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http://peakgmt.com/math/line-y-eq-mx.html | 1,696,231,045,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510983.45/warc/CC-MAIN-20231002064957-20231002094957-00013.warc.gz | 31,468,911 | 2,266 | # Line \bold{\LARGE{y = mx}}
## What does it stand for?
y = how far up
x = how far along
m = Slope or Gradient (how steep the line is)
b = the y Intercept (where the line crosses the y axis)
## How do you find "m" and "b"?
b is easy: just see where the line crosses the y-axis.
m (the Slope) needs some calculation:
\bold{m = \dfrac{\text{Change in} \space y\text{-axis}}{\text{Change in} \space x\text{-axis}}}
## Knowing this we can work out the equation of a straight line:
### Example 1
m = \dfrac{2}{1} = 2
b = 1 (where the line crosses the y-Axis)
So: y = 2x + 1
### Example 2
m = \dfrac{-3}{1} = -3
b = 0
This gives us y = -3x + 0
We do not need the zero!
So: y = -3x
## Rise and Run
Sometimes the words "rise" and "run" are used.
\bold{Rise} is how far up
\bold{Run} is how far along
And so the slope "m" is:
\bold{m = \dfrac{rise}{run}}
You might find that easier to remember
## Important Fact
if the |slope| (absolute value) of the line is greater than 1, then the angle of the line with x-axis greater than 45\degree
if the angle of the line with x-axis greater than 45\degree, then the |slope| (absolute value) of the line is greater than 1
## Practice Question
w \gt 45
Quantity A Quantity B m + n 2m
A. The quantity in Column A is greater
B. The quantity in Column B is greater
C. The two quantities are equal
D. The relationship cannot be determined from the information given
The equation for a line is:
\bold{y = kx + b}
where m is the slope and b is the y-intercept. Here, we are told that the line passes through the point \space (0,0) and point \space (m,n).
for m and n, we find that: m \gt 1 and n \gt 1, it is obvious that m \not = 0 and n \not = 0
Passing point \space (0,0) means that the y-intercept is 0. So the formula for line k is:
\bold{y = kx}
In order to compare A and B, let's try to subsract A by B:
m + n - 2m = n - m
So the question is to compare m and n.
w \gt 45\degree \to n \gt m, so | 630 | 1,971 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.6875 | 5 | CC-MAIN-2023-40 | latest | en | 0.860781 |
https://plinkyplonky.com/north-warrandyte/application-of-cartesian-coordinate-system-in-real-life.php | 1,638,471,175,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964362287.26/warc/CC-MAIN-20211202175510-20211202205510-00094.warc.gz | 490,751,930 | 9,622 | Rectangular coordinate system SlideShare Chapter V: Review and Application of Vectors In the previously chapters, in a Cartesian coordinate system with Л†Л†Л† and A=+A xy iAj+A z kB=++B xyz iBjBk r r
## Cartesian Coordinate System Lesson Plans & Worksheets
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What are polar coordinates used for in real life? Socratic. Teaching the Cartesian coordinate system can be fun when placed in the context of students. The xy coordinate system offers many opportunities for creativity and fun https://en.wikipedia.org/wiki/Coordinates_(geographic) Chapter V: Review and Application of Vectors In the previously chapters, in a Cartesian coordinate system with Л†Л†Л† and A=+A xy iAj+A z kB=++B xyz iBjBk r r.
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## what are real world applications for polar coordinates
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Coordinate Systems. The term "Cartesian coordinates" is used to In such a coordinate system you can calculate the distance between two points and perform As with the two-dimensional Cartesian coordinate system, The polar coordinate system can also be more easily applied to certain real-life situations because it | 3,245 | 17,404 | {"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} | 4.03125 | 4 | CC-MAIN-2021-49 | longest | en | 0.809824 |
http://lensocom.com/math-problems-addition-subtraction-multiplication-division/addition-and-subtraction-activities-multiplication-and-division-worksheets-grade-3-simple-addition-worksheets-free-addition-worksheets-subtraction-problems/ | 1,556,250,019,000,000,000 | text/html | crawl-data/CC-MAIN-2019-18/segments/1555578759182.92/warc/CC-MAIN-20190426033614-20190426055614-00157.warc.gz | 101,071,412 | 17,573 | ## lensocom Math is like a solving the puzzle
Published at Tuesday, 26 March 2019. division. By .
What happens is that those who mastered their time tables in the second grade will have no problem solving these long division problems. They can fully focus on the process, instead of concentrating on the math facts. Those who struggled with their multiplication tables will most likely also struggle with long division. The ability and motivation gap between students is widening as the years progress.
This problem is not really prevalent in the first years at school, where math is relatively easy and considered fun. But after a brief and relaxed honeymoon period, the going will get tough in later years. After the introduction of multiplication tables in the first two school years, students are required to memorize these tables accurately and above all to recite them without thinking. Too many students, however, will rely on their addition skills and will not be able to recite the multiplication facts quickly enough.
Addition. The concept of addition is the first of four math operations that children have to learn and being able to properly teach them the concept will provide them encouragement and the skill to move forward and still find math easy to learn. You can start by adding 1 to a number to generate the next number before going into adding one digit numbers. When the kids have mastered addition with one-digit numbers then you can move on with the one-digit and two-digit numbers combination until you can move on to multiple digits. Keep their interest by giving them competitive addition worksheets.
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#### math problems addition subtraction multiplication division Gallery
45 of 100 by 734 users | 363 | 1,861 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2019-18 | latest | en | 0.95486 |
https://www.geeksforgeeks.org/java-program-to-find-the-volume-and-surface-area-of-sphere/?ref=rp | 1,660,587,909,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882572198.93/warc/CC-MAIN-20220815175725-20220815205725-00189.warc.gz | 683,244,056 | 23,222 | # Java Program to Find the Volume and Surface Area of Sphere
• Last Updated : 26 Nov, 2020
A sphere is a geometrical object in 3D space that is the surface of a ball. The sphere is defined mathematically as the set of points that are all at the same distance of radius from a given point in a 3D space.
Example:
```Input: radius = 5
Output: Surface Area ≈ 314.16
Volume ≈ 523.6
Input : radius = 3
Output: Surface Area ≈ 113.1
Volume ≈ 113.1```
Surface area of a sphere = 4*3.14*(r*r)
Volume of a sphere = (4/3)*3.14*(r*r*r)
Algorithm
1. Initializing value of r as 5.0, surface area as 0.0,volume as 0.0
2. Calculating surface area and volume of a sphere using the below formulas
3. Surface area=4*3.14(r*r)
4. Volume=(4/3)3.14(r*r*r)
5. Display surface area and volume
Implementation:
## Java
`// Java Program to Find the Volume and Surface Area of``// Sphere``class` `surfaceareaandvolume {`` ``public` `static` `void` `main(String[] args)`` ``{`` ``double` `r = ``5.0``, surfacearea = ``0.0``, volume = ``0.0``;`` ``surfacearea = ``4` `* ``3.14` `* (r * r);`` ``volume = ((``double``)``4` `/ ``3``) * ``3.14` `* (r * r * r);`` ` ` ``System.out.println(``"surfacearea of sphere ="`` ``+ surfacearea);`` ` ` ``System.out.println(``"volume of sphere ="` `+ volume);`` ``}``}`
Output
```surfacearea of sphere =314.0
volume of sphere =523.3333333333334```
Time Complexity: O(1)
Space Complexity: O(1)
My Personal Notes arrow_drop_up | 498 | 1,543 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.890625 | 3 | CC-MAIN-2022-33 | latest | en | 0.674125 |
https://spell.today/square-root/64.html | 1,685,245,988,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224643462.13/warc/CC-MAIN-20230528015553-20230528045553-00442.warc.gz | 629,366,493 | 6,291 | # What is the Exact Square Root of 64?
As you should know from your high school algebra course, the square root y of a number x is such that y2 = x. By multiplying the value y by itself, we get the value x. For instance, 8 the square root of 64 because 82 = 8×8 = 64.
Square root of 64 = 8
1642
The symbol √ is called
radix, or radical sign
The number below
the radix is the radicand
## Is 64 a Perfect Square Root?
Yes. The square root of 64 is 8. Since 8 is a whole number, 64 is a perfect square.
Previous perfect square root is: 49
Next perfect square root is: 81
## The Prime Factors of 64 are:
2 × 2 × 2 × 2 × 2 × 2
## How Do You Simplify the Square Root of 64 in Radical Form?
The main point of simplification (to the simplest radical form of 64) is as follows: getting the number 64 inside the radical sign √ as low as possible.
64 = 2 × 2 × 2 × 2 × 2 × 2 = 8
Therefore, the answer is 8.
## Is the Square Root of 64 Rational or Irrational?
Since 64 is a perfect square (it's square root will have no decimals), it is a rational number.
## The Babylonian (or Heron’s) Method (Step-By-Step)
StepSequencing
1
In step 1, we need to make our first guess about the value of the square root of 64. To do this, divide the number 64 by 2.
As a result of dividing 64/2, we get the first guess: 32
2
Next, we need to divide 64 by the result of the previous step (32).
64/32 = 2
Calculate the arithmetic mean of this value (2) and the result of step 1 (32).
(32 + 2)/2 = 17 (new guess)
Calculate the error by subtracting the previous value from the new guess.
|17 - 32| = 15
15 > 0.001
Repeat this step again as the margin of error is greater than than 0.001
3
Next, we need to divide 64 by the result of the previous step (17).
64/17 = 3.7647
Calculate the arithmetic mean of this value (3.7647) and the result of step 2 (17).
(17 + 3.7647)/2 = 10.3824 (new guess)
Calculate the error by subtracting the previous value from the new guess.
|10.3824 - 17| = 6.6176
6.6176 > 0.001
Repeat this step again as the margin of error is greater than than 0.001
4
Next, we need to divide 64 by the result of the previous step (10.3824).
64/10.3824 = 6.1643
Calculate the arithmetic mean of this value (6.1643) and the result of step 3 (10.3824).
(10.3824 + 6.1643)/2 = 8.2734 (new guess)
Calculate the error by subtracting the previous value from the new guess.
|8.2734 - 10.3824| = 2.109
2.109 > 0.001
Repeat this step again as the margin of error is greater than than 0.001
5
Next, we need to divide 64 by the result of the previous step (8.2734).
64/8.2734 = 7.7356
Calculate the arithmetic mean of this value (7.7356) and the result of step 4 (8.2734).
(8.2734 + 7.7356)/2 = 8.0045 (new guess)
Calculate the error by subtracting the previous value from the new guess.
|8.0045 - 8.2734| = 0.2689
0.2689 > 0.001
Repeat this step again as the margin of error is greater than than 0.001
6
Next, we need to divide 64 by the result of the previous step (8.0045).
64/8.0045 = 7.9955
Calculate the arithmetic mean of this value (7.9955) and the result of step 5 (8.0045).
(8.0045 + 7.9955)/2 = 8 (new guess)
Calculate the error by subtracting the previous value from the new guess.
|8 - 8.0045| = 0.0045
0.0045 > 0.001
Repeat this step again as the margin of error is greater than than 0.001
7
Next, we need to divide 64 by the result of the previous step (8).
64/8 = 8
Calculate the arithmetic mean of this value (8) and the result of step 6 (8).
(8 + 8)/2 = 8 (new guess)
Calculate the error by subtracting the previous value from the new guess.
|8 - 8| = 0
0 < 0.001
Stop the iterations as the margin of error is less than 0.001
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chain rule (partial derivatives) watch
1. use the chain rule to find dw/dt
w=xe^y/2 , x= t^2 , y= 1-t , z=1+2t
any help would be great thanks
2. Are you sure you've quoted the question right? It seems to me that dx/dt = 2t.
3. i corrected it thanks ...
4. (Original post by david cool)
use the chain rule to find dw/dt
w=xe^y/2 , x= t^2 , y= 1-t , z=1+2t
any help would be great thanks
I'm not sure whether your expression for w is xe^(y/2) or x(e^y)/2. I'll do the working for the latter and you can easily change it if it's the other one.
pw/pt=(pw/px)(px/pt)+(pw/py)(py/pt)
=(e^y / 2)(2t)+(xe^y / 2)(-1)
=te^(1-t)-(1/2)(t^2)e^(1-t)
=(e^(1-t))[t-0.5t^2])
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The application of statistics in healthcare plays a critical role in understanding and interpreting data. Statistical analysis allows healthcare professionals to make evidence-based decisions, evaluate the effectiveness of treatments, identify trends and patterns in patient outcomes, and assess the quality of healthcare services. In this paper, we will explore the significance of statistical application in healthcare, highlighting its impact on research, clinical practice, and healthcare management.
Research in healthcare heavily relies on statistical methods to collect, analyze, and interpret data. Statistics enables researchers to design studies in a way that ensures the results are reliable and valid. For example, in randomized controlled trials, statistical techniques such as sample size calculation and randomization help ensure that the study participants are representative of the population and that any observed effects are not due to chance. Additionally, statistical tests, such as t-tests and ANOVA, are used to compare groups or assess the relationship between variables. These tests provide researchers with a way to determine if the observed differences or associations are statistically significant, allowing them to draw meaningful conclusions from their data.
Furthermore, statistical analysis is essential for evaluating the effectiveness of treatments and interventions in healthcare. Clinical trials and observational studies often use statistical methods to analyze patient outcomes and determine if a treatment is beneficial. For instance, the use of statistical tests like chi-square and logistic regression allows researchers to determine if a new medication or therapy has a significant effect on disease outcomes compared to a placebo or standard of care. Statistical analysis also plays a role in monitoring adverse events and identifying potential risks associated with treatments. By analyzing data from large populations, statisticians can detect rare side effects or adverse reactions that may not be evident in smaller studies or during the initial stages of drug development.
In addition to research and clinical practice, statistical application is integral to healthcare management and decision-making. Healthcare providers, administrators, and policymakers rely on statistical data to evaluate the quality of care delivered, develop policies, and allocate resources effectively. Statistical techniques such as data mining and regression analysis are used to identify healthcare trends, patient preferences, and service utilization patterns. This information can guide decisions on resource allocation, staffing levels, and the development of healthcare programs. For example, statistical analysis can help hospitals determine optimal staffing ratios by analyzing patient volume, acuity levels, and nursing workload. It can also assist in predicting patient readmission rates and identifying areas for improvement in healthcare delivery.
Statistical analysis in healthcare is not without its challenges and limitations. One common challenge is the presence of confounding variables, which can influence the relationship between an exposure and an outcome. Confounding variables can introduce bias into the results and undermine the validity of the findings. To mitigate this, researchers often use statistical techniques such as multivariable regression analysis to adjust for confounding factors and identify true associations. Another limitation is the reliance on data quality. Accurate and reliable data are crucial for accurate statistical analysis. However, healthcare data can be prone to errors, inconsistencies, and missing information. Careful data collection, validation, and cleaning processes are essential to minimize biases and ensure the reliability of the results.
In conclusion, statistical application is of utmost importance in healthcare. It facilitates research by providing valid and reliable results, helps evaluate treatment effectiveness, and supports decision-making in healthcare management. The use of statistical methods allows healthcare professionals to make evidence-based decisions that improve patient outcomes, enhance the quality of care, and optimize resource allocation. Despite the challenges and limitations, statistical analysis continues to be an indispensable tool in healthcare research and practice.
References:
1. Smith, J., Jones, A., & Johnson, B. (2017). The role of statistics in healthcare research. Journal of Healthcare Analysis, 10(2), 45-60.
2. Thompson, R., Johnson, K., & Brown, L. (2018). Statistical methods for evaluating treatment effectiveness in healthcare. British Journal of Healthcare Management, 24(3), 110-116.
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### AuthorTopic: Nikon D800/E Diffraction Limits (Read 43388 times)
#### Ray
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##### Re: Nikon D800/E Diffraction Limits
« Reply #40 on: July 07, 2012, 12:05:27 am »
To simplify further, the formula is max f-stop = P x 1.054. The D800 has a pixel pitch of 4.87 microns, so the corresponding f/stop is f/5.1. As one exceeds this critical f/stop, loss of contrast is often more noticeable than the loss of resolution. These considerations derive from the laws of physics and are not a defect in the D800. If you use f/16 on the D800, the results will be no worse than with the D3, which has 8.4 micron pixels.
I'm sure you know, Bill, that many mathematical formulae and principles of Physics are approximations, not only for the sake of simplicity, but because of uncertaintanties built into the fabric of reality, and the ever-present possibility that sometimes mand-made theories can be either flat out wrong or plain imprecise, or that the people using the theories in any particular instance may be misapplying them.
Having taken the trouble to compare F16 images from a couple of cropped-format cameras with a much higher pixel density than the D3, and one of which has even a slightly higher pixel density than the D800 (the Canon 50D), I am confident that it is extremely unlikely that resolution at F16 with the D3 would be as good as resolution at F16 with the D800. I suspect the resolution differences would be clearly noticeable at 100% on monitor, after appropriate sharpening for each image and upsizing of the smaller file.
Now it so happens I still have my 12.7mp Canon 5D which is very close to the pixel density of the Nikon D3. If I have the time, and time really is a problem but I might be able to find it in the interests of the pursuit of truth, I could do another comparison between my old 5D used at F16 and my new D800E used at F16, to see which is sharpest.
I could take bets on the results of the outcome (to give me an incentive), but I doubt that Michael would allow betting activities on his site.
« Last Edit: July 07, 2012, 04:08:40 am by Ray »
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#### ErikKaffehr
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##### Re: Nikon D800/E Diffraction Limits
« Reply #41 on: July 07, 2012, 04:46:12 am »
Hi,
It seems that different posters may have different requirements.
I do agree that viewing distance plays a crucial role. If viewing distance is increased the eye may not be resolve the finest detail. So sharpening may mask lack of detail. The picture looks sharp. Watching closer the lack of detail is obvious.
Best regards
Erik
That doesn't mean anything. Why would anyone want a 150DPI print these days?
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Erik Kaffehr
#### Bart_van_der_Wolf
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##### Re: Nikon D800/E Diffraction Limits
« Reply #42 on: July 07, 2012, 06:00:56 am »
I'm sure you know, Bill, that many mathematical formulae and principles of Physics are approximations, ...
So you are trying to suggest that the laws of diffraction are not accurate, despite the fact that the phenomenon can be seen (first reports go back to 1828), measured, reproduced, and accurately calculated?
I do agree that poorly executed tests can produce puzzling results, or that flawed interpretation can lead to the wrong conclusions. But blaiming a lack of understanding on the accuracy of the laws of physics ..., surely you can do better than that.
Just perform the test and, if the result seems to add something worthwhile, by all means share it.
Cheers,
Bart
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#### Ray
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##### Re: Nikon D800/E Diffraction Limits
« Reply #43 on: July 07, 2012, 10:49:54 am »
So you are trying to suggest that the laws of diffraction are not accurate, despite the fact that the phenomenon can be seen (first reports go back to 1828), measured, reproduced, and accurately calculated?
I'm trying to say in general terms that most laws are not perfectly accurate. What is often considered to be accurate is simply sufficiently accurate for the immediate purposes. Heck! We don't even know what 95% of the matter and energy in the universe is made of, despite the recent discovery of the Higgs Boson which has caused great excitement in the world of Physics.
More specifically, the Wikipedia article you refer to contains the following comment about the Airy disk.
Quote
The Airy pattern falls rather slowly to zero with increasing distance from the center, with the outer rings containing a significant portion of the integrated intensity of the pattern. As a result, the root mean square (RMS) spotsize is undefined (i.e. infinite). An alternative measure of the spot size is to ignore the relatively small outer rings of the Airy pattern and to approximate the central lobe with a Gaussian profile.
Quote
Just perform the test and, if the result seems to add something worthwhile, by all means share it.
I have done the tests. Following Emil Martinec's advice to use a banknote as a test target, I took about 100 shots in 2009, using tripod and LiveView, comparing my 10mp Canon 40D with my 15mp 50D. I took several series of shots at different distances to the target, and different apertures, and paid particular attention to the accuracy of focussing. By varying the distance to the banknote, I eventually found a distance which produced significant aliasing and moire which was clearly visible on the LiveView screen when the target was in focus. I found this method useful because the resolution of the LiveView screen on the 40D is lower resolution than the 50D screen. As a consequence, I was sometimes not totally certain I was precisely in focus with the 40D, unless I used the presence of moire as an indication of 'spot on' focussing.
The results of my tests are quite clear. At F16, the 50D (equivalent to a 38.4mp full-frame sensor) has a very slight resolution advantage compared with the 40D, equivalent to a 25.6mp full-frame sensor. However, I admit that such an advantage, apparent in terms of the greater legibility of the finest text on the banknote, would only be noticeable on very large prints viewed close up.
Nevertheless, if I wanted to make and 8"x10" print representing a 200% crop of a part of the scene as viewed on my monitor, say a rare bird on the branch of a tree, or an interesting geological pattern on a cliff face, I know that I would prefer to use the 50D shot at F16. I might even prefer the 50D shot at F16 to the 40D shot at F8. I posted a comparison on another thread at http://www.luminous-landscape.com/forum/index.php?topic=68359.60 which shows significant aliasing in the 40D shot at F8, yet no better detail.
As a result of such tests, I would be very surprised if the gap between the D800 and the D3 at F16 were not wider and more obvious, perhaps noticeable at 100% on screen.
Cheers! Ray
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#### bjanes
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##### Re: Nikon D800/E Diffraction Limits
« Reply #44 on: July 07, 2012, 02:11:25 pm »
I'm sure you know, Bill, that many mathematical formulae and principles of Physics are approximations, not only for the sake of simplicity, but because of uncertaintanties built into the fabric of reality, and the ever-present possibility that sometimes mand-made theories can be either flat out wrong or plain imprecise, or that the people using the theories in any particular instance may be misapplying them.
Having taken the trouble to compare F16 images from a couple of cropped-format cameras with a much higher pixel density than the D3, and one of which has even a slightly higher pixel density than the D800 (the Canon 50D), I am confident that it is extremely unlikely that resolution at F16 with the D3 would be as good as resolution at F16 with the D800. I suspect the resolution differences would be clearly noticeable at 100% on monitor, after appropriate sharpening for each image and upsizing of the smaller file.
Now it so happens I still have my 12.7mp Canon 5D which is very close to the pixel density of the Nikon D3. If I have the time, and time really is a problem but I might be able to find it in the interests of the pursuit of truth, I could do another comparison between my old 5D used at F16 and my new D800E used at F16, to see which is sharpest.
I could take bets on the results of the outcome (to give me an incentive), but I doubt that Michael would allow betting activities on his site.
At least two main factors are in play: the point spread function (PSP) of the lens, which is affected by diffraction and aberrations, and the PSP of the sensor, including the effects of the raw converter. One may add defocus as discussed in an excellent article by Erik Kaffer. I now have both the D3 and the D800e and can verify that optimal sharpness with both cameras with the 60mm f/2.8 AFS MicroNikkor is at f/4 to f/5.6 and this is determined by the sweet spot of the lens and not pixel size. This is confirmed in a post by Bobn2 on DPReview.
System MTF may be obtained by convolving the PSP of the lens with that of the sensor. At f/16 both cameras will be handicapped by the larger Airy disc at this aperture, but the D800 will likely have better MTF as you predict because of the better MTF of the sensor. Since MTFs multiply, the old adage that resolution is determined by the weakest link in the imaging chain is not true. See this explanation by Bobn2 and also the demonstration of convolution on the Wolfram site.
Regards,
Bill
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#### Bart_van_der_Wolf
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##### Re: Nikon D800/E Diffraction Limits
« Reply #45 on: July 07, 2012, 02:40:36 pm »
More specifically, the Wikipedia article you refer to contains the following comment about the Airy disk.
Quote
The Airy pattern falls rather slowly to zero with increasing distance from the center, with the outer rings containing a significant portion of the integrated intensity of the pattern. As a result, the root mean square (RMS) spotsize is undefined (i.e. infinite). An alternative measure of the spot size is to ignore the relatively small outer rings of the Airy pattern and to approximate the central lobe with a Gaussian profile.
The emphasis in bold that you applied to the quote, suggests that you either are unfamiliar with the terminology, or made an attempt to spin the article in your favor. "Spotsize is indetermined" is mentioned as a consequence of "i.e. infinite", it doesn't mean that it can't be quantified with high accuracy and precision. It's somewhat similar to the division 1/3 which cannot be expressed exactly as a "real number" (it can be as a rational number though), or a division by zero (result is "complex infinity").
"To approximate the central lobe with a Gaussian profile" is mentioned as a less accurate alternative to the approximation of the actual shape of part of the exact pattern. The approximation itself can be accurately calculated to any precision one desires.
The diameter of the central lobe for a circular aperture is 'approximated' by:
2.4393397825330089098530776949305103557587186615502242... x wavelength x F-number , I limited the precision (hence approximation indicated by ellipses) to some 53 decimal positions, and the amplitude can also be calculated to any precision one requires, for as many lobes as one finds useful. BTW, the F-numbers are usually also approximations of the actual dimensions but it's up to the user to use more accurate input or not. The formula is exact.
As for your image example, the screen zoom resampling by two different amounts does eliminate the size differences, but it doesn't help the comparison due to the added resampling artifacts (especially on the lower '40D' crop). Despite that, and the additional moiré on the 40D (AA-filtering+diffraction was not strong enough to prevent that), I also see light diagonal stripes on the 40D image (on the vertical dark bar just left next to the portrait) that are missing in the 50D crop. It doesn't look like aliasing, so it seems to be higher resolution (due to less diffraction?) ...
Cheers,
Bart
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#### Fine_Art
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##### Re: Nikon D800/E Diffraction Limits
« Reply #46 on: July 07, 2012, 03:27:19 pm »
Hi,
It seems that different posters may have different requirements.
I do agree that viewing distance plays a crucial role. If viewing distance is increased the eye may not be resolve the finest detail. So sharpening may mask lack of detail. The picture looks sharp. Watching closer the lack of detail is obvious.
Best regards
Erik
A typical room may be 12ft across. You probably want to give a viewer more detail than their eye can resolve at 1/2 that distance. That leaves nothing to detract from the image when they take it home.
My pictures start to look soft when my eye is 8" from the picture. I want people to feel free to get close and explore moving their head around a bit. You want to immerse them in a place and time they will never see again. That is worth buying. Anything less will hold interest for a few seconds only.
What is the distance now specified for an immersive HDTV experience? Its much closer than most living rooms are set up for. It would actually look weird to have the sofa that close to the TV on the wall. Standing close in front of a framed picture does not look weird. People will do it.
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#### Wayne Fox
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##### Re: Nikon D800/E Diffraction Limits
« Reply #47 on: July 07, 2012, 04:22:47 pm »
I want people to feel free to get close and explore moving their head around a bit. You want to immerse them in a place and time they will never see again. That is worth buying. Anything less will hold interest for a few seconds only.
I agree with this. The idea of a "normal" viewing distance has always been more about when the printing technology will fail (normal for a billboard vs normal for a magazine spread). to me a great image can pull you in as you "immerse yourself" (very good description) in the image. Sure, not everyone will look at it close ... they may have no interest in the subject matter.
I also feel that images are not hanging in a gallery but in a normal location, and often the viewer will be directed much closer because of the natural flow of the location. I've seen some very large images hanging in a multi million dollar home in Park City, Utah. They were used on a magnificent stairway about 5 feet wide, and from the living area below looked OK. Climb the stairs and suddenly they looked terrible.
I have no control of where the image will be hung or how close people may choose to be or perhaps even be forced to be. But I certainly don't want to print with such low resolution the image quality degrades very quickly as you approach the image if a viewer so chooses.
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#### stevesanacore
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##### Re: Nikon D800/E Diffraction Limits
« Reply #48 on: July 07, 2012, 06:05:14 pm »
As a long time 4x5 shooter, I was under the impression that diffraction gets worse as the focal length gets shorter. I thought it was the actual size of the aperture at given f-stop, and had nothing to do with the focal length. In other words a 150mm 4x5 lens at f16 had much less diffraction than a 50mm lens at f16 on a 35mm camera. I remember shooting at f22-32 very often with my 4x5 cameras and don't recall any sharpness issues. I recently shot a job with my 17mm TSE Canon on my 1DsMk3 and noticed a major fall off in sharpness at anything above f11. I wonder if lenses that have mediocre sharpness to begin with don't show the effect as much?
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#### texshooter
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##### Re: Nikon D800/E Diffraction Limits
« Reply #49 on: July 07, 2012, 07:16:07 pm »
I certainly don't want to print with such low resolution the image quality degrades very quickly as you approach the image if a viewer so chooses.
I agree.
Don't you get peeved when some photogs try to argue with you about how unnecessary they feel 30+ megapixel cameras are becaue of the theoretical 5-15 ft. viewing distance. I think this argument is an anachronism, a vestige of the day when the only art hung on the wall were paintings. Those days are gone. I say use however many pixels you need so the viewer cannot tell the difference in resolution whether they are standing across the room or whether they are smudging the print with their nose. If all you can afford is a 12 MP camera that's perfectly understandable, but stop suggesting D800 and medium format shooters are off-the-deep-end show-offs. I hear this all the time at local photo hobby clubs, and it makes my jaw clinch.
« Last Edit: July 07, 2012, 07:43:08 pm by texshooter »
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#### Bart_van_der_Wolf
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##### Re: Nikon D800/E Diffraction Limits
« Reply #50 on: July 07, 2012, 07:21:21 pm »
As a long time 4x5 shooter, I was under the impression that diffraction gets worse as the focal length gets shorter. I thought it was the actual size of the aperture at given f-stop, and had nothing to do with the focal length.
Hi,
It actually has to do with the angular aperture, and as such both the actual aperture size and the focal length are in play. However, since our aperture numbers (F-number) are a ratio (f/#) between focal length and aperture size, diffraction is constant (as is the angular aperture) at a given F-number.
Quote
In other words a 150mm 4x5 lens at f16 had much less diffraction than a 50mm lens at f16 on a 35mm camera. I remember shooting at f22-32 very often with my 4x5 cameras and don't recall any sharpness issues.
That is because the image (and the diffraction) requires less output magnification for a given output size. The f/16 on the 35mm image was magnified much more. The f/22 - f/32 required much less magnification so the actual diifraction patterns stayed small enough to not affect output sharpness too much.
Quote
I recently shot a job with my 17mm TSE Canon on my 1DsMk3 and noticed a major fall off in sharpness at anything above f11. I wonder if lenses that have mediocre sharpness to begin with don't show the effect as much?
For the 1DsMk3, f/11 is probably the sweetspot where corner resolution has improved enough and center resolution has fallen enough to provide even sharpness across the image-circle, I know it does on my TS-E 24mm II. Optical theory predicts that center resolution will start to be visually impacted by diffraction at apertures narrower than f/7.1 on the 1DsMk3. It's not the optical quality, which probably is second to none, but pure physics.
Cheers,
Bart
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#### Bart_van_der_Wolf
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##### Re: Nikon D800/E Diffraction Limits
« Reply #51 on: July 07, 2012, 07:36:33 pm »
Don't you get peeved when some photogs try to argue with you about how unnecessary they feel 30+ megapixel cameras are because of the theoretical 5-15 ft. viewing distance.
Hi,
Not really, but I do think, "if thinking that makes you happy, be my guest", I know better.
I agree with Wayne, assuming we managed to capture the soul (light/composition/intent) of our image, it is about the realism with which the image is rendered that delivers the knock-out punch, total submission/submersion, nothing to distract. I want the surface/material structure of the subjects/objects to become almost tangible.
Cheers,
Bart
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#### Ray
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##### Re: Nikon D800/E Diffraction Limits
« Reply #52 on: July 07, 2012, 08:28:05 pm »
I also see light diagonal stripes on the 40D image (on the vertical dark bar just left next to the portrait) that are missing in the 50D crop. It doesn't look like aliasing, so it seems to be higher resolution (due to less diffraction?) ...
Those broad, diagonal, colored stripes that are very obvious on the 40D crop occur in a number of places around the head of the Aboriginal. Not being as knowledgeable as you on such technical matters as aliasing and moire, my first reaction was that those diagonal stripes were in fact artifacts or moire. But I always like to do real-world checking, so I pulled out a \$50 banknote from my wallet and studied it carefully with a magnifying glass.
I can assert categorically that those diagonal, faintly colored stripes do not exist on the banknote. They are false detail. We have here a case of a sensor of higher resolution than a D800 producing a better and more accurate image at F16 than a lower resolution sensor at F8; better in terms of accuracy of detail; better in terms of DoF; and at least equal in terms real detail.
I rest my case.
Cheers! Ray
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#### Bart_van_der_Wolf
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##### Re: Nikon D800/E Diffraction Limits
« Reply #53 on: July 07, 2012, 09:13:18 pm »
Those broad, diagonal, colored stripes that are very obvious on the 40D crop occur in a number of places around the head of the Aboriginal. Not being as knowledgeable as you on such technical matters as aliasing and moire, my first reaction was that those diagonal stripes were in fact artifacts or moire.
They are, which tells us that there is more optical detail present than the lens+diffraction at the sensel pitch can resolve.
Quote
But I always like to do real-world checking, so I pulled out a \$50 banknote from my wallet and studied it carefully with a magnifying glass.
Good, that saves me from getting one (Australian \$50, David Unaipon (1872–1967) portrait, not the Edith Cowan (1861–1932) version) at my local bank, which I still might because it looks like an interesting test subject ...
Quote
I can assert categorically that those diagonal, faintly colored stripes do not exist on the banknote. They are false detail.
Yes, those are obviously aliasing artifacts. No problem, that's to be expected when diffraction doesn't kill all fine detail.
What seems to be missing from the more diffraction affected 50D image crop, is the diagonal area detail in the 40D crop, that I marked in Red (the vertical bar) on the attached copy of your image.
Cheers,
Bart
« Last Edit: July 07, 2012, 10:12:41 pm by BartvanderWolf »
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#### Jan Brittenson
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##### Re: Nikon D800/E Diffraction Limits
« Reply #54 on: July 07, 2012, 10:42:50 pm »
Every point in the image consists of light that has passed through every point in the aperture. It's not a pinhole. Light close to the rim diffracts. Therefore, every point in the image consists of some proportion of light spread through diffraction and some proportion of undiffracted light that passed through the center. This proportion is purely dependent on the diameter of the aperture. Since it's not a pinhole and there is half a lens in front of it and half a lens behind it, the portion of light that diffracted represents a star-shaped point spread (PSF). This is trivial to verify by stopping down and photographing a bright point (aka 'unit impulse') like a light. Star shape. No airy disk. Because an aperture is not a pinhole.
The diffracted light subject to the PSF forms a veil that sits on top of the undiffracted image, and which results in contrast loss. The spread is very broad (as can be seen in the star from a point light - the 'arms' extend quite wide) so most of the energy is scattered widely. But it does increase in intensity as you get close to the center.
The reason the PSF is a star shape is equally easy to understand. The narrower the aperture, the greater the spread. When an aperture has straight blades it's wider in the corners where the blades meet and narrower around the middle between the corners. The former diffracts less, and the latter more. This variation in intensity shows as a star. With a perfectly round aperture it would roughly approximate a gaussian, similar to spherical aberration.
The various rules of thumb were created for telescopes (which lack apertures), where stars get distorted due to diffraction - from the barrel at the entry and exit pupils of the telescope. This is still a rule a thumb and not a strict physical relationship though, because the pupils on a telescope aren't in focus. However, they probably adequately in focus to produce something similar to airy disks. (Just like the barrel can cause fuzzy vignetting at the pupils. A photographic aperture of course can't cause vignetting.) But an aperture can't project airy disks any more than closing it causes vignetting.
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#### dimapant
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##### Re: Nikon D800/E Diffraction Limits
« Reply #55 on: July 08, 2012, 01:47:05 am »
Very, very interesting discussion, with some deep analysis, properly well done and I really learned some interesting information.
But……in a picture, what cannot be seen, it does not matter at all.
In my personal opinion, there is the most important parameters which are missing in this thread: it has not been defined the word “visible” in the original question.
In other word, talking of diffraction, it has been forgotten the most important things: the definition of the print size and viewing distance.
At which print size and viewing distance we refer to answer to the original request to be “visible”?
Having not define those two parameters, defining visibility of diffraction is impossible, on a post card print size a certain picture can be seen and can be judged as perfect at a certain viewing distance, whilst the same picture could not be printable for a good quality print on A2 print size on the same, or different, viewing distance.
What you see, even terrible, on a monitor at 200 x, it could be completely irrelevant on a certain print size/viewing distance.
Many thanks to all of you for the sharing of the deep expertise and information and best regards.
Alessandro
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#### ErikKaffehr
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##### Re: Nikon D800/E Diffraction Limits
« Reply #56 on: July 08, 2012, 05:13:01 am »
Hi,
I made some tests with a Sony Alpha 77 SLT (a really small pitch sensor) and an old 100/2.8 Minolta Macro lens.
The two samples below are taken with f/16 and f/5.6, both deconvolution sharpened. About optimal sharpening to my taste.
The f/16 image may look sharp when viewed at long distance, but fine line pattern visible in the red box is very clearly lost on the f/16 image.
Sharpening (Lightroom 4.1):
f/16: 75/1.3/100/17
f/5.6: 55/0.8/100/17
Best regards
Erik
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Erik Kaffehr
#### erpman
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##### Re: Nikon D800/E Diffraction Limits
« Reply #57 on: July 08, 2012, 06:30:21 am »
Quote
Quote
In other words a 150mm 4x5 lens at f16 had much less diffraction than a 50mm lens at f16 on a 35mm camera. I remember shooting at f22-32 very often with my 4x5 cameras and don't recall any sharpness issues.
That is because the image (and the diffraction) requires less output magnification for a given output size. The f/16 on the 35mm image was magnified much more. The f/22 - f/32 required much less magnification so the actual diifraction patterns stayed small enough to not affect output sharpness too much.
Does this apply to stitching too? Or is it related to the actual size of the negative/sensor relative to the aperture size?
Let´s say I stitch 3 images taken in portrait mode so the resolution of the digital "negative" is increased to 7360x10000 pixels. In order to print it horizontally on a 110cm roll I would have to interpolate the image to about 200% (or reduce the resolution to 150ppi) whereas a regular d800 image file (7360 × 4912px) would have to be interpolated about 280-300%. Is this what is meant with magnification?
It appears to me that stitching (if that suits your shooting style) could be one way to avoid diffraction problems when you want that ultra-deep DOF??
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#### ErikKaffehr
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##### Re: Nikon D800/E Diffraction Limits
« Reply #58 on: July 08, 2012, 06:43:37 am »
Hi,
I don't think it works that way. If you stitch the focal length would be longer, so you would need to stop down more. It is possible to achieve extended depth using "focus stacking".
The reason that diffraction was not obvious in 4x4" was mainly "that you were not looking" and perhaps also the use of Tri-X film. This article illustrates it very well:
http://www.photodo.com/topic_138.html
In short: 135 at optimal aperture using TMAX-100 outperforms 4x5" at f/22 using Tri-X. Article is worth reading.
Best regards
Erik
Does this apply to stitching too? Or is it related to the actual size of the negative/sensor relative to the aperture size?
Let´s say I stitch 3 images taken in portrait mode so the resolution of the digital "negative" is increased to 7360x10000 pixels. In order to print it horizontally on a 110cm roll I would have to interpolate the image to about 200% (or reduce the resolution to 150ppi) whereas a regular d800 image file (7360 × 4912px) would have to be interpolated about 280-300%. Is this what is meant with magnification?
It appears to me that stitching (if that suits your shooting style) could be one way to avoid diffraction problems when you want that ultra-deep DOF??
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Erik Kaffehr
#### Bart_van_der_Wolf
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##### Re: Nikon D800/E Diffraction Limits
« Reply #59 on: July 08, 2012, 07:41:05 am »
Does this apply to stitching too? Or is it related to the actual size of the negative/sensor relative to the aperture size?
Let´s say I stitch 3 images taken in portrait mode so the resolution of the digital "negative" is increased to 7360x10000 pixels. In order to print it horizontally on a 110cm roll I would have to interpolate the image to about 200% (or reduce the resolution to 150ppi) whereas a regular d800 image file (7360 × 4912px) would have to be interpolated about 280-300%. Is this what is meant with magnification?
Yes.
The optical system, limited in resolution by residual lens aberrations and diffraction, projects a still very high resolution image on the sensor array. The sensel pitch sets another physical limit on how much of that can be resolved. The denser the sampling, the higher the resolution that can be utilized from the projected optical image. That results in an on sensor resolution that can be expressed in physical units such as cycles/mm.
That resolution will be proportionally reduced when the data is going to be magnified to a larger output size than the sensor array. If we can limit the required output magnification, because we have a physically larger sensor array or because we stitch (the result of) a few smaller ones together, then the resolution that was originally captured will be better preserved.
Quote
It appears to me that stitching (if that suits your shooting style) could be one way to avoid diffraction problems when you want that ultra-deep DOF??
It does help if we do not have to magnify the existing diffraction to the point that it becomes clearly visible as lost reolution. However, for a given Field of View it won't help to just stitch some more images together, because that will only increase our FOV. While that helps to reach a certain output size with a lower output magnification, it may not give us the FOV we want, it may be too wide. To counter-act that, one typically shoots with a longer focal length with a narrower FOV but unfortunately also a shallower DOF. To compensate for that one could use a narrower aperture, but that defeats the purpose of reducing the visibility of diffraction blur.
For beating diffraction and achieving deep DOF at the same time, there are no free lunches. Diffraction is a physical boundary that can only be controlled with our choice of aperture and the result can be magnified untill we have to accept visible resolution losses.
The only real solution for deep DOF with very high resolution, is focus stacking. That will allow to reduce the diffraction losses by using a wider aperture, and we can add DOF back as we increase the number of stacked focus brackets. Focus stacking does come with its own set of practical limitations though ...
Cheers,
Bart
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Home > Error Function > Error Function Gaussian
# Error Function Gaussian
## Contents
Privacy policy About Wikipedia Disclaimers Contact Wikipedia Developers Cookie statement Mobile view Algebra Applied Mathematics Calculus and Analysis Discrete Mathematics Foundations of Mathematics Geometry History and Terminology Number Theory Probability and The imaginary error function has a very similar Maclaurin series, which is: erfi ( z ) = 2 π ∑ n = 0 ∞ z 2 n + 1 n Similarly, (8) (OEIS A103979 and A103980). Sep 4 '11 at 13:42 Indeed, on page 296 of the Glaisher article, $x$ is used for both purposes. http://qwerkyapp.com/error-function/error-function-integral-gaussian.html
## Error Function Values
Comp. 23 (107): 631–637. more stack exchange communities company blog Stack Exchange Inbox Reputation and Badges sign up log in tour help Tour Start here for a quick overview of the site Help Center Detailed Glaisher published an article on definite integrals in which he comments that while there is scarcely a function that cannot be put in the form of a definite integral, for the The denominator terms are sequence A007680 in the OEIS.
Arfken, G. D: A D package[16] exists providing efficient and accurate implementations of complex error functions, along with Dawson, Faddeeva, and Voigt functions. The integrand ƒ=exp(−z2) and ƒ=erf(z) are shown in the complex z-plane in figures 2 and 3. Error Function Explained It is unfortunate that changes from Glaisher’s original $\mathrm{Erf}$: the switch of limits, names and the standardizing factor, did not apply to what Glaisher acknowledged was its most important application: the
Math. Complementary Error Function The first derivative is (28) and the integral is (29) Min Max Re Im Erf can also be extended to the complex plane, as illustrated above. Wähle deine Sprache aus. http://mathworld.wolfram.com/Erf.html Sequences A000079/M1129, A001147/M3002, A007680/M2861, A103979, A103980 in "The On-Line Encyclopedia of Integer Sequences." Spanier, J.
The denominator terms are sequence A007680 in the OEIS. Gaussian Error Function Calculator The inverse error function is usually defined with domain (−1,1), and it is restricted to this domain in many computer algebra systems. The system returned: (22) Invalid argument The remote host or network may be down. Join them; it only takes a minute: Sign up Here's how it works: Anybody can ask a question Anybody can answer The best answers are voted up and rise to the
• MR0167642.
• Despite the name "imaginary error function", erfi ( x ) {\displaystyle \operatorname 8 (x)} is real when x is real.
• H.
## Complementary Error Function
Washington, DC: Hemisphere, pp.385-393, 1987.
Continued fraction expansion A continued fraction expansion of the complementary error function is:[11] erfc ( z ) = z π e − z 2 1 z 2 + a 1 Error Function Values Wolfram|Alpha» Explore anything with the first computational knowledge engine. How To Solve Error Function Your cache administrator is webmaster.
comm., Dec.15, 2005). check my blog When should I refuel my vehicle? ISBN978-1-4020-6948-2. ^ Winitzki, Sergei (6 February 2008). "A handy approximation for the error function and its inverse" (PDF). If nothing else, it might help me remember the definition. Probability Error Function
The integrand ƒ=exp(−z2) and ƒ=erf(z) are shown in the complex z-plane in figures 2 and 3. The Q-function can be expressed in terms of the error function as Q ( x ) = 1 2 − 1 2 erf ( x 2 ) = 1 2 Close Was this topic helpful? × Select Your Country Choose your country to get translated content where available and see local events and offers. this content Similarly, the En for even n look similar (but not identical) to each other after a simple division by n!.
Kategorie Bildung Lizenz Standard-YouTube-Lizenz Mehr anzeigen Weniger anzeigen Wird geladen... Gaussian Error Function Table To use these approximations for negative x, use the fact that erf(x) is an odd function, so erf(x)=−erf(−x). Melde dich bei YouTube an, damit dein Feedback gezählt wird.
## The error and complementary error functions occur, for example, in solutions of the heat equation when boundary conditions are given by the Heaviside step function.
Press, William H.; Teukolsky, Saul A.; Vetterling, William T.; Flannery, Brian P. (2007), "Section 6.2. Indeed, Φ ( x ) = 1 2 π ∫ − ∞ x e − t 2 2 d t = 1 2 [ 1 + erf ( x 2 Referenced on Wolfram|Alpha: Erf CITE THIS AS: Weisstein, Eric W. "Erf." From MathWorld--A Wolfram Web Resource. Gaussian Error Function Excel Taylor series The error function is an entire function; it has no singularities (except that at infinity) and its Taylor expansion always converges.
I guess I'm in the reverse situation with Nate; I have to dig up Abramowitz and Stegun to remember how the normal distribution CDF is expressed in terms of $\mathrm{erf}$. –J. Cody's rational Chebyshev approximation algorithm.[20] Ruby: Provides Math.erf() and Math.erfc() for real arguments. For iterative calculation of the above series, the following alternative formulation may be useful: erf ( z ) = 2 π ∑ n = 0 ∞ ( z ∏ k have a peek at these guys Craig, A new, simple and exact result for calculating the probability of error for two-dimensional signal constellaions, Proc. 1991 IEEE Military Commun.
Incomplete Gamma Function and Error Function", Numerical Recipes: The Art of Scientific Computing (3rd ed.), New York: Cambridge University Press, ISBN978-0-521-88068-8 Temme, Nico M. (2010), "Error Functions, Dawson's and Fresnel Integrals", | 1,402 | 5,569 | {"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} | 3.421875 | 3 | CC-MAIN-2017-30 | latest | en | 0.790275 |
https://www.hammockforums.net/forum/showthread.php/11866-Hammock-structural-ridgeline-length/page3 | 1,529,895,094,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267867364.94/warc/CC-MAIN-20180625014226-20180625034226-00380.warc.gz | 840,711,132 | 16,601 | # Thread: Hammock structural ridgeline length
1. The blackbird ridge line is 101"...as posted on Brandon's website....I just made a new hammock with a whoopie sling for the RL.
My hammock is 11 feet long. My whoopie sling is 99" to 109" long....I tried it out at 109" first (83% of hammock length) and it didn't feel comfortable enough....I kept adjusting and eventually had it set at 99" which feels great for a diagonal lay.
I used the camo 1.1 that sclittlefield just sold and made mine a double layer. I am 6'3" and 235 lbs (as reference).
I think there are some general ranges, but no true one size fits all when you are making custom gear.
2. Originally Posted by Shug
That will change depending on your hang in the woods .... tree spacing, angle of suspension to tree and so forth.
The education goes on a while.....
In theory your hang shouldnt change much if at all with a structural ridgeline. As long as your structural ridgeline stays the same length between hangs and is taught in your hang, all your hangs in varoius conditions will be the same. Thus the point of the structural ridgeline. Am I wrong?
3. Originally Posted by High@HighOutdoors
In theory your hang shouldnt change much if at all with a structural ridgeline. As long as your structural ridgeline stays the same length between hangs and is taught in your hang, all your hangs in varoius conditions will be the same. Thus the point of the structural ridgeline. Am I wrong?
You are correct!
4. I know this is a somewhat older thread but am incorrect in thinking that the 83% length figure corresponds with the 30% sag factor? Or is it a different sag angle?
5. Good question.
hammock ridgeline length: start with 83% of hammock's total (stretched flat) length between where the suspension attaches to the hammock. for gathered end is just inside the two whipped areas.
30* (degree angle) is the optimal angle of the suspension from the tree to the hammock attach-point.
anyone else chime in ?
6. Originally Posted by Loki
Good question.
hammock ridgeline length: start with 83% of hammock's total (stretched flat) length between where the suspension attaches to the hammock. for gathered end is just inside the two whipped areas.
30* (degree angle) is the optimal angle of the suspension from the tree to the hammock attach-point.
anyone else chime in ?
Sorry, I think I phrased the question in an odd way. I know what each number by itself represents, but do they go along with each other? The ridgeline is used to keep the same lay angle should the suspension angle not be the same so is that 83% length for the ridgeline supposed to ensure a sag angle of 30 degrees or another angle of lay?
7. Ahh... I'd say one does not necessarily ensure the other.
It is possible to have, for example, a correct 98" hammock ridgeline (a SRL at 83%) and if your straps are adjusted too short they can still be at angles less than 30*. If your straps are set too long then they would hang at more than 30* - even though the SRL is still set to 83% of the hammock length.
This might help. Try entering a few number combinations and see what changes
http://theultimatehang.com/hammock-hang-calculator/
8. Derek's tool is good if you are an iPhone user, for android users search for Hammock tools and there is an app that gives the same results as the ultimate hang guide.
9. Originally Posted by notquitethere
Derek's tool is good if you are an iPhone user, for android users search for Hammock tools and there is an app that gives the same results as the ultimate hang guide.
that link also works on the web in your browser, so you can test before you hit the trail
10. ## Hammock structural ridgeline length
Side question? I've got an ENO, putting Whoopies in this weekend. Does it matter where the ridgeline connects to suspension in order to achieve the 83%?
I was going to attach close to gathered end. | 935 | 3,893 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2018-26 | latest | en | 0.937441 |
http://coocktastty.top/diagram-rational-numbers-worksheet/ | 1,601,096,498,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400234232.50/warc/CC-MAIN-20200926040104-20200926070104-00270.warc.gz | 27,499,417 | 31,596 | # Diagram Rational Numbers Worksheet
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Without phonics skills, it's almost impossible, especially for kids, to learn how to read new words. Techniques to Handle Attention Issues It is extremely essential that should you discover your kid is inattentive to his learning especially when it has to do with reading and writing issues you must begin working on various ways and to improve it. Use a student's name in every sentence so there's a single sentence for each kid. Because he or she learns at his own rate, there is some variability in the age when a child is ready to learn to read. Teaching your kid to form the alphabets is quite a complicated practice.
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Have faith. But just because it's possible, doesn't mean it will be easy. Know that whatever life you want, the grades you want, the job you want, the reputation you want, friends you want, that it's possible.
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# lect10 - THE CONDITIONAL RULE OF EXPECTATION SOME MORE...
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THE CONDITIONAL RULE OF EXPECTATION SOME MORE COMMON DISCRETE DISTRIBUTIONS Outline THE CONDITIONAL RULE OF EXPECTATION SOME MORE COMMON DISCRETE DISTRIBUTIONS Geometric Distributions Uniform Distributions Hypergeometric Disributions Negative Binomial Distributions An application: How long does a game of craps last? 1 / 14 Xinghua Zheng Lect 10: Expectations and variances: examples and methods
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THE CONDITIONAL RULE OF EXPECTATION SOME MORE COMMON DISCRETE DISTRIBUTIONS THE CONDITIONAL RULE OF EXPECTATION The Conditioning Rule. Let C 1 , C 2 , . . . be a collection of mutually exclusive and exhaustive cases. For each i , let E ( X | C i ) = X x x · P [ X = x | C i ] . be the conditional expectation of X given case C i . Then E ( X ) = X i P [ C i ] · E ( X | C i ) . 2 / 14 Xinghua Zheng Lect 10: Expectations and variances: examples and methods
THE CONDITIONAL RULE OF EXPECTATION SOME MORE COMMON DISCRETE DISTRIBUTIONS Geometric Distributions Uniform Distributions Hypergeometric THE GEOMETRIC DISTRIBUTIONS Let N be the number of tosses required to produce the first head, when tossing a coin that has probability p to land heads. N has a geometric distribution with success probability p What is the expected value of N ? Method 1: By definition, E ( N ) = n = 1 n · q n - 1 p = . . . . Method 2: Let H be the event that the first toss is a head, and T = H c the event that first toss is a tail. Then P [ H ] = , P [ T ] = , E ( N | H ) = , E ( N | T ) = , E ( N ) = = , ( 1 - q ) E ( N ) = , E ( N ) = . 3 / 14 Xinghua Zheng Lect 10: Expectations and variances: examples and methods
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THE CONDITIONAL RULE OF EXPECTATION SOME MORE COMMON DISCRETE DISTRIBUTIONS Geometric Distributions Uniform Distributions Hypergeometric THE GEOMETRIC DISTRIBUTIONS, ctd Similarly we can find its variance: Put X = N ( N - 1 ) = N 2 - N . Then E ( X | H ) = , E ( X | T ) = , E ( N 2 ) - E ( N ) = E ( X ) = = q ( E ( N 2 ) + E ( N )) , E ( N 2 ) = ( 1 + q ) E ( N ) / p = ( 1 + q ) / p 2 , Var ( N ) = q p 2 , SD ( N ) = q p .
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• Spring '11
• Zheu
• Probability theory, Binomial distribution, Geometric distribution, Common Discrete Distributions, Xinghua Zheng
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lect10 - THE CONDITIONAL RULE OF EXPECTATION SOME MORE...
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# What is the area of a rectangle with the dimensions of 40 by 14?
Updated: 12/18/2022
Marcaroberts
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14y ago
To find the area multiply the length by the width, in this case 40 x 14 = 560 square units (units being inches, feet, centimetres, metres etc.)
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### What is the area of a rectangle with a width of 40 inches and a length of 14 inches?
Area of RECTANGLE =length * breadth[width] m*m =40*14 m*m 560m*m
### What are the dimensions of a 26' x 40' rectangle?
2 sides = 40 feet long, 2 sides = 26 feet long and an area of 1,040 square feet.
### The length and breadth of a rectangle is of ratio 3 to 4 The area of a rectangle is 6000 what are the dimensions of the rectangle?
The dimensions work out as: length = 30 times the square root of 5 breadth = 40 times the square root of 5 check: (30 times sq rt of 5)*(40 times sq rt of 5) = 6000 square units
### The perimeter of a rectangle is 700 yards what are the dimensions of the rectangle if the length is 40 yards more than the width?
the perimeter of a rectangle is 700 yards. what are the dimensions of the rectangle if the lenght is 80 yards more than the width?
### How long is the diagonal dimension of a rectangle with dimensions of 26' 40 '?
The diagonal is 47.707'
### The area of a rectangle is 100 inches the perimeter is 40 a second rectangle has the same area but a different perimeter. Is the second rectangle a square?
the area of a rectangleis 100 square inches. The perimeter of the rectangle is 40 inches. A second rectangle has the same area but a different perimeter. Is the secind rectangle a square? Explain why or why not.
### What is the dianagle of 24x32?
If it's the dimensions of a rectangle then by using Pythagoras' theorem its diagonal is 40
### What is the area of a rectangle of 8m and 5m?
8m*5m=40 m2 The area of the rectangle is 40m2
### What is the area of the rectangle in square inches 20 inches by 40 inches?
Area rectangle: 20 times 40 = 800 square inches
### What is the area of a square 90 by 40?
A square cannot measure 90 by 40; this would be a rectangle. Anyway, the area of this rectangle is 90 x 40 = 3,600 square units.
### If the length of a rectangle is 6 in and the width is 14 in what is the perimiter?
The answer is 40 (cm?). 6 + 6 (both sides) +14+14= 40 | 676 | 2,451 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.8125 | 4 | CC-MAIN-2024-38 | latest | en | 0.892308 |
https://progassignments.com/can-someone-help-me-understand-matlab-concepts-in-my-assignment | 1,723,608,191,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641095791.86/warc/CC-MAIN-20240814030405-20240814060405-00746.warc.gz | 368,030,521 | 19,482 | # Can someone help me understand MATLAB concepts in my assignment?
Can someone help me understand MATLAB concepts in my assignment? I’m very new to MATLAB so please excuse my poor formatting Here’s what I currently have at my disposal: /*MATLAB v 1.8.2: Make a table for the’real part’, the ‘imagined part’ When we have a few cells we sort those by the scale column, matlab makes us the real part (see this picture) and Imagined is left to the left of MATLAB when the first row is an over (figure1.1). When the last row is the real part we sort this by Imagined column. Matlab keeps the order of the row order, and the row order is ‘normal’, ‘right’ with correct effects. The first row was created as an over according to where you specify where the real part, Imagined, is and MATLAB will try and show you it’s view on this row. The first 10 spaces will be to make sure we understand what the first 10 spaces are of which we shouldn’t say this. Our first 10 possible spaces will be to decide between the first 9th and last, we have the table here if we want it? If not we have a list of the possible rows, we will have a list of rows or a list of columns. We will always pick the right column value (the scale value) that maximizes the sum over the possible real rows. Matlab gets the results for all possible scales, the result can be all columns. Additionally we’ll read out everything which were in between the maximum, max, min, order, etc, if you want to know about all possible rows from back to front or wherever they’re right. In this case we start where every range of possible real cells is listed somewhere. In the list of rows go we’ve found one type (this one just like the list of columns) and the other there (this one means that we pick the one which is the row with two columns and that will do the work). If we dont find the “real” part, let it be the left half, do that. The row starting point is the normal part, it’s the right half. Matlab does the work here, it reads’real’/imagined/ideal’ when given the column values that are outside the range of the real columns. The row starting point is the image row, however it is not the right half of the left half. So when you sort the last 20 possible rows from the row starting point to rightCan someone help me understand MATLAB concepts in my assignment? I’ve considered MATLAB’s classic IEEF approach as if human beings are just a single cell in a living biological organism. It seems to me in MATLAB only the first cell does the job.
## Takemyonlineclass
This kind of theoretical work is challenging and really is rather advanced. I don’t think anyone is really new in MATLAB, but it takes some get-tuples to clearly state the answer… a fact that is most interesting to have figured out at some later date. I will finish it tomorrow and get ready for class 2 as I am entering college. i have 3 answers at least and 9 are obvious and the reason for the issue i am fixing is this: try to make my “image” not expand to the full array (containing all the data). It looks very similar to the image of a map. If i go to image3 you will get a 2nd and 3rd digit, but the numbers are not that well understood. what number??? more simple case? it is 7 and not sure if that is correct. The third digit will be easier to understand and the sign if the order is correct. ok now goes to student1 (new), student2 (yes, student2 thinks I am already at the beginning of class I think, but does not really know what “new” would be) and student3 (no, student3 thinks I am already here) and a few seconds later it says that i am having problems placing data in image3 and its not even clear between it and image1. On again note Image3 should contain at the end of the first line of my file I wrote above who can be called at the time I put in this. Now again to student1 and Student3 should be at the end of Photo3 but its not clear between them. at most of them thats 100% correct. but everything else on this file are just imwise. i think thats btw images are not that high quality. that’s it. thank you As a “test assignment”. No, i could not find a better way to solve it than to make people understand my problem.
## Where Can I Find Someone To Do My Homework
And again, i was trying to make a “file”. i put some text in the last line that i made and then wrote it in it. and now with that i get a black rectangle. I need to apply some control to it. and if that is it, your point is moot. The problem is that they don’t know enough. Does it look like a map can be seen at once? If yes, need to think about it while you apply some blocks see this page can do what you expect a map to do *not* but some other block where you can examine it. Now, I did once what they suggested I could do while not trying. But that wasn’t useful at all! Like they don’t use photoshop to draw these photos.Can someone help me understand MATLAB concepts in my assignment? Since some of you recently read my blog, please don’t hesitate to provide me click this link. Furthermore, I am company website that there will be plenty of places you’ll need to find them (or any one on the internet if you’d like help here). I was learning MATLAB yesterday, and did not know many basic concepts, so I was like “you wrote these good questions, and you wanted to find some one other to explain them…” so I compiled it correctly until it became completely obsolete. I then created a new project with MATLAB just before that, and added my two questions. I finally got the project ready for my assignment, by clicking the “add project” button and sending my assignments online where they can be addressed. As I typed out my assignments, I was soon frustrated, a bit. To fix it: I created a new click here now project and used as the root of the project. To use the “Cascade” project in a new context, I used the add project function in MATLAB (based on it’s current user interface, and through my new “configuration”).
## Online Test Takers
Subsequently, I added all other non-trivial steps (or errors that I had in my code) together into my “set” settings variable. Later that function was used to change the project path and to open a new GUI as the student did on his assignment page. In the left sidebar, the following code is viewable through the MATLAB app’s toolbar: Next, I want to reference the assignment page (the paper and its post- and final project pop over to these guys from the top. Also, I want to directly reference the assignments page of a real assignment, without creating a CTE (“created new task”) and with a separate “change assignment path”. The assignment page or | 1,512 | 6,620 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.234375 | 3 | CC-MAIN-2024-33 | latest | en | 0.919559 |
https://www.physicsforums.com/threads/the-effect-of-a-stationary-proton-on-a-stationary-electron.750324/ | 1,508,365,644,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187823153.58/warc/CC-MAIN-20171018214541-20171018234541-00259.warc.gz | 983,207,416 | 16,847 | # The effect of a stationary proton on a stationary electron
1. Apr 23, 2014
### Mindstein
1. The problem statement, all variables and given/known data
On the utterly unrealistic assumption that there are no other charged particles in the vicinity, at what distance below a proton would the upward force on an electron equal the electrons weight?
2. Relevant equations
I used Newtons universal law of gravitation as well as Coulombs law.
3. The attempt at a solution
Basically, I equated the sum of the forces on the electron (due to the proton) to the weight of the electron. The sum of the aforementioned forces correspond to the gravitational attractive force and the electrical attractive force (both forces being produced by the proton).
I got an answer of 4.85 meters. Right or wrong, I don't really care because the number will be wrong by virtue of entering the values into the calculator wrong (hopefully); I'm definitely comfortable with the methodology. My problem is the language in this exercise. The weight of the electron depends on the acceleration due to gravity (which, on earth, is 9.8 m/s^2) of some other mass.
I arrived at 4.85 meters by assuming that this experiment takes place on earth. I'm not comfortable with making that assumption, but I'm also not comfortable with having to calculate the acceleration of a mass that is in the presence of a proton (the mass, for this example, would be the electron). I feel like calculating such a quantity would be out of the scope and spirit of the problem.
In a nutshell, my question pertains to the intent of the word 'weight'. Do I calculate 'g' for an object in the presence of its gravitational field or do I assume this experiment takes place on earth?
Any guidance on this would be much appreciated. Thank you.
Last edited: Apr 23, 2014
2. Apr 23, 2014
### Staff: Mentor
If "weight"/"gravity"/... is used without any other information, the experiment always takes place on the surface of earth. This is just a convention for homework problems.
3. Apr 23, 2014
### Mindstein
Ah great, thanks mfb. I appreciate it. | 469 | 2,100 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2017-43 | longest | en | 0.938515 |
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owever, since the release of Excel 2007 users can now create as many worksheets within one workbook as the memory of the computer can handle. Even if the user does not have access to one of the newest versions of Excel such as Excel 2007 or Excel 2010, they can still make as many worksheets as they would like, but earlier versions of Excel will require more workbooks. How can I learn more about Worksheets and find Tutorials? I have created a website to teach as much about Excel as I can possibly learn. I will be offering valuable advice, knowledge and tutorials about many different features of Excel Worksheets as well as many other aspects of Excel. | 845 | 4,178 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2018-51 | longest | en | 0.963913 |
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Word Problems Involving Systems of Linear Equations Many word problems will give rise to systems of equations that is, a pair of equations like 5.3 Systems of Linear Equations in Three Variables 5.3 OBJECTIVES 1. Find ordered triples associated with three equations 2. Solve a system by the...Solving Equations by Factoring 9 - Cool Math has free online cool math lessons, cool math games and fun math activities. welcome to coolmath. We use first party cookies on our website to enhance your browsing experience, and third party cookies to provide advertising that may be of interest to you.
Chapter 6 lesson 2 homework practice
Rhiel funeral home | 997 | 4,652 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.703125 | 3 | CC-MAIN-2021-17 | latest | en | 0.91437 |
https://istopdeath.com/simplify-5-x-4-5-4x-5/ | 1,670,549,382,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446711376.47/warc/CC-MAIN-20221209011720-20221209041720-00580.warc.gz | 350,732,722 | 15,453 | # Simplify (5/x-4)/((5-4x)/5)
5x-45-4×5
Multiply the numerator by the reciprocal of the denominator.
(5x-4)55-4x
To write -4 as a fraction with a common denominator, multiply by xx.
(5x-4⋅xx)55-4x
Combine -4 and xx.
(5x+-4xx)55-4x
Combine the numerators over the common denominator.
5-4xx⋅55-4x
Cancel the common factor of 5-4x.
Cancel the common factor.
5-4xx⋅55-4x
Rewrite the expression.
1x⋅5
1x⋅5
Combine 1x and 5.
5x
Simplify (5/x-4)/((5-4x)/5) | 192 | 450 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.03125 | 4 | CC-MAIN-2022-49 | latest | en | 0.782798 |
https://askworksheet.com/simplifying-fractions-worksheet-pdf/ | 1,718,286,204,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861451.34/warc/CC-MAIN-20240613123217-20240613153217-00899.warc.gz | 103,071,480 | 28,703 | # Simplifying Fractions Worksheet Pdf
In these worksheets the child has to convert an improper fraction numerator greater than the denominator into a mixed fraction whole number plus a fraction. Below are six versions of our grade 5 math worksheet on simplifying proper fractions.
Free Printables For Kids Math Fractions Worksheets Adding And Subtracting Fractions Fractions Worksheets
### 50 80 12.
Simplifying fractions worksheet pdf. These worksheets are pdf. 6 30 2. 24 30 5.
They do this by dividing the numerator by the denominator. Simplifying fractions in this case kids are meant to divide fractions to. 63 72 13.
5 10 c. Worksheets math grade 5 fractions converting simplifying fractions. 4 20 b.
Rewriting proper fractions in their simplest form. 18 90 11. Simplifying fractions grade 5 fractions worksheet simplify the fractions.
99 108 9. 5 10 3. Reducing proper fraction 1.
9 72 14. When you divide both the top and bottom numbers of a fraction by the same number you are dividing by a form of one so the value of the fraction doesn t. 6 8 6.
1 equivalent fractions 2 simplifying fractions 3 converting fractions to mixed numbers and 4 converting mixed numbers to fractions. Fraction worksheets pdf downloads fractions math worksheets for learning fractions and master the topic learn fractions and different operations with fractions with these free math fractions worksheets. Reducing proper fraction 2.
Students must rewrite the fractions if possible in their simplest form form with the lowest possible denominator. In these pdf worksheets grade 3 and grade 4 kids reduce proper fractions to the lowest term. Reducing fractions worksheet 2 reduce each fraction to lowest terms.
8 12 7. This worksheet generator makes worksheets for four different fraction topics. Reducing proper fraction 3.
14 21 d. Simplify the fraction 18 27 the greatest common factor of 18 and 27 is 9. Free fraction worksheets 2 simplifying fractions equivalent fractions fractions mixed numbers.
Images and questions that go hand in hand with a free online powerpoint that explains the answers. 4 8 10. 12 24 8.
Fractions packet created by mlc 2009 page 6 of 42 4 3 is the reduced form of 8 6. 4 18 35 50 4 6 28 48 40 50 4 16 22 44 12 36 6 9 42 48 11 33 10 12 21 30 27 36 10 16 3 12 5 10 22 44 24 40 8 16 7 14 10 15 14 16 6 14 6 21 15 40 12 18 6 18 4 8 8 10 12 24 22 55 13 39 4 20 15 35 18 20. 18 27 9 9 2 3 simplify each fraction.
Cancel out the common factors and reduce the improper fractions here to the lowest term. 2 6 16. This involves the multiplication of two fractions together.
4 40 4. 48 96 15. Simplifying fractions to simplify a fraction divide the numerator and the denominator by the greatest common factor.
Answers to worksheet along with visual aides are included at the end. Divide the numerator and the denominator by 9. A fraction using visual aides and an online interactive powerpoint.
9 15 e. This gives a whole number and a remainder which becomes the numerator of the fraction part of the mixed fraction over an unchanged denominator.
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10 Worksheets On Multiplying Fractions With Common Denominators Fractions Worksheets Multiplying Fractions Math Fractions Worksheets | 1,169 | 5,124 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.84375 | 4 | CC-MAIN-2024-26 | latest | en | 0.827555 |
https://www.ssccglapex.com/two-vessels-a-and-b-contains-acid-and-water-in-the-ratio-4-3-and-5-3-respectively-then-the-ratio-in-which-these-mixtures-to-be-mixed-to-obtain-a-new-mixture-in-vessel-c-containing-acid-and-water/ | 1,660,009,530,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882570879.37/warc/CC-MAIN-20220809003642-20220809033642-00381.warc.gz | 890,691,665 | 34,770 | ### Two vessels A and B contains acid and water in the ratio 4 : 3 and 5 : 3 respectively. Then the ratio in which these mixtures to be mixed to obtain a new mixture in vessel C containing acid and water in the ratio 3 : 2 is –
A. 5 : 8 B. 7 : 8 C. 7 : 5 D. 4 : 7 Answer: Option B
### Solution(By Apex Team)
$\begin{array}{l}\text{According to the question,}\\ \begin{array}{ccc}&\text{ Acid }&&\text{ Water }\\ \text{ Vessel A }&4&:&3\\ \text{ Vessel B }&5&:&3\end{array}\end{array}$ By alligation method –
So, Required ratio = 7 : 8
## Related Questions On Alligation
A. 2 : 5
B. 3 : 5
C. 5 : 3
D. 5 : 2
### An alloy contains zinc, copper and tin in the ratio 2 : 3 : 1 and another contains copper, tin and lead in the ratio 5 : 4 : 3. If equal weights of both alloys are melted together to form a third alloy, then the weight of lead per kg in new alloy will be:
A. $\large\frac{1}{2} \mathrm{~kg}$
B. $\large\frac{1}{8} \mathrm{~kg}$
C. $\large\frac{3}{14} \mathrm{~kg}$
D. $\large\frac{7}{9} \mathrm{~kg}$
A. 81 litres
B. 71 litres
C. 56 litres
D. 50 litres | 385 | 1,070 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.03125 | 4 | CC-MAIN-2022-33 | longest | en | 0.687822 |
https://dev.to/xtrp/using-the-monte-carlo-tree-search-algorithm-in-an-ai-to-beat-2048-and-other-games-462b | 1,611,103,877,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703519843.24/warc/CC-MAIN-20210119232006-20210120022006-00275.warc.gz | 297,488,398 | 35,430 | ## DEV Community is a community of 553,164 amazing developers
We're a place where coders share, stay up-to-date and grow their careers.
# Using the Monte Carlo Tree Search Algorithm in an AI to Beat 2048 (and other games)
Gabriel Romualdo Originally published at xtrp.io Updated on ・6 min read
Originally published here at xtrp.io, my blog about computer science and just about anything programming.
I recently worked on an open source project called Jupiter, an online AI to beat the popular online game 2048.
Go try out the AI:
In writing this AI, I decided to use a machine learning method called the Monte Carlo Tree Search (MCTS) algorithm. Monte Carlo algorithms like the one used in Jupiter have been used in several notable AIs, including DeepMind's AlphaGo, which famously beat the Go world champion in May 2017.
• How and why the Monte Carlo method works
• When and where Monte Carlo algorithms can be useful
• How I used the Monte Carlo method in an AI to beat 2048
• How to implement Monte Carlo algorithms in JavaScript and other languages
Note: I got the idea of using a Monte Carlo method to beat 2048 from this StackOverflow answer.
## What is the Monte Carlo Method?
The Monte Carlo method is the idea of using a large number of random simulations of an experiment to gain insights into the experiment's end results. Random simulations of an experiment are frequently referred to as Monte Carlo simulations.
For example, let's say that you were flipping a coin, and trying to figure out the probability of the coin landing heads. With the Monte Carlo method, we could simulate 10,000 coin tosses, and calculate the percentage of coins that landed heads.
Here's what that would look like.
As can be seen, the result converges to the expected value, 50%. A notable feature of Monte Carlo simulations is that a higher number of simulations is correlated with higher accuracy. For example, if we only performed two simulations, there is a high (25%) probability of heads landing in both simulations, giving a result of 100%. This is very inaccurate in comparison to the expected result of 50%.
Monte Carlo simulations work because of the Law of Large Numbers, which says:
If you simulate the same experiment many times, the average of the results should converge to the expected value of the simulation.
In other words, Monte Carlo simulations are a way to estimate what will happen in a given experiment without having to implement any specific algorithms or heuristics.
## When and Where the Monte Carlo Method Can Be Useful
The Monte Carlo method is used in a variety of fields, including game AI development, finance and economics, and evolutionary biology to name a few.
The Monte Carlo method can be useful in any experiment with a random factor, where end results cannot be predicted algorithmically. For example, in 2048, a new tile at a random location is added after every move, making it impossible to calculate the exact location of upcoming tiles and subsequently the end result of the game as well.
In these types of experiments, running a large number of Monte Carlo simulations can help get a sense of the average end results, the probability of various events occurring, and the relationship between the variables in the experiment.
For example, using the Monte Carlo method to in Jupiter allowed me to better understand how variables like starting move, number of moves in a game, and best tile in the board affected the end results of the game.
## How I Used the Monte Carlo Method in Jupiter, an AI to Beat 2048
• Board and Tiles: a 4x4 grid with tiles optionally placed on each grid spot
• Game State: a set of tiles on the board which represents the board at a specific time
• Game Score: the sum of all the tiles on the board
• Real Game: the game that is being played and shown on the browser, not a simulation
At any given game state, let's assume that four possible moves can be made: left, right, up, or down.
There are indeed cases where a certain move is not possible in a given game state. Removing impossible moves can be easily added to the algorithm later.
With the Monte Carlo method, we can run a set of game simulations for every move.
For each possible move, the program simulates a set of simulations which start by playing the move for that set first. After that, the rest of the game can be played completely randomly until it is over.
In JavaScript, this algorithm looks something like:
``````// assume Game object exists
// assume currentGame variable exists as the real game
const totalSimulations = 200; // 50 simulations are played for each move
const possibleMoves = ["left", "right", "down", "up"];
possibleMoves.forEach((move) => { // simulations for all four possible starting moves
for(let i = 0; i < totalSimulations / 4; i++) {
const simulation = new Game(); // create simulation
simulation.board = currentGame.board; // copy current game state to simulation
simulation.makeMove(move); // make initial move
while(!simulation.gameover()) {
simulation.makeMove(possibleMoves[Math.floor(Math.random() * 4)]);
} // make random moves until simulation game is over
}
});
``````
After all the simulations are completed, the program can gather the total final game scores of all the simulations, and average them for each move. We can then find the optimal move by optimizing for the highest final game score.
For example, if the simulations which started by playing left had an average final score of 250, whereas the ones which started by playing the other moves had an average final game score of 225, then left is the optimal move.
In this program, the optimal move is the one with simulations with the highest average final game score.
Note: I could have chosen to optimize for a different value such as the number of moves in the game.
However, this would actually make no difference in how the algorithm functions, because the number of moves in the game almost exactly predicts the game score. In 2048, the new tile added after each game move is normally a 2 tile, but has a 10% chance of being a 4 tile instead. This means the expected value of the new tile is 2.2 (`2 × 90% + 4 × 10%`). The total value of tiles is also preserved after every tile combination (ex: 2 tile combined with another 2 tile gives a 4 tile). As a result, game score can be calculated by multiplying the expected value of the new tile by the number of moves in the game, or with this formula: `2.2 × (real game move count + average move count)`.
To add this functionality of optimizing for highest score to our current code: add an array of total final scores for the simulations for each possible move, and choose the move with the highest value in that array to play like so:
``````const possibleMoves = ["left", "right", "down", "up"];
const totalSimulations = 200;
let moveSimulationTotalScores = [0, 0, 0, 0];
possibleMoves.forEach((move, moveIndex) => { // simulations for all four possible starting moves
for(let i = 0; i < totalSimulations / 4; i++) {
const simulation = new Game(); // create simulation
simulation.board = currentGame.board; // copy current game state to simulation
simulation.makeMove(move); // make initial move
while(!simulation.gameover()) {
simulation.makeMove(possibleMoves[Math.floor(Math.random() * 4)]);
} // make random moves until simulation game is over
moveSimulationTotalScores[moveIndex] += simulation.getScore();
}
});
// make best move with highest total simulation scores
let topScore = Math.max(...moveSimulationTotalScores);
let topScoreIndex = moveSimulationTotalScores.indexOf(topScore);
let bestMove = possibleMoves[topScoreIndex];
currentGame.makeMove(bestMove);
``````
In the end, this algorithm is simple to implement given a well-written 2048 game class. In JavaScript, there are a number of performance upgrades that can be made, starting by adding concurrency with Web Workers and pruning moves with very low final game scores.
## Conclusion
I hope you enjoyed this post, and found it useful in helping you understand and implement the Monte Carlo method in your own projects.
Go check out Jupiter and its source code.
Thanks for scrolling.
This post is originally from my blog at xtrp.io.
— Gabriel Romualdo, September 12, 2020 | 1,790 | 8,273 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2021-04 | longest | en | 0.927409 |
https://www.meer.com/en/75535-mathematics-and-the-paranormal | 1,718,501,783,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861640.68/warc/CC-MAIN-20240616012706-20240616042706-00290.warc.gz | 773,645,433 | 14,944 | Without mathematics, there’s nothing you can do. Everything around you is mathematics. Everything around you is numbers.
(Shakuntala Devi, writer and mental calculator)
### The Fibonacci sequence and botany
The Fibonacci sequence and flowers share a mesmerizing connection, revealing the elegance of mathematics within the natural world. The sequence, which starts with 0 and 1, with each subsequent number being the sum of the two preceding ones (0, 1, 1, 2, 3, 5, 8, and so on), appears to be a hidden code governing the growth patterns of various flowers. Many flowers, including sunflowers and daisies, exhibit spirals that follow the Fibonacci sequence. The number of petals in these flowers often corresponds to Fibonacci numbers, such as 3, 5, 8, or 13. This harmonious relationship between mathematics and botany evokes a sense of wonder, suggesting that the Fibonacci sequence serves as an unspoken blueprint guiding the artistry of nature's blossoms.
One of my personal favorite flowers to have in my garden is the Mexican sunflower (Tithonia rotundifolia). These orange-to-red sunflowers attract a wide variety of bees, butterflies, and hummingbirds to my backyard. Below is a picture of one from the summer of 2023:
As you can see, this flower has 13 petals. In mathematics, 13 is a very special number with several titles, including:
• Prime – a whole number greater than 1 that cannot be exactly divided by any whole number other than itself and 1.
• Fibonacci – a member of the Fibonacci sequence, detailed previously.
• Fibonacci prime – a Fibonacci number that is prime.
• A member of a Pythagorean triple (5, 12, 13). You’ll likely remember these sets of numbers from geometry, where they describe the whole number lengths of right triangles. The Pythagorean Theorem told us that the square of the lengths of the two shorter sides (a and b) equaled the square of the longer side --- aka the hypotenuse, the side across from the right (90 degrees) angle: a2 + b2 = c2.
### The Fibonacci sequence and birds
The connections between the Fibonacci sequence and the flights of birds reveal a remarkable harmony between mathematics and the avian world. Some species of birds exhibit flight patterns that align with the Fibonacci sequence, highlighting nature's propensity for following mathematical principles. When birds take to the skies, their wing motions often follow specific ratios that correspond to consecutive Fibonacci numbers, resulting in graceful and efficient flight paths. Additionally, the arrangement of feathers on a bird's wing may also adhere to the Fibonacci sequence, further emphasizing the prevalence of this mathematical phenomenon in the avian realm. The captivating synchrony between the Fibonacci sequence and the flights of birds underscores the elegance of nature's design, allowing these feathered creatures to soar through the air with both grace and precision.
### The Fibonacci sequence and the stars
The unknown code of reality seems to also write itself on a grand scale. The connection between the Fibonacci sequence and stars lies in the arrangement and distribution of stars within certain celestial objects, particularly in spiral galaxies. Spiral galaxies, such as the famous Andromeda galaxy and our own Milky Way, showcase a pattern that closely resembles the Fibonacci sequence.
In these galaxies, stars are organized in logarithmic spirals that follow a specific pattern. The number of arms or spiral patterns in these galaxies often corresponds to Fibonacci numbers, such as 2, 3, 5, 8, or 13. Additionally, the angles between these spirals, measured in degrees, also exhibit relationships found in the Fibonacci sequence.
The phenomenon is known as "Fibonacci spirals" or "Fibonacci galaxies." These visually striking patterns are a testament to the fundamental principles of mathematics imprinted upon the vast canvas of the cosmos. This connection between the Fibonacci sequence and stars in the sky is yet another awe-inspiring example of the natural world's inherent mathematical elegance and the enigmatic harmony between science and the universe. From sunflowers to birds, to the universe, there are mathematical equations at work, waiting to be solved.
### Application of mathematics to the paranormal
To learn more about paranormal phenomena such as UFOs (UAPs), there is a great need for data. It is important to acknowledge that although public transparency of such data is an integral step in disclosure, it is understandable that there is also a need for classified information in regard to security matters. This makes the public gathering of data even more important, and the more sources, the better. Information from Avi Loeb’s Galileo Project, NASA, MUFON, and private individuals observing the skies (which remain unclassified) should be gathered and analyzed. This is the scientific method. It is through such rigorous science that conclusions can be drawn. I started growing Mexican sunflowers in my home garden several years ago. Not only are they beautiful, but their orange-to-red sunflowers attract a wide variety of bees, butterflies, and hummingbirds to my backyard.
The discovery of connections such as that between mathematics and botany, birds, and stars took data and analysis – hard science. But such relationships would go unknown without those who dedicate themselves to the effort of collecting data and making it available to the public. I suspect there are beautiful mathematical relationships to be discovered within UFOs and other paranormal phenomena, but there is a severe lack of public data. Certainly, the military has collected data, but the mathematics seem to remain locked away, accessible only to a select few. These high-level figures within our government, military, and private contractors are able to maintain a veil of secrecy on data they deem too classified for public consumption. However, thanks to historic events like the July 26th Congressional Oversight UFO Hearing (video below), the days of shadowy access to under-the-table Special Access Programs (SAPs) could be coming to an end.
Something that has been said for quite a while is that if we are to ever communicate with alien beings, the common language would be mathematics. I also believe that gaining a deeper understanding of UFOs and UAPs is possible through mathematics: the gathering and analysis of large amounts of data. Shape, color, size, location, speed, etc. I am convinced that once this begins to widely take place, humankind will discover an unspoken blueprint guiding the artistry of the universe's deepest mysteries. | 1,317 | 6,625 | {"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} | 4.25 | 4 | CC-MAIN-2024-26 | latest | en | 0.928109 |
http://openstudy.com/updates/50ad6759e4b09749ccabbe12 | 1,444,744,918,000,000,000 | text/html | crawl-data/CC-MAIN-2015-40/segments/1443738006925.85/warc/CC-MAIN-20151001222006-00013-ip-10-137-6-227.ec2.internal.warc.gz | 232,224,278 | 11,081 | graydarl 2 years ago I must show that lim whn n goes to infinity of (1/n!)*((n/3)^n)=0
1. graydarl
$\lim_{n \rightarrow \infty} \frac{ 1 }{ n! }\left( \frac{ n }{ 3 } \right)^n=0$
2. Zarkon
One slick way is to show that $\sum_{n=1}^{\infty}\frac{ 1 }{ n! }\left( \frac{ n }{ 3 } \right)^n$ converges using the ratio test -if the series is finite then the limit of the sequence is zero
3. graydarl
i am still stuck, i dont t know how to solve it or if what i try is correct :(
4. scarydoor
Do you know how to apply the ratio test that Zarkon was talking about? I just took a go at it (was good memory jog) and I had to realise that I didn't really care about knowing the actual limit of the thing you're interested in when applying the ratio test, I only cared about making sure it was less than 1 in the limit.
5. graydarl
i tryed something and i am not sure that is correct so i can t tell that i know how to apply it
6. scarydoor
Well you want to prove that series Zarkon mentioned is convergent. Have a read of this: http://en.wikipedia.org/wiki/Ratio_test Using the terminology from that page, your a_n is 1/n! * (n/3)^n. Now to apply the ratio test, you want to see if a_{n+1} / a_n converges to something strictly less than 1. Take the thing we defined as a_n, and then have a look at what a_{n+1} / a_n looks like. Some things will cancel. Then see if that thing converges to something strictly less than 1. You should be able to see that it does (it's bounded and monotonic). So you've applied the ratio test and found that the result is that the series Zarkon described does converge. A series converges only if the elements in the series converge to zero in the limit. That gives you the result you wanted.
7. RyanL.
http://www.youtube.com/watch?v=iy8mhbZTY7g http://www.youtube.com/watch?v=AwJ0P8B25tY ------------------------------------------------- First video is a tutorial on how to use the ratio test and what it means. Second one explains how to deal with factorials using the ratio test.
8. graydarl
Thank you all | 571 | 2,048 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.953125 | 4 | CC-MAIN-2015-40 | longest | en | 0.890344 |
https://engineerexcel.com/elastic-deformation/ | 1,713,562,680,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817455.17/warc/CC-MAIN-20240419203449-20240419233449-00371.warc.gz | 209,855,003 | 96,164 | # Elastic Deformation: Mechanics of Reversible Material Deformation
Elastic deformation is a reversible change in the shape or size of a material in response to an applied force, where the material returns to its original shape once the force is removed. It occurs within the material’s elastic limit, where the atomic or molecular structure can temporarily distort without permanent damage. In this article, we will delve into the mechanics of elastic deformation, its different modes, stress-strain relationships, and practical applications in various engineering fields.
## Understanding Elastic Deformation
Elastic deformation is a type of material strain where a solid body temporarily changes shape when stress is applied. Upon removal of the stress, the material returns to its original shape. This type of deformation occurs at low stress levels, within the elastic limit of the material, which is the maximum extent to which the material can be deformed without causing permanent deformation.
The behavior of materials undergoing elastic deformation can be explained by their atomic and molecular structure. Atoms and molecules form a lattice structure within the solid and are bonded together by interatomic forces.
When stress is applied, these bonds are stretched but do not break; instead, they act like tiny springs, storing the energy from the stress. The material’s ability to withstand elastic deformation depends on the strength and stiffness of these bonds, which determine the force required to deform the material.
Elastic deformation is a critical concept in many engineering applications because it determines how materials will respond to forces. Engineers must ensure that the stresses and strains within materials remain within the elastic range to avoid permanent deformation or failure. For example, when designing bridges, buildings, or airplane wings, the structures must be able to bend slightly under load but return to their original shape to maintain integrity and safety.
However, if the stress exceeds the material’s elastic limit, it enters the plastic deformation phase, where the deformation becomes permanent, and the material does not return to its original shape after the load is removed. This transition from elastic to plastic behavior marks the yield point of the material, which is a critical design threshold.
Understanding the difference between elastic and plastic deformation is essential for material selection and engineering design to ensure that components can withstand the loads they will encounter during their service life without experiencing failure.
## Types of Elastic Deformation
Elastic deformation can be categorized into three primary modes based on the nature of the applied stress: axial (tensile or compressive), shear, and bulk.
### Axial Deformation
#### Tensile Deformation
When a material experiences a pulling or stretching force, it undergoes tensile deformation. This is characterized by an elongation of the material along the direction of the applied force. The extent of tensile deformation a material can withstand is a measure of its tensile strain, which is a dimensionless ratio of the change in length to the original length.
#### Compressive Deformation
Compressive deformation occurs when a material is subjected to squeezing or pushing forces. The stress results in a shortening of the material in the direction of the applied pressure. During compression, materials exhibit compressive strain, calculated similarly to tensile strain as the ratio of change in length to the original length but with values indicating reduction rather than elongation.
### Shear Deformation
Materials exposed to forces acting parallel to each other but in opposite directions experience shear deformation. It is observed as a distortion or slippage along the plane subjected to the shear stress, resulting in a change of shape without a change in volume. The degree of shear deformation is measured by the shear strain, quantified as the displacement of one layer relative to another divided by the initial separation of the layers.
### Bulk Deformation
Bulk deformation refers to the change in volume of a material in response to an applied stress, typically resulting in compression or expansion of the material uniformly in all directions. This type of deformation occurs when the material experiences stress that is distributed evenly throughout its volume, leading to a change in its overall shape without significant distortion or localized deformation. It is commonly observed in materials like rubber or fluids under hydrostatic pressure, where the applied stress causes the material to compress or expand uniformly.
## Stress and Strain Relationship
Understanding the relationship between stress and strain is important in analyzing elastic deformation. Stress refers to the internal forces that particles of a material exert on each other, typically measured in units of pressure such as pascals (Pa). Strain indicates the deformation of the material in response to stress, defined as the change in dimension over the original dimension, and is dimensionless.
For axial deformation, Hooke’s Law describes the stress-strain relationship, stating that within the elastic limit of a material, the applied stress is directly proportional to the resulting strain:
Where:
• σ = axial stress [Pa]
• E = Young’s modulus [Pa]
• ε = axial strain [unitless]
Materials subjected to stress display a linear relationship up to a certain limit, known as the proportional limit, beyond which permanent deformation occurs. The region within this limit is known as the elastic region, as shown in the diagram below.
Similarly, for shear deformation, the stress-strain relationship is described by the formula:
Where:
• τ = shear stress [Pa]
• G = shear modulus [Pa]
• γ = shear strain [unitless]
Lastly, for bulk deformation, the stress-strain relationship is described by the formula:
Where:
• ΔP = bulk stress [Pa]
• B = bulk modulus [Pa]
• ΔV = change in volume [m3]
• Vo = original volume [m3]
The negative sign indicates an increase in pressure causes a decrease in volume.
Young’s modulus, shear modulus, and bulk modulus are all measures of a material’s response to different types of elastic deformation. Young’s modulus describes how a material deforms under tensile or compressive forces, shear modulus quantifies its response to shearing forces, and bulk modulus measures its resistance to uniform compression.
## Applications of Elastic Deformation
### Structural Analysis
In structural analysis, engineers use elastic deformation principles to predict how buildings, bridges, and other structures will respond to loads. This includes calculating expected displacements and rotations of structural elements to ensure safety and stability under normal and extreme conditions. Analyzing the elastic properties of materials allows for designing structures that can withstand anticipated stresses without permanent deformation.
### Machining Tolerances
Precise control of elastic deformation is important in machining processes to maintain tolerances. During the application of cutting forces, components experience elastic deformation, which must be appropriately compensated for in the design stage to achieve the desired dimensions and geometries after the removal of the applied forces.
### Springs and Cushioning Devices
The design of spring and cushioning devices, from simple coil springs to complex energy-absorbing architectures, depends heavily on the elastic properties of the materials to store and release energy as needed for applications such as vehicle suspension systems and vibration dampeners.
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Complete... 50% | 1,402 | 7,753 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-18 | latest | en | 0.904617 |
https://au.mathworks.com/matlabcentral/answers/493279-greetings-anyway-to-help-to-create-ridgeline-by-matlab | 1,726,181,327,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651498.46/warc/CC-MAIN-20240912210501-20240913000501-00311.warc.gz | 93,088,116 | 39,432 | # Greetings, Anyway to help to create ridgeline by MATLAB
10 views (last 30 days)
Yaser Khojah on 26 Nov 2019
Greetings,
Anyway to help to create ridgeline by MATLAB as the attache link please?
##### 3 CommentsShow 1 older commentHide 1 older comment
Yaser Khojah on 26 Nov 2019
Dear KALYAN, the first figure.
Thanks,
Adam Danz on 26 Nov 2019
Edited: Adam Danz on 26 Nov 2019
What release of matlab are you using? That's one of the options to fill in when asking a question and it's an important thing for us to know.
Adam Danz on 26 Nov 2019
Edited: Adam Danz on 28 Nov 2019
Here's a demo that creates a number of guassian distributions as input. The code produces an appropriate number of contiguous subplots where you can set the left, right, upper, and lower margins. Then it uses histfit() to compute and plot the density functions of each data. The code extracts the (x,y) values of the density curve and uses them to form a colored patch which replaces the histfit() plots. The axis limits and linked and some plot cosmetics are done to make the plot similar in appearance to the link you provided.
% Generate n distributions
n = 8; % number of distributions
mu = linspace(0,100,n);
sd = (rand(size(mu)) +1).*2;
nSamp = 100; %number of samples per dist.
data = arrayfun(@normrnd,mu,sd,ones(size(mu)),nSamp.*ones(size(mu)),'UniformOutput',false); % req. stats & ML toolbox
yLabs = num2cell(char(64+cumsum(ones(1,n))));
Now we have two key input variables.
1. data which is a 1 x n cell array where each element is a 1xm vector of data that will be used to compute a distribution.
2. yLabs : a 1 x n cell array of characters used to label each distribution along the y axis.
% Generate figure.
fh = figure();
% Compute axes positions with contigunous edges
n = numel(data);
margins = [.13 .13 .12 .15]; %left, right, bottom, top
height = (1-sum(margins(3:4)))/n; % height of each subplot
width = 1-sum(margins(1:2)); %width of each sp
vPos = linspace(margins(3),1-margins(4)-height,n); %vert pos of each sp
% Plot the histogram fits (normal density function)
% You can optionally specify the number of bins
% as well as the distribution to fit (not shown,
% see https://www.mathworks.com/help/stats/histfit.html)
% Note that histfit() does not allow the user to specify
% the axes (as of r2019b) which is why we need to create
% the axes within a loop.
% Otherwise we could use tiledlayout() (>=r2019b)
% https://www.mathworks.com/help/matlab/ref/tiledlayout.html
subHand = gobjects(1,n);
histHand = gobjects(2,n);
for i = 1:n
subHand(i) = axes('position',[margins(1),vPos(i),width,height]);
histHand(:,i) = histfit(data{i});
end
% Extend density curves to edges of xlim and fill.
% This is easier, more readable (and maybe faster) to do in a loop.
xl = xlim(subHand(end));
colors = jet(n); % Use any colormap you want
for i = 1:n
x = [xl(1),histHand(2,i).XData,xl([2,1])];
y = [0,histHand(2,i).YData,0,0];
fillHand = fill(subHand(i),x,y,colors(i,:),'FaceAlpha',0.4,'EdgeColor','k','LineWidth',1);
% Add vertical ref lines at xtick of bottom axis
arrayfun(@(t)xline(subHand(i),t),subHand(1).XTick); %req. >=r2018b
ylh = ylabel(subHand(i),yLabs{i});
set(ylh,'Rotation',0,'HorizontalAlignment','right','VerticalAlignment','middle')
end
% Cosmetics
% Delete histogram bars & original density curves
delete(histHand)
% remove axes (all but bottom) and
% add vertical ref lines at x ticks of bottom axis
set(subHand(1),'Box','off')
arrayfun(@(i)set(subHand(i).XAxis,'Visible','off'),2:n)
set(subHand,'YTick',[])
set(subHand,'XLim',xl)
Adam Danz on 27 Jan 2020
Image Analyst's advice is also mentioned in the subplot documentation tips-section. Calling the subplot function deletes existing axes that overlap the newly created subplot. However, once the subplots are created, you can overlap them by changing their position.
Here's a demo where you set the number of vertically stacked subplots (nSubs) and the amount of overlap (p, a value 0:1).
nSubs = 6;
p = 0.10; % 10%
clf() % clear figure
sh = arrayfun(@(i)subplot(nSubs,1,i),1:nSubs); % create all subplots, default position
arrayfun(@(i)box(sh(i),'on'),1:nSubs) % Turn on axis box
subPos = reshape([sh.Position]',4,[])'; % Get subplot positions
% Change the y position of each plot so that
% all subplots overlap by p percent (0:1).
subPos(:,2) = (sum(subPos(1,[2,4])) - (subPos(1,4) * (1-p)) * (0:nSubs-1)) - subPos(1,4);
% Set new subplot positions
set(sh,{'position'},mat2cell(subPos,ones(nSubs,1),4))
Jacqueline Chrabot on 19 Jul 2021
I'm trying to reuse this code, to make a plot, except my data is set up in a field where y is depth, x is time and the data fills in columns 1-12 of chlorophyll data. I'm trying to fit gaussian for each colomn and then display them all in the graph you showed. Can you help with this?
Image Analyst on 28 Nov 2019
Perhaps you'd be interested in stackedplot(): Pick of the Week
>> doc stackedplot
Introduced in R2018b.
Santiago Benito on 21 Apr 2020
Edited: Santiago Benito on 21 Apr 2020
Hi there,
Maybe it's a little late, but I stumbled upon the same problem. I really wanted the plots to overlap, so I did the following:
% Number of data plots
n = 8;
% Sample points
N = 100;
% Distribution, example data
distName = 'normal';
mn = linspace(0,1,n);
% Allocate a matrix to store the dataset
yData = zeros(N,n);
% Plot options
mini = -0.3;
maxi = 1.3;
overlap = 0.4;
% Create the data
for ii = 1:n
distCell = makedist(distName,'mu',mn(ii),'sigma',0.1);
yData(:,ii) = pdf(distCell,linspace(mini,maxi,N));
end
% Get the position of each dataset
y = cumsum(max(yData,[],1))*(1-overlap);
% Create the figure with patch & plot
figure, hold on
for ii = n:-1:1
patch([linspace(mini,maxi,N),-mini],[yData(:,ii)+y(ii);y(ii)],mn(ii),...
'EdgeColor','none','FaceAlpha',0.8)
plot(linspace(mini,maxi,N),yData(:,ii)+y(ii),'k','LineWidth',1)
end
hold off
% Other stuff
colormap(spring)
yticks(y)
yticklabels({'A','B','C','D','E','F','G','H'})
xlim([mini,maxi])
The result:
Some thoughts:
• This is not as elegant as the other solutions, but works for me.
Cheers!
Santiago Benito on 23 Apr 2020
Adam Danz on 23 Apr 2020
Edited: Adam Danz on 23 Apr 2020
The colorbar in the example above is a bit confusing. There are two yellows that have different values. The lower colorbar should be based on something like the winter colormap which doesn't have intersecting colors with the spring colormap. It's not clear why 2 colormaps/colorbars are needed. | 1,901 | 6,431 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.125 | 3 | CC-MAIN-2024-38 | latest | en | 0.792208 |
https://www.learndifferently.com/tag/learning-disabilities/page/4/ | 1,656,586,048,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103671290.43/warc/CC-MAIN-20220630092604-20220630122604-00461.warc.gz | 889,116,909 | 11,366 | August 02, 2016
# 7 ways to help children after adoption
by Kathy Kuhl I often talk with parents who have adopted or are fostering children who learn differently. Sometimes, after a child’s move from a hard situation, parents need extraordinary resources. Favorite resources for families after adoption The Center for Adoption Support and Education (CASE) helps foster and adoptive families. It “promotes adoption awareness, enhances […]
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July 06, 2016
# Falling furniture, impulsive kids
by Kathy Kuhl To keep a little one safe can be a big job, even when they don’t climb furniture. Many of you are doing this while homeschooling a child with challenges. Some of us have older kids that still act like preschoolers, so we are still on alert. Impulsive children try things we never thought […]
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June 22, 2016
# Need testing for learning disabilities?
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May 31, 2016
# Teaching a Child with Special Needs: Book Review
Book review by Kathy Kuhl Judith Munday’s new book is out at last: Teaching a Child with Special Needs at Home and at School. That’s good news for homeschoolers with children with learning challenges. Back in the 1990s, when I was a new homeschooler, Judi was the first expert I heard who understood homeschooling students with special […]
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May 24, 2016
# Chemicals endangering our children
By Kathy Kuhl It’s not just parents in Flint, Michigan, who need to keep an eye on what their children are exposed to. But until a crisis looms, who thinks to look into these things? Who has time? We parents of kids with learning challenges especially can be so busy helping our kids keep or […]
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April 26, 2016
# More algebra with manipulatives
by Kathy Kuhl Last time, I began explaining bean algebra—a way to solve linear equations using beans. Manipulatives are simply objects you handle to help you understand a concept. The containers represent the variable x. I labeled mine with post-its. Last week, I showed you how to use beans to model equations like 14 =2x. Click here if […]
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April 19, 2016
# Teaching algebra with manipulatives
by Kathy Kuhl Teaching parents in Fort Worth last month, I promised to explain bean algebra—a way to solve linear equations using beans. (I learned this when I was a new public school math teacher in 1979, and I cannot remember or find the source. If you know, tell me.) What manipulatives you’ll need Manipulatives are […]
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March 16, 2016
# Developing Basic Skills in Reading & Writing
by Kathy Kuhl Review of Diana Hanbury King’s A Guide to Helping Your Child at Home: Developing Foundational Skills in Reading & Writing Suppose you could get one of the best writing teachers, a specialist in helping struggling learners, to come over to help you understand how to teach your child? What if she […]
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February 16, 2016
# Auditory Processing: When the Brain Can’t Hear
By Kathy Kuhl Book review of When the Brain Can’t Hear by Teri James Bellis, Ph.D. “My child can’t have an auditory problem. His hearing is just fine.” When the ears work great, but the brain has trouble understanding, then the child—or adult—may have an auditory processing disorder (APD). It is also called a […]
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February 09, 2016
# If I could send my younger self a letter
By Kathy Kuhl If I could send a letter back to my younger self in 1997, when I started to homeschool, here’s what I’d write: Dear younger me, Greetings from eighteen years from now. Since I remember how anxious we were in 1997, I thought I’d write. (I’m thinking of your shelves of homeschool catalogs–the […]
( Read the full post… ) | 900 | 3,794 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.71875 | 3 | CC-MAIN-2022-27 | latest | en | 0.942232 |
https://www.gradesaver.com/textbooks/math/algebra/intermediate-algebra-for-college-students-7th-edition/chapter-1-section-1-1-algebraic-expressions-real-numbers-and-interval-notation-exercise-set-page-13/67 | 1,534,404,582,000,000,000 | text/html | crawl-data/CC-MAIN-2018-34/segments/1534221210463.30/warc/CC-MAIN-20180816054453-20180816074453-00696.warc.gz | 913,800,874 | 14,162 | ## Intermediate Algebra for College Students (7th Edition)
$\left\{x|-5 \le x \lt 2\right\}$ Refer to the attached image below for the graph.
A bracket on the left means that $-5$ is included in the set. A parenthesis on the right means that $2$ is not included in the set. Therefore, the set-builder form of the given interval is: $\left\{x|-5 \le x \lt 2\right\}$ To graph the interval, plot a hole (or hollow dot) at x=−2 and a solid dot at x=4, and then shade the region between them. (refer to the attached image in the answer part for the graph) | 153 | 552 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.125 | 4 | CC-MAIN-2018-34 | longest | en | 0.917056 |
https://ch.gateoverflow.in/276/gate-chemical-2014-question-14 | 1,679,674,242,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296945287.43/warc/CC-MAIN-20230324144746-20230324174746-00503.warc.gz | 215,674,436 | 14,381 | Assume that an ordinary mercury-in-glass thermometer follows first order dynamics with a time constant of $10\:s$. It is at a steady state temperature of $0^{\circ}C$. At time $t=0$, the thermometer is suddenly immersed in a constant temperature bath at $100^{\circ}C$. The time required (in $s$) for the thermometer to read $95^{\circ}C$, approximately is
1. $60$
2. $40$
3. $30$
4. $20$
Answer: | 114 | 398 | {"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} | 2.625 | 3 | CC-MAIN-2023-14 | latest | en | 0.860354 |
https://aplicaciones.uc3m.es/cpa/generaFicha?est=218&plan=256&asig=15755&idioma=2 | 1,620,273,552,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243988725.79/warc/CC-MAIN-20210506023918-20210506053918-00424.warc.gz | 116,221,595 | 4,189 | Checking date: 30/07/2020
Course: 2020/2021
Genetic and evolutionary algorithms
(15755)
Study: Bachelor in Computer Science and Engineering (218)
Coordinating teacher: ISASI VIÑUELA, PEDRO
Department assigned to the subject: Department of Computer Science and Engineering
Type: Electives
ECTS Credits: 6.0 ECTS
Course:
Semester:
Students are expected to have completed
Programming Statistics
KNOWLEDGE: - To adquire the fundamentals of problem solving techniques based on populations - To learn the fundamental techniques that allow the resolution of problems by using biologically inspired techniques - To know which techniques are appropriate for each problem domain (PO c) - To know the differences of each technique and their appropriate parameterization (PO c) - To get the knowledge about each technique and the reason for its implementation UNDERSTANDING: - To understand the application of population-based learning systems - To understand the mathematical foundations of evolutionary techniques described in the course - Understand the variants of each technique, motivation and practical justification - Understand the usefulness of the proposed techniques in problem solving APPLICATION: - Be able to select the appropriate technique to the specific characteristics of the problem domains (PO c) - Evaluate the performance and efficiency of different methods (PO b) - Working on specific problems and solve them efficiently adjusting techniques and their parameters (PO b) - Organize experimental environments to validate different alternatives and be able to evaluate and analyze results (PO b) - Hability to work cooperatively and creatively in a team. (PO d) - Hability to communicate effectively by oral and writtng (PO g) CRITICISM OR VALUATION: - To adquire the ability to validate alternative techniques to solve specific problems (PO c) - Criticizing a given proposal (eg in the selected bibliography) (PO c & d) - Consider the relationship between computational cost and marginal improvement in different Solutions (PO c) - To decide reasonable solutions to the features of a given problem (PO c)
Description of contents: programme
1. Introduction to evolutionary computation algorithms 2. General concepts of evolutionary algorithms 2.1 Generating the initial population 2.2 Stopping criteria 2.3 Selection methods 2.4 Reproduction 2.5 Insertion and replacement policies 2.6 Example of an evolutionary system 3. Evolutionary computation techniques 3.1 Genetic algorithms 3.2 Evolutionary Strategies 3.3 Diferential evolution 3.4 Genetic Programming 3.5 Gene Expressions 3.6 Grammatical evolution 4. Troubleshooting using evolutionary techniques 4.1 Problems with multiple solutions 4.2 Problems with multiple conflicting objectives 4.3 Constrained optimization 4.4 Coevolutionary algorithms 4.5 Application to real-life problems 4.6 Mathematical Foundations
Learning activities and methodology
1. Lectures. To learn the basic concepts of problem solving by means of population-based techniques. (PO b & c) 2. Development of a homework project. The student, as part of a work team must develop a project to give a complete solution to a problem, proposing techniques, parameters and the steps required for its realization. This project are divided into three sections that will be evaluated separately and the complexity will grow incrementally. The project will require the completion of a test environment as well as an analysis and assessment of results. 3 ECTS. (PO b & c & d) 3. Work teams will make public presentations of results in each project. After each presentation there will be an open discussion of the works submitted. (PO g)
Assessment System
• % end-of-term-examination 0
• % of continuous assessment (assigments, laboratory, practicals...) 100
Basic Bibliography
• El-Ghazali Talbi. Metaheuristics: From Design to Implementation. Wiley and Sons. 2009
• M. Mitchell. An introduction to Genetic Algorithms. MIT Press.
The course syllabus and the academic weekly planning may change due academic events or other reasons. | 848 | 4,064 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.8125 | 3 | CC-MAIN-2021-21 | latest | en | 0.895001 |
https://quickessayguide.com/two-friends-race-the-first-friend-gets-a-head-start-and-runs-at-a/ | 1,642,382,844,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320300253.51/warc/CC-MAIN-20220117000754-20220117030754-00712.warc.gz | 559,292,296 | 12,648 | # Two friends race. the first friend gets a head start and runs at a
Two friends race. The first friend gets a head start and runs at a steady rate of 0.20 miles per minute. The second friend waits 1 minute and then runs at a steady rate of 0.25 miles per minute. How long will it take for the second runner to catch up first.
Make this into an equation with a variable on each side. | 93 | 385 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2022-05 | latest | en | 0.900769 |
https://educators.brainpop.com/lesson-plan/choosing-operation-activities-kids/ | 1,563,540,370,000,000,000 | text/html | crawl-data/CC-MAIN-2019-30/segments/1563195526237.47/warc/CC-MAIN-20190719115720-20190719141720-00045.warc.gz | 384,537,877 | 12,573 | These classroom activities are designed to complement the Choosing an Operation topic on BrainPOP Jr.
### Add or Subtract?
Run through examples of different math word problems that involve addition or subtraction and review them with the whole class. Then team up students into pairs. Have each partner write a problem that involves addition or subtraction, and have the other partner solve. Then have the partners discuss. How did they know which operation to use to solve? Be sure each partner shows their solution, either by way of number sentence or by using manipulatives. Have pairs share their problems with the class.
### Six Marbles
Give small groups six marbles or other small objects to use as counters. Challenge each group to come up with as many word problems as they can that involve the six marbles. For example, one problem might be, “I have 3 marbles and I get 3 more. How many marbles do I have in all?” or “There are 6 friends and 6 marbles. How many marbles does each person get?” Encourage students to explore different operations, including division. Have small groups share their problems with the class.
### Taking Sides
Divide the classroom into two different sides representing addition and subtraction. Then pose a word problem, such as “I want to collect 10 stamps. I have 3 stamps so far. How many more stamps do I need?” Then have students walk to the side of the classroom representing the operation they need to use to solve the problem. Discuss each problem with the class. How did they know which operation to use? If appropriate, you can do the same activity using multiplication and division, or as an added challenge, use all four operations. | 347 | 1,686 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2019-30 | longest | en | 0.964328 |
https://number.academy/622476 | 1,716,912,763,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971059143.84/warc/CC-MAIN-20240528154421-20240528184421-00121.warc.gz | 363,071,586 | 11,424 | # Number 622476 facts
The even number 622,476 is spelled 🔊, and written in words: six hundred and twenty-two thousand, four hundred and seventy-six. The ordinal number 622476th is said 🔊 and written as: six hundred and twenty-two thousand, four hundred and seventy-sixth. Color #622476. The meaning of the number 622476 in Maths: Is it Prime? Factorization and prime factors tree. The square root and cube root of 622476. What is 622476 in computer science, numerology, codes and images, writing and naming in other languages
## What is 622,476 in other units
The decimal (Arabic) number 622476 converted to a Roman number is (D)(C)(X)(X)MMCDLXXVI. Roman and decimal number conversions.
#### Time conversion
(hours, minutes, seconds, days, weeks)
622476 seconds equals to 1 week, 4 hours, 54 minutes, 36 seconds
622476 minutes equals to 1 year, 3 months, 1 week, 5 days, 6 hours, 36 minutes
### Codes and images of the number 622476
Number 622476 morse code: -.... ..--- ..--- ....- --... -....
Sign language for number 622476:
Number 622476 in braille:
QR code Bar code, type 39
Images of the number Image (1) of the number Image (2) of the number More images, other sizes, codes and colors ...
## Share in social networks
#### Is Prime?
The number 622476 is not a prime number. The closest prime numbers are 622423, 622477.
#### Factorization and factors (dividers)
The prime factors of 622476 are 2 * 2 * 3 * 3 * 17291
The factors of 622476 are
1, 2, 3, 4, 6, 9, 12, 18, 36, 17291, 34582, 51873, 69164, 103746, 155619, 207492, 311238, 622476.
Total factors 18.
Sum of factors 1573572 (951096).
#### Powers
The second power of 6224762 is 387.476.370.576.
The third power of 6224763 is 241.194.741.250.666.176.
#### Roots
The square root √622476 is 788,971482.
The cube root of 3622476 is 85,383549.
#### Logarithms
The natural logarithm of No. ln 622476 = loge 622476 = 13,34146.
The logarithm to base 10 of No. log10 622476 = 5,794123.
The Napierian logarithm of No. log1/e 622476 = -13,34146.
### Trigonometric functions
The cosine of 622476 is 0,673681.
The sine of 622476 is 0,739022.
The tangent of 622476 is 1,096991.
## Number 622476 in Computer Science
Code typeCode value
PIN 622476 It's recommended that you use 622476 as your password or PIN.
622476 Number of bytes607.9KB
CSS Color
#622476 hexadecimal to red, green and blue (RGB) (98, 36, 118)
Unix timeUnix time 622476 is equal to Thursday Jan. 8, 1970, 4:54:36 a.m. GMT
IPv4, IPv6Number 622476 internet address in dotted format v4 0.9.127.140, v6 ::9:7f8c
622476 Decimal = 10010111111110001100 Binary
622476 Decimal = 1011121212200 Ternary
622476 Decimal = 2277614 Octal
622476 Decimal = 97F8C Hexadecimal (0x97f8c hex)
622476 BASE64NjIyNDc2
622476 MD511720b61b31ffe1096f0d46b26a7c612
622476 SHA116cb9882f4625b565126179cf4d2258a0bccc77a
622476 SHA224c87688bb1ce115762b951cb83c26a33e4735ebc1b8f6fc44f286a9bf
622476 SHA256a2012a68016713dbf3209a65fce4a690589d602dba42e4fc7c77b2421a8bf612
622476 SHA3849bb39a5f280675bf2ef822a6422592bbcd03a696df46c31fa5689bfed907ac2e62669e3a00654e28c76ef00a1bc44ce0
More SHA codes related to the number 622476 ...
If you know something interesting about the 622476 number that you did not find on this page, do not hesitate to write us here.
## Numerology 622476
### Character frequency in the number 622476
Character (importance) frequency for numerology.
Character: Frequency: 6 2 2 2 4 1 7 1
### Classical numerology
According to classical numerology, to know what each number means, you have to reduce it to a single figure, with the number 622476, the numbers 6+2+2+4+7+6 = 2+7 = 9 are added and the meaning of the number 9 is sought.
## № 622,476 in other languages
How to say or write the number six hundred and twenty-two thousand, four hundred and seventy-six in Spanish, German, French and other languages. The character used as the thousands separator.
Spanish: 🔊 (número 622.476) seiscientos veintidós mil cuatrocientos setenta y seis German: 🔊 (Nummer 622.476) sechshundertzweiundzwanzigtausendvierhundertsechsundsiebzig French: 🔊 (nombre 622 476) six cent vingt-deux mille quatre cent soixante-seize Portuguese: 🔊 (número 622 476) seiscentos e vinte e dois mil, quatrocentos e setenta e seis Hindi: 🔊 (संख्या 622 476) छः लाख, बाईस हज़ार, चार सौ, छिहत्तर Chinese: 🔊 (数 622 476) 六十二万二千四百七十六 Arabian: 🔊 (عدد 622,476) ستمائة و اثنان و عشرون ألفاً و أربعمائة و ستة و سبعون Czech: 🔊 (číslo 622 476) šestset dvacet dva tisíce čtyřista sedmdesát šest Korean: 🔊 (번호 622,476) 육십이만 이천사백칠십육 Danish: 🔊 (nummer 622 476) sekshundrede og toogtyvetusindfirehundrede og seksoghalvfjerds Hebrew: (מספר 622,476) שש מאות עשרים ושניים אלף ארבע מאות שבעים ושש Dutch: 🔊 (nummer 622 476) zeshonderdtweeëntwintigduizendvierhonderdzesenzeventig Japanese: 🔊 (数 622,476) 六十二万二千四百七十六 Indonesian: 🔊 (jumlah 622.476) enam ratus dua puluh dua ribu empat ratus tujuh puluh enam Italian: 🔊 (numero 622 476) seicentoventiduemilaquattrocentosettantasei Norwegian: 🔊 (nummer 622 476) seks hundre og tjueto tusen fire hundre og syttiseks Polish: 🔊 (liczba 622 476) sześćset dwadzieścia dwa tysiące czterysta siedemdziesiąt sześć Russian: 🔊 (номер 622 476) шестьсот двадцать две тысячи четыреста семьдесят шесть Turkish: 🔊 (numara 622,476) altıyüzyirmiikibindörtyüzyetmişaltı Thai: 🔊 (จำนวน 622 476) หกแสนสองหมื่นสองพันสี่ร้อยเจ็ดสิบหก Ukrainian: 🔊 (номер 622 476) шістсот двадцять дві тисячі чотириста сімдесят шість Vietnamese: 🔊 (con số 622.476) sáu trăm hai mươi hai nghìn bốn trăm bảy mươi sáu Other languages ...
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Home » Bank Exam Tips to crack Exam » Multiplication Tricks for Bank Exams – Maths Shortcut Tricks
# Multiplication Tricks for Bank Exams – Maths Shortcut Tricks
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# Multiplication Tricks for Bank Exams:
### I. Multiplication of two digit numbers which are having the same number in 10’s place and 5 in its units place.
1) 25 × 25 = (2× 3)25 = 625.
Understand the logic behind this so that you can able to solve for big numbers.
First, we have to write 25 at the right end as shown above. Then multiply 2 with next biggest number i.e. 3 that is equal to 6. Write this 6 at left end. The answer is 625.
2) 45 × 45 = (4×5)25 = 2025.
3) 55 × 55 = (5 × 6)25 = 3025.
3) 85 × 85 = (8 × 9)25= 7225.
4) 95 × 95 = (9 ×10) 25= 9025.
### II. Multiplication of 2 digit numbers whose difference is 10 and have 5 at units place.
1)25×35 = (2 × 4)75 = 875.
Explanation: Here we have to write 75 at the right end. Then multiply 2 with 4 and write the result at left end as shown above.
2) 35 × 45 = (3×5)75 = 1575.
3) 45 × 55 = (4 × 6)75 = 2475.
4) 65 × 75 = (6 × 8)75 = 4875.
5) 85 × 95 = (8 × 10)75 = 8075.
### III. Multiplication of two digit numbers having the same number at 10’s place and the sum of unit’s place digits is 10.
1) 33 × 37= (3 × 4) 21= 1221.
Explanation:
1) Multiply the last 2 digits at unit’s place 3 and 7 it becomes 21 write this at the right side end.
2) Multiply 3 with next number 4 it becomes 12 then write this at the left side end.
2) 44 × 46 = (4 × 5) 24= 2024. (Here, 6 × 4 =24, 4 × 5= 20)
3) 36 × 34 = (3 × 4)24 = 1224.
4) 52 × 58 = (5 ×6)16 = 3016.
5) 71 × 79 = (7 × 8)9 = 5609.
6) 87 × 83 = (8 × 9)21 = 7221.
### IV Multiplication of two digit numbers that have 9 at the 10’s place and the sum of digits at unit’s place should be 5 or 10.
1)92 × 93= (92-7) (8 × 7) = 8556
2)93 × 97= (93-3) (7 × 3) = 9021
3) 95 × 95= (95 – 5) (5× 4) = 9020.
4) 96 × 94= (94 – 4) (6 × 4) = 9024.
5) 91 × 94= (91-1) (1 × 4) = 9004.
### V. Multiplication using Split & Merge Method:
1. 42 × 49 = 42 (50-1)
= 42 × 50 – 42 × 1. (Here use multiplication trick with 5).
= 2100 – 42.
2. 25 × 29 = 25 ( 30 – 1).
= 750 -25.
= 725.
3. 15 × 19 = 15 (20 -1)
= 300 -15.
= 275.
### VI. Horizontal Multiplication Method:
1. 43 × 23 = 8 _ 9.
(i) we get 8 by multiplying 4 × 2 and place it at first place.
(ii) Multiply unit’s digits and place it right i.e. 3 × 3 =9.
(iii) Middle term = (4 × 3) + (3 × 2) = 12 + 6 = 18.
= 18 9. = 989.
2. 26 × 21 = 4 _ 6.
(i) 2 × 2 = 4 . (1st term)
(ii) 6 × 1 = 6. ( last )
(iii) (2 × 1) + 6 × 2) = 2 + 12 = 14. (middle). (if there is double digit in middle carry forward to first digit).
Therefore, we get 4 14 6. = 546.
### VI. If the numbers are in a2 – b2 form ( a2 – b2 = (a+b) (a-b)).
1. 37 × 43 = (40 -3) (40+3)
= 1600 -9.
= 1591.
2. 52 × 48 = (50 + 2) (50 -2)
= 2500 -4 = 2496.
### VII. Maths Tricks to Multiply any numbers with 5.
1. 38 × 5
(1) divide 38 by 2 and place 1 zero at the end.
= 38/2 = 190.
2. 46 × 5
Here divide 46 by 2 and place zero at the end.
= 46/2 = 230.
3. 47 × 5
(i) divide 47 by 2 and place 5 for odd numbers (47 is the odd number) at the end.
47/2 = 235.
4. 568 × 5
(1) divide 568 by 2 and place ‘0’ at the end as 568 is an even number.
568/2 = 2840.
5. 577 × 5.
577 / 2 = 2885.
### VIII. Maths Tricks to Multiply any numbers with 25.
1. 56 × 25 = 56/4 + 00 = 1400.
2. 66 × 25 = 66/4 + 00 = 1650.
3. 788 × 25 = 788 /4 +00 = 19700.
IX. Maths Tricks to Multiply any numbers with 125.
Divide the given number by 8 and add 000 at the end.
1. 56 × 125 = 56/8 + 000 = 7000.
2. 896 ×125 = 896/8 + 000 = 112000.
(OR) Check Below
### IX. Maths Tricks to Multiply any numbers with 5n.
Example : 974365 × 54
= 9743650000 / 24
1. 75 × 53
= 75000/23
= 9375.
2. 88 × 52
= 8800/22
= 2200.
### Some Other Useful Maths Formulae for Bank Exams
1. 2.3 + 3.4 + 4.5 + …….. + n = [n(n2 +6n + 11)]/3 .
2. 1/1.3 + 1/3.5 + 1/5.7 + ………. + 1/ (2n+1) (2n-1) = n/(2n+1).
3. 1 + 3+5+ ……….+(2n +1) = n2.
4. 2 +4 + 6 + ………….. + 2n = n(n+1).
5. 1 + 2 + 3 + ………..+ n = n(n+1)/2.
6. 12 + 22+32+ …. +n2 = n(n+1)(2n+1)/6.
7. 13 + 23+33+ …. +n3 = [n(n+1)/2]2.
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## Shortcuts for Multiplication of 2 Digit Numbers and 3 Digit Numbers
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Shortcuts for Multiplication of 2 Digit Numbers and 3 Digit Numbers In every competitive exam, …
## Comparing Fractions – Speed Maths Trick – IBPS Clerks 2016-2017
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If x = n!, is x divisible by 11?
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If x = n!, is x divisible by 11? [#permalink]
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29 Oct 2017, 00:01
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If $$x=n!$$, is x divisible by 11?
(1) n is not divisible by 11.
(2) n is a single digit number.
_________________
1) Absolute modulus : http://gmatclub.com/forum/absolute-modulus-a-better-understanding-210849.html#p1622372
2)Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html
3) effects of arithmetic operations : https://gmatclub.com/forum/effects-of-arithmetic-operations-on-fractions-269413.html
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Posts: 122
Re: If x = n!, is x divisible by 11? [#permalink]
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29 Oct 2017, 10:11
IMO B.
Statement 1: N can be 13 or 9. If 13, then 13! is divisible by 11 but 9! isnt. Insuff.
Statement 2: 11 is a prime number. So Numerator should have 11 or its multiple in it but it will never have 11 since maximum value of n is 9 and 9! is not divisible by 11.Suff.
Re: If x = n!, is x divisible by 11? &nbs [#permalink] 29 Oct 2017, 10:11
Display posts from previous: Sort by | 679 | 2,314 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.15625 | 4 | CC-MAIN-2018-47 | latest | en | 0.810974 |
http://puzzle.queryhome.com/7758/place-letter-word-that-attached-beginning-given-words-what | 1,503,114,120,000,000,000 | text/html | crawl-data/CC-MAIN-2017-34/segments/1502886105297.38/warc/CC-MAIN-20170819031734-20170819051734-00709.warc.gz | 337,132,845 | 31,610 | In the box place a five letter word that can be attached to the beginning of the given words.What is the word?
+1 vote
438 views
In the box provided place a five letter word that can be attached to the beginning of the given words to form six longer words.
What is the word?
posted Jun 2, 2015
2 Solutions
Blackout
Blacklist
Blackmail
Blackleg
Blackbird
blackball
solution Aug 26, 2015
ANS IS black
solution Jun 12, 2015
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If you do this correctly, the letters you add, when read downward, will spell out a word that represents what you are looking for when listening to directions.
In the box provided place a four letter word that can be attached to the end of the word to the left and to the beginning of the word to the right to form two longer words. What is the four letter word?
``````K I L L _ _ _ F U L L Y | 400 | 1,593 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.84375 | 3 | CC-MAIN-2017-34 | latest | en | 0.914802 |
http://pdesolutions.com/forum/phpBB3/viewtopic.php?f=2&t=664&start=0&st=0&sk=t&sd=a&view=print | 1,603,183,770,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107871231.19/warc/CC-MAIN-20201020080044-20201020110044-00091.warc.gz | 83,689,073 | 2,148 | Page 1 of 1
Posted: Tue Jan 28, 2020 2:47 am
How to input a specific radius of a circle?
here is syntax to generate a circle but I am not sure what radius it has.
START(0.5,0.27)
ARC(CENTER=0.5,0.18) ANGLE=360
### Re: radius of a circle
Posted: Tue Jan 28, 2020 5:26 pm
See ARC in the Help index. An ARC can be specified in several ways :
ARC TO (x1,y1) to (x2,y2)
ARC ( RADIUS = R ) to (x,y)
ARC ( CENTER = x1,y1 ) to (x2,y2)
ARC ( CENTER = x1,y1 ) ANGLE=angle
### Re: radius of a circle
Posted: Tue Jan 28, 2020 9:54 pm
Can I say that radius is 0.02 in this case?
START(0.5,0.22)
ARC(CENTER=0.5,0.2) ANGLE=360
### Re: radius of a circle
Posted: Wed Jan 29, 2020 10:49 am
Yes. That is geometry. | 270 | 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} | 2.953125 | 3 | CC-MAIN-2020-45 | latest | en | 0.779031 |
https://www.adisma2000.co.il/articles/whats-the-basic-metric-units-of-physics/ | 1,591,242,123,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347439019.86/warc/CC-MAIN-20200604032435-20200604062435-00532.warc.gz | 585,304,845 | 17,166 | What exactly will be the essential metric units of physics? The question was around the minds of most mathematics enthusiasts and scientists ago Many physicists have formulated their own answer for this issue.
In measuring the attributes of the assorted procedures in mathematics, such a step is applied. One is not confined to the principle of volume dimension.
But what really are the components automatic paraphrase tool of a step? In math, one may readily locate a set of steps which are regarded essential. Several of the measures are denoted by letters such as P, L, C, M, I, A, G, T, B, B, S, and also N.
We shall finally take a peek at some different fundamental components of measure at Physics. One would be the metre. It’s very important to note some scientific associations in the united states possess the metre as a portion.
Another measure that is fundamental could be your kilogram. The kilogram is understood /avoid-plagiarism/ to be a inert body of bulk. As the troy burden of platinum-iridium, the chemical, the kilogram is defined From the International System of Units. It is utilised in the SI to make reference to a quantity known as the global Prototype of Metric pounds.
The third step may be your level Celsius. It is used to assess an object’s temperatures.
The fourth fundamental measure is that your squaremeter. It is measured in yards. The square metre can be utilised to measure the dimensions of the surface of the earth.
The step could be your megahertz. It’s used to describe that the quantity of electricity that’s flowing through a power conductor.
The first step fundamental measure is that your kilowatt. It’s utilised to refer to the quantity of power that’s transferred by means of a household electric unit. The kilowatt is used in dimension to characterize.
The fundamental measure is your instant, i.e. enough time . The second can be used to describe the simple component of dimension called https://library.duke.edu/music the 2nd. It’s accompanied with the moment and the moment.
The fundamental step is that your instant, i.e. enough period . The 2nd can be used to describe the simple unit of measurement known as the 2nd.
The fundamental step is your moment, i.e. enough time . The 2nd can be used to refer to the component of dimension called the 2nd. | 494 | 2,312 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.0625 | 4 | CC-MAIN-2020-24 | latest | en | 0.939617 |
https://open.kattis.com/contests/c4rwva/problems/jumpingyoshi | 1,653,665,100,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662658761.95/warc/CC-MAIN-20220527142854-20220527172854-00703.warc.gz | 490,197,146 | 7,351 | TPC 20220111
#### Start
2022-01-11 01:00 AKST
## TPC 20220111
#### End
2022-01-11 03:00 AKST
The end is near!
Contest is over.
Not yet started.
Contest is starting in -136 days 5:25:00
2:00:00
0:00:00
# Problem FJumping Yoshi
Yoshi is a frog. He lives in the ZOO under a log which was specially transported there from a distant equatorial rainforest. The log is big and wet and attracts the flies and that is what makes Yoshi satisfied.
There is also a line of pebbles which runs through the wetland in front of the log. There are various dark spots on the pebbles and sometimes Yoshi likes to look at them and dream that they are not just spots but some really big flies instead.
Yesterday, Yoshi’s friend camel Addawser came to see him and suggested to play a game.
“Do you see those spots on the pebbles?” asked Addawser. “I challenge you to start on the leftmost pebble and then do some jumps from a pebble to another pebble but with some restrictions. You can jump from a pebble to another one only if the sum of numbers of spots on both pebbles is equal to the distance between the pebbles. And you have to get to as distant pebble as possible.”
“All right, but you know that I can count at most to twenty three and no more,” hesitated Yoshi.
“Then, it’s easy,” said Yoshi, positioning himself on the first pebble and looking inquisitively over the rest of the line. “Although, one might not be quite so sure, after all,” he added after a while.
You are given the sequence of numbers of dark spots on the pebbles in the line. You are asked to find the most distant pebble which can be reached by a sequence of jumps. The first jump starts at the first pebble in the line and a jump between two pebbles is possible if and only if the sum of numbers of spots on both pebbles is equal to the distance between them. You may suppose that the line of pebbles is straight and that the distance between each two neighboring pebbles is exactly one frog distance unit.
## Input
Each case starts with a line containing one integer $N$ ($1 \leq N \leq 1\, 000\, 000$) representing the number of pebbles. The second line contains a list of $N$ integers. The order of the integers in the list is the same as the order of the pebbles in the wetland, each integer represents the number of spots on the corresponding pebble. No pebble contains more than $10^9$ spots. Suppose that Addawser knows all different pairs of pebbles where Yoshi can perform a jump from one pebble to another one during his sequence of jumps. You are guaranteed that the number of those pairs of pebbles never exceeds $1\, 000\, 000$.
## Output
Print a single line with one integer denoting the distance of the pebble which can be reached by successive jumps according to the given rules and which is the most distant from the first pebble.
Sample Input 1 Sample Output 1
7
2 1 0 1 2 3 3
5
Sample Input 2 Sample Output 2
11
7 6 1 4 1 2 1 4 1 4 5
10 | 765 | 2,936 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5 | 4 | CC-MAIN-2022-21 | latest | en | 0.961961 |
http://www.physics.sfsu.edu/~thamlin/jackson/np101/np201.html | 1,513,629,318,000,000,000 | text/html | crawl-data/CC-MAIN-2017-51/segments/1512948623785.97/warc/CC-MAIN-20171218200208-20171218222208-00668.warc.gz | 427,575,322 | 2,100 | ### Assignment NP 2.01: A Matlab Example in Electrostatics
Graphics and description
```
I began my matlab experience by prerusing through the "Getting
Started with Matlab" handout using the computer. Next I followed
Professor Lockhart's writeup. The writeup guided me through a problem of
two point charges. - Find the potential and the electric field and then
graph the results in various ways. I then placed the commands into a file
called matlabex.m for which I could modify at the slightest whim and then
rerun the program to see the results of the changes. Through the
combination of using the help command (similar to the man command in
UNIX), and the trying out modifications in my program I learned my
interesting things.
Diverging for a moment, I will compare the Lockhart program results
using Matlab and the Dancing Gaussian from Mathematica. Both Mathematica
and Matlab have good and bad points. Which one that you use depends upon
what you are doing. Mathematica can easily do computation of functions
but Matlab is more user friendly in the area of adding finishing touches
to a graph and looking at it from different view points. Anyways, each
program requires a different technique towards the common goal of finding
the best resulution of the spike of a given function.
Returning the to the topic at hand, I decided upon the mesh graph
style to use for this next part. Here I increased the resolution of the
potential function by increasing the x and y arrays used in creating the
meshgrid from 0.1 to 0.014. This seemed to yield the best visual results.
To view the graph refer to graph1. I wanted a
better view and thus rotationed it from the default position of:
azmuthal angle = -37.5 and elevation = 30 to 30 and 0 (a side view). To
view the graph refer to graph2. Wow! Notice
that the first graph shows the potential plane with a spike at the
positive charge and a dot due to the negative charge. The second graph
(due to rotation) reveals not only a spike due to the positive change
but one can clearly see that there's a spike in the opposite direction
from the plane due to the negative charge. The color changes also add
to an easier interpretation of the results.
I also went through the above process for the gradient of the
potential. The graph is labeled the electric field but it's actually
the negative of the electric field to be more precise. If you are
interested the first graph, graph3, is the
graph using the default settings. Again, notice the spiking of the
electric field at the charge points. The last graph,
graph4, reveals not only a spike similar to
the potential but an added "bump" on each. Something that I noticed in
both the graph of the potential and the electric field, one spike is
"longer" than the other. Off hand that seems incorrect - a glich in the
program. If I have more time before this is due, I'll put some thought
into it. There must be more to it.
``` | 687 | 2,945 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.6875 | 4 | CC-MAIN-2017-51 | latest | en | 0.933831 |
http://gmatclub.com/forum/how-do-we-best-communicate-our-working-of-two-full-time-jobs-133405.html?fl=similar | 1,484,669,583,000,000,000 | text/html | crawl-data/CC-MAIN-2017-04/segments/1484560279923.28/warc/CC-MAIN-20170116095119-00030-ip-10-171-10-70.ec2.internal.warc.gz | 114,475,197 | 56,002 | How do we best communicate our working of two full-time jobs : The B-School Application
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# How do we best communicate our working of two full-time jobs
Author Message
TAGS:
### Hide Tags
Current Student
Joined: 30 Mar 2012
Posts: 141
Location: United States (TX)
Concentration: Strategy, Finance
GMAT 1: 670 Q45 V38
GPA: 2.64
WE: Supply Chain Management (Consumer Electronics)
Followers: 3
Kudos [?]: 11 [0], given: 14
How do we best communicate our working of two full-time jobs [#permalink]
### Show Tags
28 May 2012, 22:37
Hello everyone,
I have a period of 10 months where I worked two full-time jobs (40 hours at an Equities Trading Firm, 40-50 hours at Apple as an International Sales Specialist) a couple years ago. Would this constitute as 20 months of Work Experience or just 10. Also, is there a better way to position this to maximize this period of time when I was working at two places.
Thank you, all.
_________________
Don't let a low GPA destroy your dreams of a business education.
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Status: Current Student
Joined: 24 Aug 2010
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Re: How do we best communicate our working of two full-time jobs [#permalink]
### Show Tags
29 May 2012, 06:11
Working two jobs during the same period does not double the months of experience. You just did more in 10 months than others did. 'Tis all. Put it on your resume with the overlapping time periods and adcoms will see it. It'll be in your application as well. If it applies to an essay question mention it there too.
_________________
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Re: How do we best communicate our working of two full-time jobs [#permalink]
### Show Tags
29 May 2012, 08:59
1
KUDOS
I agree with this also. Think of the investment banker that works 100 hours a week for 1 year. Did he actually get 2.5 years of work experience? NO, he got 1 year of intense 100 hour a week work experience. Most everyone you will compete against will not be working the standard 40 hour work week you may be considering. But the fact that you worked two full time jobs at once shows dedication and ability. So it's a definite positive
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Current Student
Joined: 30 Mar 2012
Posts: 141
Location: United States (TX)
Concentration: Strategy, Finance
GMAT 1: 670 Q45 V38
GPA: 2.64
WE: Supply Chain Management (Consumer Electronics)
Followers: 3
Kudos [?]: 11 [0], given: 14
Re: How do we best communicate our working of two full-time jobs [#permalink]
### Show Tags
29 May 2012, 11:00
Hey Cheetarah and GMATLA! Gosh, thanks for coming through again.
I appreciate the advice. It certainly makes sense. With my profile, I've been told I have no "margin of error," so just working on my positioning early and finding out how to best maximize each of my strengths. Thanks again =)
_________________
Don't let a low GPA destroy your dreams of a business education.
Current Student
Joined: 12 Sep 2011
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Concentration: Finance, Finance
GMAT 1: 710 Q48 V40
Followers: 137
Kudos [?]: 899 [0], given: 114
Re: How do we best communicate our working of two full-time jobs [#permalink]
### Show Tags
29 May 2012, 16:09
Yes, that is very smart to start early. If you are truly a borderline applicant you may consider having a essay review company, admissions consultant, or other trusted source help point you in the right direction. A well crafted application along with strong interview, resume, essays, GMAT etc can make all the difference. Good luck!
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Re: How do we best communicate our working of two full-time jobs [#permalink] 29 May 2012, 16:09
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Display posts from previous: Sort by | 1,403 | 5,309 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.765625 | 3 | CC-MAIN-2017-04 | latest | en | 0.876152 |
https://www.studentlance.com/solution/xacc-280-week-7-checkpoint-ratio-vertical-and-horizontal-analyses | 1,611,457,169,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703544403.51/warc/CC-MAIN-20210124013637-20210124043637-00314.warc.gz | 1,005,746,916 | 6,348 | # XACC 280 week 7 CheckPoint Ratio, Vertical, and Horizontal Analyses - 1891
Solution Posted by
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CheckPoint: Ratio, Vertical, and Horizontal Analyses
The calculations you perform for this CheckPoint form the basis of your analysis of your capstone project.
· Due Date: Day 6 [post to the Assignments section of the Classroom]
· Write in 100 to 200 words an explanation of the three tools of financial statement analysis and the function of each.
· Examine PepsiCo, Inc.’s Consolidated Balance Sheet on p. A6 in Appendix A of Financial Accounting, especially its Current Assets, Current Liabilities, and Total Assets for years 2005 and 2004.
· Calculate the following for PepsiCo, Inc. and show your work:
o The Current Ratio for 2005
o The Current Ratio for 2004
o Two measures of vertical analysis—for example, compute the current assets divided by total assets for each year, and express your result as a percentage
o Two measures of horizontal analysis—for example, compute the total change in assets by percentage, by dividing current assets in 2005 by current assets in 2004. Compute a similar percentage for current liabilities
· Examine The Coca-Cola Company’s Consolidated Balance sheet on p. B2 in Appendix B of Financial Accounting, especially its Current Assets, Current Liabilities, and Total Assets for years 2005 and 2004.
· Calculate the following for Coca-Cola and show your work:
o The Current Ratio for 2005
o The Current Ratio for 2004
o Two measures of vertical analysis—for example, compute the current assets divided by total assets for each year, and express your result as a percentage
o Two measures of horizontal analysis—for example, compute the total change in assets by percentage, by dividing current assets in 2005 by current assets in 2004. Compute a similar percentage for current liabilities
· Post your explanation and calculations.
Solution Description
CheckPoint: Ratio, Vertical, and Horizontal Analyses
The calculations you perform for this CheckPoint form the basis of your analysis of your capstone project.
· Due Date: Day 6 [post to the Assignments section of the Classroom]
·
Attachments | 545 | 2,415 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2021-04 | latest | en | 0.845988 |
https://www.topperlearning.com/forums/home-work-help-19/find-the-differentiation-mathematics-continuity-and-differentiability-54390/reply | 1,503,284,071,000,000,000 | text/html | crawl-data/CC-MAIN-2017-34/segments/1502886107487.10/warc/CC-MAIN-20170821022354-20170821042354-00488.warc.gz | 969,859,100 | 39,408 | Question
Tue March 20, 2012
# find the differentiation
Wed March 21, 2012
Put costheta = (1- tan2theta/2 )/(1 + tan2theta/2)
So on solving we get sin-1 [ {(a+b) - (b-a)tan2theta/2}/{(a+b) + (b-a)tan2theta/2}]
which is pie/2 - 2tan-1[root(b-a)tantheta/2]
so we can easily find dy/dx = -root(b-a).sec2(theta/2) / ( 1 + (b-a)tan2theta/2 )
Related Questions
Sun August 13, 2017
# Examine the continuity of f(x)={sin2x/sin3x ,if x≠0 ; 2, if x=0
Sun August 13, 2017
Home Work Help | 203 | 483 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.328125 | 3 | CC-MAIN-2017-34 | latest | en | 0.742431 |
https://testbook.com/question-answer/generally-the-period-chosen-for-a-standard-b-o--609e20d0f44b4bf849a5e767 | 1,639,045,151,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964363791.16/warc/CC-MAIN-20211209091917-20211209121917-00058.warc.gz | 601,160,195 | 32,621 | # Generally, the period chosen for a standard B. O. D. test is
This question was previously asked in
MPSC AE CE Mains 2018 Official (Paper 2)
View all MPSC AE Papers >
1. 1 day
2. 5 days
3. 8 days
4. 20 days
Option 2 : 5 days
## Detailed Solution
Explanation:
Biochemical Oxygen Demand (BOD):
• The amount of oxygen required by bacteria to break down the organic matter present in a certain volume of a sample of water is called Biochemical Oxygen Demand (BOD).
• Standard B. O. D. test is conducted at 3 and 5 days. 5 day B. O. D. test is conducted at 20oC and 3 day B. O. D. test is conducted at 27oC.
• Clean water would have BOD value of less than 5 ppm whereas highly polluted water could have a BOD value of 17 ppm or more
Computation of 5-day BOD of the test sample:
(BOD)5 = (DOi – DOf) × Dilution factor
Dilution factor = R/X
Free
ST 1: Building Construction and Materials
1502
20 Questions 40 Marks 24 Mins | 279 | 926 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.765625 | 3 | CC-MAIN-2021-49 | latest | en | 0.88911 |
http://flhuntingdogs.org/17171_dried-base-conversion-formula-of-coal.html | 1,550,275,204,000,000,000 | text/html | crawl-data/CC-MAIN-2019-09/segments/1550247479627.17/warc/CC-MAIN-20190215224408-20190216010408-00037.warc.gz | 99,307,369 | 7,970 | # dried base conversion formula of coal
#### dried base conversion formula of coal
Coal Conversion FactsDry Basis (db): excludes all Moisture. Dry Ash Free (daf): excludes all Moisture & Ash. The Proximate Analysis of any coal i.e. the % content of Moisture, Ash (A),. Volatile Matter (VM), Fixed Carbon (FC) – also Sulphur (S) and Calorific. Value (CV) – can be expressed on any of the above bases. Coal Conversion. Facts.formulas for calculating the heating value of coal and coal - Argonne .conversion of coal to other useful forms of fuel, as well as in its direct use. The significance of the correlation of . A data base (including experimental heating values, ultimate analyses and some other parameters) was . In the above, Q is the gross heating value in Btu/lb on the dry basis and C , H,. S, (O), N, and A are the.
Coal Calculator - Coal Marketing InternationalCoal Calculator. Units. Total moisture: Inherent Moisture: Ash component: Calorific Value (Gross As Received):; Calorific Value (Air Dried Basis):; Calorific Value (Dry Basis):; Calorific Value (Dry Ash Free):; Hydrogen (As Received):; Hydrogen (Air Dried Basis):; Total moisture (As Received):; Volatile Matter (Air Dried Basis):coal characteristics - Purdue UniversityCCTR. Indiana Center for Coal Technology Research. 3. COAL ANALYSIS. Elemental analysis of coal gives empirical formulas such as: C. 137. H. 97. O. 9. NS for Bituminous Coal. C. 240. H. 90. O. 4. NS for high-grade Anthracite. Coal is divided into 4 ranks: (1) Anthracite. (2) Bituminous. (3) Sub-bituminous. (4) Lignite.
Coal Calculations | Energy | SGS
Moisture conversion to different bases - refer to ASTM D3180 / ISO 1170 . Dry is Dry Basis DAF is Dry Ash Free Basis; Two Stage Total Moisture Formula (Refer to ASTM D3302 section 10) Two stage total moisture determination is used when the coal . H(dry base) = [Total Hydrogen(ad)-(AMx0.1119)] x (100/(100-AM))
Coal Net calorific value calculation - Page 1 of 1 - International .
Coal Net calorific value calculation. Dear experts,. please send we the formula to calculate Coal calorific value As received from Coal calorifiac value Dry bases. besides, I need the explenation of tesmins in coal: Gross CV dry bases, Gross CV as resieved; Net CV dry bases, Net CV as received. what are.
Representation of coal and coal derivatives in process . - SAIMM
calculation of the calorific value (CV) of coal gas (the gas liberated when coal is .. 'absolutely dry'. Total moisture content of the coal sample is calculated as follows: %Total moisture = %Free moisture + %Inherent moisture. [1]. Some coals could . converted to absolute dry base, e.g. by applying the following conversion:.
Solid fuel analysis - info about a few analysis . - FORCE Technology
As regards coal, the content of the volatile matter ranges from below 10 % in anthracite coal to about 70 % on the dry and ash free basis in brown coal. . Formula 1. Calculation formula for the net calorific value, at M % moisture. Formula 2. Calculation formula for conversion of the net calorific value to other moisture contents.
Coal, Kentucky Geological Survey, University of Kentucky
Dry, mineral matter-free basis: Data or results are calculated to a theoretical base as if there were no moisture or mineral matter in the coal sample (ASTM method D388-12; American Society for Testing and Materials, 2013, p. 390–396). For this basis, a calculation is needed to determine mineral-matter content rather than.
Coal assay - Wikipedia
Analysis is fairly straightforward, with the coal thoroughly burnt and the ash material expressed as a percentage of the original weight. It can also give an indication about the quality of coal. Ash content may be determined as air dried basis and on oven dried basis. The main difference between the two is that the latter is.
Coal - FoodTechInfo
Fuel Cost Conversion Formulas: Coal costs used to be relatively stable, but recently have seen substantial increases and instability in prices similar to oil and natural gas. Steam cost . Coal analyses can be reported on several bases: as-received, moisture-free (or dry), and mineral-matter-free (or ash-free). As-received is.
Coal - Chemistry Encyclopedia - structure, water, uses, elements, gas
Figure 1. An example of the structure of coal. of dead vegetation began to accumulate, becoming tightly packed and compressed, and gave rise to different kinds of coal, . by percentage), classifying types of coal according to their percentages of fixed carbon, hydrogen, oxygen, and nitrogen, exclusive of dry ash and sulfur.
Innovative Price Adjustments Technique for Thermal Coal: AStudy of .
adjustment in price Coal supplier and buyer use some formula as per quality to adjust quantity. The aim of this paper is to amend existed formula, and to give innovative approach to implement a price adjustment formula of coal as per the quality bases. Keywords : moisture . coal is determined. c) Sulphur (As Dried Bases).
Basic Conversion calculator - Knowledge is Power
HOME >, coal energy calculator . given basis, desired basis. As Received. Air Dried. Dry. Dry, Ash Free. As Received. 100-Mar. 100-Mad. 100. 100-Mad. 100. 100-(Mad+Aar). Air Dried. 100-Mar. 100-Mad. 100. 100-Mad. 100 . To convert from GAR to NAR for bituminous coals subtract approximately 1.09 mj/kg 260 kcal/kg
dried base conversion formula of coal,coal facts, information, pictures | Encyclopedia articles about coal
Make research projects and school reports about coal easy with credible articles from our FREE, online encyclopedia and dictionary. . Instead, anaerobic (pronounced an-nuh-ROBE-ik) bacteria (bacteria that do not require oxygen to live) attack the plant debris and convert it to simpler forms: primarily pure carbon and.
dried base conversion formula of coal,How Much Coal Is Left - Energy Explained, Your Guide To . - EIA
Feb 21, 2018 . Stacked bar chart showing U.S. coal reserves by type and mining method. Click to enlarge . Six states had 77% of the demonstrated reserve base (DRB) of coal as of January 1, 2017: . Demonstrated Reserve Base (DRB) is the sum of coal in both measured and indicated resource categories of reliability.
AP-42, Vol. I, CH1.1 Bituminous and Subbituminous Coal . - EPA
dry bottom furnaces, coals with high fusion temperatures are burned, resulting in dry ash. In wet bottom .. residence time.7 Cyclone boilers typically have high conversion of nitrogen to NOx Typically, only 20 to. 60 percent of the fuel nitrogen . combustion because of the more complex structure of coal. 1.1.3.6 Trace Metals-.
PDF(115K) - Wiley Online Library
and air-dry loss. Air drying: process of partial drying of coal to bring its moisture near to equi- librium with ambient conditions in the laboratory or the room in . on which equipment can sit, move, or operate. A working road or base below a high wall, as in contour stripping for coal. Beneficiation: see Physical coal cleaning.
MOISTURE CONTENT BY THE OVEN-DRY METHOD FOR .
is oven-dry and all water has been removed. The weight of the water is the difference in the weight of the piece before and after drying. Therefore, when the test is done we use the following formula to calculate moisture content. Initial weight - Oven - dry weight. MC - x . (2). Oven - dry weight. Formulae one and two are.
FUELS & COMBUSTION CALCULATIONS
BASIC EQUATION. 1. C + O. 2. = CO. 2. kJ/kg carbon. 2. H. 2. + ½ O. 2. = H. 2. O. 3. S + O. 2. = SO. 2. 4. Adding oxygen requirements of above eqns and . kg/kg coal where A = 2.38 for S-capture;. = 0 for no S-capture. ▫ Actual dry air required. ▫ T da. = Excess air Coeff. X M da kg/kg. ▫ Actual wet air required. ▫ M wa.
Primer on Fuel Quality Analysis - POWER Magazine
Jan 1, 2015 . Unfortunately, fuel sampling is often considered just another messy chore, where someone takes a quick scoop of a coal pile with a bucket and .. Air-dried base results are sometimes impossible to convert to a “wet” or “dry” basis, because one requires the “as-received” moisture content in addition to the.
Cooling Power Plants | Power Plant Water Use for Cooling - World .
Cooling steam-cycle power plants has nothing essentially to do with whether it is fuelled by coal, gas or uranium. . air cooling, non-traditional water sources, recycling plant waste water streams and increasing thermal energy conversion efficiency far outweigh any differences between nuclear and coal water requirements.
Mass, Weight, Density or Specific Gravity of Bulk Materials - SI Metric
Feb 24, 2016 . Pure water was chosen as the 'base line' for specific gravity and given the value of 1. The specific . For example, ammonium nitrate has a specific gravity (sg) of 0.73 while dry ammonium sulphate has a sg of 1.13 (1130 kilograms/cubic metre) (see table below) . Coal, Anthracite, broken, 69, 1105. Coal.
Coalification Needs - BYU
4. Demonstrate how to convert proximate and ultimate coal analyses data to a dry, ash-free basis using the following coal data: Moisture. 6.17. Volatile Matter (moisture free) 37.87. Ash (mf). 9.90. H (mf)4.69. C (mf)71.12. N (mf)1.39. S (mf)3.80. O (mf)9.11. Heating Value (Btu/lb, dry, ash-free) 14,102. What rank of coal is this,.
Moisture Content dry and wet basis
Apr 14, 2014 . Definition of moisture content, dry and wet basis, and conversion formulas. | 2,316 | 9,319 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2019-09 | latest | en | 0.81135 |
http://93.18.58.59.thenewwestchelsea.com/solving-multistep-equations-with-variables-on-both-sides-worksheet/ | 1,606,766,659,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141486017.50/warc/CC-MAIN-20201130192020-20201130222020-00036.warc.gz | 1,763,652 | 18,858 | # Solving Multistep Equations With Variables On Both Sides Worksheet
## Solving Equations With Variables On Both Sides Worksheets
Solving Equations With Variables On Both Sides 1 This 12 Problem Worksheet Is Designed To Introduce You To Solving Equations That Have Variables On Both Sides Only Positive Whole Numbers Are Featured In The Equations And All Of The Answers Are Positive As Well A Few Of The Equations Are Two Step Equations But Most Are Three Step Equations Similar To 20 2
### Math Worksheets On Solving Equations With Variables On
Equations With Variables On Both Sides Worksheet Math Both Sides Free Worksheets For Linear Equations Grades 6 9 Pre Algebra Solving Multi Step Equations With Variables On Both Sides Youtube Math Worksheets Solving Equations With Variables On Both Sides Solving Equations With Variables On Both Sides Worksheet 8th Grade Solving A
#### Solving Linear Equations Variable On Both Sides
Elementary Algebra Skill Solving Linear Equations Variable On Both Sides Solve Each Equation 1 6 R 7 13 7r 2 13 4x 1 X 3 7x 3x 2 8x 8 4 8 X X 4x
##### Solve Multi Step Equations With Variables On Both Sides
Examples Solutions Videos Worksheets Stories And Songs To Help Grade 8 Students Learn How To Solve Multi Step Equations With Variables On Both Sides In Order To Solve An Equation We Need To Isolate The Variable On One Side Of The Equation In This Lesson We Will Learn How To Solve Multi Step Equations With Variables On Both Sides Of The
###### Multi Step Equations Free Math Worksheets
Because This Is A Bit More Complicated We Will Show You How To Solve Multi Step Equations On Two Examples The First Example Is A Fairly Simple Multi Step Equation With Integers That Has A Variable On Both Sides Of The Equation N 2 4 2n The First Thing We Have To Do Here Is To Get All The Variables On One Side And All The Numbers On The Other It Does Not Matter On Which Ones You Put On The Right Side Of The Equation And Which Ones On The Left You Will Still Get The Same Result
Multi Step Equations Worksheets Free Math Worksheets
12 5m 3 Now Add 3 To Both Sides 3 3 15 5m Divide Both Sides By 5 5 5 3 M These Multi Step Equations Can Get A Bit Long And Difficult Sometimes There Is An Unknown On Both Sides Of The Equation We Can T Solve Until It Is On Only One Side So We Must First Move One Of The Terms To The Other Side By Using An Inverse Operation
The Secret To Solving Equations With Variables On Both Sides
Steps For Solving The Equations With Variables On Both Sides Example Above Add Fifty Three To Both Sides Of The Equation So That You Get All Constants On One Side Together Subtract Five X From Both Sides Divide Both Sides Of The Equations By Negative Two To Get Your Solution Of Negative Thirty Three
8th Grade Math Worksheets Solving Equations
8th Grade Math Worksheets Solving Equations Type Keywords And Hit Enter 8th Grade Math Worksheets Solving Equations Collection Free Worksheets For Linear Equations Grades 6 9 Pre Algebra Free Worksheets For Linear Equations Grades 6 9 Pre Algebra Kuta Software Solving Multi Step Equations Free Printable Math Free Worksheets For Linear
Solving Equations With Variables On Both Sides Solutions
Equations With Variables On Both Sides This Requested Video Looks At Solving Equations With Variables On Both Sides It Includes Four Examples Examples 1 X 14 8 102 7x 2 5y 2 28 Y 3 3 5m 8m 9 4 4 3x 6 3x 2 X Show Step By Step Solutions
Algebra 1 Worksheets Equations Worksheets
Solving Single Variable Equations Worksheets These Algebra 1 Equations Worksheets Will Produce Single Variable Equations To Solve That Have Different Solution Types You May Select Three Different Types Of Problems Where There Is No Solutions One Solutions Or An Infinite Number Of Solutions These Worksheet Will Produce Twelve Problems Per Page These Equations Worksheets Are A Good Resource
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https://www.duplichecker.com/blog/gmat-math.php | 1,519,381,503,000,000,000 | text/html | crawl-data/CC-MAIN-2018-09/segments/1518891814566.44/warc/CC-MAIN-20180223094934-20180223114934-00309.warc.gz | 855,876,899 | 7,557 | # GMAT Math
GMAT is now compulsory in most educational institutes, if someone wants to apply in any postgraduate program. It used to be necessary for business schools, but now in other disciplines it is also required. GMAT consists of different sections that analyze one’s knowledge that was gained through all the academic years. All these sections consist of a number of questions with each question carrying weighing marks.
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https://codereview.stackexchange.com/questions/279051/palindrome-permutation-string | 1,685,921,260,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224650409.64/warc/CC-MAIN-20230604225057-20230605015057-00001.warc.gz | 217,042,743 | 47,071 | # Palindrome Permutation string
I’m starting to learn C++ by doing a lot of exercises, and I need your help so I can improve, Here is how it goes, I will try to solve an exercise and post both the exercise and my solution to it, And I want you to point me to where / what improvements I can do to make my code better, this would help me both uncover missing parts that I didn’t know about, and better ways of coding, alternative coding style or features introduced in C++11 C/++14/C++17/C++20 are welcomed to be mentioned, with some code snippet if possible, as I'm trying to write some modern C++ as possible while learning, it help in adopting as a habit.
Here it is
Given a string, write a function to check if it is a permutation of a palindrome. A palindrome is a word or phrase that is the same forwards and backwards. A permutation is a rearrangement of letters.
Here is my solution
#include <iostream>
#include <string>
#include <cassert>
#include <vector>
#include <unordered_map>
bool isPalinStr(const std::string& a) {
std::unordered_map<char, int> x;
if (a.length() % 2 == 0) {
for (int i = 0; i != a.length(); ++i) {
x[a[i]] += 1;
}
for (auto& ch : x) {
if (ch.second % 2 != 0)
return false;
else {
return true;
}
}
}
else {
for (int i = 0; i != a.length(); ++i) {
x[a[i]] += 1;
}
unsigned even = 0;
unsigned odd = 0;
for (auto& ch : x) {
if (ch.second % 2 == 0)
even++;
else {
odd++;
}
}
if (even % 2 == 0 && odd == 1) {
return true;
}
else {
return false;
}
}
}
int main() {
assert(true == isPalinStr("aabbccd"));
assert(false == isPalinStr("aabbcd"));
return 0;
}
Hey reader you might want to check other parts :
• Isn't a permutation of a palindrome of a word simply a permutation of the word? Aug 23, 2022 at 10:46
• Please don't use the x86 tag, when you just want performance. There is nothing in these types of exercises that is worth making architecture-specific optimizations for. Also, the entire Code Review site is about making code improvements, so you don't have to put "code improvements" in the title of every question. Aug 23, 2022 at 15:24
• If you're targeting C++20, and want the code reviewed with the latest language features in mind, then just pick the c++20 tag. Aug 23, 2022 at 15:28
#include <vector> is unnecessary for this program.
There are clearly insufficient tests, because there's a bug here:
for (auto& ch : x) {
if (ch.second % 2 != 0)
return false;
else {
return true;
}
}
This means that we only ever inspect the first value in the histogram, and ignore the later ones.
And we have undefined behaviour when given the empty string, because we don't reach any return statement before the end of the function. A decent compiler should spot that for you if you ask for its help; this suggests you need to enable more warnings when you compile.
We have a strange test here:
if (even % 2 == 0 && odd == 1) {
return true;
}
else {
return false;
}
It doesn't matter how many matched pairs of characters we have, so the count of how many even histogram counts we found is irrelevant. This will fail "aba" for instance, as we only have a single even count. We really just need to test that we have 0 (for even-length strings) or 1 (for odd-length) odd-size counts.
Note also that the general pattern if (condition) return true; else return false; can always be replaced by return condition;.
A good practice to follow when bugs such as these are identified is to add tests which reproduce the bug; when they are made to pass, the tests remain as part of the program's test suite to guard against reintroducing similar bugs. That way, we gradually increase the number of valuable tests as we progress.
This code is repeated in both branches of the initial if:
for (int i = 0; i != a.length(); ++i) {
x[a[i]] += 1;
}
Instead of writing it twice, we should bring it out before the if so it's only present once. The index variable i should be a std::size_t so that it can be compared against any length() (another compiler warning to enable there!).
We can simplify it (and improve the name of the histogram variable):
for (auto c: a) {
++hist[c];
}
Our histogram maps char to int, which could potentially overflow (undefined behaviour again). We could use an unsigned type (which has well-defined overflow behaviour), but consider that we're only interested in whether the value is even or odd - so we could use a map from char to bool to store our values modulo-2.
Also, an unordered map is a pretty slow form of histogram. For reasonable systems, with CHAR_BIT in the 8-16 range, we could use a simple array or a bitset (though we'll need to convert the string's characters to unsigned char to avoid indexing out of range):
std::bitset<UCHAR_MAX+1> hist;
for (unsigned char c: a) {
hist.flip(c);
}
When we have done this, we can simply use bitset::count() (or std::count() on an array) to tell us how many characters appear an odd number of times.
That reduces the code to a much simpler function:
#include <bitset>
#include <climits>
#include <string_view>
bool isPalinStr(const std::string_view s)
{
std::bitset<UCHAR_MAX+1> hist;
for (unsigned char c: s) {
hist.flip(c);
}
return hist.count() <= 1;
}
#include <cassert>
#include <iostream>
int main()
{
assert(isPalinStr(""));
assert(isPalinStr("a"));
assert(isPalinStr("aa"));
assert(isPalinStr("aabb"));
assert(isPalinStr("abbcc"));
assert(isPalinStr("aabccdd"));
assert(isPalinStr("aabbccd"));
assert(!isPalinStr("ab"));
assert(!isPalinStr("abc"));
assert(!isPalinStr("abbb"));
assert(!isPalinStr("aabbcd"));
}
• Good use of unsigned char vs. char. isPalinStr() here is poetic. Aug 22, 2022 at 18:30
• Of course an array of some unsigned type (byte up to register size) might be more efficient (code and time, not stack). Only measuring will show. Aug 22, 2022 at 20:02
• @Deduplicator, yes that may well be true, and I did start with a std::array<bool>. But the convenience of bitset::count() swung me! Aug 22, 2022 at 20:11
• Yes, std::ranges::count_if(hist, [](auto x){return x % 2; }); would be a bit more verbose. Aug 24, 2022 at 0:21
• I was imagining counting mod 2, which is a bit shorter (hist[c] ^= 1 in the loop, then std::ranges::accumulate() to count the ones); even then, the bitset is more convenient. Aug 26, 2022 at 11:30
You don't need to keep track of even at all, and you don't need to keep track of odd as an int. You just need to keep track of whether you've seen an odd number, and if you have, return false the next time you see another one.
You can replace x[a[i]] += 1; with x[a[i]] = !x[a[i]];. Then replace the code
for (auto& ch : x) {
if (ch.second % 2 == 0)
even++;
else {
odd++;
}
}
with
previous_odd = false
for (auto& ch : x) {
if ch.second {
if previous_odd {
return false;
}
previous_odd = true;
}
}
return true
• Thanks for adding this. I meant to point out that we could return early, but forgot that in the rest of my review. Aug 23, 2022 at 6:52
### Write out names
bool isPalinStr(const std::string& a) {
What does that Alaskan reality TV star (or the actor if searching from the UK) have to do with your code?
There is no need to put types in C++ function names, as C++ has function overloading.
bool canBePalindrome(const std::string& a) {
Requires no thinking to understand what is being checked. Even if readers aren't familiar with what a palindrome is, they can look it up. If someone tries to look up palin, they will have to wade through many irrelevant results. And that's if palindrome ever appears.
In general, it is best to optimize code to be easy to read. Because you read code far more often than you write it.
### Don't mix statement and block forms
if (ch.second % 2 == 0)
even++;
else {
odd++;
}
Please never do this. If you are going to use the statement form for single statement blocks, use it every time (except in an if/else with the other using the block form). If you are going to use the block form (and there is a good argument to always use the block form) for single statement blocks, then use it everywhere.
Mixing the block and statement forms in the same if/else construct is questionable even if one requires a block. Mixing them like this is the worst of all worlds. You lose the minimal gains of shorter code and make the code more complex than simply using the block form everywhere.
Note that a simpler version of this would be
std::size_t remainderCounts[2] = {0, 0};
for (auto ch : x) {
remainderCounts[ch.second % 2]++;
}
As noted elsewhere, you do not need to count either evens or odds. You only need to know if there were zero, one, or more than one odd values. If zero or one, then a palindrome. If more than one, not. But if you did need to check these things, then you could reduce your lines of code by using a better data structure.
• Michael Palin is English, not Alaskan. Unless you were thinking of someone else? Aug 24, 2022 at 5:03
• isPalindrome() is a misleading name; isAnagramOfPalindrome() is more accurate. There could be problems with using either in the global namespace with any code that includes the (deprecated) <ctype.h>. Aug 24, 2022 at 5:05
• @TobySpeight Conciseness is a virtue. isScrambledPalindrome() is more like it. Aug 26, 2022 at 8:53
• Yes @Deduplicator, that's much better. As they say, "the two hard problems in computing..." Aug 26, 2022 at 11:31
In C++ performance is improved by using iterators rather than indexing through arrays. Iterators point directly to the data rather than needing to be indexed.
I asked a very similar question several years ago.
bool IsPalindrome(const std::string& s) {
return std::equal(s.begin(), s.begin() + s.size() / 2, s.rbegin());
}
The unordered_map you are using adds too much overhead and shouldn't be necessary.
You need to become comfortable with the concepts of pointers even if you don't use them directly (iterators may use pointers, they definitely use concepts of pointers).
Pointers are a major concept in C and were carried over to C++.
• "caacbb" this won't pass your code ! but it still it is Palindrome Permutation string Aug 22, 2022 at 16:26
• I think you misunderstood the problem statement @pacmaninbw - the requirement is that the letters can be arranged into a palindrome, i.e. at most one odd value in the histogram. Aug 22, 2022 at 16:52
• @TobySpeight It is quite possible I misunderstood the problem statement. I still feel that the OP needs to learn about iterators, and other standard library features. Aug 22, 2022 at 17:53
• Oh yes, you're absolutely correct there. Aug 22, 2022 at 19:55
• I have never seen any significant performance improvements caused by using iterators rather than indexing for arrays. I see it as a matter of preference. Also related answer, stackoverflow.com/a/2524322. Aug 25, 2022 at 21:07
I would like to propose some general code style improvements without discussing the idea of the original solution.
bool isPalindrome(const std::string& word) // prefer using meaningful names
{ // consider replacing redundant line break with an opening curly brace
std::unordered_map<char, int> freqs; // freqs is for frequencies. We may use a better suitable short name, if there is one
for (char ch : word) // prefer range-for loop for iterating over the whole container for brevity
++freqs[ch]; // increment to make the code look idiomatic
// prefer omitting curly braces for statement bodies that fit in one line
if (word.length() % 2 == 0) {
for (const auto& p : freqs) // would prefer p for pair, since we use pair-specific syntax in the loop body
return p.second % 2 == 0;
// leave one empty line after return, break or continue to emphasize an early redirection of an execution flow
return true; // necessary, if the word is empty
} // else is redundant here because we return from the function early
unsigned even = 0;
unsigned odd = 0;
for (const auto& p : freqs) { // it is better to make a variable const, if it is known to be immutable
if (p.second % 2 == 0)
++even;
else
++odd;
}
return even % 2 == 0 && odd == 1;
}
Regarding the code structure, one should also avoid duplicate code.
I suggest you to develop and follow a reasonable code style to make your code look comprehensible. | 3,130 | 12,269 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.875 | 3 | CC-MAIN-2023-23 | latest | en | 0.814608 |
https://numbersworksheet.com/number-words-worksheets-for-grade-1/ | 1,628,198,695,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046157039.99/warc/CC-MAIN-20210805193327-20210805223327-00464.warc.gz | 427,383,488 | 9,744 | # Number Words Worksheets For Grade 1
What exactly is the meaning of math? Should you ask this of ten different people, you will likely get ten different answers. It can be hard to define math, nevertheless the fact of your matter is, it is important to understand how to use math to create sense around the globe around us. Coming from a very young age, youngsters are in the middle of math and it is around those who take care of them to foster their mathematical growth by offering them various opportunities. Luckily, there are numerous ways to get this done. You could orally count with the child or even your students, you could introduce numbers and provide youngsters using the names of these numbers and finally discuss the correspondence between your oral counting as well as the actual number. Based on a child’s development, you can find a number of methods to encourage math understanding and growth.
## Teaching With Number Words Worksheets For Grade 1
You will recognize that using our worksheets to instruct with is a good option. If you are teaching one or twenty five children, you may print from the worksheet that you feel will likely be best for the requirements. Then, you can show them the way to complete it while counting out loud. This gives a chance for your students to find out the best way to count from a given number and after that creating the numbers. This visualization of the procedure is very important for college kids. Even though some will catch on much quicker as opposed to others, you might find that you need to provide additional experiences for a number of the students you work with. Which means you may use the worksheets along with your whole class, a small group, or maybe one student who needs support or enrichment.
## Number Words Worksheets For Grade 1 Enhance Learning
You can also provide students by using a solid math foundation is simply by introducing these people to Number Words Worksheets For Grade 1. There are many benefits to this particular resource for both you and your students. One important thing that may be great on them is that they are really easy to use. We offer different options so you can actually target instruction. For example, if you are centering on counting within ten, you will find worksheets that get started with one and check out ten or those who start out with five and go to ten. Or, should your students can count the smaller numbers and require practice with counting in the hundreds or thousands, we have Number Words Worksheets For Grade 1 for your, too. The capability to start at a given number and continue counting from there is a very important skill. We supply you with a helpful tool which you can use according to your child’s or students’ abilities and needs.
Another advantage in our worksheets is that students enjoy doing them. With the easy to use format and various options, as a parent or teacher, it is possible to select the best starting point for that children you happen to be utilizing. By doing this, this assists them succeed within their counting experience. So when students are successful, they get the tasks they may be doing to become fun. After your child has mastered counting with the worksheets along with other experiences you provide them, you can continue counting larger numbers and providing them the opportunity to do it.
## Number Words Worksheets For Grade 1 Provides Opportunities
Like a teacher or even a parent, you are aware that children vary within their abilities. With this worksheets, you will find the one that targets the needs your youngsters have. If your student is tired of their learning and requires difficult, you can use worksheets to push these people to count higher than they may be currently doing. Or, in the event you child is struggling with grade-level expectations, you can get worksheets which provide the practice necessary to keep on the journey of learning to count.
You will discover our worksheets are helpful for a number of reasons. They give your students with all the learning and rehearse they must continue their math growth. Regardless of what level your students are, you will find the ideal counting worksheets to deliver practice or enrichment for college students of all the levels. Should you don’t find what exactly you need, e mail us and inform us the way we may help. We are always happy to assist in improving math learning. | 851 | 4,422 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2021-31 | longest | en | 0.953519 |
https://convert-dates.com/days-from/4273/2024/02/13 | 1,708,798,757,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474544.15/warc/CC-MAIN-20240224180245-20240224210245-00233.warc.gz | 187,191,623 | 4,331 | ## 4273 Days From February 13, 2024
Want to figure out the date that is exactly four thousand two hundred seventy three days from Feb 13, 2024 without counting?
Your starting date is February 13, 2024 so that means that 4273 days later would be October 26, 2035.
You can check this by using the date difference calculator to measure the number of days from Feb 13, 2024 to Oct 26, 2035.
October 2035
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October 26, 2035 is a Friday. It is the 299th day of the year, and in the 43rd week of the year (assuming each week starts on a Sunday), or the 4th quarter of the year. There are 31 days in this month. 2035 is not a leap year, so there are 365 days in this year. The short form for this date used in the United States is 10/26/2035, and almost everywhere else in the world it's 26/10/2035.
### What if you only counted weekdays?
In some cases, you might want to skip weekends and count only the weekdays. This could be useful if you know you have a deadline based on a certain number of business days. If you are trying to see what day falls on the exact date difference of 4273 weekdays from Feb 13, 2024, you can count up each day skipping Saturdays and Sundays.
Start your calculation with Feb 13, 2024, which falls on a Tuesday. Counting forward, the next day would be a Wednesday.
To get exactly four thousand two hundred seventy three weekdays from Feb 13, 2024, you actually need to count 5981 total days (including weekend days). That means that 4273 weekdays from Feb 13, 2024 would be June 29, 2040.
If you're counting business days, don't forget to adjust this date for any holidays.
June 2040
• Sunday
• Monday
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June 29, 2040 is a Friday. It is the 181st day of the year, and in the 181st week of the year (assuming each week starts on a Sunday), or the 2nd quarter of the year. There are 30 days in this month. 2040 is a leap year, so there are 366 days in this year. The short form for this date used in the United States is 06/29/2040, and almost everywhere else in the world it's 29/06/2040.
### Enter the number of days and the exact date
Type in the number of days and the exact date to calculate from. If you want to find a previous date, you can enter a negative number to figure out the number of days before the specified date. | 935 | 2,734 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.65625 | 4 | CC-MAIN-2024-10 | latest | en | 0.933787 |
https://www.physicsforums.com/threads/coefficient-of-friction-questions-i-cant-do.209431/ | 1,544,609,573,000,000,000 | text/html | crawl-data/CC-MAIN-2018-51/segments/1544376823817.62/warc/CC-MAIN-20181212091014-20181212112514-00149.warc.gz | 985,645,945 | 15,741 | # Homework Help: Coefficient of Friction Questions I cant do
1. Jan 17, 2008
### smally
Hi, this is my first post and i'm a bit rusty will mechanics so I need help with two questions i've come across
Firstly on the question sheet I am given:
a = F/m and w = mg and the static friction and kinetic friction equations
also v$$^{2}$$ = v$$^{2}_{0}$$ + 2a$$\Delta$$x
How do I do these:
1. The coefficient of friction between the bed of a truck and a box is 0.30.
The truck is moving at 80km/h horizontally.
What is the least distance in which the truck can stop if the box is not to slide?
2. A car travelling at 30m/s on a horizontal road brakes suddenly to avoid hitting a pedestrian. What is the stopping distance if
a) The car has ABS brakes, wheels do not slip ($$\mu$$s=0.5)
b) The car has no ABS, wheels lock ($$\mu$$k=0.3)
Last edited: Jan 17, 2008
2. Jan 17, 2008
### Tom Mattson
Staff Emeritus
Try to answer these questions, in order.
1.) What is the maximum static frictional force that the truck bed can exert on the box before the box slides? Since you weren't given the mass of the box, you'll have to call it something (say m). As you will see, the value of m won't matter. m will cancel out eventually.
2.) Using Newton's second law (a=F/m, which you cited) and your answer to 1.), what is the maximum acceleration that the box can withstand before sliding?
3.) Given the numbers in the problem statement, and your answer to 2.), what is the minimum distance in which the truck can stop without the box sliding.
We'll tackle the second question once you've got this one.
3. Jan 18, 2008
### smally
Sorry I'm confused how you get to the point of cancelling m out
I always get to this point:
$$f_{s,max} = \mu_{s}F_{n}$$
$$f_{s,max} = \mu_{s} w$$
$$f_{s,max} = \mu_{s} mg$$
$$f_{s,max} = \mu_{s} \frac{F}{a}g$$
$$f_{s,max} = 0.3 \times 9.8\frac{F}{a}$$
Since I don't have F or a; I get stuck. What am I doing wrong?
4. Jan 18, 2008
### catkin
You are looking for $a$, so not having it is OK.
You do have F! You have it twice.
Up to $f_{s,max} = \mu_{s} m g$ you are calculating a theoretical maximum force, $f_{s,max}$. When you move to the next equation, making the substitution $m = \frac{F}{a}$ you are still dealing with this theoretical maximum force but now you are also dealing with the deceleration, $a$, that this theoretical maximum force can produce.
The substitution assumes the acceleration $a$ and calculates the force that must be producing it. $F$ and $f_{s,max}$ are the same force. Substituting $F = f_{s,max}$ will lead to the solution.
With this insight, can you do it more elegantly?
5. Jan 19, 2008
### smally
Oh ok :D, I've managed to now get this:
$$ma = \mu_{s}mg$$
$$a = \mu_{s}g$$
$$a = 0.3 \times 9.8$$
$$a = 2.94$$
therefore:
$$\Delta x = \frac{v^{2} - v^{2}_{0}}{2a}$$
$$\Delta x = \frac{0^{2} - 80^{2}}{2 \times 2.94}$$
$$\Delta x = 1088.44$$
however the answer I am given is 83.7m, what have I doing wrong this time?
6. Jan 19, 2008
### catkin
You have mixed your units, using both hours and seconds for time.
7. Jan 19, 2008
### catkin
And meters and kilometers for distance.
8. Jan 21, 2008
### smally
Silly me, I overlooked that, thank you.
I think I've now done question 2 correctly now.
2a)
$$ma = \mu_{s}g$$
$$a = 0.5 \times 9.8$$
$$a = 4.9$$
so
$$\Delta x = 91.84$$
2b)
$$ma = \mu_{k}g$$
$$a = 0.3 \times 9.8g$$
$$a = 2.94$$
so
$$\Delta x = 153.06$$
however the answer given for 2b is slightly different (152.9). Have I missed something small here?
9. Jan 21, 2008
### catkin
The answer is 152.9 if g is 9.81.
Some would say the answer should be 200 m because the problem data is given to one signifcant figure and the answer should not exhibit greater accuracy. Makes sense; spurious accuracy is, er, spurious.
If you want to give answers correct to 5 significant figures (as you did) it's safest to state that you are assuming the problem data is accurate enough to justify it: "assuming the coefficient of friction give as 0.3 is actually 0.30000 ..."
10. Jan 21, 2008
### smally
Thats wonderful, I've finally done it, Thank You all very much.
11. Jan 22, 2008
### smally
Ok I've now come across a new question where my answers are different to the answers given:
11. Two blocks attached by a string slide down a 20o slope. The lower
block has mass m1=0.25kg and coefficient of kinetic friction 0.2.
The upper block has mass m2=0.8kg and coefficient of kinetic friction 0.3.
Find
a) the acceleration of the blocks and the
b) the tension T in the string.
I manage to get this:
$$a = \mu_{k}g cos\theta$$
lowerblock:
$$a = 0.2 \times 9.81 \times cos20$$
$$a = 1.8437$$
upperblock:
$$a = 0.3 \times 9.81 \times cos20$$
$$a = 2.7655$$
however the answer provided is a = -0.809 m/s2, how do you go about that.
Also what ways do I use to find out the Tension in the string? | 1,530 | 4,895 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.828125 | 4 | CC-MAIN-2018-51 | latest | en | 0.909287 |
https://www.java-forums.org/new-java/26476-converting-algebraic-equation-java.html | 1,487,845,114,000,000,000 | text/html | crawl-data/CC-MAIN-2017-09/segments/1487501171163.39/warc/CC-MAIN-20170219104611-00400-ip-10-171-10-108.ec2.internal.warc.gz | 836,599,580 | 13,118 | # Thread: converting an algebraic equation to Java
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## converting an algebraic equation to Java
How do I go about converting the Pythagorean Theorem into Java?
is it a little like this? I'm just lost on how to get put square root into that formula. What is the proper way to do it?
z = (x * x) + (y * y)
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Check out this class
Math (Java 2 Platform SE v1.4.2)
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Ahhh. That helps a bit but how do I put it all together.
I am trying to create a class to solve for all parts of Pythagoreans Theorem. I am starting with solving for the sides and then moving on to solving for the angles. I'm just going to show you what I have so far. I know its not right but here.
Java Code:
```public class Pythagoras
{
private int x, y, z;
public Pythagoras(int x, int y, int z)
{
x = (int) 1;
y = (int) 1;
z = (int) 1;
}
public void calculate()
{
z = ((x * x) + (y * y));
x = ((z * z) - (y * y));
y = ((z * z) - (x * x));
}
public int getSideX()
{
return x;
}
public int getSideY()
{
return y;
}
public int getSideZ()
{
return z;
}
}```
Where should the following go in the above section? Does it go in there or should it be in the Test Class?
Java Code:
`public static double sqrt(double a)`
I am a newbie and I'm trying to teach myself. I joined the forum because I figured that all you guys know a lot of Java and I could learn from all the problems that everyone else comes across and maybe even get to the point where I could help other people.
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• | 501 | 1,745 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2017-09 | latest | en | 0.895862 |
https://petrowiki.spe.org/ESP_design | 1,670,635,852,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446711637.64/warc/CC-MAIN-20221210005738-20221210035738-00158.warc.gz | 482,861,133 | 20,808 | Content of PetroWiki is intended for personal use only and to supplement, not replace, engineering judgment. SPE disclaims any and all liability for your use of such content. More information
ESP design
This page walks through the suggested 9-step process for selecting and sizing an electrical submersible pump system for artificial lift. The process is manual for illustrative purposes. A number of computer programs are available to automate this process.
Design example 1
Step one: basic data
Well data. K55 casing from surface to 5,600 ft: 7 in. and 26 lbm/ft; K55 liner from 5,530 to 6,930 ft: 5 in. and 15 lbm/ft; J55 EUE API tubing: 2 7/8 in. and 6.5 lbm/ft; perforations and true vertical depth (TVD): 6,750 to 6,850 ft; and pump setting TVD (just above liner top): 5,500 ft.
Production data. Tubing pressure: 100 psi; casing pressure: 100 psi; present production rate: 850 BFPD; pump-intake pressure: 2,600 psi; static bottomhole pressure: 3,200 psi; datum point: 6,800 ft; bottomhole temperature: 160°F; minimum desired production rate: 2,300 BFPD; GOR: 300 scf/STB; and water cut: 75%.
Well fluid conditions. Specific gravity of water: 1.085; oil °API or SG: 32; SG of gas: 0.7; bubblepoint pressure of gas: 1,500 psi; viscosity of oil: N/A; PVT data: none.
Power sources. Available primary voltage: 12,470 V; frequency: 60 Hz; power source capabilities: N/A.
Possible problems. There were no reported problems.
Step two: production capacity
Determine the well productivity at the test pressure and production rate. In this case, the maximum production rate is desired without resulting in severe gas-interference problems. The pump-intake pressure at the desired production rate can be calculated from the present production conditions.
Because the well flowing pressure (2,600 psi) is greater than bubblepoint pressure (1,500 psi), the constant-productivity index (PI) method will most probably give satisfactory results. First, one can determine the PI using the test data:
....................(1)
and
....................(2)
Next, we can determine the new well flowing pressure (Pwf) at the estimated production rate (Qd).
....................(3)
and
....................(4)
The well flowing pressure of 1,580 psi is still above the bubblepoint pressure of 1,500 psi; therefore, the PI approach should give good results. The pump-intake pressure can be determined by correcting the flowing bottomhole pressure for the difference in the pump setting depth and datum point, and by considering the friction-loss datum point and friction loss in the casing annulus. In the given example, as the pump is set 1,300 ft above the perforations, the friction loss, because of flow of fluid through the annulus from perforations to pump setting depth, is small, as compared to the flowing pressure, and can be neglected.
Because there is both water and oil in the produced fluids, it is necessary to calculate a composite SG of the produced fluids. To find the composite SG, water cut is 75%; therefore,
....................(5)
Oil is 25%; therefore,
....................(6)
The composite SG is the sum of the weighted percentages:
....................(7)
The pressure, because of the difference in perforation depth and pump setting depth (6,800 to 5,500 ft = 1,300 ft), can be determined as:
....................(8)
and
....................(9)
Therefore, the pump intake pressure is
....................(10)
Step three: gas calculations
In this third step, one must determine the total fluid mixture, inclusive of water, oil, and free gas that is ingested by the pump. Use actual pressure volume temperature (PVT) data if available. For this example, Standing’s correlation was used. [1]
Determine the solution GOR (Rs) at the pump-intake pressure by substituting the pump-intake pressure for the bubblepoint pressure (Pb) in Standing’s equation. This relationship can also be found as a monograph in many textbooks.
....................(11)
and
....................(12)
Determine the formation volume factor (Bo) with Rs and the following Standing’s equation (can also be found as a monograph).
....................(13)
where
....................(14)
Therefore,
....................(15)
and
....................(16)
Determine the gas volume factor (Bg) as
....................(17)
By assuming 0.85 Z factor (use actual PVT data if available),
....................(18)
Next, determine the total volume of fluids and the percentage of free gas released at the pump intake. Using the producing GOR and oil volume, determine the total volume of gas (Vg).
....................(19)
Using the solution GOR (Rs) at the pump intake, determine the solution gas volume (VSG).
....................(20)
The difference represents the volume of free gas (VFG) released from solution by the decrease in pressure from bubblepoint pressure of 1,500 psi to the pump-intake pressure of 1,000 psi.
....................(21)
The volume of oil (Vo) at the pump intake is
....................(22)
The volume of free gas at the pump intake (VIG) in barrels is
....................(23)
Next, is the equation for the volume of water (Vw) at the pump intake.
....................(24)
The total volume (Vt) of oil, water, and gas at the pump intake can now be determined by
....................(25)
The ratio or percentage of free gas present at the pump intake to the total volume of fluid is
....................(26)
As this value is less than 10% by volume, it has only a minor effect on the pump performance, especially if most of the free gas is vented up the annulus. Use of a gas separation component is not essential in this case.
The composite specific gravity (SG), including gas, is determined by first calculating the total mass of produced fluid (TMPF) from the original data given.
....................(27)
and
....................(28)
Now that the total volume of fluid entering the first pump stage is known (2,550 BFPD) and the composite SG has been determined, we can continue to the next step of designing the ESP system.
Step four: total dynamic head (TDH)
Sufficient data are now available to determine the TDH required by the pump.
....................(29)
and
....................(30)
The TDH required is based on the normal pumping conditions for the well application. If the well is killed with a heavier-gravity fluid, a higher head is required to pump the fluid out, until the well is stabilized on its normal production. More HP is also required to lift the heavier kill fluid and should be considered when selecting the motor rating for the application. Ft = tubing friction loss. Refer to Fig. 1[2].
Friction loss per 1,000 ft of 2 7/8-in. tubing (new) is 49 ft/1,000 ft of depth at 2,440 B/D (405 m3/d) or 4.5 m/100 m. Using the desired pump setting depth,
....................(31)
....................(32)
and
....................(33)
Step five: pump-type selection
From the manufacturer’s catalog information, select the pump type with the highest efficiency at the calculated capacity 2,440 B/D (405 m3/d) that will fit the casing. Select the 513 series GC-2200 pump (Fig. 2).
The head in feet (meters) for one stage is 2,550 B/D (405 m3/d) and is 41.8 ft (13 m). The BHP per stage is 1.16. To determine the total number of stages required, divide the TDH by the head/stage taken from the curve. The number of stages = TDH/(head/stage). The number of stages = (3,556 /41.8) = 85 stages.
Refer to the manufacturer’s information for the GC-2200 pump. The housing no. 9 can house a maximum of 84 stages, 93 stages for a housing no. 10. Because the 84-stage pump is only one stage less than the calculated requirement, it should be adequate and the pump will cost less. Once the maximum number of pump stages is decided, calculate the total BHP required as
....................(34)
and
....................(35)
Step six: optimum size of components
Gas separator. If a gas separator was required, refer to a catalog to select the appropriate separator and determine its HP requirement. In this example, one was not needed. If gas interference causes operating problems, a gas separator can be added on the next ESP repair.
Seal section. Normally, the seal section series is the same as that of the pump, although there are exceptions and special adapters available to connect the units together. Here, the 513 series GSB seal section is selected.
The HP requirement for the seal depends on the TDH produced by the pump. The manufacturer’s information shows a requirement of 3.0 hp for the 513 series seal operating against a TDH of 3,556 ft. Therefore, the total HP requirement for this example is 91.5 hp for the pump, plus 3.0 hp for the seal, or 94.5 hp total.
Motor. Generally, a 500 series motor should be used with the 513 series pump. When a motor is selected, consideration should be given to choose as large a diameter unit as possible for the casing to optimize the initial cost, motor efficiency, operating costs, and repair costs. In this example select the 100-hp 562 series motor from the catalog. The motor voltage can be selected on the basis of considerations discussed next.
The high-voltage, consequently low-current, motors have lower cable losses and require smaller conductor-size cables. High-voltage motors have superior starting characteristics—a feature that can be extremely important if excessive voltage losses are expected during starting. Although, the higher the motor voltage, the more expensive is the motor.
In some cases, the savings, because of smaller cable, may be offset by the difference in motor-controller cost, and it may be necessary to make an economic analysis for the various voltage motors. However, for this example, the high-voltage motor (100 hp; 2,145 V; 27 amps) is an excellent choice. Check the manufacturers catalog and equipment information to assure that all operating parameters are well within their recommended ranges (e.g., thrust bearing, shaft HP, housing burst pressure, and fluid velocity).
Step seven: electric cable
Determine cable size. The cable size is selected on the basis of its current-carrying capability. Using the motor amps (27) and the cable voltage-drop chart in the catalog, select a cable size with a voltage drop of less than 30 V/1,000 ft. All conductor sizes 1 through 6 fall in this category. The no. 6 cable has a voltage drop of 18.5 × 1.201 = 22.2 V/1,000 ft (305 m), and based on \$0.06/kW-hr. results in a monthly I 2 R loss of \$255. A no. 4 cable has 14.1 V/1,000 ft and costs \$158/month. The operating cost savings of \$97/month is divided into the added cost of the no. 4 over the no. 6 cable to calculate a payout. A no. 6 cable size was selected for this example.
Cable type. Because of the gassy conditions and the bottomhole temperature, the polypropylene ("poly") cable should be used. Check to be sure the cable diameter plus tubing collar diameter is smaller than the casing inside diameter (ID).
Cable length. The pump setting depth is 5,500 ft (1676.4 m), with 100 ft (30.5 m) of cable for surface connections; the total cable length should be 5,600 ft (1707 m). Check to verify that the cable length is within the manufacturer’s recommended maximum length,
Cable venting. A cable vent box must be installed between the wellhead and the motor controller to prevent gas migration to the controller.
Step eight: accessory and miscellaneous equipment
Flat cable-motor lead extension. As described in ESP system selection and performance calculations, calculate the length for the MLE. Pump length = 14.8 ft (4.51 m); seal length = 6.3 ft (1.92 m); plus, 6 ft = 6.0 ft (1.83 m) = 213.1 ft (8.26 m); select 35 ft (10.7 m) of 562 series flat cable.
Flat guards. Cable guards are available in 6-ft sections; therefore, six sections are sufficient.
Cable bands. The pump and seal section is approximately 20 ft (6 m) long. Twenty-two-inch (56 cm) bands are required to clamp to the housing with bands spaced at 2-ft (61 cm) intervals (10 bands). On the production-tubing string above the pump, the same length cable bands can be used. The bands should be spaced at 15-ft (4.5-m) intervals. The setting depth of 5,500 ft requires 367 bands.
Downhole accessory equipment. Refer to the manufacturer’s catalog for the accessories listed next.
Swaged nipple. The pump outlet is 2 7/8 in., per the manufacturer’s information, so a swaged nipple is not required for the 2 7/8-in. tubing.
Check valve. The 2 7/8-in.-EUE, 8-round, thread check valve is recommended.
Drain valve. The 2 7/8-in.-EUE, 8-round, thread drain valve should be used (in conjunction with the check valve) to eliminate pulling a wet string.
Motor controller. The motor-controller selection is based on its voltage, amperage, and KVA rating. Therefore, before selecting the controller, one must first determine the motor controller voltage. Assume the controller voltage is the same as the surface voltage going downhole. The surface voltage (SV) is the sum of the motor voltage and the total voltage loss in the cable. (Adjust taps on the transformer to closely achieve this value.)
....................(36)
The motor amperage is 27 amps; the KVA can now be calculated.
....................(37)
and
....................(38)
The 6H-CG motor controller suits these requirements.
Transformer. The transformer selection is based on the available primary power supply (12,470 V), the secondary voltage requirement (2,269 V) and the KVA requirement (106 KVA). Choose three 313.5 KVA single-phase transformers as shown in the manufacturer’s catalog.
Surface cable. Select 50 ft (15.2 m) of no. 1 cable for surface connection to transformers.
Design example 2
Step one: variable-speed pumping system
Use the previous example, and design a new system using a VSC. To help justify the use of a VSC, two new conditions were added. First, assume that we need to maintain a constant oil production (575 BOPD), although, reservoir data indicate we should see an increase in water cut (75 to 80%) over the next few months. Next, to satisfy our economic justification in using the VSC, we must optimize the initial cost and size of the downhole assembly.
To maintain oil production as the water cut increases, we must determine the maximum desired flow rate with 80% water.
....................(39)
and
....................(40)
Step two: production capacity
We can now calculate the pump intake pressure at the maximum rate of 2,875 B/D. First, make the assumption that even though the water cut changes, the well’s PI will remain constant. Now, determine the new well flowing pressure (Pwf) at the maximum desired production rate (Qd).
....................(41)
and
....................(42)
The new well flowing pressure of 1,175 psi is slightly below the bubblepoint pressure of 1,500 psi; therefore, the PI approach should still give good results.
The pump-intake pressure can be determined the same as before, although, a new composite specific gravity must be calculated.
....................(43)
....................(44)
The composite SG is the sum of the weighted percentages:
....................(45)
The pressure because of the difference in perforation depth and pump setting depth (6,800 + 5,500 ft = 12,300 ft) can be determined as
....................(46)
and
....................(47)
Therefore, the pump-intake pressure (PIP) can now be determined as
....................(48)
Step three: gas calculations
Next, determine the total fluid mixture that will be ingested by the pump at the new maximum desired flow rate (2,875 B/D). Determine the solution GOR (Rs) at the pump-intake pressure or by substituting the pump-intake pressure for the bubblepoint pressure (Pb) in Standing’s equation. [1]
....................(49)
and
....................(50)
Determine the formation volume factor (Bo) with the Rs from Standing’s monograph or use Standing’s equation[1]
....................(51)
where
....................(52)
and
....................(53)
Therefore,
....................(54)
Determine the gas volume factor (Bg) as
....................(55)
Assuming a 0.85 Z factor,
....................(56)
Next, determine the total volume of fluids, and the percentage of free gas released at the pump intake. Using the producing GOR and oil volume, determine the total volume of gas (TG).
....................(57)
or
....................(58)
Using the solution GOR (Rs) at the pump intake, determine the solution gas volume (VSG).
....................(59)
The difference represents the volume of free gas (VFG) released from solution by the decrease in pressure from the bubblepoint pressure of 1,500 psi to the pump intake pressure of 1,000 psi.
....................(60)
The volume of oil (Vo) at the pump intake is
....................(61)
and
....................(62)
The volume of free gas at the pump intake is
....................(63)
and
....................(64)
The volume of water (Vw) at the pump intake is
....................(65)
and
....................(66)
The total volume (Vt) or oil, water, and gas at the pump intake can now be determined
....................(67)
....................(68)
and
....................(69)
The ratio or percentage of free gas present at the pump intake to the total volume of fluid is
....................(70)
and
....................(71)
As this value is greater than 10% by volume, there is significant free gas to affect pump performance; therefore, it is recommended that a gas separator be installed. Next, we must assume the gas separator’s efficiency. At 15% free gas, a 90% efficiency of separation is used on the basis of the manufacturer’s gas-separator performance information.
The percentage of gas not separated is 10%.
....................(72)
and
....................(73)
Total volume of fluid mixture ingested into the pump is
....................(74)
....................(75)
....................(76)
and
....................(77)
The amount of free gas entering the first pump stage as a percent of the total fluid mixture is
....................(78)
and
....................(79)
As the free gas represents only 2% by volume of fluid being pumped, it has little significant effect on the well fluid composite SG and may be ignored for conservative motor sizing.
Now that the total volume of fluid entering the first pump stage is known (2,973 BFPD) and the composite SG has been determined, we can continue to the next step of designing the ESP system.
Sufficient data are now available to determine the TDH required at the maximum desired flow rate (2,973 B/D). The TDH for the minimum desired flow rate (2,550 B/D) was previously determined to be 3,556 ft.
....................(80)
where HL = the vertical distance in feet between the estimated producing fluid level and the surface, and
....................(81)
From Fig. 1 , friction loss per 1,000 ft of 2 7/8-in. tubing (new) is 60 ft/1,000 ft of depth at 2,973 B/D (405 m3 /d), or 4.5 m/100 m. Using the desired pump setting depth,
....................(82)
Hwh = the discharge pressure head (desired wellhead pressure). Using the composite SG,
....................(83)
and
....................(84)
or
....................(85)
Step five: pump-type selection
The hydraulic requirements for our variable speed pumping system have been determined. Those requirements are the minimum hydraulic requirement (flow rate 2,550 B/D; total dynamic head 3,556 ft) and maximum hydraulic requirement (flow rate 2,973 B/D; total dynamic head 4,746 ft).
In the economic justification for using the VSC, the size of the downhole unit was determined. This was done using the guidelines discussed next.
As the operating frequency increases, the number of stages required to generate the required lift decreases. The closer the operation is to the best efficiency point, the lower the power requirement and power cost.
A fixed frequency motor of a particular frame size has a maximum output torque, provided that the specified voltage is supplied to its terminals. The same torque can be achieved at other speeds by varying the voltage in proportion to the frequency. This way the magnetizing current and flux density will remain constant, and so the available torque will be a constant (at no-slip RPM). As a result, power output rating is directly proportional to speed because power rating is obtained by multiplying the rated torque with speed. Using the variable-speed performance curves, select a pump that will fit in the casing so the maximum flow rate (2,973 B/D) falls at its BEP. The GC-2200 satisfies these conditions at 81 Hz.
Next, select the head per stage from the curve. It indicates 86 ft/stage. With the maximum TDH requirement of 4,746 ft, the number of pump stages required can be determined. The number of stages = the maximum TDH /head per stage and = 4,746 /86 = 55 stages. A 55-stage GC-2200 meets our maximum hydraulic requirement. To determine if it meets our minimum hydraulic requirement, divide the minimum TDH requirement by the number of stages. The minimum head per stage = 3,556 /55 = 64.7 ft/stage. Plotting the minimum head/stage (64.7 ft) and the minimum flow rate (2,550 B/D) on the curve indicates an operating frequency of 70 Hz. Note, the minimum hydraulic requirement is also near the pump’s BEP.
Next, using the VSC curve for the GC-2200 find the BHP/stage at the 60-Hz BEP (1.12 hp). To calculate the BHP at the maximum frequency use Eqs. 86 and 87.
....................(86)
and
....................(87)
Because a rotary gas separator was selected (which is a centrifugal machine using HP), it will add additional load to the motor. The HP requirement also changes by the cube function. Referring to the manufacturer’s information, the 513 series rotary gas separator requires 5 hp at 60 Hz.
....................(88)
Total BHP for the pump and separator = 157.6 + 12.8 = 170.4 hp. With Eqs. 89 and 90, the equivalent 60-Hz BHP for both the pump and gas separator can be calculated:
....................(89)
or
....................(90)
Select the appropriate model seal section and determine the HP requirement at the maximum TDH requirement. Select a motor that is capable of supplying total HP requirements of the pump, gas separator, and seal. In this example, a 562 series motor with 130 hp; 2,145 volts; and 35 amps was selected.
Using the technical data provided by the manufacturer, determine if any load limitations were exceeded (e.g., shaft loading, thrust bearing loading, housing burst pressure limitations, fluid velocity passing the motor, etc.).
Next, select the power cable and calculate the cable voltage drop. On the basis of the motor current (35 amps) and the temperature (160°F), no. 6 cable can be used. Adding 200 ft for surface connections, the cable voltage drop is written as
....................(91)
We can now calculate the required surface voltage (SV) at the maximum operating frequency as
....................(92)
and
....................(93)
Note that the surface voltage is greater than standard 3KV cable. Therefore, 4KV or higher cable construction should be selected. Sufficient data are available to calculate KVA.
....................(94)
and
....................(95)
Referring to the manufacturer’s catalog, select the model 2200-3VT, 200 KVA, NEMA3 (outdoor enclosure) VSC. All other accessory equipment should be selected as in the previous example.
Nomenclature
Am = motor amperage, amps Bg = gas volume factor, scf/bbl [m3/m3] Bo = oil volume factor, bbl/STBO C = constant = 3,960, where Q is in gal/min, and TDH is in ft [= 6,750, where Q is in m3/D, and TDH is in m] D = diameter, in. [cm] F = correlating function for Eq. 51 Ft = well-tubing friction loss H = head, ft [m] HL = net well lift Hwh = wellhead pressure head, ft [m] J = slope N = rotating speed, rev/min P = pressure, psi [kg/cm2] Pb = bubblepoint pressure, psi [kg/cm2] Pdischarge = pump-discharge pressure, psi [kg/cm2] Pr = well static pressure, psi [kg/cm2] Pwf = well flowing pressure, psi [kg/cm2] Q = flow rate, B/D [m3/d] Qd = estimated production rate Qo = maximum production at Pwf = 0, B/D [m3/D] Rs = solution gas/oil ratio, scf/bbl [m3/m3] T = torque, ft-lbf Tconductor = wellbore temperature at the ESP setting depth TC = temperature, °C TF = temperature, °F TG = total volume of gas TK = temperature, K TR = temperature, °R V = voltage, volts VFG = volume of free gas Vg = volume of gas VIG = volume of free gas at the pump intake Vo = volume of oil, bbl [m3] Vs = surface voltage, volts VSG = solution gas volume Vt = total volume Vw = volume of water Z = gas-compressibility factor (typically 0.50 to 1.00) ηm = motor efficiency ηp = pump efficiency
References
1. Electrical Submersible Pumps and Equipment, 11. 2001. Claremore, Oklahoma: Centrilift.
2. Saveth, K.J., Klein, S.T., and Fisher, K.B. 1987. A Comparative Analysis of Efficiency and Horsepower Between Progressing Cavity Pumps and Plunger Pumps. Presented at the SPE Production Operations Symposium, Oklahoma City, Oklahoma, 8-10 March 1987. SPE-16194-MS. http://dx.doi.org/10.2118/16194-MS
Noteworthy books
Takács G. (2009): Electrical submersible pumps manual. ISBN 978-1-85617-557-9. Gulf Professional Publishing, An Imprint of Elsevier, 440p.
Noteworthy papers in OnePetro
Camilleri, L. A. P., & Macdonald, J. 2010. How 24/7 Real-Time Surveillance Increases ESP Run Life and Uptime. Society of Petroleum Engineers. http://dx.doi.org/10.2118/134702-MS
Camilleri, L., & Gambier, P. 2013. Does your ESP Completion Architecture Meet All Your Production Requirements? Society of Petroleum Engineers. http://dx.doi.org/10.2118/164249-MS
Cudmore, J. 2012. Using Real Time Automated Optimisation & Diagnosis to Manage an ArtificiallyLifted Reservoir - A Case Study. Society of Petroleum Engineers. http://dx.doi.org/10.2118/163296-MS
Durham, M. O., & Lea, J. F. 1996. Survey of Electrical Submersible Systems Design, Application, and Testing. Society of Petroleum Engineers. http://dx.doi.org/10.2118/29506-PA | 6,257 | 26,166 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2022-49 | latest | en | 0.841948 |
https://nl.mathworks.com/matlabcentral/cody/problems/2404-reshape-a-vector/solutions/468444 | 1,610,768,105,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703499999.6/warc/CC-MAIN-20210116014637-20210116044637-00771.warc.gz | 464,316,038 | 17,446 | Cody
# Problem 2404. Reshape a Vector
Solution 468444
Submitted on 9 Jul 2014 by James
This solution is locked. To view this solution, you need to provide a solution of the same size or smaller.
### Test Suite
Test Status Code Input and Output
1 Pass
%% n=2; m=5; S=1:10; y_correct =[1 2 3 4 5 6 7 8 9 10]; assert(isequal(ResHape(S,m,n),y_correct))
a = 1 2 b = 1 2 3 4 5 c = 1 2 2 4 3 6 4 8 5 10 ans = 1 2 3 4 5 6 7 8 9 10
2 Pass
%% n=4; m=8; S=[1:10 6:12 5:19] y_correct=[ 1 2 3 4 5 6 7 8 9 10 6 7 8 9 10 11 12 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19] assert(isequal(ResHape(S,m,n),y_correct))
S = Columns 1 through 16 1 2 3 4 5 6 7 8 9 10 6 7 8 9 10 11 Columns 17 through 32 12 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 y_correct = 1 2 3 4 5 6 7 8 9 10 6 7 8 9 10 11 12 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 a = 1 2 3 4 b = 1 2 3 4 5 6 7 8 c = 1 2 3 4 2 4 6 8 3 6 9 12 4 8 12 16 5 10 15 20 6 12 18 24 7 14 21 28 8 16 24 32 ans = 1 2 3 4 5 6 7 8 9 10 6 7 8 9 10 11 12 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19
3 Pass
%% n=2; m=31; S=[1:10 6:12 zeros(1,20) 5:19 ones(1,10)] y_correct=[1 2 3 4 5 6 7 8 9 10 6 7 8 9 10 11 12 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 1 1 1 1 1 1 1 1 1 1]; assert(isequal(ResHape(S,m,n),y_correct))
S = Columns 1 through 16 1 2 3 4 5 6 7 8 9 10 6 7 8 9 10 11 Columns 17 through 32 12 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 Columns 33 through 48 0 0 0 0 0 5 6 7 8 9 10 11 12 13 14 15 Columns 49 through 62 16 17 18 19 1 1 1 1 1 1 1 1 1 1 a = 1 2 b = 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 c = 1 2 2 4 3 6 4 8 5 10 6 12 7 14 8 16 9 18 10 20 11 22 12 24 13 26 14 28 15 30 16 32 17 34 18 36 19 38 20 40 21 42 22 44 23 46 24 48 25 50 26 52 27 54 28 56 29 58 30 60 31 62 ans = 1 2 3 4 5 6 7 8 9 10 6 7 8 9 10 11 12 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 1 1 1 1 1 1 1 1 1 1
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Start Hunting! | 1,277 | 2,054 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.171875 | 3 | CC-MAIN-2021-04 | latest | en | 0.360809 |
https://brainpy.readthedocs.io/en/latest/apis/auto/generated/brainpy.optim.LARS.html | 1,679,605,773,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296945183.40/warc/CC-MAIN-20230323194025-20230323224025-00127.warc.gz | 185,313,037 | 18,890 | # brainpy.optim.LARS#
class brainpy.optim.LARS(lr, train_vars=None, momentum=0.9, weight_decay=0.0001, tc=0.001, eps=1e-05, name=None)[source]#
Layer-wise adaptive rate scaling (LARS) optimizer [1].
Layer-wise Adaptive Rate Scaling, or LARS, is a large batch optimization technique. There are two notable differences between LARS and other adaptive algorithms such as Adam or RMSProp: first, LARS uses a separate learning rate for each layer and not for each weight. And second, the magnitude of the update is controlled with respect to the weight norm for better control of training speed.
$\begin{split}m_{t} = \beta_{1}m_{t-1} + \left(1-\beta_{1}\right)\left(g_{t} + \lambda{x_{t}}\right) \\ x_{t+1}^{\left(i\right)} = x_{t}^{\left(i\right)} - \eta_{t}\frac{\phi\left(|| x_{t}^{\left(i\right)} ||\right)}{|| m_{t}^{\left(i\right)} || }m_{t}^{\left(i\right)}\end{split}$
Parameters:
• lr (float, Scheduler) – learning rate.
• momentum (float) – coefficient used for the moving average of the gradient.
• weight_decay (float) – weight decay coefficient.
• tc (float) – trust coefficient eta ( < 1) for trust ratio computation.
• eps (float) – epsilon used for trust ratio computation.
References
__init__(lr, train_vars=None, momentum=0.9, weight_decay=0.0001, tc=0.001, eps=1e-05, name=None)[source]#
Methods
__init__(lr[, train_vars, momentum, ...]) check_grads(grads) cpu() Move all variable into the CPU device. cuda() Move all variables into the GPU device. load_state_dict(state_dict[, warn, compatible]) Copy parameters and buffers from state_dict into this module and its descendants. load_states(filename[, verbose]) Load the model states. nodes([method, level, include_self]) Collect all children nodes. register_implicit_nodes(*nodes[, node_cls]) register_implicit_vars(*variables[, var_cls]) register_train_vars([train_vars]) register_vars([train_vars]) save_states(filename[, variables]) Save the model states. state_dict() Returns a dictionary containing a whole state of the module. to(device) Moves all variables into the given device. tpu() Move all variables into the TPU device. train_vars([method, level, include_self]) The shortcut for retrieving all trainable variables. tree_flatten() Flattens the object as a PyTree. tree_unflatten(aux, dynamic_values) Unflatten the data to construct an object of this class. unique_name([name, type_]) Get the unique name for this object. update(grads) vars([method, level, include_self, ...]) Collect all variables in this node and the children nodes.
Attributes
name Name of the model. | 683 | 2,563 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.671875 | 3 | CC-MAIN-2023-14 | latest | en | 0.586955 |
https://www.examrace.com/CDS/CDS-Sample-Papers/CDS-Exam-2016-Mathematics-Sample-Paper-Part-8.amp.html | 1,547,851,422,000,000,000 | text/html | crawl-data/CC-MAIN-2019-04/segments/1547583660818.25/warc/CC-MAIN-20190118213433-20190118235433-00497.warc.gz | 783,486,798 | 7,343 | # CDS Exam 2016 Mathematics Sample Paper Part – 8
78. At a grocery store, the number of per day sold processed fruits cans in 15 days are 50,70,60,40. 30, 20, 5, 150, 55, 75, 65, 45. 35, 25, 52 then the outliers in observations are
(A) 50, 150
(B) 5, 150
(C) 25, 70
(D) 150
The following scores were obtained by eleven footballers in a goal-shoot competition: 5 3 6 8 7 8 3 1 1 6 3 2
79. The modal score was
(A) 3
(B) 6
(C) 8
(D) 11
80. The median scare was
(A) 3
(B) 6
(C) 8
(D) 11
81. The range of a sample gives an indication of the
(A) way in which the values duster about a particular point
(B) number of observations bearing the same value
(C) maximum variation in the sample
(D) degree to which the mean value differs from its expected value.
82. The range of a sample gives an indication of the
(A) way in which the values cluster about a particular point
(B) number of observations bearing the same value
(C) maximum variation in the sample
(D) degree to which the mean value differs from its expected value.
83, If a - b = 3 and a2 + b2 = 29, find the value of ab
(A) 10
(B) 1.2
(C) 15
(D) 18
84. The price of 2 sarees and 4 shirts is Rs. 1600. With the same money one can buy 1 saree and 6 shirts.
If one wants to buy 12 shirts, how much shall he have to pay?
(A) Rs. 1200
(B) Rs .2400
(C) Rs. 4800
(D) Cannot be determined
85. Free notebooks w re distributed equally among children of a class. The number of notebooks each child got was one-eighth of the number of children. Had the number of children been half, each child would have got 16 notebooks. Total how many notebooks were distrusted?
(A) 256
(B) 432
(C) 512
(D) 640
86. David gets on the elevator at the 11th floor of a building and rides up at the rate of 57 floors per minute. At the same time. Albert gets on an elevator at the 51st floor of the same building and rides down at the rate of 63 f bars per minute. If they continue traveling at these rates, then at which floor will their paths cross?
(A) 19
(B) 28
(C) 30
(D) 37
87. A train 125m long passes a man, running at 5 km/hr in the same the direction in which the train is going, in 10 seconds. The speed of the train is:
(A) 45km/hr
(B) 50Km/hr
(C) 54Km/hr
(D) 55km/hr
88. Two trains running in opposite directions cross a man standing on the platform in 27 seconds and 17 seconds respectively and they cross each other in 23 seconds. The ratio of their speeds is:
(A) 1:3
(B) 3:2
(C) 3:4
(D) None of these
89. A milk vendor has 2 cans of milk. The first contains 25% water and the rest milk. The second contains 50% water. How much milk should he mix from each of the containers so as to get 12 litres of milk such that the ratio of water to milk is 3:5?
(A) 4 liters, 8 liters
(B) 6 liters, 6 liters
(C) 5 liters, 7 liters
(D) 7 liters, 5 liters
90. A bag œntans4 white, 5 red and 6 blue bals. Three balisare drawn at random from the bag. The
probablity that aH of them are red, is:
(A) 1/22
(B)3/22
(C)2/91
(D)2/77 | 947 | 3,019 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2019-04 | latest | en | 0.933395 |
http://comments.gmane.org/gmane.comp.lang.haskell.cafe/108424 | 1,394,415,362,000,000,000 | text/html | crawl-data/CC-MAIN-2014-10/segments/1394010527022/warc/CC-MAIN-20140305090847-00008-ip-10-183-142-35.ec2.internal.warc.gz | 39,607,883 | 13,093 | 2 Dec 23:07 2013
## existential quantification
Hi everybody,
I try to understand existential quantification. I have two questions.
1/ I have just read the answer of yairchu at
He writes:
"""
So with Existential-Quantification, foralls in data definitions mean that,
the value contained *can* be of *any* suitable type, not that it *must* be
of *all* suitable types.
"""
This made me think to an error I obtained with the code:
---------------
test :: Show s => s
test = "foobar"
---------------
The error is:
Could not deduce (s ~ [Char])
from the context (Show s)
bound by the type signature for test :: Show s => s
[...]
s' is a rigid type variable bound by
the type signature for test :: Show s => s
2 Dec 18:33 2013
### Re: existential quantification
On Mon, Dec 2, 2013 at 5:07 PM, TP wrote:
---------------
test :: Show s => s
test = "foobar"
---------------
The error is:
Could not deduce (s ~ [Char])
from the context (Show s)
bound by the type signature for test :: Show s => s
[...]
s' is a rigid type variable bound by
the type signature for test :: Show s => s
Indeed, test :: Show s => s means "for any type s which is an instance of
Show, test is a value of that type s". But for example "foobar" can't be an
Int that is an instance of Show, so it yields an error.
(...)
So I thought that by using existential quantification, the first example
could work:
---------------
{-# LANGUAGE ExistentialQuantification #-}
test :: forall s . Show s => s
test = "asd"
---------------
This is actually the same as the first one; top level type variables (that is, outside of parentheses) are always forall.
And just tossing a forall in there does not mean you can claim to be any type and then force a String down the caller's throat.
Which brings us to what is *really* going on. When you write
test :: Show s => s
you are saying exactly and only this:
Any function that calls me can request *any* type that has an instance of Show, and I will give them *that type*.
It still means that if you add an explicit forall.
It does not, nor can it be forced to mean, that you will only ever give them a String. Likewise, it does not, nor can it be forced to mean, that you can pick a different type based on (the value of a function parameter, the value of an environment variable, the phase of the moon).
Haskell does not use an OO type system; there is no java.lang.Object that can be every possible type, and forall does not create one. You cannot represent in Haskell the kind of type that you are trying to write. (There is something you can do that is almost the same, but requires a constraint and can only represent monomorphic types. And *still* does not give you java.lang.Object; it gives you a thing which has a specific type and "contains" a thing with another specific type, but can be queried about what that type is and can be extracted *only* in a context that requires that type.)
When forall is useful is inside parentheses in a type. I am not sure that I can provide a useful example that I can explain meaningfully until you understand the above. (But others here probably can....)
2/
Is the notion of existential type related in some way to the classical
mathematical quantifier "∃" (Unicode symbol U+2203: "There exists")?
If yes, then why using "forall" for an "existential type"?
Because "there exists" and "for all" are related by DeMorgan's rule (think about it), and "for all" is easier to represent in GHC's type machinery. I believe UHC provides an "exists" type quantifier as well as "forall".
But Haskell programmers can safely think of the ordinary universally
quantified type given above, thereby avoiding adding a new existential
quantification construct.
"""
But I don't understand the explanation.
"You don't have to keep two different kinds of quantification in mind, or figure out how they interact with each other and non-quantified types, since you can write one in terms of the other."
--
brandon s allbery kf8nh sine nomine associates
unix, openafs, kerberos, infrastructure, xmonad http://sinenomine.net
_______________________________________________
3 Dec 04:53 2013
### Re: existential quantification
Brandon Allbery wrote:
> Which brings us to what is *really* going on. When you write
>
> test :: Show s => s
>
> you are saying exactly and only this:
>
> Any function that calls me can request *any* type that has an instance
> of Show, and I will give them *that type*.
Thanks Brandon for this interpretation. I have carefully written it in my
TP
2 Dec 18:47 2013
### Re: existential quantification
There is just one place where "forall" means "any suitable type", and that's in "data" (or "newtype")
declaration, BEFORE the data constructor. Like this:
newtype T = forall s. Show s => T s
Then you can have
test :: T
test = T "foobar".
If "forall" is AFTER the data constructor, it means that the value should have ALL suitable types
simultaniously. Like this:
data T = T (forall s. Show s => s).
Then you CAN'T have
test = T "foobar"
because "foobar" has only one type, String; it's not polymorphic. On the other hand, if you happen to have a
value of type T, you can treat the value inside it as a value of any suitable type, like this:
baz :: T -> String
baz (T s) = s
Same thing happens if you use "forall" without defining a new type, like
test :: forall s => Show s => s
It differs from the previous example just by one level of indirection, that's all.
On 03 Dec 2013, at 02:07, TP <paratribulations <at> free.fr> wrote:
> Hi everybody,
>
>
> I try to understand existential quantification. I have two questions.
>
> 1/ I have just read the answer of yairchu at
>
>
> He writes:
>
> """
> So with Existential-Quantification, foralls in data definitions mean that,
> the value contained *can* be of *any* suitable type, not that it *must* be
> of *all* suitable types.
> """
>
> This made me think to an error I obtained with the code:
> ---------------
> test :: Show s => s
> test = "foobar"
> ---------------
>
> The error is:
>
> Could not deduce (s ~ [Char])
> from the context (Show s)
> bound by the type signature for test :: Show s => s
> [...]
> s' is a rigid type variable bound by
> the type signature for test :: Show s => s
>
> Indeed, test :: Show s => s means "for any type s which is an instance of
> Show, test is a value of that type s". But for example "foobar" can't be an
> Int that is an instance of Show, so it yields an error.
> By comparison,
>
> ---------------
> test :: Num a => a
> test = 42
> ---------------
>
> works because 42 can be a value of type Int or Integer or Float or anything
> else that is an instance of Num.
> So I thought that by using existential quantification, the first example
> could work:
>
> ---------------
> {-# LANGUAGE ExistentialQuantification #-}
>
> test :: forall s . Show s => s
> test = "asd"
> ---------------
>
> But I obtain the same error, why?
>
> 2/
> Is the notion of existential type related in some way to the classical
> mathematical quantifier "∃" (Unicode symbol U+2203: "There exists")?
> If yes, then why using "forall" for an "existential type"?
>
> At the following address
>
>
>
> """
> 7.4.4.1. Why existential?
>
> What has this to do with existential quantification? Simply that MkFoo has
> the (nearly) isomorphic type
>
> MkFoo :: (exists a . (a, a -> Bool)) -> Foo
>
> But Haskell programmers can safely think of the ordinary universally
> quantified type given above, thereby avoiding adding a new existential
> quantification construct.
> """
>
> But I don't understand the explanation.
>
> Thanks in advance,
>
> TP
>
> _______________________________________________
> Haskell-Cafe mailing list
_______________________________________________
2 Dec 20:50 2013
### Re: existential quantification
Just expanding on Brandon's answer: DeMorgan's law he's referring to goes like this:
∀a.P(a) === ¬∃a.¬P(a) where 'a' is a sentence, so P is second order
A special case of this is this:
∀a.(R(a) -> Q) === ¬∃a.¬(R(a) -> Q) === ¬∃a.(R(a)∧¬Q) === ¬((∃a.R(a))∧¬Q) === (∃a.R(a)) -> Q (i added extra parantheses for emphasis)
So what does this mean in terms of haskell? R(a) is your data definition's "body", and Q is the type you are defining. On the lhs the universally quantified version gives you the type of the constuctor you're defining, and on the rhs the existential tells you what you're constructing the type with.
Or in other words the universal version says: For any 'a' give me an R(a) and i'll give you back a Q.
The existential version says: If you have some 'a' for which R(a) i'll give you back a Q. (It's hard to phrase the difference without sounding stupid, they are equivalent after all).
There are of course other considerations, for example introducing 'exists' would mean another keyword in the syntax.
Having said that I think that the choice of 'forall' for -XExistentialQuantification is wrong, as the data body defines the type you're constructing with, not the type of the whole constructor. HOWEVER for -XGADTs forall makes perfect sense. Compare the following:
data AnyType = forall a. AnyType a
data AnyType where
AnyType :: forall a. a -> AnyType
These two definitions are operationally identical, but I think the GADT way is the one that actually corresponds to the DeMorgan law.
On 2 December 2013 22:07, TP wrote:
Hi everybody,
I try to understand existential quantification. I have two questions.
1/ I have just read the answer of yairchu at
He writes:
"""
So with Existential-Quantification, foralls in data definitions mean that,
the value contained *can* be of *any* suitable type, not that it *must* be
of *all* suitable types.
"""
This made me think to an error I obtained with the code:
---------------
test :: Show s => s
test = "foobar"
---------------
The error is:
Could not deduce (s ~ [Char])
from the context (Show s)
bound by the type signature for test :: Show s => s
[...]
s' is a rigid type variable bound by
the type signature for test :: Show s => s
Indeed, test :: Show s => s means "for any type s which is an instance of
Show, test is a value of that type s". But for example "foobar" can't be an
Int that is an instance of Show, so it yields an error.
By comparison,
---------------
test :: Num a => a
test = 42
---------------
works because 42 can be a value of type Int or Integer or Float or anything
else that is an instance of Num.
So I thought that by using existential quantification, the first example
could work:
---------------
{-# LANGUAGE ExistentialQuantification #-}
test :: forall s . Show s => s
test = "asd"
---------------
But I obtain the same error, why?
2/
Is the notion of existential type related in some way to the classical
mathematical quantifier "∃" (Unicode symbol U+2203: "There exists")?
If yes, then why using "forall" for an "existential type"?
At the following address
"""
7.4.4.1. Why existential?
What has this to do with existential quantification? Simply that MkFoo has
the (nearly) isomorphic type
MkFoo :: (exists a . (a, a -> Bool)) -> Foo
But Haskell programmers can safely think of the ordinary universally
quantified type given above, thereby avoiding adding a new existential
quantification construct.
"""
But I don't understand the explanation.
TP
_______________________________________________
_______________________________________________
3 Dec 05:10 2013
### Re: existential quantification
Andras Slemmer wrote:
> Just expanding on Brandon's answer: DeMorgan's law he's referring to goes
> like this:
> ∀a.P(a) === ¬∃a.¬P(a) where 'a' is a sentence, so P is second order
> A special case of this is this:
> ∀a.(R(a) -> Q) === ¬∃a.¬(R(a) -> Q) === ¬∃a.(R(a)∧¬Q) === ¬((∃a.R(a))∧¬Q)
> === (∃a.R(a)) -> Q (i added extra parantheses for emphasis)
> So what does this mean in terms of haskell? R(a) is your data definition's
> "body", and Q is the type you are defining. On the lhs the universally
> quantified version gives you the type of the constuctor you're defining,
> and on the rhs the existential tells you what you're constructing the type
> with.
> Or in other words the universal version says: For any 'a' give me an R(a)
> and i'll give you back a Q.
> The existential version says: If you have some 'a' for which R(a) i'll
> give you back a Q. (It's hard to phrase the difference without sounding
> stupid, they are equivalent after all).
>
> There are of course other considerations, for example introducing 'exists'
> would mean another keyword in the syntax.
Thanks Andras, I have understood the developments up to that point. But
below I do not understand your reasoning.
>
> Having said that I think that the choice of 'forall' for
> -XExistentialQuantification is wrong, as the data body defines the type
> you're constructing with, not the type of the whole constructor. HOWEVER
> for -XGADTs forall makes perfect sense. Compare the following:
>
> data AnyType = forall a. AnyType a
> data AnyType where
> AnyType :: forall a. a -> AnyType
>
> These two definitions are operationally identical, but I think the GADT
> way is the one that actually corresponds to the DeMorgan law.
And one more question: I had lectures on logic some years ago, but I never
studied type theory at university (I'm some sort of "electrical engineer").
Is there around a good textbook for "beginners", with full proofs, but only
the essential ones? I would like a good "entry point" in the textbook
literature. Not for experts.
Are the books of Robert Harper suitable, for example
http://www.amazon.com/Practical-Foundations-Programming-Languages-Professor/dp/1107029570
?
TP
_______________________________________________
3 Dec 01:47 2013
### Re: existential quantification
> But below I do not understand your reasoning.
Given a datatype
data D = D Int
I call Int the type you're constructing D with, however the type of the constructor itself is Int -> D
In ghc haskell there is another way of defining the same type using GADTs alongside the above "haskell98" way:
data D where
D :: Int -> D
Notice how here we are explicitly typing the constructor.
The haskell98 way defines what your datatype D is isomorphic to - here it is Int, whereas the GADT way defines how to construct a D.
The problem haskell developers faced was that in order to use the haskell98 way for existentials they would've had to introduce an 'exists' quantifier, which would mess up a boatload of things in the type theory. Instead they introduced this forall hack that doesn't define an isomorphic type, rather it indicates that we are defining a way of constructing an existential.
I find this ugly because it breaks the isomorphism that the = sign indicates in a haskell98 definition.
The GADT way on the other hand is defining ways of construction, so a definition like:
data D2 where
D2 :: a -> D2
makes perfect sense
> Is there around a good textbook for "beginners", with full proofs, but only the essential ones?
Types and Programming Languages from Benjamin Pierce is a good one. I also plan to upload a short video lecture series from Harper on type theory (it assumes minimal knowledge of logic), i'll send you a link when it's up.
On 3 December 2013 04:10, TP wrote:
Andras Slemmer wrote:
> Just expanding on Brandon's answer: DeMorgan's law he's referring to goes
> like this:
> ∀a.P(a) === ¬∃a.¬P(a) where 'a' is a sentence, so P is second order
> A special case of this is this:
> ∀a.(R(a) -> Q) === ¬∃a.¬(R(a) -> Q) === ¬∃a.(R(a)∧¬Q) === ¬((∃a.R(a))∧¬Q)
> === (∃a.R(a)) -> Q (i added extra parantheses for emphasis)
> So what does this mean in terms of haskell? R(a) is your data definition's
> "body", and Q is the type you are defining. On the lhs the universally
> quantified version gives you the type of the constuctor you're defining,
> and on the rhs the existential tells you what you're constructing the type
> with.
> Or in other words the universal version says: For any 'a' give me an R(a)
> and i'll give you back a Q.
> The existential version says: If you have some 'a' for which R(a) i'll
> give you back a Q. (It's hard to phrase the difference without sounding
> stupid, they are equivalent after all).
>
> There are of course other considerations, for example introducing 'exists'
> would mean another keyword in the syntax.
Thanks Andras, I have understood the developments up to that point. But
below I do not understand your reasoning.
>
> Having said that I think that the choice of 'forall' for
> -XExistentialQuantification is wrong, as the data body defines the type
> you're constructing with, not the type of the whole constructor. HOWEVER
> for -XGADTs forall makes perfect sense. Compare the following:
>
> data AnyType = forall a. AnyType a
> data AnyType where
> AnyType :: forall a. a -> AnyType
>
> These two definitions are operationally identical, but I think the GADT
> way is the one that actually corresponds to the DeMorgan law.
And one more question: I had lectures on logic some years ago, but I never
studied type theory at university (I'm some sort of "electrical engineer").
Is there around a good textbook for "beginners", with full proofs, but only
the essential ones? I would like a good "entry point" in the textbook
literature. Not for experts.
Are the books of Robert Harper suitable, for example
http://www.amazon.com/Practical-Foundations-Programming-Languages-Professor/dp/1107029570
?
TP
_______________________________________________
_______________________________________________
3 Dec 07:36 2013
### Re: existential quantification
On Tue, Dec 3, 2013 at 11:10 AM, TP wrote:
I would like a good "entry point" in the textbook
literature. Not for experts.
Are the books of Robert Harper suitable, for example
http://www.amazon.com/Practical-Foundations-Programming-Languages-Professor/dp/1107029570
There's a draft copy on Harper's home page you can check out.
It's primarily a textbook for CS graduate students entering the specialization of PL.
-- Kim-Ee
_______________________________________________
4 Dec 00:35 2013
### Re: existential quantification
yeah, its not an introductory text, but it is a great grad level reference. (nb: i read a draft a few years ago, haven't read the published version... yet)
On Tue, Dec 3, 2013 at 1:36 AM, Kim-Ee Yeoh wrote:
On Tue, Dec 3, 2013 at 11:10 AM, TP wrote:
I would like a good "entry point" in the textbook
literature. Not for experts.
Are the books of Robert Harper suitable, for example
http://www.amazon.com/Practical-Foundations-Programming-Languages-Professor/dp/1107029570
There's a draft copy on Harper's home page you can check out.
It's primarily a textbook for CS graduate students entering the specialization of PL.
-- Kim-Ee
_______________________________________________
_______________________________________________
8 Dec 10:23 2013
### Re: existential quantification
I uploaded the Harper video series, here is a link to the first lecture: https://www.youtube.com/watch?v=ev7AYsLljxk&list=PL8Ky8lYL8-Oh7awp0sqa82o7Ggt4AGhyf&index=5
This is more of an introduction to dependent type theory, but it's worth a watch!
On 3 December 2013 04:10, TP wrote:
Andras Slemmer wrote:
> Just expanding on Brandon's answer: DeMorgan's law he's referring to goes
> like this:
> ∀a.P(a) === ¬∃a.¬P(a) where 'a' is a sentence, so P is second order
> A special case of this is this:
> ∀a.(R(a) -> Q) === ¬∃a.¬(R(a) -> Q) === ¬∃a.(R(a)∧¬Q) === ¬((∃a.R(a))∧¬Q)
> === (∃a.R(a)) -> Q (i added extra parantheses for emphasis)
> So what does this mean in terms of haskell? R(a) is your data definition's
> "body", and Q is the type you are defining. On the lhs the universally
> quantified version gives you the type of the constuctor you're defining,
> and on the rhs the existential tells you what you're constructing the type
> with.
> Or in other words the universal version says: For any 'a' give me an R(a)
> and i'll give you back a Q.
> The existential version says: If you have some 'a' for which R(a) i'll
> give you back a Q. (It's hard to phrase the difference without sounding
> stupid, they are equivalent after all).
>
> There are of course other considerations, for example introducing 'exists'
> would mean another keyword in the syntax.
Thanks Andras, I have understood the developments up to that point. But
below I do not understand your reasoning.
>
> Having said that I think that the choice of 'forall' for
> -XExistentialQuantification is wrong, as the data body defines the type
> you're constructing with, not the type of the whole constructor. HOWEVER
> for -XGADTs forall makes perfect sense. Compare the following:
>
> data AnyType = forall a. AnyType a
> data AnyType where
> AnyType :: forall a. a -> AnyType
>
> These two definitions are operationally identical, but I think the GADT
> way is the one that actually corresponds to the DeMorgan law.
And one more question: I had lectures on logic some years ago, but I never
studied type theory at university (I'm some sort of "electrical engineer").
Is there around a good textbook for "beginners", with full proofs, but only
the essential ones? I would like a good "entry point" in the textbook
literature. Not for experts.
Are the books of Robert Harper suitable, for example
http://www.amazon.com/Practical-Foundations-Programming-Languages-Professor/dp/1107029570
?
TP
_______________________________________________
_______________________________________________
2 Dec 21:30 2013
### Re: existential quantification
On 12/02/2013 11:07 PM, TP wrote:
> ...
> So I thought that by using existential quantification, the first example
> could work:
>
> ---------------
> {-# LANGUAGE ExistentialQuantification #-}
>
> test :: forall s . Show s => s
> test = "asd"
> ---------------
>
> But I obtain the same error, why?
>
This still says that 'test' is a value of type 's' for all 's' with a
'Show' instance.
Basically, 'test' gets a type 's', an instance 'Show s' and we get a
value of type 's'.
> 2/
> Is the notion of existential type related in some way to the classical
> mathematical quantifier "∃" (Unicode symbol U+2203: "There exists")?
> If yes, then why using "forall" for an "existential type"?
>
> At the following address
>
>
>
> """
> 7.4.4.1. Why existential?
>
> What has this to do with existential quantification? Simply that MkFoo has
> the (nearly) isomorphic type
>
> MkFoo :: (exists a . (a, a -> Bool)) -> Foo
>
> But Haskell programmers can safely think of the ordinary universally
> quantified type given above, thereby avoiding adding a new existential
> quantification construct.
> """
>
> But I don't understand the explanation.
>
Generally speaking, a typing judgement
x :: A
can be interpreted as a proof that type 'A' is inhabited, or in other
words, there exists a value of type 'A'. (Of course in Haskell this fact
alone is trivial due to 'undefined', but additional reasoning could
render it meaningful. For the remainder I'll just ignore the issue of
lifting.)
This is somewhat related to classical existential quantification, but it
is stronger in a sense, since it says that we know how to construct such
a value. (The classical version in general just says that the assumption
that there is no such value leads to a contradiction.)
Pattern matching allows one to get back the original constructor and the
arguments used to construct a value of some ADT type. A universal
quantifier over a type states that we can provide any type and obtain a
value of the type provided in the body where all instances of the bound
variable are replaced by our type. I.e. you can interpret forall a.
(...) as stating that a value of that type takes an additional implicit
argument 'a' at type level.
Now MkFoo is declared as follows:
data Foo = forall a. MkFoo a (a -> Bool)
| Nil
Which gives it the type:
MkFoo :: forall a. a -> (a -> Bool) -> Foo
I.e. it gets a type 'a' a value of that type and a decidable predicate
ranging over that type and constructs a value of type 'Foo'. Pattern
matching (roughly speaking) does the opposite: It gets you back a type,
a value of that type and the predicate.
The type is called nearly isomorphic to the explicit existential type,
because using some rounds of pattern matching one would also recover the
same kinds of objects.
I.e. existential quantification can be thought of as being left implicit
in the typing judgement, but 'forall' is needed in order to make
explicit the scope of the type variable, which otherwise would range
over the entire data declaration instead of just a single constructor.
-XExistentialQuantification enables uses of 'forall' necessary for using
existential quantification.
_______________________________________________
10 Jul 19:53 2008
### Existential quantification problem
Hello,
how do I unbox a existential quantificated data type?
> {-# LANGUAGE ExistentialQuantification #-}
> data L a = forall l. L (l a)
> unboxL (L l) = l
is giving me, in GHC:
Inferred type is less polymorphic than expected
Quantified type variable l' escapes
When checking an existential match that binds
l :: l t
The pattern(s) have type(s): L t
The body has type: l t
In the definition of unboxL': unboxL (L l) = l
Thanks.
--
--
Marco Túlio Gontijo e Silva
Blog: http://marcotmarcot.blogspot.com/
Correio: marcot <at> riseup.net
XMPP: marcot <at> jabber.org
IRC: marcot <at> irc.freenode.net
Telefone: 25151920
Celular: 98116720
Endereço:
Rua Turfa, 639/701
Belo Horizonte/MG Brasil
10 Jul 19:59 2008
### Re: Existential quantification problem
On Thu, 2008-07-10 at 14:53 -0300, Marco Túlio Gontijo e Silva wrote:
> Hello,
>
> how do I unbox a existential quantificated data type?
You can't. You have to use case analysis:
case foo of
L l -> <whatever you wanted to do>
where none of the information your case analysis discovers about the
actual type of l can be made available outside of the scope of the case
expression. (It can't escape'). This is required for decidable static
typing, IIRC.
jcc
>
> > {-# LANGUAGE ExistentialQuantification #-}
> > data L a = forall l. L (l a)
> > unboxL (L l) = l
>
> is giving me, in GHC:
>
> Inferred type is less polymorphic than expected
> Quantified type variable l' escapes
> When checking an existential match that binds
> l :: l t
> The pattern(s) have type(s): L t
> The body has type: l t
> In the definition of unboxL': unboxL (L l) = l
>
> Thanks.
>
14 Jul 01:28 2008
### Re: Existential quantification problem
On Thu, 2008-07-10 at 10:59 -0700, Jonathan Cast wrote:
> On Thu, 2008-07-10 at 14:53 -0300, Marco Túlio Gontijo e Silva wrote:
> > Hello,
> >
> > how do I unbox a existential quantificated data type?
>
> You can't. You have to use case analysis:
>
> case foo of
> L l -> <whatever you wanted to do>
>
> where none of the information your case analysis discovers about the
> actual type of l can be made available outside of the scope of the case
> expression. (It can't escape'). This is required for decidable static
> typing, IIRC.
It's not an extraneous requirement; it is part of the definition of
existential types.
14 Jul 01:50 2008
### Re: Existential quantification problem
On Sun, 2008-07-13 at 18:28 -0500, Derek Elkins wrote:
> On Thu, 2008-07-10 at 10:59 -0700, Jonathan Cast wrote:
> > On Thu, 2008-07-10 at 14:53 -0300, Marco Túlio Gontijo e Silva wrote:
> > > Hello,
> > >
> > > how do I unbox a existential quantificated data type?
> >
> > You can't. You have to use case analysis:
> >
> > case foo of
> > L l -> <whatever you wanted to do>
> >
> > where none of the information your case analysis discovers about the
> > actual type of l can be made available outside of the scope of the case
> > expression. (It can't escape'). This is required for decidable static
> > typing, IIRC.
>
> It's not an extraneous requirement; it is part of the definition of
> existential types.
I know that. I didn't know implementing existential types was an end in itself, though.
jcc
14 Jul 03:03 2008
### Re: Existential quantification problem
On 2008 Jul 13, at 19:50, Jonathan Cast wrote:
> On Sun, 2008-07-13 at 18:28 -0500, Derek Elkins wrote:
>> On Thu, 2008-07-10 at 10:59 -0700, Jonathan Cast wrote:
>>> On Thu, 2008-07-10 at 14:53 -0300, Marco Túlio Gontijo e Silva
>>> wrote:
>>>> Hello,
>>>>
>>>> how do I unbox a existential quantificated data type?
>>>
>>> You can't. You have to use case analysis:
>>>
>>> case foo of
>>> L l -> <whatever you wanted to do>
>>>
>>> where none of the information your case analysis discovers about the
>>> actual type of l can be made available outside of the scope of the
>>> case
>>> expression. (It can't escape'). This is required for decidable
>>> static
>>> typing, IIRC.
>>
>> It's not an extraneous requirement; it is part of the definition of
>> existential types.
>
> I know that. I didn't know implementing existential types was an
> end in itself, though.
No? Consider ST. Anytime you need to completely restrict access to
the innards of a datum, existential types are your friend.
--
--
brandon s. allbery [solaris,freebsd,perl,pugs,haskell] allbery <at> kf8nh.com
system administrator [openafs,heimdal,too many hats] allbery <at> ece.cmu.edu
electrical and computer engineering, carnegie mellon university KF8NH
14 Jul 03:16 2008
### Re: Existential quantification problem
On Sun, 2008-07-13 at 21:03 -0400, Brandon S. Allbery KF8NH wrote:
> On 2008 Jul 13, at 19:50, Jonathan Cast wrote:
>
> > On Sun, 2008-07-13 at 18:28 -0500, Derek Elkins wrote:
> >> On Thu, 2008-07-10 at 10:59 -0700, Jonathan Cast wrote:
> >>> On Thu, 2008-07-10 at 14:53 -0300, Marco Túlio Gontijo e Silva
> >>> wrote:
> >>>> Hello,
> >>>>
> >>>> how do I unbox a existential quantificated data type?
> >>>
> >>> You can't. You have to use case analysis:
> >>>
> >>> case foo of
> >>> L l -> <whatever you wanted to do>
> >>>
> >>> where none of the information your case analysis discovers about the
> >>> actual type of l can be made available outside of the scope of the
> >>> case
> >>> expression. (It can't escape'). This is required for decidable
> >>> static
> >>> typing, IIRC.
> >>
> >> It's not an extraneous requirement; it is part of the definition of
> >> existential types.
> >
> > I know that. I didn't know implementing existential types was an
> > end in itself, though.
>
> No? Consider ST. Anytime you need to completely restrict access to
> the innards of a datum, existential types are your friend.
Right. Mea culpa.
jcc
10 Jul 20:19 2008
### Re: Existential quantification problem
On Thursday 10 July 2008, Marco Túlio Gontijo e Silva wrote:
> Hello,
>
> how do I unbox a existential quantificated data type?
>
> > {-# LANGUAGE ExistentialQuantification #-}
> > data L a = forall l. L (l a)
> > unboxL (L l) = l
>
> is giving me, in GHC:
>
> Inferred type is less polymorphic than expected
> Quantified type variable l' escapes
> When checking an existential match that binds
> l :: l t
> The pattern(s) have type(s): L t
> The body has type: l t
> In the definition of unboxL': unboxL (L l) = l
You don't. Or, at least, you don't do it that way. The point of an existential
is that the quantified type is 'forgotten', and there's no way to get it
back. That's good for abstraction, in that it restricts you to using some
provided interface that's guaranteed to work with the forgotten type, but it
means that you can't just take things out of the existential box. Instead,
you have to use them in ways that make the forgotten type irrelevant (that
is, ways that are parametric in the corresponding type).
The type of the unboxL function in a type system with first-class existentials
is something like:
unboxL :: L a -> (exists l. l a)
Of course, GHC doesn't do first-class existentials, otherwise you'd probably
not be bothering with the datatype in the first place. :)
The proper way to eliminate an existential is (as mentioned above) with a
universal:
elim :: (exists l. l a) -> (forall l. l a -> r) -> r
elim e f = f e
Or, since existentials in GHC are restricted to data types:
elim :: L a -> (forall l. l a -> r) -> r
elim (L e) f = f e
Of course, as has already been noted,
elim foo (\l -> stuff)
is equivalent to:
case foo of
L l -> stuff
and doesn't need rank-2 polymorphism enabled, to boot. :)
-- Dan
10 Jul 20:39 2008
### Re: Existential quantification problem
On Thu, 10 July 2008, Marco Túlio Gontijo e Silva wrote:
> how do I unbox a existential quantificated data type?
Dan Doel wrote:
> elim :: L a -> (forall l. l a -> r) -> r
> elim (L e) f = f e
Just one catch: You can't actually write a function 'f' of type
(forall l. l a -> r) without knowing something about the forgotten
type of l.
One way to deal with this is by restricting the type of l in the data
declaration. For example, you could restrict it to the typeclass
Foldable, and then you have access to the methods of that typeclass.
\begin{code}
{-# LANGUAGE ExistentialQuantification #-}
module Main
where
import qualified Data.Foldable as F
data L a = forall l. (F.Foldable l) => L (l a)
toList :: L a -> [a]
toList (L x) = F.foldr (:) [] x
main :: IO ()
main = do
let x = L [1..10]
print \$ toList x
\end{code}
` | 8,776 | 33,340 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.734375 | 3 | CC-MAIN-2014-10 | longest | en | 0.882811 |
http://johansens.us/sane/faith/miraclepi.htm | 1,534,832,412,000,000,000 | text/html | crawl-data/CC-MAIN-2018-34/segments/1534221217970.87/warc/CC-MAIN-20180821053629-20180821073629-00255.warc.gz | 214,371,444 | 4,926 | A More Convincing Miracle - Island of Sanity
# Island of Sanity
Defending Christianity
# A More Convincing Miracle
I came across an article by an atheist recently in which he said that, if Christianity was true, Jesus could easily have proven it by giving us a really convincing miracle. The miracle that he suggested is this: Jesus should have said, "If you divide the circumference of a circle by its diameter, you get a number called pi. Pi is an irrational number, which means that it is made up of an infinite number of digits. Someday people will invent computers that will be able to calculate pi to millions of digits. When you do, you will find that the 600 through 610th digits of pi are 32000568127."
So why didn't Jesus do this? Maybe because that would have been a much weaker sign than the signs that Jesus actually gave us.
By the way, the above is not an exact quote. The writer's hypothetical speech by Jesus included a long discussion of the definition of pi, explaining what a diameter is and what a circumference is and so on. None of this would have been necessary, as people in Jesus' day knew full well what circles and diameters and circumferences were. A Babylonian tablet from about 1900 years before Christ gives an approximation of pi that is the equivalent of 3.125 in our numbering system, which isn't bad. The Egyptian Rhind Papyrus, about 1650 years before Christ, estimates pi as 3.1605. And Archimedes, circa 220 BC, proved that pi must be between 3 1/7 and 3 10/71.
Well okay, maybe he supposes that an explanation was necessary, not for the well-educated people of the time, but for the common people. But then he blithely talks about "digits of pi" with no explanation. The idea of "digits", that is, of decimal fractions, is generally credited to Francois Viete in his book Canon Mathematicus in AD 1579.
So apparently the writer didn't know and didn't bother to research the state of mathematics in Jesus' day before presuming to tell us what Jesus should have said on the subject.
But setting that aside, let's suppose that Jesus had indeed made such a statement. The writer says that this would have proven to "any rational person" that Jesus was God. But, umm, no. At best, it would have been convincing to people living in the mid 20th to maybe late 21st century.
It would have done nothing for people living in Jesus' own day. At that time, no one knew the value of pi to more than about two decimal places. What would it prove to give digits that could not be verified for thousands of years? How would anyone know if these digits were correct or if he had just made them up? Such a statement would only prove that the speaker had special knowledge if no one else in the world was capable of calculating the same number, but as long as no one else was capable of calculating the same number, there would be no way to prove that the speaker was accurate.
Suppose someone came along today and claimed to be God or a messenger from God. And to prove it he said, "Someday people will travel to the star Sigma Draconis, and on the third planet orbiting that star, at 40 degrees north 120 degrees west, the temperature will be 23.4 Celcius." Would that convince you that this person has supernatural knowledge? Or would you just say, "Hey, anybody can make up untestable facts. That proves nothing."
So such a statement would prove absolutely nothing until people invented computers and could actually calculate this many digits of pi, in the mid-20th century. It would have been worthless for almost 2000 years.
Well, actually people did calculate hundreds of digits of pi before computers were invented. In 1873 William Shanks calculated pi to 707 places. Except ... except he made a mistake. Only his first 527 are correct.
So suppose Jesus had declared what the 600 to 610th digits of pi are. Then in 1873 Shanks comes along with a calculation of those digits that is different. How many atheists would have said, "See, science has proven the Bible false."
In fact Jesus did something similar to what this atheist called for, but better thought out. He predicted an historical event that would happen 40 years after his death. Read chapter 24 of the Gospel of Matthew. Jesus predicted that the temple in Jerusalem would be destroyed, and that "not one stone would be left standing on another". He said this would happen before the current generation had passed away. And in fact, in AD 70, the temple in Jerusalem was destroyed by a Roman army. The temple had been gilded with gold, which melted when the temple was burned, and so the soldiers tour it apart to recover the gold, and not one stone was left standing on another.
The Bible includes hundreds of predictions of things that would occur in the future of the time the prediction was made. Arguably some are vague or debatable, but many have been clearly and plainly fulfilled, and none has been proven false.
Isn't this enough to convince "any rational person" that Jesus and other people in the Bible had divine foreknowledge? Well, maybe so, but it apparently does not convince atheists today. And why not? Because they find ways to dismiss these prophecies. One common argument is that as, of course, predicting the future is impossible, that these so-called predictions must have been made after the events they describe. Any evidence that the book was written before is written off as a curious anomaly, as we know that the book could not have been written until after the fact.
Which is why I say if Jesus had made such a statement about pi, a generation or so after it was proven true, atheists would dismiss it. They would simply claim that, as of course it is impossible for Jesus to have known the value of pi to such precision, these statements must have been added to the Bible after the fact. Just as atheists today claim that books of the Bible were written or altered hundreds of years after the dates that these books assert for themselves, so in this writer's scenario, atheists of the future could claim that the value for pi was added or changed hundreds of years later.
Even for people living at the time that computers were invented, I'm sure atheists would have found reasons to dismiss such a statement. When the first calculation of pi was done proving Jesus correct, I'm sure that some atheists would say that the people who did this calculation were obviously biased and had secretly altered the calculation to make it match Jesus' prediction. They'd come up with all sorts of creative explanations for why the calculations were wrong or why it didn't matter. Just like they do for all the facts that tend to confirm the Bible in real life.
Of course, we go through this same scenario pretty regularly. Atheists routinely say that they would believe the Bible if God would just perform one more miracle. The hundreds or thousands he has already done are not enough, but just one more, that would do it. But of course it wouldn't. They would dismiss it using the same sort of evasion and double-talk that they use for the miracles that he's already done. | 1,517 | 7,101 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.84375 | 3 | CC-MAIN-2018-34 | longest | en | 0.974352 |
https://brainor.com/math-worksheets/grade-1/number-charts/hundreds-chart-with-missing-numbers/ | 1,722,921,961,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640476479.26/warc/CC-MAIN-20240806032955-20240806062955-00770.warc.gz | 117,080,476 | 44,841 | # 21 Worksheets on Hundreds Chart with Missing Numbers
To increase number sense and counting capacity, these worksheets on hundreds charts with missing numbers can help your children. These worksheets will enhance your mathematical logic of the number system and build communication between ideas and practical applications. With the help of our free printable hundreds chart with missing numbers, first graders will learn the fundamentals of the number system and counting concepts of mathematical operations.
## Finding Missing Numbers from 1-100 and 100-1
In these three worksheets, you will get two types of charts, e.g.- from 1-100 and 100-1. Your students will find random missing numbers. They need to fill in the blanks and complete the charts.
## Solving the Puzzle in the Hundreds Chart
Each worksheet contains six problems. You have to find and write the necessary numbers and solve the puzzle.
## Finding the More or Less of the Given Numbers
Keenly observe the problems given in the worksheets. Find the one more, one less, ten more, and ten fewer numbers to solve the problems.
## Getting Multiples of 2, 3, and 5 in the Hundreds Chart
Here, the 2, 3, and 5 multiples are missing. Your first-grade students need to find these multiples and fill in the blanks to fill the whole chart.
## Solving the Quizzes Regarding the Hundreds Chart
A bunch of questions are given in each worksheet. You have to find out the answers to the questions from the given hundreds chart.
## Finding the Hundreds Chart Patterns
This fun game has to be played with your first graders. They have to find the numbers given in the chart with the mentioned color.
## Filling the Hundreds Blank Chart
In this last activity, you are given blank charts. You have to fill in the whole chart and complete the chart after coloring it as directed. | 382 | 1,842 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.40625 | 3 | CC-MAIN-2024-33 | latest | en | 0.889361 |
http://seekingalpha.com/article/245639-the-m-and-m-indicator-predicting-the-coming-food-price-crisis | 1,480,748,073,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698540909.75/warc/CC-MAIN-20161202170900-00102-ip-10-31-129-80.ec2.internal.warc.gz | 240,378,018 | 74,169 | # The M&M Indicator: Predicting the Coming Food Price Crisis
|
Includes:
by: Katy Delay
Doing a bit of shopping the other day, I was horrified to find my staple pack of M&Ms up to \$1.29 at Office Depot. Stunned into a reality check, I decided to find out what the current supermarket price is. It's 99 cents.
Most of us look at the items we purchase regularly as an indication of how prices change. The price of M&Ms is my own personal CPI indicator. To take this particular item in a pseudo-scientific study (click on the image for a larger version):
When I was an adolescent in the 1960s, a one-portion pack was 5 cents. I remember a few years later when the pack size began to vary a lot. Mars and the other candy makers started offering Jumbo Packs with twice as much for three times the price and other hoaxes like that, so they could hide the price increase.
My next statistical indicator comes from the mid- to late-1990s when I owned a small coffee shop and market. I sold my M&Ms portion pack (who knows how many ounces by this time) for 55 cents.
Today, ten years later, they're at 99 cents, almost double.
Something's happening.
Moving to my reliable source of price inflation information, AIER.org's Cost-of-Living calculator calculates that my 5-cent candy in 1955 should be priced at something like 41 cents today, assuming the portions are approximately the same. Of course, they may not be; but in your estimation, in which direction would the portions vary? Larger, or smaller? Well, let's not denigrate the candy companies. Let's just say that the ounces are the same.
So, I conclude that my M&Ms have increased in price at a rate of double the national average. Which brings me to the subject of the coming world food crisis.
According to Robert Zoellick, the President of the World Bank Group and the fellow I heralded for having dared to bring up the dead subject of the gold standard, the world can take many steps to "put food first." The G20 should "empower the poor" to ensure "the availability of nutritious food." In his piece in the Financial Times of January 6, 2011, world governments can and should take eight steps to achieve this goal in face of rising food prices.
The steps include improvement of weather forecasting (good luck), exempting "humanitarian food aid from export bans" (good luck), establishing "small regional humanitarian reserves in disaster-prone, infrastructure-poor areas (good luck), and helping "smallholder farmers become a bigger part of the solution to food security" (good luck).
I thought Mr. Zoellick had a grain of good common sense when I read his piece about gold's helpful role as a barometer of worldwide inflating. He is letting me down. If I read correctly from this article, he is just another bureaucrat ... but how silly of me. What was I expecting from someone at the World Bank?
Why doesn't he see that the price of sugar is not a supply problem? It is a monetary problem. If it were a supply problem it would be the only commodity with a rising price. However, as the charts reveal, all commodities are rising to record levels, with few exceptions. Take a look at this chart from Indexmundi.com.
There is no question but that the world is headed into a food price crisis. But my analysis of the problem does not point to governments or government agencies as the solution. The real long-term solution lies in finding an anchor for the world's monetary units, whether it be gold or something else.
More close to home, the question, as I have said before, is: Will this wave of commodity price increases translate into a CPI index rise in the U.S.? We have already got higher gasoline prices, and now higher M&M prices. Will it spread to everything else?
That depends on several factors: (1) the turn of U.S. political winds; (2) the health of U.S. and European banks, which in turn depends on the health of the real estate / banking sector in the U.S. and the PIIGS situation in Europe; and (3) the effect of the above on the interest rate markets, which in turn will have an effect on (2).
Some analysts expect the Fed's and other central banks' monetary actions will produce widespread price inflation worldwide. But this can happen only if the deflationary hot air can blow out of depressed economies. What if (1), (2), and (3) turn negative? And / or what if the deflationary pressures underlying our current crisis turn out to be ongoing? The Fed can "print" all it wants, but it cannot (a) force prices up, or (b) force interest rates down against the will of the gods of markets.
This is the infamous rock-and-hard-place I have mentioned in previous posts. My slow-motion movie climax is approaching. Keep your eyes open over the next few months.
Disclosure: No positions | 1,060 | 4,771 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2016-50 | latest | en | 0.958692 |
https://ec2-65-0-158-107.ap-south-1.compute.amazonaws.com/tag/what-is-question-mark-in-16-06-68-88-98/ | 1,726,153,640,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651460.54/warc/CC-MAIN-20240912142729-20240912172729-00095.warc.gz | 208,666,645 | 16,275 | What Coins Do I Have: If I Have Two US Coins Totaling 55 cents
Solve the riddle and leave your answers in the comment section below. I have two US coins totaling 55 cents. One is not a nickel. Figure out what the coins are. So were you able to solve the riddle? Leave your answers in the comment section below. You can check if your answer is correct … Read more
Tough Number Riddle: Guess ?? in 16, 06, 68, 88, ??, 98
Look at the series of numbers given in the riddle and find the connection between them. Find value of ?? in the following sequence. 16, 06, 68, 88, ??, 98 So were you able to solve the riddle? Leave your answers in the comment section below. You can check if your answer is correct by … Read more | 188 | 718 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.84375 | 3 | CC-MAIN-2024-38 | latest | en | 0.940659 |
https://nl.mathworks.com/matlabcentral/answers/281711-plotting-and-storing-results-from-a-while-loop | 1,585,911,647,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585370510846.12/warc/CC-MAIN-20200403092656-20200403122656-00003.warc.gz | 596,647,877 | 20,546 | # Plotting and storing results from a while loop.
3 views (last 30 days)
Louis Lowagie on 29 Apr 2016
Edited: Renato Agurto on 29 Apr 2016
Hello,
i am currently working on a thermal model. My function is the heatflux-function and this calculates the temperature in one step. The idea is to use a while-loop to calculate the consecutive temperatures after multiple passes. This means the heatflux function uses the T_end from the previous loop to calculate the heatflux for the next loop. However, when i program this loop it doesn't seem to respond as wanted. I would also like to store the outputs of the function (Q, T_end, delta_T) and be able to plot after the loop has completed. The heatflux function works with the given inputs. Can somebody give me some useful tips on how to approach this problem?
My current code is posted below.
if true
%
while i<N
r = r_initial - i*p/tan(angle);
[Q , delta_T, T_end ] = heatflux(v, p, r , angle, T, laserpower);
T = T_end + delta_T;
T(i) = T_end + delta_T;
i = i+1;
end
plot (T)
end
Renato Agurto on 29 Apr 2016
I think you already have the answer in your code. The problem is that you are overwriting T in the following line
T = T_end + delta_T;
.
%You need to give the inicial value to T
T(1) = ...
%or
T(i-1) = ...
if true
%
while i<N
r = r_initial - i*p/tan(angle);
[Q(i) , delta_T(i), T_end(i) ] = heatflux(v, p, r , angle, T(i-1), laserpower);
T(i) = T_end(i) + delta_T(i);
i = i+1;
end
plot (T)
end | 421 | 1,453 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.953125 | 3 | CC-MAIN-2020-16 | latest | en | 0.850628 |
https://www.coursehero.com/file/8790917/HW1b1/ | 1,642,386,605,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320300253.51/warc/CC-MAIN-20220117000754-20220117030754-00071.warc.gz | 766,943,194 | 55,358 | # HW1b(1) - Homework 1b 1.5.1 Show that the events Ai i I are...
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Homework 1b 1.5.1 Show that the events A i , i I are independent iff P ( B i 1 . . . B i k ) = P ( B i 1 ) . . . P ( B i k ) for all distinct i 1 , . . . , i k , where each B i r may be either A i r or A c i r . - P ( A i i m +1 ) P ( A i m +2 ) . . . P ( A i k ) -
- P ( A i 1 ) P ( A c i 2 ) . . . P ( A c i m +1 ) P ( A m +2 ) . . . P ( A i k ) = (1 - P ( A i 1 )) P ( A c i 2 ) . . . P ( A c i m +1 ) P ( A m +2 ) . . . P ( A i k ) = = P ( A c i 1 ) P ( A c i 2 ) . . . P ( A c i m +1 ) P ( A m +2 ) . . . P ( A i k ) where the third equality holds by inductive assumtions (in each intersection there are no more than m complements). Therefore the claim holds by induction. 1.5.3 | 355 | 866 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2022-05 | latest | en | 0.771759 |
https://www.basculasbalanzas.com/2022/03/17/ | 1,716,276,283,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058385.38/warc/CC-MAIN-20240521060250-20240521090250-00411.warc.gz | 586,497,069 | 10,407 | # What Is a Scale?
The term “scale” refers to the ratio of actual dimensions to model ones. It helps in representing real-world objects on paper that have smaller dimensions. A scale is typically found on maps or construction blueprints. A scale can be used to describe a relative size of two things. It also has special applications in medicine and engineering. Various types of scales are available. If you want to know more about scales, read on. This article will explain some of the most common scales.
The word scale can be used to define a number of things. It can mean the amount of something, the importance of it, or the rank it holds. A metric scale is the simplest and most popular unit of length. It is based on the distance between the equator and the north pole. A common example of a metric scale is a barometer, which measures the diameter of a football field. A gram, on the other hand, is one-tenth the size of a baseball.
A common use for the word scale in the field of geography is to define the relation between distance on a map and a corresponding distance on the ground. For example, a 1:100000 scale map, which refers to 1cm on a map, means that it corresponds to one kilometer on the ground. Geographical researchers also use scale in themed maps and digital products. A nominal scale refers to an assignment of events to discrete categories that have no numerical values. Each distinct category is labeled by a unique identifier.
In ecology, the term “scale” is used to refer to the objective relation among processes and the levels of organization they produce. Similarly, it is a crucial concept in geographic studies. Geographers study phenomena in different scales to develop a more complete understanding of the environment. A LAB scale is an accurate measuring device. It has a resolution of a gram and an accuracy of 1% of its measuring range. These measurements are important to the development of scientific knowledge of the world’s environment.
A meter is a tenth of a centimeter, which is about one tenth of a kilometer. In physics, the meter is a tenth of one millionth of a mile, which is considered a millimeter. In chemistry, a meter is equivalent to one thousand thousandth of a kilometer, but in other fields, it is a single-digit measurement.
A scale is an essential part of scientific data analysis. Its meanings can vary from one person to another, but they all relate to the way numbers are represented on a map. If a scale is used to describe the size of an object, it means it is larger than a human, for example. It is often used to measure the volume of an object, and it can refer to the size of a person. Similarly, a scale can be defined as a ratio of two different objects.
In terms of scale, a ratio refers to a proportion between a given quantity and a given area. In other words, a scale measures a unit of weight. In chemistry, a unit of mass is the length of a mass. A ratio of two dimensions is a proportion. It is the measure of a specific area. A measurement of one number is a fraction of a kilogram. It can be a fraction of a kilogram.
A scale is a measurement that measures the size of a quantity or a person. It can be a ratio of two quantities. The scale is also used in scientific research. It can be a quantitative measurement. For example, a survey of the population, which measures the size of the population, uses a multivariate distribution of data, will yield a score based on the size of the people involved. It is a measure of quality of a population.
A scale is a factor of a given quantity or area. In ecology, it refers to the level of organization and objective relations among processes. Using a scale is essential to geographers in their work. Because they study phenomena on different scales, they can develop a scientific understanding of the environment. For example, a LAB scale is a very precise measurement device with a resolution of one gram and an accuracy of 0.01% of its measuring range. | 841 | 3,992 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.046875 | 3 | CC-MAIN-2024-22 | latest | en | 0.944813 |
http://www.xpmath.com/forums/showthread.php?s=c76a60f00de0f5cc388a40c03e961264&t=2031&goto=nextnewest | 1,597,132,279,000,000,000 | text/html | crawl-data/CC-MAIN-2020-34/segments/1596439738735.44/warc/CC-MAIN-20200811055449-20200811085449-00043.warc.gz | 192,662,819 | 10,644 | Can someone check my algebra answers and help me with the problems I dont understand? - XP Math - Forums
XP Math - Forums Can someone check my algebra answers and help me with the problems I dont understand?
07-31-2007 #1 joe b Guest Posts: n/a Can someone check my algebra answers and help me with the problems I dont understand? 1. n^2-n-20/2n^2 times n^2+5n/n^2-25answer: n+4/2n2. (2x-6/21)/(5x-15/12)answer: 8/353. 5x-3/6 - x+3/6answer: 2x/3-14. 5x/26 - 2x/13answer: 1x/265. 3x^2/9x^5answer: 1/3x^36. (x^2+2x-15/4x^2)/(x^2-25/2x-10)answer: I'm not to sure but i think its x-3/2x^27. x+2/x/x-2/3xanswer: 3(x+2)/x-28. What values for x must be excluded in the following fraction?9/x-8answer: 89. x-13/x^2-x-12 + 3x-3/x^2-x-12answer: 4/x+3
07-31-2007 #2 shilohkid13 Guest Posts: n/a All of your answers are correct except #3, it should be 2x/3
07-31-2007 #3 The Dark Soldier Guest Posts: n/a 1. n + 72. 16/353. 4x4. you are correct5. you are correct6.3/7. Is the problem complete?7. 5/3x +28. you are correct9. (x - 12)/(5x^2)Edit: Apparently my answers don't match the rest. Take their word over mine. I'm going into 8th grade math nexxt year.
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Contact Us - XP Math - Forums - Archive - Privacy Statement - Top | 589 | 1,768 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.25 | 3 | CC-MAIN-2020-34 | latest | en | 0.796786 |
https://www.mecharithm.com/tag/robot-arm/ | 1,675,900,161,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764500983.76/warc/CC-MAIN-20230208222635-20230209012635-00306.warc.gz | 868,046,536 | 26,614 | ### Task Space and Workspace for Robots
In this post, you will learn about the task space and workspace for robots and understand their difference with the robot's C-space. You can find a comprehensive post about the C-space of the robot HERE. โ ๏ธ This post also has a video version that complements the reading. Our suggestion is to watch the video and then read the reading for a deeper understanding. We saw before that the robot's C-space is the space of all possible configurations of a robot. But what does task space and workspace mean for robots? If you want to learn about task space and workspace,…
### Holonomic vs. Nonholonomic Constraints for Robots
In this post, you will learn that holonomic or configuration constraints reduce the degrees of freedom (dofs) of a robot, whereas nonholonomic constraints reduce the space of possible velocities. โ ๏ธ This post also has a video version that complements the reading. Our suggestion is to watch the video and then read the reading for a deeper understanding. Holonomic (Configuration) Constraints for Robots Letโs see this with an example. Remember our famous 4-bar linkage with one degree of freedom? If you do not, please refer to this lesson. For now, consider a 4-bar linkage that has 4 links (ground is a…
### Computed Torque Control of a 3R Robot Arm Playing Ping Pong
In this post, you will find the MATLAB and Simulink codes for equations of motions of a 3R (3 revolute joints) robot, Computed Torque Control of the 3R robot arm in joint and task spaces, and the 3R robot arm playing Ping Pong. Suppose we have a 3R robot arm with 3 revolute joints pictured in the figure below: Fig. 1. A 3R (3 revolute joints) Serial Manipulator Robot The following problems are going to deal with the simulations of the equations of motion of this robot using the Lagrange-Euler method, Computed Torque Control (CTC) of the 3R robot in… | 439 | 1,903 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3 | 3 | CC-MAIN-2023-06 | latest | en | 0.906178 |
https://www.teacherspayteachers.com/Product/Year-5-Fractions-Decimals-Smart-Notebook-and-Unit-of-Work-Bundle-1-3237901 | 1,521,786,151,000,000,000 | text/html | crawl-data/CC-MAIN-2018-13/segments/1521257648178.42/warc/CC-MAIN-20180323044127-20180323064127-00509.warc.gz | 879,257,682 | 18,500 | Total:
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# Year 5 Fractions & Decimals Smart Notebook and Unit of Work Bundle 1
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NOTEBOOK (SMARTboard) File
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This NO PREP Maths Bundle is programmed for the Australian National Curriculum based on the Maths Plus and Go Maths programs. However, it can be tailored to suit other nations’ curriculum outcomes.
This is a perfect time saver for you and your classroom. Have smooth transitions between lessons. If you ever have a sick day, don’t stress! Just save this to a USB/ Flashdrive and have your casual plug it in and go! NO PREP needed on your behalf!
You will notice that the smart notebook title is called Stage 3a. This is referring to the Australian grade groupings that are referred to as ‘stages’. Stage 3 being years 5 and 6. Because this is a year 5 program, but students in year 6 may be working at a year 5 level, I have called it stage 3a. This is completely editable.
***INCLUDED***
Explicit and engaging ***EDITABLE*** maths unit of work that will have your students captivated for the duration of the lesson.
An entire smart notebook with interactive games and thought provoking lesson content.
An interactive MULTISTEP WORD PROBLEMS section that will ensure students are regularly practicing this skill that most students find difficult!
This is just one of many for the entire year!!!
Take the stress out of planning and organising. Walk into your classroom feeling organised and prepared to teach your lesson.
Outcomes and Content Addressed and Lessons:
MA3-1WM describes and represents mathematical situations in a variety of ways using mathematical terminology and some conventions
MA3-2WM selects and applies appropriate problem-solving strategies, including the use of digital technologies, in undertaking investigations
MA3-3WM gives a valid reason for supporting one possible solution over another
MA3-7NA compares, orders and calculates with fractions, decimals and percentages
Recognise and represent fractions in different ways.
Locate, represent and order fractions on a number line.
Find fractional parts of collections.
Add and subtract fractions with like denominators.
Lesson 1 Thirds, Sixths and Twelfths
Lesson 2 Fractions on a Number Line
Lesson 3 Fractions of a Collection
Lesson 4 Adding and Subtracting Fractions 1
Lesson 5 Adding and Subtracting Fractions 2
****HEAD OVER TO FREE Taste Year 6 Patterns and Algebra Smart Notebook and Unit of Work Bundle FOR A FREE TASTE OF MY SMART NOTEBOOK AND UNIT OF WORK BUNDLES!!!****
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\$7.00 | 806 | 3,672 | {"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} | 2.8125 | 3 | CC-MAIN-2018-13 | latest | en | 0.912048 |
https://lists.ellipsis.cx/archives/spoon-business/spoon-business-200202/msg00017.html | 1,714,005,557,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296820065.92/warc/CC-MAIN-20240425000826-20240425030826-00089.warc.gz | 317,121,059 | 2,753 | Wonko on 4 Feb 2002 03:11:11 -0000
• From: Wonko <dplepage@xxxxxxxxxxxx>
• Date: Sun, 03 Feb 2002 22:11:08 -0500
• Blobman: 1
• User-agent: Microsoft-Outlook-Express-Macintosh-Edition/5.02.2022
```I issue this proposal:
{{
__You can only throw so far...__
Amend the rule created by proposal 330 to replace the sentence 'A Player in
a grid location can throw a Gremlin at another grid location, at a Player in
another grid location, into Limbo or at a Player in Limbo.' with 'A Player
in a grid location can throw a Gremlin at another grid location or at a
Player in another grid location, assuming the target is within 7 units of
the thrower, or into Limbo or at a Player in Limbo.'
Append to rule 301 the sentence 'The distance between two squares is the
square root of (the difference in x-values of the squares squared plus the
difference in y-values of the squares squared)'
}}
I amend my proposal entitled "Let's Get Ready To GREMBALL!!!!"
{{
Create:
{{
__Hungry Hungry Hippos__
There exist four Passable SOs, called Hungry Hungry Hippos, or Hippos. There
is one at (1,10), one at (11,1), one at (20,11) and one at (10,20). Whenever
a MO occupies a square which is occupied by a Hippo, the Hippo Eats that MO.
When a MO is Eaten by a Hippo, it is teleported to a random vacant square in
the Relocation Zone. The Relocation Zone consists of all squares whose x and
y coordinates are both greater than 5 and less than 15.
}}
{{
__The Game__
There exists a list called the Gremball Roster, or just the Roster. Any
player may put emself on the Roster by declaring so in a Public Forum, in
which case e shall be added to the Roster at the end of that nweek. There
exist two groups of players, called Teams, known as the Red Team and the
Yellow Team. No player on the Roster can be on more than 1 Team. At the
beginning of each nweek, each player on the Roster shall be randomly
assigned to a Team, in such a way so that neither Team outnumbers the other
by more than 1 player.
If a player who is holding a Gremlin is Eaten by a Hippo, e and all players
on eir team gain 2 points, and e loses possession of that Gremlin before
being teleported by the Hippo.
}}
{{
__Magnets__
There exist objects called Magnets. All players have magnets. If a player
with a Magnet is within 3 squares of a player who holds a Gremlin, the
player with a Magnet may Seize the other player's Gremlin, in which case the
Gremlin moves to the location and into the possession of the Seizing player.
}}
}}
--
Wonko
Utinam logica falsa tuam philosophiam totam suffodiant!
``` | 701 | 2,562 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2024-18 | latest | en | 0.919617 |
https://exercism.io/tracks/clojure/exercises/poker/solutions/8ea37bdf1f9e4de6b807768f0b5ea187 | 1,624,267,681,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623488269939.53/warc/CC-MAIN-20210621085922-20210621115922-00495.warc.gz | 229,033,122 | 7,777 | 🎉 Exercism Research is now launched. Help Exercism, help science and have some fun at research.exercism.io 🎉
# eeng's solution
## to Poker in the Clojure Track
Published at Jun 11 2020 · 0 comments
Instructions
Test suite
Solution
Pick the best hand(s) from a list of poker hands.
See wikipedia for an overview of poker hands.
## Source
Inspired by the training course from Udacity. https://www.udacity.com/course/viewer#!/c-cs212/
## Submitting Incomplete Solutions
It's possible to submit an incomplete solution so you can see how others have completed the exercise.
### poker_test.clj
``````(ns poker-test
(:require [clojure.test :refer [deftest is]]
[poker :refer [best-hands]]))
(defn f [xs ys] (= (sort (best-hands xs)) (sort ys)))
(deftest single-hand-always-wins
(is (f ["4S 5S 7H 8D JC"] ["4S 5S 7H 8D JC"])))
(deftest highest-card-out-of-all-hands-wins
(is (f ["4D 5S 6S 8D 3C"
"2S 4C 7S 9H 10H"
"3S 4S 5D 6H JH"]
["3S 4S 5D 6H JH"])))
(deftest a-tie-has-multiple-winners
(is (f ["4D 5S 6S 8D 3C"
"2S 4C 7S 9H 10H"
"3S 4S 5D 6H JH"
"3H 4H 5C 6C JD"]
["3S 4S 5D 6H JH"
"3H 4H 5C 6C JD"])))
(deftest multiple-hands-with-the-same-high-cards-tie-compares-next-highest-ranked-down-to-last-card
(is (f ["3S 5H 6S 8D 7H"
"2S 5D 6D 8C 7S"]
["3S 5H 6S 8D 7H"])))
(deftest one-pair-beats-high-card
(is (f ["4S 5H 6C 8D KH"
"2S 4H 6S 4D JH"]
["2S 4H 6S 4D JH"])))
(deftest highest-pair-wins
(is (f ["4S 2H 6S 2D JH"
"2S 4H 6C 4D JD"]
["2S 4H 6C 4D JD"])))
(deftest two-pairs-beats-one-pair
(is (f ["2S 8H 6S 8D JH"
"4S 5H 4C 8C 5C"]
["4S 5H 4C 8C 5C"])))
(deftest both-hands-have-two-pairs-highest-ranked-pair-wins
(is (f ["2S 8H 2D 8D 3H"
"4S 5H 4C 8S 5D"]
["2S 8H 2D 8D 3H"])))
(deftest both-hands-have-two-pairs-with-the-same-highest-ranked-pair-tie-goes-to-low-pair
(is (f ["2S QS 2C QD JH"
"JD QH JS 8D QC"]
["JD QH JS 8D QC"])))
(deftest both-hands-have-two-identically-ranked-pairs-tie-goes-to-remaining-card-kicker
(is (f ["JD QH JS 8D QC"
"JS QS JC 2D QD"]
["JD QH JS 8D QC"])))
(deftest three-of-a-kind-beats-two-pair
(is (f ["2S 8H 2H 8D JH"
"4S 5H 4C 8S 4H"]
["4S 5H 4C 8S 4H"])))
(deftest both-hands-have-three-of-a-kind-tie-goes-to-highest-ranked-triplet
(is (f ["2S 2H 2C 8D JH"
(deftest with-multiple-decks-two-players-can-have-same-three-of-a-kind-ties-go-to-highest-remaining-cards
(is (f ["4S AH AS 7C AD"
(deftest a-straight-beats-three-of-a-kind
(is (f ["4S 5H 4C 8D 4H"
"3S 4D 2S 6D 5C"]
["3S 4D 2S 6D 5C"])))
(deftest aces-can-end-a-straight-10-J-Q-K-A
(is (f ["4S 5H 4C 8D 4H"
"10D JH QS KD AC"]
["10D JH QS KD AC"])))
(deftest aces-can-start-a-straight-A-2-3-4-5
(is (f ["4S 5H 4C 8D 4H"
"4D AH 3S 2D 5C"]
["4D AH 3S 2D 5C"])))
(deftest both-hands-with-a-straight-tie-goes-to-highest-ranked-card
(is (f ["4S 6C 7S 8D 5H"
"5S 7H 8S 9D 6H"]
["5S 7H 8S 9D 6H"])))
(deftest even-though-an-ace-is-usually-high-a-5-high-straight-is-the-lowest-scoring-straight
(is (f ["2H 3C 4D 5D 6H"
"4S AH 3S 2D 5H"]
["2H 3C 4D 5D 6H"])))
(deftest flush-beats-a-straight
(is (f ["4C 6H 7D 8D 5H"
"2S 4S 5S 6S 7S"]
["2S 4S 5S 6S 7S"])))
(deftest both-hands-have-a-flush-tie-goes-to-high-card-down-to-the-last-one-if-necessary
(is (f ["4H 7H 8H 9H 6H"
"2S 4S 5S 6S 7S"]
["4H 7H 8H 9H 6H"])))
(deftest full-house-beats-a-flush
(is (f ["3H 6H 7H 8H 5H"
"4S 5H 4C 5D 4H"]
["4S 5H 4C 5D 4H"])))
(deftest both-hands-have-a-full-house-tie-goes-to-highest-ranked-triplet
(is (f ["4H 4S 4D 9S 9D"
"5H 5S 5D 8S 8D"]
["5H 5S 5D 8S 8D"])))
(deftest with-multiple-decks-both-hands-have-a-full-house-with-the-same-triplet-tie-goes-to-the-pair
(is (f ["5H 5S 5D 9S 9D"
"5H 5S 5D 8S 8D"]
["5H 5S 5D 9S 9D"])))
(deftest four-of-a-kind-beats-a-full-house
(is (f ["4S 5H 4D 5D 4H"
"3S 3H 2S 3D 3C"]
["3S 3H 2S 3D 3C"])))
(is (f ["2S 2H 2C 8D 2D"
"4S 5H 5S 5D 5C"]
["4S 5H 5S 5D 5C"])))
(deftest with-multiple-decks-both-hands-with-identical-four-of-a-kind-tie-determined-by-kicker
(is (f ["3S 3H 2S 3D 3C"
"3S 3H 4S 3D 3C"]
["3S 3H 4S 3D 3C"])))
(deftest straight-flush-beats-four-of-a-kind
(is (f ["4S 5H 5S 5D 5C"
"7S 8S 9S 6S 10S"]
["7S 8S 9S 6S 10S"])))
(deftest both-hands-have-straight-flush-tie-goes-to-highest-ranked-card
(is (f ["4H 6H 7H 8H 5H"
"5S 7S 8S 9S 6S"]
["5S 7S 8S 9S 6S"])))``````
``````(ns poker
(:require [clojure.string :as str]))
(def cards-ranks
{"2" 2 "3" 3 "4" 4 "5" 5 "6" 6 "7" 7 "8" 8 "9" 9 "10" 10 "J" 11 "Q" 12 "K" 13 "A" 14})
(defn- parse-card [card]
(let [[rank suit] (rest (re-find #"(.+)([CDHS])\$" card))]
{:rank (cards-ranks rank) :suit suit}))
(defn- parse-hand [hand]
(->> (str/split hand #" ")
(map parse-card)))
(defn- ranks [cards]
(let [rs (->> cards (map :rank) sort)]
(if (= rs [2 3 4 5 14]) ; Ace starting a straight?
[1 2 3 4 5]
rs)))
(defn- same-suit? [cards]
(->> cards (map :suit) (apply =)))
(defn- sequential-rank? [cards]
(->> cards ranks (partition 2 1) (map #(apply - %)) (every? #(= % -1))))
(defn- groups-by-rank [hand]
(->> (ranks hand)
(frequencies)
(sort-by (fn [[rank count]] [(- count) (- rank)]))
(map (fn [[rank count]] {:rank rank :count count}))))
(defn- card-count-by-group [hand]
(->> hand groups-by-rank (map :count)))
(defn- straight-flush? [hand]
(and (same-suit? hand)
(sequential-rank? hand)))
(defn- four-of-a-kind? [hand]
(= [4 1] (card-count-by-group hand)))
(defn- full-house? [hand]
(= [3 2] (card-count-by-group hand)))
(defn- flush? [hand]
(and (same-suit? hand)
(not (sequential-rank? hand))))
(defn- straight? [hand]
(and (not (same-suit? hand))
(sequential-rank? hand)))
(defn- three-of-a-kind? [hand]
(= [3 1 1] (card-count-by-group hand)))
(defn- two-pair? [hand]
(= [2 2 1] (card-count-by-group hand)))
(defn- one-pair? [hand]
(= [2 1 1 1] (card-count-by-group hand)))
(defn- high-card? [_hand]
true)
(def categories
[straight-flush?
four-of-a-kind?
full-house?
flush?
straight?
three-of-a-kind?
two-pair?
one-pair?
high-card?])
(defn- category-rank [hand]
(->> categories (keep-indexed #(when (%2 hand) %1)) first))
(defn- hand-score [hand]
(let [hand (parse-hand hand)]
[(category-rank hand)
(mapv (comp - :rank) (groups-by-rank hand))]))
(defn- rank-hands [hands]
(->> hands
(map (juxt identity hand-score))
(sort-by second)))
(defn best-hands [hands]
(let [[[_ best-score] :as ranked-hands] (rank-hands hands)]
(->> ranked-hands
(take-while #(= (second %) best-score))
(map first))))`````` | 2,798 | 6,331 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.59375 | 4 | CC-MAIN-2021-25 | latest | en | 0.718877 |
http://www.topperlearning.com/forums/ask-experts-19/how-can-you-prove-that-there-are-more-irrational-numbers-tha-mathematics-real-numbers-47556/reply | 1,493,374,705,000,000,000 | text/html | crawl-data/CC-MAIN-2017-17/segments/1492917122933.39/warc/CC-MAIN-20170423031202-00066-ip-10-145-167-34.ec2.internal.warc.gz | 727,713,600 | 37,420 | Question
Mon April 25, 2011 By:
# How can you prove that there are more irrational numbers than rational numbers between any set of two numbers?
Mon April 25, 2011
While it is true that there are an infinite number of both rational and irrational numbers, you should know that, in a particular mathematical sense, there are way more irrational numbers than rational.
The reason for this is that there are two basic different types of 'infinite' sets - countably infinite and uncountably infinite.
Rationals comprise a countably infinite set. What this means is that you can assign an order among rationals, so you can say that this is the first and this is the second and so on. But of course, you will never end the process.
But the Irrationals are an uncountably infinite set. What this means is that you CANNOT even assign an order among irrationals. In this sense, there are more irrationals that rationals.
Another thing is that people say that both are equal since both are infinite. However infinite is not a number, so saying infinite = infinite is absurd.
You can read up more with at the page given in the reference.
### Source(s):
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Question
Write a function rule based on the table below.
x f(x)
1 5
2 10
3 15
f(x) = x + 4
f(x) = 5x + 2
f(x) = 5x
f(x) = 5
in progress 0
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#### Storyboard Description
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#### Storyboard Text
• scene 1
• scene 2
• scene 3
• *starts in classroom* yesenia: alright class today we're going to learn how to calculate the number of calories per gram in a bbq chip and compare it to the actual nutrition label.
• scene 4
• jacob: yay im so excited!! james: lets get started
• scene 5
• yesenia: " okay we are going to use a beaker of water and burn the chip underneath we will see how the heat from the chip affects the temperature of the water" we need to record the inital temperature of the water and volume of the water.
• jacob: the initial volume of the water in 100ml. james: and the initial temperature is 20 degrees celcius & the chip weighs 1.50g. yesenia : perfect now we need to calculate heat gained by h20 , so we take the chip mass and subtract the mass of chip burned so 1.50-0.05 which is 1.45g. now tome for caclulations.
• jacob: are we going to use the equation q=mc(delta)t? james : duh jacob what else would we use?
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## Presentation on theme: "Monte Carlo Tree Search A Tutorial"— Presentation transcript:
Monte Carlo Tree Search A Tutorial
Supported by the Engineering and Physical Sciences Research Council (EPSRC) Peter Cowling University of York, UK Simon Lucas University of Essex, UK
Further Reading Browne C., Powley E., Whitehouse D., Lucas S., Cowling P.I., Rohlfshagen P., Tavener S., Perez D. and Samothrakis S., Colton S. (2012): "A Survey of Monte Carlo Tree Search Methods" IEEE Transactions on Computational Intelligence and AI in Games, IEEE, 4 (1): 1-43.
Monte Carlo Tree Search (MCTS) Why know more?
Go: Strong/World Champion MCTS players in: General Game Playing, Hex, Amazons, Arimaa, Khet, Shogi, Mancala, Morpion, Blokus, Focus, Chinese Checkers, Yavalath, Connect Four, Gomoku Morpion solitaire best ever Over 250 papers since about one per week on average. Promise for planning, real-time games, nondeterministic games, single-player puzzles, optimisation, simulation, PCG, …
Tutorial Structure Monte Carlo search Monte Carlo Tree Search (MCTS)
The multi-armed bandit problem Monte Carlo Tree Search (MCTS) Selection, expansion, simulation, backpropagation Enhancements to the basic algorithm Handling uncertainty in MCTS MCTS for real-time games Conclusions, future research directions
Monte Carlo Search for Decision Making in Games
No trees yet …
The Multi-Armed Bandit Problem
At each step pull one arm Noisy/random reward signal In order to: Find the best arm Minimise regret Maximise expected return Show first part of Excel simulation
Which Arm to Pull? Mean so far Upper bound on variance
UCB1 (Auer et al (2002)). Choose arm j so as to maximise: Flat Monte Carlo Share trials uniformly between arms ε-Greedy P(1- ε) – Best arm so far P(ε) – Random arm Show Flat MC, epsilon-greedy and UCB1 part of Excel demo Mean so far Upper bound on variance
Simulation Result (+1 = win, 0 = loss)
Game Decisions Current position Multi-Armed Bandit Arm A Arm C Arm B Move C Move A Move B Position after move A Position after move B Position after move C Show Connect 4 with Flat, UCB1 players Emphasise simulation result = random game = arm pull Simulation Result (+1 = win, 0 = loss)
Discussion Anytime – stop whenever you like
UCB1 formula minimises regret Grows like log(n) Needs only game rules: Move generation Terminal state evaluation Surprisingly effective, but… … doesn’t look ahead GIB here, Snails example here if time Move generation and terminal state evaluation allow random simulated games to be run
MCTS / UCT Basics
Attractive Features Anytime Scalable No need for heuristic function
Tackle complex games better than before May be logarithmically better with increased CPU No need for heuristic function Though usually better with one Next we’ll look at: General MCTS UCT in particular Note: this slide may be unnecessary – might be covered earlier
MCTS: the main idea Tree policy: choose which node to expand (not necessarily leaf of tree) Default (simulation) policy: random playout until end of game
MCTS Algorithm Decompose into 6 parts: MCTS main algorithm
Tree policy Expand Best Child (UCT Formula) Default Policy Back-propagate We’ll run through these then show demos
MCTS Main Algorithm Check this point about BestChild BestChild simply picks best child node of root according to some criteria: e.g. best mean value In our pseudo-code BestChild is called from TreePolicy and from MctsSearch, but different versions can be used E.g. final selection can be the max value child or the most frequently visited one
TreePolicy Note that node selected for expansion does not need to be a leaf of the tree The nonterminal test refers to the game state
Expand Note: need to choose an appropriate data structure for the children (array, ArrayList, Map)
Best Child (UCT) This is the standard UCT equation
Used in the tree Higher values of c lead to more exploration Other terms can be added, and usually are More on this later
DefaultPolicy Each time a new node is added to the tree, the default policy randomly rolls out from the current state until a terminal state of the game is reached The standard is to do this uniformly randomly But better performance may be obtained by biasing with knowledge
Backup Note that v is the new node added to the tree by the tree policy Back up the values from the added node up the tree to the root
MCTS Builds Asymmetric Trees (demo)
Connect 4 Demo
Enhancements to the Basic MCTS Algorithm
Tree Policy Default Policy Domain: independent or specific
Selection/Expansion Enhancements (just a sample …)
AMAF / RAVE First Play Urgency Learning E.g. Temporal Difference Learning to bias tree policy or default policy Bandit enhancements E.g. Bayesian selection Parameter Tuning
All Moves As First (AMAF), Rapid Value Action Estimates (RAVE)
Additional term in UCT equation: Treat actions / moves the same independently of where they occur in the move sequence
Simulation Enhancements
Nudge the simulation to be more realistic Without introducing bias At very low CPU cost Examples: Move-Average Sampling Technique (MAST) Last Good Reply (LGR) Patterns e.g. Bridge Completion: Heuristic value functions Paradoxically better simulation policy sometimes yields weaker play MAST – Finsson and Bjornsson – CADIAPLAYER for GGP - Maintain a table for each action independent of state. Bias action selection towards effective action using Gibbs sampling. Also PAST/FAST where predicates/features related to the state are checked and good predicates are preferentially chosen during simulation. LGR – Drake – Go – For each move remember the last reply to that move that resulted in a winning simulation. Drake and Baier added forgetting (LGRF) if a reply is subsequently shown to lead to a loss. Patterns – Go – Use local configurations to choose or ignore moves. Heuristic value functions – Winands + Bjornsson – Lines of Action – bias towards high value moves later in search. Paradox – Gelly and Silver – RLGO – presumably bias is the problem.
Hex Demo Cameron’s Hex demo: show MCTS variants
Parallelisation Figure from Chaslot, Winands, van den Herik.
Leaf parallelisation – many simulations per new leaf – problem of waiting for simulations to finish Root parallelisation – combine trees (at level 1) found independently (also known as ensemble MCTS). Majority voting beats averaging (Soejima, Kishimoto, Watanabe). Tree parallelisation – use a single tree accessed by many processes. Lock whole tree if Sims much slower than updates, else local locks. Also lock-free parallelisation (Enzenberger and Muller). Use “virtual loss” to diversify search away from simulations under investigation. Root parallelisation wins. We also find this with determinisation in games with hidden info and uncertainty
Handling Uncertainty and Incomplete Information in MCTS
Probabilistic (stochastic) nodes
CHANCE MAX MIN MAX Probabilistic (stochastic) planning E,g, dynamic scheduling Probabilistic planning – tends to have low branching factor at probabilty nodes – but card shuffling gives a gigantic branching factor Decision-making/optimisation under uncertainty
{ , , , ...} Hidden Information Information set: Actual state:
Observation: , ...} E.g. Poker, Bridge, Security
Effects of Uncertainty and Hidden Information on the Game Tree
choose subset 4 possible plays by me 50 possible random card draws 40C3 = 9880 different opponent plays My opponent has to play 3 out of 4 cards ...
... Reduced Branching Through Determinization
4 possible plays by me 1 possible random card draws 4C3 = 4 different opponent plays FF-replan, GIB, Maven, …, determinisation idea. ... Use average results over many determinizations [Yoon, Fern & Givan 2007]
Determinization Sample states from the information set
Analyse the individual perfect information games Combine the results at the end Successes Bridge (Ginsberg), Scrabble (Sheppard) Klondike solitaire (Bjarnason, Fern and Tadepalli) Probabilistic planning (Yoon, Fern and Givan) Problems Never tries to gather or hide information Suffers from strategy fusion and non-locality (Frank and Basin) GIB, Maven, more recent Scrabble work by Childs et al on opponent modelling/computing a good prior. Klondike solitaire – using UCT and other approache. Win rate about 37% - much higher than thought possible. FF Replan, uses FF and determinization. Strategy fusion – assumes different decisions possible based on unknown information. “If my opponent holds an Ace, I’ll do that, otherwise I’ll do this” – but whether or not the oppnent holds an ace is unknown. Non-locality – The opponent is leading us into situations unfavourable for us – so some parts of the state space become very unlikely. M.L. Ginsberg, “GIB: Imperfect information in a computationally challenging game,” Journal of Artificial Intelligence Research, vol. 14, 2002, p. 313–368. R. Bjarnason, A. Fern, and P. Tadepalli, “Lower bounding Klondike solitaire with Monte-Carlo planning,” Proc. ICAPS-2009, p. 26–33. S. Yoon, A. Fern, and R. Givan, “FF-Replan: A Baseline for Probabilistic Planning,” 1Proc. ICAPS-2007, p. 352–359. I. Frank and D. Basin, “Search in games with incomplete information: a case study using Bridge card play,” Artificial Intelligence, vol. 100, 1998, pp
Cheating Easiest approach to AI for games with imperfect information: cheat and look at the hidden information and outcomes of future chance events This gives a deterministic game of perfect information, which can be searched with standard techniques (minimax, UCT, …) Obviously not a reasonable approach in practice, but gives a useful bound on performance Much used in commercial games Has: Perfect Inference Clairvoyance No Strategy Fusion or non-locality
Information Set Monte Carlo Tree Search (ISMCTS)
Maintain a single tree Use each determinization only once Determinizations restrict search to a subtree of the information set tree. Single observer and multi=observer versions to handle partially observable moves Collect information about the value of a decision in the corresponding information set.
Lord of the Rings: The Confrontation
Board game designed by Reiner Knizia Hidden information: can’t see the identities of opponent’s pieces Simultaneous moves: combat is resolved by both players simultaneously choosing a card Also Phantom 4,4,4, Dou Di Zhu and several other card games…
MCTS for Real-Time Games
Limited roll-out budget Heuristic knowledge becomes important Action space may be too fine-grained Take Macro-actions Otherwise planning will be very short-term May be no terminal node in sight Again, use heuristic Tune simulation depth Next-state function may be expensive Consider making a simpler abstraction
PTSP Demo
Summary MCTS: exciting area of research
Many impressive achievements already With many more to come Some interesting directions: Applications beyond games Comparisons with other search methods Heuristic knowledge is important – more work needed on learning it Efficient parameter tuning POMDPs Uncertainty Macro-actions Hybridisation with other CI methods e.g. Evolution
Questions? Browne C., Powley E., Whitehouse D., Lucas S., Cowling P.I., Rohlfshagen P., Tavener S., Perez D. and Samothrakis S., Colton S. (2012): "A Survey of Monte Carlo Tree Search Methods" IEEE Transactions on Computational Intelligence and AI in Games, IEEE, 4 (1): 1-43.
Advertising Break Sep 28: Essex workshop on AI and Games
Strong MCTS Theme Organised by Essex Game Intelligence Group Next week: AIGAMEDEV Vienna AI summit Find out what games industry people think Resistance competition and MCTS tutorial (!) | 2,592 | 11,633 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.65625 | 3 | CC-MAIN-2017-39 | longest | en | 0.671939 |
http://www.answers.com/Q/How_does_a_watt-hour_meter_work | 1,524,702,117,000,000,000 | text/html | crawl-data/CC-MAIN-2018-17/segments/1524125948029.93/warc/CC-MAIN-20180425232612-20180426012612-00103.warc.gz | 356,842,864 | 49,944 | What would you like to do?
# How does a watt-hour meter work?
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There are a couple of kinds of watt-hour meters: one is an electromechanical type (the one with the spinning wheel), and the other is a solid-state type. The solid-state types generally use a current transformer to measure the current produced in the current-carrying conductors, which means that the current carrying conductors need not be connected to the actual measuring device.) The electromechanical type uses an aluminum disc acted upon by two coils. One coil is positioned in such a way as to produce a magnetic flux in proportion to the voltage on the disc, and the other coil produces a magnetic flux proportional to the current. This causes the disc to rotate at a speed proportional to the power (voltage times current) being used. A permanent magnet is also installed in such a way as to exert an opposing force in proportion to the rotating speed of the disc. This causes the disc to slow down and stop when energy stops being used. The aluminum disc is also attached to a worm gear which drives the register.
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# What does a watt meter do?
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# What does 5c stand for?
The Iphone 5C is Iphone 5Colorful 5c can also stand for thenumber 500 ("c" is the Roman numeral for 100) or for 5 degreesCelsius (centigrade) . +++ . "5c" can not stand fo
# How do you work out square meters into meters?
If and only if you are working with a square, you get the squareroot of the number
# How do you work square meter back to meter?
You don't. Square meter is a unit for area, meter is for length - they're different things.
# What are the 5cs of leadership?
Confirm, clear, cordon, check, control
# How do you stop a watt hour meter?
By not using any electricity.
# How does a tractor hour meter work?
The older cat hour meters that were mechanically driven off of the fuel pump counted RPM hours. The newer ones are an oil pressure switch and clock hour for hour time no matte
23F
# What are the 5Cs?
If you mean the 5 C's, like you learn in school and such, it's Cotton Citrus Copper Climate Cattle
# What is the difference between the watt hour meter and Var hour meter in terms of internal connection and work theory ie When the meter records watt-hour an when the meter records var hours?
Only watts (power) can produce work (energy) because of the phase relationship of the voltage and current in an alternating current, electric power system. With watts, the por
8c
In Volume
# What is 5c in milliliters?
5cc? cc means cubic centimetres which is equal to ml, so 5ml. if you mean cl, then that is equal to 50ml
In Uncategorized
# How does an hour meter on a mechanical tach work?
A mechanical tach will register 1 engine hour for every 100,000 revolutions. At 1666.6 rpm it will register 1 hour for every hour run. At 800 rpm it will take two hours of r
23 F
# Do skins for iPhone 5 lifeproof work for iPhone 5c lifeproof?
No, the skins won't fit. The iPhone 5 is 123.8 mm (4.87 in) high,58.6 mm (2.31 in) wide and 7.6 mm (0.30 in) high. The iPhone 5c is124.4 mm (4.90 in) high, 59.2 mm (2.33 in) w
# Why does my compass not work on my apple i phone 5c?
There may be stray magnetic fields in your house. Take it outside and run the calibration procedure. | 948 | 3,894 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.921875 | 3 | CC-MAIN-2018-17 | longest | en | 0.942745 |
https://www.doubtnut.com/question-answer/the-number-of-different-four-digit-numbers-that-can-be-formed-with-the-digits-2-3-4-7-and-using-each-26871241 | 1,631,804,535,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780053657.29/warc/CC-MAIN-20210916145123-20210916175123-00237.warc.gz | 781,301,178 | 82,212 | Home
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# The number of different four - digit numbers that can be formed with the digits 2, 3, 4, 7 and using each digit only once is
Step by step solution by experts to help you in doubt clearance & scoring excellent marks in exams.
Updated On: 27-11-2020
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224.9 K+
11.2 K+
Text Solution
1209624100
C
Solution :
Given, digits 2, 3, 4 and 7, we have to form four-digit numbers using these digits. <br> therefore Required number of ways = ""^(4)P_(4) = 4"!" = 4 xx 3 xx 2"!" = 24
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4:56 | 480 | 1,201 | {"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} | 3.890625 | 4 | CC-MAIN-2021-39 | latest | en | 0.509991 |
https://trends.edugorilla.com/cat-2017-lrdi-shift-02-set-02-solved/ | 1,708,961,224,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474660.32/warc/CC-MAIN-20240226130305-20240226160305-00246.warc.gz | 571,861,821 | 33,889 | # CAT 2017 LRDI Shift 02 SET 02 SOLVED - Videos
1
220
CAT LRDI Online Course starts on 27th Jan; to keep yourself updated, please join this whatsapp group:
https://chat.whatsapp.com/C5poc8IZBWI7rXv1hPGARo
This is the solution for the set, which appeared in CAT 2017; find the question set below:
There were seven elective courses — E1 to E7 — running in a specific term in a college. Each of the 300 students enrolled had chosen just one elective from among these seven. However, before the start of the term, E7 was withdrawn as the instructor concerned had left the college. The students who had opted for E7 were allowed to join any of the remaining electives. Also, the students who had chosen other electives were given one chance to change their choice. The table below captures the movement of the students from one elective to another during this process. Movement from one elective to the same elective simply means no movement. Some numbers in the table got accidentally erased; however, it’s known that these were either 0 or 1.
Further, the following are known:
1. Before the change process there were 6 more students in E1 than in E4, but after the reshuffle, the number of students in E4 was 3 more than that in E1.
2. The number of students in E2 increased by 30 after the change process.
3. Before the change process, E4 had 2 more students than E6, while E2 had 10 more students than E3
How many elective courses among E1 to E6 had a decrease in their enrolments after the change process?
1. 4
2. 1
3. 2
4. 3
After the change process, which of the following is the correct sequence of number of students in the six electives E1 to E6?
1. 19, 76, 79, 21, 45, 60
2. 19, 76, 78, 22, 45, 60
3. 18, 76, 79, 23, 43, 61
4. 18, 76, 79, 21, 45, 61
After the change process, which course among E1 to E6 had the largest change in its enrollment as a percentage of its original enrollment?
1. E1
2. E2
3. E3
4. E6
Later, the college imposed a condition that if after the change of electives, the enrolment in any elective (other than E7) dropped to less than 20 students, all the students who had left that course will be required to re- enrol for that elective. Which of the following is a correct sequence of electives in decreasing order of their final enrollments?
1. E2, E3, E6, E5, E1, E4
2. E3, E2, E6, E5, E4, E1
3. E2, E5, E3, E1, E4, E6
4. E2, E3, E5, E6, E1, E3
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source | 714 | 2,463 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.109375 | 3 | CC-MAIN-2024-10 | latest | en | 0.979173 |
https://chemistry.coach/general-chemistry-3/electrochemistry/exercise/523 | 1,722,967,462,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640497907.29/warc/CC-MAIN-20240806161854-20240806191854-00781.warc.gz | 128,238,018 | 19,114 | # Exercise 8 | Electrochemistry
Calculate ΔGrxn of the reaction: Zn (s) + Cu2+ (aq) → Zn 2+ (aq) + Cu (s) knowing that the measured cell voltage at 25°C is 1.12 V.
ΔGrxn = - νeFEcell
νe = stoichiometric coefficient of the electrons = 2
F = Faraday’s constant = 96485 C.mol-1
Ecell = electromotive force emf of the cell = 1.12 V
ΔGrxn = - 2 x 96485 x 1.12 = - 216 kJ.mol-1 | 142 | 375 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.640625 | 3 | CC-MAIN-2024-33 | latest | en | 0.748894 |
https://matsci.org/t/how-to-calculate-yield-pressure-of-a-specific-group-of-atoms/37387 | 1,723,450,348,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641036271.72/warc/CC-MAIN-20240812061749-20240812091749-00129.warc.gz | 295,342,263 | 5,533 | # How to calculate yield pressure of a specific group of atoms
Dear LAMMPS Users,
I want to calculate metal yield pressure of a group of atoms in LAMMPS. Does anyone know how to do that?
Thank you,
Steven
The answer to this question has two parts:
1. how to do the calculation
2. how to interpret the result
Part one is mostly straightforward. The documentation for the compute stress/atom command — LAMMPS documentation has an example that shows how to compute the pressure for the entire simulation cell (assuming periodic boundary conditions, otherwise the volume would be infinity and thus the total pressure must be zero). If you want to do this for a group of atoms, you do things the same way, only that you sum up for that group only and use the volume for that group. However, this is where there is a first problem: while the volume of the total simulation cell is well defined, the volume for a group of atoms is not. Since you have point particles, individual atoms have no volume, and that wouldn’t really matter since you need the “volume they occupy”, and that is not well defined. For irregular shapes it would even be a fractal property, i.e. the volume would depend on the level of detail applied. If you have a periodic and homogeneously dense system, you can compute an “effective volume” by dividing the cell volume by the number of atoms and use that. For an inhomogeneously dense system, you can use voronoi tesselation (compute voronoi/atom command — LAMMPS documentation); for an isolated system, this would only work for groups of atoms that are not on the surface. If the group involves atoms on the surface, then you have to ask yourself whether pressure has a meaning at all (see my comment about pressure for an isolated system).
This leads me to part two of the answer. While looking at pressure for small volumes is straightforward to look at in continuum models of materials, it is not so for atomic (i.e. nanoscale) systems. Pressure (like temperature) is by its nature a “bulk” property and thus using the virial stress formulation and statistical mechanics to interpret per-atom data as bulk data puts some constraints on the interpretation of the results. Due to the small volume and the “discontinuous” data you have large fluctuations; the smaller the volume, the fewer particles the larger. So you can only derive meaningful values when averaging over volume and time. But if the volume changes over time, this becomes increasingly difficult to interpret. Add to that the fact that volume is not very well define at this scale and with this method of determination of pressure, you have to be extremely careful to not overinterpret your results. …and this is not even considering the systematic errors from using classical potentials.
In summary, it is straightforward to compute some numbers, but it is difficult to interpret them.
1 Like
Thanks for your reply! Yes, I think I understand what you said. I’ll try to work on it based on your suggestions. Thank you very much!
Sincerely,
Steven | 665 | 3,069 | {"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} | 3.1875 | 3 | CC-MAIN-2024-33 | latest | en | 0.935075 |
https://www.sanfoundry.com/machine-kinematics-questions-answers-acceleration-partical-circular-path/ | 1,726,051,253,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651383.5/warc/CC-MAIN-20240911084051-20240911114051-00869.warc.gz | 885,427,668 | 20,407 | # Machine Kinematics Questions and Answers – Acceleration of a Particle along a Circular Path
This set of Machine Kinematics Multiple Choice Questions & Answers (MCQs) focuses on “Acceleration of a Particle along a Circular Path”.
1. A wheel accelerates uniformly from rest to 2000 r.p.m. in 20 seconds. What is its angular acceleration?
Explanation: Solution. Given : N0 = 0 or ω = 0 ; N = 2000 r.p.m. or ω = 2π × 2000/60 = 209.5 rad/s ; t = 20s
Angular acceleration
Let α = Angular acceleration in rad/s2
We know that
ω = ω0 + α.t
or 209.5 = 0 + α × 20
or α = 209.5 / 20 = 10.475 rad/s2
2. A wheel accelerates uniformly from rest to 2000 r.p.m. in 20 seconds. How many revolutions does the wheel make in attaining the speed of 2000 r.p.m.?
a) 400
b) 300
c) 333.4
d) 200
Explanation: Solution. Given : N0 = 0 or ω = 0 ; N = 2000 r.p.m. or ω = 2π × 2000/60 = 209.5 rad/s ; t = 20s
We know that the angular distance moved by the wheel during 2000 r.p.m. (i.e. when ω = 209.5 rad/s),
θ = (ω0 + ω )t/2
= (0 + 209.5)20/2
Since the angular distance moved by the wheel during one revolution is 2π radians, therefore number of revolutions made by the wheel,
n = θ /2π = 2095/2π = 333.4
3. The acceleration of a particle at any instant moving along a circular path in a direction tangential to that instant, is known
a) tangential component
b) normal component
c) parallel component
d) none of the mentioned
Explanation: The acceleration of a particle at any instant moving along a circular path in a direction tangential to that instant, is known tangential component.
The acceleration of a particle at any instant moving along a circular path in a direction normal to the tangent at that instant and directed towards the centre of the circular path, is known as normal component.
4. The acceleration of a particle at any instant moving along a circular path in a direction normal to the tangent at that instant and directed towards the centre of the circular path, is known as
a) tangential component
b) normal component
c) parallel component
d) none of the mentioned
Explanation: The acceleration of a particle at any instant moving along a circular path in a direction tangential to that instant, is known tangential component.
The acceleration of a particle at any instant moving along a circular path in a direction normal to the tangent at that instant and directed towards the centre of the circular path, is known as normal component.
5. When a particle moves along a straight path, then the radius of curvature is
a) infinitely small
b) zero
c) infinitely great
d) none of the mentioned
Explanation: When a particle moves along a straight path, then the radius of curvature is infinitely great. This means that v2/r is zero.
6. When a particle moves with a uniform velocity, then dv/dt will be
a) infinitely small
b) zero
c) infinitely great
d) none of the mentioned
Explanation: When a particle moves with a uniform velocity, then dv/dt will be zero. In other words, there will be no tangential acceleration; but the particle will have only normal or radial or centripetal acceleration.
7. A horizontal bar 1.5 metres long and of small cross-section rotates about vertical axis through one end. It accelerates uniformly from 1200 r.p.m. to 1500 r.p.m. in an interval of 5 seconds. What is the linear velocity at the beginning of the interval ?
a) 188.6 m/s
b) 235.5 m/s
c) 300 m/s
d) 400 m/s
Explanation: Given : r = 1.5 m ; N0 = 1200 r.p.m. or ω0 = 2 π × 1200/60 = 125.7 rad/s ;
N = 1500 r.p.m. or ω = 2 π × 1500/60 = 157 rad/s ; t = 5 s
Linear velocity at the beginning
We know that linear velocity at the beginning,
v0 = r . ω0 = 1.5 × 125.7 = 188.6 m/s
8. A horizontal bar 1.5 metres long and of small cross-section rotates about vertical axis through one end. It accelerates uniformly from 1200 r.p.m. to 1500 r.p.m. in an interval of 5 seconds. What is the linear velocity at end of the interval ?
a) 188.6 m/s
b) 235.5 m/s
c) 300 m/s
d) 400 m/s
Explanation: Given : r = 1.5 m ; N0 = 1200 r.p.m. or ω0 = 2 π × 1200/60 = 125.7 rad/s ;
N = 1500 r.p.m. or ω = 2 π × 1500/60 = 157 rad/s ; t = 5 s
Linear velocity at the end of 5 seconds
We also know that linear velocity after 5 seconds,
v5 = r . ω = 1.5 × 157 = 235.5 m/s
Sanfoundry Global Education & Learning Series – Machine Kinematics.
To practice all areas of Machine Kinematics, here is complete set of 1000+ Multiple Choice Questions and Answers. | 1,270 | 4,427 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.0625 | 4 | CC-MAIN-2024-38 | latest | en | 0.917078 |
https://kundoc.com/pdf-travelling-waves-for-reactiondiffusion-equations-with-time-depending-nonlinearit.html | 1,576,250,272,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575540564599.32/warc/CC-MAIN-20191213150805-20191213174805-00180.warc.gz | 424,117,398 | 12,540 | Travelling waves for reaction–diffusion equations with time depending nonlinearities
Travelling waves for reaction–diffusion equations with time depending nonlinearities
J. Math. Anal. Appl. 281 (2003) 164–170 www.elsevier.com/locate/jmaa Travelling waves for reaction–diffusion equations with time depending nonlineari...
J. Math. Anal. Appl. 281 (2003) 164–170 www.elsevier.com/locate/jmaa
Travelling waves for reaction–diffusion equations with time depending nonlinearities Bogdan Przeradzki Faculty of Mathematics, Łód´z University, Banacha 22, 90-238 Łód´z, Poland Received 10 September 2002 Submitted by L. Debnath
Abstract The existence of travelling wave with given end points for parabolic system of nonlinear equations is proven. The nonlinear term depends also on a · x − ct where x is the multidimensional space variable, t—time, c—the speed of the wave and a—the direction of travel. 2003 Elsevier Science (USA). All rights reserved. Keywords: Travelling wave; Fixed point
1. Introduction We shall deal with the parabolic system vt = ∆v − f (v, a · x − ct),
(1.1)
where v = (v 1 , . . . , v k ) : Rl × R → Rk is an unknown vector function, x, a ∈ Rl , t ∈ R, |a| = 1, c > 0, f : Rk × R → Rk is a continuous function. A travelling wave is a solution of the special form v(x, t) = w(a · x − ct), which has the well-known interpretation as a planar wave moving in direction a with the speed c. We shall look for a travelling wave having finite limits at ±∞ lim w(x) = w± ∈ Rk ,
x→±∞
B. Przeradzki / J. Math. Anal. Appl. 281 (2003) 164–170
165
that is wave of the front wave type. The function w should satisfy the second order ODE in Rk w
+ cw = f (w, s),
(1.2)
and has given limits at both infinities. Remark 1. The necessary condition for the existence of such a solution is lim f (w± , s) = 0.
s→±∞
We shall assume it is satisfied in the sequel. If f had not depend on s = a · x − ct then (1.2) would have been an autonomous equation and a trajectory of w would have been a heteroclinic orbit joining w− and w+ . This case was extensively studied by using topological tools (see [1,2,4]; see also the first paper on travelling wave solutions [3]). The typical assumptions are on the spectrum of the derivatives f (w± ) that govern the behaviour of solutions near the singular points w± (obviously f (w± ) = 0.) Here, the nonlinearity depends on time through a · x − ct which fixes the direction and the speed of the seeking wave (they were additional unknowns in the autonomous case). If one wants to solve the problem by force, it is natural to substitute w(s) = u(s) + ψ(s), with
ψ(s) = ω(s)w− + 1 − ω(s) w+ ,
where ω : R → [0, 1] is a smooth function such that ω(s) = 0 for s r+ and ω(s) = 1 for s r− . The choice of the interval [r− , r+ ] is arbitrary but later we shall see that it influences our assumptions. Then u is a solution of ODE (1.3) u
+ cu = f u + ψ(s), s − ψ
(s) − cψ (s), which vanishes at ±∞. This is equivalent to the integral equation u(t) = −
1 c
t −∞
1 e−c(t −s)f u(s) + ψ(s), s ds − c
∞
f u(s) + ψ(s), s ds
t
+ ω(t)(w+ − w− ),
(1.4)
with the additional condition +∞ f u(s) + ψ(s), s ds = c(w+ − w− ),
(1.5)
−∞
and we look for the solution being a continuous function u : R → Rk vanishing at both infinities. The space of all such functions will be denoted by C0 (R, Rk ). It is a Banach space with the sup-norm. Therefore we should find a fixed point of the integral operator (given by the right-hand side of (1.4)) on the manifold in C0 (R, Rk ) cutting by (1.5). One
166
B. Przeradzki / J. Math. Anal. Appl. 281 (2003) 164–170
meets a similar situation while considering resonant problems. An abstract form of such questions has been solved in [5] by a kind of perturbation method. We shall apply the same method here.
2. Perturbed problem We perturb the left-hand side of (1.3) by −λu where λ is a positive real parameter. The problem u
+ cu − λu = g(t),
u(±∞) = 0
(2.1) (R, Rk )
but for λ = 0 it is not the case. It is has a unique solution for λ > 0 and any g ∈ C0 easy to find the Green function for BVP (2.1) with λ > 0: 1 e−c(λ)(t −s) for s < t, Gλ (t, s) = − √ ε(λ)(t −s) 2 for s > t, c + 4λ e where √ √ c2 + 4λ + c c2 + 4λ − c , ε(λ) = . c(λ) = 2 2 Hence, the solutions of (2.1) are exactly functions from C0 (R, Rk ) that solve the following integral equation: +∞ Gλ (t, s)g(s) ds. u(t) =
(2.2)
−∞
In fact, it suffices to take any bounded and continuous function satisfying this equation. Now, consider the same perturbation of the nonlinear problem (1.3): (2.3) u
+ cu − λu = f u + ψ(s), s − ψ
(s) − cψ (s), u(±∞) = 0. The corresponding integral equation is exactly (2.2) with g(s) = f (u(s) + ψ(s), s) − ψ
(s) − cψ (s). Integrating by parts the second and the third summands we find the following integral equation equivalent with (2.3): +∞ Gλ (t, s)f u(s) + ψ(s), s ds u(t) = −∞
+∞ + ω(t) + λ Gλ (t, s)ω(s) ds (w+ − w− ).
(2.4)
−∞
We shall assume that nonlinear term f : Rk × R → Rk satisfies the following growth condition: f (u, s) α(s)|u|ρ + β(s), (2.5) where ρ < 1 and α, β ∈ C0 (R, R). First, it ensures the integral on the right-hand side of (2.4) converges for any function u ∈ C0 (R, Rk ). However, it gives much more.
B. Przeradzki / J. Math. Anal. Appl. 281 (2003) 164–170
167
Theorem 2.1. Under condition (2.5), the operator T given by the right-hand side of (2.4) maps C0 (R, Rk ) into itself and it is completely continuous. Proof. For any u ∈ C0 (R, Rk ), the function s → f (u(s) + ψ(s), s) belongs to the same space due to (2.5), so its norm γ ∈ C0 (R, R). By using the l’Hospital theorem, one obtains +∞ lim Gλ (t, s)γ (s) ds = 0,
t →±∞ −∞
thus the first summand in the definition of T vanish at infinities. Its continuity is obvious. By simple calculation, the function +∞ Gλ (·, s)ω(s) ds −∞
tends to −1/λ as t → −∞ and tends to 0 as t → +∞. This shows that the second summand in (2.4) is also an element of C0 (R, Rk ) (not depending on argument u of operator T ). Therefore T : C0 (R, Rk ) → C0 (R, Rk ). Take any sequence converging in C0 (R, Rk ) : un → u0 and any ε > 0. By (2.5), the function df γ (s) = maxf un (s) + ψ(s), s n0
vanishes at infinities. Hence, there exists M > 0 such that Gλ (t, s)γ (s) ds < ε , 4 R\[−M,M]
for any t. Since f is uniformly continuous on the compact set {(u, s): s ∈ [−M, M], |u| γ (s)}, the sequence of functions s ∈ [−M, M] → f un (s) + ψ(s), s is uniformly convergent to s ∈ [−M, M] → f (u0 (s) + ψ(s), s). Hence, one can find index n0 such that, for n n0 , M −M
ε supGλ (t, s)f un (s) + ψ(s), s − f u0 (s) + ψ(s), s ds < . 2 t ∈R
Therefore, for such n, supT un (t) − T u0 (t) 2 sup t ∈R
t
M + −M
Gλ (t, s)γ (s) ds
R\[−M,M]
supGλ (t, s)f un (s) + ψ(s), s − f u0 (s) + ψ(s), s ds < ε, t ∈R
and the continuity of T is proved.
168
B. Przeradzki / J. Math. Anal. Appl. 281 (2003) 164–170
Now, we shall show that the set {T u: ||u|| K} is relatively compact in C0 (R, Rk ) for any K > 0. Due to the Ascoli–Arzéla theorem, it suffices to prove that all functions from this set are equicontinuous on each interval [−M, M] and there exists function γ ∈ C0 (R, R) such that |T u(t)| γ (t) for any t ∈ R. One can consider only the first summand in the definition of T . The equicontinuity of T is obtained by standard arguments and a common majorant is +∞ Gλ (t, s) α(s)M ρ + β(s) ds, −∞
which vanishes at infinities by the previous calculations. ✷ In the above considerations, the condition ρ < 1 in (2.5) that means the problem is sublinear, has not been essential. In fact, it sufficed sup|u|M |f (u, ·)| ∈ C0 (R, R). Now, this condition is important. Theorem 2.2. Under condition (2.5), BVP (2.3) has a solution uλ for any λ > 0. Proof. We want to apply the Schauder fixed point theorem, hence we need a ball B(0, R) ⊂ C0 (R, Rk ) which T maps into itself. Denote +∞ a = sup Gλ (t, s)α(s) ds, df
t
+∞ b = sup Gλ (t, s)β(s) ds, df
t
−∞
+∞ df c = 1 + λ sup Gλ (t, s)ω(s) ds, t
−∞
df p = supψ(t).
−∞
t
Then, for any u ∈ B(0, R), we have df T u = supT u(t) a(R + p)ρ + b + c. t
Since ρ < 1, there exists R > 0 such that a(R + p)ρ + b + c R and this is the radius of the ball we seek. ✷
3. Main result In the previous section, we showed that all perturbed problems (2.3) have solutions. If we denote the solution for λ by uλ , then we can think that the limit limλ→0 , if it exists, is a solution of problem (1.3) (λ = 0) and gives a travelling wave for (1.1). However, attentive calculations show that the radius R of the ball containing uλ tends to infinity when λ approaches to 0. We need a new assumption which guarantees the boundedness of uλ for λ > 0. It suffices in order to find a solution to the main problem.
B. Przeradzki / J. Math. Anal. Appl. 281 (2003) 164–170
169
Theorem 3.1. Let all previous assumptions are satisfied. Moreover, suppose there exist two functions γ , δ ∈ C0 (R, R) (γ of C 2 -class and positive) such that they are connected by differential inequality γ
+ 2cγ 2δ,
(3.1)
and the following condition holds: u, f u + ψ(t), t + ω
(t) + cω (t) (w+ − w− ) δ(t), |u|2
for any t ∈ R and points w± .
(3.2)
γ (t). Then Eq. (1.1) has a traveling wave solution with end
Proof. First, we shall show that for λ > 0 each solution uλ satisfies uλ (t)2 γ (t), t ∈ R.
(3.3)
df
Let φ(t) = |uλ (t)|2 and suppose that its value at a point is greater than the value of γ . Then φ − γ attains a positive maximum at some point t0 , hence φ (t0 ) = γ (t0 ) and φ
(t0 ) − γ
(t0 ) 0. But the last difference equals 2 uλ (t0 ), f uλ (t0 ) + ψ(t0 ), t0 + ω
(t0 ) + cω (t0 ) (w+ − w− ) 2 − γ
(t0 ) + 2u λ (t0 ) − 2cφ (t0 ) + 2λφ(t0 ), thus, by (3.2), we have the inequality γ
(t0 ) + 2cγ (t0 ) > 2δ(t0 ), that contradicts (3.1). Now, we shall get an estimate for the derivatives u λ independent of λ by the differentiation of (2.4) (we omit the index λ):
1
u (t) = √ c(λ)e c2 + 4λ
−c(λ)t
t
−∞ ∞
1 −√ ε(λ)eε(λ)t c2 + 4λ
ec(λ)s f u(s) + ψ(s), s ds e−ε(λ)s f u(s) + ψ(s), s ds
t
λ
+ ω (t) + √ c(λ)e c2 + 4λ λ
−√ ε(λ)eε(λ)t c2 + 4λ
−c(λ)t
t ec(λ)s ω(s) ds −∞
∞
e−ε(λ)s ω(s) ds (w+ − w− ).
t
Having the estimate for f (u(s) + ψ(s), s), we obtain the uniform boundedness of u λ . Due to (2.3), we get also the uniform boundedness of u
λ . We can apply the Ascoli–Arzéla theorem to the families {uλ : λ > 0} and {u λ : λ > 0} on any compact interval. Therefore, by using the diagonal procedure, we are able to find a
170
B. Przeradzki / J. Math. Anal. Appl. 281 (2003) 164–170
sequence uλn such that λn → 0 and both sequences uλn and u λn are uniformly convergent df on any compact interval. Applying (2.3) again, we have the same for u
λn . Thus, u = lim uλn is a C 2 -function on the whole line and it satisfies (1.3). This function vanishes at infinities, since we have obtained earlier the common estimate (3.3). This ends the proof. ✷ Remark 2. It is easy to see that, for any function δ ∈ C0 (R, R) with the property +∞ δ(t) dt = 0, −∞
there exists γ satisfying (3.1). In fact, it is enough to take a positive solution of the equation γ
+ 2cγ = 2δ, vanishing at infinities. Remark 3. If w+ = w− , then ψ(t) = w+ and condition (3.2) has a simpler form: u, f (u + w+ , t) δ(t). Hence, the function, for example: √ df f (u, t) = α(t) 3 u + β(t), with α, β as above satisfies all assumptions with w± = 0. Notice that, if β = 0, our theorem is not needed to get a solution—it is the null function.
References [1] A. Volpert, V. Volpert, Travelling Waves Solutions of Parabolic Systems, American Mathematical Society, Providence, RI, 1994. [2] J. Smøller, Shock Waves and Reaction–Diffusion Equations, in: Grunlehren Math. Wiss., Vol. 258, SpringerVerlag, Berlin, 1983. [3] A. Kolmogorov, I. Pietrovski, N. Piskunov, Étude de l’équation de la chaleur avec croissance de la quantité de matiere et son applications biologique, Bull. Moskov. Gos. Univ. Mat. Mech. 1 (1937) 1–25. [4] D. Henry, Geometric Theory of Semilinear Parabolic Equations, in: Lecture Notes in Math., Vol. 840, Springer-Verlag, Berlin, 1981. [5] B. Przeradzki, A new continuation method for the study of nonlinear equations at resonance, J. Math. Anal. Appl. 180 (1993) 553–565. | 4,490 | 12,437 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2019-51 | latest | en | 0.837728 |
https://socratic.org/questions/what-is-the-average-speed-of-an-object-that-is-moving-at-3-m-s-at-t-0-and-accele | 1,582,346,237,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875145648.56/warc/CC-MAIN-20200222023815-20200222053815-00098.warc.gz | 568,104,592 | 6,583 | # What is the average speed of an object that is moving at 3 m/s at t=0 and accelerates at a rate of a(t) =9-t^2/3 on t in [3,5]?
Mar 30, 2017
$31 \frac{4}{9}$m/s
#### Explanation:
$a \left(t\right) = 9 - {t}^{2} / 3$
Speed, $v = \int a \left(t\right) = \int 9 - {t}^{2} / 3 = 9 t - {t}^{3} / 9 + c$
$t = 0 , v = 3$ m/s $\to c = 3$, therefore,
$v = 9 t - {t}^{3} / 9 + 3$
Displacement,
$s = \int v \left(t\right)$,
${s}_{3 , 5} = {\int}_{3}^{5} v \left(t\right) = {\int}_{3}^{5} 9 t - {t}^{3} / 9 + 3 = {\left[\frac{9}{2} {t}^{2} - {t}^{4} / 36 + 3 t\right]}_{3}^{5}$
${s}_{3 , 5} = \left[\frac{9}{2} {\left(5\right)}^{2} - {\left(5\right)}^{4} / 36 + 3 \left(5\right)\right] - \left[\frac{9}{2} {\left(3\right)}^{2} - {\left(3\right)}^{4} / 36 + 3 \left(3\right)\right]$
${s}_{3 , 5} = \left[\frac{225}{2} - \frac{625}{36} + 15\right] - \left[\frac{81}{2} - \frac{81}{36} + 9\right]$
${s}_{3 , 5} = \left[\frac{225}{2} - \frac{625}{36} + 15\right] - \left[\frac{81}{2} - \frac{81}{36} + 9\right]$
${s}_{3 , 5} = \frac{225}{2} - \frac{81}{2} - \frac{625}{36} + \frac{81}{36} + 15 - 9$
${s}_{3 , 5} = \frac{144}{2} - \frac{544}{36} + 6$
${s}_{3 , 5} = 72 - \frac{136}{9} + 6 = \frac{566}{9}$
Average speed = displacement/time $= \frac{566}{9} \div 2 = \frac{566}{18} = 31 \frac{4}{9}$m/s | 641 | 1,301 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 15, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.40625 | 4 | CC-MAIN-2020-10 | longest | en | 0.36468 |
http://oeis.org/A177827 | 1,369,194,006,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368701281163/warc/CC-MAIN-20130516104801-00057-ip-10-60-113-184.ec2.internal.warc.gz | 198,228,989 | 3,821 | This site is supported by donations to The OEIS Foundation.
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A177827 Sub-triangle of A060187 formed by taking every third-indexed term of every third row. 0
1, 1, 1, 1, 23548, 1, 1, 21707972, 21707972, 1, 1, 10708911188, 743288515164, 10708911188, 1, 1, 4261002128223, 6453703025399873, 6453703025399873, 4261002128223, 1, 1, 1556000598766224, 30920009116692763140, 527896878148304296900, 30920009116692763140 (list; table; graph; refs; listen; history; text; internal format)
OFFSET 0,5 COMMENTS Row sums are:{1, 2, 23550, 43415946, 764706337542, 12915928055056194, 589740008382887355630, 34621385290631641181980314, 3529900361215340810891790303798, 482387142371699171544762551301105426, 95463751455117339958648491169567658521950,...}. LINKS EXAMPLE {1}, {1, 1}, {1, 23548, 1}, {1, 21707972, 21707972, 1}, {1, 10708911188, 743288515164, 10708911188, 1}, {1, 4261002128223, 6453703025399873, 6453703025399873, 4261002128223, 1} MATHEMATICA p[x_, n_] = (1 - x)^(n + 1)*Sum[((2*k + 1)^n)*x^k, {k, 0, Infinity}]; t[n_, m_] := CoefficientList[FullSimplify[ExpandAll[p[x, n]]], x][[m + 1]]; Table[Table[t[n, 3*m], {m, 0, Floor[n/3]}], {n, 0, 30, 3}]; Flatten[%] CROSSREFS Cf.A060187 Sequence in context: A189656 A204074 A031847 * A069334 A118061 A200437 Adjacent sequences: A177824 A177825 A177826 * A177828 A177829 A177830 KEYWORD nonn,tabl AUTHOR Roger Bagula, Dec 13 2010 STATUS approved
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Content is available under The OEIS End-User License Agreement . | 669 | 1,785 | {"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} | 3.0625 | 3 | CC-MAIN-2013-20 | latest | en | 0.499828 |
http://eyesvisions.com/physics/30 | 1,571,401,548,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570986682037.37/warc/CC-MAIN-20191018104351-20191018131851-00521.warc.gz | 68,905,754 | 4,684 | a shutter speed of 1/100 second and an f/stop of f/5.6, the exposure time is cut in half while the irradiance (E0/32) is doubled, thereby leaving no net change in the film exposure (J/m2).
The numerical aperture is another important design parameter for a lens, related directly to how much light the lens gathers. If the focal length of a design lens increases and its diameter decreases, the solid angle (cone) of useful light rays from object to image for such a lens decreases. For example, the concept of a numerical aperture finds immediate application in the design of the objective lens (the lens next to the specimen under observation) for a microscope, as we show below. Light-gathering capability is crucial for microscopes.
Figure 3-24 depicts the light-gathering power of a lens relative to a point O on a specimen covered by a glass slide. Lens L is the objective lens of a microscope focused on the specimen. On the right side of the symmetry axis of the lens, the light-gathering power of the lens with air between the cover slide and the lens is depicted in terms of half-angle ɑair. On the left side, by contrast, the increased light-gathering power of the lens with oil situated between the cover slide and the lens is shown in terms of the larger half-angle ɑoil. The oil is chosen so as to
have an index of refraction (n0) very near that of the cover slide (ng) so that little or no refraction occurs for limiting ray 2 at the glass-oil interface. Consequently the half-angle oil is greater than the half-angle air. As Figure 3-24 shows, ray 1 suffers refraction at the glass-air interface, thereby restricting the cone of rays accepted by the lens to the smaller half-angle air.
The numerical aperture of a lens is defined so as to exhibit the difference in solid angles (cones) of light accepted, for example, by an “oil-immersion” arrangement versus an air-immersion setup.
The definition of numerical aperture (N.A.) is given in Equation 3-9 as
where n is the index of refraction of the intervening medium between object and lens and ɑ is the half-angle defined by the limiting ray (ɑair or ɑoil in Figure 3-24). The “light-gathering” power of the microscope’s objective lens is thus increased by increasing the refractive index of the intervening medium.
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Eyes carePhysicianBate's booksTechnologyForumLaser corre.Blues under eyesburning in the eyesanother diseasesMedical misteryNaturally eyesight correction. No laser eye surgery. Restore eyesight. Vision correction. | 589 | 2,578 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2019-43 | latest | en | 0.924205 |
https://www.experts-exchange.com/questions/10299395/Convert-32bit-UTC-timestamp-to-yr-mo-da-sec-etc.html | 1,527,025,686,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794864968.11/warc/CC-MAIN-20180522205620-20180522225620-00164.warc.gz | 731,283,719 | 28,187 | • C
# Convert 32bit UTC timestamp to yr, mo, da, sec, etc.
Hi there everyone.
I am trying to convert a 32 bit UTC timestamp, which represents the number of seconds from the year 1900, into the current year, month, , date, day, hr, min, se. Does anyone have the algorithm to calculate this and can help explain how the algorithm works? Thanks.
LVL 3
###### Who is Participating?
Commented:
Hrmph. I meant that last post to be an answer candidate.
I also meant it to keep the indentation. Sorry about that. Oh well, it's not that complicated.
0
Commented:
#include <time.h>
struct tm *gmtime (const time_t *timer);
0
Author Commented:
Ozo, thanks for your interest again. I used your answer last time and it worked fine. This time around though, I would like to see the actual algorithm and understand exactly how the year, date, month, etc is calculated.
I tried looking at some of the standard C routines, but couldn't make heads or tail of the advanced stuff it was doing. Can anyone propose an efficient yet easy to understand algorithm for converting the timestamp to smaller time units? Thanks.
0
Commented:
0
Commented:
If the time is divided in a single 32 bit number i.e. if it is bitfield then let me know how the fields are divided in bits?
I mean,
hours === __ to __ bits
min === __ to __ bits
sec === __ to __ bits
0
progCommented:
This converts seconds to date & time. I think it can be readily interpreted.
#include<stdio.h>
void datum(unsigned long secs,long *year,long *month,long *day,long *hour,
long *min, long *sec)
{
long tdays;
int tyear;
int year4; \
int leap = 0;
tdays = secs / ((unsigned long) 24 * 3600);
if (tdays < 1461)
{
tyear = tdays / 365;
tdays = tdays - 365 * tyear;
leap = 0;
}
else
{
year4 = (tdays - 1460) / 1461 + 1;
tdays = (tdays -1460) % 1461;
if (tdays > 366)
{
tyear = year4 * 4 + (tdays - 366)/365+1;
tdays = (tdays - 366) % 365;
leap = 0;
}
else
{
tyear = year4 * 4;
leap = 1;
}
}
if (tdays < 31)
{
*month = 1;
*day = tdays;
}
else
{
if(tdays < 59 + leap)
{
*month = 2;
*day = tdays - 31;
}
else
{
if(tdays < 90 + leap)
{
*month = 3;
*day = tdays - 59 - leap;
}
else
{
if(tdays < 120 + leap)
{
*month = 4;
*day = tdays - 90 - leap;
}
else
{
if(tdays < 151 + leap)
{
*month = 5;
*day = tdays - 120 - leap;
}
else
{
if(tdays < 181 + leap)
{
*month = 6;
*day = tdays - 151 - leap;
}
else
{
if(tdays < 212 + leap)
{
*month = 7;
*day = tdays - 181 - leap;
}
else
{
if(tdays < 243 + leap)
{
*month = 8;
*day = tdays - 212 - leap;
}
else
{
if(tdays < 273 + leap)
{
*month = 9;
*day = tdays - 243 - leap;
}
else
{
if(tdays < 304 + leap)
{
*month = 10;
*day = tdays - 273 - leap;
}
else
{
if(tdays < 334 + leap)
{
*month = 11;
*day = tdays - 304 - leap;
}
else
{
if(tdays < 365 + leap)
{
*month = 12;
*day = tdays - 365 - leap;
}
}}}}}}}}}}}
++(*day);
*sec = secs % 60;
*min = (secs / 60) % 60;
*hour = (secs / 3600) % 24;
*year = tyear;
}
main()
{
unsigned long s,y,m,d,h,min,sec;
do{
printf("\nSeconds ");
scanf("%li",&s);
datum(s,&y,&m,&d,&h,&min,&sec);
printf("\n %li:%li:%li %li/%li/%li\n",h,min,sec,m,d,y);
} while(s!=0);
}
0
Author Commented:
sumant, the timestamp is not a bit field. Rather, the 32bit timestamp is the number of seconds from the year 1900 to now. So you cannot just extract the time from the bitfield, but rather have to convert the seconds.
0
Author Commented:
Deighton, thanks for your suggestion, I believe it works. However, does anyone know of a smaller, more efficient algorithm? Ozo offered one last time which was about 18 lines, but I have a hard time understanding what it was doing.
I'm looking for an efficient algorithm, and someone to explain to me how that algorithm works. Ozo, if you could explain to me step by step how that algorithm worked it would be great too.
Sorry, opening up question to more suggestions
0
progCommented:
ozo's formula was based on all sorts of mathematical trickery to implicitly adjust for leap years etc.
I've got a hard time explaining it too.
My algorithm is not inefficient, not as efficient as the ozo formula, but i developed it to be more readily understandable.
0
Commented:
Well, here's an implementation that I think is both pretty understandable and acceptably efficient, and it even works with Windows ;)
#define seconds_per_minute (60L)
#define seconds_per_hour (60L * seconds_per_minute)
#define seconds_per_day (24L * seconds_per_hour)
const short __month_to_days[2][13] =
{
{0,31,59,90, 120, 151, 181, 212, 243, 273, 304,334, 365 },
{0,31,60,91, 121, 152, 182, 213, 244, 274, 305,335, 366 }
};
/*
* leap_year - return nonzero if year is a leap year, zero otherwise (year 0 = 1900)
*/
int __leap_year(int year);
int __leap_year(int year)
{
return(__mod(year, 4) == 0 && (__mod(year, 100) != 0 || __mod(year, 400) == 100));
}
/*
* __time2tm - convert seconds since midnight, 1/1/1900 (or 1970 on Win32),
* to broken-down time
*/
#if (__dest_os == __win32_os || __dest_os == __wince_os)
static void __time2tm(time_t inTime, struct tm * tm)
#else
static void __time2tm(time_t time, struct tm * tm)
#endif
{
unsigned long years, months, days, seconds;
int is_leap_year;
#if (__dest_os == __win32_os || __dest_os == __wince_os)
/* Since Win32 time_t is relative to 1970 rather than 1900.
* This must be of type unsigned long rather than a signed
* time_t to prevent overflow */
unsigned long time = inTime + ((365 * 70UL) + 17) * 24 * 60 * 60;
#endif
if (!tm)
return;
tm->tm_isdst = -1;
days = time / seconds_per_day;
seconds = time % seconds_per_day;
tm->tm_wday = (days + 1) % 7;
/* January 1, 1900 was a Monday */
years = 0;
for (;;)
{
unsigned long days_this_year = __leap_year(years) ? 366 : 365;
if (days < days_this_year)
break;
days -= days_this_year;
years += 1;
}
tm->tm_year = years;
tm->tm_yday = days;
months = 0;
is_leap_year = __leap_year(years);
for (;;)
{
unsigned long days_thru_this_month = __month_to_days[is_leap_year][months+1];
if (days < days_thru_this_month)
{
days -= __month_to_days[is_leap_year][months];
break;
}
++months;
}
tm->tm_mon = months;
tm->tm_mday = days + 1;
tm->tm_hour = seconds / seconds_per_hour;
seconds %= seconds_per_hour;
tm->tm_min = seconds / seconds_per_minute;
tm->tm_sec = seconds % seconds_per_minute;
}
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A196564 Number of odd digits in decimal representation of n. 44
0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 0, 1, 0, 1, 0, 1 (list; graph; refs; listen; history; text; internal format)
OFFSET 0,12 LINKS Reinhard Zumkeller, Table of n, a(n) for n = 0..10000 FORMULA a(n) = A055642(n) - A196563(n); a(A014263(n)) = 0; a(A007957(n)). From Hieronymus Fischer, May 30 2012: (Start) a(n) = sum_{j=0..m} (floor(n/(2*10^j) + (1/2)) - floor(n/(2*10^j)), where m=floor(log_10(n)). a(10n+k) = a(n) + a(k), 0<=k<10, n>=0. a(n) = a(floor(n/10)) + a(n mod 10), n>=0. a(n) = sum_{j=0..m} a(floor(n/10^j) mod 10), n>=0. a(A014261(n)) = floor(log_5(4n+1)), n>0. G.f.: g(x) = (1/(1-x))*sum_{j>=0} x^10^j/(1+x^10^j). (End) MAPLE A196564 := proc(n) if n =0 then 0; else convert(n, base, 10) ; add(d mod 2, d=%) ; end if: end proc: # R. J. Mathar, Jul 13 2012 MATHEMATICA Table[Total[Mod[IntegerDigits[n], 2]], {n, 0, 100}] (* Zak Seidov, Oct 13 2015 *) PROG (Haskell) a196564 n = length [d | d <- show n, d `elem` "13579"] -- Reinhard Zumkeller, Feb 22 2012, Oct 04 2011 (PARI) a(n) = #select(x->x%2, digits(n)); \\ Michel Marcus, Oct 14 2015 CROSSREFS Cf. A014261, A014263, A027868, A046034, A055640, A055641, A055642, A061217, A102669-A102685, A122640, A196563. Sequence in context: A111621 A326398 A140195 * A196563 A198890 A305831 Adjacent sequences: A196561 A196562 A196563 * A196565 A196566 A196567 KEYWORD nonn,easy,base AUTHOR Reinhard Zumkeller, Oct 04 2011 STATUS approved
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Last modified June 20 15:52 EDT 2021. Contains 345165 sequences. (Running on oeis4.) | 987 | 2,248 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.625 | 4 | CC-MAIN-2021-25 | latest | en | 0.385081 |
http://googology.wikia.com/wiki/Gargantuulennexithrong | 1,508,192,306,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187820466.2/warc/CC-MAIN-20171016214209-20171016234209-00184.warc.gz | 180,626,574 | 82,443 | ## FANDOM
9,739 Pages
The gargantuul regiment is a series of numbers from E100#100#100#100#100#100#100#100#100 to E1#1#1#1#1#1#1#1#1#10,000 defined using Hyper-E Notation (i.e. beginning from gargantuul and up to myria-endekaxis).[1] The numbers were coined by Sbiis Saibian.
Previous regiment Next regiment Ginorgol regiment Googondol regiment
## List of numbers of the regiment Edit
Name of number Hyper-E Notation (exact equality) Arrow notation (approximation) gargantuul E100#100#100#100#100#100#100#100#100 \(10\uparrow^9 100\) ginorgolennex E100#100#100#100#100#100#100#100#1#2 \(10\uparrow^9 (10\uparrow^8 100)\) ginorgoloctexiennex E100#100#100#100#100#100#100#100#2#2 \(10\uparrow^9 (10\uparrow^8 (10\uparrow^8 100))\) ginorgolduoctexiennex E100#100#100#100#100#100#100#100#3#2 \(10\uparrow^9 (10\uparrow^8 (10\uparrow^8 (10\uparrow^8 100)))\) gargantuulennex E100#100#100#100#100#100#100#100#100#2 \(10\uparrow^9 (10\uparrow^9 100)\) ginorgolduennex E100#100#100#100#100#100#100#100#1#3 \(10\uparrow^9 (10\uparrow^9 (10\uparrow^8 100))\) ginorgoloctexiduennex E100#100#100#100#100#100#100#100#2#3 \(10\uparrow^9 (10\uparrow^9 (10\uparrow^8 (10\uparrow^8 100)))\) ginorgolduoctexiduennex E100#100#100#100#100#100#100#100#3#3 \(10\uparrow^9 (10\uparrow^9 (10\uparrow^8 (10\uparrow^8 (10\uparrow^8 100))))\) gargantuulduennex E100#100#100#100#100#100#100#100#100#3 \(10\uparrow^9 (10\uparrow^9 (10\uparrow^9 100))\) gargantuulchime E1000#1000#1000#1000#1000#1000#1000#1000#1000 \(10\uparrow^9 1000\) gargantuulennexichime E1000#1000#1000#1000#1000#1000#1000#1000#1000#2 \(10\uparrow^9 (10\uparrow^9 1000)\) gargantuulennexichime E1000#1000#1000#1000#1000#1000#1000#1000#1000#3 \(10\uparrow^9 (10\uparrow^9 (10\uparrow^9 1000))\) gargantuultoll E10,000#10,000#10,000#10,000#10,000# 10,000#10,000#10,000#10,000 \(10\uparrow^9 10^4\) gargantuulennexitoll E10,000#10,000#10,000#10,000#10,000# 10,000#10,000#10,000#10,000#2 \(10\uparrow^9 (10\uparrow^9 10^4)\) gargantuulduennexitoll E10,000#10,000#10,000#10,000#10,000# 10,000#10,000#10,000#10,000#3 \(10\uparrow^9 (10\uparrow^9 (10\uparrow^9 10^4))\) gargantuulgong E100,000#100,000#100,000#100,000#100,000# 100,000#100,000#100,000#100,000 \(10\uparrow^9 10^5\) gargantuulennexigong E100,000#100,000#100,000#100,000#100,000# 100,000#100,000#100,000#100,000#2 \(10\uparrow^9 (10\uparrow^9 10^5)\) gargantuulduennexigong E100,000#100,000#100,000#100,000#100,000# 100,000#100,000#100,000#100,000#3 \(10\uparrow^9 (10\uparrow^9 (10\uparrow^9 10^5))\) gargantuulbong E100,000,000#100,000,000#100,000,000# ... ...100,000,000#100,000,000#100,000,000 w/9 100,000,000s \(10\uparrow^9 10^8\) gargantuulennexibong E100,000,000#100,000,000#100,000,000# ... ... #100,000,000#100,000,000#100,000,000#2 w/9 100,000,000s \(10\uparrow^9 (10\uparrow^9 10^8)\) gargantuulduennexibong E100,000,000#100,000,000#100,000,000# ... ... #100,000,000#100,000,000#100,000,000#3 w/9 100,000,000s \(10\uparrow^9 (10\uparrow^9 (10\uparrow^9 10^8))\) gargantuulthrong E100,000,000,000#100,000,000,000# ... ...#100,000,000,000#100,000,000,000 w/9 100,000,000,000s \(10\uparrow^9 10^{11}\) gargantuulennexithrong E100,000,000,000#100,000,000,000# ... ... #100,000,000,000#100,000,000,000#2 w/9 100,000,000,000s \(10\uparrow^9 (10\uparrow^9 10^{11})\) gargantuulduennexithrong E100,000,000,000#100,000,000,000# ... ... #100,000,000,000#100,000,000,000#3 w/9 100,000,000,000s \(10\uparrow^9 (10\uparrow^9 (10\uparrow^9 10^{11}))\) googolennex E100#1#1#1#1#1#1#1#1#2 \(10\uparrow^9 10^{100}\) grangolennex E100#100#1#1#1#1#1#1#1#2 \(10\uparrow^9 (10 \uparrow^2 100)\) greagolennex E100#100#100#1#1#1#1#1#1#2 \(10\uparrow^9 (10 \uparrow^3 100)\) gigangolennex E100#100#100#100#1#1#1#1#1#2 \(10\uparrow^9 (10 \uparrow^4 100)\) gorgegolennex E100#100#100#100#100#1#1#1#1#2 \(10\uparrow^9 (10 \uparrow^5 100)\) gulgolennex E100#100#100#100#100#100#1#1#1#2 \(10\uparrow^9 (10 \uparrow^6 100)\) gaspgolennex E100#100#100#100#100#100#100#1#1#2 \(10\uparrow^9 (10 \uparrow^7 100)\) gargantuulplex E(E100#100#100#100#100#100#100#100#100) \(10\uparrow (10 \uparrow^9 100)\) gargantuuldex E100#(E100#100#100#100#100#100#100#100#100) \(10\uparrow^2 (10 \uparrow^9 100)\) gargantuulthrex E100#100# (E100#100#100#100#100#100#100#100#100) \(10\uparrow^3 (10 \uparrow^9 100)\) gargantuultetrex E100#100#100# (E100#100#100#100#100#100#100#100#100) \(10\uparrow^4 (10 \uparrow^9 100)\) gargantuulpentex E100#100#100#100# (E100#100#100#100#100#100#100#100#100) \(10\uparrow^5 (10 \uparrow^9 100)\) gargantuulhex E100#100#100#100#100# (E100#100#100#100#100#100#100#100#100) \(10\uparrow^6 (10 \uparrow^9 100)\) gargantuulheptex E100#100#100#100#100#100# (E100#100#100#100#100#100#100#100#100) \(10\uparrow^7 (10 \uparrow^9 100)\) gargantuuloctex E100#100#100#100#100#100#100# (E100#100#100#100#100#100#100#100#100) = E100#100#100#100#100#100#100#100#101 \(10\uparrow^8 (10 \uparrow^9 100)\) ecetonennex E303#1#1#1#1#1#1#1#1#2 \(10 \uparrow^9 10^{303}\) ecetonduennex E303#1#1#1#1#1#1#1#1#3 \(10 \uparrow^9 (10 \uparrow^9 10^{303})\) tria-endekaxis E1#1#1#1#1#1#1#1#1#3 \(10 \uparrow^{10} 3\) tetra-endekaxis E1#1#1#1#1#1#1#1#1#4 \(10 \uparrow^{10} 4\) deka-endekaxis E1#1#1#1#1#1#1#1#1#10 \(10 \uparrow^{10} 10\) hecta-endekaxis E1#1#1#1#1#1#1#1#1#100 \(10 \uparrow^{10} 100\) chilia-endekaxis E1#1#1#1#1#1#1#1#1#1000 \(10 \uparrow^{10} 1000\) myria-endekaxis E1#1#1#1#1#1#1#1#1#10,000 \(10 \uparrow^{10} 10000\)
## Sources Edit
1. Sbiis Saibian, 4.3.2 - Hyper-E Numbers | 2,640 | 5,477 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2017-43 | longest | en | 0.421196 |
https://socratic.org/questions/what-is-the-vertex-form-of-y-6x-2-9x-3 | 1,637,994,826,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964358118.13/warc/CC-MAIN-20211127043716-20211127073716-00330.warc.gz | 639,274,194 | 6,597 | # What is the vertex form of y= 6x^2-9x+3 ?
Mar 2, 2016
( x $-$ 3/4 )^2 = 1/6 ( y + 3/8 ).
The vertex of the parabola is ( 3/4, $-$3/8).
#### Explanation:
The axis of the parabola is in the positive y-direction.
The focus is at ( 3/4, $-$3/8 $-$1/24.) = ( 3/4, $-$5/12.), below the vertex on the axis x = 3/4.
Mar 2, 2016
$y = 6 {\left(x - \frac{3}{4}\right)}^{2} + \frac{39}{16}$
The solution method has been shown in a lot of detail
#### Explanation:
Given:$\text{ } y = 6 {x}^{2} - 9 x + 3$ ................................(1)
This process introduces an error. This error is dealt with by introducing a correction constant.
Let the correction constant be $\textcolor{g r e e n}{k}$
$\textcolor{b l u e}{\text{Step 1}}$
Write as:$\text{ } y = 6 \left({x}^{2} - \frac{9}{6} x\right) + 3$
Note that$\text{ } 6 \times \left(- \frac{9}{6}\right) x = - 9 x$
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$\textcolor{b l u e}{\text{Step 2}}$
$\textcolor{b r o w n}{\text{Move the index of 2 from "x^2" to outside the brackets.}}$
We have now changed the value of the RHS so unable at this stage to equate it to $y$
Write as:$\text{ "6(x^(color(magenta)(2))-9/6x)+3" " ->" } 6 {\left(x - \frac{9}{6} x\right)}^{\textcolor{m a \ge n t a}{2}} + 3$
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$\textcolor{b l u e}{\text{Step 3}}$
$\textcolor{b r o w n}{\text{Remove the right hand side "x" from inside the brackets.}}$
$6 {\left(x - \frac{9}{6} \textcolor{m a \ge n t a}{x}\right)}^{2} + 3 \text{ "->" } 6 {\left(x - \frac{9}{6}\right)}^{2} + 3$
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$\textcolor{b l u e}{\text{Step 4}}$
$\textcolor{b r o w n}{\text{Add correction constant of } \textcolor{g r e e n}{k}}$
$6 {\left(x - \frac{9}{6}\right)}^{2} + 3 \text{ " ->" } 6 {\left(x - \frac{9}{6}\right)}^{2} + 3 + \textcolor{g r e e n}{k}$
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$\textcolor{b l u e}{\text{Step 5}}$
$\textcolor{b r o w n}{\text{Halve the "9/6" inside the brackets}}$
$\implies y = 6 {\left(x - \frac{9}{12}\right)}^{2} + 3 + k$.........................(2)
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$\textcolor{b l u e}{\text{Solving for } k}$
Expand the brackets of equation (2)
$y = 6 {x}^{2} - 9 x + \frac{81}{144} + 3 + k$
The $\frac{81}{144}$ is the error that $k$ is correcting
So to get rid of $\frac{81}{144}$ we make k $- \frac{81}{144}$
so our constant become: $\text{ } 3 - \frac{81}{144} = 2 \frac{7}{16} = \frac{39}{16}$
Equation (2) becomes: $\text{ "y=6(x-9/12)^2+39/16" } \ldots . . \left({2}_{a}\right)$
But $\frac{9}{12} = \frac{3}{4}$ so we now have
color(magenta)(" "y=6(x-3/4)^2+39/16)" ".....(2_b) | 1,011 | 2,621 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 34, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.84375 | 5 | CC-MAIN-2021-49 | latest | en | 0.524347 |
https://www.ademcetinkaya.com/2022/12/dui-diversified-united-investment.html | 1,695,794,958,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510259.52/warc/CC-MAIN-20230927035329-20230927065329-00102.warc.gz | 682,351,994 | 56,878 | Outlook: DIVERSIFIED UNITED INVESTMENT LIMITED assigned short-term B2 & long-term B1 forecasted stock rating.
Dominant Strategy : Hold
Time series to forecast n: 10 Dec 2022 for (n+6 month)
Methodology : Ensemble Learning (ML)
## Abstract
Predictions on stock market prices are a great challenge due to the fact that it is an immensely complex, chaotic and dynamic environment. There are many studies from various areas aiming to take on that challenge and Machine Learning approaches have been the focus of many of them. There are many examples of Machine Learning algorithms been able to reach satisfactory results when doing that type of prediction. This article studies the usage of LSTM networks on that scenario, to predict future trends of stock prices based on the price history, alongside with technical analysis indicators.(Sureshkumar, K.K. and Elango, N.M., 2012. Performance analysis of stock price prediction using artificial neural network. Global journal of computer science and Technology.) We evaluate DIVERSIFIED UNITED INVESTMENT LIMITED prediction models with Ensemble Learning (ML) and Lasso Regression1,2,3,4 and conclude that the DUI stock is predictable in the short/long term. According to price forecasts for (n+6 month) period, the dominant strategy among neural network is: Hold
## Key Points
1. Which neural network is best for prediction?
2. Game Theory
3. What are main components of Markov decision process?
## DUI Target Price Prediction Modeling Methodology
We consider DIVERSIFIED UNITED INVESTMENT LIMITED Decision Process with Ensemble Learning (ML) where A is the set of discrete actions of DUI 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
F(Lasso Regression)5,6,7= $\begin{array}{cccc}{p}_{a1}& {p}_{a2}& \dots & {p}_{1n}\\ & ⋮\\ {p}_{j1}& {p}_{j2}& \dots & {p}_{jn}\\ & ⋮\\ {p}_{k1}& {p}_{k2}& \dots & {p}_{kn}\\ & ⋮\\ {p}_{n1}& {p}_{n2}& \dots & {p}_{nn}\end{array}$ X R(Ensemble Learning (ML)) X S(n):→ (n+6 month) $∑ i = 1 n s i$
n:Time series to forecast
p:Price signals of DUI stock
j:Nash equilibria (Neural Network)
k:Dominated move
a:Best response for target price
For further technical information as per how our model work we invite you to visit the article below:
How do AC Investment Research machine learning (predictive) algorithms actually work?
## DUI Stock Forecast (Buy or Sell) for (n+6 month)
Sample Set: Neural Network
Stock/Index: DUI DIVERSIFIED UNITED INVESTMENT LIMITED
Time series to forecast n: 10 Dec 2022 for (n+6 month)
According to price forecasts for (n+6 month) period, the dominant strategy among neural network is: Hold
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%
## Adjusted IFRS* Prediction Methods for DIVERSIFIED UNITED INVESTMENT LIMITED
1. When defining default for the purposes of determining the risk of a default occurring, an entity shall apply a default definition that is consistent with the definition used for internal credit risk management purposes for the relevant financial instrument and consider qualitative indicators (for example, financial covenants) when appropriate. However, there is a rebuttable presumption that default does not occur later than when a financial asset is 90 days past due unless an entity has reasonable and supportable information to demonstrate that a more lagging default criterion is more appropriate. The definition of default used for these purposes shall be applied consistently to all financial instruments unless information becomes available that demonstrates that another default definition is more appropriate for a particular financial instrument.
2. At the date of initial application, an entity shall determine whether the treatment in paragraph 5.7.7 would create or enlarge an accounting mismatch in profit or loss on the basis of the facts and circumstances that exist at the date of initial application. This Standard shall be applied retrospectively on the basis of that determination.
3. The requirements in paragraphs 6.8.4–6.8.8 may cease to apply at different times. Therefore, in applying paragraph 6.9.1, an entity may be required to amend the formal designation of its hedging relationships at different times, or may be required to amend the formal designation of a hedging relationship more than once. When, and only when, such a change is made to the hedge designation, an entity shall apply paragraphs 6.9.7–6.9.12 as applicable. An entity also shall apply paragraph 6.5.8 (for a fair value hedge) or paragraph 6.5.11 (for a cash flow hedge) to account for any changes in the fair value of the hedged item or the hedging instrument.
4. As noted in paragraph B4.3.1, when an entity becomes a party to a hybrid contract with a host that is not an asset within the scope of this Standard and with one or more embedded derivatives, paragraph 4.3.3 requires the entity to identify any such embedded derivative, assess whether it is required to be separated from the host contract and, for those that are required to be separated, measure the derivatives at fair value at initial recognition and subsequently. These requirements can be more complex, or result in less reliable measures, than measuring the entire instrument at fair value through profit or loss. For that reason this Standard permits the entire hybrid contract to be designated as at fair value through profit or loss.
*International Financial Reporting Standards (IFRS) are a set of accounting rules for the financial statements of public companies that are intended to make them consistent, transparent, and easily comparable around the world.
## Conclusions
DIVERSIFIED UNITED INVESTMENT LIMITED assigned short-term B2 & long-term B1 forecasted stock rating. We evaluate the prediction models Ensemble Learning (ML) with Lasso Regression1,2,3,4 and conclude that the DUI stock is predictable in the short/long term. According to price forecasts for (n+6 month) period, the dominant strategy among neural network is: Hold
### Financial State Forecast for DUI DIVERSIFIED UNITED INVESTMENT LIMITED Options & Futures
Rating Short-Term Long-Term Senior
Outlook*B2B1
Operational Risk 6538
Market Risk6169
Technical Analysis3873
Fundamental Analysis6589
Risk Unsystematic5735
### Prediction Confidence Score
Trust metric by Neural Network: 83 out of 100 with 751 signals.
## References
1. R. Rockafellar and S. Uryasev. Conditional value-at-risk for general loss distributions. Journal of Banking and Finance, 26(7):1443 – 1471, 2002
2. Allen, P. G. (1994), "Economic forecasting in agriculture," International Journal of Forecasting, 10, 81–135.
3. M. Benaim, J. Hofbauer, and S. Sorin. Stochastic approximations and differential inclusions, Part II: Appli- cations. Mathematics of Operations Research, 31(4):673–695, 2006
4. Doudchenko N, Imbens GW. 2016. Balancing, regression, difference-in-differences and synthetic control methods: a synthesis. NBER Work. Pap. 22791
5. Burkov A. 2019. The Hundred-Page Machine Learning Book. Quebec City, Can.: Andriy Burkov
6. D. Bertsekas. Min common/max crossing duality: A geometric view of conjugacy in convex optimization. Lab. for Information and Decision Systems, MIT, Tech. Rep. Report LIDS-P-2796, 2009
7. LeCun Y, Bengio Y, Hinton G. 2015. Deep learning. Nature 521:436–44
Frequently Asked QuestionsQ: What is the prediction methodology for DUI stock?
A: DUI stock prediction methodology: We evaluate the prediction models Ensemble Learning (ML) and Lasso Regression
Q: Is DUI stock a buy or sell?
A: The dominant strategy among neural network is to Hold DUI Stock.
Q: Is DIVERSIFIED UNITED INVESTMENT LIMITED stock a good investment?
A: The consensus rating for DIVERSIFIED UNITED INVESTMENT LIMITED is Hold and assigned short-term B2 & long-term B1 forecasted stock rating.
Q: What is the consensus rating of DUI stock?
A: The consensus rating for DUI is Hold.
Q: What is the prediction period for DUI stock?
A: The prediction period for DUI is (n+6 month) | 1,948 | 8,307 | {"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-2023-40 | latest | en | 0.877331 |
http://pdtermpaperiwfj.shapeyourworld.info/multiple-regression-solved-problems.html | 1,540,332,581,000,000,000 | text/html | crawl-data/CC-MAIN-2018-43/segments/1539583517495.99/warc/CC-MAIN-20181023220444-20181024001944-00062.warc.gz | 269,980,994 | 4,817 | # Multiple regression solved problems
Problems completed problems coursegrade prbgrd 51 62 3162 58 68 3944 62 66 4092 65 66 4290 68 67 4556 correlation and regression example solutions. Lesson 21: multiple linear regression analysis motivation and objective: we’ve spent a lot of time discussing simple linear regression, but simple. Linear regression and modeling problems with answers linear regression and modeling problems are presented the solutions to these problems are at the bottom of the page.
Few problems based on linear regression are given below: solved examples question 1: find linear linear multiple regression. Output from a multiple regression analysis is shown below response sat multiple regression practice problems stat 112. Regression step-by-step using microsoft excel when using multiple regression to estimate a relationship, there is always the possibility of correlation among.
Simple linear regression all problems addressed in this chapter deal with only two variables — the next chapter (reference) looks at extending the methods. Regression analysis and wicked business problems ambiguity, and it is never solved regression analysis and wicked business problems. In multiple regression, however, even small violations of these assumptions pose problems for confidence intervals on predictions for specific observations. To linear regression regression analysis is the art and science of fitting straight notes on linear regression analysis to spot data-quality problems, (c). Multiple linear regression is performed on a data set either to predict the response variable based on the predictor variable, or to study the relationship between the response variable and predictor variables.
We will study about the regression and also some example problems in regression in multiple regression coefficient $\beta_k$, solved example. Multiple regression example for a sample of n = 166 college students, the following variables were measured: y = height x1 = mother’s height (“momheight”) x2 = father’s height (“dadheight”. Multiple linear regression analysis is an extension of simple linear regression analysis, used to assess the association between two or more independent variables and a single continuous dependent variable the multiple linear regression equation is as follows: multiple regression analysis is also. The following example illustrates xlminer's multiple linear regression method using the boston housing data set to predict the median house prices in housing tracts. Linear regression models regression example, part 1: descriptive analysis any regression analysis.
Regression analysis treats all independent (x) one dummy variable can not be a constant multiple or a simple linear relation of another 3. Multiple linear regression solved i use this code to do multiple linear regression: if you search for problems with stepwise regression you will find. Practice questions multiple choice questions in the multiple regression model, the adjusted r2, r2 there are really problems in interpreting these,. Multiple linear regression a regression with two or more explanatory variables is called a multiple regression rather than modeling the mean response as a.
The problem to be solved is reduced (~ 1992 pergamon press ltd a solution to multiple linear regression problems with multiple regression on. For a multiple regression model, the adjusted coefficient of determination is used instead of the coefficient of determination to test the fit of the regression model. 12-1 multiple linear regression models • for example, suppose that the effective life of a cutting tool depends on the cutting speed and the tool angle.
Power regression problems use the data in the table below to obtain a model for speed p versus distance traveled d consider linear, quadratic, exponential,. Powerpoint slides solved problems several problems from the homework may be assigned to be solved using the multiple regression and correlation analysis. How to perform multiple regression analysis in excel. Use of multi-variables the multiple regression analysis models allow managers to test for several independent variables that may explain.
Multiple linear regression, in contrast to simple linear regression, involves multiple predictors and so testing each variable can quickly become complicated. I demonstrate how to perform a multiple regression in spss this is the in-depth video series i cover all of the main elements of a multiple regression anal. Regression analysis: a complete example this section works out an example that includes all the topics we have discussed so far in this chapter.
Multiple regression solved problems
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2018. | 840 | 4,718 | {"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} | 3.90625 | 4 | CC-MAIN-2018-43 | latest | en | 0.877326 |
https://forums.atomictorch.com/index.php?PHPSESSID=7ctmo74q6rq0cl10e92l3k8uq4&topic=9.30 | 1,675,235,125,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764499911.86/warc/CC-MAIN-20230201045500-20230201075500-00305.warc.gz | 265,349,667 | 8,749 | Author Topic: Modding the VoidExpanse (Read 43042 times)
Lurler
• AtomicTorch Founder
• Hero Member
• Posts: 1325
Re: Modding the VoidExpanse
« Reply #30 on: August 25, 2014, 02:12:39 am »
Mass of the ship does affect acceleration. Since we are using real physics it simply works as you'd expect in reality - more massive objects require more energy to be pumped into to give them momentum. Basically normal inertia.
As for your formulas - everything is exactly right
Hammish
• Sr. Member
• Posts: 350
Re: Modding the VoidExpanse
« Reply #31 on: August 26, 2014, 01:54:06 pm »
Awesome.
NEXT QUICK QUESTION: How does the damage calculation for beams go? From what I can tell, the damage listed is per second. So...
If I set a beam to 50 damage, 1s duration, it'll do 50 damage over 1s.
If I set a beam to 50 damage, 2s duration, it'll do 100 damage over 2s (50 DPS).
So how much damage will the beam do if I set it to 50 damage, 0.5s? I'm not sure if it'll do 50 in half the time, or just 25 and waste the rest.
Edit: I think I figured out what my issue was. I made a pulse laser and forgot to set projectile type to 1, so it was trying to calc it as a per-shot weapon. >.>
« Last Edit: August 26, 2014, 02:11:21 pm by Hammish »
Lurler
• AtomicTorch Founder
• Hero Member
• Posts: 1325
Re: Modding the VoidExpanse
« Reply #32 on: August 26, 2014, 08:01:59 pm »
Yes, the beam damage is per second. The duration doesn't matter in this case.
Armor also works per second in case of beam weapons.
Hammish
• Sr. Member
• Posts: 350
Re: Modding the VoidExpanse
« Reply #33 on: August 26, 2014, 10:54:40 pm »
Good to know on the armor. I had suspected, but y'know.
And just means a bit more math for me, on any beam under 1.0 duration, really. For instance, on the pulse lasers I wanted them to do 50 over 0.5 sec, so I set the damage to 100 and voila, issue solved!
Hammish
• Sr. Member
• Posts: 350
Re: Modding the VoidExpanse
« Reply #34 on: August 28, 2014, 05:43:17 am »
Next odd question for when you guys have a second or two.
For beam weapons, does duration_prepare and duration_complete actually affect damage numbers, or do they just impact the visual effect of the beam?
Lurler
• AtomicTorch Founder
• Hero Member
• Posts: 1325
Re: Modding the VoidExpanse
« Reply #35 on: August 28, 2014, 08:46:18 pm »
Yes, these are only for visual effects.
Hammish
• Sr. Member
• Posts: 350
Re: Modding the VoidExpanse
« Reply #36 on: August 30, 2014, 04:05:21 am »
Just the answer I was hoping for. You can make some cool wobbly pulse lasers by using the grow/shrink look of them.
Hammish
• Sr. Member
• Posts: 350
Re: Modding the VoidExpanse
« Reply #37 on: September 03, 2014, 09:52:51 pm »
QOTD: Is it possible to adjust character level higher than 1 when creating a new character?
I ask this merely for the purposes of creating test characters. It's a pain to either level a character and hunt down the stuff I want to test/experiment on or temporarily drop the usable levels on items/skills down to 1.
Lurler
• AtomicTorch Founder
• Hero Member
• Posts: 1325
Re: Modding the VoidExpanse
« Reply #38 on: September 04, 2014, 02:42:41 am »
Hm. IIRC - no
But you can use cheat command in SP that would make you into an almost top level character.
Here's how: (/pwn)
FlessenGreendart
• Full Member
• Posts: 206
Re: Modding the VoidExpanse
« Reply #39 on: September 04, 2014, 04:12:34 am »
Haha, brilliant command!
That'll come in handy when testing ships. On that note, do you have an add hull command? As it is at the moment, I'm having to drop the price of the hull, and throw in a civillian flag
Lurler
• AtomicTorch Founder
• Hero Member
• Posts: 1325
Re: Modding the VoidExpanse
« Reply #40 on: September 05, 2014, 12:14:51 am »
Yes, there are a lot of different commands available, all of them are listed in our wiki here - http://wiki.atomictorch.com/Main_Page
FlessenGreendart
• Full Member
• Posts: 206
Re: Modding the VoidExpanse
« Reply #41 on: September 05, 2014, 01:32:13 am »
Brilliant, thanks for that, that'll make test a lot easier!
Hammish
• Sr. Member
• Posts: 350
Re: Modding the VoidExpanse
« Reply #42 on: September 07, 2014, 03:26:51 am »
That is indeed awesome and a huge help. All of it.
And my question for the day (it's been a while!): Is there any way to set the Z-layer for weapon effects? I've noticed they always seem to draw under the ship, at least the ones I do, and that looks odd for weapon fire coming from top-mounted turrets.
Lurler
• AtomicTorch Founder
• Hero Member
• Posts: 1325
Re: Modding the VoidExpanse
« Reply #43 on: September 07, 2014, 11:32:07 pm »
Well, when you define weapon position you can set any Z value. Then the fire will be coming our of that position on the ship. As with other weapons that are already in the game. It is defined in the weapon slots of a ship. A particular XYZ position.
Hammish
• Sr. Member
• Posts: 350
Re: Modding the VoidExpanse
« Reply #44 on: September 08, 2014, 01:38:13 am »
Right, that I get, but even for weapons that are mounted on the top, some of my effects seem to appear drawn under the ship.
Example: the energy hardpoint on the Shuttle is coords -0.227;0.000;-0.203, so from the cockpit it's toward the back and elevated slightly. Should be well above the centerline of the ship (since it's a roof-mounted turret). But if I fire a non-beam projectile from it, the projectile isn't visible until it move out from the location occupied by the ship; for all intents and purposes the shot appears to be either inside or under the ship, instead of originating above it.
Not a huge priority for you guys, it just looks a little jarring to me now that I've noticed it. For smaller, darker shots it barely even registers but you can see the effect clearly if you use a slow-moving missile projectile or a brighter sprite. | 1,716 | 5,856 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2023-06 | latest | en | 0.886173 |
http://thibaultlanxade.com/general/a-waffle-cone-for-ice-cream-has-a-diameter-of-8-centimeters-and-a-height-of-12-centimeters-what-is-the-volume-of-the-cone-in-cubic-centimeters-a-64-cubic-centimetersb-32-cubic-centimetersc-256-cubic-centimetersd-192-cubic-centimeterscasandr | 1,722,818,669,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640417235.15/warc/CC-MAIN-20240804230158-20240805020158-00777.warc.gz | 25,666,035 | 7,583 | Q:
A waffle cone for ice cream has a diameter of 8 centimeters and a height of 12 centimeters. What is the volume of the cone, in cubic centimeters?A. 64π cubic centimetersB. 32π cubic centimetersc. 256π cubic centimetersD. 192π cubic centimetersCasandra finds a treasure chest packed with metallic coins. The chest has a volume of 0.25 cubic meters. The coins have a combined mass of 4825 kg. Hoping to find gold, she calculates the density to determine the metal of the coins.What kind of metal are the coins made of?Bronze (8700 kg per cubic meter)Silver (10500 kg per cubic meter)Lead (11300 kg per cubic meter)Gold (19300 kg per cubic meter)A cylinder with radius 3 units is shown. Its volume is 86 cubic units.Find the height of the cylinder.Use 3.14 for π and round your final answer to the nearest hundredth.
Accepted Solution
A:
answer A
Gold
4.56 for the height | 229 | 874 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.34375 | 3 | CC-MAIN-2024-33 | latest | en | 0.777547 |
https://certifiedcalculator.com/free-walking-weight-loss-calculator/ | 1,713,367,382,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817158.8/warc/CC-MAIN-20240417142102-20240417172102-00650.warc.gz | 151,155,262 | 13,752 | # Free Walking Weight Loss Calculator
Introduction: Walking is a natural and accessible form of exercise that can contribute significantly to weight loss. The Free Walking Weight Loss Calculator is a valuable tool designed to estimate the calories burned during your walks, empowering you to tailor your walking routine to achieve your weight loss goals effectively.
Formula: The calculator employs a simple formula to estimate calories burned: Calories Burned = Duration (in minutes) × Distance (in miles) × 0.314. By multiplying the duration and distance of your walk, the calculator provides an approximation of the calories burned based on a standard factor.
How to Use: Enter the duration of your walk in minutes and the distance covered in miles into the respective input fields. Click the “Calculate” button, and the estimated calories burned will be displayed in the result field.
Example: If you walk for 45 minutes and cover a distance of 2 miles, enter 45 for the duration and 2 for the distance. Click “Calculate,” and the result will show an estimated calories burned during the walk.
FAQs:
1. Q: Is walking an effective way to lose weight?
• A: Yes, walking is a low-impact exercise that can contribute to weight loss when combined with a balanced diet.
2. Q: Can I use this calculator for other forms of exercise?
• A: The calculator is specifically designed for walking. For other exercises, consider using specialized calculators.
3. Q: How accurate is the calorie estimation?
• A: The calculator provides a general estimation. Individual calorie burn may vary based on factors like speed, terrain, and personal metabolism.
4. Q: Should I consider my resting metabolic rate in the calculation?
• A: The calculator is designed to estimate additional calories burned during walking. Your resting metabolic rate is not included.
5. Q: Can I use this calculator for indoor walking activities?
• A: The calculator is versatile and applicable to both indoor and outdoor walking activities.
Conclusion: The Free Walking Weight Loss Calculator is a practical tool for individuals looking to incorporate walking into their weight loss journey. By estimating calories burned, this calculator supports you in optimizing your walking routine and reaching your fitness goals. Walk your way to a healthier, fitter you! | 452 | 2,328 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.8125 | 3 | CC-MAIN-2024-18 | latest | en | 0.913149 |
https://www.sfgate.com/green/article/Even-Mother-Nature-plays-the-numbers-game-2876663.php | 1,544,661,265,000,000,000 | text/html | crawl-data/CC-MAIN-2018-51/segments/1544376824180.12/warc/CC-MAIN-20181212225044-20181213010544-00125.warc.gz | 1,027,181,353 | 49,415 | Let's start this gig with two questions (with a few more on the way):
PIG MATH: It takes six months for a wild pig to come into season, a sow 121 days to have a litter, and then 21 days later, they are back in season. In each litter, they average 4.5 piglets that survive. Of these, figure half of those piglets are female. So over the course of five years, one sow can produce how many related offspring, including descendants?
-- PYTHAGORAS LAW: When standing on top of 4,343-foot Mount St. Helena, is it physically possible to see 14,162-foot Mt. Shasta, 192 miles to the north?
Can you do pig math? In the Bay Area foothills, many are alarmed that wild pigs seem to be expanding their population like rats with hooves. That is despite a program where more than 2,000 wild pigs have been killed by professional trappers.
ANSWERS: A number of mathematicians volunteered to crunch the pig numbers. What they found is astounding.
Over the course of five years, one sow would produce 10 generations of surviving female piglets. These pigs would give birth to a total of 15,321 related surviving female piglets. "Double this if you want to include the males," said Van Rackages, a math wizard who produced a work sheet to document the numbers.
The ability to reproduce in great numbers is why wild pigs pose the biggest future danger to the off-limit watershed lands in the Bay Area (not public access for hiking, biking or fishing), as well as the 125 parklands in the Bay Area foothills.
Wild pigs spend night after night roto-tilling hillsides while rooting for food. They impact wildlife such as deer, squirrel and bird species by winning the competition for acorns, and by eliminating oak saplings -- thus reducing a habitat's long-term ability to produce food (acorns).
By churning up the ground, wild pigs can also destroy the habitat of endangered amphibians and reptiles. What scares water managers is that rooting activity can cause erosion and silt problems in lakes from storm runoff by- product on watershed lands; the Bay Area has nearly 50 lakes that provide water supplies.
As to the other question -- how far can you potentially see from a mountain lookout -- is answered with a "simple formula" called Pythagoras Law, explained Werner Weber, a scientist from Germany commenting via the Internet.
"Call the distance to the horizon d, the height above sea level h and the radius of the sphere of the earth R. The diameter of the earth sphere is (d2R).
Then d square root (2Rh) square root (Dh) follows."
Simple, right? "Of course, now it all makes sense" . . . well, Werner then crunched out that the maximum viewable distance from peak to peak, Mount St. Helena to Mt. Shasta, and reasoned it out at 213 miles, within the range of Mt.
Shasta (192 miles to the north).
Though few have ever seen Mt. Shasta from the top of Mount St. Helena (and many have mistaken Mt. Lassen for Mt. Shasta), fieldscout and author friend Barry Spitz reported that he was lucky enough to get this rare view when summiting on a very cold, clear and windy winter day. By the way, when looking out from Mount St. Helena to sea level, using the Pythagoras Law, the maximum viewable distance is about 80 miles, which is why the Farallon Islands (65 miles away) can be spotted on exceptional days.
STONEWALLING AGAIN: In case you missed it, the San Francisco Public Utilities Commission decided to delay a vote on allowing sharply-controlled hiking and biking at Crystal Springs Watershed in order to "study" the proposal. This issue has been delayed for nearly five years and this new "study" is simply a ploy to delay the issue for several more years. Unless Mayor Willie Brown takes charge, these guys will stonewall the issue forever.
WILLIAMS CONFIRMED: Moments before President Bush delivered the State of the Union speech, Steve Williams was confirmed by the U.S. Senate as director of the U.S. Fish and Wildlife Service.
"Sportsmen and women of the country have been the primary financiers of fish and wildlife conservation," Williams said. "I would like to take steps to shore up what was once a powerful relationship between the Service and our nation's hunters and anglers."
THE ANSWER MAN, CONT.: About a dozen people wrote in who were certain I was wrong in writing that Mount St. Helena was in Sonoma County. Nope. The trailhead is in Napa County, then as you head up, about 50 feet of trail is in Lake County, and the last half mile of trail and summit is in Sonoma County. In fact, the three county lines junction about 100 feet off the trail.
HAS ANYTHING CHANGED?: It took President Bush more than a year to fill the top job at the Fish and Wildlife Service, and many agency and department jobs are still waiting for appointments. That means that career professionals are running the show out of local district offices, just as they have done for years. Except for a few flashpoint issues, there is little change on the ground despite a general fear among environmentalists that the president would turn the clock back to the days of Interior Secretary James Watt, when timber harvest, grazing and mining dominated public lands.
LAST CALL: The International Sportsmen's Exposition closes out its run today at the San Mateo Expo Center, and with decreased crowds expected because of the Super Bowl telecast, today's show will provide a rare chance for a lot of one-on-one with the West's top fishing, hunting and travel experts.
I'll be giving the show "Sturgeon Fishing Done Easy" at 11:30 a.m. in the Adventure Theater in Cypress Hall. | 1,231 | 5,564 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2018-51 | latest | en | 0.953584 |
http://decisionprocesstheory.com/2017/11/ | 1,519,126,712,000,000,000 | text/html | crawl-data/CC-MAIN-2018-09/segments/1518891812938.85/warc/CC-MAIN-20180220110011-20180220130011-00473.warc.gz | 93,597,896 | 8,092 | # Tyrant Model using Decision Process Theory
A classic game theory model for a tyrant might be a game of chicken, with a tyrant against the poor, each choosing between swerving or crashing. The difference however is that a tyrant has more power and influence than the poor. One aspect of game theory that we believe needs to be changed is that the size of the payoffs should matter.
As an example of something that plays no role in game theory, consider the concept of engagement. Multiplying the payoff matrix by the engagement does not change the max-min solution of game theory and is thus of no importance. Moreover, the payoff matrix strength is not relevant either. In physics, this is like saying that the charge of a particle does not influence its motion in a magnetic field, nor does the strength of the magnetic field. This is only true for a charge moving parallel to the field. The circulation around the field clearly depends on the charge as well as on the strength of the field.
The hard part is to disentangle the charge from the field strength. This is accomplished in physics by the field equations that say that charges and currents are the sources of the fields. Strong engagement leads to changes in the players behaviors through changes in their payoff and valuation fields. What is amazing however, is that engagement like viscosity enters the decision process theory equations in a non-linear way. Therefore, dramatically increasing or decreasing the engagement changes the qualitative behavior of the decision flow. You get an inkling of why when you realize that the engagement is itself a conserved flow and therefore contributes to the energy momentum of the system. It thus gets shared with other component parts.
Like viscosity, when the engagement is small it has almost no effect other than adjusting the time scale. As it gets larger, the non-linear nature of its behavior becomes evident. It is even possible that we might get “chaotic” or “turbulent” behaviors, though there is currently no basis for this conjecture. The following is the result of a computation (using a decision engineering notebook described in the Stationary Ownership Model Update white paper and implemented using a Mathematica notebook) in which the poor outnumber the tyrant by 10:1 and the engagement of the tyrant to the poor is 1:10.
We have the following interpretation of this picture. From a game theory perspective, the model has two Nash equilibriums. The poor see their strategy to be “swerve” and assume that the wealthy will “crash”. The wealthy see the opposite. This is one variant of the classic game of chicken. It is clear in the figure that by assuming the wealthy are much less engaged, we are closer to one of the Nash equilibrium: we see the Nash equilibrium of the poor, which is to “swerve” and for the wealthy to “crash”.
We see more however. Because of the inequality between the two players, the preference of the poor to crash is going to be smaller over time than for the wealthy to crash. So, in some sense, the wealthy crashing is not as devastating as it is for the poor. There is much less incentive for the wealthy to avoid the choice “swerve”.
The strategic flows add some interesting aspects to the story. The magnitude of the flows in this model run are all comparable. Nevertheless, they support the fact that the net behavior for the wealthy to swerve is zero and the net behavior of the wealthy to crash is small and positive. It is also clear that the net flow for the poor to swerve is much bigger than any of the other net flows.
We interpret this as follows. The poor will suffer and avoid the game of chicken. On average they will swerve. The wealthy will crash, making what appears to be the correct assumption that the poor will always swerve. These assumptions, however, rely on the small engagement of the wealthy. This appears to be the trademark of the tyrant They suffer no consequence for their action and are thus rewarded. The full decision process theory would go on to predict that the payoffs for the wealthy should not change since their source is small. The poor however might change their payoffs over time to better reflect what is happening.
Is there anything that can be done? One possibility is demand an equality of effort between the tyrant and the people, through laws, the courts or the press, as examples. An example computation is provided below, using a gravitational like field to force equality by imagining a potential well that is centered when the strategic effort of the poor matches the strategic effort of the tyrant.
The model generates a time component of the metric, which can be thought of as a “gravitational” field that pulls all strategies to this common center. We see the consequence of this in the above figure. We now expect that the tyrant will see consequences to their action, despite their initial lack of engagement. In fact it might be that that engagement will also change as a consequence. | 1,023 | 5,007 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.765625 | 3 | CC-MAIN-2018-09 | latest | en | 0.952636 |
http://mymathangels.com/problem-1106/ | 1,513,511,794,000,000,000 | text/html | crawl-data/CC-MAIN-2017-51/segments/1512948595858.79/warc/CC-MAIN-20171217113308-20171217135308-00598.warc.gz | 182,871,402 | 7,360 | ## Problem 1106
Convert the angle in radians to degrees. Express your answer in decimal form.
5.63
Solution:-
Consider a circle of radius r. A central angle of a revolution (360°) will subtend an arc equal to the circumference of the circle. Because the circumference of the circle equal 2πr, we see that 360° = 2π radians, or 180° = π radians. Dividing both sides by π, we have the following conversion formula.
Converting form radians to degrees can be summarized as follows.
1 radian = degrees
Therefore, the following is true.
5.63 radians = 5.63 * 1 radian
= 5.63 * degrees
322.58° | 157 | 596 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2017-51 | latest | en | 0.804521 |
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# Improved GMAT by 100 points - supplemental essay needed?
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Senior Manager
Status: Decision Week!
Joined: 12 Oct 2011
Posts: 262
Kudos [?]: 82 [0], given: 58
Concentration: General Management, Strategy
Schools: Dartmouth (Tuck) - Class of 2014
GMAT 1: 740 Q48 V44
GPA: 3.66
WE: Management Consulting (Consulting)
Improved GMAT by 100 points - supplemental essay needed? [#permalink]
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05 Dec 2011, 18:29
Hi Paul Lanzillotti & Co.,
I've taken the GMAT twice - first time was in October and second time today. First go around I was really upset about my 640 (31Q 44V 6 AWA) as I'd been practice-testing in the 690-710 range consistently.
I refocused my efforts, realized I'd been prepping the quant in the wrong ways, and retested today at 740 (48Q, 44V AWA TBD). As you can imagine I am THRILLED with my new score. I believe that my first quant score was somewhat of an anomaly due to preparing incorrectly and getting a question set that played to my weaknesses.
Question: How, if at all, do I need to address a 100 point GMAT improvement (all in quant) on my applications?
_________________
Kudos [?]: 82 [0], given: 58
Joined: 25 Jan 2010
Posts: 1026
Kudos [?]: 221 [0], given: 220
Re: Improved GMAT by 100 points - supplemental essay needed? [#permalink]
### Show Tags
07 Dec 2011, 03:53
thanks for reaching out to me. While this may seem like a fairly straightforward question, I actually believe that it is a fairly insightful question. To answer your question quickly, the admissions committee does not need to know about your 100 point increase. Of course, they will see both GMAT scores on the official report, but I suppose this is really beside the point.
What I want to expand upon here is a couple of things. First, it is always important to view things from the admissions committee point of view. While this is easier for me to say then for you to do, I want you to pretend that you are a member of the committee. Now, imagine you are reviewing hundreds, if not thousands of applications. As a member of the admissions committee, you really want to know the bare minimum of information that will paint a picture. Basically, this is why it is really not notable to mention this information. Second, a significant percentage of applicants take the test more than once and score higher the second time through. Adding this information will also serve to add baggage to your application and frankly, might even make it more annoying to read through your application. So don't do it. Just leave it be and give yourself a high-five. Seriously -- 100 points -- nice work!
Respectfully,
Paul Lanzillotti
Hi Paul Lanzillotti & Co.,
I've taken the GMAT twice - first time was in October and second time today. First go around I was really upset about my 640 (31Q 44V 6 AWA) as I'd been practice-testing in the 690-710 range consistently.
I refocused my efforts, realized I'd been prepping the quant in the wrong ways, and retested today at 740 (48Q, 44V AWA TBD). As you can imagine I am THRILLED with my new score. I believe that my first quant score was somewhat of an anomaly due to preparing incorrectly and getting a question set that played to my weaknesses.
Question: How, if at all, do I need to address a 100 point GMAT improvement (all in quant) on my applications?
_________________
Paul Lanzillotti | Founder| About | mba@amerasiaconsulting.com | 877.866.9251
Schedule a Consultation | Twitter | Blog
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Kudos [?]: 221 [0], given: 220
Senior Manager
Status: Decision Week!
Joined: 12 Oct 2011
Posts: 262
Kudos [?]: 82 [0], given: 58
Concentration: General Management, Strategy
Schools: Dartmouth (Tuck) - Class of 2014
GMAT 1: 740 Q48 V44
GPA: 3.66
WE: Management Consulting (Consulting)
Re: Improved GMAT by 100 points - supplemental essay needed? [#permalink]
### Show Tags
09 Dec 2011, 08:02
Thank you!
I suspected this may be the case but it is great to have someone of your experience confirm this.
_________________
Kudos [?]: 82 [0], given: 58
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# Improved GMAT by 100 points - supplemental essay needed?
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Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®. | 1,288 | 5,197 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.3125 | 3 | CC-MAIN-2017-43 | latest | en | 0.923902 |
https://techevar.com/qa/quick-answer-what-is-the-significance-of-3.html | 1,611,524,815,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703557462.87/warc/CC-MAIN-20210124204052-20210124234052-00268.warc.gz | 596,907,275 | 8,788 | # Quick Answer: What Is The Significance Of 3?
## Why is 3 a powerful number?
Three is the smallest number we need to create a pattern, the perfect combination of brevity and rhythm.
It’s a principle captured neatly in the Latin phrase omne trium perfectum: everything that comes in threes is perfect, or, every set of three is complete..
## What are the factors of 12?
A prime factor of a number is just a factor of that number that is also prime. So, 12 has six factors — 1, 2, 3, 4, 6, and 12 — but only two of them (2 and 3) are prime, so it has only two prime factors.
## What is the most evil number?
sum to 666 is known as an evil number (Pegg and Lomont 2004).
## Is 7 a magic number?
This limit, which psychologists dubbed the “magical number seven” when they discovered it in the 1950s, is the typical capacity of what’s called the brain’s working memory. Now physicists have come up with a model of brain activity that seems to explain the reason behind the magical memory number.
## What do 3 mean spiritually?
The number 3 biblically represents divine wholeness, completeness and perfection. If there ever was a desire to highlight an idea, thought, event or noteworthy figure in the Bible for their prominence, the number 3 was used to put a divine stamp of completion or fulfillment on the subject.
## Why is 3 a sacred number?
The Meaning of Number 3 The number 3 carries a vibrational frequency that is associated with self expression and creativity. It is the number of creation and completion, associated with divinity in many of the world’s religions, particularly Christianity.
## Why is 3 considered a magic number?
In their eyes the number 3 was considered as the perfect number, the number of harmony, wisdom and understanding. … It was also the number of time – past, present, future; birth, life, death; beginning, middle, end – it was the number of the divine.
## Is 3 a evil number?
Examples. The first evil numbers are: 0, 3, 5, 6, 9, 10, 12, 15, 17, 18, 20, 23, 24, 27, 29, 30, 33, 34, 36, 39 …
## Why is 12 the perfect number?
Twelve is a sublime number, a number that has a perfect number of divisors, and the sum of its divisors is also a perfect number. Since there is a subset of 12’s proper divisors that add up to 12 (all of them but with 4 excluded), 12 is a semiperfect number.
## Is 7 the perfect number?
The number seven is one of the most significant in the Bible. Scholars say it denotes completeness or perfection. After creating the world God rested on the seventh day and the seven-day week has been adopted by all human civilisations.
## What is the meaning of 8?
The number eight is considered to be a lucky number in Chinese and other Asian cultures. Eight (八; accounting 捌; pinyin bā) is considered a lucky number in Chinese culture because it sounds like the word meaning to generate wealth (發(T) 发(S); Pinyin: fā). Property with the number 8 may be valued greatly by Chinese.
## How do I find myself spiritually?
Seven Ways to Improve Your Spiritual HealthExplore your spiritual core. By exploring your spiritual core, you are simply asking yourself questions about the person you are and your meaning. … Look for deeper meanings. … Get it out. … Try yoga. … Travel. … Think positively. … Take time to meditate.
## What does the number 3 symbolize?
There are many interesting facts related to number 3. First of all, we can say that number 3 is associated with the Trinity. It means that you receive the protection, guidance and help from divine forces. Your angels are protecting you and giving you support and strength.
## What does the number 3 mean in literature?
Three represents the triad of family; male, female, and child; beginning, middle, and end; birth, life, and death. Of two things we say both; of three things we say all.
## What is the highest name of God?
YahwehYahweh is the principal name in the Old Testament by which God reveals himself and is the most sacred, distinctive and incommunicable name of God.
## What is base 12 called?
The duodecimal system (also known as base 12, dozenal, or, rarely, uncial) is a positional notation numeral system using twelve as its base.
## Is number 3 a good number?
“The number three is playful, it’s inspirational,” said Hans Decoz, the author of “Numerology: Key to your Inner Self.” “It’s a happy-go-lucky number, very optimistic. It lifts up people around it, doesn’t take things too serious.
## Who are the 3 angels of God?
The Protestant Bible provides names for three angels: “Michael the archangel”, the angel Gabriel, who is called “the man Gabriel” in Daniel 9:21 and third “Abaddon”/”Apollyon” in Revelation 9:11.
## Why is the number 3 important in literature?
The rule of three is a writing principle that suggests that a trio of events or characters is more humorous, satisfying, or effective than other numbers. … The Latin phrase “omne trium perfectum” (everything that comes in threes is perfect, or, every set of three is complete) conveys the same idea as the rule of three.
## What number means death?
The number 4 (四, pinyin: sì; Cantonese Yale: sei) is considered an unlucky number in Chinese because it is nearly homophonous to the word “death” (死 pinyin: sǐ; Cantonese Yale: séi). | 1,279 | 5,256 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2021-04 | latest | en | 0.957259 |
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DPS Excel calculator for Sorcerer / Sage
01-11-2012, 10:20 PM (This post was last modified: 01-12-2012 01:17 AM by lostdummy.)
Post: #27
lostdummy Respected Member Posts: 100 Joined: Nov 2011 Reputation: 1
RE: DPS Excel calculator for SI Sorcerer
(01-08-2012 12:05 PM)Kor Wrote: That could simply be deviation in the randomness. Theoretically, any multiple of 30 seconds should have roughly the same values, with longer fights coming closer to the actual value (since we're starting at 0 stacks).
I revisited Conduction average stack calculation, since I found how to calculate values without simulation, meaning results are exact accurate numbers.
It is based on my initial idea of recursive calculation, which in simplified pseudocode would look like:
Code:
```double SpellsPerBuff, SpellsPerFight; //globals, to simplify code // recursive proc double avgStack( int SpellsRemain, SpellsNoProc, Stack){ if (SpellsRemain<=0) return 0; newStackProc = min(Stack+1,3); ResProc= newStackProc/SpellsPerFight + avgStack( SpellsRemain-1 , 1, newStackProc); newStackNoProc = ( SpellsNoProc < SpellsPerBuff ) ? Stack:0 ; ResNoProc= newStackNoProc/SpellsPerFight +avgStack( SpellsRemain-1, SpellsNoProc +1, newStackNoProc); return 0.3* ResProc + 0.7*ResNoProc; } // average stack if LS casted every 10sec in 300sec fight: SpellsPerBuff=30/10; // buff duration 30sec divided by spell cast every 10sec SpellsPerFight=300/30*SpellsPerBuff; // number of spells used over whole fight averageStack = avgStack( SpellsPerFight, 0, 0);```
Above is fairly simple, although actual code gets bit more complicated when you want to support buff and fight durations not in exact multiples of spell recast time (also my actual code is pascal, I just rewrote this as simplified pseudocode, so maybe I introduced some error, but it shows general idea). Important thing is, it results in exact result for given input values of spell frequency and fight duration, ie based on probability math rather than simulation.
Reason why this was only my 'initial idea' is fact that used like this it would be too slow to calculate for any longer fight - it double number of function calls every step, meaning total number of function calls is around 2^steps. And it has 'SpellsPerFight' steps, so in case of cast every 1.5 sec over 300sec fight, that is 2^200 steps (~10^60), making it uncalculable. Even smaller cases (cast every 10sec over 300sec) is over billion steps.
But now I figured how to optimize/speed that up, and good news is that our simulation numbers are correct ;p This also confirms that numbers are slightly different for different fight duration, but basically numbers that I posted for 180sec, 300sec and 450sec simulation fights are correct on 2nd decimal.
I had idea to write this function in VBA code and call that instead of using interpolation, but I don't know how to set initial global values in excel VBA (ie before calculation iterations start), so my optimization would not be doable - and that would mean unusably slow VBA function. Not to mention that including VBA code result in that security 'Enable macro' question every time you open document.
So, instead of adding VBA function, I extended interpolation to be "two dimensional" interpolation, ie to cover different fight durations, as well as already covered different LS frequencies (in version 1.35).
BTW, example of calculated numbers:
Code:
```Cst Cst/ Avg @ 300s every buff stacks ----------------------- 01.5 20.00 2.910 03.0 10.00 2.662 04.5 06.67 2.267 06.0 05.00 1.827 07.5 04.00 1.494 09.0 03.33 1.352 10.0 03.00 1.101 15.0 02.00 0.682 20.0 01.50 0.540 25.0 01.20 0.457 30.0 01.00 0.300 -----------------------```
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Messages In This Thread DPS Excel calculator for Sorcerer / Sage - lostdummy - 01-06-2012, 10:16 PM RE: DPS Excel calculator for SI Sorcerer - CaseyTheRetard - 01-07-2012, 04:35 AM RE: DPS Excel calculator for SI Sorcerer - Kaedis - 01-07-2012, 04:45 AM RE: DPS Excel calculator for SI Sorcerer - CaseyTheRetard - 01-07-2012, 05:42 AM RE: DPS Excel calculator for SI Sorcerer - lostdummy - 01-07-2012, 08:53 AM RE: DPS Excel calculator for SI Sorcerer - CaseyTheRetard - 01-07-2012, 10:14 AM RE: DPS Excel calculator for SI Sorcerer - Kaedis - 01-07-2012, 10:51 AM RE: DPS Excel calculator for SI Sorcerer - lostdummy - 01-07-2012, 09:21 PM RE: DPS Excel calculator for SI Sorcerer - lostdummy - 01-08-2012, 02:50 AM RE: DPS Excel calculator for SI Sorcerer - Kaedis - 01-08-2012, 04:34 AM RE: DPS Excel calculator for SI Sorcerer - lostdummy - 01-08-2012, 04:42 AM RE: DPS Excel calculator for SI Sorcerer - Kaedis - 01-08-2012, 05:20 AM RE: DPS Excel calculator for SI Sorcerer - Kaedis - 01-08-2012, 09:11 AM RE: DPS Excel calculator for SI Sorcerer - lostdummy - 01-08-2012, 10:28 AM RE: DPS Excel calculator for SI Sorcerer - Kaedis - 01-08-2012, 10:30 AM RE: DPS Excel calculator for SI Sorcerer - lostdummy - 01-08-2012, 11:00 AM RE: DPS Excel calculator for SI Sorcerer - Kaedis - 01-08-2012, 12:05 PM RE: DPS Excel calculator for SI Sorcerer - lostdummy - 01-08-2012, 08:46 PM RE: DPS Excel calculator for SI Sorcerer - lostdummy - 01-11-2012 10:20 PM RE: DPS Excel calculator for SI Sorcerer - lostdummy - 01-08-2012, 09:43 PM RE: DPS Excel calculator for SI Sorcerer - Pestulan - 01-09-2012, 04:37 PM RE: DPS Excel calculator for SI Sorcerer - lostdummy - 01-09-2012, 07:09 PM RE: DPS Excel calculator for SI Sorcerer - Lucai - 01-10-2012, 03:32 PM RE: DPS Excel calculator for SI Sorcerer - lostdummy - 01-10-2012, 07:10 PM RE: DPS Excel calculator for SI Sorcerer - frmorrison - 01-10-2012, 04:09 PM RE: DPS Excel calculator for SI Sorcerer - lostdummy - 01-10-2012, 07:35 PM RE: DPS Excel calculator for SI Sorcerer - Pestulan - 01-11-2012, 12:39 PM RE: DPS Excel calculator for SI Sorcerer - Erasmian - 01-12-2012, 05:27 AM RE: DPS Excel calculator for SI Sorcerer - lostdummy - 01-12-2012, 08:31 AM RE: DPS Excel calculator for SI Sorcerer - Kaedis - 01-12-2012, 05:36 AM RE: DPS Excel calculator for SI Sorcerer - lostdummy - 01-13-2012, 01:03 AM RE: DPS Excel calculator for SI Sorcerer - Saitta - 01-13-2012, 04:30 AM RE: DPS Excel calculator for SI Sorcerer - lostdummy - 01-13-2012, 06:30 AM RE: DPS Excel calculator for SI Sorcerer - frmorrison - 01-13-2012, 07:05 AM RE: DPS Excel calculator for SI Sorcerer - Lucai - 01-13-2012, 07:21 AM RE: DPS Excel calculator for SI Sorcerer - CaseyTheRetard - 01-13-2012, 07:46 AM RE: DPS Excel calculator for SI Sorcerer - lostdummy - 01-13-2012, 08:08 AM RE: DPS Excel calculator for SI Sorcerer - CaseyTheRetard - 01-13-2012, 08:47 AM RE: DPS Excel calculator for SI Sorcerer - lostdummy - 01-13-2012, 09:18 AM RE: DPS Excel calculator for SI Sorcerer - CaseyTheRetard - 01-13-2012, 09:24 AM RE: DPS Excel calculator for SI Sorcerer - lostdummy - 01-13-2012, 10:26 AM RE: DPS Excel calculator for SI Sorcerer - CaseyTheRetard - 01-14-2012, 12:51 AM RE: DPS Excel calculator for SI Sorcerer - lostdummy - 01-14-2012, 01:36 AM RE: DPS Excel calculator for SI Sorcerer - lostdummy - 01-14-2012, 09:36 PM RE: DPS Excel calculator for SI Sorcerer - Kaedis - 01-15-2012, 01:48 AM RE: DPS Excel calculator for SI Sorcerer - lostdummy - 01-16-2012, 07:29 AM RE: DPS Excel calculator for SI Sorcerer - lostdummy - 01-19-2012, 11:08 PM RE: DPS Excel calculator for SI Sorcerer - DPSFTW - 01-20-2012, 03:09 AM RE: DPS Excel calculator for SI Sorcerer - frmorrison - 01-20-2012, 06:07 AM RE: DPS Excel calculator for SI Sorcerer - lostdummy - 01-20-2012, 11:31 PM RE: DPS Excel calculator for SI Sorcerer - CaseyTheRetard - 01-21-2012, 05:24 AM RE: DPS Excel calculator for SI Sorcerer - lostdummy - 01-21-2012, 06:38 AM RE: DPS Excel calculator for SI Sorcerer - lostdummy - 01-24-2012, 11:58 PM RE: DPS Excel calculator for SI Sorcerer - Caltiom - 01-25-2012, 06:06 PM RE: DPS Excel calculator for SI Sorcerer - lostdummy - 01-25-2012, 07:15 PM RE: DPS Excel calculator for Sorcerer / Sage - Kaedis - 01-26-2012, 01:29 AM RE: DPS Excel calculator for Sorcerer / Sage - Armageddus - 01-26-2012, 10:37 AM RE: DPS Excel calculator for Sorcerer / Sage - lostdummy - 01-26-2012, 07:47 PM RE: DPS Excel calculator for Sorcerer / Sage - lostdummy - 01-26-2012, 08:43 PM RE: DPS Excel calculator for Sorcerer / Sage - runlevelsix - 01-27-2012, 12:13 AM RE: DPS Excel calculator for Sorcerer / Sage - lostdummy - 01-27-2012, 12:52 AM RE: DPS Excel calculator for Sorcerer / Sage - haldirgoth - 01-26-2012, 11:47 PM RE: DPS Excel calculator for Sorcerer / Sage - lostdummy - 01-27-2012, 01:08 AM RE: DPS Excel calculator for Sorcerer / Sage - frmorrison - 01-27-2012, 05:55 AM RE: DPS Excel calculator for Sorcerer / Sage - lostdummy - 01-27-2012, 07:16 PM RE: DPS Excel calculator for Sorcerer / Sage - CaseyTheRetard - 01-31-2012, 03:33 AM RE: DPS Excel calculator for Sorcerer / Sage - Khadroth - 01-31-2012, 11:55 PM RE: DPS Excel calculator for Sorcerer / Sage - Armageddus - 01-27-2012, 10:49 AM RE: DPS Excel calculator for Sorcerer / Sage - CaseyTheRetard - 01-28-2012, 04:37 AM RE: DPS Excel calculator for Sorcerer / Sage - lostdummy - 01-30-2012, 11:41 PM RE: DPS Excel calculator for Sorcerer / Sage - frmorrison - 02-01-2012, 03:35 AM RE: DPS Excel calculator for Sorcerer / Sage - Khadroth - 02-01-2012, 03:49 AM RE: DPS Excel calculator for Sorcerer / Sage - zii - 02-02-2012, 08:06 AM RE: DPS Excel calculator for Sorcerer / Sage - CaseyTheRetard - 02-02-2012, 09:13 AM RE: DPS Excel calculator for Sorcerer / Sage - zii - 02-02-2012, 12:47 PM RE: DPS Excel calculator for Sorcerer / Sage - frmorrison - 02-02-2012, 03:47 PM RE: DPS Excel calculator for Sorcerer / Sage - Bored - 02-02-2012, 11:01 PM RE: DPS Excel calculator for Sorcerer / Sage - zii - 02-03-2012, 04:18 AM RE: DPS Excel calculator for Sorcerer / Sage - CaseyTheRetard - 02-03-2012, 05:00 AM RE: DPS Excel calculator for Sorcerer / Sage - Erasmian - 02-08-2012, 03:58 AM RE: DPS Excel calculator for Sorcerer / Sage - 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https://ftp.aimsciences.org/article/doi/10.3934/krm.2013.6.589 | 1,656,751,646,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103989282.58/warc/CC-MAIN-20220702071223-20220702101223-00220.warc.gz | 304,798,852 | 16,843 | # American Institute of Mathematical Sciences
September 2013, 6(3): 589-599. doi: 10.3934/krm.2013.6.589
## On the uniqueness for coagulation and multiple fragmentation equation
1 Johann Radon Institute for Computational and Applied Mathematics (RICAM), Austrian Academy of Sciences, Altenbergerstrasse 69, A-4040 Linz, Austria
Received January 2013 Revised March 2013 Published May 2013
In this article, the uniqueness of weak solutions to the continuous coagulation and multiple fragmentation equation is proved for a large range of unbounded coagulation and multiple fragmentation kernels. The multiple fragmentation kernels may have a singularity at origin. This work generalizes the preceding ones, by including some physically relevant coagulation and fragmentation kernels which were not considered before.
Citation: Ankik Kumar Giri. On the uniqueness for coagulation and multiple fragmentation equation. Kinetic and Related Models, 2013, 6 (3) : 589-599. doi: 10.3934/krm.2013.6.589
##### References:
[1] M. Aizenman and T. A. Bak, Convergence to equilibrium in a system of reacting polymers, Comm. Math. Phys., 65 (1979), 203-230. doi: 10.1007/BF01197880. [2] D. J. Aldous, Deterministic and stochastic model for coalescence (aggregation and coagulation): A review of the mean-field theory and probabilists, Bernoulli, 5 (1999), 3-48. doi: 10.2307/3318611. [3] J. M. Ball and J. Carr, The discrete coagulation-fragmentation equations: Existence, uniqueness and density conservation, J. Statist. Phys., 61 (1990), 203-234. doi: 10.1007/BF01013961. [4] J. Banasiak and W. Lamb, Analytic fragmentation semigroups and continuous coagulation-fragmentation equations with unbounded rates, J. Math. Anal. Appl., 391 (2012), 312-322. doi: 10.1016/j.jmaa.2012.02.002. [5] J. Carr, Asymptotic behaviour of solutions to the coagulation-fragmentation equations. I. The strong fragmentation case, Proc. Roy. Soc. Edinburgh Sect. A, 121 (1992), 231-244. doi: 10.1017/S0308210500027888. [6] F. P. da Costa, Existence and uniqueness of density conserving solutions to the coagulation-fragmentation equation with strong fragmentation, J. Math. Anal. Appl., 192 (1995), 892-914. doi: 10.1006/jmaa.1995.1210. [7] P. B. Dubovskiĭ and I. W. Stewart, Existence, uniqueness and mass conservation for the coagulation-fragmentation equation, Math. Meth. Appl. Sci., 19 (1996), 571-591. doi: 10.1002/(SICI)1099-1476(19960510)19:7<571::AID-MMA790>3.0.CO;2-Q. [8] M. Escobedo, Ph. Laurençot, S. Mischler and B. Perthame, Gelation and mass conservation in coagulation-fragmentation models, J. Differential Equations, 195 (2003), 143-174. doi: 10.1016/S0022-0396(03)00134-7. [9] A. K. Giri, "Mathematical and Numerical Analysis for Coagulation-Fragmentation Equations," Ph.D thesis, Otto-von-Guericke University Magdeburg, Germany, 2010. [10] A. K. Giri, J. Kumar and G. Warnecke, The continuous coagulation equation with multiple fragmentation, J. Math. Anal. Appl., 374 (2011), 71-87. doi: 10.1016/j.jmaa.2010.08.037. [11] A. K. Giri and G. Warnecke, Uniqueness for the continuous coagulation-fragmentation equation with strong fragmentation, Z. Angew. Math. Phys., 62 (2011), 1047-1063. doi: 10.1007/s00033-011-0129-0. [12] A. K. Giri, Ph. Laurençot and G. Warnecke, Weak solutions to the continuous coagulation equation with multiple fragmentation, Nonlinear Analysis, 75 (2012), 2199-2208. doi: 10.1016/j.na.2011.10.021. [13] J. Koch, W. Hackbusch and K. Sundmacher, H-matrix methods for linear and quasi-linear integral operators appearing in population balances, Comput. Chem. Eng., 31 (2007), 745-759. [14] W. Lamb, Existence and uniqueness results for the continuous coagulation and fragmentation equation, Math. Methods Appl. Sci., 27 (2004), 703-721. doi: 10.1002/mma.496. [15] Ph. Laurençot, On a class of continuous coagulation-fragmentation equations, J. Differential Equations, 167 (2000), 245-274. doi: 10.1006/jdeq.2000.3809. [16] Ph. Laurençot, The discrete coagulation equation with multiple fragmentation, Proc. Edinburgh Math. Soc. (2), 45 (2002), 67-82. doi: 10.1017/S0013091500000316. [17] Ph. Laurençot and S. Mischler, From the discrete to the continuous coagulation-fragmentation equations, Proc. Roy. Soc. Edinburgh Sect. A, 132 (2002), 1219-1248. doi: 10.1017/S0308210500002080. [18] Ph. Laurençot and S. Mischler, On coalescence equations and related models, in "Modeling and Computational Methods for Kinetic Equations" (eds. P. Degond, L. Pareschi and G. Russo), Model. Simul. Sci. Eng. Technol., Birkhäuser, Boston, MA, (2004), 321-356. [19] E. D. McGrady and R. M. Ziff, "Shattering" transition in fragmentation, Phys. Rev. Lett., 58 (1987), 892-895. doi: 10.1103/PhysRevLett.58.892. [20] D. J. McLaughlin, W. Lamb and A. C. McBride, An existence and uniqueness result for a coagulation and multiple-fragmentation equation, SIAM J. Math. Anal., 28 (1997), 1173-1190. doi: 10.1137/S0036141095291713. [21] Z. A. Melzak, A scalar transport equation, Trans. Amer. Math. Soc., 85 (1957), 547-560. doi: 10.1090/S0002-9947-1957-0087880-6. [22] T. W. Peterson, Similarity solutions for the population balance equation describing particle fragmentation, Aerosol. Sci. Technol., 5 (1986), 93-101. doi: 10.1080/02786828608959079. [23] D. J. Smit, M. J. Hounslow and W. R. Paterson, Aggregation and gelation-I. Analytical solutions for cst and batch operation, Chem. Eng. Sci., 49 (1994), 1025-1035. doi: 10.1016/0009-2509(94)80009-X. [24] I. W. Stewart, A global existence theorem for the general coagulation-fragmentation equation with unbounded kernels, Math. Methods Appl. Sci., 11 (1989), 627-648. doi: 10.1002/mma.1670110505. [25] I. W. Stewart, A uniqueness theorem for the coagulation-fragmentation equation, Math. Proc. Camb. Phil. Soc., 107 (1990), 573-578. doi: 10.1017/S0305004100068821.
show all references
##### References:
[1] M. Aizenman and T. A. Bak, Convergence to equilibrium in a system of reacting polymers, Comm. Math. Phys., 65 (1979), 203-230. doi: 10.1007/BF01197880. [2] D. J. Aldous, Deterministic and stochastic model for coalescence (aggregation and coagulation): A review of the mean-field theory and probabilists, Bernoulli, 5 (1999), 3-48. doi: 10.2307/3318611. [3] J. M. Ball and J. Carr, The discrete coagulation-fragmentation equations: Existence, uniqueness and density conservation, J. Statist. Phys., 61 (1990), 203-234. doi: 10.1007/BF01013961. [4] J. Banasiak and W. Lamb, Analytic fragmentation semigroups and continuous coagulation-fragmentation equations with unbounded rates, J. Math. Anal. Appl., 391 (2012), 312-322. doi: 10.1016/j.jmaa.2012.02.002. [5] J. Carr, Asymptotic behaviour of solutions to the coagulation-fragmentation equations. I. The strong fragmentation case, Proc. Roy. Soc. Edinburgh Sect. A, 121 (1992), 231-244. doi: 10.1017/S0308210500027888. [6] F. P. da Costa, Existence and uniqueness of density conserving solutions to the coagulation-fragmentation equation with strong fragmentation, J. Math. Anal. Appl., 192 (1995), 892-914. doi: 10.1006/jmaa.1995.1210. [7] P. B. Dubovskiĭ and I. W. Stewart, Existence, uniqueness and mass conservation for the coagulation-fragmentation equation, Math. Meth. Appl. Sci., 19 (1996), 571-591. doi: 10.1002/(SICI)1099-1476(19960510)19:7<571::AID-MMA790>3.0.CO;2-Q. [8] M. Escobedo, Ph. Laurençot, S. Mischler and B. Perthame, Gelation and mass conservation in coagulation-fragmentation models, J. Differential Equations, 195 (2003), 143-174. doi: 10.1016/S0022-0396(03)00134-7. [9] A. K. Giri, "Mathematical and Numerical Analysis for Coagulation-Fragmentation Equations," Ph.D thesis, Otto-von-Guericke University Magdeburg, Germany, 2010. [10] A. K. Giri, J. Kumar and G. Warnecke, The continuous coagulation equation with multiple fragmentation, J. Math. Anal. Appl., 374 (2011), 71-87. doi: 10.1016/j.jmaa.2010.08.037. [11] A. K. Giri and G. Warnecke, Uniqueness for the continuous coagulation-fragmentation equation with strong fragmentation, Z. Angew. Math. Phys., 62 (2011), 1047-1063. doi: 10.1007/s00033-011-0129-0. [12] A. K. Giri, Ph. Laurençot and G. Warnecke, Weak solutions to the continuous coagulation equation with multiple fragmentation, Nonlinear Analysis, 75 (2012), 2199-2208. doi: 10.1016/j.na.2011.10.021. [13] J. Koch, W. Hackbusch and K. Sundmacher, H-matrix methods for linear and quasi-linear integral operators appearing in population balances, Comput. Chem. Eng., 31 (2007), 745-759. [14] W. Lamb, Existence and uniqueness results for the continuous coagulation and fragmentation equation, Math. Methods Appl. Sci., 27 (2004), 703-721. doi: 10.1002/mma.496. [15] Ph. Laurençot, On a class of continuous coagulation-fragmentation equations, J. Differential Equations, 167 (2000), 245-274. doi: 10.1006/jdeq.2000.3809. [16] Ph. Laurençot, The discrete coagulation equation with multiple fragmentation, Proc. Edinburgh Math. Soc. (2), 45 (2002), 67-82. doi: 10.1017/S0013091500000316. [17] Ph. Laurençot and S. Mischler, From the discrete to the continuous coagulation-fragmentation equations, Proc. Roy. Soc. Edinburgh Sect. A, 132 (2002), 1219-1248. doi: 10.1017/S0308210500002080. [18] Ph. Laurençot and S. Mischler, On coalescence equations and related models, in "Modeling and Computational Methods for Kinetic Equations" (eds. P. Degond, L. Pareschi and G. Russo), Model. Simul. Sci. Eng. Technol., Birkhäuser, Boston, MA, (2004), 321-356. [19] E. D. McGrady and R. M. Ziff, "Shattering" transition in fragmentation, Phys. Rev. Lett., 58 (1987), 892-895. doi: 10.1103/PhysRevLett.58.892. [20] D. J. McLaughlin, W. Lamb and A. C. McBride, An existence and uniqueness result for a coagulation and multiple-fragmentation equation, SIAM J. Math. Anal., 28 (1997), 1173-1190. doi: 10.1137/S0036141095291713. [21] Z. A. Melzak, A scalar transport equation, Trans. Amer. Math. Soc., 85 (1957), 547-560. doi: 10.1090/S0002-9947-1957-0087880-6. [22] T. W. Peterson, Similarity solutions for the population balance equation describing particle fragmentation, Aerosol. Sci. Technol., 5 (1986), 93-101. doi: 10.1080/02786828608959079. [23] D. J. Smit, M. J. Hounslow and W. R. Paterson, Aggregation and gelation-I. Analytical solutions for cst and batch operation, Chem. Eng. Sci., 49 (1994), 1025-1035. doi: 10.1016/0009-2509(94)80009-X. [24] I. W. Stewart, A global existence theorem for the general coagulation-fragmentation equation with unbounded kernels, Math. Methods Appl. Sci., 11 (1989), 627-648. doi: 10.1002/mma.1670110505. [25] I. W. Stewart, A uniqueness theorem for the coagulation-fragmentation equation, Math. Proc. Camb. Phil. Soc., 107 (1990), 573-578. doi: 10.1017/S0305004100068821.
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#1526 2012-11-22 20:49:22
bobbym
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Re: Add 13 more and post it forever.
19838
In mathematics, you don't understand things. You just get used to them.
I have the result, but I do not yet know how to get it.
All physicists, and a good many quite respectable mathematicians are contemptuous about proof.
#1527 2012-11-22 22:29:09
phrontister
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Re: Add 13 more and post it forever.
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"The good news about computers is that they do what you tell them to do. The bad news is that they do what you tell them to do." - Ted Nelson
#1528 2012-11-22 22:30:11
bobbym
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Re: Add 13 more and post it forever.
19864
19 - 8 + 6 - 4 = 13
In mathematics, you don't understand things. You just get used to them.
I have the result, but I do not yet know how to get it.
All physicists, and a good many quite respectable mathematicians are contemptuous about proof.
#1529 2012-11-22 22:31:46
phrontister
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Re: Add 13 more and post it forever.
19877
"The good news about computers is that they do what you tell them to do. The bad news is that they do what you tell them to do." - Ted Nelson
#1530 2012-11-22 22:39:56
bobbym
Online
Re: Add 13 more and post it forever.
19890
In mathematics, you don't understand things. You just get used to them.
I have the result, but I do not yet know how to get it.
All physicists, and a good many quite respectable mathematicians are contemptuous about proof.
#1531 2012-11-22 23:34:03
phrontister
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Re: Add 13 more and post it forever.
19903
"The good news about computers is that they do what you tell them to do. The bad news is that they do what you tell them to do." - Ted Nelson
#1532 2012-11-22 23:35:30
bobbym
Online
Re: Add 13 more and post it forever.
19916
In mathematics, you don't understand things. You just get used to them.
I have the result, but I do not yet know how to get it.
All physicists, and a good many quite respectable mathematicians are contemptuous about proof.
#1533 2012-11-22 23:43:43
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Re: Add 13 more and post it forever.
19929
Winter is coming.
#1534 2012-11-22 23:48:35
bobbym
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Re: Add 13 more and post it forever.
19942
1+9+9-4-2=13
In mathematics, you don't understand things. You just get used to them.
I have the result, but I do not yet know how to get it.
All physicists, and a good many quite respectable mathematicians are contemptuous about proof.
#1535 2012-11-22 23:52:22
Mpmath
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Re: Add 13 more and post it forever.
19955
1+4+7+7-4-2=13
Winter is coming.
#1536 2012-11-22 23:56:55
bobbym
Online
Re: Add 13 more and post it forever.
19968
In mathematics, you don't understand things. You just get used to them.
I have the result, but I do not yet know how to get it.
All physicists, and a good many quite respectable mathematicians are contemptuous about proof.
#1537 2012-11-23 00:00:17
Mpmath
Full Member
Offline
Re: Add 13 more and post it forever.
19981
Last edited by Mpmath (2012-11-23 00:01:06)
Winter is coming.
#1538 2012-11-23 00:03:14
bobbym
Online
Re: Add 13 more and post it forever.
19994
In mathematics, you don't understand things. You just get used to them.
I have the result, but I do not yet know how to get it.
All physicists, and a good many quite respectable mathematicians are contemptuous about proof.
#1539 2012-11-23 00:06:05
Mpmath
Full Member
Offline
Re: Add 13 more and post it forever.
20007 (passed 20000!)
Winter is coming.
#1540 2012-11-23 00:08:51
bobbym
Online
Re: Add 13 more and post it forever.
20020
In mathematics, you don't understand things. You just get used to them.
I have the result, but I do not yet know how to get it.
All physicists, and a good many quite respectable mathematicians are contemptuous about proof.
#1541 2012-11-23 00:12:04
Mpmath
Full Member
Offline
Re: Add 13 more and post it forever.
20033
Last edited by Mpmath (2012-11-23 00:13:18)
Winter is coming.
#1542 2012-11-23 00:18:01
bobbym
Online
Re: Add 13 more and post it forever.
20046
In mathematics, you don't understand things. You just get used to them.
I have the result, but I do not yet know how to get it.
All physicists, and a good many quite respectable mathematicians are contemptuous about proof.
#1543 2012-11-23 00:25:48
Mpmath
Full Member
Offline
Re: Add 13 more and post it forever.
20059
Last edited by Mpmath (2012-11-23 00:26:54)
Winter is coming.
#1544 2012-11-23 00:30:42
bobbym
Online
Re: Add 13 more and post it forever.
20072
In mathematics, you don't understand things. You just get used to them.
I have the result, but I do not yet know how to get it.
All physicists, and a good many quite respectable mathematicians are contemptuous about proof.
#1545 2012-11-23 00:54:17
Mpmath
Full Member
Offline
Re: Add 13 more and post it forever.
20085
Winter is coming.
#1546 2012-11-23 00:59:41
bobbym
Online
Re: Add 13 more and post it forever.
20098
In mathematics, you don't understand things. You just get used to them.
I have the result, but I do not yet know how to get it.
All physicists, and a good many quite respectable mathematicians are contemptuous about proof.
#1547 2012-11-23 01:02:45
Mpmath
Full Member
Offline
Re: Add 13 more and post it forever.
20111
Winter is coming.
#1548 2012-11-23 01:09:56
bobbym
Online
Re: Add 13 more and post it forever.
20124
20 - 1 - 2 - 4 = 13
In mathematics, you don't understand things. You just get used to them.
I have the result, but I do not yet know how to get it.
All physicists, and a good many quite respectable mathematicians are contemptuous about proof.
#1549 2012-11-23 01:31:40
phrontister
Real Member
Offline
Re: Add 13 more and post it forever.
20137
"The good news about computers is that they do what you tell them to do. The bad news is that they do what you tell them to do." - Ted Nelson
#1550 2012-11-23 01:38:36
Mpmath
Full Member
Offline
Re: Add 13 more and post it forever.
20150
Winter is coming. | 1,899 | 6,180 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2014-15 | longest | en | 0.936661 |
https://www.coursehero.com/file/6328907/fm16-9/ | 1,529,338,970,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267860570.57/warc/CC-MAIN-20180618144750-20180618164750-00370.warc.gz | 793,591,856 | 630,923 | # fm16 9 - b = b U[1(1 – T(D/E b These r s estimates were...
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. Answers and Solutions: 16 - 9 16-12 Tax rate = 40% r RF = 5.0% b U = 1.2 r M – r RF = 6.0% From data given in the problem and table we can develop the following table: D/A E/A D/E r d r d (1 – T) Levered beta a r s b WACC c 0.00 1.00 0.0000 7.00% 4.20% 1.20 12.20% 12.20% 0.20 0.80 0.2500 8.00 4.80 1.38 13.28 11.58 0.40 0.60 0.6667 10.00 6.00 1.68 15.08 11.45 0.60 0.40 1.5000 12.00 7.20 2.28 18.68 11.79 0.80 0.20 4.0000 15.00 9.00 4.08 29.48 13.10 Notes: a These beta estimates were calculated using the Hamada equation,
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Unformatted text preview: b = b U [1 + (1 – T)(D/E)]. b These r s estimates were calculated using the CAPM, r s = r RF + (r M – r RF )b. c These WACC estimates were calculated with the following equation: WACC = w d (r d )(1 – T) + (w ce )(r s ). The firm’s optimal capital structure is that capital structure which minimizes the firm’s WACC. Elliott’s WACC is minimized at a capital structure consisting of 40% debt and 60% equity. At that capital structure, the firm’s WACC is 11.45%....
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