776352377 402 777777777777777777777777777777777 6600
884149845715851922583839509122 355 55555555555555555555555555555555555 6090
10943672915503901419394377140858 143 111111111111111111111111111111111111 210
2178649237472766884531590413943357 18 333333333333333333333333333333333333 48
2959580585151361407069169626247251820 73 7777777777777777777777777777777777777777 72
3265092891892774349430241290364710878 83 11111111111111111111111111111111111111111 205
3630844752340079442883181200938210286 429 333333333333333333333333333333333333333333 318
74681483472987707427820346223357380773 173 1111111111111111111111111111111111111111111 903
25972676744065243363981091891330320502833 93 111111111111111111111111111111111111111111111 330
4357298474945533769063180827886710239651418 18 666666666666666666666666666666666666666666666 48
103662238808180431531091267196824973714220 123 777777777777777777777777777777777777777777777 60
411305012045361067042716963393853927962867 62 777777777777777777777777777777777777777777777 60
408040686552937566510631908128823341235 1953 777777777777777777777777777777777777777777777 180
254200666005744935051729835532169094283029 94 1111111111111111111111111111111111111111111111 690
65662037493023408516366262845136084572704293 143 666666666666666666666666666666666666666666666666 210
39067230797479382268946630256007563415882394 239 1111111111111111111111111111111111111111111111111 336
Table 3. All fancy primitive repdigit polygonal numbers less than 1050.
As an example, we derive the entry in Table 1 which says that there are 9 manifestations of r = 11111 11111 11111 11111 11111 11111 11111 1 (36 1's) as an RP number. First, there is the simple primitive solution (k,n) = (r, 2). Looking at Table 2, we see that the simple primitive solution (11,2) has period 2, so it generates solutions with any even number of digits, including 36. The 6-digit simple primitive solution (111111,2), which is just beyond the end of Table 2, has, as described earlier, period 6 x 5 = 30, so it also generates a 36-digit solution. There is also the obvious simple primitive solution (2, r).
To complete the list, look in Table 3 for RP numbers consisting of 1's with the correct period so that a 36-digit solution can be generated. We find five possibilities: (38,3) and (9,6) with length 3 and period 3, (531,21) with length 6 and period 6, (131,42) with length 6 and period 30, and the n=143 36-digit solution. Thus all nine solutions can be found from Tables 2 and 3.
Two of the simple primitive solutions (the ones marked with **: (2,9) and (2,33)) in Table 2 do not have finite ring periods, and so do not generate any additional solutions. For example, for (2,9) the recurrence gets stuck in a cycle of length one at the value 6. Are there other solutions with this property?
Note from Table 3 that one of the "fancy" things about the fancy primitives is the progression of k values, which also form an interesting sequence, A033706 (as do the values of n, A033707):
3, 4, 5, 3, 3, 16, 38, 9, 75, 11, 149, 186, 3, 260, 297, 9, 1589, 531, 131, 1475, 514, 41139, 21604940, ...
It has been noted since 1979 (see [2]) that all primitive RP numbers with k>2 tend to have large k and small n. (If a primitive solution has this property, then all RP numbers derived from it will also. So this is equivalent to saying that all RP numbers with k>2 tend to have large k and small n.)
In particular, the following conjecture, made in [2], is now strongly supported by the numerical evidence in Table 3 (although we still do not have a proof).
Conjecture. The only RP numbers with k > 2 and n > k are P(3,10), P(3,11), P(3,36), and P(9,44).
A related conjecture comes from defining the wickedness of an RP number to be the value of n/k.
Conjecture. The most wicked RP number with k>2 is the "Beast number", 666 = (3,36), with n/k = 12.
Another remarkable solution in Table 3 is (8944083, 27302) = 3333333333333333, whose k value is the largest among the fancy primitives so far.
Finally, define a simple RP number to be one generated from a simple (or trivial) primitive solution, and a fancy RP number to be one generated from a fancy primitive solution. What is the distribution of simple versus fancy RP numbers? Here are the two sequences (number of RP numbers of m digits, A033708, and their partial sums, A033709) for simple:
Numbers with exactly m digits:
15 21 21 24 21 22 24 24 22 24 21 22 24 24 22 25 21 22 24 25 23 24 21 22 25
24 22 25 21 22 24 24 22 24 21 23 24 25 23 25 21 22 24 26 23 24 21 22 25 24
Numbers with m or fewer digits:
15 36 57 81 102 124 148 172 194 218
239 261 285 309 331 356 377 399 423 448
471 495 516 538 563 587 609 634 655 677
701 725 747 771 792 815 839 864 887 912
933 955 979 1005 1028 1052 1073 1095 1120 1144
and fancy (A033710 and A033711) RP numbers :
Numbers with exactly m digits:
2 6 11 4 5 15 2 5 13 4 6 16 5 5 12 8 5 15 2 6 12 3 5 14 2
7 14 4 7 15 2 6 13 3 6 18 2 6 11 4 6 16 3 5 17 5 6 15 3 5
Numbers with m or fewer digits:
2 8 19 23 28 43 45 50 63 67
73 89 94 99 111 119 124 139 141 147
159 162 167 181 183 190 204 208 215 230
232 238 251 254 260 278 280 286 297 301
307 323 326 331 348 353 359 374 377 382
As can be seen from these sequences, there are many fewer fancy RP numbers than simple ones. Of the 1526 RP numbers with 50 digits or less, only 382 (= 25.03%) are fancy ones.
## References
[1] D. W. Ballew and R. C. Weger, "Repdigit Triangular Numbers", Journal of Recreational Mathematics, Vol. 8, No. 2, p. 96, 1975.
[2] M. Keith, "Repdigit Polygonal Numbers", Journal of Recreational Mathematics, Vol. 12, No. 1, p. 9, 1979.
Received May 20, 1998; published in Journal of Integer Sequences May 25, 1998.
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posted by .
A flat thin non-conducting ring has an inner radius R1 and an outer radius R2 . The disk is uniformly charged with charge per unit area σ(>0) .
(a) What is the electric field at the center, P, of the ring? Express your answer in terms of the following variables, if necessary, R1, R2, σ and the constant ϵ0 (if needed, enter pi for π, epsilon_0 for ϵ0, sigma for σ, R_1 for R1 .. etc).
(b)What is the magnitude of the electric field at point A which is a distance x above the point P? The line AP is perpendicular to the plane of the ring. Express your answer in terms of the following variables, if necessary, R1, R2, σ, x and the constant ϵ0 (if needed, enter pi for π, epsilon_0 for ϵ0, sigma for σ, R_1 for R1 .. etc).
• Physics -
a) 0
b) ((sigma*x)/(2*epsilon_0))*(1/(sqrt(x^2+R_1^2))-1/(sqrt(x^2+R_2^2)))
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Voting power distribution (Posted on 2005-01-24)
The five owners of Plexus and Co. are voting on a very important decision (Top secret!). Each must vote for or against the decision. They don't necessarily own equal shares of the company, so they don't necessarily have equal voting power. For example, one person might have 5 votes and the other four have 1 vote each. However, it is distributed in a way that a tie is impossible. Obviously, everyone has positive voting power.
There are 2^5=32 different ways that the five people can vote (such as YYNNY, YNNYY, NNNNN, ...). Each way will result in favor or against the decision, depending on how the voting power is distributed.
There are 2^32 different combinations of the 32 outcomes, but not every combination is possible. For example, it is impossible for YYYNN to be in favor of the decision while YYYYN is against the decision, no matter how the voting power is distributed.
Out of the 2^32 different combinations, how many are possible, remembering that combinations where a tie is possible are not allowed?
See The Solution Submitted by Tristan Rating: 3.7143 (7 votes)
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Monotone functions Comment 13 of 13 |
81. The answer involves monotone functions.
Posted by Math Man on 2013-07-28 22:00:57
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# LCR Meters
LCR Meters are electronic test equipment used to measure the inductance (L), capacitance (C) and resistance (R) of an electronic component.
When there is a change in the current flowing through a conductor, a corresponding change is induced in the voltage in it and in conductors surrounding it. This property is known as inductance. The ability of a body or a conductor to store electrical charge is known as capacitance. The opposition that a conductor offers to passaging electric current through it is called resistance.
Most LCR meters use AC signals to measure the impedance of components. When using an LCR meter make sure you are using the correct settings as the wrong settings can cause poor measurement accuracy. Both analogue and digital LCR meters are available. The analogue is less expensive, and the digital variety are seen to be more accurate.
## What are the uses of an LCR Meter?
A digital LCR meter is used to measure the impedance flowing through a Device Under Test (DUT). It measures the voltage (V) across it, the current (I) flowing through it, and the phase angle between current and voltage. Subsequently, you can determine all the impedance parameters from these three factors.
Thus, an LCR meter measures the following parameters related to a circuit:
• Inductance
• Capacitance
• Resistance
• Dissipation factor
• Quality factor
• Current
• Voltage
• Phase angle between the current and voltage
• Conductance
## What are the types of LCR Meters?
Handheld LCR meters: Small, lightweight and portable. They have multiple test frequency and the data it captures can be transferred to a PC via a USB port. Typically used in field operations.
Benchtop LCR meters: Big and bulky. They can be operated on programmable frequencies and can be controlled via a computer.
## Techniques used with LCR Meters
Bridge Method: This method is used for measuring frequencies below 100 kHz.
LCR Measurement by Current-Voltage Technique: Using this method, the LCR measurement of a component is done by measuring the current and voltage. Then the impedance values are found out from these two quantities.
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https://codeshive.com/product/ve311-electronic-circuits-homework-03-solved/ | 1,718,335,980,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861520.44/warc/CC-MAIN-20240614012527-20240614042527-00576.warc.gz | 160,996,026 | 22,532 | # VE311 Electronic Circuits Homework 03 solved
\$35.00
## Description
5/5 - (1 vote)
• Diodes and Rectifiers
1. A diode is doped with NA = 1019 /cm3 on the p-type side and ND = 1018/cm3
on the n-type side. (a) What is the depletion-layer width wdo ? (b) What are
the values of xp and xn? (c) What is the value of the built-in potential of the
junction? (d) What is the value of EMAX? Use Eq. (3.3) and Fig. 3.5.
2. A diode has wdo = 0.4 µm and φj = 0.85 V. (a) What reverse bias is required
to triple the depletion-layer width? (b) What is the depletion region width if a
reverse bias of 7 V is applied to the diode?
3. Suppose a drift current density of 5000 A/cm2
exists in the neutral region on the
p-type side of a diode that has a resistivity of 2.5 Ω·cm. What is the electric field
needed to support this drift current density?
4. Suppose that NA(x) = No exp(−x/L) in a region of silicon extending from x =
0 to x = 12 µm, where No is a constant. Assume that p(x) = NA(x). Assuming
that jp must be zero in thermal equilibrium, show that a built-in electric field
must exist and find its value for L = 1 µm and No = 1018/cm3
5. A diode has n=1.05 at T =320 K. What is the value of n·VT ? What temperature
would give the same value of n·VT if n=1.00?
6. A diode has IS = 10−17A and n = 1.07. (a) What is the diode voltage if the diode
current is 70 µA? (b) What is the diode voltage if the diode current is 5 µA? (c)
What is the diode current for vD = 0 V? (d) What is the diode current for vD =
-0.075 V? (e) What is the diode current for vD = -5 V?
7. The saturation current for diodes with the same part number may vary widely.
Suppose it is known that 10−14 A ≤ IS ≤ 10−12 A. What is the range of forward
voltages that may be exhibited by the diode if it is biased with iD=1mA?
8. A diode with IS = 2.5×10−16A at 30◦C is biased at a current of 1 mA. (a) What
is the diode voltage? (b) If the diode voltage temperature coefficient is -2 mV/K,
what will be the diode voltage at 50◦C?
1
9. A diode has a doping of ND = 1020/cm3 on the n-type side and NA = 1018/cm3
on the p-type side. What are the values of wdo and φj? What is the value of wd
at a reverse bias of 5 V? At 25 V?
10. (a) Plot the load line and find the Q-point for the diode circuit in Fig.P3.57 if V
=10 V and R = 5 kΩ . Use the i − v characteristic in Fig. P3.42. (b) Repeat for
V = -10 V and R = 5 kΩ . (c) Repeat for V = -2 V and R = 2 kΩ
11. Use the i − v characteristic in Fig.1a. (a) Plot the load line and find the Q-point
for the diode circuit in Fig. 1 if V = 6V and R = 4kΩ. (b) For V = -6 V and R
= 3 kΩ.(c) For V = -3 V and R = 3 kΩ. (d) For V = 12 V and R = 8 kΩ. (e)
For V = -25 V and R = 10kΩ
Figure 1: i − v diode characteristics and schematic diagram for the diode…
12. Simulate the behavior of the half-wave rectifier in Fig.2 for vI = 10 sin 120πt,
R = 0.025 Ω and C = 0.5 F. (Use IS = 10−10 A, RS=0 and RELTOL = 10−6
.)
Compare the simulated values of dc output voltage, ripple voltage, and peak diode
current to hand calculations. Repeat simulation with RS =0.02Ω.
Figure 2: Half-wave rectifier
2
13. What are the dc output voltages V1 and V2 for the rectifier circuit in Fig. 3 if
vI = 40 sin 377t and C = 20 000 µF?
Figure 3: Full-wave rectifier
3 | 1,128 | 3,246 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-26 | latest | en | 0.900055 |
https://community.smartsheet.com/discussion/65203/please-advise-if-you-can-help-me-convert-this-excel-formula-to-a-smartsheet-formula | 1,718,285,431,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861451.34/warc/CC-MAIN-20240613123217-20240613153217-00018.warc.gz | 166,615,883 | 110,196 | # Please advise if you can help me convert this Excel formula to a Smartsheet formula.
Options
I keep getting #UNPARSEABLE when using the following formula:
=IF([CUSTODY DATE]7>1/1/2000,IF(\$MILEAGE\$2<=([CUSTODY DATE]7+2),(125+75+40),(\$MILEAGE\$2-([CUSTODY DATE]7+2)*15+(125+75+40)), )
I was told to make this correction, however, I am still having the same trouble:
Based on the IF formula provided, \$ MILEAGE \$ 2 seems to be causing the #UNPARSEABLE error. If this is a column, you will need to change it to [MILEAGE](row number or @row).
Any help is appreciated.
• ✭✭✭✭✭✭
Options
Try changing your date to fit the DATE function. Smartsheet does not recognize "1/1/2000" as a date. It recognizes it as 1 divided by 1 divided by 2000. To use dates in Smartsheet formulas, the DATE function is required.
DATE(yyyy, mm, dd)
=IF([CUSTODY DATE]7>DATE(2000, 1, 1),IF(\$MILEAGE\$2<=[CUSTODY DATE]7+2,125+75+40,(\$MILEAGE\$2-([CUSTODY DATE]7+2)*15+(125+75+40)), )
There is also an extra comma towards the end that is causing the actual #UNPARSEABLE error.
Here is the formula with the DATE function included and the comma (and some unnecessary parenthesis) removed:
=IF([CUSTODY DATE]7 > DATE(2000, 1, 1), IF(\$MILEAGE\$2 <= [CUSTODY DATE]7 + 2,125 + 75 + 40, \$MILEAGE\$2 - ([CUSTODY DATE]7 + 2) * 15 + (125 + 75 + 40)))
• ✭✭
Options
You do have to correct your reference to the MILEAGE column.
Looks like you're using a nested IF formula (an IF within an IF) and didn't complete the first IF structure.
Your value if true is another IF which appears complete, it's got an expression, a value if true, and a value if false.
=IF([CUSTODY DATE]2 > 1 / 1 / 2000, IF(MILEAGE2 <= ([CUSTODY DATE]2 + 2), (125 + 75 + 40), (MILEAGE2 - ([CUSTODY DATE]2 + 2) * 15 + (125 + 75 + 40)),
Your value if False on the first IF is missing:
=IF([CUSTODY DATE]2 > 1 / 1 / 2000, IF(MILEAGE2 <= ([CUSTODY DATE]2 + 2), (125 + 75 + 40), (MILEAGE2 - ([CUSTODY DATE]2 + 2) * 15 + (125 + 75 + 40)), 12)), #Value if False should be here)
From the help site:
IF Function
Evaluates a logical expression and returns one value when true or another when false.
Sample Usage
IF([Due Date]1 > [Due Date]2, "Date 1 is Larger", "Date 2 is Larger")
Syntax
IF(logical_expressionvalue_if_true[value_if_false])
• logical_expression—The expression to evaluate. Must be true or false.
• value_if_true—The value or formula to return if the logical expression is true.
• value_if_false—[optional] The value or formula to return if the logical expression is false. If omitted, a blank value is returned.
• ✭✭✭✭✭✭
Options
@andyrami Having an IF function nested in the "value if true" portion of another IF function is similar to using "IF(AND(..................),"
I do this quite frequently if I am using multiple variables associated to the same variable. Instead of
=IF(AND(variable 1 = 1, variable 2 = 2), 2, IF(AND(variable 1 = 1, variable 3 = 3), 3))
I just use
=IF(variable 1 = 1, IF(variable 2 = 2, 2, IF(variable 3 = 3, 3)))
Because I am repeating "variable 1" for multiple portions of the remaining IF statements, doing it this way saves a few key strokes. It doesn't seem like much of a difference here, but when you start getting into more complex nested IF's, it can actually be a huge lifesaver.
For example...
I have a very long series of nested IF's, but I only want them to run if a certain cell contains a date. Instead of going through and changing all of my IF statements to have an AND statement and adding that in, I can just wrap the whole thing in a single IF formula that says
=IF(ISDATE([Date Column]@row), run this other long and complex nested IF)
By leaving the third portion of this new IF statement out, it will remain blank if that first IF is not true.
• Options
Here is a screenshot of the sheet, as I am still getting the error.
• ✭✭✭✭✭✭
Options
@Yolanda Brown-Mccutchen Can you also post the exact formula that you are now using?
• Options
Hi Paul,
I tried the formula you gave but it's not working. I must be doing something wrong. Please be patient with me as this is my first time working with Smartsheet, as well as formulas.
• ✭✭✭✭✭✭
Options
No worries. Welcome to Smartsheet! Formulas can be rather daunting especially when first starting out.
What is the column type for the MILEAGE column?
• Options
Text/Number?
• ✭✭✭✭✭✭
Options
Ok. And what type of column is the [Custody Date] column?
• Options
Date
• ✭✭✭✭✭✭
Options
Ok. So this next step goes back to the DATE function. Because the data in the \$MILEAGE\$2 cell is if the text/number type, we need to convert it into a date for your formula to run calculations off of.
I am going to assume the format for that data is always going to be mm/dd/yy. If there is a possibility of single digit day or month or four digit year, please let me know so we can rewrite to accommodate. I will also assume that the year is 2000 or greater. Again... Please let me know if we need to accommodate anything else.
To get the year...
=VALUE("20" + RIGHT(\$MILEAGE\$2, 2))
To get the month...
=VALUE(LEFT(\$MILEAGE\$2, 2))
And the day...
=VALUE(MID(\$MILEAGE\$2, 4, 2))
Now we drop these into the appropriate portions of the DATE function...
=DATE(VALUE("20" + RIGHT(\$MILEAGE\$2, 2)), VALUE(LEFT(\$MILEAGE\$2, 2)), VALUE(MID(\$MILEAGE\$2, 4, 2)))
Now we take the above and use it to replace ever reference to \$MILEAGE\$2...
=IF([CUSTODY DATE]7 > DATE(2000, 1, 1), IF(DATE(VALUE("20" + RIGHT(\$MILEAGE\$2, 2)), VALUE(LEFT(\$MILEAGE\$2, 2)), VALUE(MID(\$MILEAGE\$2, 4, 2))) <= [CUSTODY DATE]7 + 2,125 + 75 + 40, DATE(VALUE("20" + RIGHT(\$MILEAGE\$2, 2)), VALUE(LEFT(\$MILEAGE\$2, 2)), VALUE(MID(\$MILEAGE\$2, 4, 2))) - ([CUSTODY DATE]7 + 2) * 15 + (125 + 75 + 40)))
• Options
Thank you. I will try this shortly and let you know if it works.
## Help Article Resources
Want to practice working with formulas directly in Smartsheet?
Check out the Formula Handbook template! | 1,828 | 5,974 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-26 | latest | en | 0.797158 |
https://learnmechanical.com/properties-of-fluid/ | 1,579,766,600,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579250609478.50/warc/CC-MAIN-20200123071220-20200123100220-00074.warc.gz | 527,528,754 | 50,171 | # Properties of Fluid-Density or Mass Density, Specific weight, Specific Volume, Gravity, PDF
Are you looking for the Properties of Fluid? So today we will study the Properties of Fluid-Mass Density, Specific weight, Volume, Gravity, PDF. Here you will get the articles of Mechanical Engineering in brief with some key points and you will get to know an enormous amount of knowledge from It. So If you find this articles helpful, please let us know in the comment box, either if any correction required too, also let us know in the comment box.
Before going to study properties of fluid let’s have an overview of What is fluid and Types of Fluid.
What is Fluid:
Generally, a Fluid definition is,
A fluid is any substance that flows or deforms under applied shear stress. or
The substances which have the tendency to flow is also called as fluid.
#### Different types of fluids are:
Basically, there are 5 different types of fluid has been classified:
1. Ideal fluid
2. Real fluid
3. Newtonian fluid
4. Non-Newtonian fluid, and
5. Ideal plastic fluid
1. Ideal Fluid:
The fluid which cannot be compressed and have no viscosity falls in the category of an ideal fluid. Ideal fluid is not found in actual practice but
It is an imaginary fluid because all the fluid that exists in the environment have some viscosity. So, there is no ideal fluid in reality.
2. Real Fluid:
A fluid which is having some viscosity is called real fluid. Actually, all the fluids existing or present in the environment are called real fluids.
3. Newtonian Fluid:
The Fluid which obeys Newton’s law of viscosity (i.e the shear stress is directly proportional to the shear strain) then it is known as the Newtonian fluid.
4. Non-Newtonian Fluid:
The Fluid which does not obey Newton’s law of viscosity then it is called Non-Newtonian fluid.
5. Ideal Plastic Fluid:
A fluid having the value of shear stress more than the yield value and shear stress is proportional to the shear strain (velocity gradient) is known as ideal plastic fluid.
## Properties of Fluids:
1. Density or Mass Density
2. Weight Density or Specific Weight:
3. Specific Volume and
4. Specific Gravity
properties of fluids
### 1. Density or Mass Density:
The mass density or Density of a fluid is defined as the ratio of a mass of fluid to its a volume of the fluid.
Density is called a Mass per unit volume of a fluid. This is denoted by symbol ρ (rho) and the unit of mass density is (kg/m3).
The density of liquid may be constant but the density of gases changes with the variation of temperature and pressure.
(The Density of water is 1000 g/m3) or (we can say 1 g/cm3).
Mathematically,
ρ (rho) = (Mass of Fluid) / (Volume of Fluid)
### 2. Weight Density or Specific Weight:
Weight density or specific density of a fluid is defined as the ratio of the weight of fluid to its volume of the fluid.
Weight density is called Weight per unit volume of a fluid. This is denoted by symbol ‘w‘ and the unit of mass density is (N/m3).
Mathematically,
w = (Weight of Fluid) / (Volume of fluid)
=(Mass of fluid) * (Acceleration due to gravity) / (Volume of fluid)
And we know from the previous formula of Density. So this becomes,
w = ρ * g
The value of weight density or specific weight for water is 9.81*1000 N/m3
### 3. Specific Volume:
The specific volume of a fluid is defined as the ratio of the volume of fluid to the mass of fluid.
or
The volume of a fluid occupied by a unit mass or volume per unit mass of a fluid is called Specific volume.
The unit of Specific volume is m³/kg and This is commonly applied to Gases.
Mathematically,
Specific Volume = (Volume of fluid) / (Mass of fluid)
When we divide by volume of fluid to the Numerator and Denominator the fraction we get is,
= (1) / (ρ)
Specific volume is the reciprocal of Mass Density as seen from the formula we got.
### 4. Specific Gravity:
This is defined as the ratio of the Density or weight density of a fluid to density or weight density of a standard fluid.
We know the Standard fluid is water so for liquid the water is standard fluid and
For gases, the Standard fluid is taken as Air.
Most important The specific gravity is called Relative Density. This is denoted by the symbol ‘S’ and this is dimensionless because the upper unit and lower units get canceled.
Mathematically,
(liquid) = ( Weight Density of liquid)/(Weight Density of water)
Here liquid cannot be the same as water it can be some oil and other.
(Gases) = ( Weight Density of Gases / (Weight Density of Air)
Now,
Weight density of liquid = S * Weight Density of water
= S * 9.81 * 1000 N/m3
Density of liquid = S * Density of water = S * 1000 kg/m3
I have covered the topics Types of Fluid Flow in details.
Today we have learned about Properties of Fluid-Mass Density, Specific weight, Specific Volume, Specific Gravity, PDF. If I have missed something, kindly inform me through commenting. If you like this article, don’t forget to share it on social networks. Subscribe our website for more informative articles. Thanks for reading it. Thank you. | 1,177 | 5,091 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2020-05 | longest | en | 0.931615 |
https://www.quantinsti.com/blog/usdinr-option-payoff-excel-model/ | 1,537,537,669,000,000,000 | text/html | crawl-data/CC-MAIN-2018-39/segments/1537267157203.39/warc/CC-MAIN-20180921131727-20180921152127-00152.warc.gz | 843,875,444 | 43,383 | # USDINR Option Payoff [EXCEL MODEL]
29 Shares
In this post we are considering trading option position on USD/INR currency pair. The strike price here is in terms of rupee against the dollar. We are holding different option positions (call and put options) on this underlying (USD/INR pair). The excel workbook shows the payoff of holding different positions simultaneously on the same underlying. It also provides with single option position for a given combination across the spot price range.
It is highly useful for people involved in the option products structuring in bank/corporate treasuries as well as in fund houses for their clients. This workbook can also be used for other currency pairs (like EUR/USD) as well as other asset class options (like Nifty or SPY options) by changing the initial and the incremental value on the Sheet1 below the final spot (cell A9 and A10 onwards).
The excel sheet consists of 4 sheets. We shall explore all the sheets one at a time.
Sheet 1
The snapshot of the table data at the beginning of the sheet is provided here.
Row 2: type of position i.e. call or put denoted by C or P respectively,
Row 3: Whether the respective option is being bought or sold denoted by B or S respectively,
Row 4: strike price of the respective option,
Row 5: notional amount in USD (millions), and
Row 6: the net position. Net position calculation is explained below:
Net position is the effective notional amount for the respective option if we were to replicate it with a future contract. So if we are buying a call option or selling a put option, net notional for that option would be considered positive. Similarly if we are buying a put option or selling a call option, net notional for that option would be considered negative.
Column A from row 8 onwards represents increasing “Final Spot Price” i.e. the exchange rate of dollar against rupee on expiry. The model considers the increment in price at the price interval of 1 paisa (i.e. 0.01 Rupee) with thousand such increments resulting in a range of Rs 10 in total. If you want to cover a different range, you can put the starting price of the range that you want to consider in cell A9.
Column B (similarly Column C, D, etc) from row 8 onwards represent the profit/loss for the respective option in that column at the final spot price as mentioned in the Column A from row 8 onwards. Let us explore how the Pay-off is calculated for each option position.
Column B formula
The profit/loss for cell B9 (and hence for the entire column B thereafter) is calculated in the following manner:
Cell number B2 (where we have ‘P’ for put as the value) and B3 (where we have ‘B’ for buy as the value) suggest that we have bought a Put option. In a long Put, lower is the final spot price from the strike price, higher is the profit. So, in cell B9 we have the highest Pay-off value as it has the lowest value of the final spot price. Please notice that the final spot price is increasing from B9 on wards hence the payoff is decreasing. When the final spot price is higher than the strike price the long Put expires worthless. Hence the profit is zero (as we’ve bought the put option and we have the right to sell but no obligation to sell). This is evident from the cell B335 onwards.
The same formula is extended for other options mentioned in Column C, D, etc. For example in Column C9, we’ve the following formula:
The pay-off for cell C9 (and hence for the entire column C thereafter) is calculated in the following manner:
C2 and C3 suggest that we have sold a Call option. In a short Call, we do not have any positive pay-off (as we only receive premium which we are not considering here). Since we have ignored any premium payment or transaction costs, the maximum pay-off is zero. When the final spot price starts increasing above the strike price, the short call incurs loss. This is evident from the cell C335 on wards.
Column AG: The Total Payoff
Column AG denotes the total profit/loss from the three option positions combined at that final spot price.
The total payoff sheet displays the net payoff from all the option positions on the left.
Column AI: The status
Column AI denotes the status of change in the total payoff difference. The formula is as follows:-
If the difference between two payoffs is zero then the entry in column AI is null.
The next part can be broken into two steps.
• The ratio of difference between total payoff to the difference of final spot is calculated.
• The absolute value of the differences of the ratio calculated above is computed.
If the absolute value is greater than 0.000001 then the entry is “Change” else if the absolute value is less than 0.000001 then the entry is “Ok”.
Column AJ: The signal
Column AI denotes the status of buy or sell signal.
If the absolute value of the difference between successive total payoffs is less than 0.000001 then the entry is “Fixed”. If the total payoff is greater than 0.000001 then the signal is “Buy” else if the total payoff is less than 0.000001 then the signal is “Sell”.
Column AK: The option type
Column AK denotes the option position that is held.
If the corresponding entry in column AJ is “Fixed”, then the entry in column AK is “Fixed Payout”. Else the following conditions are tested:-
1. If the ratio of difference of the payoffs and the difference of final spot is greater than 0.000001 and if the corresponding entry in AJ is “Sell” then it is “Put” else it is “Call”.
2. If the ratio of difference of the payoffs and the difference of final spot is less than 0.000001 and if the corresponding entry in AJ is “Sell” then it is “Call” else it is “Put”.
Column AL: Strike price
Column AL calculates the strike price as follows:-
If the corresponding entry in column AK is “Fixed Payout”, then the entry in the column AL is “Fixed”.
Else the strike price is calculated as follows:-
The ratio of difference in the total payoffs to the difference in final spot is calculated. This entity is used to divide the total final payoff from the AG column. The resulting value is subtracted from the final spot price to obtain the strike price.
Column AM: Net position
Column AM denotes the net position and is calculated as follows:-
We will see in a moment what the contents of the sheet3 are, as of now the notional amount is the sum of the product of corresponding cell values from sheet1 and sheet3 for the ranges $B$6:$AE$6 and B9:AE9 respectively.
The values from the sheet1 for the range $B$6:$AE$6 is shown in the screen shot below
Only the first three values will be considered since the rest of the values are zeros.
The values from the sheet3 for the range B9:AE9 is shown in the screen shot below.
Now the sum of products is calculated as follows:-
(-3*1) +(-6*0) + (3*0)=-3
Now we will turn our attention to sheet3.
Sheet 3 is essentially the binary representation of sheet 1 for the payoff columns. For example column B from B8 onwards shows the payoff for the “Put” and “Buy” combination. The payoff is positive from the cell B9 to B334. So the entry in the column B of sheet 3 from B9 onwards is 1 if the corresponding absolute value in sheet 1 is at least 0.000001. The screen shot below shows the formula.
Similar logic is applied for all other columns of sheet 3.
Let us explore sheet 2.
Sheet 2
The macro written to fill the cells of sheet 2 is shown below.
With the initial parameters of i=11 and j=11, the Do While loop fills the cells of sheet 2 with the contents of sheet 1 as follows:-
Worksheet(2).Cells(3,1).Value= Worksheet(1).Cells(14,36).Value
All this means is the value of cell defined by row 13 and column 36 (Sheet1!AJ13) of sheet 1 is placed in the cell defined by row 3 and column 1(Sheet2!A2) of sheet 2.
Similarly for the cells B2, C2 and D2.
Cell E5 of sheet 2 is filled on the following logic:-
If the cell AJ13 of sheet 1 has the entry as “Fixed”, then the entry in E5 of sheet 2 should be of AG13 of sheet 1. Otherwise it should be “-“.
Cell F5 of the sheet 2 is the sum of products of range of cell values from sheet 1 and sheet 3 described in the formula below.
Sheet 4 (named Total Payoff) shows the plot for the payoff. The X axis shows the rupee exchange against dollar (the final spot). Y axis shows the final payoff (column AG of sheet 10). As you can see, for this combination of options, payoff increases as the final spot decreases and payoff decreases as the final spot increases. In other words, payoff increases as the rupee strengthens and decreases as rupee weakens against the greenback. | 2,001 | 8,576 | {"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-2018-39 | latest | en | 0.915816 |
https://community.tes.com/threads/the-maths-of-a-jury-and-a-courtroom.456620/ | 1,603,218,840,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107874026.22/warc/CC-MAIN-20201020162922-20201020192922-00298.warc.gz | 266,501,205 | 19,002 | The Maths of a Jury and a Courtroom
Discussion in 'Mathematics' started by tango78, Jan 2, 2011.
1. tango78New commenter
Hi all,
I am teaching Maths in the context of Citizenship this half term and am really interested in the probabilities and associated Maths of the Judicial System. Only problem is that I know that there is a lesson hidden in it somewhere but cannot think of how! I'm thinking along the lines of expert witnesses giving probability evidence that is then misinterpreted, eg the Sally Clark trial. Has anyone done this? Or have any idea where I can go with this?
3. tango78New commenter
Thanks, I'll have a good readthrough of that article. Ha ha, you never know about the dance approach, could count as Healthy Schools!
4. KarvolOccasional commenter
I cannot remember the exact title of the book that I read this in, but there is something to do with probablities and evidence.
I think the name of the book was "Knowledge Problems: A Intoroduction to Epistemology" or something along those lines. It looked at probabilities to do with what happens when we say we saw something, and what are the real probabilities involved in that. It came under the heading of Gettier problems.
There are also court room scenes in the film "My Cousin Vinny" which look at evidence in a light hearted manner using mathematics.
5. ColinWilson
In the Sally Clark trial it was the expert witness who was misinterpreting the probabilities, the tragety was that no one questioned how Professor Sir Roy Meadows had calculated the probalities in the first case.
Meadows was given a Knighthood his research which ignored conditional probabilities. He cliamed that the probabity of both twins dying from cot death was 1/73000000. But he had arrived at this number by squaring, 1/8500, the probability of a single cot death.
A lesson based on Game Theory looking at 'The Prisoner's Dilemma' might be worth looking into.
6. Maths_MikeNew commenter
Yes a classic where he simply assumed the deaths to be independent - something which even a *** could see to be wrong.
You could be genetically likely to lose children to cot death so the probability of a death given that you already had a previous death is likely to be much lower.
He also coined the
one is unlikely
two is suspicious
and three is murder
I feel so sorry for the women punished as a result of this supposedly intelligent mans stupidity.
How the defense lawyers were unable to discredit his theories is beyond me.
7. tango78New commenter
This all sounds great, I have found a reference to John Moriarty appearing on the BBC's The One Show in January of last year talking about the probability of being called for Jury Service but cannot find the actual clip yet, if I do I will let you all know.
The prisoner's dilemma sounds really good too and could be great with probability trees, it would also be something that would interest the kids!
The 'expert witness' has so many probabilities also, I know that after Sally was cleared the Appeals system was filled with cases who had been convicted on the evidence of an 'expert' witness so there could be more in that.]
Thanks so much for your help so far, any more ideas...
8. ColinWilson
You could try this as a game between pairs over a number of rounds using a payoff matrix. The overall winner is the player with the most points at the end. This has been simulated using computers with different statergies.
9. alabasterNew commenter
this https://www.ncetm.org.uk/resources/20330 resource is really good - examining evidence to work out who the murderer is. What I like is that the maths isn't contrived, working out the percentage of arsenic in different samples to find the samples that match the one found at the murder scene for exanple, actually seems like a real life example of maths being used, rather than made up to give the kids practice. My classes that I have used it with have really enjoyed it.
10. DMNew commenter
I have used this too alabaster and I agree it went down well. I can't agree the maths is not contrived however - poisons were found at all the suspects houses etc - really?!
11. florapost
i'm dargging this out of an old memory - you could maybe get the figure from the sids charity website, but i think i remember them saying that the actual probability of a second cot death in a family given a first is something like 1 in 3000
not that teaching numeracy matters, of course....
12. ColinWilson
There is no such thing as an actual probability in such cases, but this is missing the point. If Bayes Theorem had been understood Sally Clark might still be alive.
The probability that Dover will beat Stevenage in this years FA Cup final might be estimated at 73000000 to 1, it would be unbelievable. But, in the very unlikley event. if they did both reach the final the odds on a Dover win might be around say 5 to 1.
The cot death of twins is very unlikely, murder of young twins by their mother is very unlikely. But both twins died, so one event could be taken as true. Bayers Theorem should have be used to estimate the conditional probability orf murder given that both twins died.
13. tango78New commenter
That looks great, thanks for the tip. How long did it take your classes?
14. tango78New commenter
ooh that's an interesting statistic, if I was feeling brave enough I could even repeat the survey in my class, hmm...
thanks... | 1,198 | 5,404 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2020-45 | latest | en | 0.968979 |
https://pedagogue.app/activities-to-teach-students-division-facts-up-to-12-find-the-missing-number-3/ | 1,721,734,322,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763518029.87/warc/CC-MAIN-20240723102757-20240723132757-00871.warc.gz | 399,600,121 | 29,539 | # Activities to Teach Students Division Facts Up to 12: Find the Missing Number
Division is an essential skill that students need to master. As teachers, it is our responsibility to make learning division a fun and engaging activity for our students. One way to teach division facts up to 12 is by using the “Find the Missing Number” game.
The “Find the Missing Number” game involves presenting students with a division problem where one of the factors is missing. The goal of the game is for students to find the missing number that makes the equation true. This activity is an excellent way to help students think critically and improve their problem-solving skills.
Here are some activities you can use to teach students division facts up to 12 using the “Find the Missing Number” game:
1. Division Bingo
Division Bingo is an exciting way to teach students division facts while also reinforcing their ability to find the missing number. In this activity, create bingo cards with different division problems on them, leaving one of the factors missing. Students will solve the equation and mark the answer on their bingo cards. The first student to get a straight or diagonal line marked on their bingo card wins the game.
Task cards are a versatile tool that can be used to teach a variety of skills, including division facts. Create task cards with different division problems on them, leaving one of the factors blank. Provide students with a recording sheet where they can write down the missing number for each problem. This activity helps students practice their division skills and keeps them engaged.
3. Digital Games
There are many digital games available that use the “Find the Missing Number” concept. Students can play these games on their computers or tablets, making it a fun and engaging activity for them. Digital games are a great way to supplement traditional teaching methods and keep students motivated.
4. Division War Card Game
The Division War Card Game is a fun and competitive way to practice division facts. In this game, students draw two cards with division equations on them and race to find the missing number first. The first student to solve the problem correctly gets to keep the cards. The student with the most cards at the end of the game wins. This game is a fun way to practice division skills while also promoting healthy competition among students.
In conclusion, teaching division facts up to 12 can be an enjoyable and engaging experience for students. Using the “Find the Missing Number” game is an effective way to help students build critical thinking and problem-solving skills. By incorporating these activities into your curriculum, you can ensure that your students are equipped with the tools they need to succeed in math. | 527 | 2,785 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.71875 | 4 | CC-MAIN-2024-30 | latest | en | 0.95456 |
https://hackage.haskell.org/package/sbv-8.1/src/Documentation/SBV/Examples/Puzzles/MagicSquare.hs | 1,620,276,660,000,000,000 | text/plain | crawl-data/CC-MAIN-2021-21/segments/1620243988725.79/warc/CC-MAIN-20210506023918-20210506053918-00511.warc.gz | 288,982,206 | 1,923 | ----------------------------------------------------------------------------- -- | -- Module : Documentation.SBV.Examples.Puzzles.MagicSquare -- Copyright : (c) Levent Erkok -- License : BSD3 -- Maintainer: erkokl@gmail.com -- Stability : experimental -- -- Solves the magic-square puzzle. An NxN magic square is one where all entries -- are filled with numbers from 1 to NxN such that sums of all rows, columns -- and diagonals is the same. ----------------------------------------------------------------------------- module Documentation.SBV.Examples.Puzzles.MagicSquare where import Data.List (genericLength, transpose) import Data.SBV -- | Use 32-bit words for elements. type Elem = SWord32 -- | A row is a list of elements type Row = [Elem] -- | The puzzle board is a list of rows type Board = [Row] -- | Checks that all elements in a list are within bounds check :: Elem -> Elem -> [Elem] -> SBool check low high = sAll \$ \x -> x .>= low .&& x .<= high -- | Get the diagonal of a square matrix diag :: [[a]] -> [a] diag ((a:_):rs) = a : diag (map tail rs) diag _ = [] -- | Test if a given board is a magic square isMagic :: Board -> SBool isMagic rows = sAnd \$ fromBool isSquare : allEqual (map sum items) : distinct (concat rows) : map chk items where items = d1 : d2 : rows ++ columns n = genericLength rows isSquare = all (\r -> genericLength r == n) rows columns = transpose rows d1 = diag rows d2 = diag (map reverse rows) chk = check (literal 1) (literal (n*n)) -- | Group a list of elements in the sublists of length @i@ chunk :: Int -> [a] -> [[a]] chunk _ [] = [] chunk i xs = let (f, r) = splitAt i xs in f : chunk i r -- | Given @n@, magic @n@ prints all solutions to the @nxn@ magic square problem magic :: Int -> IO () magic n | n < 0 = putStrLn \$ "n must be non-negative, received: " ++ show n | True = do putStrLn \$ "Finding all " ++ show n ++ "-magic squares.." res <- allSat \$ (isMagic . chunk n) `fmap` mkExistVars n2 cnt <- displayModels disp res putStrLn \$ "Found: " ++ show cnt ++ " solution(s)." where n2 = n * n disp i (_, model) | lmod /= n2 = error \$ "Impossible! Backend solver returned " ++ show n ++ " values, was expecting: " ++ show lmod | True = do putStrLn \$ "Solution #" ++ show i mapM_ printRow board putStrLn \$ "Valid Check: " ++ show (isMagic sboard) putStrLn "Done." where lmod = length model board = chunk n model sboard = map (map literal) board sh2 z = let s = show z in if length s < 2 then ' ':s else s printRow r = putStr " " >> mapM_ (\x -> putStr (sh2 x ++ " ")) r >> putStrLn "" | 672 | 2,540 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.0625 | 3 | CC-MAIN-2021-21 | latest | en | 0.54301 |
https://math.answers.com/Q/How_many_laps_are_in_a_.75k_swim | 1,701,704,115,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100531.77/warc/CC-MAIN-20231204151108-20231204181108-00780.warc.gz | 447,915,020 | 45,233 | 0
# How many laps are in a .75k swim?
Updated: 10/22/2022
Wiki User
11y ago
.75k = 820.209974 yards.
If you are swimming a 25 yard pool:
A lap (technically defined as ONE length of a pool not two): 32.8 laps is the answer.
So if you're swimming a sprint tri, swim 33 laps.
Wiki User
11y ago
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Q: How many laps are in a .75k swim?
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Related questions
16 laps
### How many laps in a 25 meter pool to swim a mile?
64 laps to swim a mile in a 25 meter pool
### How many laps do you need to swim to make a kilometer?
it depends on the length of the pool. if the pool is 25 meters, you have to swim 40 laps. For 50 meters, 20 laps. For 100 m, 10 laps. For 200 meters, 5 laps.
### Alyssa swims 3 laps in 12 minutes At this rate how many laps will she swim in 10 minutes?
Two and a half laps
### How many swim laps 1300 meters?
52 laps in a 25-meter per lap pool.
16
### What is the difference in number of laps to swim a mile between a 25-yard pool and a 25-meter pool?
first, how many laps to swim a mile in a 25 yard pool. Second, how many laps to swim a mile in a 25 meter pool. Third, subtract the lowest number to get your answer.
### How many laps of the pool do the athletes swim in the 1500 meter race?
each lap is 50m so its 30 laps if you add 2 more laps you got a mile witch is a 1600 meter swim
43
### How many laps did they swim in 2008?
'They' is not defined as a specific group of people.
76
### How many laps do you have to swim to swim 0.5 mile?
more than 10 but less than 20 | 451 | 1,554 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.421875 | 3 | CC-MAIN-2023-50 | latest | en | 0.965936 |
https://codegolf.stackexchange.com/questions/100552/japanese-mahjong-score-calculator | 1,718,539,159,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861659.47/warc/CC-MAIN-20240616105959-20240616135959-00060.warc.gz | 143,621,564 | 44,916 | # Japanese Mahjong Score Calculator
I'm new to Japanese Mahjong and have trouble calculating the scores. Please help me write a program to calculate it.
# Introduction
(aka you can skip this boring part)
Mahjong is a popular tile game in Asia. Different rules have evolved and systematically established in various regions. Japan, being an isolated island state, have its own mahjong variant including rules and scoring.
In Japanese Mahjong, 4 players participate in the game at the same time. They take turn to draw a tile from wall and discard a tile from hand. Occasionally then can take other's discarded tiles for use. Each round 1 of the 4 players will be the dealer. Independent of that, each round may have a winner if he/she can assemble a winning hand. There could be no winners in a round but we don't care about that in this challenge.
When a player become the winner of that round, the change in score is calculated depending on 4 things: (1) Han and (2) fu depends on the winning hand, (3) whether the winner is also a dealer, and (4) whether he win by self-draw or from other's discard. The winner get his points by taking the points from the losers. So points lost by all losers added should equal to points gained by winner. Moreover, sometimes only one loser lose his points while the other two don't lose any, and sometimes one of the three losers lose more points. (See below)
# Objective
Write a program or function that accepts the 4 parameters mentioned above, and return (1) how many points the winner will gain, and (2-4) how many points each of the three losers will lose.
## Details of inputs
Han: An integer from 1 to 13 inclusive.
Fu: An integer that is either 25, or an even number from 20 to 130 inclusive. If han is 1 or 2, fu is further capped at 110.
dealer or not: A truthy/falsy value. Though if you are a mahjong lover you can instead accept the winner's seat-wind :-)
self-draw or discard: A two-state value. Truthy/falsy value will suffice but other reasonable values are acceptable.
## Steps of calculating the points gained/lost:
1. Calculate the base score
The base score is calculated based on 2 cases.
1a) Han ≤ 4
If fu is 25, then keep it as-is. Otherwise round it up to nearest 10.
Then base score is calculated by
base_score = fu×2han+2
And it's capped at 2000 points.
1b) Han ≥ 5
Fu is not used here. The base score is obtained by direct lookup of han in the following table:
Han Base score
5 2000
6-7 3000
8-10 4000
11-12 6000
13 8000
2. Calculate the points lost by losers
2a) Dealer wins by self-draw
Each loser pays an amount equal to double of the base score, then rounded up to nearest 100.
2b) Dealer wins by discard
The discarding loser pays an amount equal to 6x the base score, then rounded up to nearest 100. Other two losers don't pay.
2c) Non-dealer wins by self-draw
Dealer (one of the loser) pays an amount equal to double of the base score, then rounded up to nearest 100. The other two losers each pays an amount equal to base score, also rounded up to 100.
2d) Non-dealer wins by discard
The discarding loser pays an amount equal to 4x the base score, then rounded up to 100. Other two losers don't need to pay. Scoring isn't affected by whether the discarding loser is dealer or not.
3. Calculate the points gained by winner
Winner gains the total points lost by the three losers. Not a point less and not a point more.
# Example
(Some examples are shamelessly taken/modified from Japanese Wikipedia page of calculating mahjong score)
4 han 24 fu, dealer win by self-draw
24 fu rounded up to 30 fu
Base score = 30 × 2 4+2 = 1920
Each loser pays 1920×2 = 3840 => 3900 (rounded)
Winner takes 3900×3 = 11700 points
Output: +11700/-3900/-3900/-3900
3 han 25 fu, dealer win by discard
25 fu no need to round up
Base score = 25 × 2 3+2 = 800
Discarding loser pays 800×6 = 4800
Non-discarding loser pays 0 points
Winner takes 4800 points
Output: +4800/-4800/0/0
4 han 40 fu, non-dealer win by self-draw
40 fu no need to round up
Base score = 40 × 2 4+2 = 2560 => 2000 (capped)
Dealer pays 2000×2 = 4000
Non-dealer pays 2000
Winner takes 4000+2000×2 = 8000 points
Output: +8000/-4000/-2000/-2000
9 han 50 fu, non-dealer win by discard
Fu not used to calculate score
Base score = 4000 (table lookup)
Discarding loser pays 4000×4 = 16000
Non-discarding loser pays 0 points
Winner takes 16000 points
Output: +16000/-16000/0/0
## Output
Output 4 numbers: The point that the winner is gaining (as positive number), and the points that each of the three losers is losing (as negative number).
You may output the numbers as 4 individual numerical values in your language's native format, an array/list of numbers, 4 separate strings (or array/list of characters), or a string contatining the numbers that are properly delimited. If you use string, the positive sign is optional but negative sign is mandatory.
The order of the numbers doesn't matter, since they can be inferred from sign and magnitude of the numbers. e.g. 0/-16000/+16000/0 is acceptable.
## Format
• Full program or function
• Any reasonable input/output format
• Default rules apply. Standard loopholes forbidden. so fewest bytes wins.
# Notes
• All outputs are guaranteed to be integers, more specifically multiples of 100 in the range [-48000,+48000]
• Rounding is applied only to (1) fu and (2) loser's points. It is not applied to base score.
• Theoretically winning by self-draw or discard should gain the same points (6x or 4x base points). However due to rounding, this may not be true at low han. (See 1st and 2nd test cases)
# Test cases
(Also see example)
Inputs Outputs
Han Fu Dealer Self Winner Loser1 Loser2 Loser3
1 22 N Y +1100 -500 -300 -300
1 22 N N +1000 -1000 0 0
3 106 Y N +12000 -12000 0 0
4 20 N Y +5200 -2600 -1300 -1300
4 25 Y N +9600 -9600 0 0
5 88 Y Y +12000 -4000 -4000 -4000
7 25 Y Y +18000 -6000 -6000 -6000
12 64 N N +24000 -24000 0 0
13 130 Y N +48000 -48000 0 0
• Just spent a few hours for a C#6 answer but it's over 200 bytes. I will further golf it and post it after others have a chance to. Commented Nov 22, 2016 at 6:04
# 05AB1E, 79 bytes
05AB1E uses CP-1252 encoding.
Input order:
Han
Fu
Dealer = 1, Non-dealer = 0
Self = 2, Discard = 0
Code:
3‹i6b‚ï{0è}¹5‹iÐT/îT*‚sÈè¹Ìo*3°·‚{0èë\•j¤È°•¹è3°*}U•é2TˆÞ•S3ôŠ+èX*Tn/îTn*(DOĸì
Try it online!
Explanation to come.
## Python 2, 128 bytes
def m(h,f,o,t):C=100;b=[24/(50/h)*1000,min(2000,[-f/10*~9,f][f%2]<<h+2)][h<5];x=(o|2-t)*-2*b/C*C;y=~o*t*b/C*C;return-x-y-y,x,y,y
h is han, f is fu, o is dealer (1/0), t is self-draw (1/0). Outputs in the same order as the test cases. ideone link.
Expanded slightly:
def m(h,f,o,t):
C = 100
# Base score
b = [24/(50/h)*1000, min(2000,[-f/10*~9,f][f%2]<<h+2)][h<5]
# Player scores - uses the -x/100*-100 approach to round up
x = (o|2-t)*-2*b/C*C
y = ~o*t*b/C*C
return -x-y-y, x, y, y
Indirect thanks to @MartinEnder.
## C#6, 211 bytes
using static System.Math;(h,f,x,y)=>{int b=h>5?24/(50/h)*100:Min(f%2<1?(int)Ceiling(f*.1)<<h+2:10<<h,200),c=-(int)Ceiling(b*.1*(y?2:x?6:4))*100,d=-(int)Ceiling(b*.1*(y?x?2:1:0))*100;return new[]{-c-d*2,c,d,d};};
Contains an idea shamelessly stolen from @Sp3000.
Assign to Func<int,int,bool,bool,int[]>
h=han
f=fu
x=true if dealer wins, false if non-dealer wins
y=true if self draw, false if discard
returns int[], [0]=winner, [1]-[3]:loser
repl.it demo
Ungolfed + explanations
// b is base score / 10
int b=h>5
// han is [6,13], use Sp3000's idea
?24/(50/h)*100
// han <= 5, calculate base score (/ 10) from formula...
// Yes this formula works for han==5 also
:Min(
f%2<1
// round up to 10, then times 2^(h+2)
?(int)Ceiling(f*.1)<<h+2
// b=2.5*2^(h+2)=10*2^h=10<<h
:10<<h
// Cap at 200, thus base score is capped at 2000
,200),
// Multiply 0.1b by 2/4/6 according to input, then multiply by 100 to get actual score
c=-(int)Ceiling(b*.1*(y?2:x?6:4))*100,
d=-(int)Ceiling(b*.1*(y?x?2:1:0))*100;
// Array type implied from arguments
return new[]{-c-d*2,c,d,d}; | 2,590 | 8,377 | {"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} | 3.078125 | 3 | CC-MAIN-2024-26 | latest | en | 0.952665 |
http://www.stata.com/statalist/archive/2012-10/msg01211.html | 1,500,692,311,000,000,000 | text/html | crawl-data/CC-MAIN-2017-30/segments/1500549423842.79/warc/CC-MAIN-20170722022441-20170722042441-00399.warc.gz | 566,255,496 | 4,046 | Notice: On April 23, 2014, Statalist moved from an email list to a forum, based at statalist.org.
# Re: st: Subtract Closest Cell Which has A Value
From Nick Cox To statalist@hsphsun2.harvard.edu Subject Re: st: Subtract Closest Cell Which has A Value Date Fri, 26 Oct 2012 09:49:39 +0100
```This can be solved using the same techniques as for your previous problem.
See my previous posting.
You need a variable that is the previously observed value and one that
is the next observed value. You get those variables by copying down in
a cascade in the usual (panel id, time) order and then again after
reversing time.
Then you use the previous value if the timeline date is positive and
the next value if it is negative.
FAQ . . . . . . . . . . . . . . . . . . . . . . . Replacing missing values
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . N. J. Cox
2/03 How can I replace missing values with previous or
following nonmissing values?
http://www.stata.com/support/faqs/data/missing.html
(Reversing time is one of the top tricks many people need occasionally
in Stata but often don't seem to know about.)
NIck
On Fri, Oct 26, 2012 at 3:41 AM, Lisa Wang <lhwang0925@gmail.com> wrote:
> Dear all,
>
> I have another question. I have two columns of values - variables A
> and B to make it simple. I would like to A-B in a new column C;
> however, column B has a lot of missing values (.), so it doesn't seem
> that simple. Variable A is complete, so no missing data. I would like
> Stata to find the nearest cell (data is already sorted by column D)
> with a value in column B to subtract off for each panel that I have.
>
> For instance,
>
> 51 . 31
> 50 . 30
> 47 20 27
> 51 . 31
> 25 . 5
> 49 . 29
> 45 . 25
> 24 . 4
> 51 . 1
> 22 . -28
> 27 . -23
> 57 . 7
> 20 . -30
> 57 50 7
> 59 . 9
> 33 . -17
>
> Would this be possible?
>
> Many thanks,
> Lisa
> *
> * For searches and help try:
> * http://www.stata.com/help.cgi?search
> * http://www.stata.com/support/faqs/resources/statalist-faq/
> * http://www.ats.ucla.edu/stat/stata/
*
* For searches and help try:
* http://www.stata.com/help.cgi?search
* http://www.stata.com/support/faqs/resources/statalist-faq/
* http://www.ats.ucla.edu/stat/stata/
``` | 737 | 2,423 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2017-30 | latest | en | 0.85024 |
https://www.teacherspayteachers.com/Browse/Search:ESL/PreK-12-Subject-Area/Geometry/Type-of-Resource/Video-Files?ref=filter/resourcetype/all | 1,550,517,188,000,000,000 | text/html | crawl-data/CC-MAIN-2019-09/segments/1550247487624.32/warc/CC-MAIN-20190218175932-20190218201932-00363.warc.gz | 970,799,209 | 30,783 | You Selected:
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Math Geometry Common Core Bundle - Animations of real-world applications. Use this geometry bundle to introduce or review math concepts for grades 6-7. It includes 9 story-based animations of real life scenarios. It will help you show the concepts visually and engage your students in learning from
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Use engaging animation of math word problem and worksheet to review area of quadrilaterals concept. In this animation, challenge your kids to help George, Ann, Rick and Jeff till their farmland by finding the total area of their farms. You can use this activity as warm up/bell ringers, review,
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Use engaging animation of Math problem and worksheet to review Common Core Math skills. The local church is being painted. The painters need your help to find out how much is it going to cost to paint the frontage of the church, which is a rectangle with a triangle on top of it. You can use thi
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Use engaging animation of Math problem and worksheet to review Common Core Math skills. Charles is cleaning up his messy room but does he have a box big enough to pack all his 325 dice? Challenge your students to find that out using this engaging animation. You can use this activity as warm up
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Use engaging animation of Math problem and worksheet to review Common Core Math skills. Challenge your students to use knowledge of areas to help Grandma cut down the size of the pancakes she has been baking for Sammy. You can use this activity as warm up/bell ringers, review, group activity,
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Use engaging animation of Math problem and worksheet to review Common Core Math skills. Use knowledge of areas to tackle this math problem that involves paving the paths running through a beautiful city park. You can use this activity as warm up/bell ringers, review, group activity, and remedia
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Use engaging animation of Math problem and worksheet to review Algebra and Geometry Math skills A customer provides limited information to Paul about a fence she needs around her garden. Help him determine the length and width of the fence. You can use this activity as warm up/bell ringers, re
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Use engaging animation of a math problem and a worksheet to review common core math skills. Challenge your students to help Max figure out the quantity of color paper needed for the kaleidoscope he is building. You can use this activity as warm up/bell ringers, review, group activity, and rem
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Use engaging animation of Math problem and worksheet to review Common Core Math skills. Rita is feeling geeky today! Join her in finding a size of a cookie before she gobbles it up. You can use this activity as warm up/bell ringers, review, group activity, and remedial-activity. The voice-over
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A math music video that explains what congruent shapes are. This song is intended to lower students' affective filter prior to a video lesson and information gap activities. Those components will be available for download shortly. The lessons that follow this song use similarity and congruence to
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A math music video to help students understand what makes a quadrilateral a quadrilateral, and to identify different types of quadrilaterals.
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This lesson accompanies the song, "Quadrilaterals," which has been posted for free download. It is the first of five lessons that accompany that song. This lesson directly addresses the properties of squares and rectangles, but is also meant to foster academic vocabulary growth that will make futur
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This lesson accompanies the song, "Quadrilaterals," which has been posted for free download. It is the fifth of five lessons that accompany that song. This lesson directly addresses the properties of squares and rectangles, but is also meant to foster academic vocabulary growth that will make futur
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This lesson accompanies the song, "Quadrilaterals," which has been posted for free download. It is the second of five lessons that accompany that song. This lesson directly addresses the properties of squares and rectangles, but is also meant to foster academic vocabulary growth that will make futu
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This lesson accompanies the song, "Quadrilaterals," which has been posted for free download. It is the third of five lessons that accompany that song. This lesson directly addresses the properties of squares and rectangles, but is also meant to foster academic vocabulary growth that will make futur
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This lesson accompanies the song, "Quadrilaterals," which has been posted for free download. It is the fourth of five lessons that accompany that song. This lesson directly addresses the properties of squares and rectangles, but is also meant to foster academic vocabulary growth that will make futu
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Teachers Pay Teachers is an online marketplace where teachers buy and sell original educational materials. | 1,386 | 5,838 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2019-09 | latest | en | 0.90068 |
https://oeis.org/A188675 | 1,600,894,218,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400212039.16/warc/CC-MAIN-20200923175652-20200923205652-00347.warc.gz | 516,323,404 | 4,079 | The OEIS Foundation is supported by donations from users of the OEIS and by a grant from the Simons Foundation.
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A188675 Partial sums of the binomial coefficients binomial(3*n,n) (A005809). 6
1, 4, 19, 103, 598, 3601, 22165, 138445, 873916, 5560741, 35605756, 229142476, 1480820176, 9603245620, 62463474700, 407330900284, 2662179813931, 17433248900656, 114359597479261, 751343566800961, 4943188072606456 (list; graph; refs; listen; history; text; internal format)
OFFSET 0,2 LINKS G. C. Greubel, Table of n, a(n) for n = 0..1000(terms 1..100 from Vincenzo Librandi) FORMULA a(n) = Sum_{k=0..n} binomial(3*k,k). Recurrence: 2*(n+2)*(2n+3)*a(n+2)-(31*n^2+95*n+72)*a(n+1)+3*(3*n+4)(3*n+5)*a(n)=0. G.f.: 2*cos((1/3)*arcsin(3*sqrt(3*x)/2))/((1-x)*sqrt(4-27*x)). a(n) ~ sqrt(3)*27^(n+1)/(46*4^n*sqrt(Pi*n)). - Vaclav Kotesovec, Oct 20 2012 MATHEMATICA Table[Sum[Binomial[3k, k], {k, 0, n}], {n, 0, 20}] PROG (Maxima) makelist(sum(binomial(3*k, k), k, 0, n), n, 0, 20); (PARI) for(n=0, 25, print1(sum(k=0, n, binomial(3*k, k)), ", ")) \\ G. C. Greubel, Jan 27 2017 CROSSREFS Cf. A001764, A005809, A104859, A188676, A188678 - A188687. Cf. A263134: Sum_{k=0..n} binomial(3*k+1,k). Cf. A087413: Sum_{k=0..n} binomial(3*k+2,k). Sequence in context: A307678 A151382 A234958 * A199876 A225029 A078940 Adjacent sequences: A188672 A188673 A188674 * A188676 A188677 A188678 KEYWORD nonn,easy AUTHOR Emanuele Munarini, Apr 08 2011 STATUS approved
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Last modified September 23 16:41 EDT 2020. Contains 337315 sequences. (Running on oeis4.) | 722 | 1,839 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2020-40 | latest | en | 0.490676 |
https://unizor.blogspot.com/2015/03/unizor-probability-event-arithmetic.html | 1,701,977,766,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100686.78/warc/CC-MAIN-20231207185656-20231207215656-00462.warc.gz | 657,799,338 | 10,634 | ## Monday, March 16, 2015
### Unizor - Probability - Event Arithmetic
Let's discuss how logical operations on events are related to their probabilities. This part of the theory is heavily dependent on the representation of events as subsets of elements (each element is an elementary event) and probabilities as an additive measure defined on these subset.
Starting from the simplest case, consider a finite set of elementary events (say, outcomes of rolling a pair of dice) of equal probabilities (in case of a pair of dice each elementary event has probability 1/36).
Since this probability is an additive measure, any combination of elementary events into one bigger event has a probability equal to a sum of probabilities of all elementary events that compose this bigger event (for example, the probability of rolling a pair of dice with a sum equal to 5 is a sum of probabilities of the composing elementary events - rolling 1/4, 2/3, 3/2 and 4/1 on two dice, each having the probability 1/36 and, therefore, the combined probability of having a sum of 5 on two dice equals to 4·1/36=1/9).
Consider now a logical operation OR between two events X and Y (this operation is usually signified in the set theory by a union sign ∪):
Z = X OR Y = X ∪ Y.
First, consider these two events to be mutually exclusive, that is having no common elementary events. Since they are mutually exclusive, elementary events constituting event X are completely different from those that constitute event Y. As we know, event Z consists of elementary events that form a union of elementary events included into event X and those included into event Y. Therefore, the probability measure of event Z is a sum of probabilities of events X and Y. We can, therefore, state that the probability of an event that is the result of an operation OR between two mutually exclusive events equals to a sum of their probabilities.
Now let's establish a connection between intersection of certain type of events and an operation of multiplication.
Consider now a logical operation AND between two events X and Y (this operation is usually signified in the set theory by an intersection sign ∩):
Z = X AND Y = X ∩ Y.
This is an operation of taking only common elementary events between X and Y to form a resulting event Z.
Let's start with an example of rolling two dice (we will call them #1 and #2). Consider an event X{#1 is equal or greater than 5} and an event Y{#2 is equal or less than 4}. From the common sense perspective these events are independent because they apply to two different dice with no connection between them. Let's examine the probabilities of X, Y and X∩Y.
Elementary events in our experiment are still all the possible pairs of numbers from 1 to 6 each - 36 different combinations with equal probability of 1/36 each.
Event X consists of elementary events 5/1, 5/2, 5/3, 5/4, 5/5, 5/6, 6/1, 6/2, 6/3, 6/4, 6/5, 6/6. Therefore, it's probability is 12/36=1/3.
Event Y consists of elementary events 1/4, 2/4, 3/4, 4/4, 5/4, 6/4, 1/3, 2/3, 3/3, 4/3, 5/3, 6/3, 1/2, 2/2, 3/2, 4/2, 5/2, 6/2, 1/1, 2/1, 3/1, 4/1, 5/1, 6/1. Therefore, it's probability is 24/36=2/3.
Event X∩Y consists of only one elementary events 5/4, 6/4, 5/3, 6/3, 5/2, 6/2, 5/1, 6/1. Therefore, it's probability is 8/36=2/9.
Notice that
P(X∩Y) = P(X)·P(Y)
and this is not an accident. The important consideration here is the independence of X and Y, which we did not define precisely, but just mentioned it in a casual common sense. It was indeed natural to consider the results of rolling two different dice as independent events.
As you see, logical operation AND between independent events results in the multiplication of probabilities.
This fact is a basis for using multiplication sign · instead of logical AND between independent events. The above equality looks now as
P(X · Y) = P(X) · P(Y).
To summarize, this lecture was about two logical operations on events, OR (∪) and AND (∩).
We have shown that the OR of mutually exclusive events is similar to addition, so we are justified replacing sign ∪ between two events with sign of addition +:
P(X+Y)=P(X)+P(Y)
and the AND of independent events is similar to multiplication, so we are justified replacing sign ∩ between two events with sign of multiplication ·:
P(X·Y)=P(X)·P(Y)
Finally, using the operation of intersection, we can express the probability of a union between any two (not necessarily mutually exclusive) events as
P(X∪Y)=P(X)+P(Y)−P(X∩Y).
Indeed, summing together probability measures of events X and Y, we counted twice the probability measures of all elementary events common for them. That's why we subtracted the probability measure of their intersection. | 1,216 | 4,702 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.21875 | 4 | CC-MAIN-2023-50 | latest | en | 0.967733 |
https://brilliant.org/problems/three-marbles-in-polyas-urn-2/ | 1,495,551,881,000,000,000 | text/html | crawl-data/CC-MAIN-2017-22/segments/1495463607647.16/warc/CC-MAIN-20170523143045-20170523163045-00007.warc.gz | 731,842,452 | 18,513 | # Three Marbles in Pólya's Urn
You have an urn with three marbles in it, a red one a yellow one and a blue one. Every ten seconds you pull out one marble at random and replace it together with one of the same color. e.g. If you pick out a red marble, you will replace it with two red marbles.
So, now the question...
The probability that at some point you will have 8 red marbles, 12 yellow marbles, and 3 blue ones after the corresponding number of moves is $$\dfrac{a}{b}$$, where $$a$$ and $$b$$ are coprime positive integers.
What is $$a + b$$?
× | 151 | 555 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.296875 | 3 | CC-MAIN-2017-22 | longest | en | 0.8984 |
https://www.teacherspayteachers.com/Product/I-Spy-Divide-Practice-Division-Facts-Puzzles-FUN-NO-PREP-Gr-3-CORE-MATH-993181 | 1,513,329,198,000,000,000 | text/html | crawl-data/CC-MAIN-2017-51/segments/1512948567785.59/warc/CC-MAIN-20171215075536-20171215095536-00088.warc.gz | 805,906,624 | 27,729 | Total:
\$0.00
# I Spy Divide | Practice Division Facts | Puzzles | FUN |NO PREP |Gr 3 CORE MATH
Common Core Standards
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Product Description
Division Facts . . . Early Finishers Worksheet Activity
I Spy Divide Puzzles CORE MATH ... Grade 3
I Spy Divide
is a great way to reinforce division facts with your students. The puzzles work just like a word search puzzle except they use numbers instead of words.
. . . NO PREP . . . DIVISION FUN . . . NO PREP . . . PUZZLES . . .
♥ ♥ A note from the authors ♥ ♥
In creating the puzzles and answer keys, we noticed we were repeating the division patterns over and over, 36 ÷6 = 6, 36 ÷ 6 = 6; 25 ÷ 5 = 5, 25 ÷ 5 = 5, and realized how this repetition will help students reinforce the basic division facts as they search the puzzles for each set of dividends, divisors, and quotients.
• The word searches are a great way to practice division. I can actually hear my students whispering the facts over and over as they look for them--certainly a fun way to memorize division facts.
• These are great practice!
• Great Resource! Thanks.
• Love this!!!!
Thank you!!! The kids love it!
© Jean VanDerford and Linda Schwartz at Teaching Stuff Place
Art by Corbin Hillam: corbinhillamdesign.weebly.com
Also check out other books I’ve written with my friend and co-author,
Linda Schwartz.
OTHER DIVISION PRACTICE:
Click here for Division in Designs Basic division skills practice in a fun format.
Click here DINOSAUR DIVISION to continue math practice in the same format.
MULTIPLICATION PRACTICE:
Click here for MULTIPLICATION SPORTS RIDDLES: Begins with multiplication of a 1-digit number by a 1-digit number and progresses to a 3-digit number by a 2-digit number.
Look for MONSTER MULTIPLICATION to continue multiplication practice
Our goal is to make learning fun for kids. ♥ ♥ ♥
Please take time to review I SPY DIVIDE
NEWEST PRODUCTS with SKILLS IN FUN FORMATS
♥ ♥ ♥ Please click here to check out NEW PRODUCTS in my SHOP ♥ ♥ ♥
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\$6.00 | 774 | 3,028 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.21875 | 4 | CC-MAIN-2017-51 | longest | en | 0.836988 |
https://www.doubtnut.com/qna/647476348 | 1,716,029,264,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971057379.11/warc/CC-MAIN-20240518085641-20240518115641-00810.warc.gz | 675,321,094 | 33,731 | # Find the angel between any two diagonals of a cube.
A
45
B
60
C
30
D
cos1(13)
Video Solution
Text Solution
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Step by step video, text & image solution for Find the angel between any two diagonals of a cube. 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
## Find the angel between any two diagonals of a cube.
A45
B60
C30
Dcos1(13)
• Question 1 - Select One
## The vertices of a cube are (0,0,0),(2,0,0),(0,0,2),(0,2,0),(2,0,2),(0,2,2),(2,2,0),(2,2,2)respectively What is the angle between any two diagonals of the cube?
AA) cos1(12)
BB) cos1(13)
CC) cos1(13)
DD) cos1(22)
• Question 1 - Select One
## The angle between two diagonals of a cube İs
Acos1(13)
B30
Ccos1(13)
D45
• Question 1 - Select One
## The angle between two diagonals of a cube will be
Asin1(13)
Bcos1(13)
Cvariable
Dnone of these
• Question 1 - Select One
## The angle between two diagonals of a cube will be
Asin1(13)
Bcos1(13)
Cvariable
Dnone of these
### Similar Questions
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Doubtnut is the perfect NEET and IIT JEE preparation App. Get solutions for NEET and IIT JEE previous years papers, along with chapter wise NEET MCQ solutions. Get all the study material in Hindi medium and English medium for IIT JEE and NEET preparation | 616 | 2,084 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-22 | latest | en | 0.857076 |
https://cs.lth.se/edan55/ | 1,721,063,048,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514713.2/warc/CC-MAIN-20240715163240-20240715193240-00694.warc.gz | 165,014,811 | 8,534 | lunduniversity.lu.se
Computer Science
Faculty of Engineering, LTH
We have selected a number of course-related programming exercises at the self-checking Kattis server.
(This system works very much like the ACM programming contest: You can upload your solution to a server, which checks them against known input-output pairs.)
This activity is optional, but useful.
* Register on kattis: https://lu.kattis.com/
* Register on the course offering aa17: https://lu.kattis.com/courses/EDAN55/aa17
* Go to the problem group "aa-exercises": https://lu.kattis.com/problemgroups/609
* Solve the 10 problems.
Course Format
The course consists of lectures and mandatory labs.
For 2017, labs are completed in groups of 4 students. You can now register at
https://sam.cs.lth.se/LabsSelectSession?occasionId=506
Informal Description
This is an advanced course in algorithms, targeted at students who are familiar with fundamental data structures (stacks, queues, trees, hash tables), algorithms (sorting, graph connectivity, dynamic programming, network flow), and computation complexity (NP-hardness). At LTH, this currently corresponds to the courses EDAA01 and EDAF05.
The focus of the course is “Algorithms for Hard Problems,” which could be a good alternative title for this course. Here, “hard” means one of two things. Either the instance is huge compared to what was seen in previous courses, so we need algorithms and data structures that work on massive data sets. Or the problem is hard in the sense of computational complexity (for example NP-hard), so no good algorithm is known even for tiny instances.
Technically, the glue that keeps this course together is randomisation: many of our troubles will be solved by randomised algorithms—a large part of the course could equally well be called “Randomised Algorithms”. That’s why the course also requires a solid background in discrete probability theory (events, probabilities, conditional probabilities, stochastic variables, mean, variance), corresponding to the LTH course FMS012. This allows us to visit (or revisit) a bunch of famous, useful, and important algorithms like Quicksort, Hashing or Google’s PageRank algorithm.
Finally, the course attempts to include a significant amount of bleeding edge current research of the highest international caliber, so that parts of the course material will (I hope) be research articles where the ink is still fresh.
Materials
The required textbook for this course is
• Kleinberg and Tardos, Algorithm Design, Addison Wesley, 1st edition (2005)
We’ll use the second half of this book. Many LTH students own this book from EDAF05, which is why I chose it. We’ll supplement this book with additional material from various sources (single chapters, online course notes, etc.). This will be announced. If you have too much money, you can buy two other excellent books:
• Mitzenmacher and Upfal, Probability and Computing: Randomized Algorithms and Probabilistic Analysis, Cambridge University Press (2005)
• Hromkovic, Algorithmics for Hard Problems: Introduction to Combinatorial Optimization, Randomization, Approximation, and Heuristics , 2nd edition, Springer (2010)
However, I won’t be using enough from either of these books to justify elevating them to required course books. View them as secondary references.
Assessment
The course contains of a number of mandatory exercises that are typically a mixture of implementation (coding) and analysis (maths). These are graded pass/fail; successful completion accounts for 3 of the 7.5 credits for this course.
The course ends with a written exam. The format is open book (you can bring the course book and your notes and whatever other printed material you want). The exam accounts for 4.5 of the course’s 7.5 credits. The final grade for EDAN55 is entirely determined by the performance at the exam.
Course Facts
Credits: 7.5 hp
Course period: HT1 2014 (3rd quarter)
Optional for: D4, E4, F4, Pi4
Course coordinator:
Thore Husfeldt
e-post: Thore.Husfeldt@ (lägg till cs.lth.se)
Schedule: Schedule for HT1 2014 via TimeEdit
Offcial course plan: Course plan for 2014 | 956 | 4,144 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.625 | 3 | CC-MAIN-2024-30 | latest | en | 0.894308 |
https://www.ercankoclar.com/en/2016/12/mathematical-solution-of-touch-panel-calibration/ | 1,722,926,869,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640476915.25/warc/CC-MAIN-20240806064139-20240806094139-00809.warc.gz | 595,948,817 | 26,582 | # MATHEMATICAL SOLUTION OF TOUCH PANEL CALIBRATION
## TOUCH PANEL CALIBRATION
• Touch panels are used by attaching them to the screens. However, the coordinates of the panel and the coordinates of the screen do not match. That’s why they have to be calibrated.
• Previously I had to calibrate through the function in the library I wrote for 4-wire resistive type panes. However, I learned the mathematical solution of the 2 point calibration method with the aim of learning the calibration mathematics of users of different types of touch panel.
• Do not forget to watch the video lesson.
### 2 POINT CALIBRATION METHOD
• There are many calibration methods. (2-4-6-12-16 point methods are available.)
• In general, large screens are calibrated using 2 denier values to provide angular calibration.
• Even though it is a 2 point method, touch 4 points. This is because X1 is X2 and Y1 is Y2.
• To do this, first of all, on the screen where the touch panel is installed, pixels are formed in certain places.
• The picture above is a template. Here 4 red dots are our calibration points.
• These points are created in the GLCD where the pixel rendering function corresponds to the above metrics. This is just to show where the user will touch.
• For a 240×128 GLCD, the coordinates that need to be created are points that need to be touched according to the template image above;
• 1.POINT x=%20 ve y=%80 so x=240*20/100= 48 y=128*80/100=102 so (48,102)
• 2.POINT x=%20 ve y=%20 so x=240*20/100=48 y=128*20/100=25 so (48,25)
• 3.POINT x=%80 ve y=%20 so x=240*80/100=192 y=128*20/100=25 so (192,25)
• 4.POINT x=%80 ve y=%80 so x=240*80/100=192 y=128*80/100=102 so (192,102)
• We have calculated above in which coordinates the 4 calibration points on the 240×128 screen should be created on the screen. Pixels will be created at these points and the user will be asked to touch the place.
• When the user touches this point, the raw x and y values coming from the touch panel will be read. (see circuit diagram and other details for “resistive touch panel library”)
• For example (the raw values are the values used in the lesson, and the value that each panel gives may vary).
• The raw value read when the first POINT is touched is x = 230 and y = 730. That is, when the point on the GLCD is touched (48,102), the read value is (230,730).
• The raw value read when the second POINT is touched is x = 228 and y = 237. That is, the value read when the point (48,25) on the GLCD is touched (228,237) is.
• The raw value read when the third POINT is touched is x = 780 and y = 232. That is, when the point (192,25) is touched on the GLCD, the read value is (780,232).
• The raw value read when the fourth POINT is touched is x = 773 and y = 704. That is, when the point (192,102) on the GLCD is touched, the read value is (773,704).
• We record the points touched in the above manner and the raw data from these points (the data comes from the touch panel).
• For screen control, you can use the “SAP1024B_GRAFIK_LCD” library I wrote earlier.
Touched Points in GLCD Values read from Touch Panel XD YD X Y 1.POINT 48 102 230 730 2.POINT 48 25 228 237 3.POINT 192 25 780 232 4.POINT 192 102 773 704
• The values read when this point is touched are our raw values.
• After these values, the mathematical operation sequence came up.
• With this process we will arrive at the solution by calculating our deviation shares.
As can be understood from its name, the deviation share calculates the deviation between the GLCD and the touch panel. Correcting the values accordingly enables the GLCD to have the same point in the coordinate plane in the touch panel.
• Example = If you touch the point of the screen (100,100) this time you will read it (100,100) through the touch panel instead of the different value.
### MATHEMATICAL SOLUTION OF CALIBRATION
#### Calculation of Deviation Values
• Before going to the end with the last formula, I need to calculate 4 values which I call “deviation value“.
• We will calculate the deviation values by using the table above.
• Four deviation values will be calculated. These are “a” value, “b” value, “c” value and “d” value.
• Each value will be calculated as “(a1 + a2) / 2 = averaging”.
• For all deviation values, 4 formulas will be used as shown in the picture above.
#### Calculation of Deviation Values “a1 and b1“
FORMULA1: YD1=(a1*Y1)+(b1*1) YD3=(a1*Y3)+(b1*1)
• Let’s put into place by reading the values from the table above.
102=(a1*730)+b1
25=(a1*232)+b1
———————–
102= (730a1) + b1
–25(232a1– b1
———————–
77=498a1
———————–
a1=(77/498) = 0,154
• Let us find the value of b1 based on the value of a1.
25=(0,154*232)+b1
———————–
25=(35,728)+b1
———————–
b1=(2537,728)= -10,728
#### Calculation of Deviation Values “a2 and b2“
FORMULA2: YD2=(a2*Y2)+(b2*1) YD4=(a2*Y4)+(b2*1)
• Let’s put into place by reading the values from the table above.
25=(a2*237)+b2
102=(a2*704)+b2
———————–
25= (237a2) b2
102= (704a2b2
———————–
77=467a2
———————–
a2=(77/467) = 0,164
• Let us find the value of b2 over this value as we find the value of a2.
25=(0,164*237)+b2
———————–
25=(38,868)+b2
———————–
b2=(2538,868)= -13,868
#### Calculation of Deviation Values “c1 and d1“
FORMULA3: XD1=(c1*X1)+(d1*1) XD3=(c1*X3)+(d1*1)
• Let’s put into place by reading the values from the table above.
48=(c1*230)+d1
192=(c1*780)+d1
———————–
48= (230c1) d1
192= (780c1+ d1
———————–
144=550c1
———————–
c1=(144/550) = 0,261
• Let’s find the value of d1 based on this value as we find the value of c1.
48=(0,261*230)+d1
———————–
48=(60,03)+d1
———————–
d1=(4860,03)= -12,03
#### Calculation of Deviation Values “c2 and d2“
FORMULA4: XD2=(c2*X2)+(d2*1) XD4=(c2*X4)+(d2*1)
• Let’s put into place by reading the values from the table above.
48=(c2*228)+d2
192=(c2*773)+d2
———————–
48= (228c2) d2
192= (773c2+ d2
———————–
144=545c2
———————–
c2=(144/545) = 0,264
• Since we find the value of c2, let’s find the value of d2 over this value.
48=(0,264*228)+d2
———————–
48=(60,192)+d2
———————–
d2=(4860,192)= -12,192
#### Deviation Values Result Table
1 2 Average a 0,154 0,164 0,159 b -10,728 -13,868 -12,298 c 0,261 0,264 0,262 d -12,03 -12,192 -12,111
• The results from the above tablature will be used in the final form.
• The averaging method is (a1 + a2) / 2.
#### Calibration by Using Result Formula
RESULT FORMULA YD=a*Y(b) XD=c*X(d)
• After all values have been calculated, calibration can now be done.
• The “Result Formula” above is the last point in the software. With this formula, the software calibrates. See how it works with examples.
• As an example I will use the values given in the table above.
• For example, I read 730 on the Y axis. But the point I clicked is actually 102 dir. (see the table at the beginning)
Let’s place them in the form;
YD=(0,159*730)-12,298=103,772
• I saw that the point I touched when I made transactions based on the deviations I have found should be 103. Actually, the point I actually touched is 102. Such deviations can be caused by touching places and fractions. However, it should not be taken into consideration. Because the software is processed with the value of the float, a large part of the fractions are accounted for, and such mistakes are eliminated.
• If we had to do an X-axis calculation:
• For example, when I touch the X axis, I read 230, but the point I clicked is actually 48. (see the table at the beginning)
Put your place in the form;
XD=(0,262*230)-12,111=48,149
• I have seen that the point to be touched in the calculations made according to the deviation values I have found should be 48. The point I already touched was the 48 axes on the X axis, that is, the calibration was done correctly.
## RESULT
• Calibration of the touch panels is done by calculations as described above. Random number determination is unscientific and not true.
• Regardless of the screen size, the above method takes you to the right conclusion because it deals with the percentiles.
• When calibrating, the raw “X” and “Y” values of each point touch are substituted in the “Conclusion Formula” and the result is taken.
Be sure to watch the video lesson below for action and other details. | 2,507 | 8,201 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.1875 | 4 | CC-MAIN-2024-33 | latest | en | 0.828205 |
http://www.answers.com/Q/Formula_to_determining_the_connections_in_a_full_mesh_network | 1,542,499,628,000,000,000 | text/html | crawl-data/CC-MAIN-2018-47/segments/1542039743913.6/warc/CC-MAIN-20181117230600-20181118012600-00505.warc.gz | 390,752,449 | 51,087 | # Formula to determining the connections in a full mesh network?
Would you like to merge this question into it?
#### already exists as an alternate of this question.
Would you like to make it the primary and merge this question into it?
#### exists and is an alternate of .
Where n = number of nodes The number of connections in a full mesh = n(n - 1) / 2
7 people found this useful
# What are the reasons for using mesh networking?
If one route to a certain computer goes down, theres another route which can be used. Just really to make a network less susceptible to downtime.
# 'do i need a wireless lan card on my mesh computer to connect to broadband with bt hub i have network ready integrated 10100 Ethernet?
I also can not get my Mesh computer to connect to the internet. I have bought a netgear router and wireless adaptor. In addition, i have connected the router via ethernet cabl
# Formula in a full mesh network?
Where n = number of nodes. The number of connections in a full mesh = n(n - 1) / 2
# What are the advantages of Mesh Network Topology?
Mesh network in which multiple redundant links exist between multiple nodes (computers/hosts/servers/distribution switches, etc) so that the network has a web-like topology.
secure: because each node is diretly connected to every other node robustness:if one link break down then it does not effect on whole network fast response:because each node i
# How many links are required to connect six computers in a full mesh topology?
let suppose total number of nodes/computers = n the formula will be = n(n-1)/2 e.g = 6(6-1)/2 =15 links.
# Find the number of connections needed to be used in Mesh topology in network of 8 nodes and also find that how many IO ports are needed for each device Also write formulas used to solve this question?
Find the number of connections needed to be used in Mesh topology in network of 8 nodes and also find that how many IO ports are needed for each device Also write formulas use
# What is mesh networking?
In a mesh networking, every node of the network is connected with all other nodes of the network. It is normally used for high redundancy of the network.
# What is mesh network topology?
A mesh topology is where each node or computer has a direct link to every other node that it needs to talk to. This topology is okay for small computer set ups (e.g. upto 6 c
# What are the reasons for not using mesh networking?
there a lot of reasons. on top my mind, exceptionally high cost of implementation. that is because the enormous length of cables that'll be required, the equipment needed to s
# What is mesh network or topology?
A network in which there are multiple network links betweencomputers to provide multiple paths for data to travel is called amesh network or topology.
# If 20 Routers are Networked with full mesh topology. How many Virtual connections between routers?
For a full mesh between 20 devices, you need a total of 20 (20-1) / 2 connections.\n For a full mesh between 20 devices, you need a total of 20 (20-1) / 2 connections.\n For
# What will happen when a SINGLE connection to one of the network hosts in a full mesh network fails?
Nothing should happen; that is reason you have fully redundant links, in case a link fails.
redundancy
# What is the full meaning of brc mesh?
Steel Reinforcement fabrics
In Linux
# How do you determine if a Linux computer has connected to a wireless network?
Generally speaking, look for a radiowave-like icon where the network connection area in the task bar should be. If it has radio waves, you're connected to wifi. The more te
In Uncategorized
# What are the benefits of a wireless mesh network?
A wireless mesh network is like a mini internet that can be used for cities, colleges, and large businesses. The advantages are less expensive due no need to install wires an | 853 | 3,871 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.203125 | 3 | CC-MAIN-2018-47 | latest | en | 0.936935 |
https://ftp.aimsciences.org/article/doi/10.3934/dcdsb.2018185 | 1,660,686,831,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882572581.94/warc/CC-MAIN-20220816211628-20220817001628-00651.warc.gz | 241,779,597 | 19,037 | # American Institute of Mathematical Sciences
July 2018, 23(5): 1873-1893. doi: 10.3934/dcdsb.2018185
## Generalized network transport and Euler-Hille formula
1 Department of Mathematics and Applied Mathematics, University of Pretoria, Pretoria, South Africa 2 Institute of Mathematics, Łódź University of Technology, Łódź, Poland 3 Advanced System Analysis Program, International Institute for Applied System Analysis, Laxenburg, Austria
* Corresponding author: Jacek Banasiak
The research was partially conducted during the scholarship of A. P. at the International Institute for Applied System Analysis and supported by a grant for young scientists of the Institute of Mathematics of Lódź University of Technology. J. B. was partially supported by the Incentive Funding of the National Research Foundation of South Africa.
Received April 2017 Revised August 2017 Published July 2018 Early access May 2018
In this article we consider asymptotic properties of network flow models with fast transport along the edges and explore their connection with an operator version of the Euler formula for the exponential function. This connection, combined with the theory of the regular convergence of semigroups, allows for proving that for fast transport along the edges and slow rate of redistribution of the flow at the nodes, the network flow semigroup (or its suitable projection) can be approximated by a finite dimensional dynamical system related to the boundary conditions at the nodes of the network. The novelty of our results lies in considering more general boundary operators than that allowed for in previous papers.
Citation: Jacek Banasiak, Aleksandra Puchalska. Generalized network transport and Euler-Hille formula. Discrete and Continuous Dynamical Systems - B, 2018, 23 (5) : 1873-1893. doi: 10.3934/dcdsb.2018185
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The research was partially conducted during the scholarship of A. P. at the International Institute for Applied System Analysis and supported by a grant for young scientists of the Institute of Mathematics of Lódź University of Technology. J. B. was partially supported by the Incentive Funding of the National Research Foundation of South Africa.
##### References:
[1] H. Amann and J. Escher, Analysis II, Birkhäuser, Basel, 2008. [2] F. M. Atay and L. Roncoroni, Lumpability of linear evolution equations in Banach spaces, Evolution Equation and Control Theory, 6 (2017), 15-34. doi: 10.3934/eect.2017002. [3] P. Auger, R. Bravo de la Parra, J.-C. Poggiale, E. Sánchez and T. Nguyen-Huu, Aggregation of variables and applications to population dynamics, Journal of Evolution Equations, 11 (2011), 121-154. doi: 10.1007/978-3-540-78273-5_5. [4] J. Banasiak, A. Goswami and S. Shindin, Aggregation in age and space structured population models: An asymptotic analysis approach, Journal of Evolution Equations, 11 (2011), 121-154. doi: 10.1007/s00028-010-0086-7. [5] J. Banasiak, A. Falkiewicz and P. Namayanja, Semigroup approach to diffusion and transport problems on networks, Semigroup Forum, 93 (2016), 427-443. doi: 10.1007/s00233-015-9730-4. [6] J. Banasiak and A. Falkiewicz, Some Transport and diffusion processes on networks and their graph realizability, Applied Mathematical Letters, 45 (2015), 25-30. doi: 10.1016/j.aml.2015.01.006. [7] J. Banasiak and A. Falkiewicz, A singular limit for an age structured mutation problem, Mathematical Biosciences and Engineering, 14 (2017), 17-30. doi: 10.3934/mbe.2017002. [8] J. Banasiak, A. Falkiewicz and P. Namayanja, Asymptotic state lumping in transport and diffusion problems on networks with application to population problems, Mathematical Models and Methods in Applied Sciences, 26 (2016), 215-247. doi: 10.1142/S0218202516400017. [9] J. Banasiak and M. Lachowicz, Methods of Small Parameter in Mathematical Biology, Birkhaüser/Springer, Cham, 2014. doi: 10.1007/978-3-319-05140-6. [10] J. Banasiak and P. Namayanja, Asymptotic behaviour of flows on reducible networks, Networks and Heterogeneous Media, 9 (2014), 197-216. doi: 10.3934/nhm.2014.9.197. [11] J. Bang-Jensen and G. Gutin, Digraphs. Theory, Algorithms and Applications, 2nd ed., Springer Verlag, London, 2009. doi: 10.1007/978-1-84800-998-1. [12] A. Bobrowski, On Hille-type approximation of degenerate semigroups of operators, Linear Algebra and its Applications, 511 (2016), 31-53. doi: 10.1016/j.laa.2016.08.036. [13] A. Bobrowski, Convergence of One-parameter Operator Semigroups in Models of Mathematical Biology and Elsewhere, Cambridge University Press, Cambridge, 2016. doi: 10.1017/CBO9781316480663. [14] A. Bobrowski, From diffusions on graphs to Markov chains via asymptotic state lumping, Annales Henri Poincare, 13 (2012), 1501-1510. doi: 10.1007/s00023-012-0158-z. [15] B. Dorn, Semigroups for flows in infinite networks, Semigroup Forum, 76 (2008), 341-356. doi: 10.1007/s00233-007-9036-2. [16] K.-J. Engel, R. Nagel, One-parameter Semigroups for Linear Evolution Equations, Springer Verlag, New York Berlin Heidelberg, 2000. [17] Y. Iwasa, V. Andreasen and S. Levin, Aggregation in model ecosystems. Ⅰ. Perfect aggregation, Ecological Modelling, 37 (1987), 287-302. [18] Y. Iwasa, V. Andreasen and S. Levin, Aggregation in model ecosystems. Ⅱ. Approximate aggregation, IMA Journal of Mathematics Applied in Medicine and Biology, 6 (1989), 1-23. doi: 10.1093/imammb/6.1.1-a. [19] T. Kato, Perturbation Theory for Linear Operators, Springer-Verlag, Berlin, 1995. [20] M. Kimmel and D. N. Stivers, Time-continuous branching walk models of unstable gene amplification, Bull. Math. Biol., 50 (1994), 337-357. [21] M. Kimmel, A. Świerniak and A. Polański, Infinite-dimensional model of evolution of drug resistance of cancer cells, J. Math. Systems Estimation Control, 8 (1998), 1-16. [22] M. Kramar and E. Sikolya, Spectral properties and asymptotic periodicity of flows in networks, Mathematische Zeitschrift, 249 (2005), 139-162. doi: 10.1007/s00209-004-0695-3. [23] J. L. Lebowitz and S. I. Rubinow, A theory for the age and generation time distribution of a microbial population, J. Theor. Biol., 1 (1974), 17-36. doi: 10.1007/BF02339486. [24] C. D. Meyer, Matrix Analysis and Applied Linear Algebra, SIAM, Philadelphia, 2000. doi: 10.1137/1.9780898719512. [25] D. Mugnolo, Semigroup Methods for Evolution Equations on Networks, Springer, Cham, 2014. doi: 10.1007/978-3-319-04621-1. [26] M. Rotenberg, Transport theory for growing cell population, Journal of Theoretical Biology, 103 (1983), 181-199. doi: 10.1016/0022-5193(83)90024-3.
Commutativity of the aggregation diagram
The graph G representing the canal network in Example 1
The line graph of the graph shown on Fig. 2
Graphical representation of the Kimmel–Stivers model
Kimmel–Stievers model with vital dynamics
Discrete Lebowitz–Rubinow–Rotenberg model
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2021 Impact Factor: 1.497 | 5,008 | 15,857 | {"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} | 2.546875 | 3 | CC-MAIN-2022-33 | latest | en | 0.845376 |
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A painless guide to crc error detection algorithms Painless Grammar (Painless Series) · Read more Software Error Detection through Testing and Analysis. A PAINLESS GUIDE TO CRC ERROR DETECTION ALGORITHMS INDEX V (9/24/96). Contents: Table of Contents · 1. Preface · ) About the Author &. A Painless Guide to CRC Error Detection Algorithms – gentooinit/crc.
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For example, if we chose a checksum function which was simply the sum of the bytes in the message mod i. Suppose that we drive the next 8 iterations using the calculated values which we could perhaps store in a single byte register and shift out to pick off each bit.
See Tanenbaum for a clearer explanation of this; I’m a little fuzzy on this one. It should be clear that the above algorithm will work for any width W. For example, in the second example above, the summing register could be a Megabyte wide, and the error would still go undetected. XOR this The point of this is that you can XOR constant values into a register to your heart’s delight, and in the end, there will exist a value which when XORed in with the original register will have the same effect as all the other XORs.
Actually, I’m being rather hard on whoever cooked this up because it seems that hardware implementations of the CRC algorithm used the reflected checksum value and so producing a reflected CRC was just right. Some popular polys are: If you’re reading this document in a sequential scroller, you can skip this code by searching for the string “Roll Your Own”. While addition is clearly not strong enough to form an effective checksum, it turns out that division is, so long as the divisor is about as wide as the checksum register.
LIBRO LA CARRERA ELIYAHU GOLDRATT PDF
And they’re consistent with the results of the reference code given in the Guide. Then we note three things: The most important aspect of the model algorithm is that it focusses exclusively on functionality, ignoring all implementation details. To ensure that the following code is working, configure it for the CRC and CRC algorithms given above and ensure that they produce the specified “check” checksum when fed the test string “” see earlier.
The steps of the algorithm are very simple: This is the header.
Suppose that we use a checksum register one-byte wide and use a constant divisor of gude, then the checksum is the remainder after is divided by Painlews code can be made even more unreadable as follows: While it seems clear that is greater than 10, it is no longer the case that can be considered to be greater than This section attempts to do that. Any multiple of G will be constructed using shifting and adding and it is impossible to construct a value with a single bit by shifting an adding a single value with more than one bit set, as the two end bits will always persist.
Typically, widths of 16 or 32 are chosen so as to simplify implementation on modern computers. In this situation, a normal sane software engineer would simply reflect each byte before processing it. An extra parameter allows the algorithm’s register to be initialized to a particular value.
## The Boston Diaries
And from that description, the code pretty much follows: An effect of this convention is that hardware engineers constructing hardware CRC calculators that operate at the bit level took to calculating CRCs of bytes streams with each of the bytes reflected within itself. However, at the further pinless of clarity which, you must admit, is already a pretty scare commodity in this code we can reorganize this small loop further so as to avoid the need to either augment the message with zero bytes, or to explicitly process zero bytes at the end as above.
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Thus, the result becomes: The upshot of all this is that a reflected algorithm is not equivalent to the original algorithm with the poly reflected.
Choosing A Poly 8. However, permission is granted to make and distribute verbatim copies of this document provided that this information block and copyright notice is included. So what’s the point?
### A painless guide to crc error detection algorithms – PDF Free Download
The only trick is that W zero bits are appended to the message before the CRC is calculated. Except this time, it’s in binary: However, in the next section, we will be assuming option b because it is marginally mathematically cleaner.
Unfortunately, this possibility is completely unavoidable and the detectionn that can be done is to minimize its probability by increasing the amount of information in the checksum e. End The register now contains the remainder. For this reason, the next section attempts to provide a well-defined parameterized model for CRC algorithms.
This is a name given to the algorithm.
I didn’t think the code I wrote for reflected CRCs was that unreasonable based upon the information in the Guide, but I guess I was wrong for some of dettection.
However, we do not need to go so far; the next arithmetic step suffices. This parameter should be specified as a hexadecimal number. That is, E consists of all zeros except for a run of 1s somewhere inside. | 1,174 | 5,479 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2021-25 | latest | en | 0.86767 |
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It seems a general guideline because that measuring tea is to usage 1 rounded teaspoon for 1 cup of water. However, some world say 1 cup is 6oz and also some say it is 8oz. What is the basic consensus?
Well, its as with they say around case law...it depends. And as frustrating as it sounds, the depends. The confident is, you room the variable. Play roughly with how it tastes and make adjustments because that what girlfriend like. (or whoever you room making it for)
"You desire the taste of dried leaves in boiled water?""Er, yes. V milk.""Squirted the end of a cow?""Well, in a path of speaking I suppose ..."
I read somewhere that a typical cup that tea means 6oz if a common cup the coffee method 8oz. However I can"t remember whereby I check out it or if it to be reputable. For me, the size of my cup the tea relies on the cup I"m utilizing for the tea. Yet I with simply 2 oz difference, ns think a teaspoon per 6 or 8 oz is more than likely okay. Also using a scale, there is very small difference in the amount.
You are watching: How many ounces in a cup of tea
The book "The Story of Tea" defines that the measurement come from the 1920s, when they (modern food pioneers, according to the book) to be trying come determine continuous parameters for impending tea and also coffee. They discovered that 1 measure teaspoon coincided to 2 grams the tea, which was the approximate amount in a teabag. The tea tasters determined that 6 oz. The water to 2 grams the tea produced the finest flavor and had the appropriate level of "soluble solids taken into consideration to be the perfect cup that brewed tea."I have made a cheat sheet for the weights that tea for every my favorite cups, none of them take place to be 6 oz.!
semi-official modern source because that conversions:http://unit-converter.org/en/volume/usteacup.htmlUS teacup is 6 oz.US cup is 8 ozSeveral of my vintage English china teapots are 4 teacups, 24+ oz, conforming come the virtually uniform size of the Staffordshhire teacup that came to be standardized in the early on 20th century (I think this corresponded with standardization of the size of mattresses and beds, furniture favor dining table (to to the right linens) and also chairs and probably room dimensions on affordable homes, at the beginning of mass-production era of residential goods in the UK (Brit Empire) and also Europe and North America.
I agree with the above posters that 6oz is what i see many often however I think its important to ask castle how plenty of oz in there cup because there room so plenty of differences. You have the right to have as many oz in her cup as you want
due to the fact that I use a coffee mug many of the time my cup is in between 8-12oz.
See more: What Does It Mean When You Dream About A Tiger Dream Meaning And Interpretations
I"ve seen some Japanese tea instructions for a cup wherein a "cup" is 4oz. Well, that was yes, really something favor 3.7oz. Convert from mL, I"m sure.
My cup are mostly 8 oz. I usage a teaspoonful for 8 oz. And I mix 2 infusions with each other for 16 ounces the end of every teaspoonful. Steve
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U.S. Army Research Laboratory experts have found out a way to leverage emerging mind-like personal computer architectures for an age-old range-theoretic difficulty recognized as integer factorization.
By mimicking the brain capabilities of mammals in computing, Army experts are opening up a new remedy house that moves absent from traditional computing architectures and toward devices that are capable to function within just extreme dimension-, body weight-, and energy-constrained environments.
“With far more computing electric power in the battlefield, we can system info and address computationally-really hard issues more quickly,” reported Dr. John V. “Vinnie” Monaco, an ARL computer system scientist. “Programming the variety of units that fit this criteria, for illustration, brain-inspired personal computers, is tough, and cracking crypto codes is just 1 software that shows we know how to do this.”
The dilemma alone can be stated in very simple conditions. Get a composite integer N and specific it as the product of its key components. Most persons have done this undertaking at some level in grade college, often an exercise in elementary arithmetic. For case in point, 55 can be expressed as 5*11 and 63 as 3*3*7. What lots of did not notice is they were being accomplishing a process that if accomplished rapidly sufficient for large quantities, could split significantly of the fashionable day net.
Community essential encryption is a approach of secure interaction utilized extensively nowadays, centered on the RSA algorithm formulated by Rivest, Shamir, and Adleman in 1978. The safety of the RSA algorithm relies on the problem of factoring a massive composite integer N, the community vital, which is distributed by the receiver to any one who needs to mail an encrypted information. If N can be factored into its key factors, then the private crucial, needed to decrypt the information, can be recovered. Having said that, the problems in factoring big integers rapidly gets obvious.
As the dimensions of N raises by a single digit, the time it would consider to issue N by trying all achievable combinations of key components is roughly doubled. This signifies that if a selection with 10 digits will take 1 moment to variable, a variety with 20 digits will take about 17 several hours and a selection with 30 digits about two yrs, an exponential advancement in energy. This problem underlies the safety of the RSA algorithm.
Tough this, Monaco and his colleague Dr. Manuel Vindiola, of the lab’s Computational Sciences Division, shown how brain-like personal computers lend a speedup to the presently greatest recognized algorithms for factoring integers.
The staff of researchers have devised a way to element huge composite integers by harnessing the significant parallelism of novel laptop or computer architectures that mimic the working of the mammalian mind. So named neuromorphic computers work beneath vastly unique rules than standard computer systems, these kinds of as laptops and mobile products, all centered on an architecture described by John von Neumann in 1945.
In the von Neumann architecture, memory is individual from the central processing device, or CPU, which will have to read and compose to memory in excess of a bus. This bus has a minimal bandwidth, and a lot of the time, the CPU is ready to entry memory, usually referred to as the von Neumann bottleneck.
Neuromorphic desktops, on the other hand, do not put up with from a von Neumann bottleneck. There is no CPU, memory, or bus. In its place, they incorporate several unique computation models, a great deal like neurons in the mind.
These models are related by actual physical or simulated pathways for passing knowledge around, analogous to synaptic connections concerning neurons. Several neuromorphic gadgets operate based mostly on the actual physical response qualities of the fundamental substance, these as graphene lasers or magnetic tunnel junctions. Because of this, these products consume orders of magnitude a lot less energy than their von Neumann counterparts and can operate on a molecular time scale. As this sort of, any algorithm able of functioning on these devices stands to gain from their capabilities.
The speedup acquired by the ARL researchers is due to the formulation of a method for integer factorization with the enable of a neuromorphic co-processor. The recent speediest algorithms for factoring integers consist generally of two levels, sieving and a matrix reduction, and the sieving stage contains most of the computational hard work.
Sieving will involve browsing for several integers that fulfill a particular home referred to as B-clean, integers that don’t consist of a key variable larger than B. Monaco and Vindiola ended up capable to construct a neural network that discovers B-sleek figures quicker and with better precision than on a von Neumann architecture. Their algorithm leverages the enormous parallelism of brain-impressed pcs and the innate capability of unique neurons to accomplish arithmetic operations, these as addition. As neuromorphic architectures keep on to raise in size and speed, not restricted by Moore’s Legislation, their capacity to deal with bigger integer factorization complications also grows. In their get the job done, it is estimated that 1024-little bit keys could be damaged in about a calendar year, a process as soon as believed to be out of access. For comparison, the current history, a 232 decimal digit selection (RSA-768) took about 2,000 yrs of computing time around the course of various many years.
From a broader point of view, this discovery pushes us to issue how a change in computing paradigm may well have an affect on some of our most simple safety assumptions. As rising units shift to include enormous parallelism and harness product physics to compute, the computational hardness underlying some security protocols might be challenged in strategies not earlier imagined. This get the job done also opens the doorway to new analysis regions of rising personal computer architectures, in phrases of algorithm layout and operate illustration, alongside very low-electricity device studying and synthetic intelligence applications.
“Encrypted messages in warfare frequently have an expiration day, when their contents become un-actionable,” Monaco reported. “There is an urgency to decrypt enemy communications, especially all those at the discipline level, considering the fact that these expire the quickest, as opposed to conversation at higher echelons. In industry circumstances, electrical power and connectivity are extremely confined. This is a robust motivating issue for working with a mind-inspired laptop for these types of a process the place typical desktops are not functional.” | 1,317 | 6,882 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2019-43 | latest | en | 0.944608 |
https://math.stackexchange.com/questions/3784520/looking-at-nature-for-answers-to-math-problems/3784526 | 1,610,738,291,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703495936.3/warc/CC-MAIN-20210115164417-20210115194417-00616.warc.gz | 504,679,778 | 36,530 | # Looking at nature for answers to math problems
I recently watched a video about the Gömböc, a shape with one stable and one unstable point of equilibrium. Dr. Gábor Domokos proved the existence of this shape. Domokos originally looked at pebbles when trying to find the shape, although this approach was ultimately unsuccessful. Later on, some turtles were found to have a similar shape to the Gömböc.
I am looking for some other examples of when researchers have drawn inspiration (or completely copied) from nature in order to find solutions to math/engineering problems. To clarify, I am not looking for examples of explaining nature using math, as there are many examples of that (e.g. the Fibonacci spiral, arrangement of electrons in atoms/molecules, normal distribution, etc), but instead for examples of explaining math/engineering using nature.
I wasn't able to come up with or find any more examples myself, only the Gömböc.
Edit: I feel like there must have been some instance of when researchers have used the natural positioning of electrons to model something else, but I can't think of anything off the top of my head.
• You're welcome! I also only heard about it recently, and was fascinated by it (although I still don't understand the physics behind "unstable equilibrium points"). – Varun Vejalla Aug 8 '20 at 23:26
I remember reading about a research paper in which the researchers used an amoeba to gain approximate solutions to the traveling salesman problem. Although that is a computer science problem, it is not entirely irrelevant to mathematics.
Amoeba finds approximate solutions to NP-hard problem in linear time. In the new study, the researchers found that an amoeba can find reasonable (nearly optimal) solutions to the TSP in an amount of time that grows only linearly as the number of cities increases from four to eight.
I also found the following book that might be of interest to you. A look at the table of contents will provide you with a lot of interesting examples.
Nature-Inspired Optimization Algorithms by Xin-She Yang.
Hope this helps.
• Neat - this is an interesting example! Thanks for this. Maybe we'll have living computer chips made out of amoeba one day :) – Varun Vejalla Aug 9 '20 at 0:01
• @VarunVejalla You are very welcome. :$)$ – Khashayar Baghizadeh Aug 9 '20 at 0:02
• @VarunVejalla Check out the edit. – Khashayar Baghizadeh Aug 9 '20 at 18:45
• Wow, there's a lot of examples in the book. I guess a lot of machine learning and artificial intelligence uses nature. – Varun Vejalla Aug 9 '20 at 18:55
• I think we can also learn from the chipmuks about how they solve their TSP. Chipmunks love peanuts so I feed them peanuts so they can survive the Canadian winter. It's really interesting to see how they collect the peanuts you throw at them really fast. The advantage of observing chipmunks instead of amoeba is obvious. Chipmunks are much bigger and can be observed an video-recorded. One can also feed peanuts to the squirrels, the chipmunks cousins to learn how they solve their TSP. But it will cost more to feed a squirrel. – user25406 Sep 6 '20 at 21:53
This is a cool question. One example I know of is that of the wholeness axiom of Paul Corazza. Roughly, the axiom is a large cardinal axiom which states something about the reflection properties of the universe $$V$$ by way of large cardinals. Paul Corazza is said to have had access to this axiom after meditating. Not exactly looking at nature, but I guess looking inside oneself.
https://en.wikipedia.org/wiki/Wholeness_axiom
More corny is the classical story of Newton and the apple.
• I'm not sure the wholeness axiom fits, since "looking inside oneself" could just be considered as thinking intensely (or self-surgery, but that's another matter). Newton and the apple might work, although I think it fits better into the "using math to explain nature" category. – Varun Vejalla Aug 8 '20 at 23:13
• Yes Varun. Maybe another way to look at the question would be to see how the entire beginings of Differential Equations be seen as humans looking at nature and seeing it change. As it changes, humans want to understand what governs that change. We have access to how fast nature changes, hence rate of change. So maybe just looking at a river or leaves shaking under the action of the wind, makes you think about the laws upon which all of this rests. Is that what you had in mind? – Rachid Atmai Aug 8 '20 at 23:21
• As a decades-long meditator and former math professor I assure you that meditating is the exact OPPOSITE of looking to nature as asked in the question. I can't even imagine a method that is more the opposite of what the questioner had in mind. – David G. Stork Aug 8 '20 at 23:25
• @RachidAtmai That's somewhat what I'm looking for, but not completely. I guess it is the "direction of influence" that matters - I'm not looking for when humans have tried explaining nature using math, but explaining math using nature. – Varun Vejalla Aug 8 '20 at 23:28
• @DavidG.Stork You seem pretty riled up for someone who meditates haha... Peace man, hope you find some equanimity. Although aren't we ourselves part of nature though? Think about it... – Rachid Atmai Aug 8 '20 at 23:39 | 1,247 | 5,235 | {"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": 1, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.890625 | 3 | CC-MAIN-2021-04 | latest | en | 0.961291 |
http://stackoverflow.com/questions/7565675/enumerating-all-possible-strings-of-length-k-from-an-alphabet-in-python?answertab=oldest | 1,406,351,707,000,000,000 | text/html | crawl-data/CC-MAIN-2014-23/segments/1405997894983.24/warc/CC-MAIN-20140722025814-00068-ip-10-33-131-23.ec2.internal.warc.gz | 344,825,356 | 16,865 | # enumerating all possible strings of length K from an alphabet in Python [duplicate]
Possible Duplicate:
is there any best way to generate all possible three letters keywords
how can I enumerate all strings of length K from an alphabet L, where L is simply a list of characters? E.g. if `L = ['A', 'B', 'C']` and `K = 2`, I'd like to enumerate all possible strings of length 2 that can be made up with the letters `'A'`, `'B'`, `'C'`. They can be reused, so `'AA'` is valid.
This is essentially permutations with replacement, as far as I understand. if theres a more correct technical term for this, please let me know.... its essentially all strings of length K that you can make by choosing ANY letter from the alphabet L, and possibly reusing letters, in a way that is sensitive to order (so `AB` is NOT identical to `BA` according to this.) is there a clearer way to state this?
in any case i believe the solution is:
``````[ ''.join(x) for x in product(L, repeat=K) ]
``````
but i am interested in other answers to this, esp. naive approaches versus fast Pythonic ones, and discussions of speed considerations.
-
## marked as duplicate by agf, JMax, Jeff Atwood♦Sep 27 '11 at 9:03
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
Please write some code and / or tell us specifically what problem you're facing trying to do that. – agf Sep 27 '11 at 7:39
You can do it by using recursion. – tgmath Sep 27 '11 at 7:40
I agree it's a duplicate, apologies for missing it. would still like to know about recursive way to do this since that is not mentioned in other thread in detail – user248237dfsf Sep 27 '11 at 8:03
Your edit has made this a discussion / polling question -- it no longer has a specific correct answer, or even a specific question ("I am interested in..."). That's not on-topic for this site. – agf Sep 27 '11 at 8:07
## 2 Answers
this is part of the Python Documentation
EDIT2: of course the right answer is the product, thanks for the comment
``````print [''.join(x) for x in product('ABC', repeat=3)]
``````
prints 27 elements
``````['AAA', 'AAB', 'AAC', 'ABA', 'ABB', 'ABC', 'ACA', 'ACB', 'ACC', 'BAA', 'BAB',
'BAC', 'BBA', 'BBB', 'BBC', 'BCA', 'BCB', 'BCC', 'CAA', 'CAB', 'CAC', 'CBA',
'CBB', 'CBC', 'CCA', 'CCB', 'CCC']
``````
@agf gave the right answer before
-
Doesn't look right - if the alphabet size (iterable) is length 3, and r = 3, you'd expect 3^3 sequences, but len([x for x in combinations_with_replacement(['A', 'B', 'C'], 3)]) returns 10. – user248237dfsf Sep 27 '11 at 7:48
see my edit, which element would be wrong? :-) – knitti Sep 27 '11 at 7:53
none are wrong, but it misses some. there should be 3^3, or 27 elements: [ ''.join(x) for x in product('ABC', repeat=3) ] ['AAA', 'AAB', 'AAC', 'ABA', 'ABB', 'ABC', 'ACA', 'ACB', 'ACC', 'BAA', 'BAB', 'BAC', 'BBA', 'BBB', 'BBC', 'BCA', 'BCB', 'BCC', 'CAA', 'CAB', 'CAC', 'CBA', 'CBB', 'CBC', 'CCA', 'CCB', 'CCC'] – user248237dfsf Sep 27 '11 at 7:58
Where is `'ABA'`, `'ACA'` etc? Please read your link "Combinations are emitted in lexicographic sort order." The right answer is given (by me) in stackoverflow.com/questions/7074051/… – agf Sep 27 '11 at 7:59
thanks, your right. linked to you answer – knitti Sep 27 '11 at 8:06
You can use `itertools`:
``````n = 3
itertools.product(*['abc']*n)
``````
This gives you 27 elements as you expected.
-
Read the `product` docs more carefully and see my answer in the linked duplicate stackoverflow.com/questions/7074051/… – agf Sep 27 '11 at 8:01 | 1,084 | 3,605 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2014-23 | latest | en | 0.939425 |
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