url stringlengths 6 1.61k | fetch_time int64 1,368,856,904B 1,726,893,854B | content_mime_type stringclasses 3
values | warc_filename stringlengths 108 138 | warc_record_offset int32 9.6k 1.74B | warc_record_length int32 664 793k | text stringlengths 45 1.04M | token_count int32 22 711k | char_count int32 45 1.04M | metadata stringlengths 439 443 | score float64 2.52 5.09 | int_score int64 3 5 | crawl stringclasses 93
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https://forums.cpanel.net/threads/understanding-top-and-load-averages.271441/ | 1,675,579,501,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764500250.51/warc/CC-MAIN-20230205063441-20230205093441-00649.warc.gz | 285,474,260 | 17,852 | # Understanding top and load averages
#### czerdrill
##### Well-Known Member
I'm just trying to understand the top output as I'm a little confused at what I'm looking at here. I'm running a quad core box, and when I run top the load averages are 7.33, 7.43, 7.10
From what I understand #cores=max load right? So doesn't this mean that my server is under some pretty bad strain? Or does it mean I have to divide each average by 4 to get the actual results so it would be 1.83, 1.86, 1.78?
Just trying to understand if I have something to worry about or nothing at all...
#### ruzbehraja
##### Well-Known Member
To put it simply, Load Average shows you the Number of processes, waiting to use the CPU(s) of your machine.
This is not a % of the CPU usage.
The CPU can either be 100% used or 100% unused, it cant be half used or 20% used logically.
But the Load Average shows you how many processes will be using the CPU shortly.
CPU Load Average != CPU % Used.
Load Averages show you the value for an entire machine and not per CPU. They do not know the number of CPUs you have in your machine.
Assuming you had a single CPU:
This means that they are high and if you had a processor 7.33 times faster, it would have done your job instantly.
The load of 7.33 means you made it work 633% more times than what it could have done.
If you had 4 CPU's, then you can say that they are being overworked. Ideally it should have been 4.00
If you had 7 CPU's, then you can say that all are being used at their full potential.
To check out the actual CPU% used for the day use the command | 406 | 1,586 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2023-06 | latest | en | 0.962205 |
https://algebra-answer.com/algebra-help/unlike-denominators/worksheets-dividing-monomials.html | 1,558,858,335,000,000,000 | text/html | crawl-data/CC-MAIN-2019-22/segments/1558232258862.99/warc/CC-MAIN-20190526065059-20190526091059-00196.warc.gz | 376,339,609 | 7,099 | worksheets dividing monomials
Related topics:
solving subtraction equations 6th grade | Solving Inequalities In Math | College Algebra Problem Solver | Calculator Common Least Multiple | Story Problems Intermediate Algebra Tutor | Homework Solutions For Rational Expressions | Manipulating Algebra
Author Message
BaneCxrif
Registered: 22.10.2003
From: Everywhere
Posted: Saturday 24th of May 11:53 worksheets dividing monomials is my worst problem. Can any one help me understand it? I am a beginner at it. What book or other resource do you suggest for beginning this subject?
espinxh
Registered: 17.03.2002
From: Norway
Posted: Monday 26th of May 07:01 Algebra Helper is a real treasure that can help you with Basic Math. Since I was not strong in Intermediate algebra, one of my class instructors recommended me to check out the Algebra Helper and based on his advice, I looked for it online, purchased it and began using it. It was just remarkable. If you sincerely follow each and every example offered there on Algebra 1, you would definitely master the fundamentals of inequalities and lcf within hours.
Ashe
Registered: 08.07.2001
From:
Posted: Tuesday 27th of May 09:05 It would really be great if you could let us know about a utility that can provide both. If you could get us a resource that would give a step-by-step solution to our assignment, it would really be good. Please let us know the reliable links from where we can get the software.
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Registered: 11.07.2001
From: Earth
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ORDER NOW! | 869 | 3,937 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2019-22 | latest | en | 0.945594 |
https://www.nagwa.com/en/videos/697164284184/ | 1,600,900,604,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400212959.12/warc/CC-MAIN-20200923211300-20200924001300-00127.warc.gz | 970,283,263 | 7,556 | # Video: Solving Word Problems by Dividing 5-Digit Numbers by 2-Digit Numbers
An airport registered 838,500 passengers traveling to Africa over a period of 12 months. Find the average number of passengers that traveled to Africa in one month.
07:38
### Video Transcript
An airport registered 838,500 passengers traveling to Africa over a period of 12 months. Find the average number of passengers that traveled to Africa in one month.
This problem is all about the number of people that travel from an airport to Africa. To begin with, we’re told the total number of passengers that the airport registered. And that’s 838,500. But how long did the airport measure this amount for? Is this the number of people in a day, a week? We’re told that the airport registered this amount over a period of a year, 12 months. And the problem asks us to use this information to find the average number of passengers that traveled to Africa in one of those 12 months. So how are we going to calculate this amount?
This is a division problem. We’re going to need to divide 838,500 by 12. And because these numbers are so large, we’re going to need to use long division. How many lots of 12 are there in 838,500? If we just look at the first digit, there are no 12s in eight. So we need to begin by looking at the first two digits. How many 12s are there in 83?
Before we start, it might be worth just jotting down all the multiples of 12 that we know already. These are multiples of 12 that we don’t need to think about. We already know them. Obviously, one times 12 is 12. We can double that to get two lots of 12 are 24. We could double again quickly to find four lots of 12 are 48. Then 10 times a number is always a good one. 10 multiplied by 12 is 120. We can then go backwards and find half of 10 lots of 12 to find what five 12s are worth. Half of 120 is 60. So there are one or two 12 times tables facts there that could help us.
Let’s get back to the question. How many 12s are there in 83? Now instead of starting right at the beginning with 12, because we’ve already written out some number facts that we knew, we can start with five 12s are 60. It looks like we can fit one maybe two lots of 12 into 83. If we add 12 to 60, we’ll find out what six 12s are. 60 plus 12 equals 72.
Let’s just see whether seven 12s fit into our 83. We need to add another lot of 12. 72 plus 12 equals 84. This is too large. We can see that only six lots of 12 fit into 83. So we’ll write the number six at the top. And six lots of 12 as we’ve said are 72. And if we subtract this from 83, we can find the remainder. Three take way two, is one, and eight take away seven is one. We have 11 left. Now there are no lots of 12 in 11. It’s too small. So let’s include an extra digit, and we’re gonna have to bring down this digit from the top number.
The next digit along is an eight. And it’s going to turn our number 11 into 118. How many lots of 12 are there in 118? Again, our facts that we knew already can be helpful here. We know that 10 lots of 12 are 120, which is slightly too much. But if we subtract one lot of 12 from 120, we’ll be left with nine 12s. And this is going to fit into 118. We know that 20 take away 12 leaves us with eight. So 120 take away 12 is 108. So we can write the digit nine at the top as part of our answer.
To calculate the remainder, we need to subtract nine lots of 12, which we said was 108, from the number we were dividing, which was 118. We can work this answer out mentally. The difference between 108 and 118 is 10. Again, 10 is too small to be dividing by 12. So let’s include another digit.
And the next digit along to bring down is going to be the digit five. It’s going to turn our number 10 into 105. How many lots of 12 are there in 105? As we’ve just seen, nine lots of 12 is 108. But this is too large. It’s only too large by three. So we know the answer is going to be eight lots of 12. And to calculate eight lots of 12, we need to subtract another lot of 12 from 108. 108 take away two is 106. And if we take away 10, we have the answer 96.
We could’ve found the same answer by adding 12 to 84. So eight 12s are 96. And this is as close as we’re going to get to 105. To find the remainder, we need to subtract our eight 12s, which is 96, from 105. Again, a quick way to do this would be mentally. We can count up from 96. If we count from 96 to 100, that’s four. And then from 100 onto 105 is another five. And four plus five equals nine. There’s a remainder of nine. Nine is too small to be dividing by 12. So we need to bring down another digit from the top number.
This time, it’s a zero. And it’s going to turn our nine into 90. How many 12s are there in 90? This is where we can see the benefits of writing out all the multiples of 12 that we’ve done as we’ve gone along. We’ve got quite an extensive list now. And we just need to glance across and see how close we can get to 90. The closest multiple of 12 that we can get to 90 without going above 90 is 84. And this is equal to seven 12s. So we can say 90 divided by 12 equals seven. Seven 12s are 84 as we’ve just said. And if we subtract this from 90, we can find the remainder. Again, it’s one we can do in our heads. The difference between 84 and 90 is six.
There are no 12s in six. So we need to bring down our final digit, another zero. And so six becomes 60. Do you recognize the number 60? It’s in one of our facts. There are five 12s in 60. So we can complete our answer by writing the digit five at the top. And we can prove that there’s nothing left over by subtracting five lots of 12, which is 60, from the number that we wanted to divide, which was also 60, to show there’s no remainder.
And so we can see the answer to our problem at the top of the calculation. We used long division to find the answer. If an airport registered 838,500 passengers traveling to Africa over a period of 12 months, we knew that the way to find the average number of passengers that travel to Africa in one month was to divide 838,500 by 12. And so the answer to the problem and the average number of passengers is 69,875. | 1,615 | 6,120 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.75 | 5 | CC-MAIN-2020-40 | latest | en | 0.94913 |
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Aaron Meurer
@asmeurer
something that can help here is minpoly. minpoly((sqrt(x)+1)/(sqrt(x)*(x+sqrt(x)), y) gives x*y - 1 meaning that the expression equals 1/x
I've mentioned it on some issues before, but we should have a simplification function that uses minpoly. If the minimal polynomial is linear or constant then the simplification is trivial. It's harder if the degree is higher because you have to make sure you keep the same root, but it's doable.
Aaron Meurer
@asmeurer
I didn't see any dedicated issue for it so I opened sympy/sympy#18907
Megan Ly
@meganly
The minpoly idea is cool. Is there a canonical form for algebraic functions? If so, what is it?
Aaron Meurer
@asmeurer
I'm not sure if there's a straightforward way to define one.
I think you can do something using the fact that algebraic extensions form a vector space over the extended field. So if you have a basis, then every element of the extension is a linear combination of those basis elements. But there's the question of how to chose that basis in a canonical way.
Megan Ly
@meganly
That's something you could do, but I don't think that works for quotients of algebraic expressions. As far as I know there isn't really a canonical form for algebraic functions like there is for rational
Saanidhya
@Saanidhyavats
After making some changes in pull request #18681 I think it's ready to be merged. It would be helpful if someone could take a look init.
Dhruv Mendiratta
@dhruvmendiratta6
@jksuom @Arpan612 I too am running wsl with the sympy repo in windows. I am not able to run profilers(eg: vprof) I installed in ubuntu on the repo. Also local tests using ./bin/test do not work on ubuntu. So is the solution here simply cloning a wsl version of the sympy repo and keeping its functioning separate from windows?
Arpan Chattopadhyay
@Arpan612
Yes, I think this solution will be best for now.
If I get any alternative, I will tell you.
@asmeurer Thank you, Sir.
Maarten van der Velde - (Ideogram)
@ideogram
@jksuom I tried it, and it doesn't seem to make any difference. Everythink works fine, until I introduce a variable I. It then says:
TypeError: can't convert expression to float
Kalevi Suominen
@jksuom
Does the same error occur if you change the name to something else?
Maarten van der Velde - (Ideogram)
@ideogram
No, I tried that. It works fine with X, Y, Z, W, H and L
Kalevi Suominen
@jksuom
What is the difference then? Do you begin by defining I in the same way as I = Symbol('I')? That should be done before creating the expression to which evalf is to be applied so that the expression will involve the symbol I and not the imaginary unit I.
danil179
@danil179
I don't know what is the current state of my PR #17745 as it seems 5 months passed already and I don't see any specific problems with the code right now. In my PR I tried to keep the original printing where possible (and hence my PR keeps the original printing with changes only where necessary and doesn't require code changes in the tests). I think that my solution is near optimal in doing that.
Arpan Chattopadhyay
@Arpan612
@asmeurer Sir, could you please go through my query in issue #18882?
BasileiosKal
@BasileiosKal
Hi everyone. I made the pr #18875 a few days ago. I was wondering if anyone could review it? Thank you!
Saanidhya
@Saanidhyavats
I have made some changes in pr #18681 and #18559. It would be helpful if someone could review it
Aaron Meurer
@asmeurer
@danil179 sometimes PRs fall by the wayside. Just ping people if no one has said anything for a while.
Saanidhya
@Saanidhyavats
@asmeurer thank you for suggesting the changes, I have made the changes in pr #18681. Can you review it
Daniel Shapero
@danshapero
Hi all -- I'm looking at how to differentiate expressions involving IndexedBase objects. When I run the following code:
import sympy
n = sympy.symbols('n', integer=True)
M = sympy.IndexedBase('M', shape=(n, n))
x = sympy.IndexedBase('x', shape=n)
i = sympy.Idx('i', n)
j = sympy.Idx('j', n)
k = sympy.Idx('k', n)
energy = 0.5 * M[i, j] * x[i] * x[j]
print(sympy.simplify(energy.diff(x[k])))
I get an expression involving lots of Kronecker deltas. I feel it should be possible to simplify these out but I don't know how. Thoughts?
Maq_Owais
@MaqOwais
@MaqOwais
Sir, as I came to know that ' Handling modular equations ' already integrated with _solveset then why it showed
in gsoc project under solvers module?
do we again have to reintegrate under _transolve ?
and remove that part from _solveset ? to have a reply : )
waiting for reply
OmarWagih1
@OmarWagih1
Hello All, i just posted my application on the github wiki for some reviews and comments, could you please look at it? Would very much appreciate it. Thanks!
https://github.com/sympy/sympy/wiki/GSoC-2020-Application-Omar-Wagih:-Benchmarks-and-performance
Hello All, i just posted my application on the github wiki for some reviews and comments, could you please look at it? Would very much appreciate it. Thanks!
Gagandeep Singh
@czgdp1807
Please add the links to your applications here. See, https://github.com/sympy/sympy/wiki/GSoC-2019-Current-Applications for reference.
Arpan Chattopadhyay
@Arpan612
Yes sir, I saw it now. I will do it.
@czgdp1807 Could you please tell me who are the potential mentors for symengine project?
Saanidhya
@Saanidhyavats
I have made the changes in pr #18681. It would be helpful if someone could review it
Arpan Chattopadhyay
@Arpan612
@Saanidhyavats Could you please go through my proposal and suggest changes?
I am reviewing your changes. I will put up my comments on Github.
Saanidhya
@Saanidhyavats
Sure
BasileiosKal
@BasileiosKal
Hi all! I have made the changes the reviewer's proposed in my pr #18875. Could you please review it?
Mohit Balwani
@Mohitbalwani26
In my PR #18881 travis has already passed on when i checked on travis offcial websote but it is showing still pending in GitHub. What could be the reason?
mohit
@mohitacecode
Hello Arpan,
I have left comment on the issue.
mohit
@mohitacecode
the files mentioned in the issues are already created you have to identify the test in test_matrices.py and test_commonmatrix.py and move them to their related test file. for example all the solvers related test should be transferred to test_solvers.py.
just a opinion the discussion related to pr should be kept there.I will also post the previous comment there.
Saanidhya
@Saanidhyavats
I have made a pr #18932, it would be helpful if someone could guide me further on it
Arpan Chattopadhyay
@Arpan612
I am on it.
Julian
@cod3monk
Is there a simple way to find and divide all terms of relationals by the gcd, if one exists?
This should turn 64*M*N - 104*N - 56 <= 11796480 into 8*M*N - 13*N - 7 <= 1474560. simplify does not do the trick.
mohit
@mohitacecode
maybe something like this.
>>> a = 64*M*N - 104*N - 56 - 11796480 <= 0
>>> factor(a)
8*(8*M*N - 13*N - 1474567) <= 0
@cod3monk
Stefan Corneliu Petrea
@wsdookadr
sss
Aaron Meurer
@asmeurer
@danshapero how should it simplify?
Daniel Shapero
@danshapero
@asmeurer the result I get is
>>> 0.5*(KroneckerDelta(i, k)*x[j] + KroneckerDelta(j, k)*x[i])*M[i, j]
which mathematically I know to be equal to 0.5 * (M[k, j] * x[j] + M[i, k] * x[i]) at the very least
of course I haven't put in any assumptions about M being self-adjoint so I'm not expecting anything there | 2,427 | 8,207 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2022-05 | latest | en | 0.829732 |
http://www.encyclopedia4u.com/m/minor-graph-theory-.html | 1,371,606,042,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368707439012/warc/CC-MAIN-20130516123039-00073-ip-10-60-113-184.ec2.internal.warc.gz | 425,888,133 | 3,823 | ENCYCLOPEDIA 4U .com
Web Encyclopedia4u.com
# Minor (graph theory)
In graph theory, a graph H is called a minor of the graph G if H is isomorphic to a graph that results from a subgraph of G by zero or more edge contractions. Here, "contracting an edge" means removing the edge and identifying its two endpoints, keeping all other edges.
For example, the graph
``` *
|
*--*--*
|
*
```
is a minor of
``` *
/|
*-*--*-*-*
|/
*
```
(the outer edges are removed, the long middle edge is contracted).
The relation "being a minor of" is a partial order on the isomorphism classes of graphs.
Many classes of graphs can be characterized by "forbidden minors": a graph belongs to the class if and only if it does not have a minor from a certain specified list. The best-known example is Kuratowski's theorem for the characterization of planar graphs. The general situation is described by the Robertson-Seymour theorem.
Another deep result by Robertson-Seymor states that if any infinite list G1, G2,... of finite graphs is given, then there always exists two indices i < j such that Gi is a minor of Gj.
In linear algebra, there is a different unrelated meaning of the word minor. See minor (linear algebra).
Content on this web site is provided for informational purposes only. We accept no responsibility for any loss, injury or inconvenience sustained by any person resulting from information published on this site. We encourage you to verify any critical information with the relevant authorities. | 340 | 1,514 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2013-20 | latest | en | 0.896 |
https://www.physicsforums.com/threads/finding-tension-acb.771988/ | 1,531,941,533,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676590314.29/warc/CC-MAIN-20180718174111-20180718194111-00426.warc.gz | 955,881,120 | 14,139 | # Homework Help: Finding tension acb
1. Sep 20, 2014
### CivilSigma
For problem 2.53, they are asking us for the tension in cable ACB and the answer is 1213N.
Now to my understanding to find tension ACB, I need to find tension CB and CA then find the resultant of them, However I know that is wrong because when I calculated tension CB I got the answer of 1213N.
Now my question is, why is tension ACB = tension CB ?
Thank you in advance.
http://[Imgur](http://i.imgur.com/pOgJai9) [Broken]
Last edited by a moderator: May 6, 2017
2. Sep 22, 2014
### pongo38
If the traction cable were not there, what would happen? Is the system stable? What difference would it make if the pulley were not frictionless? | 198 | 713 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2018-30 | latest | en | 0.94655 |
http://en.wikipedia.org/wiki/Generalized_Clifford_algebra | 1,419,208,779,000,000,000 | text/html | crawl-data/CC-MAIN-2014-52/segments/1418802772897.141/warc/CC-MAIN-20141217075252-00010-ip-10-231-17-201.ec2.internal.warc.gz | 90,165,079 | 13,341 | # Generalized Clifford algebra
In mathematics, a Generalized Clifford algebra (GCA) is an associative algebra that generalizes the Clifford algebra, and goes back to the work of Hermann Weyl,[1] who utilized and formalized these clock-and-shift operators introduced by J. J. Sylvester (1882),[2] and organized by Cartan (1898)[3] and Schwinger.[4]
Clock and shift matrices find routine applications in numerous areas of mathematical physics, providing the cornerstone of quantum mechanical dynamics in finite-dimensional vector spaces.[5][6][7] The concept of a spinor can further be linked to these algebras.[6]
## Definition and properties
### Abstract definition
The n-dimensional generalized Clifford algebra is defined as an associative algebra over a field F, generated by[8]
$e_j e_k = \omega_{jk} e_k e_j \,$
$\omega_{jk} e_l = e_l \omega_{jk} \,$
$\omega_{jk} \omega_{lm} = \omega_{lm} \omega_{jk} \,$
and
$e_j^{N_j} = 1 = \omega_{jk}^{N_j} = \omega_{jk}^{N_k} \,$
j,k,l,m = 1,...,n.
Moreover, in any irreducible matrix representation, relevant for physical applications, it is required that
$\omega_{jk} = \omega_{kj}^{-1} = e^{2\pi i \nu_{kj}/N_{kj}}$
j,k = 1,...,n, and $N_{kj} =$gcd$(N_j,N_k)$. The field F is usually taken to be the complex numbers C.
### More specific definition
In the more common cases of GCA,[6] the n-dimensional generalized Clifford algebra of order p has the property ωkj = ω, $N_k=p$ for all j,k, and $\nu_{kj}=1$. It follows that
$e_j e_k = \omega \, e_k e_j \,$
$\omega e_l = e_l \omega \,$
and
$e_j^{p} = 1 = \omega^{p} \,$
for all j,k,l = 1,...,n, and
$\omega = \omega^{-1} = e^{2\pi i /p}$
is the pth root of 1.
There exist several definitions of a Generalized Clifford Algebra in the literature.[9]
Clifford algebra
In the (orthogonal) Clifford algebra, the elements follow an anticommutation rule, with ω = −1, and p = 2.
## Matrix representation
The Clock and Shift matrices can be represented[10] by n×n matrices in Schwinger's canonical notation as
$V = \begin{pmatrix} 0&1&0&\cdots&0\\ 0&0&1&\cdots&0\\ 0&0&\cdots&1&0\\ \cdots&\cdots&\cdots&\cdots&\cdots\\ 1&0&0&\cdots&0 \end{pmatrix}$ , $U = \begin{pmatrix} 1&0&0&\cdots&0\\ 0&\omega&0&\cdots&0\\ 0&0&\omega^2&\cdots&0\\ \cdots&\cdots&\cdots&\cdots&\cdots\\ 0&0&0&\cdots&\omega^{(n-1)} \end{pmatrix}$ , $W = \begin{pmatrix} 1&1&1&\cdots&1\\ 1&\omega&\omega^2&\cdots&\omega^{n-1}\\ 1&\omega^2&(\omega^2)^2&\cdots&\omega^{2(n-1)}\\ \cdots&\cdots&\cdots&\cdots&\cdots\\ 1&\omega^{n-1}&\omega^{2(n-1)}&\cdots&\omega^{(n-1)^2} \end{pmatrix}$ .
Notably, Vn = 1, VU = ωUV (the Weyl braiding relations), and W−1VW = U (the Discrete Fourier transform). With e1 = V , e2 = VU, and e3 = U, one has three basis elements which, together with ω, fulfil the above conditions of the Generalized Clifford Algebra (GCA).
These matrices, V and U, normally referred to as "shift and clock matrices", were introduced by J. J. Sylvester in the 1880s. (Note that the matrices V are cyclic permutation matrices that perform a circular shift; they are not to be confused with upper and lower shift matrices which have ones only either above or below the diagonal, respectively).
### Specific examples
Case n = p = 2.
In this case, we have ω = −1, and
$V = \begin{pmatrix} 0&1\\ 1&0 \end{pmatrix}$ , $U = \begin{pmatrix} 1&0\\ 0&-1 \end{pmatrix}$ , $W = \begin{pmatrix} 1&1\\ 1&-1 \end{pmatrix}$
thus
$e_1 = \begin{pmatrix} 0&1\\ 1&0 \end{pmatrix}$ , $e_2 = \begin{pmatrix} 0&-1\\ 1&0 \end{pmatrix}$ , $e_3 = \begin{pmatrix} 1&0\\ 0&-1 \end{pmatrix}$ ,
which constitute the Pauli matrices.
Case n = p = 4,
In this case we have ω = i, and
$V = \begin{pmatrix} 0&1&0&0\\ 0&0&1&0\\ 0&0&0&1\\ 1&0&0&0 \end{pmatrix}$ , $U = \begin{pmatrix} 1&0&0&0\\ 0&i&0&0\\ 0&0&-1&0\\ 0&0&0&-i \end{pmatrix}$ , $W = \begin{pmatrix} 1&1&1&1\\ 1&i&-1&-i\\ 1&-1&1&-1\\ 1&-i&-1&i \end{pmatrix}$
and e1, e2, e3 may be determined accordingly.
## References
1. ^ Weyl, H., "Quantenmechanik und Gruppentheorie", Zeitschrift für Physik, 46 (1927) pp. 1–46, doi:10.1007/BF02055756. Weyl, H., The Theory of Groups and Quantum Mechanics (Dover, New York, 1931)
2. ^ Sylvester, J. J., (1882), Johns Hopkins University Circulars I: 241-242; ibid II (1883) 46; ibid III (1884) 7–9. Summarized in The Collected Mathematics Papers of James Joseph Sylvester (Cambridge University Press, 1909) v III . online and further.
3. ^ Cartan, E. (1898). "Les groupes bilinéaires et les systèmes de nombres complexes." Annales de la faculté des sciences de Toulouse 12.1 B65-B99. online
4. ^ Schwinger, J. (1960), "Unitary operator bases", Proc Natl Acad Sci U S A, April; 46(4): 570–579, PMCID: PMC222876; ibid, "Unitary transformations and the action principle", 46(6): 883–897, PMCID: PMC222951
5. ^ Santhanam, T. S.; Tekumalla, A. R. (1976). "Quantum mechanics in finite dimensions". Foundations of Physics 6 (5): 583. doi:10.1007/BF00715110.
6. ^ a b c See for example: A. Granik, M. Ross: On a new basis for a Generalized Clifford Algebra and its application to quantum mechanics, in: Rafal Ablamowicz, Joseph Parra, Pertti Lounesto (eds.): Clifford Algebras with Numeric and Symbolic Computation Applications, Birkhäuser, 1996, ISBN 0-8176-3907-1, p. 101–110
7. ^ A. K. Kwaśniewski: On Generalized Clifford Algebra C4(n) and GLq(2;C) quantum group
8. ^ For a serviceable review, see Vourdas A. (2004), "Quantum systems with finite Hilbert space", Rep. Prog. Phys. 67 267. doi: 10.1088/0034-4885/67/3/R03.
9. ^ See for example the review provided in: Tara L. Smith: Decomposition of Generalized Clifford Algebras
10. ^ Alladi Ramakrishnan: Generalized Clifford Algebra and its applications – A new approach to internal quantum numbers, Proceedings of the Conference on Clifford algebra, its Generalization and Applications, January 30 – February 1, 1971, Matscience, Madras 20, pp. 87–96 | 2,092 | 5,891 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 25, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.015625 | 3 | CC-MAIN-2014-52 | latest | en | 0.830322 |
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[–] 1 point2 points * (5 children)
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Yes, I meant his 1988 doctoral thesis "Bayesian Spectrum Analysis and Parameter Estimation" (I'm too lazy to google all these references), my mistake.
And yes, Jaynes does give the impression that uncountably infinite sets do not exist or should be ignored (he even says explicitely that "we need never depart from finite sets")-- if you read Appendix B of his book, however, he just advocates that the limiting process be postponed until after everything has been worked out mathematically in the finite or countably infinite case, and only if you want to. The result is equivalent to having probs on infinite sets from the get-go, as is the case with kolmogorov. The difference is kolmogorov is a math geek, Jaynes was a working scientist.
Could you elaborate a bit on that last part? I don't think anyone is implying the converse that "if we had all the information we'd know for certainty"... only that we consistently assign our probabilities on the information we do have. Unfortunately I haven't studied signal analysis since college so not sure what specific examples you have in mind.
[–] 1 point2 points (0 children)
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[–] 1 point2 points * (3 children)
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Sure; imagine I have a cosine with frequency function w1(t), and during the interval t1 to t2 I add another cosine to it with frequency function w2(t). Throw some white noise with known variance on top of it.
It's a bit tedious but not terribly hard to show that any analysis that I do cannot identify the functions w1, w2, and the interval t1 to t2 uniquely. This isn't a matter of not having dense enough sampling, or anything like that, it's a mathematical constraint.
When I read something like "...both the prior and posterior distributions represent, not any measurable property of the parameter, but only our own state of knowledge about it", that suggests to me that the assumption is that if we only had more data points or better analytic methods, we could identify the parameters, when in fact we can't. An intrinsic property of the parameter set is that it is effectively unmeasurable past a certain joint precision.
As for the use of Kolmogorov's formalism... without using it, is there any way to find the probability that a single realization of U~Uniform[0,1] is rational?
[–] 0 points1 point * (0 children)
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if we only had more data points or better analytic methods, we could identify the parameters, when in fact we can't.
That simply doesn't follow from "the prior and posterior distributions represent, not any measurable property of the parameter, but only our own state of knowledge about it".
You've described a problem in which there is a fundamental limit to what we can know about the parameters of interest. Congratulations, Bayesians love a problem like that.
As for the use of Kolmogorov's formalism... without using it, is there any way to find the probability that a single realization of U~Uniform[0,1] is rational?
There's no need to rule out measure theory if it helps solve some problems. The point is that it's not necessary to bring it into play just to get the ball rolling.
[–] 0 points1 point * (1 child)
sorry, this has been archived and can no longer be voted on
Would it help if I rephrased that quote as "bayesians don't model the real world-- they model data and information we have about the real world"? There's an impedance mismatch between us that I'm not seeing...
As creeping_feature said, yes, you'd need kolmogorov or some other equivalent system to handle single realizations. In practice I've never needed it. In fact I don't know of a single non-mathematician scientist or statistician outside academia that ever needed to invoke measure theory in practice... do you?
[–] 0 points1 point (0 children)
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I concede the point. I've often heard (self-proclaimed) Bayesians take the position that it's incorrect to state that we can't know what the joint parameter is to infinite precision, and we should merely state that we don't know and once we throw enough data at it and iterate Bayes rule enough somehow magic will happen and it's all cleared up. I'm happy to hear that this isn't Bayesian "orthodoxy".
As far as measure theory, it's rarely used in the actual analysis process, but builds a bridge to the nice, general results that can be conveniently specialized and then used in a particular analysis. It allows you to avoid having to reinvent the wheel for each new kind of problem you end up working with.
I'm familiar with this explicitly in the area stochastic processes -- evaluating the stability of process driven by different 'colors' or even types of noise, to pick an arbitrary example that I saw this morning -- but I imagine you can find examples of this in most other areas.
The degree to which non-mathematicians can avoid it is, in many cases, the degree to which a more mathematically inclined researcher has done the 'grunt work' to get the result ready for consumption. | 1,180 | 5,317 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.796875 | 3 | CC-MAIN-2014-41 | latest | en | 0.960159 |
https://uniontestprep.com/hiset-test/study-guide/mathematics/pages/1 | 1,685,546,173,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224646937.1/warc/CC-MAIN-20230531150014-20230531180014-00163.warc.gz | 672,454,638 | 18,478 | # Page 1-Mathematics Study Guide for the HiSET Test
## General Information
This test requires you to answer 55 questions in 90 minutes. The questions involve math content from the following areas, with approximately the noted level of emphasis:
• Numbers and Operations on Numbers (19%)
• Measurement/Geometry (18%)
• Data Analysis/Probability/Statistics (18%)
• Algebraic Concepts (45%)
Additionally, you’ll need to be fluent in these process skills to answer the questions correctly:
• Understand Mathematical Concepts and Procedures
• Analyze and Interpret Information
• Synthesize Data and Solve Problems
In this study guide, we will focus on the content first. You will not be able to get all the information you need in this guide, but you will have an idea of the topics and concepts you’ll need to know. Following that, we will outline the types of mathematical processes you’ll need to know. Be sure to seek extra help and practice with any content or procedures that give you trouble. Use our practice tests and flashcards for more information about just where you stand with mathematics before taking the test.
When taking the HiSET test, you will have access to a formula sheet. However, you must memorize the following three formulas because you may need them on the test and they are not on the formula sheet.
The Pythagorean Theorem:
${a^2} + {b^2} = {c^2}$
The Distance Formula:
$d = \sqrt {{{\left( {{x_2} - {x_1}} \right)}^2} + {{\left( {{y_2} - {y_1}} \right)}^2}}$
$x = \frac{-b \pm \sqrt{b^{2} - 4ac}}{2a}$
The content covered by the questions on this test is wide and varied, touching several areas of mathematics. This study guide takes you through the types of questions you will see and explains the procedures necessary to solve problems and find answers. There are also example problems and situations within the text.
## Numbers and Operations
In mathematics, we have numbers such as $$4,\; - 3,\; - \frac{3}{4},\;\sqrt 7 ,\;16,\;\pi ,\;{\rm{and}}\;0.$$ These numbers are real numbers, and we can add them together, subtract them, multiply them, and divide them (as long as we do not divide by $$0$$) In this section, we will explore various properties and concepts of all different types of numbers, and how to perform operations using these properties and concepts.
### Properties of Operations
When we are performing operations with numbers, there is a special order in which we must perform the operations. Thus, we must remember to follow the order of operations (PEMDAS):
Do everything in parentheses (P), left to right.
Evaluate any exponents (E), left to right.
Do all multiplication and division MD), in order, left to right.
Then do all addition and subtraction (AS), in order, from left to right.
A good way to remember this order is:
Please Excuse My Dear Aunt Sally
Example:
Simplify this expression.
$$6 + 3(5 + 2) - 2 \cdot 2$$
Add $$5 + 2$$
$$6 + 3(7) - 2 \cdot 2$$
Multiply $$3(7)$$ and $$2 \cdot 2$$
$$6 + 21 - 4$$
Add or subtract from left to right to get $$23.$$
Numbers belong to sets. The most common set of numbers is called the real numbers. These numbers are actually used in the “real world.” The set includes natural numbers, whole numbers, integers, rational numbers, and irrational numbers. An important note about the set of real numbers is that we can find them, or find where they would be, on the real number line.
The set of natural numbers is sometimes called the counting numbers. These numbers are the set: $$\{ 1,\,2,\,3,\,4,...\}$$ This set does not include negative numbers, fractions, or decimals.
The set of whole numbers is the natural numbers plus $$0.$$ These numbers are the set: $$\{ 0,\,1,\,2,\,3,\,4,...\}$$ This set does not include negative numbers, fractions, or decimals.
The set of integers is the set of all positive numbers, negative numbers, zero, with no fractions or decimals. These numbers are the set: $$\{ ... - 3, \,- 2,\, - 1,\,0,\,1,\,2,\,3,\,4,...\}$$
The set of rational numbers is the set of all numbers that can be written as a fraction. This set includes the previous sets of numbers, all terminating decimals, non-terminating repeating decimals, and radicals that have exact values. An example of this set of numbers is: $$\left\{ { - 5,\,\frac{2}{3},\,0,7,\,\sqrt {25},\,2.\bar 4,\,\sqrt {0.04} } \right\}$$
The set of irrational numbers is the set of all numbers that cannot be written as a fraction. This set includes many radicals (that do not have exact values), and the two commonly used irrational numbers $$\pi$$ and $$e.$$ Examples of these numbers are in the set: $$\left\{ {\sqrt 5 ,\,\sqrt {26} ,\,\pi ,\,e, \,- \sqrt 8 ,\, - \sqrt[3]{{25}}} \right\}$$
There are four basic mathematics operations, upon which all other processes are based. These operations are:
• Finding a sum is the result of adding numbers, such as $$6 + 15 = 21$$
• Finding a difference is the result of subtracting one number from another, such as $$18 - 7 = 11$$
• Finding a product is the result of multiplying numbers, called factors of the product. Multiplication is sometimes shown by a raised dot $$\cdot$$ instead of the times sign $$\times.$$ Examples are $$5 \cdot 7 = 35$$ or $$5 \times 7 = 35$$
• Finding a quotient is the result of dividing one number by another number. Division is indicated by the symbol $$\div$$, a horizontal bar, or a slanted fraction bar. Examples are $$8 \div 4 = 2$$, $$\frac{8}{4} = 2,$$ or $$8/4 = 2.$$ It is important to note that you cannot divide by $$0$$ because the result is undefined.
### Properties of Exponents
Exponents are used as a shortcut for writing repeated multiplication. For example: $$4 \cdot 4 \cdot 4 \cdot 4 \cdot 4 \cdot 4 \cdot 4$$ is written as $${4^7}$$ using exponents. In this example, $$4$$ is the base, and $$7$$ is the exponent.
Several points about exponents to remember are:
• If an expression does not have a visible exponent, that expression is understood to have an exponent of $$1.$$ For example, $$a = {a^1}$$.
• An exponent applies only to the term before it. For example, in the expression $$3{a^n},$$ only $$a$$ is raised to the $$n$$th power.
• If a negative expression is raised to a power, the negative is not raised to the power unless it is enclosed in parenthesis with the numeral being raised. For example: $$- {3^2} = - 9$$ but $${\left( { - 3} \right)^2} = 9.$$
Let’s review the properties of exponents. These rules apply to all exponents, whether they are integers, fractions, or decimals:
• When multiplying terms with exponents, the terms must have the same base. The property, known as the product rule of exponents, tells us to add the exponents.
Example:
${a^m} \cdot {a^n} = {a^{m + n}}$
Using numbers:
$\begin{array}{l}{3^2} \cdot {3^5} &=& {3^{2 + 5}}\\ &=& {3^7}\end{array}$
• When dividing terms with exponents, the terms, again, must have the same base. The property, known as the quotient rule of exponents, tells us to subtract the exponent in the denominator from the exponent in the numerator.
Example:
$\frac{{{a^m}}}{{{a^n}}} = {a^{m - n}}$
Using numbers:
$\begin{array}{l}\frac{{{7^6}}}{{{7^3}}} &=& {7^{6 - 3}}\\ &=& {7^3}\end{array}$
• When raising a product of two numbers to a power, we use the power of a product rule. This rule tells us to raise all parts of the product to the power.
Example:
${\left( {abc} \right)^n} = {a^n}{b^n}{c^n}$
Using numbers:
$\begin{array}{l}{\left( {3xyz} \right)^2} &=& {3^2}{x^2}{y^2}{z^2}\\ &=& 9{x^2}{y^2}{z^2}\end{array}$
• When raising a term with an exponent to a power, we use the power of a power rule. This rule tells us to multiply the two exponents.
Example:
${\left( {{a^m}} \right)^n} = {a^{mn}}$
Using numbers:
$\begin{array}{l}{\left( {{2^3}} \right)^4} &=& {2^{3 \cdot 4}}\\ &=& {2^{12}}\end{array}$
• When raising a term to the power of $$0$$, we use the power property of 0. This rule tells us that anything raised to the power of $$0$$ results in $$1.$$ However, keep in mind other exponent rules that must be followed when we use this rule.
Examples:
${5^0} = 1,\; - {5^0} = - 1,\;2{a^0} = 2 \cdot 1 = 2,\;{\left( {ab} \right)^0} = 1$
Using numbers:
${\left( {\frac{{32}}{5}} \right)^0} = 1$
• If we have a negative exponent, we can remove the negative exponent using the negative exponent rule and then continue the process using the other exponent rules. This rule tells us that if we have a negative exponent, we can remove that negative on the exponent by moving the term with the exponent from the numerator to the denominator or from the denominator to the numerator.
Example:
$${a^{ - m}} = \frac{1}{{{a^m}}}$$ or $$\frac{1}{{{a^{ - m}}}} = {a^m}$$
• Radicals can be written using exponents. The general radical expression $$\sqrt[m]{a}$$ means the $$m$$th root of $$a.$$ The $$m$$ is called the index of the radical, and the $$a$$ is called the radicand of the radical. This radical can be rewritten using the rational exponent rule, which tells us that the radical expression can be written with the radicand as the base with an exponent of $$1$$ divided by the index as the exponent.
Example:
$\sqrt[m]{a} = {a^{\frac{1}{m}}}$
Using numbers:
$\sqrt[5]{{625}} = {625^{\frac{1}{5}}}$
• If the radical has exponents inside it, then the exponent inside the radical is in the numerator of the rational exponent:
Example:
$\sqrt[m]{{{a^n}}} = {a^{\frac{n}{m}}}$
Using numbers:
$\sqrt[4]{{{7^3}}} = {7^{\frac{3}{4}}}$
• The radical expression $$\sqrt a$$ is read as the square root of $$a$$. This radical expression has no visible index. It is understood that the index is $$2$$ so the denominator of the rational exponent is $$2.$$
Example:
$\sqrt a = {a^{\frac{1}{2}}}$
Using numbers:
$\sqrt{17} = {17^{\frac{1}{2}}}$
As mentioned earlier, once a radical expression is converted to a rational expression, we can perform operations using the exponent rules.
Example:
$\begin{array}{l}\sqrt[3]{6} \cdot \sqrt[4]{6} &=& {6^{\frac{1}{3}}} \cdot {6^{\frac{1}{4}}}\\ &=& {6^{\frac{1}{3} + \frac{1}{4}}}\\ &=& {6^{\frac{7}{{12}}}}\end{array}$
### Scientific Notation
Scientific notation is a method of writing numbers that are very large or very small. Such numbers might be very difficult to visualize and perform operations with. The method of writing a number using scientific notation follows this format:
$$a \times {10^n}$$ where $$1 \le a < 10$$ and $$n$$ is an integer
The integer $$n$$ represents the number of places the decimal point is moved to create the correct $$a.$$ If the number is larger than $$10,$$ the decimal place is moved to the left, and $$n$$ is positive. If the number is smaller than $$1,$$ the decimal place is moved to the right, and $$n$$ is negative.
Lastly, before we do some examples, please remember that scientific notation typically estimates very large and very small numbers. Thus, if we wanted to use the number $$9,274,258,752$$ using scientific notation, we would probably change it to $$9,270,000,000$$ depending on how accurate we want our calculations.
Examples:
$7,600,000 = 7.6 \times {10^6}$
The requirement for $$a$$ $$1 \le a < 10$$ so we need $$a=7.6.$$ This requires us to move the decimal point $$6$$ places to the left, so the exponent of $$10$$ is $$6.$$
$0.000000045 = 4.5 \times {10^{-8}}$
The requirement for $$a$$ $$1 \le a < 10$$ so we need $$a=4.5.$$ This requires us to move the decimal point $$8$$ places to the left, so the exponent of $$10$$ is $$-8.$$
To convert from scientific notation to standard notation, we do the opposite.
Example:
$9.2 \times {10^8} = 920,000,000$
The exponent is positive $$8$$, so we move the decimal point $$8$$ places to the right.
$5.3 \times {10^{ - 6}} = 0.0000053$
The exponent is negative $$6$$, so we move the decimal point $$6$$ places to the left.
We can multiply numbers in scientific notation. We do this using multiplication and the product rule for exponents.
Example:
$\begin{array}{l}\left( {6.4 \times {{10}^4}} \right) \cdot \left( {1.2 \cdot {{10}^5}} \right) &=& \left( {6.4 \cdot 1.2} \right) \times {10^{4 + 5}}\\ &=& 7.68 \times {10^9}\end{array}$
When we multiply the $$a$$ terms, we need to remember that $$a$$ has the restriction $$1 \le a < 10$$ so typically, we must adjust the decimal point after multiplying.
Examples:
$\begin{array}{l}\left( {5.7 \times {{10}^8}} \right) \cdot \left( {6.1 \times {{10}^7}} \right) &=& \left( {5.7 \cdot 6.1} \right) \times {10^{8 + 7}}\\ &=& 34.77 \times {10^{15}}\\ &=& 3.477 \times {10^{16}}\end{array}$
Round to the nearest $$2$$ decimal places:
$\begin{array}{l}\left( {9.47 \times {{10}^6}} \right) \cdot \left( {6.15 \times {{10}^{17}}} \right) &=& \left( {9.47 \cdot 6.15} \right) \times {10^{6 + 17}}\\ &=& 58.2405 \times {10^{23}}\\ &=& 5.82 \times {10^{24}}\end{array}$
We can divide numbers in scientific notation. We do this using division and the quotient rule for exponents.
Example:
$\begin{array}{*{20}{l}}{\frac{{8.4 \times {{10}^{15}}}}{{4.2 \times {{10}^6}}}}& = &{\frac{{8.4}}{{4.2}} \times {{10}^{15 - 6}}}\\{}& = &{2 \times {{10}^9}}\end{array}$
When we divide the $$a$$ terms, we need to remember that $$a$$ has the restriction $$1 \le a < 10$$ so typically, we must adjust the decimal point after dividing.
Example:
$\begin{array}{l}\frac{{1.43 \times {{10}^{13}}}}{{7.15 \times {{10}^{19}}}} &=& \frac{{1.43}}{{7.15}} \times {10^{13 - 19}}\\ &=& 0.2 \times {10^{ - 6}}\\ &=& 2 \times {10^{ - 7}}\end{array}$
When we divided the $$a$$ values, the result was less than $$1.$$ Therefore, we had to move the decimal to the left.
We can also multiply and divide in the same operation.
Example:
$\begin{array}{l}\frac{{\left( {5 \times {{10}^7}} \right)\left( {1.8 \times {{10}^5}} \right)}}{{3 \times {{10}^8}}} &=& \left( {\frac{{5 \cdot 1.8}}{3}} \right) \times {10^{7 + 5 - 8}}\\ &=& 3 \times {10^4}\end{array}$
### Problem Solving
We use mathematics to solve problems algebraically. The concepts we learn in mathematics transfer directly to real-world problems. To solve real-world problems, we should always use this strategy:
Step 1Read and understand the problem. Read the problem carefully a few times. Decide what numbers are asked for and what information is given. Making a sketch may be helpful.
Step 2Choose a variable and use it with the given facts to represent the number(s) described in the problem. Labeling your sketch or arranging the given information in a chart may help.
Step 3Reread the problem. Then write an equation that represents the relationships among the numbers in the problem.
Step 4Solve the equation and find the required number(s).
Step 5Check your results with the original statement of the problem. Did you answer the correct question? Give the answer using units.
Example 1:
At noon a plane leaves Tampa Airport and heads north at $$180$$ miles per hour. Its destination is Chicago, $$1,174$$ miles away. To the nearest half-hour, what time does the plane arrive in Chicago. Note that Tampa is in Eastern Time Zone and Chicago is in Central Time Zone.
Answer: $$7:30$$ p.m.
Explanation:
To find the length of time the plane takes to fly from Tampa to Chicago, write and solve the equation $$t = 1174 \div 180.$$ Divide, resulting in $$t=6.52222.$$ This number equates to $$6.5$$ hours. Next, determine the time the plane took off in Central Time Zone. Central Time Zone is $$1$$ hours behind Eastern Time Zone. The plane took off at $$11:00$$ a.m. Last, add $$6.5$$ hours to $$11:00$$ a.m., giving $$5.5$$ p.m. The plane arrived in Chicago at $$5:30$$ p.m. Central Time.
Example 2:
A petroleum tanker is fully loaded with a capacity of $$5.3 \times {10^6}$$ barrels. The ship begins offloading immediately upon arrival in port and inspection. The offloading time takes $$160$$ hours. To the nearest $$1,000$$, what is the pumping rate offloading the tanker?
Answer: $$33,000$$ barrels per hour
Explanation:
Divide the capacity by the number of hours. $$\frac{{5.3 \times {{10}^6}}}{{160}} = 33,125.$$ Next, round the answer to the nearest thousand: $$33,000$$. Add units to the answer, resulting in $$33,000$$ barrels per hour.
Example 3:
The mass of the Sun, $$2.0 \times {10^{30}}$$ kg, is $$3.33 \times {10^5}$$ times the mass of the Earth. What is the mass of the Earth? Give your answer using scientific notation rounded to one decimal place.
Answer: and $$6.0 \times {10^{24}}$$ kg
Explanation:
Divide the mass of the Sun by the multiplier. $$\frac{{2.0 \times {{10}^{30}}}}{{3.33 \times {{10}^5}}}.$$ The result is $$6.0 \times {10^{24}}.$$ Add units to the answer, resulting in $$6.0 \times {10^{24}}$$ kg.
### Level of Accuracy
When solving a problem, we must determine which level of accuracy is appropriate. Different problems require different levels of accuracy. A common mistake many students make is treating every problem the same and trying to get an exact answer every time. Let’s look at a few examples to decide whether an exact answer is needed or if an estimate is okay.
Example 1:
Determine how many pizzas are needed to feed the high school football team.
Answer: Since we don’t know the exact number of slices each player will eat, or how many players will actually be there, an estimate is okay.
Example 2:
Ask each band member to reimburse the bandleader for the airfare on a recent trip.
Answer: We need an exact amount of money so everyone pays their actual cost.
Example 3:
Determine the diameter to manufacture the shaft for hydraulic cylinders.
Answer: Although there might be a tolerance, the hydraulic cylinder must not leak fluid. Therefore, we want an exact measurement of the diameter.
Example 4:
Each Friday morning Chef Bob purchases enough chicken for his restaurant to serve on the weekend. Based on the average from the past three weekends, he expects to serve $$85$$ orders of dark meat and $$65$$ orders of white meat. He orders chicken in cases that hold $$25$$ orders of each type of meat. Is he going to buy the exact amount he expects to serve or an estimate?
Answer: Since he buys the chicken in bulk, he will have to buy the closest estimate of what he thinks he needs.
Example 5:
Jason and Deidra buy some new furniture. The total cost is $$\ 2,656.88.$$ They paid $$\500$$ down and will pay the balance in $$24$$ equal payments with no interest. How much is their monthly payment? Is this an exact answer or an estimate?
Answer: $$\89.87$$ This is an exact answer because they must pay the entire amount, and they do not want to pay more than the entire amount. They will calculate the payment as follows:
$\begin{array}{c}\ 2656.88 - \ 500 &=& \ 2156.88\\\ 2156.88 \div 24 &=& \ 89.87\end{array}$
Example 5:
Jack wants to add built-in bookshelves along a certain wall in his office. The wall is $$8$$ feet tall and he plans to install $$4$$ evenly spaced $$1\frac{1}{2}$$ thick shelves. How much material should he buy? Does this problem involve exact measurements or approximate measurements?
Answer: Both. He probably doesn’t want to run out of material while he is building the shelves, so he will most likely round up, estimating the amount of material he needs. Then, when he is building the shelves, he will want them to be spaced correctly, so he will make exact measurements both cutting the material and positioning the material on the wall. | 5,355 | 19,324 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.25 | 4 | CC-MAIN-2023-23 | longest | en | 0.884694 |
https://matheducators.stackexchange.com/questions/9993/advice-needed-to-teach-a-problem-about-speed | 1,652,750,805,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662515466.5/warc/CC-MAIN-20220516235937-20220517025937-00114.warc.gz | 478,070,395 | 66,216 | I am currently helping a friend's child with his schoolwork. He is currently in primary school and being taught the topic of speed. I would like to give him the following problem as practice but I am not sure how to explain the solution to him at his level:
A and B competed in a race. Every time A ran $7$ m, B ran $3$ m. If B took $14$ minutes to complete the race, how long did A take?
My proposed solution: the ratio of A's speed to B's speed is $7:3$. So A will take a shorter time than B. Since distance $=$ speed $\times$ time, we have distance covered by A $=$ A's speed $\times$ A's time $=$ distance covered by B $=$ B's speed $\times$ B's time, i.e. $7$ $\times$ A's time is equal to $3$ $\times$ B's time. Hence the ratio of A's time to B's time is $3:7$. So, $7u=14,\ 1u=2$ and hence $3u=6$.
How do I rephrase my solution to make it more student-friendly? I am trying not to use any formula.
I think many students have difficulty grappling with the fact that speed is inversely proportional to time. If A's speed is $x$ times B's speed, where $x$ is whole number, then most students can easily see that A's time is $1/x$ times of B's time. But if $x$ is not a whole number, most will get stuck.
I appreciate all advice. Thank you.
• Are you opposed to giving him the problem to see how he broaches it - without having a pre-planned digestable solution? Alternatively: Can you give an example of his work to indicate the current stage of his proportional reasoning? I am not sure what would be "student-friendly" without more info on what the student knows/can do. Nov 14, 2015 at 22:44
• You say you gave this problem. Are you sure the problem is at his level? There are a wide range of problems related to speed and I think this may be harder than what he is learning. When my 11 year old students learned speed, a typical problem was given two of the three paremeters (speed, time, distance), find the third parameter. This is much harder than that. Nov 15, 2015 at 21:28
We are not given the speeds of A and of B. But note that the speed of A is $7$ meters for some unknown but fixed unit of time, and the speed of B is $3$ meters for the same unit of time. Thus, the distance traveled by A is $\frac{7\mathsf{~meters}}{\mathsf{unit~of~time}}\times\mathsf{?~minutes}$; and the distance traveled by B is $\frac{3\mathsf{~meters}}{\mathsf{unit~of~time}}\times 14\mathsf{~minutes}$.
Noting that $3\times 14=3\times 2\times 7=6\times 7=7\times 6$, it can be seen that A would take $6$ minutes. (For brevity, I skipped a lot of steps here. You should supply the missing steps.) | 721 | 2,595 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.34375 | 4 | CC-MAIN-2022-21 | longest | en | 0.977692 |
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# On Soduku
VIEWS: 7 PAGES: 12
• pg 1
``` Variants on SUDOKU
On the following pages you find a collection of puzzles all being variants of SUDOKU. The
rules are as follows:
• Every puzzle consists of an n x n square of cells.
• All cells have to be filled by numbers from 1 to n.
• There are at leastthe following blocks: the n rows, the n columns and the indicated groups
of cells bounded by fat lines.
• In case the diagonals are indicated there are two more blocks: the two diagonals.
• The cells have to be filled in such a way that every block contains no duplicates. For most
puzzles this means that every block contains every number exactly once. Later on, however,
this is not always the case as there are blocks with less than n cells.
• In case the line between two neighboring cells contains a circle, the corresponding numbers
should be consecutive.
• In case the line between two neighboring cells contains a cross-line, the corresponding
numbers should not be consecutive.
• In case the line between two neighboring cells contains neither a circle nor a cross-line,
there is no information on the corresponding numbers.
• There is exactly one solution.
The example below shows a solution with all possible circles and cross-lines:
All puzzles have been designed by a program written in Dephi by starting with a solution and
repeatingly removing information and checking (by back-tracking) whether the solution is still
unique. All puzzles were solved by hand. The hardest are marked by a ‘*’.
Good luck!
Hans Zantema, April / May, 2009.
Note: the next one has width 9 and height 8. This means that it should be filled by numbers 1 to 9, and
the columns should contain no duplicates.
```
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• 1.
• 2.
The odometer of Meghan’s car reads 56,193 at the start of her trip and 56,547 when she reaches her destination. She drove the car for a total of six hours. Based on her odometer readings, what was Meghan’s average driving speed during the trip?
• A.
57 mph
• B.
58 mph
• C.
60 mph
• D.
59 mph
• E.
56 mph
D. 59 mph
Explanation
Word Problem, Arithmetic
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• 3.
Which of the equations below represents the line that that passes through (-2, -1) and (-3, -4) in the standard (x, y) coordinate plane?
• A.
Y = -x + 5
• B.
Y = x + 3
• C.
Y = -3x − 5
• D.
Y = 3x + 5
• E.
Y = 3x − 5
D. Y = 3x + 5
Explanation
Multiple Choice, Coordinate Geometry
Rate this question:
• 4.
If set A includes {2,3,4,5,6,7}, what is the probability that a number chosen at random from set A will be a prime number?
• A.
0
• B.
1/6
• C.
1/3
• D.
1/2
• E.
2/3
E. 2/3
Explanation
Multiple Choice, Statistics
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• 5.
What is the volume of a cube with each side measuring 5cm?
C.
Explanation
Multiple Choice, Geometry
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• 6.
George’s Cookie Shop is considering lowering the price of its \$1.25 gingerbread cookies. George estimates that for every \$.05 he lowers the price, he will gain three new customers in addition to his regular 92 gingerbread cookie customers. Which of the following equations models George’s gingerbread cookie sales in terms of x?
• A.
(1.25 + 5x) (92 + .3x)
• B.
(1.25 − .05x) (92 + 3x)
• C.
(1.25 + 3x) (92 + .05x)
• D.
(1.25 − .3x) (92 − 5x)
• E.
(1.25 − 3x) (92 + .05x)
B. (1.25 − .05x) (92 + 3x)
Explanation
Word Problem, Functions
Rate this question:
• 7.
In Δ ABC shown above, ∠A = , ∠C = , and has a measure of 10. Which of the following best approximates the length of ?
• A.
2.9
• B.
4.0
• C.
5.3
• D.
6.2
• E.
8.5
E. 8.5
Explanation
Multiple Choice, Trigonometry
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• 8.
If a and b are positive real numbers, is equal to which of the following?
E.
Explanation
Multiple Choice, Logarithms
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• 9.
Which of the expressions below is equivalent to
D.
Explanation
Multiple Choice, Algebra
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• 10.
If set A includes {2,3,4,5,6,7}, what is the probability that a number chosen at random from set A will be a prime number?
• A.
0
• B.
1/6
• C.
1/3
• D.
1/2
• E.
2/3
E. 2/3
Explanation
Multiple Choice, Statistics
Rate this question:
• 11.
Timothy makes a model of his garden using a scale of 1" = 5 feet. If his model measures 2.5" by 3.5", what are the dimensions of the actual garden?
• A.
12.5 x 17
• B.
12 x 17
• C.
12.5 x 17.5
• D.
12 x 17.5
• E.
12.75 x 17.75
C. 12.5 x 17.5
Explanation
Word Problem, Arithmetic
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• 12.
The Tabula Rasa Home Improvement Company has been hired to paint the living room of a local residence. The living room is 14 feet wide by 16 feet long with a ceiling height of 9 feet. A single window measures 4 feet by 6 feet. Two doors measure 8 feet by 3.5 feet and 8 feet by 3 feet respectively. Based on the conservative estimate that one gallon of paint will cover 200 square feet of wall space, how many gallons of paint will be needed to coat the living room walls?
• A.
1 gallon
• B.
2 gallons
• C.
3 gallons
• D.
4 gallons
• E.
5 gallons
C. 3 gallons
Explanation
Word Problem, Geometry
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• 13.
What is the quotient of in scientific notation?
A.
Explanation
Multiple Choice, Algebra
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• 14.
The equation C = 15 + 0.23k approximates the cost of electricity for residential customers in kilowatt hours when k is the number of kilowatt hours for 0 ≤ k ≤ 100, and C is the total cost. Based on this equation, which of the following statements is true?
• A.
Each kilowatt hour of electricity costs \$0.23
• B.
Each kilowatt hour of electricity costs \$0.38
• C.
Each kilowatt hour of electricity costs \$1.03
• D.
Each kilowatt hour of electricity costs \$1.73
• E.
Each kilowatt hour of electricity costs \$15.23
A. Each kilowatt hour of electricity costs \$0.23
Explanation
Word Problem, Functions
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• 15.
Line segment lies on the standard (x,y) coordinate plane. The coordinates of endpoint A are (1, 2), and the point halfway between A and B is . Which of the following are the coordinates of endpoint B ?
• A.
(3,4)
• B.
(4,3)
• C.
(9,8)
• D.
(-3,-4)
• E.
(8,-9)
B. (4,3)
Explanation
Word Problem, Coordinate Geometry
Rate this question:
• 16.
• A.
9
• B.
10
• C.
57
• D.
43
• E.
22 | 1,661 | 5,640 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.1875 | 4 | CC-MAIN-2023-50 | latest | en | 0.902722 |
https://www.12000.org/my_notes/kamek/mma_12_maple_2019/KERNELsubsection271.htm | 1,721,500,172,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763517515.18/warc/CC-MAIN-20240720174732-20240720204732-00128.warc.gz | 543,439,787 | 2,987 | #### 2.271 ODE No. 271
$\left (x^2+y(x)^2\right ) y'(x)+2 x (y(x)+2 x)=0$ Mathematica : cpu = 0.174085 (sec), leaf count = 370
$\left \{\left \{y(x)\to \frac {\sqrt [3]{\sqrt {-8 e^{3 c_1} x^3+e^{6 c_1}+20 x^6}+e^{3 c_1}-4 x^3}}{\sqrt [3]{2}}-\frac {\sqrt [3]{2} x^2}{\sqrt [3]{\sqrt {-8 e^{3 c_1} x^3+e^{6 c_1}+20 x^6}+e^{3 c_1}-4 x^3}}\right \},\left \{y(x)\to \frac {\left (1+i \sqrt {3}\right ) x^2}{2^{2/3} \sqrt [3]{\sqrt {-8 e^{3 c_1} x^3+e^{6 c_1}+20 x^6}+e^{3 c_1}-4 x^3}}-\frac {\left (1-i \sqrt {3}\right ) \sqrt [3]{\sqrt {-8 e^{3 c_1} x^3+e^{6 c_1}+20 x^6}+e^{3 c_1}-4 x^3}}{2 \sqrt [3]{2}}\right \},\left \{y(x)\to \frac {\left (1-i \sqrt {3}\right ) x^2}{2^{2/3} \sqrt [3]{\sqrt {-8 e^{3 c_1} x^3+e^{6 c_1}+20 x^6}+e^{3 c_1}-4 x^3}}-\frac {\left (1+i \sqrt {3}\right ) \sqrt [3]{\sqrt {-8 e^{3 c_1} x^3+e^{6 c_1}+20 x^6}+e^{3 c_1}-4 x^3}}{2 \sqrt [3]{2}}\right \}\right \}$ Maple : cpu = 0.163 (sec), leaf count = 352
$\left \{ y \left ( x \right ) ={1 \left ( {\frac {1}{2}\sqrt [3]{4-16\,{x}^{3}{{\it \_C1}}^{3/2}+4\,\sqrt {20\,{{\it \_C1}}^{3}{x}^{6}-8\,{x}^{3}{{\it \_C1}}^{3/2}+1}}}-2\,{\frac {{\it \_C1}\,{x}^{2}}{\sqrt [3]{4-16\,{x}^{3}{{\it \_C1}}^{3/2}+4\,\sqrt {20\,{{\it \_C1}}^{3}{x}^{6}-8\,{x}^{3}{{\it \_C1}}^{3/2}+1}}}} \right ) {\frac {1}{\sqrt {{\it \_C1}}}}},y \left ( x \right ) =-{\frac {1}{4} \left ( \left ( 4\,i{\it \_C1}\,{x}^{2}+i \left ( 4-16\,{x}^{3}{{\it \_C1}}^{3/2}+4\,\sqrt {20\,{{\it \_C1}}^{3}{x}^{6}-8\,{x}^{3}{{\it \_C1}}^{3/2}+1} \right ) ^{{\frac {2}{3}}} \right ) \sqrt {3}-4\,{\it \_C1}\,{x}^{2}+ \left ( 4-16\,{x}^{3}{{\it \_C1}}^{3/2}+4\,\sqrt {20\,{{\it \_C1}}^{3}{x}^{6}-8\,{x}^{3}{{\it \_C1}}^{3/2}+1} \right ) ^{{\frac {2}{3}}} \right ) {\frac {1}{\sqrt [3]{4-16\,{x}^{3}{{\it \_C1}}^{3/2}+4\,\sqrt {20\,{{\it \_C1}}^{3}{x}^{6}-8\,{x}^{3}{{\it \_C1}}^{3/2}+1}}}}{\frac {1}{\sqrt {{\it \_C1}}}}},y \left ( x \right ) ={\frac {1}{4} \left ( 4\,i\sqrt {3}{\it \_C1}\,{x}^{2}+i\sqrt {3} \left ( 4-16\,{x}^{3}{{\it \_C1}}^{3/2}+4\,\sqrt {20\,{{\it \_C1}}^{3}{x}^{6}-8\,{x}^{3}{{\it \_C1}}^{3/2}+1} \right ) ^{{\frac {2}{3}}}+4\,{\it \_C1}\,{x}^{2}- \left ( 4-16\,{x}^{3}{{\it \_C1}}^{3/2}+4\,\sqrt {20\,{{\it \_C1}}^{3}{x}^{6}-8\,{x}^{3}{{\it \_C1}}^{3/2}+1} \right ) ^{{\frac {2}{3}}} \right ) {\frac {1}{\sqrt [3]{4-16\,{x}^{3}{{\it \_C1}}^{3/2}+4\,\sqrt {20\,{{\it \_C1}}^{3}{x}^{6}-8\,{x}^{3}{{\it \_C1}}^{3/2}+1}}}}{\frac {1}{\sqrt {{\it \_C1}}}}} \right \}$ | 1,425 | 2,423 | {"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} | 3.28125 | 3 | CC-MAIN-2024-30 | latest | en | 0.334842 |
http://devmaster.net/forums/topic/5674-c-casting-float-to-int-precision/page__st__20 | 1,371,622,676,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368708142617/warc/CC-MAIN-20130516124222-00098-ip-10-60-113-184.ec2.internal.warc.gz | 68,508,769 | 10,292 | # C++ casting float to int precision
20 replies to this topic
### #21poita
Senior Member
• Members
• 322 posts
Posted 07 September 2006 - 05:02 PM
.oisyn said:
I don't think the approach should be to work with rounded numbers, but rather with exact numbers. The input are digits from text so you have finite precision, perfectly representable by rational numbers. The problem was square roots were hard to do on rational numbers. But converting it to float will not be a problem if the square root is the last calculation you'll do, as whole numbers can be represented exactly by floats and doubles (well, as long as their precision is within respectively 24 and 53 bits) - you'll throw away the bits after the decimal dot anyway.
trunc(sqrt(x) / 0.1) = trunc(sqrt(x) / sqrt(0.01)) = trunc(sqrt(x / 0.01))
In your example, if the 2.1 was the result of squarerooting 4.41 (also unrepresentable by float), you would avoid the problem if you have 4.41 represented as a rational number: 441/100. Dividing by 0.01 gives the rational number 441. The square root of 441.f is 21.f. Truncating will reveal 21.
You'll never get into trouble because the bordercase lies on the whole number. But if the result of a squareroot is a whole number, it's square is also a whole number and thus representable by a float or double.
Nice :yes: I didn't think about that last thing you said about the bordercase. I just discarded rational numbers as a solution because of the squareroot but as you said, in the cases where I'm having trouble the squareroot will be rational.
I think I'll right a rational number class before the competition just in case I need it. | 408 | 1,655 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2013-20 | latest | en | 0.952454 |
https://capital.com/en-au/learn/technical-analysis/fibonacci-retracement-strategy | 1,726,243,354,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651523.40/warc/CC-MAIN-20240913133933-20240913163933-00712.warc.gz | 126,860,225 | 44,293 | # How to trade using Fibonacci retracements
## What are Fibonacci retracements?
Fibonacci retracement levels are horizontal lines that indicate where price reversals are likely to occur and are part of technical analysis. They are based around the Fibonacci sequence, a pattern of numbers, each representing the sum of the previous two. A Fibonnaci trading strategy would incorporate these levels to analyse price behaviour.
The Fibonacci sequence starts 1,2,3,5,8,13,21,34,55 and continues to infinity, with each number being the sum of the two preceding numbers. If you want to find the next Fibonacci number, you multiply it by 1.61 to go up, or by 0.61 to go down.
## Highlights
• Fibonacci retracement levels are points on a price chart where price reversals are likely to take place.
• They are based around the Fibonacci sequence.
• Fibonacci retracement levels can be used to identify entry and exit points as well as support and resistance levels.
Although the Fibonacci sequence was first identified by the Italian mathematician after whom they are named in 1202, the first person to develop the idea of using Fibonacci numbers in finance was Charles Dow, founder of the Dow Jones Industrial Average.
He noted that, after moving along with the main trend, a price often retraces before resuming its prior movement. He concluded that the latitude of this retracement was between 33% to 66%.
The concept was later refined by stock market analyst Ralph Nelson Elliott, who brought into play more accurate retracement levels:
• Based on the Fibonacci sequence: 38.2%, 61.8%
• Based on the stock price tendencies: 50%
## Understanding Fibonacci retracements
The reason why some people use the Fibonacci sequence in trading is that markets have a tendency to reverse the direction they’re going in at various points on the price chart. In percentage terms, the suggested number is 61.8%, also known as the Golden Ratio
Fibonacci retracements can be found by:
1. Taking the highest and lowest points on a price chart and marking those as 100% and 0%.
2. Mark 61.8% and two other Fibonacci based percentages: 23.6%, the sum of a number in the sequence divided by one three places higher and 38.2%, the sum of a Fibonacci number divided by one two places higher.
3. Then, there will need to be points at 50% and 78.6%, numbers which, while not correlating to the Fibonacci sequence, often see price reversals.
### Fibonacci extensions
It is worth noting that some traders and analysts use further projections of the sequence, known as Fibonacci extensions, to look at how to take things further.
The key percentage points of these extensions are:
• 161.8%
• 200%
• 261.8%
Traders can use Fibonacci retracement levels as part of their trading strategies. Here are some examples of how they can potentially utilise a Fibonacci retracement strategy.
### Entry and exit points
A trader could use Fibonacci retracement levels as potential entry and exit points for trades. Traders could look for a price to retrace to one of the levels on the chart and then look for a price action signal, such as a reversal candlestick pattern or a bullish/bearish divergence, to confirm the potential trade setup.
### Support and resistance levels
Traders could use Fibonacci retracement levels as potential levels of support and resistance in the market. Traders could look for a price to retrace to one of these levels and then look for signs, to confirm the potential trade setup. Traders could then enter a trade in the direction of the trend.
### Technical indicators
In conjunction with Fibonacci retracements, indicators could potentially provide a more comprehensive view of the market. For example, traders may use moving averages to identify the trend and then use Fibonacci retracements to find potential levels of support and resistance within that trend. They can also use oscillators, such as the relative strength index (RSI), to confirm potential trade setups identified by the retracements.
### Price analysis
Another way Fibonacci retracements could be used with other indicators is by combining them with price analysis. Traders may use Fibonacci retracements to identify potential levels of support and resistance, and then use price action analysis, such as candlestick patterns or chart patterns, to confirm the trade setup.
## Incorporating Fibonacci retracements into your strategy
Here are some approaches for using Fibonacci retracements in trading.
• Trending markets: Fibonacci retracements could be effective in trending markets, as they can help to identify potential levels of support and resistance within the trend. In range-bound markets, they may not be as useful, as the price tends to move sideways rather than retracing.
• Use multiple time frames: It may be valuable to use multiple timeframes when using Fibonacci retracements to get a better understanding of the market. Traders could use higher time frames to identify the overall trend and lower time frames to spot potential trade setups.
• Waiting for confirmation: It may be useful to wait for confirmation before entering a trade based on Fibonacci retracements. This could include waiting for a price action signal or a confirmation from another technical analysis tool.
• Risk management: As with any trading strategy, it's important to use proper risk management when trading with Fibonacci retracements. You should always conduct your own research, remember that markets can move against you, and never trade with more money than you can afford to lose.
## Conclusion
In conclusion, Fibonacci retracements could be useful tools for traders. Trading with Fibonacci retracements could be applied to different financial instruments, such as commodities, stocks and shares, forex pairs and indices. They could also be valuable when trading contracts for difference (CFDs). Retracement levels could provide helpful information that may assist a trader decide whether to go short or long on an asset while designing their Fibonnaci trading strategy.
Fibonacci retracements are key levels in a price chart, coming in at 0%, 23.6%, 38.2%, 50%, 61.8%, 78.6% and 100%. It is at these levels that price reversals are common, according to the strategy. These points could be used as entry and exit points, as well as support and resistance levels. Traders could use them in conjunction with technical indicators such as moving averages and the relative strength index (RSI), as well as with price analysis.
However, remember that they cannot predict the future. Therefore, traders will have to remember to do their own research, understand that markets can move against them and never trade with more money than they can afford to lose.
## FAQs
#### What are Fibonacci retracements in simple terms?
Put simply, Fibonacci retracements are lines on a price chart showing levels where price reversals are likely to occur. They are a technical analysis tool.
#### Why are Fibonacci retracements important?
Fibonacci retracements are important because they represent levels on a price chart where reversals could take place. This could help traders make more informed decisions. Note, however, that they may not always be accurate.
#### How do Fibonacci retracements work?
Fibonacci retracements are a technical analysis tool used to identify potential levels of support and resistance in a market trend. These levels are based on key percentages derived from the Fibonacci sequence, and are often used by traders to help identify potential entry and exit points for trades. | 1,556 | 7,591 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.875 | 4 | CC-MAIN-2024-38 | latest | en | 0.949961 |
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Partner - Creator III
## Hierarchy and percentage distribution?
Hi,
I would like to know if this is possible to be done with QV. Let's hope I make myself clear.
Asume I have the following records:
EmployeeID idScheme iTree iPercentDistribution 73506 13 0 80 74251 19 0 20 75226 18 0 100 72416 15 13 50 75226 14 13 50
The iTree column is a reference to its parent located in column idScheme. Each record has a percentage (iPercentDistribution). As you can see the 4th and 5th records (idScheme = 15 & 14) have as parent the 1st row (IdScheme = 13). Records one throug three (IdScheme = 13, 19, 18) have no parents.
The last two records have a distribution percentage of 50% each, but since it's parent only has 80% their percentage must be 40% each. It's there a way to get the following result using Hierarchy?:
EmployeeID EmployeeID1 EmployeeID2 idScheme iTree iPercentDistribution 73506 73506 13 0 80 75226 73506 75226 14 13 40 72416 73506 72416 15 13 40 75226 75226 18 0 100 74251 74251 19 0 20
Or it should be done using Set Analysis? Honestly, I'm at a loss of ideas here.
Any help would be appreciated.
Regards.
12 Replies
Master III
Hi,
I tried like this..
```T1:
Hierarchy(idScheme,iTree,EmployeeID)
EmployeeID, idScheme, iTree, iPercentDistribution
73506, 13, 0, 80
74251, 19, 0, 20
75226, 18, 0, 100
72416, 15, 13, 50
75226, 14, 13, 50
];
Left Join(T1)
LOAD idScheme as iTree,Sum(iPercentDistribution) as Temp_iPercent Resident T1 Group by idScheme;
Left Join(T1)
LOAD iTree,Count(iTree) as Temp_Count Resident T1 where iTree>0 Group by iTree ;
NoConcatenate
Final:
EmployeeID1,
EmployeeID2,
idScheme,
iTree,
if(Len(Trim(Temp_Count))>0 and Len(Trim(Temp_Count))>0,Temp_iPercent/Temp_Count,iPercentDistribution) as iPercentDistribution
Resident T1 Order by idScheme;
DROP Table T1;
```
Partner - Champion III
Unbalanced, n-level hierarchies
Logic will get you from a to b. Imagination will take you everywhere. - A Einstein
Luminary Alumni
Hi,
Try like this
Temp:
EmployeeID, idScheme, iTree, iPercentDistribution
73506, 13, 0, 80
74251, 19, 0, 20
75226, 18, 0, 100
72416, 15, 13, 50
75226, 14, 13, 50
];
Temp1:
idScheme AS iTree,
iPercentDistribution
REsident Temp
WHERE iTree = 0;
Left Join(Temp1)
iTree,
Count(iTree) AS Count
RESIDENT Temp
WHERE iTree <> 0
GROUP BY iTree;
LEFT JOIN(Temp)
iTree,
iPercentDistribution/Count AS Percent2
RESIDENT Temp1;
DROP TABLE Temp1;
Data:
NoConcatenate
EmployeeID, idScheme, iTree, if(iTree = 0, iPercentDistribution, Percent2) AS iPercentDistribution
RESIDENT Temp;
DROP TABLE Temp;
Hope this helps you.
Regards,
jagan.
MVP
if the distribution is always equal among the child elements you would follow the settu.periyasamy approach...if into the case than we might need to think for the alternate approach ...let me know your inputs on the same
Partner - Creator III
Author
Thanks settu_periasamy, this is clever, I believe this works well if the distribution is always equal between it's children but unfortunate this isn't the case. I my data a parent node could have more than two children and the distribution could be different for each of them.
Partner - Creator III
Author
Thank you for your answer. As settu's answer I belive this would work great if the distribution is equal among the children nodes which isn't my case sadly . But I appreciate your answer.
Partner - Creator III
Author
Hi Avinash, you're right, settu's approach works good if the distribution is equal, but in my data this isn't the case unfortunately. Do you have any ideas how to make this work for uneven percentage distributions?
Regards
Master III
Hi,
Can you give more data to test which is more than two children and the expected output? Let's try..
Edit:
I think the below script will work, if there uneven percentage distribution..
T1:
Hierarchy(idScheme,iTree,EmployeeID
EmployeeID, idScheme, iTree, iPercentDistribution
73506, 13, 0, 80
74251, 19, 0, 20
75226, 18, 0, 100
72416, 15, 13, 50
75226, 14, 13, 50
]
Left Join(T1)
LOAD idScheme as iTree,Sum(iPercentDistribution) as Temp_iPercent Resident T1 Group by idScheme
NoConcatenate
Final:
EmployeeID1
EmployeeID2
idScheme
iTree
if(Len(Trim(Temp_iPercent))>0,
iPercentDistribution/100)*Temp_iPercent,
iPercentDistribution) as iPercentDistribution
Resident T1 Order by idScheme
DROP Table T1;
Luminary Alumni
Hi,
Can you provide some more data and you logic for distribution and your expected output.
Regards,
Jagan.
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A girl has as many brothers as sisters, but each brother has only half as many brothers as sisters. How many brothers and sisters are there in the family?
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There is a cryptic organization called Cicada 3301 that posts challenging puzzles online, possibly to recruit codebreakers and linguists. | 719 | 2,841 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.140625 | 3 | CC-MAIN-2024-30 | latest | en | 0.976377 |
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### Ohm's and Joule's Laws
#### Joule's Law
Where,
E = Voltage in volts
I = Current in amperes (amps)
R = Resistance in omhs
P = Power in watts
NOTE: the symbol "V" ("U" in Europe) is sometimes used to represent voltage instead of "E". In some cases, an author or circuit designer may choose to exclusively use "V" for voltage, never using the symbol "E." Other times the two symbols are used interchangeably, or "E" is used to represent voltage from a power source while "V" is used to represent voltage across a load (voltage "drop").
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"The algebraic sum of all voltages in a loop must equal zero."
Kirchhoff's Voltage Law (KVL)
"The algebraic sum of all currents entering and exiting a node must equal zero."
Kirchhoff's Current Law (KCL)
##### Lessons In Electric Circuits copyright (C) 2000-2020 Tony R. Kuphaldt, under the terms and conditions of the CC BY License.
See the Design Science License (Appendix 3) for details regarding copying and distribution.
Revised July 25, 2007
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May 22, 2013, 08:02 Average value of expression in transient flow analysis #1 New Member Jacob Fogh Knudsen Join Date: May 2013 Location: Aarhus, Denmark Posts: 1 Rep Power: 0 Hi, I am currently trying to set up a transient flow analysis of an aerofoil. I want to extract the average drag coefficient as a parameter for a study, but have run into some problems. Currently i am so far: I have set up the basic model and it runs. Created an expression for the drag coefficient: 2*force_x()@AntennaSurface/(areaAve(Density)@Inlet*(areaAve(Velocity u)@Inlet)^2*FrontalArea). Created an additional variable with: Type: Unspecified, Unit: [ ], Tensor Type: Scalar. Included the additional variable in my domain and set its value to the expression for the drag coefficient. In the "Trn Stats" tab under "Output control" I have created a Transient Statistics Object and added the additional value and set the "Option" to "Arithmetic Average". When I run the model I get an instant error saying: | PROBLEM ENCOUNTERED WHEN EXECUTING CFX EXPRESSION LANGUAGE | | | | The CFX expression language was evaluating: | | Additional Variable Value | | | | The problem was: | | DIVIDE-BY-ZERO | Have I missed some small detail or am I doing it completely wrong? Please help. Thanks!
May 22, 2013, 08:52 #2 Senior Member Lance Join Date: Mar 2009 Posts: 669 Rep Power: 21 I've done almost the same thing as you, but used two AVs and the .Trnavg operator instead. It worked for me. edit: what was the initial condition for your AV? If it was set to zero it could explain your error. Code: ``` LIBRARY: ADDITIONAL VARIABLE: Additional Variable 1 Boundary Only Field = On Option = Definition Tensor Type = SCALAR Units = [Pa] Variable Type = Unspecified END ADDITIONAL VARIABLE: Additional Variable 2 Boundary Only Field = On Option = Definition Tensor Type = SCALAR Units = [Pa] Variable Type = Unspecified END and FLUID MODELS: ADDITIONAL VARIABLE: Additional Variable 1 Additional Variable Value = areaAve(Pressure)@boundary <--- your expression Option = Algebraic Equation END ADDITIONAL VARIABLE: Additional Variable 2 Additional Variable Value = Additional Variable 1.Trnavg <--- time average Option = Algebraic Equation END``` mvoss likes this. | 565 | 2,361 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2023-40 | latest | en | 0.896547 |
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## NEET & AIIMS 12th PCB-Chemistry-Solution and Colligative Properties 2 Demo videos
Hello, dear students, we are studying the chapter, Solution and Colligative Properties, so in this module let me take you through the understanding of vapour pressure versus the temperature, so let us go with that.
Now we will see the distribution of molecular speeds in liquid, as I said that it has similarities to that of gases, so let’s see that first. Well, liquid also has the random motion like gases as well as their distribution is similar to that of velocity distribution of gases. So we have learnt that in earlier modules also. So, and we learnt that this is because of the Maxwell-Boltzmann’s Distribution which is nothing but a plot between a number of molecules versus the kinetic energy. So if we plot that let us see what I can see here, number of molecules versus kinetic energy. Now at a certain temperature T1, this is how the graph varies, this was a similar graph for gases also. Similarly, if at all I increase the temperature, let me take it to T2 temperature, then the graph will vary in this way, that is it will slowly peak will start coming down and it will start becoming more uniform. So in that case now what I could suggest here is let us take a point or kinetic energy after which let’s say liquid particles are starting to escape, right, so this is the kinetic energy, what you can call is the minimum kinetic energy required for molecules to escape from the liquid state. So that is what it is, so it is the minimum kinetic energy. Now if this minimum kinetic energy if at all being drawn towards that curve they will start touching at two different points, then they will generate an area under the curve. So let us see what exactly it represents, so at temperature T1, area under the curve represents the number of particles which are escaping out to the vapour phase. So that yellow coloured is the area which shows that at temperature T1 how many number of particles have escaped. Similarly at temperature T2 you could see this blue coloured shaded part as well as the yellow coloured part, both of them combined are the total number of particles escaping out to the vapour state. So it is very clear that in vapour phase molecules are more at higher temperature, right, as we learnt this earlier also, as the temperature is increased vapour pressure will also increase, right. So what does that mean, you can see evidently here that at temperature T1 the number of molecules escaping are very less but at temperature T2, you could see number of particles escaping are more. So therefore what I could conclude here is low temperature is equivalent to say that it is less likely to vaporize that means it is low vapour pressure. While high temperature it is equivalent to say more likely to vaporize therefore it is at high vapour pressure. So you can now relate how vapour pressure and temperature are co-related to each other, right. So let us try to derive that, now we know that vapour pressure will only be defined when liquid-vapour equilibrium is existing so therefore let me consider that liquid vapour equilibrium is there where K1 is the equilibrium constant and P1 is the vapour pressure. Now therefore K1 is the equilibrium constant at temperature T1 and P1 is the vapour pressure of liquid at temperature at T2.
Now what I can say according to Vant Hoff’s equation we can always relate that as log of K2 by K1 is equal to delta H vaporization by 2.303R into 1 by T1 minus 1 by T2. This is Vant Hoff’s equation where Delta H vaporization is the enthalpy of vaporization, right.
Now observe it carefully, can we write K1 is equal to P1, because yes, liquid does not have any vapour content so therefore vapour phase as P1 so equilibrium constant K1 will be equal to P1. Similarly, if at temperature T2 I can write K2 is equal to P2. So can I change that log of K2 by K1 as P2 by P1. So in that case this Vant Hoff’s equation will be converted to Clausius-Clapeyron’s Equation. So, students, that is how we derive at Clausius-Calpeyron’s equation which is the relation between the vapour pressure of the liquid versus the temperature for a volatile solvent.
So as we go into the next module we will see its applications and more importantly how it will derive towards the Rour’s Law.
Thank you.
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2017-04-18T04:55:12+00:00 Categories: XII-NEET & AIIMS||0 Comments
Hello | 1,042 | 4,632 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.96875 | 3 | CC-MAIN-2024-38 | latest | en | 0.936212 |
http://www.jiskha.com/display.cgi?id=1267394054 | 1,493,523,921,000,000,000 | text/html | crawl-data/CC-MAIN-2017-17/segments/1492917124297.82/warc/CC-MAIN-20170423031204-00228-ip-10-145-167-34.ec2.internal.warc.gz | 567,998,561 | 3,872 | chemistry
posted by on .
If 4.57 grams of AgBr was isolated in a gravimetric analysis of a 9.30 gram sample, what is the % Br- by weight in the samples?
• chemistry - ,
What is the mass of Br?
moles AgBr = 4.57 grams AgBr x (1 mol AgBr/molar mass AgBr).
There is 1 mole Br in 1 mole AgBr; therefore, you now have moles Br.
How many grams Br do you have?
grams Br = moles Br x atomic mass Br
What percent is that of the original sample?
%Br = (grams Br/mass sample)*100 = ??
• chemistry - ,
1.94% | 152 | 504 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2017-17 | latest | en | 0.894845 |
https://www.gradesaver.com/textbooks/math/algebra/college-algebra-10th-edition/chapter-3-section-3-6-mathematical-models-building-functions-3-6-assess-your-understanding-page-266/27 | 1,537,300,259,000,000,000 | text/html | crawl-data/CC-MAIN-2018-39/segments/1537267155676.21/warc/CC-MAIN-20180918185612-20180918205612-00470.warc.gz | 775,381,344 | 12,176 | ## College Algebra (10th Edition)
$x=0$ or $x=3$
We solve: $|2x-3|-5=-2$ $|2x-3|=5-2=3$ $2x-3=-3$ or $2x-3=3$ $2x=-3+3$ or $2x=3+3$ $2x=0$ or $2x=6$ $x=0$ or $x=3$ | 104 | 164 | {"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} | 3.921875 | 4 | CC-MAIN-2018-39 | latest | en | 0.478102 |
https://cal8.cs.fau.de/redmine/projects/cool/repository/revisions/de84f40dd3d152e91065c99a8ab82dda322beff3/entry/src/lib/CoAlgFormula.ml | 1,566,098,012,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027313589.19/warc/CC-MAIN-20190818022816-20190818044816-00281.warc.gz | 415,718,371 | 38,952 | ### Profile
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## cool / src / lib / CoAlgFormula.ml @ de84f40d
1 ```(** This module implements coalgebraic formulae ``` ``` (e.g. parsing, printing, (de-)constructing, ...). ``` ``` @author Florian Widmann ``` ``` *) ``` ```module HC = HashConsing ``` ```module A = AltGenlex ``` ```module L = List ``` ```module Str = String ``` ```(** A general exception for all kinds of errors ``` ``` that can happen in the tableau procedure. ``` ``` More specific information is given in the argument. ``` ``` *) ``` ```exception CoAlgException of string ``` ```(** Indicates the sort of a sorted formula ``` ``` *) ``` ```type sort = int ``` ```type rational = { nominator: int; denominator: int } ``` ```let string_of_rational r = ``` ``` (string_of_int r.nominator)^"/"^(string_of_int r.denominator) ``` ```let rational_of_int n d = { nominator = n; denominator = d } ``` ```(** Defines (unsorted) coalgebraic formulae. ``` ``` *) ``` ```type formula = ``` ``` | TRUE ``` ``` | FALSE ``` ``` | AP of string (* atomic proposition *) ``` ``` | NOT of formula ``` ``` | AT of string * formula ``` ``` | OR of formula * formula ``` ``` | AND of formula * formula ``` ``` | EQU of formula * formula ``` ``` | IMP of formula * formula ``` ``` | EX of string * formula ``` ``` | AX of string * formula ``` ``` | ENFORCES of int list * formula ``` ``` | ALLOWS of int list * formula ``` ``` | MIN of int * string * formula ``` ``` | MAX of int * string * formula ``` ``` | MORETHAN of int * string * formula (* diamond of GML *) ``` ``` | MAXEXCEPT of int * string * formula (* box of GML *) ``` ``` | ATLEASTPROB of rational * formula (* = {>= 0.5} C ---> C occurs with *) ``` ``` (* probability of at least 50% *) ``` ``` | LESSPROBFAIL of rational * formula (* = [1/2] C = {< 1/2} ¬ C ---> C fails with *) ``` ``` (* probability of less than 50% *) ``` ``` | CONST of string (* constant functor with value string *) ``` ``` | CONSTN of string (* constant functor with value other than string *) ``` ``` | ID of formula (* modality of the identity functor *) ``` ``` | NORM of formula * formula (* default implication *) ``` ``` | NORMN of formula * formula (* \neg NORM (\neg A, \neg B) *) ``` ``` | CHC of formula * formula (* Choice is self-dual *) ``` ``` | FUS of bool * formula ``` ``` | MU of string * formula ``` ``` | NU of string * formula ``` ``` | VAR of string ``` ``` | AF of formula ``` ``` | EF of formula ``` ``` | AG of formula ``` ``` | EG of formula ``` ``` | AU of formula * formula ``` ``` | EU of formula * formula ``` ```exception ConversionException of formula ``` ```(** Defines sorted coalgebraic formulae. ``` ``` *) ``` ```type sortedFormula = sort * formula ``` ```(** Determines whether a name indicates a nominal. ``` ``` @param A string s. ``` ``` @return True iff s contains a prime character. ``` ``` *) ``` ```let isNominal s = String.contains s '\'' ``` ```(** Determines the size of a formula. ``` ``` @param f A formula. ``` ``` @return The size of f. ``` ``` *) ``` ```let rec sizeFormula f = ``` ``` match f with ``` ``` | TRUE ``` ``` | FALSE ``` ``` | AP _ -> 1 ``` ``` | NOT f1 ``` ``` | AT (_, f1) -> succ (sizeFormula f1) ``` ``` | OR (f1, f2) ``` ``` | AND (f1, f2) ``` ``` | EQU (f1, f2) ``` ``` | IMP (f1, f2) -> succ (sizeFormula f1 + sizeFormula f2) ``` ``` | EX (_, f1) ``` ``` | AX (_, f1) ``` ``` | ENFORCES (_, f1) ``` ``` | ALLOWS (_, f1) -> succ (sizeFormula f1) ``` ``` | MIN (_, _, f1) ``` ``` | MAX (_, _, f1) ``` ``` | ATLEASTPROB (_, f1) ``` ``` | LESSPROBFAIL (_, f1) ``` ``` | MORETHAN (_, _, f1) ``` ``` | MAXEXCEPT (_, _, f1) -> succ (sizeFormula f1) ``` ``` | ID (f1) -> succ (sizeFormula f1) ``` ``` | NORM(f1, f2) ``` ``` | NORMN(f1, f2) -> succ (sizeFormula f1 + sizeFormula f2) ``` ``` | CONST _ ``` ``` | CONSTN _ -> 1 ``` ``` | CHC (f1, f2) -> succ (sizeFormula f1 + sizeFormula f2) ``` ``` | FUS (_, f1) -> succ (sizeFormula f1) ``` ``` | MU (_, f1) ``` ``` | NU (_, f1) -> succ (succ (sizeFormula f1)) ``` ``` | VAR _ -> 1 ``` ``` | AF _ | EF _ ``` ``` | AG _ | EG _ ``` ``` | AU _ | EU _ -> ``` ``` raise (CoAlgException ("sizeFormula: CTL should have been removed at this point")) ``` ```(** Determines the size of a sorted formula. ``` ``` @param f A sorted formula. ``` ``` @return The size of f. ``` ``` *) ``` ```let sizeSortedFormula f = sizeFormula (snd f) ``` ```(* think of func: (formula -> unit) -> formula -> unit as identity. ``` ``` iterate over all subformulae and collect side effects. *) ``` ```let rec iter func formula = ``` ``` func formula; ``` ``` let proc = iter func in (* proc = "process" *) ``` ``` match formula with ``` ``` | TRUE | FALSE | AP _ | VAR _ -> () ``` ``` | CONST _ ``` ``` | CONSTN _ -> () ``` ``` | NOT a | AT (_,a) ``` ``` | EX (_,a) | AX (_,a) -> proc a ``` ``` | OR (a,b) | AND (a,b) ``` ``` | EQU (a,b) | IMP (a,b) -> (proc a ; proc b) ``` ``` | ENFORCES (_,a) | ALLOWS (_,a) ``` ``` | MIN (_,_,a) | MAX (_,_,a) ``` ``` | ATLEASTPROB (_, a) | LESSPROBFAIL (_, a) ``` ``` | MORETHAN (_,_,a) | MAXEXCEPT (_,_,a) -> proc a ``` ``` | ID(a) -> proc a ``` ``` | NORM(a, b) | NORMN(a, b) -> (proc a; proc b) ``` ``` | CHC (a,b) -> (proc a ; proc b) ``` ``` | FUS (_,a) -> proc a ``` ``` | MU (_, f) | NU (_, f) -> proc f ``` ``` | AF f | EF f | AG f | EG f -> proc f ``` ``` | AU (f1, f2) | EU (f1, f2) -> (proc f1; proc f2) ``` ```let rec convert_post func formula = (* run over all subformulas in post order *) ``` ``` let c = convert_post func in (* some shorthand *) ``` ``` (* replace parts of the formula *) ``` ``` let formula = (match formula with ``` ``` (* 0-ary constructors *) ``` ``` | TRUE | FALSE | AP _ | VAR _ -> formula ``` ``` | CONST _ ``` ``` | CONSTN _ -> formula ``` ``` (* unary *) ``` ``` | NOT a -> NOT (c a) ``` ``` | AT (s,a) -> AT (s,c a) ``` ``` (* binary *) ``` ``` | OR (a,b) -> OR (c a, c b) ``` ``` | AND (a,b) -> AND (c a, c b) ``` ``` | EQU (a,b) -> EQU (c a, c b) ``` ``` | IMP (a,b) -> IMP (c a, c b) ``` ``` | EX (s,a) -> EX (s,c a) ``` ``` | AX (s,a) -> AX (s,c a) ``` ``` | ENFORCES (s,a) -> ENFORCES (s,c a) ``` ``` | ALLOWS (s,a) -> ALLOWS (s,c a) ``` ``` | MIN (n,s,a) -> MIN (n,s,c a) ``` ``` | MAX (n,s,a) -> MAX (n,s,c a) ``` ``` | ATLEASTPROB (p,a) -> ATLEASTPROB (p, c a) ``` ``` | LESSPROBFAIL (p,a) -> LESSPROBFAIL (p, c a) ``` ``` | MORETHAN (n,s,a) -> MORETHAN (n,s,c a) ``` ``` | MAXEXCEPT (n,s,a) -> MAXEXCEPT (n,s,c a) ``` ``` | ID(a) -> ID (c a) ``` ``` | NORM(a, b) -> NORM(c a, c b) ``` ``` | NORMN(a, b) -> NORMN(c a, c b) ``` ``` | CHC (a,b) -> CHC (c a, c b) ``` ``` | FUS (s,a) -> FUS (s, c a) ``` ``` | MU (n, f1) -> MU (n, c f1) ``` ``` | NU (n, f1) -> NU (n, c f1) ``` ``` | AF f1 -> AF (c f1) ``` ``` | EF f1 -> EF (c f1) ``` ``` | AG f1 -> AG (c f1) ``` ``` | EG f1 -> EG (c f1) ``` ``` | AU (f1, f2) -> AU (c f1, c f2) ``` ``` | EU (f1, f2) -> AU (c f1, c f2))in ``` ``` func formula ``` ```let convertToK formula = (* tries to convert a formula to a pure K formula *) ``` ``` let helper formula = match formula with ``` ``` | ENFORCES _ | ALLOWS _ -> raise (ConversionException formula) ``` ``` | MORETHAN (0,s,a) | MIN (1,s,a) -> EX (s,a) ``` ``` | MAX (0,s,a) -> AX (s, NOT a) ``` ``` | MAXEXCEPT (0,s,a) -> AX (s, a) ``` ``` | TRUE | FALSE ``` ``` | EQU _ | IMP _ ``` ``` | AND _ | OR _ ``` ``` | AP _ | NOT _ ``` ``` | AX _ | EX _ ``` ``` | CHC _ | FUS _ -> formula ``` ``` | _ -> raise (ConversionException formula) ``` ``` in ``` ``` convert_post helper formula ``` ```let convertToGML formula = (* tries to convert a formula to a pure GML formula *) ``` ``` let helper formula = match formula with ``` ``` | ENFORCES _ | ALLOWS _ -> raise (ConversionException formula) ``` ``` | MORETHAN _ | MIN _ | MAX _ | MAXEXCEPT _ -> formula ``` ``` | TRUE | FALSE ``` ``` | EQU _ | IMP _ ``` ``` | AND _ | OR _ ``` ``` | AP _ | NOT _ ``` ``` | CHC _ | FUS _ -> formula ``` ``` | AX (r,a) -> MAXEXCEPT (0,r,a) ``` ``` | EX (r,a) -> MORETHAN (0,r,a) ``` ``` | _ -> raise (ConversionException formula) ``` ``` in ``` ``` convert_post helper formula ``` ```let gensym = Stream.from (fun i -> Some (Printf.sprintf "#gensym%x#" i)) ``` ```let convertToMu formula = ``` ``` let helper formula = ``` ``` match formula with ``` ``` | AF f1 -> ``` ``` MU ("#AF#", (OR (f1, (AX ("", (VAR "#AF#")))))) ``` ``` | EF f1 -> ``` ``` MU ("#EF#", (OR (f1, (EX ("", (VAR "#EF#")))))) ``` ``` | AG f1 -> ``` ``` NU ("#AG#", (AND (f1, (AX ("", (VAR "#AG#")))))) ``` ``` | EG f1 -> ``` ``` NU ("#EG#", (AND (f1, (EX ("", (VAR "#EG#")))))) ``` ``` | AU (f1, f2) -> ``` ``` MU ("#AU#", (OR (f2, (AND (f1, (AX ("", (VAR "#AU#")))))))) ``` ``` | EU (f1, f2) -> ``` ``` MU ("#EU#", (OR (f2, (AND (f1, (EX ("", (VAR "#EU#")))))))) ``` ``` | _ -> formula ``` ``` in ``` ``` convert_post helper formula ``` ```(** Appends a string representation of a formula to a string buffer. ``` ``` Parentheses are ommited where possible ``` ``` where the preference rules are defined as usual. ``` ``` @param sb A string buffer. ``` ``` @param f A formula. ``` ``` *) ``` ```let rec exportFormula_buffer sb f = ``` ``` let negate = function ``` ``` | NOT f -> f ``` ``` | f -> NOT f ``` ``` in ``` ``` let prio n lf = ``` ``` let prionr = function ``` ``` | EQU _ -> 0 ``` ``` | IMP _ -> 1 ``` ``` | OR _ -> 2 ``` ``` | AND _ -> 3 ``` ``` | _ -> 4 ``` ``` in ``` ``` if prionr lf < n then begin ``` ``` Buffer.add_char sb '('; ``` ``` exportFormula_buffer sb lf; ``` ``` Buffer.add_char sb ')' ``` ``` end ``` ``` else exportFormula_buffer sb lf ``` ``` in ``` ``` match f with ``` ``` | TRUE -> Buffer.add_string sb "True" ``` ``` | FALSE -> Buffer.add_string sb "False" ``` ``` | AP s -> Buffer.add_string sb s ``` ``` | NOT f1 -> ``` ``` Buffer.add_string sb "~"; ``` ``` prio 4 f1 ``` ``` | AT (s, f1) -> ``` ``` Buffer.add_string sb "@ "; ``` ``` Buffer.add_string sb s; ``` ``` Buffer.add_string sb " "; ``` ``` prio 4 f1 ``` ``` | OR (f1, f2) -> ``` ``` prio 2 f1; ``` ``` Buffer.add_string sb " | "; ``` ``` prio 2 f2 ``` ``` | AND (f1, f2) -> ``` ``` prio 3 f1; ``` ``` Buffer.add_string sb " & "; ``` ``` prio 3 f2 ``` ``` | EQU (f1, f2) -> ``` ``` prio 0 f1; ``` ``` Buffer.add_string sb " <=> "; ``` ``` prio 0 f2 ``` ``` | IMP (f1, f2) -> ``` ``` prio 2 f1; ``` ``` Buffer.add_string sb " => "; ``` ``` prio 2 f2 ``` ``` | EX (s, f1) -> ``` ``` Buffer.add_string sb "<"; ``` ``` Buffer.add_string sb s; ``` ``` Buffer.add_string sb ">"; ``` ``` prio 4 f1 ``` ``` | AX (s, f1) -> ``` ``` Buffer.add_string sb "["; ``` ``` Buffer.add_string sb s; ``` ``` Buffer.add_string sb "]"; ``` ``` prio 4 f1 ``` ``` | ALLOWS (s, f1) -> ``` ``` Buffer.add_string sb "<{"; ``` ``` Buffer.add_string sb ( ``` ``` match s with ``` ``` | [] -> " " ``` ``` | _ ->(Str.concat " " (L.map string_of_int s)) ``` ``` ); ``` ``` Buffer.add_string sb "}>"; ``` ``` prio 4 f1 ``` ``` | ENFORCES (s, f1) -> ``` ``` Buffer.add_string sb "[{"; ``` ``` Buffer.add_string sb ( ``` ``` match s with ``` ``` | [] -> " " ``` ``` | _ ->(Str.concat " " (L.map string_of_int s)) ``` ``` ); ``` ``` Buffer.add_string sb "}]"; ``` ``` prio 4 f1 ``` ``` | ATLEASTPROB (p, f1) -> ``` ``` Buffer.add_string sb "{>="; ``` ``` Buffer.add_string sb (string_of_rational p); ``` ``` Buffer.add_string sb "}"; ``` ``` prio 4 f1 ``` ``` | LESSPROBFAIL (p, f1) -> ``` ``` Buffer.add_string sb "{<"; ``` ``` Buffer.add_string sb (string_of_rational p); ``` ``` Buffer.add_string sb "} ~ "; ``` ``` prio 4 f1 ``` ``` | MIN (n, s, f1) -> ``` ``` Buffer.add_string sb "{>="; ``` ``` Buffer.add_string sb (string_of_int n); ``` ``` Buffer.add_string sb " "; ``` ``` Buffer.add_string sb s; ``` ``` Buffer.add_string sb "}"; ``` ``` prio 4 f1 ``` ``` | MAX (n, s, f1) -> ``` ``` Buffer.add_string sb "{<="; ``` ``` Buffer.add_string sb (string_of_int n); ``` ``` Buffer.add_string sb " "; ``` ``` Buffer.add_string sb s; ``` ``` Buffer.add_string sb "}"; ``` ``` prio 4 f1 ``` ``` | MORETHAN (n, s, f1) -> ``` ``` Buffer.add_string sb "{>"; ``` ``` Buffer.add_string sb (string_of_int n); ``` ``` Buffer.add_string sb " "; ``` ``` Buffer.add_string sb s; ``` ``` Buffer.add_string sb "}"; ``` ``` prio 4 f1 ``` ``` | MAXEXCEPT (n, s, f1) -> ``` ``` Buffer.add_string sb "{<="; ``` ``` Buffer.add_string sb (string_of_int n); ``` ``` Buffer.add_string sb " ~ "; ``` ``` Buffer.add_string sb s; ``` ``` Buffer.add_string sb "}"; ``` ``` prio 4 f1 (* actually is prio of ~ and not of <= *) ``` ``` | ID (f1) -> ``` ``` Buffer.add_string sb "O"; ``` ``` prio 4 f1 ``` ``` | NORM(f1, f2) -> ``` ``` Buffer.add_string sb "("; ``` ``` exportFormula_buffer sb f1; ``` ``` Buffer.add_string sb " =o "; ``` ``` exportFormula_buffer sb f2; ``` ``` Buffer.add_string sb ")" ``` ``` | NORMN(f1, f2) -> ``` ``` Buffer.add_string sb "~("; ``` ``` exportFormula_buffer sb (negate f1); ``` ``` Buffer.add_string sb " =o "; ``` ``` exportFormula_buffer sb (negate f2); ``` ``` Buffer.add_string sb ")" ``` ``` | CHC (f1, f2) -> ``` ``` Buffer.add_string sb "("; ``` ``` exportFormula_buffer sb f1; ``` ``` Buffer.add_string sb " + "; ``` ``` exportFormula_buffer sb f2; ``` ``` Buffer.add_string sb ")" ``` ``` | CONST s -> Buffer.add_string sb "="; ``` ``` Buffer.add_string sb s ``` ``` | CONSTN s -> Buffer.add_string sb "~="; ``` ``` Buffer.add_string sb s ``` ``` | FUS (first, f1) -> ``` ``` Buffer.add_string sb (if first then "[pi1]" else "[pi2]"); ``` ``` prio 4 f1 ``` ``` | MU (s, f1) -> ``` ``` Buffer.add_string sb "μ"; ``` ``` Buffer.add_string sb s; ``` ``` Buffer.add_string sb "."; ``` ``` prio 4 f1 ``` ``` | NU (s, f1) -> ``` ``` Buffer.add_string sb "ν"; ``` ``` Buffer.add_string sb s; ``` ``` Buffer.add_string sb "."; ``` ``` prio 4 f1 ``` ``` | VAR s -> ``` ``` Buffer.add_string sb s ``` ``` | AF f1 -> ``` ``` Buffer.add_string sb "AF "; ``` ``` prio 4 f1 ``` ``` | EF f1 -> ``` ``` Buffer.add_string sb "EF "; ``` ``` prio 4 f1 ``` ``` | AG f1 -> ``` ``` Buffer.add_string sb "AG "; ``` ``` prio 4 f1 ``` ``` | EG f1 -> ``` ``` Buffer.add_string sb "EG "; ``` ``` prio 4 f1 ``` ``` | AU (f1, f2) -> ``` ``` Buffer.add_string sb "A ("; ``` ``` prio 4 f1; ``` ``` Buffer.add_string sb " U "; ``` ``` prio 4 f2; ``` ``` Buffer.add_string sb ")" ``` ``` | EU (f1, f2) -> ``` ``` Buffer.add_string sb "E ("; ``` ``` prio 4 f1; ``` ``` Buffer.add_string sb " U "; ``` ``` prio 4 f2; ``` ``` Buffer.add_string sb ")" ``` ```(** Converts a formula into a string representation. ``` ``` Parentheses are ommited where possible ``` ``` where the preference rules are defined as usual. ``` ``` @param f A formula. ``` ``` @return A string representing f. ``` ``` *) ``` ```let exportFormula f = ``` ``` let sb = Buffer.create 128 in ``` ``` exportFormula_buffer sb f; ``` ``` Buffer.contents sb ``` ```let string_of_formula = exportFormula ``` ```(** export (CL)-formula suitable for tatl-inputs *) ``` ```let rec exportTatlFormula_buffer sb f = ``` ``` let prio n lf = ``` ``` let prionr = function ``` ``` | EQU _ -> 0 ``` ``` | IMP _ -> 1 ``` ``` | OR _ -> 2 ``` ``` | AND _ -> 3 ``` ``` | _ -> 4 ``` ``` in ``` ``` if prionr lf < n then begin ``` ``` Buffer.add_char sb '('; ``` ``` exportTatlFormula_buffer sb lf; ``` ``` Buffer.add_char sb ')' ``` ``` end ``` ``` else exportTatlFormula_buffer sb lf ``` ``` in ``` ``` match f with ``` ``` | TRUE -> Buffer.add_string sb "(p \\/ ~p)" ``` ``` | FALSE -> Buffer.add_string sb "(p /\\ ~p)" ``` ``` | AP s -> Buffer.add_string sb s ``` ``` | NOT f1 -> ``` ``` Buffer.add_string sb "~"; ``` ``` prio 4 f1 ``` ``` | OR (f1, f2) -> ``` ``` prio 2 f1; ``` ``` Buffer.add_string sb " \\/ "; ``` ``` prio 2 f2 ``` ``` | AND (f1, f2) -> ``` ``` prio 3 f1; ``` ``` Buffer.add_string sb " /\\ "; ``` ``` prio 3 f2 ``` ``` | EQU (f1, f2) -> ``` ``` prio 0 (AND (IMP (f1,f2), IMP (f2,f1))) ``` ``` | IMP (f1, f2) -> ``` ``` prio 2 f1; ``` ``` Buffer.add_string sb " -> "; ``` ``` prio 2 f2 ``` ``` | ALLOWS (s, f1) -> ``` ``` Buffer.add_string sb "<<"; ``` ``` Buffer.add_string sb ( ``` ``` match s with ``` ``` | [] -> " " ``` ``` | _ ->(Str.concat "," (L.map string_of_int s)) ``` ``` ); ``` ``` Buffer.add_string sb ">>X "; ``` ``` prio 4 f1 ``` ``` | ENFORCES (s, f1) -> ``` ``` Buffer.add_string sb "~<<"; ``` ``` Buffer.add_string sb ( ``` ``` match s with ``` ``` | [] -> " " ``` ``` | _ ->(Str.concat "," (L.map string_of_int s)) ``` ``` ); ``` ``` Buffer.add_string sb ">>X ~ "; ``` ``` prio 4 f1 ``` ``` | EX _ ``` ``` | AX _ ``` ``` | MIN _ ``` ``` | MAX _ ``` ``` | MORETHAN _ ``` ``` | MAXEXCEPT _ ``` ``` | AT _ ``` ``` | CONST _ ``` ``` | CONSTN _ ``` ``` | CHC _ ``` ``` | ATLEASTPROB _ ``` ``` | LESSPROBFAIL _ ``` ``` | ID _ ``` ``` | NORM _ ``` ``` | NORMN _ ``` ``` | FUS _ ``` ``` | MU _ ``` ``` | NU _ ``` ``` | VAR _ ``` ``` | AF _ ``` ``` | EF _ ``` ``` | AG _ ``` ``` | EG _ ``` ``` | AU _ ``` ``` | EU _ -> raise (CoAlgException ("export to tatl: Not connectives of CL")) ``` ```let exportTatlFormula f = ``` ``` let sb = Buffer.create 128 in ``` ``` exportTatlFormula_buffer sb f; ``` ``` Buffer.contents sb ``` ```(** Appends a string representation of a sorted formula to a string buffer. ``` ``` Parentheses are ommited where possible ``` ``` where the preference rules are defined as usual. ``` ``` @param sb A string buffer. ``` ``` @param (s, f) A sorted formula. ``` ``` *) ``` ```let rec exportSortedFormula_buffer sb (s, f) = ``` ``` Buffer.add_string sb (string_of_int s); ``` ``` Buffer.add_string sb ": "; ``` ``` exportFormula_buffer sb f ``` ```(** Converts a sorted formula into a string representation. ``` ``` Parentheses are ommited where possible ``` ``` where the preference rules are defined as usual. ``` ``` @param f A sorted formula. ``` ``` @return A string representing f. ``` ``` *) ``` ```let exportSortedFormula f = ``` ``` let sb = Buffer.create 128 in ``` ``` exportSortedFormula_buffer sb f; ``` ``` Buffer.contents sb ``` ```(** Converts a (sorted) formula query into a string representation. ``` ``` @param tbox A list of sorted formulae representing a TBox. ``` ``` @param f A sorted formula. ``` ``` @return A string representing tbox |- f. ``` ``` *) ``` ```let exportQuery tbox f = ``` ``` let sb = Buffer.create 1000 in ``` ``` let rec expFl = function ``` ``` | [] -> () ``` ``` | h::tl -> ``` ``` exportSortedFormula_buffer sb h; ``` ``` if tl <> [] then Buffer.add_string sb "; " else (); ``` ``` expFl tl ``` ``` in ``` ``` expFl tbox; ``` ``` Buffer.add_string sb " |- "; ``` ``` exportSortedFormula_buffer sb f; ``` ``` Buffer.contents sb ``` ```(** Converts a (sorted) formula query into a string representation. Such that ``` ``` coalg can read it again using importQuery ``` ``` @param tbox A list of sorted formulae representing a TBox. ``` ``` @param f A sorted formula. ``` ``` @return A string representing tbox |- f. ``` ``` *) ``` ```let exportQueryParsable tbox (_,f) = ``` ``` let sb = Buffer.create 1000 in ``` ``` let rec expFl = function ``` ``` | [] -> () ``` ``` | (_,h)::tl -> ``` ``` exportFormula_buffer sb h; ``` ``` if tl <> [] then Buffer.add_string sb "; " else (); ``` ``` expFl tl ``` ``` in ``` ``` expFl tbox; ``` ``` Buffer.add_string sb " |- "; ``` ``` exportFormula_buffer sb f; ``` ``` Buffer.contents sb ``` ```(* NB: True and False are the propositional constants. Lower case ``` ``` true/false are regardes as atomic propositions and we emit a warning ``` ```*) ``` ```let lexer = A.make_lexer ``` ``` [":";";";"|-";"(";")";"=>";"<=>";"|";"&";"~";"@";"True";"False";"true";"false";"<";">";"[";"]";"{<=";"{>=";"}";"+";"[pi1]";"[pi2]" ``` ``` ;"[{";"}]";"<{";"}>";",";"/";"{<";"=";"=o";"O" ``` ``` ;"μ";"ν";"." ``` ``` ;"AX";"EX";"AF";"EF";"AG";"EG";"A";"E";"U" ``` ``` ] ``` ```let mk_exn s = CoAlgException s ``` ```(** Process from inside out. cons all variables seen, remove them when ``` ``` binding fixpoint is found. Fixpoint type may only change if last ``` ``` (inner) fixpoint didn't include any free variables. ``` ``` *) ``` ```let rec verifyMuAltFree formula = ``` ``` let proc = verifyMuAltFree in ``` ``` match formula with ``` ``` | TRUE | FALSE | AP _ -> ("none", []) ``` ``` | CONST _ ``` ``` | CONSTN _ -> ("none", []) ``` ``` | AT (_,a) | NOT a ``` ``` | EX (_,a) | AX (_,a) -> proc a ``` ``` | OR (a,b) | AND (a,b) ``` ``` | EQU (a,b) | IMP (a,b) -> ``` ``` let (atype, afree) = proc a ``` ``` and (btype, bfree) = proc b in ``` ``` if (compare atype "μ" == 0 && compare btype "ν" == 0) || ``` ``` (compare atype "ν" == 0 && compare btype "μ" == 0) then ``` ``` raise (CoAlgException ("formula not alternation-free")); ``` ``` if compare atype "none" == 0 then ``` ``` (btype, List.flatten [afree; bfree]) ``` ``` else ``` ``` (atype, List.flatten [afree; bfree]) ``` ``` | ENFORCES (_,a) | ALLOWS (_,a) ``` ``` | MIN (_,_,a) | MAX (_,_,a) ``` ``` | ATLEASTPROB (_, a) | LESSPROBFAIL (_, a) ``` ``` | MORETHAN (_,_,a) | MAXEXCEPT (_,_,a) -> proc a ``` ``` | ID(a) -> proc a ``` ``` | NORM(a, b) | NORMN(a, b) ``` ``` | CHC (a,b) -> ``` ``` let (atype, afree) = proc a ``` ``` and (btype, bfree) = proc b in ``` ``` if (compare atype "μ" == 0 && compare btype "ν" == 0) || ``` ``` (compare atype "ν" == 0 && compare btype "μ" == 0) then ``` ``` raise (CoAlgException ("formula not alternation-free")); ``` ``` if compare atype "none" == 0 then ``` ``` (btype, List.flatten [afree; bfree]) ``` ``` else ``` ``` (atype, List.flatten [afree; bfree]) ``` ``` | FUS (_,a) -> proc a ``` ``` | MU (s, f) -> ``` ``` let (fptype, free) = proc f in ``` ``` (if (compare fptype "ν" == 0) then ``` ``` raise (CoAlgException ("formula not alternation-free"))); ``` ``` let predicate x = compare x s != 0 in ``` ``` let newfree = List.filter predicate free in ``` ``` if newfree = [] then ``` ``` ("none", []) ``` ``` else ``` ``` ("μ", newfree) ``` ``` | NU (s, f) -> ``` ``` let (fptype, free) = proc f in ``` ``` (if (compare fptype "μ" == 0) then ``` ``` raise (CoAlgException ("formula not alternation-free"))); ``` ``` let predicate x = compare x s != 0 in ``` ``` let newfree = List.filter predicate free in ``` ``` if newfree = [] then ``` ``` ("none", []) ``` ``` else ``` ``` ("ν", newfree) ``` ``` | VAR s -> ("none", [s]) ``` ``` | AF _ | EF _ | AG _ | EG _ | AU _ | EU _ -> ``` ``` raise (CoAlgException ("verifyMuAltFree: CTL should have been removed at this point")) ``` ```(** Process from outside in. For every variable bound keep the tuple ``` ``` (name, negations). For every negation crossed map a +1 over snd on ``` ``` that list. For every variable met check that the matching ``` ``` negations is even. ``` ``` *) ``` ```let rec verifyMuMonotone negations formula = ``` ``` let proc = verifyMuMonotone negations in ``` ``` match formula with ``` ``` | TRUE | FALSE | AP _ -> () ``` ``` | CONST _ ``` ``` | CONSTN _ -> () ``` ``` | AT (_,a) ``` ``` | EX (_,a) | AX (_,a) -> proc a ``` ``` | OR (a,b) | AND (a,b) ``` ``` | EQU (a,b) | IMP (a,b) -> (proc a ; proc b) ``` ``` | ENFORCES (_,a) | ALLOWS (_,a) ``` ``` | MIN (_,_,a) | MAX (_,_,a) ``` ``` | ATLEASTPROB (_, a) | LESSPROBFAIL (_, a) ``` ``` | MORETHAN (_,_,a) | MAXEXCEPT (_,_,a) -> proc a ``` ``` | ID(a) -> proc a ``` ``` | NORM(a, b) | NORMN(a, b) -> (proc a; proc b) ``` ``` | CHC (a,b) -> (proc a ; proc b) ``` ``` | FUS (_,a) -> proc a ``` ``` | MU (s, f) ``` ``` | NU (s, f) -> ``` ``` let newNeg = (s, 0) :: negations in ``` ``` verifyMuMonotone newNeg f ``` ``` | VAR s -> ``` ``` let finder (x, _) = compare x s == 0 in ``` ``` let (_, neg) = List.find finder negations in ``` ``` if ((neg land 1) != 0) then raise (CoAlgException ("formula not monotone")) ``` ``` | NOT a -> ``` ``` let increment (s, n) = (s, n+1) in ``` ``` let newNeg = List.map increment negations in ``` ``` verifyMuMonotone newNeg a ``` ``` | AF _ | EF _ | AG _ | EG _ | AU _ | EU _ -> ``` ``` raise (CoAlgException ("verifyMuMonotone: CTL should have been removed at this point")) ``` ```let rec verifyMuGuarded unguarded formula = ``` ``` let proc = verifyMuGuarded unguarded in ``` ``` match formula with ``` ``` | TRUE | FALSE | AP _ -> () ``` ``` | CONST _ ``` ``` | CONSTN _ -> () ``` ``` | AT (_,a) | NOT a -> proc a ``` ``` | EX (_,a) | AX (_,a) -> verifyMuGuarded [] a ``` ``` | OR (a,b) | AND (a,b) ``` ``` | EQU (a,b) | IMP (a,b) -> (proc a ; proc b) ``` ``` | ENFORCES (_,a) | ALLOWS (_,a) ``` ``` | MIN (_,_,a) | MAX (_,_,a) ``` ``` | ATLEASTPROB (_, a) | LESSPROBFAIL (_, a) ``` ``` | MORETHAN (_,_,a) | MAXEXCEPT (_,_,a) -> proc a ``` ``` | ID(a) -> proc a ``` ``` | NORM(a, b) | NORMN(a, b) -> (proc a; proc b) ``` ``` | CHC (a,b) -> (proc a ; proc b) ``` ``` | FUS (_,a) -> proc a ``` ``` | MU (s, f) ``` ``` | NU (s, f) -> ``` ``` let newUnguard = s :: unguarded in ``` ``` verifyMuGuarded newUnguard f ``` ``` | VAR s -> ``` ``` let finder x = compare x s == 0 in ``` ``` if List.exists finder unguarded then ``` ``` raise (CoAlgException ("formula not guarded")) ``` ``` | AF _ | EF _ | AG _ | EG _ | AU _ | EU _ -> ``` ``` raise (CoAlgException ("verifyMuGuarded: CTL should have been removed at this point")) ``` ```let verifyFormula formula = ``` ``` verifyMuAltFree formula; ``` ``` verifyMuMonotone [] formula; ``` ``` verifyMuGuarded [] formula ``` ```(** These functions parse a token stream to obtain a formula. ``` ``` It is a standard recursive descent procedure. ``` ``` @param ts A token stream. ``` ``` @return The resulting (sub-)formula. ``` ``` @raise CoAlgException if ts does not represent a formula. ``` ``` *) ``` ```let rec parse_formula (symtab: 'a list) ts = ``` ``` let formula = (parse_imp symtab ts) in ``` ``` let f1 = convertToMu formula in ``` ``` match Stream.peek ts with ``` ``` | None -> f1 ``` ``` | Some (A.Kwd "<=>") -> ``` ``` Stream.junk ts; ``` ``` let f2 = parse_formula symtab ts in ``` ``` EQU (f1, f2) ``` ``` | _ -> f1 ``` ```(** These functions parse a token stream to obtain a formula. ``` ``` It is a standard recursive descent procedure. ``` ``` @param ts A token stream. ``` ``` @return The resulting (sub-)formula. ``` ``` @raise CoAlgException if ts does not represent a formula. ``` ``` *) ``` ```and parse_imp symtab ts= ``` ``` let f1 = parse_or symtab ts in ``` ``` match Stream.peek ts with ``` ``` | None -> f1 ``` ``` | Some (A.Kwd "=>") -> ``` ``` Stream.junk ts; ``` ``` let f2 = parse_imp symtab ts in ``` ``` IMP (f1, f2) ``` ``` | _ -> f1 ``` ```(** These functions parse a token stream to obtain a formula. ``` ``` It is a standard recursive descent procedure. ``` ``` @param ts A token stream. ``` ``` @return The resulting (sub-)formula. ``` ``` @raise CoAlgException if ts does not represent a formula. ``` ``` *) ``` ```and parse_or symtab ts = ``` ``` let f1 = parse_and symtab ts in ``` ``` match Stream.peek ts with ``` ``` | None -> f1 ``` ``` | Some (A.Kwd "|") -> ``` ``` Stream.junk ts; ``` ``` let f2 = parse_or symtab ts in ``` ``` OR (f1, f2) ``` ``` | _ -> f1 ``` ```(** These functions parse a token stream to obtain a formula. ``` ``` It is a standard recursive descent procedure. ``` ``` @param ts A token stream. ``` ``` @return The resulting (sub-)formula. ``` ``` @raise CoAlgException if ts does not represent a formula. ``` ``` *) ``` ```and parse_and symtab ts = ``` ``` let f1 = parse_cimp symtab ts in ``` ``` match Stream.peek ts with ``` ``` | None -> f1 ``` ``` | Some (A.Kwd "&") -> ``` ``` Stream.junk ts; ``` ``` let f2 = parse_and symtab ts in ``` ``` AND (f1, f2) ``` ``` | _ -> f1 ``` ```(** These functions parse a token stream to obtain a formula. ``` ``` It is a standard recursive descent procedure. ``` ``` @param ts A token stream. ``` ``` @return The resulting (sub-)formula. ``` ``` @raise CoAlgException if ts does not represent a formula. ``` ``` *) ``` ```and parse_cimp symtab ts = ``` ``` let f1 = parse_rest symtab ts in ``` ``` match Stream.peek ts with ``` ``` | None -> f1 ``` ``` | Some (A.Kwd "=o") -> ``` ``` Stream.junk ts; ``` ``` let f2 = parse_cimp symtab ts in ``` ``` NORM (f1, f2) ``` ``` | _ -> f1 ``` ```(** These functions parse a token stream to obtain a formula. ``` ``` It is a standard recursive descent procedure. ``` ``` @param ts A token stream. ``` ``` @return The resulting (sub-)formula. ``` ``` @raise CoAlgException if ts does not represent a formula. ``` ``` *) ``` ```and parse_rest symtab ts = ``` ``` let boxinternals noNo es = ``` ``` let n = ``` ``` if noNo then 0 ``` ``` else ``` ``` match Stream.next ts with ``` ``` | A.Int n when n >= 0 -> n ``` ``` | t -> A.printError mk_exn ~t ~ts " expected: " ``` ``` in ``` ``` let (s,denominator) = ``` ``` match Stream.peek ts with ``` ``` | Some (A.Ident s1) -> Stream.junk ts; (s1,0) ``` ``` | Some (A.Kwd c) when c = es -> ("", 0) ``` ``` | Some (A.Kwd "/") -> begin ``` ``` Stream.junk ts; ``` ``` match Stream.next ts with ``` ``` | A.Int denom when denom > 0 -> ("", denom) ``` ``` | t -> A.printError mk_exn ~t ~ts " (the denominator) expected: " ``` ``` end ``` ``` | _ -> A.printError mk_exn ~ts ("role name or \"" ^ es ^ "\" expected: ") ``` ``` in ``` ``` if (denominator < n) then begin ``` ``` let explanation = ``` ``` ("nominator="^(string_of_int n)^" but denominator="^(string_of_int denominator)) ``` ``` in ``` ``` A.printError mk_exn ~ts ("Nominator must not be larger than the denominator " ``` ``` ^"("^explanation^") at: " ``` ``` ) ``` ``` end; ``` ``` let _ = ``` ``` match Stream.next ts with ``` ``` | A.Kwd c when c = es -> () ``` ``` | t -> A.printError mk_exn ~t ~ts ("\"" ^ es ^ "\" expected: ") ``` ``` in ``` ``` (n, denominator, s) ``` ``` in ``` ``` let rec agentlist es = ``` ``` let allAgents = CoolUtils.cl_get_agents () in ``` ``` match Stream.next ts with ``` ``` | A.Int n -> if CoolUtils.TArray.elem n allAgents ``` ``` then n::(agentlist es) ``` ``` else A.printError mk_exn ~ts ("agent name \"" ^ (string_of_int n) ^ "\" unknonwn, see --agents: ") ``` ``` | A.Kwd c when c = es -> [] ``` ``` | _ -> A.printError mk_exn ~ts ("agent name or " ^ es ^ "\" expected: ") ``` ``` in ``` ``` match Stream.next ts with ``` ``` | A.Kwd "true" -> ``` ``` print_endline "*** Warning: \"true\" used as propositional variable."; ``` ``` AP "true" ``` ``` | A.Kwd "false" -> ``` ``` print_endline "*** Warning: \"false\" used as propositional variable."; ``` ``` AP "false" ``` ``` | A.Kwd "True" -> TRUE ``` ``` | A.Kwd "False" -> FALSE ``` ``` | A.Ident s -> ``` ``` (try ``` ``` let finder (x, _) = compare x s == 0 in ``` ``` let (_, symbol) = List.find finder symtab in ``` ``` VAR symbol ``` ``` with Not_found -> AP s) ``` ``` | A.Kwd "~" -> ``` ``` let f = parse_rest symtab ts in ``` ``` NOT f ``` ``` | A.Kwd "@" -> ``` ``` let s = ``` ``` match Stream.next ts with ``` ``` | A.Ident s when isNominal s -> s ``` ``` | t -> A.printError mk_exn ~t ~ts ("nominal expected: ") ``` ``` in ``` ``` let f = parse_rest symtab ts in ``` ``` AT (s, f) ``` ``` | A.Kwd "<" -> ``` ``` let (_, _, s) = boxinternals true ">" in ``` ``` let f1 = parse_rest symtab ts in ``` ``` EX (s, f1) ``` ``` | A.Kwd "[" -> ``` ``` let (_, _, s) = boxinternals true "]" in ``` ``` let f1 = parse_rest symtab ts in ``` ``` AX (s, f1) ``` ``` | A.Kwd "[{" -> ``` ``` let ls = agentlist "}]" in ``` ``` let f1 = parse_rest symtab ts in ``` ``` ENFORCES (ls, f1) ``` ``` | A.Kwd "<{" -> ``` ``` let ls = agentlist "}>" in ``` ``` let f1 = parse_rest symtab ts in ``` ``` ALLOWS (ls, f1) ``` ``` | A.Kwd "{>=" -> ``` ``` let (n, denom, s) = boxinternals false "}" in ``` ``` let f1 = parse_rest symtab ts in ``` ``` if (denom <> 0) ``` ``` then ATLEASTPROB (rational_of_int n denom, f1) ``` ``` else MIN (n, s, f1) ``` ``` | A.Kwd "{<=" -> ``` ``` let (n, denom, s) = boxinternals false "}" in ``` ``` let f1 = parse_rest symtab ts in ``` ``` if (denom <> 0) ``` ``` then A.printError mk_exn ~ts "Can not express {<= probability}" ``` ``` else MAX (n, s, f1) ``` ``` | A.Kwd "{<" -> ``` ``` let (n, denom, s) = boxinternals false "}" in ``` ``` let f1 = parse_rest symtab ts in ``` ``` if (denom <> 0) ``` ``` then LESSPROBFAIL (rational_of_int n denom, NOT f1) ``` ``` else A.printError mk_exn ~ts "The \"Less than\" < is not implemented yet" ``` ``` | A.Kwd "=" -> begin ``` ``` match Stream.next ts with ``` ``` (* | A.Int s *) ``` ``` | A.Kwd s ``` ``` | A.Ident s -> CONST s ``` ``` | _ -> A.printError mk_exn ~ts "constant = expects an identifier afterwards" ``` ``` end ``` ``` | A.Kwd "(" -> begin ``` ``` let f = parse_formula symtab ts in ``` ``` match Stream.next ts with ``` ``` | A.Kwd ")" -> f ``` ``` | A.Kwd "+" -> begin ``` ``` let f2 = parse_formula symtab ts in ``` ``` match Stream.next ts with ``` ``` | A.Kwd ")" -> CHC (f, f2) ``` ``` | t -> A.printError mk_exn ~t ~ts "\")\" expected: " ``` ``` end ``` ``` | t -> A.printError mk_exn ~t ~ts "\")\" or \"+\" expected: " ``` ``` end ``` ``` | A.Kwd "O" -> ``` ``` let f = parse_rest symtab ts in ``` ``` ID f ``` ``` | A.Kwd "[pi1]" -> ``` ``` let f = parse_rest symtab ts in ``` ``` FUS (true, f) ``` ``` | A.Kwd "[pi2]" -> ``` ``` let f = parse_rest symtab ts in ``` ``` FUS (false, f) ``` ``` | A.Kwd "μ" -> ``` ``` let (_, _, s) = boxinternals true "." in ``` ``` let symbol = Stream.next gensym in ``` ``` let newtab = (s, symbol) :: symtab in ``` ``` let f1 = parse_rest newtab ts in ``` ``` MU (symbol, f1) ``` ``` | A.Kwd "ν" -> ``` ``` let (_, _, s) = boxinternals true "." in ``` ``` let symbol = Stream.next gensym in ``` ``` let newtab = (s, symbol) :: symtab in ``` ``` let f1 = parse_rest newtab ts in ``` ``` NU (symbol, f1) ``` ``` | A.Kwd "AF" -> ``` ``` let f = parse_rest symtab ts in ``` ``` AF f ``` ``` | A.Kwd "EF" -> ``` ``` let f = parse_rest symtab ts in ``` ``` EF f ``` ``` | A.Kwd "AG" -> ``` ``` let f = parse_rest symtab ts in ``` ``` AG f ``` ``` | A.Kwd "EG" -> ``` ``` let f = parse_rest symtab ts in ``` ``` EG f ``` ``` | A.Kwd "A" -> ``` ``` assert (Stream.next ts = A.Kwd "("); ``` ``` let f1 = parse_formula symtab ts in ``` ``` assert (Stream.next ts = A.Kwd "U"); ``` ``` let f2 = parse_formula symtab ts in ``` ``` assert (Stream.next ts = A.Kwd ")"); ``` ``` AU (f1, f2) ``` ``` | A.Kwd "E" -> ``` ``` assert (Stream.next ts = A.Kwd "("); ``` ``` let f1 = parse_formula symtab ts in ``` ``` assert (Stream.next ts = A.Kwd "U"); ``` ``` let f2 = parse_formula symtab ts in ``` ``` assert (Stream.next ts = A.Kwd ")"); ``` ``` EU (f1, f2) ``` ``` | A.Kwd "AX" -> ``` ``` let f1 = parse_rest symtab ts in ``` ``` AX ("", f1) ``` ``` | A.Kwd "EX" -> ``` ``` let f1 = parse_rest symtab ts in ``` ``` EX ("", f1) ``` ``` | t -> A.printError mk_exn ~t ~ts ``` ``` "\"<\", \"[\", \"{>=\", \"{<=\", \"@\", \"~\", \"(\", ``` ``` \"True\", \"False\", \"=\", \"=o\", \"O\" or expected: " ``` ```(** Parses a sorted formula. ``` ``` @param ts A token stream. ``` ``` @return A sorted formula. ``` ``` @raise CoAlgException If ts does not represent a sorted formula. ``` ``` *) ``` ```let parse_sortedFormula ts = ``` ``` let nr = ``` ``` match Stream.peek ts with ``` ``` | Some (A.Int n) -> ``` ``` if n >= 0 then begin ``` ``` Stream.junk ts; ``` ``` let () = ``` ``` match Stream.next ts with ``` ``` | A.Kwd ":" -> () ``` ``` | t -> A.printError mk_exn ~t ~ts ("\":\" expected: ") ``` ``` in ``` ``` n ``` ``` end else ``` ``` A.printError mk_exn ~ts (" expected: ") ``` ``` | _ -> 0 ``` ``` in ``` ``` let f = parse_formula [] ts in ``` ``` (nr, f) ``` ```(** Parses a list of sorted formulae separated by ";". ``` ``` @param ts A token stream. ``` ``` @param acc The list of sorted formulae parsed so far. ``` ``` @return The resulting list of sorted formula. ``` ``` @raise CoAlgException if ts does not represent a list of sorted formulae. ``` ``` *) ``` ```let rec parse_sortedFormulaList ts acc = ``` ``` let f1 = parse_sortedFormula ts in ``` ``` match Stream.peek ts with ``` ``` | Some (A.Kwd ";") -> ``` ``` Stream.junk ts; ``` ``` parse_sortedFormulaList ts (f1::acc) ``` ``` | _ -> List.rev (f1::acc) ``` ```(** A common wrapper for import functions. ``` ``` @param fkt An import function. ``` ``` @param s A string. ``` ``` @return The object imported from s using fkt. ``` ``` @raise CoAlgException If ts is not empty. ``` ``` *) ``` ```let importWrapper fkt s = ``` ``` let ts = lexer s in ``` ``` try ``` ``` let res = fkt ts in ``` ``` let _ = ``` ``` match Stream.peek ts with ``` ``` | None -> () ``` ``` | Some _ -> A.printError mk_exn ~ts "EOS expected: " ``` ``` in ``` ``` res ``` ``` with Stream.Failure -> A.printError mk_exn ~ts "unexpected EOS" ``` ```(** Parses a string to obtain a formula. ``` ``` @param s A string representing a formula. ``` ``` @return The resulting formula. ``` ``` @raise CoAlgException if s does not represent a formula. ``` ``` *) ``` ```let importFormula s = importWrapper (parse_formula []) s ``` ```(** Parses a string to obtain a sorted formula. ``` ``` @param s A string representing a sorted formula. ``` ``` @return The sorted formula. ``` ``` @raise CoAlgException If s does not represent a sorted formula. ``` ``` *) ``` ```let importSortedFormula s = importWrapper parse_sortedFormula s ``` ```(** Parses a string to obtain a (sorted) formula query. ``` ``` @param s A string representing a formula query. ``` ``` @return A pair (tbox, f) where ``` ``` tbox is a list of sorted formulae representing the TBox; and ``` ``` f is a sorted formula. ``` ``` @raise CoAlgException if s does not represent a formula query. ``` ``` *) ``` ```let importQuery s = ``` ``` let fkt ts = ``` ``` match Stream.peek ts with ``` ``` | Some (A.Kwd "|-") -> ``` ``` Stream.junk ts; ``` ``` let f = parse_sortedFormula ts in ``` ``` ([], f) ``` ``` | _ -> ``` ``` let fl = parse_sortedFormulaList ts [] in ``` ``` match Stream.peek ts with ``` ``` | Some (A.Kwd "|-") -> ``` ``` Stream.junk ts; ``` ``` let f = parse_sortedFormula ts in ``` ``` (fl, f) ``` ``` | _ -> ``` ``` if List.length fl = 1 then ([], List.hd fl) ``` ``` else A.printError mk_exn ~ts "\"|-\" expected: " ``` ``` in ``` ``` importWrapper fkt s ``` ```(** Converts the negation of a formula to negation normal form ``` ``` by "pushing in" the negations "~". ``` ``` The symbols "<=>" and "=>" are substituted by their usual definitions. ``` ``` @param f A formula. ``` ``` @return A formula in negation normal form and not containing "<=>" or "=>" ``` ``` that is equivalent to ~f. ``` ``` *) ``` ```let rec nnfNeg f = ``` ``` match f with ``` ``` | TRUE -> FALSE ``` ``` | FALSE -> TRUE ``` ``` | AP _ -> NOT f ``` ``` | VAR _ -> f ``` ``` | NOT f1 -> nnf f1 ``` ``` | AT (s, f1) -> AT (s, nnfNeg f1) ``` ``` | OR (f1, f2) -> AND (nnfNeg f1, nnfNeg f2) ``` ``` | AND (f1, f2) -> OR (nnfNeg f1, nnfNeg f2) ``` ``` | EQU (f1, f2) -> OR (AND (nnf f1, nnfNeg f2), AND (nnf f2, nnfNeg f1)) ``` ``` | IMP (f1, f2) -> AND (nnf f1, nnfNeg f2) ``` ``` | EX (s, f1) -> AX (s, nnfNeg f1) ``` ``` | AX (s, f1) -> EX (s, nnfNeg f1) ``` ``` | ENFORCES (s, f1) -> ALLOWS (s, nnfNeg f1) ``` ``` | ALLOWS (s, f1) -> ENFORCES (s, nnfNeg f1) ``` ``` | MIN (n, s, f1) -> if n = 0 then FALSE else MAXEXCEPT (n-1, s, nnfNeg f1) ``` ``` | MAX (n, s, f1) -> MORETHAN (n, s, nnf f1) ``` ``` | MORETHAN (n, s, f1) -> MAXEXCEPT (n, s, nnfNeg f1) ``` ``` | MAXEXCEPT (n, s, f1) -> MORETHAN (n, s, nnfNeg f1) ``` ``` | ATLEASTPROB (p, f1) -> LESSPROBFAIL (p, nnfNeg f1) ``` ``` | LESSPROBFAIL (p, f1) -> ATLEASTPROB (p, nnfNeg f1) ``` ``` | CONST s -> CONSTN s ``` ``` | CONSTN s -> CONST s ``` ``` | ID (f1) -> ID (nnfNeg f1) ``` ``` | NORM(f1, f2) -> NORMN(nnfNeg f1, nnfNeg f2) ``` ``` | NORMN(f1, f2) -> NORM (nnfNeg f1, nnfNeg f2) ``` ``` | CHC (f1, f2) -> CHC (nnfNeg f1, nnfNeg f2) ``` ``` | FUS (first, f1) -> FUS (first, nnfNeg f1) ``` ``` | MU (s, f1) -> NU(s, nnfNeg f1) ``` ``` | NU (s, f1) -> MU(s, nnfNeg f1) ``` ``` | AF _ ``` ``` | EF _ ``` ``` | AG _ ``` ``` | EG _ ``` ``` | AU _ ``` ``` | EU _ -> raise (CoAlgException ("nnfNeg: CTL should have been removed at this point")) ``` ```(** Converts a formula to negation normal form ``` ``` by "pushing in" the negations "~". ``` ``` The symbols "<=>" and "=>" are substituted by their usual definitions. ``` ``` @param f A formula. ``` ``` @return A formula in negation normal form and not containing "<=>" or "=>" ``` ``` that is equivalent to f. ``` ``` *) ``` ```and nnf f = ``` ``` match f with ``` ``` | TRUE ``` ``` | FALSE ``` ``` | AP _ ``` ``` | NOT (AP _) ``` ``` | CONST _ ``` ``` | CONSTN _ ``` ``` | VAR _ -> f ``` ``` | NOT f1 -> nnfNeg f1 ``` ``` | AT (s, f1) -> ``` ``` let ft1 = nnf f1 in ``` ``` if ft1 == f1 then f else AT (s, ft1) ``` ``` | OR (f1, f2) -> ``` ``` let ft1 = nnf f1 in ``` ``` let ft2 = nnf f2 in ``` ``` if ft1 == f1 && ft2 == f2 then f else OR (ft1, ft2) ``` ``` | AND (f1, f2) -> ``` ``` let ft1 = nnf f1 in ``` ``` let ft2 = nnf f2 in ``` ``` if ft1 == f1 && ft2 == f2 then f else AND (ft1, ft2) ``` ``` | EQU (f1, f2) -> AND (OR (nnfNeg f1, nnf f2), OR (nnfNeg f2, nnf f1)) ``` ``` | IMP (f1, f2) -> OR (nnfNeg f1, nnf f2) ``` ``` | EX (s, f1) -> ``` ``` let ft = nnf f1 in ``` ``` if ft == f1 then f else EX (s, ft) ``` ``` | AX (s, f1) -> ``` ``` let ft = nnf f1 in ``` ``` if ft == f1 then f else AX (s, ft) ``` ``` | ENFORCES (s, f1) -> ``` ``` let ft = nnf f1 in ``` ``` if ft == f1 then f else ENFORCES (s, ft) ``` ``` | ALLOWS (s, f1) -> ``` ``` let ft = nnf f1 in ``` ``` if ft == f1 then f else ALLOWS (s, ft) ``` ``` | MIN (n, s, f1) -> ``` ``` if n = 0 then TRUE ``` ``` else ``` ``` let ft = nnf f1 in ``` ``` MORETHAN (n-1,s,ft) ``` ``` | MAX (n, s, f1) -> ``` ``` let ft = nnfNeg f1 in ``` ``` MAXEXCEPT (n,s, ft) ``` ``` | MORETHAN (n,s,f1) -> ``` ``` let ft = nnf f1 in ``` ``` if ft = f1 then f else MORETHAN (n,s,ft) ``` ``` | MAXEXCEPT (n,s,f1) -> ``` ``` let ft = nnf f1 in ``` ``` if ft = f1 then f else MAXEXCEPT (n,s,ft) ``` ``` | ATLEASTPROB (p, f1) -> ``` ``` let ft = nnf f1 in ``` ``` if ft == f1 then f else ATLEASTPROB (p, ft) ``` ``` | LESSPROBFAIL (p, f1) -> ``` ``` let ft = nnf f1 in ``` ``` if ft == f1 then f else LESSPROBFAIL (p, ft) ``` ``` | ID (f1) -> ``` ``` let ft = nnf f1 in ``` ``` if ft == f1 then f else ID(ft) ``` ``` | NORM (f1, f2) -> ``` ``` let ft1 = nnf f1 in ``` ``` let ft2 = nnf f2 in ``` ``` if ft1 == f1 && ft2 == f2 then f else NORM (ft1, ft2) ``` ``` | NORMN (f1, f2) -> ``` ``` let ft1 = nnf f1 in ``` ``` let ft2 = nnf f2 in ``` ``` if ft1 == f1 && ft2 == f2 then f else NORMN (ft1, ft2) ``` ``` | CHC (f1, f2) -> ``` ``` let ft1 = nnf f1 in ``` ``` let ft2 = nnf f2 in ``` ``` if ft1 == f1 && ft2 == f2 then f else CHC (ft1, ft2) ``` ``` | FUS (first, f1) -> ``` ``` let ft = nnf f1 in ``` ``` if ft == f1 then f else FUS (first, ft) ``` ``` | MU (s, f1) -> ``` ``` let ft = nnf f1 in ``` ``` if ft == f1 then f else MU (s, ft) ``` ``` | NU (s, f1) -> ``` ``` let ft = nnf f1 in ``` ``` if ft == f1 then f else NU (s, ft) ``` ``` | AF _ ``` ``` | EF _ ``` ``` | AG _ ``` ``` | EG _ ``` ``` | AU _ ``` ``` | EU _ -> raise (CoAlgException ("nnf: CTL should have been removed at this point")) ``` ```(** Simplifies a formula. ``` ``` @param f A formula which must be in negation normal form. ``` ``` @return A formula in negation normal form that is equivalent to f. ``` ``` @raise CoAlgException if f is not in negation normal form. ``` ``` *) ``` ```let rec simplify f = ``` ``` match f with ``` ``` | TRUE ``` ``` | FALSE ``` ``` | AP _ ``` ``` | NOT (AP _) ``` ``` | VAR _ ``` ``` | NOT (VAR _) -> f ``` ``` | AT (s, f1) -> ``` ``` let ft = simplify f1 in ``` ``` begin ``` ``` match ft with ``` ``` | FALSE -> FALSE ``` ``` | TRUE -> TRUE ``` ``` | AT _ -> ft ``` ``` | AP s1 when s = s1 -> TRUE ``` ``` | NOT (AP s1) when s = s1 -> FALSE ``` ``` | _ -> if ft == f1 then f else AT (s, ft) ``` ``` end ``` ``` | OR (f1, f2) -> ``` ``` let ft1 = simplify f1 in ``` ``` let ft2 = simplify f2 in ``` ``` begin ``` ``` match ft1, ft2 with ``` ``` | TRUE, _ -> TRUE ``` ``` | FALSE, _ -> ft2 ``` ``` | _, TRUE -> TRUE ``` ``` | _, FALSE -> ft1 ``` ``` | _, _ -> ``` ``` if (f1 == ft1) && (f2 == ft2) then f ``` ``` else OR (ft1, ft2) ``` ``` end ``` ``` | AND (f1, f2) -> ``` ``` let ft1 = simplify f1 in ``` ``` let ft2 = simplify f2 in ``` ``` begin ``` ``` match ft1, ft2 with ``` ``` | TRUE, _ -> ft2 ``` ``` | FALSE, _ -> FALSE ``` ``` | _, TRUE -> ft1 ``` ``` | _, FALSE -> FALSE ``` ``` | _, _ -> ``` ``` if (f1 == ft1) && (f2 == ft2) then f ``` ``` else AND (ft1, ft2) ``` ``` end ``` ``` | NOT _ ``` ``` | EQU _ ``` ``` | IMP _ -> raise (CoAlgException "Formula is not in negation normal form.") ``` ``` | EX (s, f1) -> ``` ``` let ft = simplify f1 in ``` ``` begin ``` ``` match ft with ``` ``` | FALSE -> FALSE ``` ``` | _ -> if ft == f1 then f else EX (s, ft) ``` ``` end ``` ``` | AX (s, f1) -> ``` ``` let ft = simplify f1 in ``` ``` begin ``` ``` match ft with ``` ``` | TRUE -> TRUE ``` ``` | _ -> if ft == f1 then f else AX (s, ft) ``` ``` end ``` ``` | ENFORCES (s, f1) -> ``` ``` let ft = simplify f1 in ``` ``` begin ``` ``` match ft with ``` ``` (* Simplification rules are checked with dirks Generic.hs ``` ``` let enforce ls = M (C ls) ``` ``` let allow ls = Neg . M (C ls) . Neg ``` ``` provable \$ F <-> enforce [1,2] F ``` ``` *) ``` ``` | TRUE -> TRUE ``` ``` | FALSE -> FALSE ``` ``` | _ -> if ft == f1 then f else ENFORCES (s, ft) ``` ``` end ``` ``` | ALLOWS (s, f1) -> ``` ``` let ft = simplify f1 in ``` ``` begin ``` ``` match ft with ``` ``` (* Simplification rules are checked with dirks Generic.hs ``` ``` let enforce ls = M (C ls) ``` ``` let allow ls = Neg . M (C ls) . Neg ``` ``` provable \$ F <-> enforce [1,2] F ``` ``` *) ``` ``` | TRUE -> TRUE ``` ``` | FALSE -> FALSE ``` ``` | _ -> if ft == f1 then f else ALLOWS (s, ft) ``` ``` end ``` ``` | MIN (n, s, f1) -> ``` ``` if n = 0 then TRUE ``` ``` else ``` ``` let ft = simplify f1 in ``` ``` begin ``` ``` match ft with ``` ``` | FALSE -> FALSE ``` ``` | _ -> if ft == f1 then f else MIN (n, s, ft) ``` ``` end ``` ``` | MORETHAN (n,s,f1) -> ``` ``` let ft = simplify f1 in ``` ``` if ft == f1 then f else MORETHAN (n,s,ft) ``` ``` | MAXEXCEPT (n,s,f1) -> ``` ``` let ft = simplify f1 in ``` ``` if ft == f1 then f else MAXEXCEPT (n,s,ft) ``` ``` | MAX (n, s, f1) -> ``` ``` let ft = simplify f1 in ``` ``` begin ``` ``` match ft with ``` ``` | FALSE -> TRUE ``` ``` | _ -> if ft == f1 then f else MAX (n, s, ft) ``` ``` end ``` ``` | LESSPROBFAIL (p,f1) -> ``` ``` let ft1 = simplify f1 in ``` ``` if (ft1 == f1) then f else LESSPROBFAIL (p,ft1) ``` ``` | ATLEASTPROB (p,f1) -> ``` ``` let ft1 = simplify f1 in ``` ``` if (ft1 == f1) then f else ATLEASTPROB (p,ft1) ``` ``` | CONST _ ``` ``` | CONSTN _ -> f (* no simplifications possible *) ``` ``` | ID (f1) -> ``` ``` let ft1 = simplify f1 in ``` ``` begin ``` ``` match ft1 with ``` ``` | TRUE -> TRUE ``` ``` | FALSE -> FALSE ``` ``` | _ -> if (ft1 == f1) then f else ID (ft1) ``` ``` end ``` ``` (* todo: more simplifications for KLM? *) ``` ``` | NORM (f1, f2) -> ``` ``` let ft1 = simplify f1 in ``` ``` let ft2 = simplify f2 in ``` ``` begin ``` ``` match ft2 with ``` ``` | TRUE -> TRUE ``` ``` | _ -> ``` ``` if (f1 == ft1) && (f2 == ft2) then f ``` ``` else NORM (ft1, ft2) ``` ``` end ``` ``` | NORMN (f1, f2) -> ``` ``` let ft1 = simplify f1 in ``` ``` let ft2 = simplify f2 in ``` ``` begin ``` ``` match ft2 with ``` ``` | FALSE -> FALSE ``` ``` | _ -> ``` ``` if (f1 == ft1) && (f2 == ft2) then f ``` ``` else NORMN (ft1, ft2) ``` ``` end ``` ``` | CHC (f1, f2) -> ``` ``` let ft1 = simplify f1 in ``` ``` let ft2 = simplify f2 in ``` ``` begin ``` ``` match ft1, ft2 with ``` ``` | TRUE, TRUE -> TRUE ``` ``` | FALSE, FALSE -> FALSE ``` ``` | _, _ -> ``` ``` if (f1 == ft1) && (f2 == ft2) then f ``` ``` else CHC (ft1, ft2) ``` ``` end ``` ``` | FUS (first, f1) -> ``` ``` let ft = simplify f1 in ``` ``` begin ``` ``` match ft with ``` ``` | FALSE -> FALSE ``` ``` | TRUE -> TRUE ``` ``` | _ -> if ft == f1 then f else FUS (first, ft) ``` ``` end ``` ``` | MU (s, f1) -> ``` ``` let ft = simplify f1 in ``` ``` begin ``` ``` match ft with ``` ``` | TRUE -> TRUE ``` ``` | _ -> if ft == f1 then f else MU (s, ft) ``` ``` end ``` ``` | NU (s, f1) -> ``` ``` let ft = simplify f1 in ``` ``` begin ``` ``` match ft with ``` ``` | TRUE -> TRUE ``` ``` | _ -> if ft == f1 then f else NU (s, ft) ``` ``` end ``` ``` | AF _ ``` ``` | EF _ ``` ``` | AG _ ``` ``` | EG _ ``` ``` | AU _ ``` ``` | EU _ -> ``` ``` raise (CoAlgException ("nnf: CTL should have been removed at this point")) ``` ```(** Destructs an atomic proposition. ``` ``` @param f An atomic proposition. ``` ``` @return The name of the atomic proposition. ``` ``` @raise CoAlgException if f is not an atomic proposition. ``` ``` *) ``` ```let dest_ap f = ``` ``` match f with ``` ``` | AP s when not (isNominal s) -> s ``` ``` | _ -> raise (CoAlgException "Formula is not an atomic proposition.") ``` ```(** Destructs a nominal. ``` ``` @param f A nominal. ``` ``` @return The name of the nominal ``` ``` @raise CoAlgException if f is not a nominal. ``` ``` *) ``` ```let dest_nom f = ``` ``` match f with ``` ``` | AP s when isNominal s -> s ``` ``` | _ -> raise (CoAlgException "Formula is not a nominal.") ``` ```(** Destructs a negation. ``` ``` @param f A negation. ``` ``` @return The immediate subformula of the negation. ``` ``` @raise CoAlgException if f is not a negation. ``` ``` *) ``` ```let dest_not f = ``` ``` match f with ``` ``` | NOT f1 -> f1 ``` ``` | _ -> raise (CoAlgException "Formula is not a negation.") ``` ```(** Destructs an @-formula. ``` ``` @param f An @-formula. ``` ``` @return The name of the nominal and the immediate subformula of the @-formula. ``` ``` @raise CoAlgException if f is not an @-formula. ``` ``` *) ``` ```let dest_at f = ``` ``` match f with ``` ``` | AT (s, f1) -> (s, f1) ``` ``` | _ -> raise (CoAlgException "Formula is not an @-formula.") ``` ```(** Destructs an or-formula. ``` ``` @param f An or-formula. ``` ``` @return The two immediate subformulae of the or-formula. ``` ``` @raise CoAlgException if f is not an or-formula. ``` ``` *) ``` ```let dest_or f = ``` ``` match f with ``` ``` | OR (f1, f2) -> (f1, f2) ``` ``` | _ -> raise (CoAlgException "Formula is not an or-formula.") ``` ```(** Destructs an and-formula. ``` ``` @param f An and-formula. ``` ``` @return The two immediate subformulae of the and-formula. ``` ``` @raise CoAlgException if f is not an and-formula. ``` ``` *) ``` ```let dest_and f = ``` ``` match f with ``` ``` | AND (f1, f2) -> (f1, f2) ``` ``` | _ -> raise (CoAlgException "Formula is not an and-formula.") ``` ```(** Destructs an equivalence. ``` ``` @param f An equivalence. ``` ``` @return The two immediate subformulae of the equivalence. ``` ``` @raise CoAlgException if f is not an equivalence. ``` ``` *) ``` ```let dest_equ f = ``` ``` match f with ``` ``` | EQU (f1, f2) -> (f1, f2) ``` ``` | _ -> raise (CoAlgException "Formula is not an equivalence.") ``` ```(** Destructs an implication. ``` ``` @param f An implication. ``` ``` @return The two immediate subformulae of the implication. ``` ``` @raise CoAlgException if f is not an implication. ``` ``` *) ``` ```let dest_imp f = ``` ``` match f with ``` ``` | IMP (f1, f2) -> (f1, f2) ``` ``` | _ -> raise (CoAlgException "Formula is not an implication.") ``` ```(** Destructs an EX-formula. ``` ``` @param f An EX-formula. ``` ``` @return The role name and the immediate subformula of the EX-formula. ``` ``` @raise CoAlgException if f is not an EX-formula. ``` ``` *) ``` ```let dest_ex f = ``` ``` match f with ``` ``` | EX (s, f1) -> (s, f1) ``` ``` | _ -> raise (CoAlgException "Formula is not an EX-formula.") ``` ```(** Destructs an AX-formula. ``` ``` @param f An AX-formula. ``` ``` @return The role name and the immediate subformula of the AX-formula. ``` ``` @raise CoAlgException if f is not an AX-formula. ``` ``` *) ``` ```let dest_ax f = ``` ``` match f with ``` ``` | AX (s, f1) -> (s, f1) ``` ``` | _ -> raise (CoAlgException "Formula is not an AX-formula.") ``` ```(** Destructs a MIN-formula. ``` ``` @param f A MIN-formula. ``` ``` @return The number restriction, role name, ``` ``` and the immediate subformula of the MIN-formula. ``` ``` @raise CoAlgException if f is not a MIN-formula. ``` ``` *) ``` ```let dest_min f = ``` ``` match f with ``` ``` | MIN (n, s, f1) -> (n, s, f1) ``` ``` | _ -> raise (CoAlgException "Formula is not a MIN-formula.") ``` ```(** Destructs a MAX-formula. ``` ``` @param f A MAX-formula. ``` ``` @return The number restriction, role name, ``` ``` and the immediate subformula of the MAX-formula. ``` ``` @raise CoAlgException if f is not a MAX-formula. ``` ``` *) ``` ```let dest_max f = ``` ``` match f with ``` ``` | MAX (n, s, f1) -> (n, s, f1) ``` ``` | _ -> raise (CoAlgException "Formula is not a MAX-formula.") ``` ```(** Destructs a choice formula. ``` ``` @param f A choice formula. ``` ``` @return The two immediate subformulae of the choice formula. ``` ``` @raise CoAlgException if f is not a choice formula. ``` ``` *) ``` ```let dest_choice f = ``` ``` match f with ``` ``` | CHC (f1, f2) -> (f1, f2) ``` ``` | _ -> raise (CoAlgException "Formula is not a choice formula.") ``` ```(** Destructs a fusion formula. ``` ``` @param f A fusion formula. ``` ``` @return A pair (first, f1) where ``` ``` first is true iff f is a first projection; and ``` ``` f1 is the immediate subformulae of f. ``` ``` @raise CoAlgException if f is not a fusion formula. ``` ``` *) ``` ```let dest_fusion f = ``` ``` match f with ``` ``` | FUS (first, f1) -> (first, f1) ``` ``` | _ -> raise (CoAlgException "Formula is not a fusion formula.") ``` ```(** Tests whether a formula is the constant True. ``` ``` @param f A formula. ``` ``` @return True iff f is the constant True. ``` ``` *) ``` ```let is_true f = ``` ``` match f with ``` ``` | TRUE -> true ``` ``` | _ -> false ``` ```(** Tests whether a formula is the constant False. ``` ``` @param f A formula. ``` ``` @return True iff f is the constant False. ``` ``` *) ``` ```let is_false f = ``` ``` match f with ``` ``` | FALSE -> true ``` ``` | _ -> false ``` ```(** Tests whether a formula is an atomic proposition. ``` ``` @param f A formula. ``` ``` @return True iff f is an atomic proposition. ``` ``` *) ``` ```let is_ap f = ``` ``` match f with ``` ``` | AP s when not (isNominal s) -> true ``` ``` | _ -> false ``` ```(** Tests whether a formula is a nominal. ``` ``` @param f A formula. ``` ``` @return True iff f is a nominal. ``` ``` *) ``` ```let is_nom f = ``` ``` match f with ``` ``` | AP s when isNominal s -> true ``` ``` | _ -> false ``` ```(** Tests whether a formula is a negation. ``` ``` @param f A formula. ``` ``` @return True iff f is a negation. ``` ``` *) ``` ```let is_not f = ``` ``` match f with ``` ``` | NOT _ -> true ``` ``` | _ -> false ``` ```(** Tests whether a formula is an @-formula. ``` ``` @param f A formula. ``` ``` @return True iff f is an @-formula. ``` ``` *) ``` ```let is_at f = ``` ``` match f with ``` ``` | AT _ -> true ``` ``` | _ -> false ``` ```(** Tests whether a formula is an or-formula. ``` ``` @param f A formula. ``` ``` @return True iff f is an or-formula. ``` ``` *) ``` ```let is_or f = ``` ``` match f with ``` ``` | OR _ -> true ``` ``` | _ -> false ``` ```(** Tests whether a formula is an and-formula. ``` ``` @param f A formula. ``` ``` @return True iff f is an and-formula. ``` ``` *) ``` ```let is_and f = ``` ``` match f with ``` ``` | AND _ -> true ``` ``` | _ -> false ``` ```(** Tests whether a formula is an equivalence. ``` ``` @param f A formula. ``` ``` @return True iff f is an equivalence. ``` ``` *) ``` ```let is_equ f = ``` ``` match f with ``` ``` | EQU _ -> true ``` ``` | _ -> false ``` ```(** Tests whether a formula is an implication. ``` ``` @param f A formula. ``` ``` @return True iff f is an implication. ``` ``` *) ``` ```let is_imp f = ``` ``` match f with ``` ``` | IMP _ -> true ``` ``` | _ -> false ``` ```(** Tests whether a formula is an EX-formula. ``` ``` @param f A formula. ``` ``` @return True iff f is an EX-formula. ``` ``` *) ``` ```let is_ex f = ``` ``` match f with ``` ``` | EX _ -> true ``` ``` | _ -> false ``` ```(** Tests whether a formula is an AX-formula. ``` ``` @param f A formula. ``` ``` @return True iff f is an AX-formula. ``` ``` *) ``` ```let is_ax f = ``` ``` match f with ``` ``` | AX _ -> true ``` ``` | _ -> false ``` ```(** Tests whether a formula is a MIN-formula. ``` ``` @param f A formula. ``` ``` @return True iff f is a MIN-formula. ``` ``` *) ``` ```let is_min f = ``` ``` match f with ``` ``` | MIN _ -> true ``` ``` | _ -> false ``` ```(** Tests whether a formula is a MAX-formula. ``` ``` @param f A formula. ``` ``` @return True iff f is a MAX-formula. ``` ``` *) ``` ```let is_max f = ``` ``` match f with ``` ``` | MAX _ -> true ``` ``` | _ -> false ``` ```(** Tests whether a formula is a choice formula. ``` ``` @param f A formula. ``` ``` @return True iff f is a choice formula. ``` ``` *) ``` ```let is_choice f = ``` ``` match f with ``` ``` | CHC _ -> true ``` ``` | _ -> false ``` ```(** Tests whether a formula is a fusion formula. ``` ``` @param f A formula. ``` ``` @return True iff f is a fusion formula. ``` ``` *) ``` ```let is_fusion f = ``` ``` match f with ``` ``` | FUS _ -> true ``` ``` | _ -> false ``` ```(** The constant True. ``` ``` *) ``` ```let const_true = TRUE ``` ```(** The constant False. ``` ``` *) ``` ```let const_false = FALSE ``` ```(** Constructs an atomic proposition. ``` ``` @param s The name of the atomic proposition. ``` ``` @return The atomic proposition with name s. ``` ``` @raise CoAlgException if s is a nominal name. ``` ``` *) ``` ```let const_ap s = ``` ``` if isNominal s then raise (CoAlgException "The name indicates a nominal.") ``` ``` else AP s ``` ```(** Constructs a nominal. ``` ``` @param s The name of the nominal. ``` ``` @return A nominal with name s. ``` ``` @raise CoAlgException if s is not a nominal name. ``` ``` *) ``` ```let const_nom s = ``` ``` if isNominal s then AP s ``` ``` else raise (CoAlgException "The name does not indicate a nominal.") ``` ```(** Negates a formula. ``` ``` @param f A formula. ``` ``` @return The negation of f. ``` ``` *) ``` ```let const_not f = NOT f ``` ```(** Constructs an @-formula from a name and a formula. ``` ``` @param s A nominal name. ``` ``` @param f A formula. ``` ``` @return The formula AT (s, f). ``` ``` *) ``` ```let const_at s f = AT (s, f) ``` ```(** Constructs an or-formula from two formulae. ``` ``` @param f1 The first formula (LHS). ``` ``` @param f2 The second formula (LHS). ``` ``` @return The formula f1 | f2. ``` ``` *) ``` ```let const_or f1 f2 = OR (f1, f2) ``` ```(** Constructs an and-formula from two formulae. ``` ``` @param f1 The first formula (LHS). ``` ``` @param f2 The second formula (LHS). ``` ``` @return The formula f1 & f2. ``` ``` *) ``` ```let const_and f1 f2 = AND (f1, f2) ``` ```(** Constructs an equivalence from two formulae. ``` ``` @param f1 The first formula (LHS). ``` ``` @param f2 The second formula (LHS). ``` ``` @return The equivalence f1 <=> f2. ``` ``` *) ``` ```let const_equ f1 f2 = EQU (f1, f2) ``` ```(** Constructs an implication from two formulae. ``` ``` @param f1 The first formula (LHS). ``` ``` @param f2 The second formula (LHS). ``` ``` @return The implication f1 => f2. ``` ``` *) ``` ```let const_imp f1 f2 = IMP (f1, f2) ``` ```(** Constructs an EX-formula from a formula. ``` ``` @param s A role name. ``` ``` @param f A formula. ``` ``` @return The formula EX (s, f). ``` ``` *) ``` ```let const_ex s f = EX (s, f) ``` ```(** Constructs an AX-formula from a formula. ``` ``` @param s A role name. ``` ``` @param f A formula. ``` ``` @return The formula AX (s, f). ``` ``` *) ``` ```let const_ax s f = AX (s, f) ``` ```(** Constructs a MIN-formula from a formula. ``` ``` @param n A non-negative integer. ``` ``` @param s A role name. ``` ``` @param f A formula. ``` ``` @return The formula MIN f. ``` ``` @raise CoAlgException Iff n is negative. ``` ``` *) ``` ```let const_min n s f = ``` ``` if n < 0 then raise (CoAlgException "Negative cardinality constraint.") ``` ``` else MIN (n, s, f) ``` ```(** Constructs a MAX-formula from a formula. ``` ``` @param n A non-negative integer. ``` ``` @param s A role name. ``` ``` @param f A formula. ``` ``` @return The formula MAX f. ``` ``` @raise CoAlgException Iff n is negative. ``` ``` *) ``` ```let const_max n s f = ``` ``` if n < 0 then raise (CoAlgException "Negative cardinality constraint.") ``` ``` else MAX (n, s, f) ``` ```let const_enforces p f = ``` ``` ENFORCES (p,f) ``` ```let const_allows p f = ``` ``` ALLOWS (p,f) ``` ```(** Constructs a choice formula from two formulae. ``` ``` @param f1 The first formula (LHS). ``` ``` @param f2 The second formula (LHS). ``` ``` @return The formula (f1 + f2). ``` ``` *) ``` ```let const_choice f1 f2 = CHC (f1, f2) ``` ```(** Constructs a fusion formula. ``` ``` @param first True iff the result is a first projection. ``` ``` @param f1 A formula. ``` ``` @return The formula [pi1] f1 or [pi2] f1 (depending on first). ``` ``` *) ``` ```let const_fusion first f1 = FUS (first, f1) ``` ```(** Hash-consed formulae, which are in negation normal form, ``` ``` such that structural and physical equality coincide. ``` ``` Also CTL has been replaced by the equivalent μ-Calculus ``` ``` constructs ``` ``` *) ``` ```type hcFormula = (hcFormula_node, formula) HC.hash_consed ``` ```and hcFormula_node = ``` ``` | HCTRUE ``` ``` | HCFALSE ``` ``` | HCAP of string ``` ``` | HCNOT of string ``` ``` | HCAT of string * hcFormula ``` ``` | HCOR of hcFormula * hcFormula ``` ``` | HCAND of hcFormula * hcFormula ``` ``` | HCEX of string * hcFormula ``` ``` | HCAX of string * hcFormula ``` ``` | HCENFORCES of int list * hcFormula ``` ``` | HCALLOWS of int list * hcFormula ``` ``` | HCMORETHAN of int * string * hcFormula (* GML Diamond *) ``` ``` | HCMAXEXCEPT of int * string * hcFormula (* GML Box *) ``` ``` | HCATLEASTPROB of rational * hcFormula ``` ``` | HCLESSPROBFAIL of rational * hcFormula ``` ``` | HCCONST of string ``` ``` | HCCONSTN of string ``` ``` | HCID of hcFormula ``` ``` | HCNORM of hcFormula * hcFormula ``` ``` | HCNORMN of hcFormula * hcFormula ``` ``` | HCCHC of hcFormula * hcFormula ``` ``` | HCFUS of bool * hcFormula ``` ``` | HCMU of string * hcFormula ``` ``` | HCNU of string * hcFormula ``` ``` | HCVAR of string ``` ```(** Determines whether two formula nodes are structurally equal. ``` ``` @param f1 The first formula node. ``` ``` @param f2 The second formula node. ``` ``` @return True iff f1 and f2 are structurally equal. ``` ``` *) ``` ```let equal_hcFormula_node f1 f2 = ``` ``` match f1, f2 with ``` ``` | HCTRUE, HCTRUE -> true ``` ``` | HCFALSE, HCFALSE -> true ``` ``` | HCCONST s1, HCCONST s2 ``` ``` | HCCONSTN s1, HCCONSTN s2 ``` ``` | HCAP s1, HCAP s2 ``` ``` | HCNOT s1, HCNOT s2 -> compare s1 s2 = 0 ``` ``` | HCAT (s1, f1), HCAT (s2, f2) -> compare s1 s2 = 0 && f1 == f2 ``` ``` | HCOR (f1a, f1b), HCOR (f2a, f2b) ``` ``` | HCAND (f1a, f1b), HCAND (f2a, f2b) -> f1a == f2a && f1b == f2b ``` ``` | HCEX (s1, f1), HCEX (s2, f2) ``` ``` | HCAX (s1, f1), HCAX (s2, f2) -> compare s1 s2 = 0 && f1 == f2 ``` ``` | HCENFORCES (s1, f1), HCENFORCES (s2, f2) ``` ``` | HCALLOWS (s1, f1), HCALLOWS (s2, f2) -> compare s1 s2 = 0 && f1 == f2 ``` ``` | HCMORETHAN (n1, s1, f1), HCMORETHAN (n2, s2, f2) ``` ``` | HCMAXEXCEPT (n1, s1, f1), HCMAXEXCEPT (n2, s2, f2) -> ``` ``` n1 = n2 && compare s1 s2 = 0 && f1 == f2 ``` ``` | HCATLEASTPROB (p1, f1), HCATLEASTPROB (p2,f2) -> ``` ``` p1 = p2 && f1 == f2 ``` ``` | HCLESSPROBFAIL (p1, f1), HCLESSPROBFAIL (p2,f2) -> ``` ``` p1 = p2 && f1 == f2 ``` ``` | HCID f1, HCID f2 -> f1 == f2 ``` ``` | HCNORM (f1a, f1b), HCNORMN (f2a, f2b) -> f1a == f2a && f1b == f2b ``` ``` | HCCHC (f1a, f1b), HCCHC (f2a, f2b) -> f1a == f2a && f1b == f2b ``` ``` | HCFUS (first1, f1), HCFUS (first2, f2) -> first1 = first2 && f1 == f2 ``` ``` | HCMU (name1, f1), HCMU(name2, f2) -> compare name1 name2 = 0 && f1 == f2 ``` ``` | HCNU (name1, f1), HCNU(name2, f2) -> compare name1 name2 = 0 && f1 == f2 ``` ``` | HCVAR name1, HCVAR name2 -> compare name1 name2 = 0 ``` ```(* The rest could be shortened to | _ -> false, ``` ``` but then the compiler would not warn if the formula type is extended ``` ``` and this function is not updated accordingly.*) ``` ``` | HCTRUE, _ ``` ``` | HCFALSE, _ ``` ``` | HCAP _, _ ``` ``` | HCNOT _, _ ``` ``` | HCAT _, _ ``` ``` | HCOR _, _ ``` ``` | HCAND _, _ ``` ``` | HCEX _, _ ``` ``` | HCAX _, _ ``` ``` | HCENFORCES _, _ ``` ``` | HCALLOWS _, _ ``` ``` | HCMORETHAN _, _ ``` ``` | HCMAXEXCEPT _, _ ``` ``` | HCATLEASTPROB _, _ ``` ``` | HCLESSPROBFAIL _, _ ``` ``` | HCCONST _, _ ``` ``` | HCCONSTN _, _ ``` ``` | HCID _, _ ``` ``` | HCNORM _, _ ``` ``` | HCNORMN _, _ ``` ``` | HCCHC _, _ ``` ``` | HCFUS _, _ ``` ``` | HCMU _, _ ``` ``` | HCNU _, _ ``` ``` | HCVAR _, _ -> false ``` ```(** Computes the hash value of a formula node. ``` ``` @param f A formula node. ``` ``` @return The hash value of f. ``` ``` *) ``` ```let hash_hcFormula_node f = ``` ``` let base = 23 in (* should be larger than the no of constructors *) ``` ``` match f with ``` ``` | HCTRUE -> 0 ``` ``` | HCFALSE -> 1 ``` ``` | HCAP s ``` ``` | HCNOT s ``` ``` | HCVAR s -> base * Hashtbl.hash s + 1 ``` ``` | HCAT (s, f1) -> base * (base * (Hashtbl.hash s) + f1.HC.hkey) + 2 ``` ``` | HCOR (f1, f2) -> base * (base * f1.HC.hkey + f2.HC.hkey) + 5 ``` ``` | HCAND (f1, f2) -> base * (base * f1.HC.hkey + f2.HC.hkey) + 6 ``` ``` | HCEX (s, f1) -> base * (base * (Hashtbl.hash s) + f1.HC.hkey) + 3 ``` ``` | HCAX (s, f1) -> base * (base * (Hashtbl.hash s) + f1.HC.hkey) + 4 ``` ``` | HCMORETHAN (n, s, f1) -> base * (base * (base * (Hashtbl.hash s) + n) + f1.HC.hkey) + 7 ``` ``` | HCMAXEXCEPT (n, s, f1) -> base * (base * (base * (Hashtbl.hash s) + n) + f1.HC.hkey) + 8 ``` ``` | HCATLEASTPROB (p,f1) -> base * (base * (base * p.nominator + p.denominator) + f1.HC.hkey) + 9 ``` ``` | HCLESSPROBFAIL (p,f1) -> base * (base * (base * p.nominator + p.denominator) + f1.HC.hkey) + 10 ``` ``` | HCCHC (f1, f2) -> base * (base * f1.HC.hkey + f2.HC.hkey) + 11 ``` ``` | HCFUS (first, f1) -> base * (base * (Hashtbl.hash first) + f1.HC.hkey) + 13 ``` ``` | HCENFORCES (s, f1) -> base * (base * (Hashtbl.hash s) + f1.HC.hkey) + 14 ``` ``` | HCALLOWS (s, f1) -> base * (base * (Hashtbl.hash s) + f1.HC.hkey) + 15 ``` ``` | HCCONST s -> Hashtbl.hash s + 16 ``` ``` | HCCONSTN s -> Hashtbl.hash s + 17 ``` ``` | HCNORM (f1, f2) -> base * (base * f1.HC.hkey + f2.HC.hkey) + 18 ``` ``` | HCNORMN (f1, f2) -> base * (base * f1.HC.hkey + f2.HC.hkey) + 19 ``` ``` | HCID (f1) -> base * f1.HC.hkey + 20 ``` ``` | HCMU (s, f1) -> base * (base * (Hashtbl.hash s) + f1.HC.hkey) + 21 ``` ``` | HCNU (s, f1) -> base * (base * (Hashtbl.hash s) + f1.HC.hkey) + 22 ``` ```(** Determines the "real" formula of a formula node. ``` ``` @param f A formula node. ``` ``` @return The formula (in negation normal form) which corresponds to f. ``` ``` *) ``` ```let toFml_hcFormula_node f = ``` ``` match f with ``` ``` | HCTRUE -> TRUE ``` ``` | HCFALSE -> FALSE ``` ``` | HCAP s -> AP s ``` ``` | HCVAR s -> VAR s ``` ``` | HCNOT s -> NOT (AP s) ``` ``` | HCAT (s, f1) -> AT (s, f1.HC.fml) ``` ``` | HCOR (f1, f2) -> OR (f1.HC.fml, f2.HC.fml) ``` ``` | HCAND (f1, f2) -> AND (f1.HC.fml, f2.HC.fml) ``` ``` | HCEX (s, f1) -> EX (s, f1.HC.fml) ``` ``` | HCAX (s, f1) -> AX (s, f1.HC.fml) ``` ``` | HCENFORCES (s, f1) -> ENFORCES (s, f1.HC.fml) ``` ``` | HCALLOWS (s, f1) -> ALLOWS (s, f1.HC.fml) ``` ``` | HCMORETHAN (n, s, f1) -> MORETHAN (n, s, f1.HC.fml) ``` ``` | HCMAXEXCEPT (n, s, f1) -> MAXEXCEPT (n, s, f1.HC.fml) ``` ``` | HCATLEASTPROB (p, f1) -> ATLEASTPROB (p, f1.HC.fml) ``` ``` | HCLESSPROBFAIL (p, f1) -> LESSPROBFAIL (p, f1.HC.fml) ``` ``` | HCCONST s -> CONST s ``` ``` | HCCONSTN s -> CONSTN s ``` ``` | HCID (f1) -> ID(f1.HC.fml) ``` ``` | HCNORM (f1, f2) -> NORM(f1.HC.fml, f2.HC.fml) ``` ``` | HCNORMN (f1, f2) -> NORMN(f1.HC.fml, f2.HC.fml) ``` ``` | HCCHC (f1, f2) -> CHC (f1.HC.fml, f2.HC.fml) ``` ``` | HCFUS (first, f1) -> FUS (first, f1.HC.fml) ``` ``` | HCMU (var, f1) -> MU (var, f1.HC.fml) ``` ``` | HCNU (var, f1) -> NU (var, f1.HC.fml) ``` ```(** Determines the negation (in negation normal form) of a formula node. ``` ``` @param f A formula node. ``` ``` @return The negation (in negation normal form) of f. ``` ``` *) ``` ```let negNde_hcFormula_node f = ``` ``` match f with ``` ``` | HCTRUE -> HCFALSE ``` ``` | HCFALSE -> HCTRUE ``` ``` | HCAP s -> HCNOT s ``` ``` | HCNOT s -> HCAP s ``` ``` | HCVAR s -> f ``` ``` | HCAT (s, f1) -> HCAT (s, f1.HC.neg) ``` ``` | HCOR (f1, f2) -> HCAND (f1.HC.neg, f2.HC.neg) ``` ``` | HCAND (f1, f2) -> HCOR (f1.HC.neg, f2.HC.neg) ``` ``` | HCEX (s, f2) -> HCAX (s, f2.HC.neg) ``` ``` | HCAX (s, f2) -> HCEX (s, f2.HC.neg) ``` ``` | HCENFORCES (s, f2) -> HCALLOWS (s, f2.HC.neg) ``` ``` | HCALLOWS (s, f2) -> HCENFORCES (s, f2.HC.neg) ``` ``` | HCMORETHAN (n, s, f1) -> HCMAXEXCEPT (n, s, f1.HC.neg) ``` ``` | HCMAXEXCEPT (n, s, f1) -> HCMORETHAN (n, s, f1.HC.neg) ``` ``` | HCATLEASTPROB (p, f1) -> HCLESSPROBFAIL (p, f1.HC.neg) ``` ``` | HCLESSPROBFAIL (p, f1) -> HCATLEASTPROB (p, f1.HC.neg) ``` ``` | HCCONST s -> HCCONSTN s ``` ``` | HCCONSTN s -> HCCONST s ``` ``` | HCID (f1) -> HCID(f1.HC.neg) ``` ``` | HCNORM (f1, f2) -> HCNORMN(f1.HC.neg, f2.HC.neg) ``` ``` | HCNORMN (f1, f2) -> HCNORM(f1.HC.neg, f2.HC.neg) ``` ``` | HCCHC (f1, f2) -> HCCHC (f1.HC.neg, f2.HC.neg) ``` ``` | HCFUS (first, f1) -> HCFUS (first, f1.HC.neg) ``` ``` | HCMU (name, f1) -> HCNU (name, f1.HC.neg) ``` ``` | HCNU (name, f1) -> HCMU (name, f1.HC.neg) ``` ```(** An instantiation of hash-consing for formulae. ``` ``` *) ``` ```module HcFormula = HC.Make( ``` ``` struct ``` ``` type nde = hcFormula_node ``` ``` type fml = formula ``` ``` let equal (n1 : nde) n2 = equal_hcFormula_node n1 n2 ``` ``` let hash (n : nde) = hash_hcFormula_node n ``` ``` let negNde (n : nde) = negNde_hcFormula_node n ``` ``` let toFml (n : nde) = toFml_hcFormula_node n ``` ``` end ``` ``` ) ``` ```(** Converts a formula into its hash-consed version. ``` ``` @param hcF A hash-cons table for formulae. ``` ``` @param f A formula in negation normal form. ``` ``` @return The hash-consed version of f. ``` ``` @raise CoAlgException if f is not in negation normal form. ``` ```*) ``` ```let rec hc_formula hcF f = ``` ``` match f with ``` ``` | TRUE -> HcFormula.hashcons hcF HCTRUE ``` ``` | FALSE -> HcFormula.hashcons hcF HCFALSE ``` ``` | AP s -> HcFormula.hashcons hcF (HCAP s) ``` ``` | VAR s -> HcFormula.hashcons hcF (HCVAR s) ``` ``` | NOT (AP s) -> HcFormula.hashcons hcF (HCNOT s) ``` ``` | AT (s, f1) -> ``` ``` let tf1 = hc_formula hcF f1 in ``` ``` HcFormula.hashcons hcF (HCAT (s, tf1)) ``` ``` | OR (f1, f2) -> ``` ``` let tf1 = hc_formula hcF f1 in ``` ``` let tf2 = hc_formula hcF f2 in ``` ``` HcFormula.hashcons hcF (HCOR (tf1, tf2)) ``` ``` | AND (f1, f2) -> ``` ``` let tf1 = hc_formula hcF f1 in ``` ``` let tf2 = hc_formula hcF f2 in ``` ``` HcFormula.hashcons hcF (HCAND (tf1, tf2)) ``` ``` | NOT _ ``` ``` | EQU _ ``` ``` | MIN _ ``` ``` | MAX _ ``` ``` | IMP _ -> raise (CoAlgException "Formula is not in negation normal form.") ``` ``` | EX (s, f1) -> ``` ``` let tf1 = hc_formula hcF f1 in ``` ``` HcFormula.hashcons hcF (HCEX (s, tf1)) ``` ``` | AX (s, f1) -> ``` ``` let tf1 = hc_formula hcF f1 in ``` ``` HcFormula.hashcons hcF (HCAX (s, tf1)) ``` ``` | ENFORCES (s, f1) -> ``` ``` let tf1 = hc_formula hcF f1 in ``` ``` HcFormula.hashcons hcF (HCENFORCES (s, tf1)) ``` ``` | ALLOWS (s, f1) -> ``` ``` let tf1 = hc_formula hcF f1 in ``` ``` HcFormula.hashcons hcF (HCALLOWS (s, tf1)) ``` ``` | MORETHAN (n, s, f1) -> ``` ``` let tf1 = hc_formula hcF f1 in ``` ``` HcFormula.hashcons hcF (HCMORETHAN (n, s, tf1)) ``` ``` | MAXEXCEPT (n, s, f1) -> ``` ``` let tf1 = hc_formula hcF f1 in ``` ``` HcFormula.hashcons hcF (HCMAXEXCEPT (n, s, tf1)) ``` ``` | ATLEASTPROB (p, f1) -> ``` ``` HcFormula.hashcons hcF (HCATLEASTPROB (p, hc_formula hcF f1)) ``` ``` | LESSPROBFAIL (p, f1) -> ``` ``` HcFormula.hashcons hcF (HCLESSPROBFAIL (p, hc_formula hcF f1)) ``` ``` | CONST s -> ``` ``` HcFormula.hashcons hcF (HCCONST s) ``` ``` | CONSTN s -> ``` ``` HcFormula.hashcons hcF (HCCONSTN s) ``` ``` | ID f1 -> ``` ``` let tf1 = hc_formula hcF f1 in ``` ``` HcFormula.hashcons hcF (HCID tf1) ``` ``` | NORM (f1, f2) -> ``` ``` let tf1 = hc_formula hcF f1 in ``` ``` let tf2 = hc_formula hcF f2 in ``` ``` HcFormula.hashcons hcF (HCNORM (tf1, tf2)) ``` ``` | NORMN (f1, f2) -> ``` ``` let tf1 = hc_formula hcF f1 in ``` ``` let tf2 = hc_formula hcF f2 in ``` ``` HcFormula.hashcons hcF (HCNORMN (tf1, tf2)) ``` ``` | CHC (f1, f2) -> ``` ``` let tf1 = hc_formula hcF f1 in ``` ``` let tf2 = hc_formula hcF f2 in ``` ``` HcFormula.hashcons hcF (HCCHC (tf1, tf2)) ``` ``` | FUS (first, f1) -> ``` ``` let tf1 = hc_formula hcF f1 in ``` ``` HcFormula.hashcons hcF (HCFUS (first, tf1)) ``` ``` | MU (var, f1) -> ``` ``` let tf1 = hc_formula hcF f1 in ``` ``` HcFormula.hashcons hcF (HCMU (var, tf1)) ``` ``` | NU (var, f1) -> ``` ``` let tf1 = hc_formula hcF f1 in ``` ``` HcFormula.hashcons hcF (HCNU (var, tf1)) ``` ``` | AF _ ``` ``` | EF _ ``` ``` | AG _ ``` ``` | EG _ ``` ``` | AU _ ``` ``` | EU _ -> ``` ``` raise (CoAlgException ("nnf: CTL should have been removed at this point")) ``` ```(* Replace the Variable name in f with formula formula ``` ``` hc_replace foo φ ψ => ψ[foo |-> φ] ``` ``` *) ``` ```let rec hc_replace hcF name (formula: hcFormula) (f: hcFormula) = ``` ``` let func = hc_replace hcF name formula in ``` ``` let gennew = HcFormula.hashcons hcF in ``` ``` match f.HC.node with ``` ``` | HCTRUE ``` ``` | HCFALSE ``` ``` | HCAP _ ``` ``` | HCNOT _ ``` ``` | HCCONST _ ``` ``` | HCCONSTN _ -> f ``` ``` | HCVAR s -> ``` ``` if compare s name == 0 ``` ``` then formula ``` ``` else f ``` ``` | HCAT (s, f1) -> ``` ``` let nf1 = func f1 in ``` ``` if nf1 == f1 then f else gennew (HCAT(s, nf1)) ``` ``` | HCOR (f1, f2) -> ``` ``` let nf1 = func f1 in ``` ``` let nf2 = func f2 in ``` ``` if nf1 == f1 && nf2 == f2 then f else gennew (HCOR(nf1, nf2)) ``` ``` | HCAND (f1, f2) -> ``` ``` let nf1 = func f1 in ``` ``` let nf2 = func f2 in ``` ``` if nf1 == f1 && nf2 == f2 then f else gennew (HCAND(nf1, nf2)) ``` ``` | HCEX (s, f1) -> ``` ``` let nf1 = func f1 in ``` ``` if nf1 == f1 then f else gennew (HCEX(s, nf1)) ``` ``` | HCAX (s, f1) -> ``` ``` let nf1 = func f1 in ``` ``` if nf1 == f1 then f else gennew (HCAX(s, nf1)) ``` ``` | HCENFORCES (s, f1) -> ``` ``` let nf1 = func f1 in ``` ``` if nf1 == f1 then f else gennew (HCENFORCES(s, nf1)) ``` ``` | HCALLOWS (s, f1) -> ``` ``` let nf1 = func f1 in ``` ``` if nf1 == f1 then f else gennew (HCALLOWS(s, nf1)) ``` ``` | HCMORETHAN (n, s, f1) -> ``` ``` let nf1 = func f1 in ``` ``` if nf1 == f1 then f else gennew (HCMORETHAN(n, s, nf1)) ``` ``` | HCMAXEXCEPT (n, s, f1) -> ``` ``` let nf1 = func f1 in ``` ``` if nf1 == f1 then f else gennew (HCMAXEXCEPT(n, s, nf1)) ``` ``` | HCATLEASTPROB (p, f1) -> ``` ``` let nf1 = func f1 in ``` ``` if nf1 == f1 then f else gennew (HCATLEASTPROB(p, nf1)) ``` ``` | HCLESSPROBFAIL (p, f1) -> ``` ``` let nf1 = func f1 in ``` ``` if nf1 == f1 then f else gennew (HCLESSPROBFAIL(p, nf1)) ``` ``` | HCID f1 -> ``` ``` let nf1 = func f1 in ``` ``` if nf1 == f1 then f else gennew (HCID(nf1)) ``` ``` | HCNORM (f1, f2) -> ``` ``` let nf1 = func f1 in ``` ``` let nf2 = func f2 in ``` ``` if nf1 == f1 && nf2 == f2 then f else gennew (HCNORM(nf1, nf2)) ``` ``` | HCNORMN (f1, f2) -> ``` ``` let nf1 = func f1 in ``` ``` let nf2 = func f2 in ``` ``` if nf1 == f1 && nf2 == f2 then f else gennew (HCNORMN(nf1, nf2)) ``` ``` | HCCHC (f1, f2) -> ``` ``` let nf1 = func f1 in ``` ``` let nf2 = func f2 in ``` ``` if nf1 == f1 && nf2 == f2 then f else gennew (HCCHC(nf1, nf2)) ``` ``` | HCFUS (first, f1) -> ``` ``` let nf1 = func f1 in ``` ``` if nf1 == f1 then f else gennew (HCFUS(first, nf1)) ``` ``` | HCMU (var, f1) -> ``` ``` if compare var name != 0 then ``` ``` let nf1 = func f1 in ``` ``` if nf1 == f1 then f else gennew (HCMU(var, nf1)) ``` ``` else ``` ``` f ``` ``` | HCNU (var, f1) -> ``` ``` if compare var name != 0 then ``` ``` let nf1 = func f1 in ``` ``` if nf1 == f1 then f else gennew (HCNU(var, nf1)) ``` ``` else ``` ``` f ``` ```let rec hc_freeIn variable (f: hcFormula) = ``` ``` match f.HC.node with ``` ``` | HCTRUE ``` ``` | HCFALSE ``` ``` | HCAP _ ``` ``` | HCNOT _ ``` ``` | HCCONST _ ``` ``` | HCCONSTN _ -> false ``` ``` | HCVAR s -> ``` ``` if compare variable s == 0 ``` ``` then true ``` ``` else false ``` ``` | HCAT (s, f1) -> ``` ``` hc_freeIn variable f1 ``` ``` | HCOR (f1, f2) ``` ``` | HCAND (f1, f2) -> ``` ``` hc_freeIn variable f1 || hc_freeIn variable f2 ``` ``` | HCEX (_, f1) ``` ``` | HCAX (_, f1) ``` ``` | HCENFORCES (_, f1) ``` ``` | HCALLOWS (_, f1) ``` ``` | HCMORETHAN (_, _, f1) ``` ``` | HCMAXEXCEPT (_, _, f1) ``` ``` | HCATLEASTPROB (_, f1) ``` ``` | HCLESSPROBFAIL (_, f1) ``` ``` | HCID f1 -> ``` ``` hc_freeIn variable f1 ``` ``` | HCNORM (f1, f2) ``` ``` | HCNORMN (f1, f2) ``` ``` | HCCHC (f1, f2) -> ``` ``` hc_freeIn variable f1 || hc_freeIn variable f2 ``` ``` | HCFUS (first, f1) -> ``` ``` hc_freeIn variable f1 ``` ``` | HCMU (var, f1) ``` ``` | HCNU (var, f1) -> ``` ``` (* Do we need to exclude bound variables here? *) ``` ``` hc_freeIn variable f1 ``` ```(** An instantiation of a hash table (of the standard library) ``` ``` for hash-consed formulae. The test for equality ``` ``` and computing the hash value is done in constant time. ``` ``` *) ``` ```module HcFHt = Hashtbl.Make( ``` ``` struct ``` ``` type t = hcFormula ``` ``` let equal (f1 : t) f2 = f1.HC.tag = f2.HC.tag ``` ``` let hash (f : t) = f.HC.tag ``` ``` end ``` ``` ) ``` | 24,824 | 77,633 | {"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} | 2.734375 | 3 | CC-MAIN-2019-35 | latest | en | 0.736628 |
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# How do you express sinx2 in terms of cosx using the double angle identity?
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## Express sin(x/2) in terms of cosx.sin(x2)=√1−cosx2Explanation:By applying the trig identity: cos2a=1−sin2a we getcosx=1−2sin2(x2)2sin2(x2)=1−cosxsin2(x2)=1−cosx2sin(x2)=±√1−cosx2
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# MATH135 - Assignment6 Solutions - MATH 135 Assignment#6...
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MATH 135 Fall 2007 Assignment #6 Due: Wednesday 31 October 2007, 8:20 a.m. Hand-In Problems 1. In each part, explain how you got your answer. (a) What is the remainder when 14 585 is divided by 3? (b) Is 8 24 + 13 12 divisible by 7? (c) What is the last digit in the base 6 representation of 8 24 ? (d) Determine the remainder when 42 2007 + 2007 10 is divided by 17. 2. Suppose that p is a prime number. (a) Prove that if x y (mod p ), then x n y n (mod p ) for every n P by induction on n . (This proof is not difficult mathematically. We will be looking for a very carefully written proof here.) (b) Using the definition of congruence, prove that if x 2 y 2 (mod p ), then x ≡ ± y (mod p ). (c) Determine the number of integers a with 0 a < 4013 with the property that there exists an integer m with m 2 a (mod 4013). (Note that 4013 is a prime number.) (d) Disprove the statement “If x 4 y 4 (mod p ), then x ≡ ± y (mod p )”. 3. (a) Prove that 5 n 7 + 7 n 5 + 23 n 0 (mod 7) for all n P . (b) Prove that 35 | 5 n 7 + 7 n 5 + 23 n for all n P 4. Let p be a prime number. Fermat’s Little Theorem tells us that if p | a , then a p - 1 1. But p - 1 might not be the smallest positive integer k for which a k 1 (mod p ). (a) Find a prime p > 5, a positive integer b > 1 that is not divisible by p and a positive integer k < p - 1 for which b k 1 (mod p ). (b) Suppose that p is a prime number, a is a positive integer not divisible by p , and s is the smallest positive integer for which a s 1 (mod p ). Prove that s | p - 1. (Hint: Start by dividing p - 1 by s , giving quotient q and remainder r .) 5. For each of the following congruences, determine if there exists a positive integer k that makes the congruence true. If so, determine the smallest integer k that works and justify why the integer that you’ve found is indeed the smallest such integer. If not, explain why not. (a) 2 k 1 (mod 18) (b) 8 k 1 (mod 17) 6. Suppose that A ( a, a 2 ) and B ( b, b 2 ) are points on the parabola y = x 2 , with a = b . (a) Write down the coordinates of the midpoint M of AB . (b) The line is tangent to the parabola at P so that is parallel to AB . Determine the coordinates of point P . (c) Determine the equations of the tangent lines to the parabola at A and B , and the coordi- nates of their point of intersection, Q . (d) Prove that M , P and Q are collinear (that is, lie on the same straight line). (This problem is not directly related to the course material, but is included to keep your problem solving skills sharp.)
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Recommended Problems 1. Text, page 82, #4 2. Text, page 83, #31 3. Text, page 85, #74 4. Text, page 85, #76 5. Suppose that a = ( r n r n - 1 · · · r 2 r 1 r 0 ) 10 . (a) Prove that a is divisible by 8 if and only if ( r 2 r 1 r 0 ) 10 is divisible by 8.
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NPTEL Data Science For Engineers Week 8 Assignment Solutions
## NPTEL Data Science For Engineers Week 8 Assignment Answers 2023
1. According to the built model, the within cluster sum of squares for each cluster is ________ (the order of values in each option could be different):-
8.316061 11.952463 16.212213 19.922437
7.453059 12.158682 13.212213 21.158766
8.316061 13.952463 15.212213 19.922437
None of the above
`Answer :-For Answer Click Here`
2. According to the built model, the size of each cluster is _ (the order of values in each option could be different):-
• 13 13 7 14
• 11 18 25 24
• 8 13 16 13
• None of the above
`Answer :- For Answer Click Here`
3. The Between Cluster Sum-of-Squares (BCSS) value of the built K-means model is ______ (Choose the appropriate range)
• 100 – 200
• 200 – 300
• 300 ā 350
• None of the above
`Answer :- For Answer Click Here`
4. The Total Sum-of-Squares value of the built k-means model is _ (Choose the appropriate range)
• 100 – 200
• 200 – 300
• 300 ā 350
• None of the above
`Answer :- For Answer Click Here`
5. Which of the statement is INCORRECT about KNN algorithm?
• KNN works ONLY for binary classification problems
• If k=1, then the algorithm is simply called the nearest neighbour algorithm
• Number of neighbours (K) will influence classification output
• None of the above
`Answer :- For Answer Click Here`
6. K means clustering algorithm clusters the data points based on:-
• Dependent and independent variables
• The eigen values
• Distance between the points and a cluster centre
• None of the above
`Answer :- For Answer Click Here`
7. The method / metric which is NOT useful to determine the optimal number of clusters in unsupervised clustering algorithms is
• Scatter plot
• Elbow method
• Dendrogram
• None of the above
`Answer :- For Answer Click Here`
8. The unsupervised learning algorithm which aims to partition n observations into k clusters in which each observation belongs to the cluster with the nearest centroid is
• Hierarchical clustering
• K-means clustering
• KNN
• None of the above
`Answer :- For Answer Click Here` | 583 | 2,191 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.984375 | 3 | CC-MAIN-2024-26 | latest | en | 0.779757 |
http://mathhelpforum.com/calculus/159262-find-equation-plane-perdendicular-another-plane.html | 1,513,423,446,000,000,000 | text/html | crawl-data/CC-MAIN-2017-51/segments/1512948587577.92/warc/CC-MAIN-20171216104016-20171216130016-00160.warc.gz | 168,332,716 | 11,256 | # Thread: Find equation of a plane perdendicular to another plane
1. ## Find equation of a plane perdendicular to another plane
As the title states, going over a review here and I can't figure this one out.
It asks to give the equation of a plane perpendicular to the plane x+y-2z=1
Any insight as to how I should start this.
2. There are infinitely many answers to that question.
Two planes are perpendicular if their normals are perpendicular.
3. I'm only a n00b, so I may be wrong, but this is how I'd attempt it:
Firstly, what you have there is the Cartesian form of a plane. When you have a plane in Cartesian form, the co-efficients amazingly form the normal of the plane. In this case, the vector x = 1, y = 1, z = -2 or (1, 1, -2).
Of course, planes are linear combinations of two vectors. We want a plane perpendicular to the given plane. We have the normal, which by definition, is perpendicular. So, we can use that to construct a perpendicular plane. If you convert the given plane into parametric form, you'll easily see two vectors that make up its span. Basically, take one of those vectors, and create a linear combination of that and the normal.
For instance, (-1, 1, 0) is a vector which lies on the given plane. And the normal is perpendicular to this. Thus a linear combination of the two produces a perpendicular plane.
4. The easiest way to do this is to find a direction vector of the plane (which is perpendicular to the normal) and make this you normal - bada boom bada bing you're done.
From Inspection it can be seen that the points (1,0,0) and (0,1,0) satisfy the equation of the plane.
Therefore a direction vector of the plane is (1,0,0)-(0,1,0) = (1,-1,0)
Therefore the perpendicular plane has a normal of (1,-1,0), so a set of perpendicular planes are in the form of x - y = d.
Taking the dot product of the normals of these planes is (1,-1,0) dot (1,1,-2) = 1 -1+0 = 0. So they are perpendicular.
There are heaps though. Another way to do this even quicker is see that the plane has a normal of (1,1,-2) so a perpendicular vector would be (1,-1,0) or (0,2,1) or (2,0,1)
Cheers,
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# eqn of plane perpendicular to other plane
Click on a term to search for related topics. | 587 | 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} | 4.46875 | 4 | CC-MAIN-2017-51 | longest | en | 0.932067 |
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Create a graph (bar graph, line graph, pie chart, or scatter plot) based on a given data set. Title the graph, label the axes, and choose a scale. Adjust the graph to fit the data, and then check your accuracy. The Gizmo can also be used to create a data table based on a given graph.
### LEARNING OBJECTIVES:
• Identify bar graphs, line graphs, pie charts and scatter plots.
• Explain how each type of graph is used.
• Learn to title and label graphs properly.
• Choose the optimal scale for graph axes.
• Plot bars, points, or regions accurately on a graph.
• Interpret what is shown on a graph.
• Create a data table that corresponds to a graph. (Extension)
• Recognize positive and negative relationships from scatter plots. (Extension)
##### VOCABULARY:
• bar graph, line graph, negative relationship, pie chart, positive relationship, scale, scatter plot, variable | 201 | 891 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2021-31 | latest | en | 0.829573 |
https://brilliant.org/problems/chess-and-a-pawn/ | 1,534,451,732,000,000,000 | text/html | crawl-data/CC-MAIN-2018-34/segments/1534221211167.1/warc/CC-MAIN-20180816191550-20180816211550-00479.warc.gz | 661,987,629 | 11,797 | # Chess and a Pawn
Logic Level 3
Given a chessboard with sides $$n \times n$$. A pawn is placed on the bottom left box. Two players, Jihyun and Ziying, took turns moving the pawn with Jihyun getting their first turn. At each turn, the pawn can only be moved to a neighboring box (two boxes are said to be neighbors if and only if the two boxes are different and have only one same side) and have not been previously occupied. Players who can not move the pawn are declared defeated. Determine all $$n$$ values such that Jihyun has a winning strategy.
× | 133 | 557 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.6875 | 3 | CC-MAIN-2018-34 | latest | en | 0.959282 |
http://mathhelpforum.com/differential-geometry/181612-double-limit.html | 1,498,711,823,000,000,000 | text/html | crawl-data/CC-MAIN-2017-26/segments/1498128323864.76/warc/CC-MAIN-20170629033356-20170629053356-00250.warc.gz | 255,078,131 | 10,537 | 1. ## Double limit.
Prove the following:
$\lim_{n\to\infty}\lim_{k\to\infty}\cos^{2k}(\pi n! x)=\begin{cases} 1 \text{ if }x \in\mathbb{Q} \\ 0 \text{ if } x\in \mathbb{R}/\mathbb{Q}\end{cases}$ and show that:
$\lim_{n\to\infty}\lim_{k\to\infty}\cos^{2k}(\pi n! e)\neq \lim_{k\to\infty}\lim_{n\to\infty}\cos^{2k}(\pi n! e)$
2. Hint if
$x \in \mathbb{Q}$
Then there exists an N such that for all n > N
$xn! \in \mathbb{Z}$
the above fails if x is irrational. See where this takes you.
3. Originally Posted by TheEmptySet
Hint if
$x \in \mathbb{Q}$
Then there exists an N such that for all n > N
$xn! \in \mathbb{Z}$
the above fails if x is irrational. See where this takes you.
Yes, I know this part. My trouble is writing the proof or the "explanation" in a right way.
Is the following will consider a mathematical proof:
When x is rational the double limit can be written as:
$\{cos((2t-1)\pi)cos((2t-1)\pi)\}\cdot \{cos((2t-1)\pi)cos((2t-1)\pi)\}\cdot ...=\{(-1)(-1)\}\{(-1)(-1)\}...=1\cdot 1\cdot 1 ...=1$
or:
$\{cos(2t\pi)cos(2t\pi)\}\cdot \{cos(2t\pi)cos(2t\pi)\}\cdot ...=1\cdot 1\cdot 1\cdot ... =1$
and if x irrational:
The double limit can be understood like that(?):
$\prod_{n = 0}^\infty x_n^2$ when $|x_i|<1$ for all $i$, which is equals (the product) to 0.
$\lim_{k\to\infty}\lim_{n\to\infty}cos^{2k}(\pi n!e)=?$ | 536 | 1,348 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 12, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2017-26 | longest | en | 0.551768 |
https://www.nationalbraille.org/forums/reply/21793/ | 1,726,647,038,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651886.88/warc/CC-MAIN-20240918064858-20240918094858-00098.warc.gz | 814,801,398 | 12,099 | Home Forums Nemeth Code for Math and Science Aligned Expression Reply To: Aligned Expression
#21793
Lindy Walton
Moderator
Hi Patty. It's good to come up with a format for this right away, and check to see if you can continue to apply it as the guided practices vary throughout the book.
Since we cannot easily apply the usual Nemeth formats to this one, I am going to go out on a limb and suggest a format that reflects some of the new layouts given to us in Braille Formats 2011. I think you will find the result to be easy to read, with the illustration clearly belonging to the itemized problem, and can be confidently applied to similar problems that may be more involved later in the book.
The bigger picture is that, in 1-3 (NC itemized), braille the numbered problem and its conclusion.
In this example, that will be "1. Evaluate {28+[(2x4^2)/8]}. So, {28+[(2x4^2)/8]} = ____."
Interrupt before "So, ..." with the guided practice portion, using the ideas behind displayed material (with indented right margin) and nested lists from BF2011. I think this is what your colleague was suggesting. Like this:
With a blank line before and after the five-step guided practice (BF), put "Write the expression. Find 4^2. Multiply. Divide. Add." each in cell 5;
put each of the five math problems in cell 7;
put the comments in cell 9.
This particular example has no runovers in the guided portion, but I expect you will come across some that do, in which case *all runovers to the displayed portion will go in cell 11 (BF). Ignore the purple lines drawn below the steps.
After a blank line (BF), "So, ..." will go in cell 3--the runover cell for the itemized problem (NC).
I have attached a print rendition of this format, to --hopefully-- clarify what I am suggesting.
Your ideas: Alignment as in print is reserved for problems that need to be solved vertically (spatially). This example does not meet that criterion. Reserve the use of guide dots for tables.
edited by Lindy on 11/28/2012
edited by Lindy on 11/28/2012
edited by NBAStaff on 11/28/2012
edited by Lindy on 11/28/2012 | 518 | 2,089 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.40625 | 3 | CC-MAIN-2024-38 | latest | en | 0.932473 |
https://www.teacherspayteachers.com/Store/Lindsay-Perro/Type-of-Resource/Fun-Stuff?ref=filter/resourcetype/top | 1,534,807,683,000,000,000 | text/html | crawl-data/CC-MAIN-2018-34/segments/1534221217354.65/warc/CC-MAIN-20180820215248-20180820235248-00023.warc.gz | 961,841,314 | 32,441 | # Lindsay Perro
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TEACHING EXPERIENCE
After spending 8 years in the classroom teaching Middle School Math, Math Intervention and 4th grade, I am currently working from home as a curriculum designer creating resources for teachers across the country! I absolutely LOVE my job and feel so grateful for the opportunity to help teachers and students near and far.
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Presenter at the 1st Teachers Pay Teachers Conference in Las Vegas - July 2014 Presenter at the 2nd Teachers Pay Teachers Conference in Las Vegas - July 2015 Panelist at the 3rd Teachers Pay Teachers Conference in Orlando - July 2016
MY OWN EDUCATIONAL HISTORY
I have a B.S. in Elementary Education (K-8) with a Minor in History. I am highly qualified to teach Middle School Mathematics.
I'm a wife, mom, teacher and writer. I love creating fun materials for the classroom that guide teachers away from the textbook! | 1,648 | 7,240 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.28125 | 3 | CC-MAIN-2018-34 | latest | en | 0.895663 |
https://allebestekredite.info/?post=199 | 1,632,289,168,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780057329.74/warc/CC-MAIN-20210922041825-20210922071825-00025.warc.gz | 133,272,004 | 8,466 | # Quantum cryptography will completely replace classical cryptography
## Quantum cryptography - security through quantum effects
Whether correspondence, telephone conversation or e-mail: every communication channel can be tapped. The aim of cryptography - the science of encryption - is to make life as difficult as possible for spies. Modern encryption methods work with randomly generated one-time keys. But the exchange of such a key can also be eavesdropped. Quantum effects offer a remedy: if an eavesdropper is listening in, he disrupts the transmitted information and is thereby exposed.
The issue of security is playing an increasingly important role in our lives - from protection against terrorism to securing energy supplies to safety in road traffic. Information can also be valuable and must be protected - not just industrial and military secrets, but also, for example, access data to bank accounts. But documents can fall into the wrong hands, telephones can be tapped and Internet connections tapped. In many cases, the spy leaves no trace. It is bad enough when a secret has been betrayed - but it is even worse when we have no knowledge of the betrayal.
It is therefore hardly surprising that the science of encrypting information, known as cryptography, is almost as old as humanity itself. The Spartans used the Skytale method in ancient Greece, Julius Caesar used the substitution and in World War II played the famous one Enigma cipher machine plays an important role.
Symmetrical encryption with random key
Modern cryptographic systems consist of two parts, the key and the algorithm. The algorithm describes how the key must be applied to the original information. In the case of a text, for example, the algorithm could be the exchange of letters and the key would then indicate which letter is exchanged for which.
An information-theoretic "unbreakable" encryption is obtained if you use a randomly generated key that is exactly as long as the information to be encrypted itself (one time pad, One-time encryption).
Today it is generally assumed that the spy knows the encryption process, i.e. the algorithm. The security of the encryption is therefore based on the confidentiality of the key used. If it is possible to distribute an identical key to two communication partners, from now on called "Alice" and "Bob", a message can be securely encrypted and decrypted again. With such a "symmetrical key distribution" it must of course be ensured that the key does not fall into the hands of a spy. From now on we want to call this possible listener "Eve" (from "Eavesdrop", engl. for eavesdropping).
The search for an absolutely secure encryption process is reduced to the secure symmetrical key distribution. This is exactly where quantum cryptography, or more precisely: the quantum key distribution, comes into play. quantum key distribution): It offers a secure solution to the problem of key distribution.
### The basic idea
The main difference between quantum cryptography and classic key distribution is that the information is encoded on individual quantum systems, the so-called qubits. Individual photons are used as carriers. In conventional forms of transmission, on the other hand, the information is encoded in pulses made up of many photons, with all photons of a pulse carrying the same information. In this way, Eve can take out some of the photons of the strong pulse in order to read out the information. In quantum cryptography, each bit of information sits on a single photon. Eve cannot simply branch off these photons, because then they would never get to Bob and could no longer become part of the key.
Instead, Eve could try to measure the photon and thus obtain the information. But this is where quantum physics throws a spanner in the works. Because the quantum laws do not allow measurements to be carried out on quantum states without influencing them. The same laws also forbid Eve to generate an identical copy of the photon (“no-cloning theorem”) and to carry out the measurements on this copy in order to remain undetected. Whenever Eve tries to eavesdrop on the key, she inevitably changes the quantum state of a qubit. These changes show up as errors in the transmission between Alice and Bob. Alice and Bob therefore know whether the key has been tapped - even before they use it to encrypt the message.
This is the new thing about quantum cryptography: It enables secure key distribution based on quantum physical laws. In the event of eavesdropping, the communication partners are warned and can simply discard the key. Since only one key was tapped, the secret information itself is still secure.
### The BB84 protocol
How does quantum cryptography work in detail? In order to store the information on individual photons, one often uses the state of polarization, i.e. the direction of the oscillations of the photon's electric field.
Alice encodes the classic bit values 0 and 1 of the key in the form of certain polarization states. Only orthogonal - i.e. mutually perpendicular - states are allowed to be used, because only these can be completely differentiated by Bob with a measurement and thus read out the values 0 and 1 again. A pair of such orthogonal states are, for example, the horizontal and vertical polarization (H / V basis).
Schematic mode of operation of the BB84 protocol
So Alice codes 0 as H and 1 as V and sends these photons to Bob. This can measure in the H / V basis and always receives the same result of the polarization and thus the values coded by Alice. In principle, an identical classic bit key could thus be exchanged between Alice and Bob. But this transmission is not yet secure against eavesdropping. Eve could simply carry out a measurement in the base H / V, determine the polarization, then generate and transmit a photon of the same polarization and thus receive the encoded information.
In order to make the transmission of the key secure against eavesdropping, Alice has to use a second base rotated 45 degrees from H / V. In this base +/-, 0 and 1 correspond to a polarization at angles of +45 and -45 degrees. When generating the key, Alice now randomly chooses the base for each individual bit of the key and sends the photon prepared in this way to Bob.
When measuring, Bob has to choose between the bases H / V or +/-. In half the cases, Bob measures on the same basis that Alice coded the information. In this case its results are correct and a bit for the key can be saved. In the other cases, Bob measures on a basis that does not match that chosen by Alice. Since a horizontally prepared photon is measured with the same probability at +45 or -45 degrees, there are no correlations in this case and the photon cannot be used for the key.
Of course, in order to know which photons can actually be used for the key, Alice and Bob have to exchange their base choice for each photon. After the key data has been transferred, this can even be done publicly. Because the information about the base alone does not allow any conclusions to be drawn as to which bit value was encoded with it. This means that eavesdroppers cannot find out anything about the secret key if they only listen in on the basic comparison.
### Bugging!
The security of this "BB84" protocol, developed in 1984 by Charles H. Bennett and Gilles Brassard, relies on the fact that the rules of quantum physics make it impossible for Eve to determine in which base (H / V or +/-) Alice has encoded the individual photons of the key. So Eve doesn't know what basis to measure. If it randomly selects the correct base, it actually receives the full information of this photon and also goes unnoticed. But if Eve chooses incorrectly, her measurement changes the polarization state. Because the polarization of the photon naturally corresponds to the measurement result after the measurement. A photon horizontally polarized by Alice is repolarized by Eves measurement in the +/- base to +45 or -45 degrees. But since Bob measures a photon polarized in +/- direction when measuring in H / V with the same probability in H or V, in half of these cases he receives a wrong result that no longer matches the information sent by Alice. An eavesdropper therefore causes errors in the transmission.
Experimental setup for the generation of entangled photons
On the basis of these errors it is now possible to determine whether or not an eavesdropping attack has taken place. To do this, Alice and Bob simply exchange a number of bits - again quite publicly - and use this to estimate the error rate. Since there can also be natural transmission errors, the problem is now to find a limit value above which an eavesdropping attack must be assumed. A possible eavesdropper could use the system's technical flaws to disguise his attack. Therefore, all errors that have occurred must always be attributed to an eavesdropper. In order to achieve a high key rate, all technical errors must be reduced to a minimum.
In order to guarantee “permanent security”, the strongest possible attack must be accepted. Eve has all the options that are allowed within the framework of quantum mechanics. Detectors with one hundred percent efficiency and quantum memories are just as much a part of their tools as quantum computers. Nevertheless, Eve will leave mistakes by which one can measure her gain in information. For the BB84 protocol, these considerations result in a limit of the error of eleven percent. If the error is above this, it is no longer a question of secure transmission. On the other hand, if the error is smaller, Alice and Bob share more information than Eve does with Alice or Bob. With a special procedure, the "privacy amlification", even Eve's information can be filtered out of the faulty, insecure key and a guaranteed secure key can be created for Alice and Bob.
### Entanglement as an alternative
In 1991 Artur Ekert developed an alternative protocol based on the quantum mechanical entanglement of states. Pairs of photons are used, the polarization of which is parallel to one another due to the mechanism of their formation. The entanglement has the result that when the polarization is measured in a randomly selected base, the polarization of the second photon instantly assumes a state in the same base.
If Alice and Bob each share a photon of such an entangled pair, then their measurement results must always be identical if they measure in the same basis. As in the case of the BB84 protocol, this results in a key. Instead of the active coding as with BB84, Alice carries out a passive measurement of the polarization state in the E91 protocol.
Although the E91 protocol is very elegant, the BB84 protocol is used almost exclusively in practice today. The reason for this is that for BB84 there is theoretically sound evidence for the safety of the procedure with given errors. With the E91 protocol, on the other hand, the connections between the eavesdropping attack and its consequences for the transmitted data have not yet been fully clarified.
### realization
Currently, the biggest problem in the technical implementation of quantum cryptography systems is the measurement of individual photons. The key rates achieved so far are rather modest and are mostly in the range of a few kilobits per second. This rate is also limited by the length of the transmission path. In contrast to a classic pulse made up of many photons, individual photons cannot be sent through an amplifier because of the no-cloning theorem. Losses in transmission therefore have a direct effect on the key: the loss of a photon means the loss of a bit.
The systems tested in laboratories today still achieve a transmission length of over a hundred kilometers in glass fibers with key rates of around a hundred bits per second. Shorter distances, such as those found in inner-city fiber optic networks, can be supplied with ten kilobits per second.
If the photons are not transmitted through the air through glass fibers but rather through a telescope, the attenuation is reduced and longer distances can be bridged. A quantum cryptography connection over 144 kilometers between the Canary Islands Tenerife and La Palma could be realized in experiments.
For use in practice, however, it is not only the distance that can be bridged and the transmission rate that are important. It is also important that the systems are fail-safe and easy to operate. In recent years, the leap from room-filling laboratory setups to compact and reliable devices has actually been made. Spin-off companies from various university groups have already started offering their systems to external customers. Quantum cryptography is the first commercially usable application of the fundamental but not always intuitive laws of quantum physics. | 2,542 | 12,834 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2021-39 | latest | en | 0.937175 |
https://physics.stackexchange.com/tags/integrable-systems/new | 1,603,572,006,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107884755.46/warc/CC-MAIN-20201024194049-20201024224049-00625.warc.gz | 488,825,140 | 19,470 | # Tag Info
For rigor, you might try to ask on Math.SE or MO.SE. In this answer, we will give a heuristic derivation via discretization. We will use a slightly different notation to connect to usual time-evolution in QM, but the idea is the same: \begin{align} U_{\lambda}(t_f,t_i)~=~&T e^{-\frac{i}{\hbar}\int_{t_i}^{t_f}\! dt~ H_{\lambda}(t)}\cr ~=~&\lim_{N\... 0 T is the ordered product of many infinitesimal factors of V(x,t) = \exp\{\epsilon U(x,t)\} \approx 1+ \epsilon U(x,t). $$Apply Liebniz' product rule to differentiate. If you insist on a rigorous derivation then some properties of the operator U(x,t) need to be adduced. 1 Such analytical expressions were derived for special cases by multiple authors. More general results were obtained by Yang and Yang in C. N. Yang and C. P. Yang One-Dimensional Chain of Anisotropic Spin-Spin Interactions. II. Properties of the Ground-State Energy Per Lattice Site for an Infinite System, Phys. Rev. 150, 327 (1966). For example, the ground ... 0 A single autonomous (possibly nonlinear) 2nd-order ODE$$F(x,\dot{x},\ddot{x})~=~0. \tag{1}$$can in principle be written as a couple of autonomous (possibly nonlinear) 1st-order ODEs of the form$$\dot{x}~=~f(x,y), \qquad \dot{y}~=~g(x,y). \tag{2} One may show that there always exists an integral of motion/first integral for the latter system (2), at ... | 390 | 1,355 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2020-45 | longest | en | 0.856325 |
https://www.physicsforums.com/threads/legendre-transform.362476/ | 1,519,488,611,000,000,000 | text/html | crawl-data/CC-MAIN-2018-09/segments/1518891815843.84/warc/CC-MAIN-20180224152306-20180224172306-00012.warc.gz | 949,310,183 | 14,610 | # Legendre transform
1. Dec 11, 2009
### Niles
Hi guys
I am looking at f(x) = (|x|+1)2. I write this as
$$f(x) = \left\{ {\begin{array}{*{20}c} {x^2 + 1 + 2x\,\,\,\,\,\,\,\,\,\,\,\,\,for\,\,\,\,x > 0} \\ {x^2 + 1 - 2x\,\,\,\,\,\,\,\,\,\,\,\,\,for\,\,\,\,x < 0} \\ {1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,for\,\,\,\,x = 0} \\ \end{array}} \right.$$
I want to find the Legendre transform of this function. For x>0 I get the Legendre transform
f*(p) = p2/4-p-1/2.
I am a little unsure of how this works. Because I need to find the Legendre transform of f for x<0 and x=0. But how do these solutions get "patched" together?
2. Dec 11, 2009
### HallsofIvy
I don't get that at all. For x> 0, f is differentiable so f*(p) is just -f(x) where x is such that f'(x)= p. Here, f'(x)= 2x+ 2= p when x= (p-2)/2. So f*(p)= -f((p-2)/2)= -((p-2)2/4+ 1+ 2(p-1)/2)= -((1/4)p2- p+ 1+ 1+ p- 2)= -(1/4)p^2.
Similarly, for x< 0, f'(x)= 2x- 2= p when x= (p+2)/2. So F*(p)= -f((p+2)/2)= -)(p+2)2/4+ 1+ 2(p+2)/2)= -((1/4)p2+ p+ 1+ 1+ p+ 2)= -(1/4)p2- 2p- 4.
f is not differentiable at x= 0 so have to go back to the basic definition: f*(0)= max(0(x)-f(x))= max -f(x). Of course, that maximum is 0: f*(0)= 0. Since f is not differentiable at 0, it Legendre transform is not continuous there and they cannot be "patched" together.
3. Dec 11, 2009
### Niles
Thanks. Ok, I've done it a slight different way. First, I must admit that I have not heard of a theorem stating that if f is differentiable, then f*(p) is just -f(x).
I used that since f is strictly convex (we look at x>0 for now), then f*(p)=xp-f(x(p)), where x is such that f'(x)= p.
But OK: Let us look at the solutions found in #2 (I am mostly trying to learn the method). So
for x>0: f*(p) = -(1/4)p^2
for x<0: f*(p) = -(1/4)p^2- 2p- 4
x=0 : f*(p) = 0 (by the way, shouldn't this be -1 rather than 0?)
What solutions are "valid" for the different intervals of p? I mean, we have 3 solutions, one for each interval of x. But how do I find out what intervals of p they correspond to?
Last edited: Dec 11, 2009 | 846 | 2,085 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.21875 | 4 | CC-MAIN-2018-09 | longest | en | 0.889564 |
https://physics.stackexchange.com/questions/782446/the-difference-in-path-lengths-for-waves-in-the-double-slit-experiment | 1,723,609,301,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641095791.86/warc/CC-MAIN-20240814030405-20240814060405-00587.warc.gz | 346,603,782 | 42,207 | # The difference in path lengths for waves in the double slit experiment
Fig.1
I don't quite understand the diagram, because it shows $$L_1$$ and $$L_2$$ as parallel, even though they are supposed to meet at the same point. I believe the idea is that $$\Delta L$$ approaches $$d \sin \theta$$ as the distance to the screen increases. However, that is not exactly the case, and here is my proof:
Fig.2
I want to get a formula for $$AF - BF$$ in terms of $$AF$$, $$\theta_1$$ and $$\theta_2$$, where these angles measure the angle between the horizontal gridlines and $$AF$$ and $$BF$$, respectively.
As we can see, we have dotted line that is parallel to $$AF$$, and a dotted line between them that is perpendicular to both of them. We have an even more finely dotted line $$CD$$ that is perpendicular to $$BF$$.
Now, first off, we can establish the following:
\begin{align} AF - BF &= AF - BC - CF \\ &= AF - BC - \sqrt{AF^2 + AC^2} \\ &= AF - BC - \sqrt{AF^2 + \cos(\theta_1)^2d^2} \end{align}
So, now we must get another expression for $$BC$$. As we know from fig. 1 (using some trigonometry), $$BE = d\sin(\theta_1)$$. In fig. 2, we see that $$BC = BE - DE$$.
The angle between $$BF$$ and the dotted line is $$\theta_2 - \theta_1$$. Therefore, we have the following two identities:
$$CE= \sin(\theta_2 - \theta_1)\cdot BE = \sin(\theta_2 - \theta_1) d\sin(\theta_1)$$
$$CD = \sin(\theta_2 - \theta_1)\cdot BC$$
The line $$DE$$ can be given a new expression by the Pythagorean theorem:
$$DE = \sqrt{CE^2 -DC^2} = \sqrt{(\sin(\theta_2 - \theta_1) d\sin(\theta_1))^2 - (\sin(\theta_2 - \theta_1)\cdot BC)^2 }$$
As already established, $$BC = BE - DC$$. So, we have the following expression for $$BC$$:
$$BC = d\sin(\theta_1) - \sqrt{(\sin(\theta_2 - \theta_1) d\sin(\theta_1))^2 - (\sin(\theta_2 - \theta_1)\cdot BC)^2}$$
Thus, we have our final expression for the difference:
$$AF - BF = AF - d\sin(\theta_1) + \sqrt{(\sin(\theta_2 - \theta_1) d\sin(\theta_1))^2 - (\sin(\theta_2 - \theta_1)\cdot BC)^2} - \sqrt{AF^2 + \cos(\theta_1)^2d^2}$$
Now, as the distance to the screen, denoted as $$d_s$$, increases relative to the distance between the slits, denoted as $$d$$, the difference in $$\theta_1$$ and $$\theta_2$$ decreases, which is fairly intuitive and simple to prove. Thusly, asking what $$AF - BF$$ approaches as $$d/d_s$$ approaches zero is the sake as asking what $$AF - BF$$ approaches as $$\theta_2 - \theta_1$$ approaches zero. Thusly:
\begin{align} \lim_{\theta_2 - \theta_1 = 0} \Delta L &= \lim_{\theta_2 - \theta_1 = 0}AF - BF \\[2ex] &= \lim_{\theta_2 - \theta_1 = 0}AF - d\sin(\theta_1) + \sqrt{(\sin(\theta_2 - \theta_1) d\sin(\theta_1))^2 - (\sin(\theta_2 - \theta_1)\cdot BC)^2} - \sqrt{AF^2 + \cos(\theta_1)^2d^2} \\[2ex] &= AF - d\sin(\theta_1) - \sqrt{AF^2 + \cos(\theta_1)^2d^2} \end{align}
So, what gives? Is it just my trig that is off, or have I misunderstood what the diagram is trying to say to begin with?
• For $d$ of 100 microns and $L$ of 5 meters, what is the difference between the two? Commented Sep 29, 2023 at 18:07
• Yeah, maybe like complete the square under the root, then take the square root, and try to expand in some small parameter....the approximation makes sense in the large $L/d$ limit, and I don't immediately see something wrong in your approach, so I think it's just massaging the last line. Commented Sep 29, 2023 at 18:30
• Apply a binomial formula, binomial expansion, and you get $$AF-\sqrt{AF^2-\cos^2(\theta_1)d^2}=\frac{\cos^2(\theta_1)d^2}{AF+\sqrt{AF^2-\cos^2(\theta_1)d^2}},$$ so the whole limit expression is dominated by $-d\sin\theta_1$. Commented Sep 30, 2023 at 5:46
Let $$R_b=\bar {BF}$$ and $$R_a=\bar {AF}$$, and also the angle $$\alpha = \measuredangle [ \bar {AB}, \bar {BF}]$$. Then using the Taylor exapansion of $$\sqrt{1+x}=1+\frac{1}{2}x+\mathcal O(x^2)$$ we get $$R_a = \sqrt{R_b^2+d^2-2dR_b \cos(\alpha)} = R_b\sqrt{1+\frac{d^2}{R_b^2}-2\frac{d}{R_b} \cos(\alpha)}\\ =R_b\left(1-\frac{d}{R_b} \cos(\alpha) + \mathcal O \left(\frac{d^2}{R_b^2}\right)\right)$$ in other words, asymptotically the path difference $$\delta L$$ is $$\delta L = R_b - R_a = d \cos(\alpha) + \mathcal O \left(\frac{d^2}{R_b^2}\right)$$ showing, to a 1st order error, that it is independent of the distance between the point of interference and the slit. | 1,480 | 4,335 | {"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": 43, "wp-katex-eq": 0, "align": 2, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.3125 | 4 | CC-MAIN-2024-33 | latest | en | 0.872034 |
http://mathathome.org/LessonPlans/WhatsInAName/LessonPlanWhatsInANameWeb.php | 1,498,294,266,000,000,000 | text/html | crawl-data/CC-MAIN-2017-26/segments/1498128320243.11/warc/CC-MAIN-20170624082900-20170624102900-00137.warc.gz | 248,948,617 | 9,460 | Math Access for Teachers and Home Child Care Providers
What's In a Name?
Children will count, compare, contrast, and graph the number of letters in their names.
Content Area Standard Target
• Number and Operations
• Algebra
• Measurement
• Data Analysis and Probability
• Understand numbers, ways of representing numbers, relationships among numbers, and number systems
• Understand measurable attributes of objects and the units, systems and processes of measurement
• Understand patterns, relations, and functions
• Formulate questions that can be addressed with data, and collect, organize, and display relevant data to answer them
• Count with understanding and recognize "how many" in sets of objects
• Sort, classify, and order objects by size, number and other properties
• Connect number words and numerals to the quantities they represent, using various physical models and representations
• Identify whether the number of objects in one group is greater than, less than, or equal to the
number of objects in another group
• Directly compare two objects with a measureable attribute (letters in a name) in common, to see which object has “more of”/ “less of" the attribute, and describe the difference
Obtain the Materials
• The book, Chrysanthemum by Kevin Henkes
• 5x8 Index cards
• Colored markers
• Easel w/ large chart graphing paper to record collected data
Note: Small parts create a choking hazard for children. Make sure that all materials you choose to use for an activity or lesson with children meet safety requirements. Small parts are not appropriate for children who are 5 years of age or younger.
Introduce the Activity
1. Explain to the children that today they are going to read a story about a little mouse who has a very unique name. Explain that each of us has a special name and that they are going to use their names to solve some math problems.
• “Is there someone else in your family who has the same name that you do?"
3. Explain that the one thing that all of our names have in common is that all of our names are composed of letters. Explain that while we can all have a different amount of letters in our name all of our names are composed of letters.
4. Say: “My name is Stephanie" (use your own name) and I have 9 letters in my name. Say:"S – t – e – p – h – a – n – i – e.” While you spell out your name, hold up a finger to represent each of the letters in your name so that when you are finished spelling out your name, you are holding up 9 fingers.
5. Distribute the index cards. Give one to each child. Ask them to print their name on the card. Some children might need help with the spelling of their names or with the formation of the letters in their names. For younger children, you can already have the child’s name printed on the index card and then ask the child to look at his/her name on the card.
6. Ask the children what they notice about their names. By asking this question you are looking for quantitative answers, “I have 3 letter S’s in my name,” “My name has 4 letters.” Later you will be analyzing the data supplied by the letters in each of their names.
7. Introduce the book Chrysanthemum by Kevin Henkes. Explain that you are going to read a book about a little mouse with a big name. Explain that Chrysanthemum loves her name until she starts school. There are many themes imbedded in this book, the most obvious being about bullying and respect. There are many cross-curricular teaching opportunities as well, but for the sake of our math focus, try to keep the children’s focus on the length of the children’s name. Ask: “Have you noticed that the name Chrysanthemum has A LOT of letters?”, “I wonder if it was difficult to learn how to spell Chrysanthemum?”
8. Read the book. Pause when you come to the part of the book where the name is written out on an envelope. When book reads “Chrysanthemum loved the way her name looked when it was written with ink on an envelope. She loved the way it looked when it was written with icing on her birthday cake.” Say: “Let’s see just how many letters Chrysanthemum has in her name. Ask: Can anyone make a guess before we start to count?” As you count the letters, point to each letter, reinforcing one-to-one correspondence. Say: “Wow! 13 letters in her name. Ask: Do any of you have 13 letters in your name? Can anyone think of a name that also has 13 letters?”
Engage the Children
1. Explain to the children that they are now going to investigate their own names. Say: “Everyone look at your names." Ask: How many letters does your name have?” Have the children write the number of letters on the other side of their index card so that the name is on one side of the card and the number of letters is on the other side.
2. Categorize the children into groups according to the number of letters in their name. Say: “Everyone who has 2 letters in their name, please stand up and sit in a group over here.” Repeat this process until all of the children are in a group. There may be only one child in a group and you may not have any children for certain numbers (for example, you may not have any children who have 5 letters in their name).
3. Chart their names. On the graph paper on your easel (this activity also makes a nice bulletin board!), graph the children’s names according to the letters in their names. Title the graph, Our Name Graph. Across the bottom label with numbers going from 0 to the largest number in your class. Then, have the children come up as you call the numbers and tape their name going up the chart. Say: “All of you who have 2 letters in your name, please come up with your index cards and place your card in the space above the number 2.” Again, continue this until all of the children’s name index cards are up on the graph.
4. Compare and examine the collected data. Look at the graph and ask questions that use the vocabulary the most, more than, less than, the least, the same
• "Whose name has the most letters?"
• “Does David have more or less letters than Amy?”
• “Which names have the same amount of letters?”
5. Extend the Activity by identifying, counting and sorting by letters. For example, you can count the number of vowels each person has in his/her name. Or you could examine the amount of syllables each person has in his/her name. Again, all of this data can be recorded and displayed in the classroom. This is an activity that you can build upon as the children’s skills increase and it ties into many other subject areas.
• Once the children are able to identify vowels within the alphabet, they can count, compare and record the number of vowels each person has in his/her name.
• This activity is a wonderful introduction to Syllabication. You can have the children clap out the number of syllables in their name and again, compare and record.
• This can also be a special Home Project. Children can interview their parents, siblings and other relatives asking them the correct spelling of their names and then together, they can count the number of letters in each of their family member’s names. The children bring this information back to school and construct a Family Tree and organize their family members in numerical order according to the amount of letters in their name.
Encourage Vocabulary
• More – A value that is higher or greater in number (e.g., "Sally has more letters in her name than Ted, Jane and Amy.")
• Fewer – A value that is smaller in number (e.g., "Ed has fewer letters in his name than Sally.")
• Greatest Amount – Largest amount; the one with the most (e.g., "Chrysanthemum has the greatest amount of letters in her name.")
• Equal – To be the same in number or amount (e.g., "Jane and Noah have an equal amount of letters in their name.")
• Numeral – The symbol used to represent a number of "how many" (e.g., "The numeral '5' represents how many letters there are in the name Brian.")
• Graph – A diagram that exhibits a relationship, often functional, between two sets of numbers as a set of points having coordinates determined by the relationship
• Data – Facts and statistics collected together for reference or analysis
Glossary of MATH vocabulary
Supporting Children at Different Levels
Toddlers Pre-K
Toddlers may:
• Not know how to spell their own names.
• Not have one-to-one correspondence.
• Not be able to recognize all their numbers or letters.
Pre-K Children may:
• Be able to identify vowels in the alphabet.
• Be able to identify syllables within words.
• Have a working knowledge of letters and how to form their letters and may be able to use invented spelling to write the names of family members and friends.
Home child care providers may:
• Provide assistance when children are counting the letters in their name.
• Write the number of letters that are in the child’s name on the back of their index card.
• Help the child read and recognize letters in their name.
Home child care providers may:
• Provide opportunities for the children to sort and count the letters in their name based on vowels and consonants.
• Help the children to identify the syllabication patterns in their names by clapping out the beats in their names and allowing the children to notice that each of the beats is a syllable.
• Help the children to count the syllables in not only their names but in other objects in their environement.
Books
• Chrysanthemum by Kevin Henkes (New York: Greenwillow Books, 2007)
• Cristanemo by Kevin Henkes (New York: Greenwillow Books, 2008 [Spanish version of the book. Opportunities to teach counting in Spanish])
Music and Movement
Outdoor Connections
This activity can be taken outside to a place where there are a variety of flowers. The flower does not necessarily need to be a Chrysanthemum.
• Looking at several types of flowers, children can count the petals on the flowers and graph that data. Rose has 16 petals, Daisy has 10 petals…. Compare and contrast the flowers using the recorded data.
• Children can also sort flowers according to color and then graph that information. There are 4 yellow flowers, 7 red flowers….. Compare and contrast the flowers using the recorded data.
Web Resources
Comment on this lesson
To report a problem with the site, please email us. © 2011. M.A.T.H. | 2,316 | 10,299 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.53125 | 5 | CC-MAIN-2017-26 | latest | en | 0.925497 |
https://cs.stackexchange.com/questions/156009/how-to-reduce-3sat-to-twoormoresat | 1,719,247,882,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198865401.1/warc/CC-MAIN-20240624151022-20240624181022-00079.warc.gz | 160,369,174 | 40,807 | # How to reduce 3SAT to TwoOrMoreSAT?
I want to prove, that 2OrMoreSAT is NP-complete. It's defined as follows:
A formula is considered strongly satisfiable if there exists a model such that two or more different literals in every clause evaluate to true.
$$\textit{2OrMoreSAT} = \{\langle \phi' \rangle \mid \phi' \text{ is a boolean formula in CNF which is strongly satisfiable} \}$$
I want to solve this by reducing 3SAT to 2OrMoreSAT. My first thought was to duplicate all of the literals in a clause, which would ensure 2 or more literals being true - but I don't really know how to limit the literals after. Say $${\phi}$$ ∈ 3SAT and $${\phi = X ∨ Y ∨ Z}$$, then $${\phi'}$$ could look something like this $${\phi' = (X_1 ∨ X_2 ∨ Y_1 ∨ Y_2 ∨ Z_1 ∨ Z_2) ∧ (...)}$$
• Uni. We've been told to assume it is a NP-complete problem and try to prove it. The definition above was a little confusing, I've made it more accurate and easier to understand. Commented Dec 7, 2022 at 20:33
• I suggest you keep working on it. What are the criteria for the reduction to be correct? Does your example meet those criteria? If not, can you fix it up? FYI, you might receive some resistance to questions that post an exercise and ask for the solution.
– D.W.
Commented Dec 7, 2022 at 20:42
As pointed out by Yuval Filmus in the comments, I assumed there that the formula had to be in 3CNF, which is not necessarily the case. However, I will let my answer, maybe it can still help.
I may be wrong, but I think this problem is in $$\mathsf{P}$$, there is a reduction to $$\texttt{2SAT}$$.
Consider a clause $$C = (x\lor y \lor z)$$. From this clause, define:
$$\phi_C = (x\lor a) \land (y \lor b) \land (z\lor c) \land (x\lor \overline{b}) \land(x\lor \overline{c}) \land(y\lor \overline{a})\land (y\lor \overline{c})\land (z\lor \overline{a}) \land(z\lor \overline{b})$$
Then I claim that $$C$$ is satisfiable with at least two true litterals if and only if $$\phi_C$$ is satisfiable, and moreover, the truth values of $$x$$, $$y$$ and $$z$$ are the same in both truth assignments.
The idea is that $$a\equiv y\land z$$, $$b\equiv x\land z$$ and $$c\equiv x\land y$$.
Now if $$\varphi = \bigwedge\limits_{i=1}^n C_i$$ is a formula in 3CNF, then $$\psi = \bigwedge\limits_{i=1}^n\phi_{C_i}$$ is a formula in 2CNF, and $$\varphi$$ is satisfiable with at least two true litterals per clause if and only if $$\psi$$ is satisfiable (this is due to the fact that in the previous claim, the truth values of $$x$$, $$y$$ and $$z$$ are the same, so a satisfying truth assignment does not change values between clauses).
Since this construction is done in polynomial time, $$\texttt{2orMore3SAT}\leqslant_m^p\texttt{2SAT}$$, which is known to be in $$\mathsf{P}$$.
What would be a $$\mathsf{NP}$$-complete problem, though, is the following problem:
Input: a boolean formula $$\varphi$$ in 3CNF.
Question: is $$\varphi$$ satisfiable with exactly two true litterals per clause?
The reduction from $$\texttt{3SAT}$$ transforms a clause $$C = (x\lor y \lor z)$$ into: $$\phi_C = (x \lor a \lor b) \land (y \lor c \lor d) \land (z\lor e \lor f) \land (a \lor c\lor e)$$
• if $$C$$ is satisfied, distinguish depending on the number of true litterals:
• if exactly one litteral is true, say $$x$$ WLOG, then assign $$b, c, d, e$$ and $$f$$ to true and $$a$$ to false;
• if two litterals are true, say $$x$$ and $$y$$ WLOG, assign $$a, d, e, f$$ to true and $$b, c$$ to false;
• if all three litterals are true, assign $$a, c, f$$ to true and $$b, d, e$$ to false.
In all cases, $$\phi_C$$ is satisfied with exactly two true litterals per clause, and the truth values of $$x$$, $$y$$ and $$z$$ are unchanged.
• Conversely, if $$\phi_C$$ is satisfied with exactly two true litterals per clause, then exactly two among $$a$$, $$c$$ and $$e$$ are true, say $$a$$ and $$c$$. That means that $$z$$ and $$f$$ are true, so $$C$$ is satisfiable. Other cases are similar.
The rest of the reduction is similar to the problem above.
• The input doesn't have to be a 3CNF. Commented Dec 8, 2022 at 16:38
• @YuvalFilmus Indeed, you are right! Commented Dec 8, 2022 at 16:52
• "if two litterals are true, say x and y WLOG, assign a,c,e,f to true and b,d to false;" is incorrect, because $a,c,e$ cannot all be true. I suppose, it is $a,d,e,f$ to true and $b,c$ to false.
– sas
Commented Mar 18 at 12:08
• @sas Thank you! Commented Mar 18 at 12:48 | 1,388 | 4,421 | {"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": 56, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.859375 | 4 | CC-MAIN-2024-26 | latest | en | 0.932635 |
https://codeforces.com/blog/entry/89080 | 1,657,047,270,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656104597905.85/warc/CC-MAIN-20220705174927-20220705204927-00052.warc.gz | 219,640,734 | 18,475 | ### coder_of_india.'s blog
By coder_of_india., history, 15 months ago,
*****************************************************TLE CODE *************************************************************
int dx[4] = {0 , -1 , 0 , 1} ; . int dy[4] = {-1 , 0 , 1 , 0} ;
bool checker (vector<vector> grid , int i , int j , int n , int m )
{
if(i >= 0 && i < n && j >= 0 && j < m && ( grid[i][j] == '1'))
return true;
return false;
}
void dfs(vector<vector> &grid , int i , int j ,int r , int c) {
grid[i][j] = '0';
for(int k = 0 ; k < 4 ; k ++ )
{
int newx = i + dx[k] ;
int newy = j + dy[k] ;
if( checker(grid , newx , newy , r , c) )
{
grid[newx][newy] = '0';
dfs(grid , newx , newy , r , c);
}
}
} class Solution { public: int numIslands(vector<vector>& grid) {
int r = grid.size() ;
int c = grid[0].size() ;
int cnt = 0 ;
for(int i = 0 ; i < r ; i ++ )
{
for(int j = 0 ; j < c ; j ++ )
{
if(grid[i][j] == '1')
{
cnt ++ ;
dfs(grid , i , j ,r , c);
}
}
}
return cnt ;
}
};
************************************************** AC CODE *******************************************************************
int dx[4] = {0 , -1 , 0 , 1} ; int dy[4] = {-1 , 0 , 1 , 0} ;
void dfs(vector<vector> &grid , int i , int j ,int r , int c) {
grid[i][j] = '0';
for(int k = 0 ; k < 4 ; k ++ )
{
int newx = i + dx[k] ;
int newy = j + dy[k] ;
if(newx >= 0 && newx < r && newy >= 0 && newy < c && ( grid[newx][newy] == '1'))
{
grid[newx][newy] = '0';
dfs(grid , newx , newy , r , c);
}
}
} class Solution { public: int numIslands(vector<vector>& grid) {
int r = grid.size() ;
int c = grid[0].size() ;
int cnt = 0 ;
for(int i = 0 ; i < r ; i ++ )
{
for(int j = 0 ; j < c ; j ++ )
{
if(grid[i][j] == '1')
{
cnt ++ ;
dfs(grid , i , j ,r , c);
}
}
}
return cnt ;
}
}; ******************************************************************************************************************************
Both codes are same but there is one difference I used cheacker function that's why i am getting tle why this is happening ? please tell me anyone.
» 15 months ago, # | +5 1) Plz, use spoiler tag or smth to more readable code in blogs. bool readable = false; 2) Your checker function copying vector. You need to use reference to vector instead of copy; bool checker(vector& grid, int i, int j, int n, int m) | 721 | 2,289 | {"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} | 2.921875 | 3 | CC-MAIN-2022-27 | latest | en | 0.252623 |
https://www.atozexams.com/mcq/quantitative-aptitude/54.html | 1,720,957,318,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514564.41/warc/CC-MAIN-20240714094004-20240714124004-00694.warc.gz | 590,448,761 | 10,594 | # If both the pipes are opened, how many hours will be taken to fill the tank I. The capacity of the tank is 400 litres. II. The pipe A fills the tank in 4 hours. III. The pipe B fills the tank in 6 hours.
1. Only I and II
2. Only II and III
3. All I, II and III
4. Any two of the three
4
Only II and III
Explanation :
No Explanation available for this question
# One ship goes along the stream direction 28 km and in opposite direction 13 km in 5 hrs for each direction.What is the velocity of stream
1. 1.5 kmph
2. 2.5 kmph
3. 1.8 kmph
4. 2 kmph
4
1.5 kmph
Explanation :
No Explanation available for this question
# In a 200 metres race A beats B by 35 m or 7 seconds. A's time over the course is:
1. 40 sec
2. 47 sec
3. 33 sec
4. None of these
4
33 sec
Explanation :
No Explanation available for this question
# A can have a piece of work done in 8 days, B can work three times faster than the A, C can work five times faster than A. How many days will they take to do the work together
1. 3 days
2. 8/9 days
3. 4 days
4. None of the above
4
8/9 days
Explanation :
No Explanation available for this question
# 7 Pink, 5 Black, 11 Yellow balls are there. Minimum no. atleast to get one black and yellow ball
1. 17
2. 13
3. 15
4. 19
4
17
Explanation :
No Explanation available for this question
# (1/10)18 – (1/10)20 =
1. 99/1020
2. 99/10
3. 0.9
4. none of these
4
99/1020
Explanation :
No Explanation available for this question
# Three friends divided some bullets equally. After all of them shot 4 bullets the total number of bullets remaining is equal to the bullets each had after division. Find the original number divided
1. 18
2. 20
3. 54
4. 8
4
18
Explanation :
No Explanation available for this question
# The simple interest on a sum of money is Rs. 50. What is the sum I. The interest rate is 10% p.a. II. The sum earned simple interest in 10 years.
1. I alone sufficient while II alone not sufficient to answer
2. II alone sufficient while I alone not sufficient to answer
3. Both I and II are necessary to answer
4. Both I and II are not sufficient to answer
4
Both I and II are necessary to answer
Explanation :
No Explanation available for this question
# A sum of Rs. 427 is to be divided among A, B and C in such a way that 3 times A’s share, 4 times B’s share and 7 times C’s share are all equal. The share of C is
1. Rs.84
2. Rs.76
3. Rs.98
4. RS.34
4
Rs.84
Explanation :
No Explanation available for this question
# A hall is 15 m long and 12 m broad. If the sum of the areas of the floor and the ceiling is equal to the sum of the areas of four walls, the volume of the hall is:
1. 720
2. 900
3. 1200
4. 1800
4
1200
Explanation :
No Explanation available for this question
## Interview Questions
MongoDB Java Script Node JS PHP JQuery Python | 883 | 2,888 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.59375 | 4 | CC-MAIN-2024-30 | latest | en | 0.77513 |
https://www.physicsforums.com/threads/modeling-boiling-in-a-closed-container-with-a-small-hole.802620/ | 1,531,830,415,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676589710.17/warc/CC-MAIN-20180717105812-20180717125812-00456.warc.gz | 951,221,325 | 14,757 | # Modeling boiling in a closed container with a small hole
Tags:
1. Mar 11, 2015
### Hang11
I'm trying to build a mathematical model of something like Heron's Aeolipile:
http://en.wikipedia.org/wiki/Aeolipile
I'd like to know, based on a known heat flux, the pressure and temperature attained in the container.
I assume as water boils, the control volume loses mass and energy, the pressure and temperature of the vapors will stabilize.
2. Mar 11, 2015
### Staff: Mentor
The exit velocity of the gas (and therefore mass flow) will have some relation to the pressure difference. Equilibrium happens where you can boil enough water (transfer enough heat) to maintain that mass flow. Temperature is given by the vapour pressure - I would expect no significant deviations from standard pressure and 100°C.
3. Mar 12, 2015
### Hang11
but what kind of pressure increase in the vessel and what steam mass outflow should I get once it reaches a steady state boil?
How would I calculate that?
4. Mar 13, 2015
### Staff: Mentor
You'll need the rate of heat flow to the container, everything else will follow from that. That rate will depend on your heating mechanism.
5. Mar 14, 2015
### Hang11
Assuming pool boiling, nucleate regime, the equation of the heat flux contains both the saturation temperature and the surface temperature. However, as more heat is pumped into the system (the exhaust is small), the Tsat changes. So , even knowing the heat flux, I still have two unknowns, the surface temperature and the saturation temperature.
6. Mar 14, 2015
### Staff: Mentor
If the approximation "all water is at 100°C" is not precise enough for your model then you'll have to model heat and gas flow in the water. Every new detail you add in the model gives the same number of unknown quantities as it gives equations to calculate those, as long as you include all relevant material properties and so on.
7. Mar 14, 2015
### Staff: Mentor
40 years ago, I derived a closed form solution for dP/dt, dm(water)/dt, and dm(steam)/dt in a setup very similar to this. It was great fun deriving it from mass balances of water and steam, and energy balances of water and steam, and the properties of water and steam. I used the result in simulators for nuclear power plants.
Give it a try. You'll succeed if you persevere. | 548 | 2,326 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5 | 4 | CC-MAIN-2018-30 | latest | en | 0.94417 |
https://math.answers.com/math-and-arithmetic/If_the_length_of_a_rectangle_is_six_times_the_width_and_area_is_37_square_meters_what_is_the_length_and_width. | 1,685,938,563,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224650620.66/warc/CC-MAIN-20230605021141-20230605051141-00682.warc.gz | 440,779,410 | 50,609 | 0
# If the length of a rectangle is six times the width and area is 37 square meters what is the length and width.?
Wiki User
2009-12-01 05:07:05
Length = 6* width
Area = 37
Area = length * width
Area = 6* width * width
37 = 6w^2
w = square root of 37/6
W = square root(37 / 6) = 2.4832774 meters
Wiki User
2009-12-01 05:07:05
Study guides
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https://primewriters.org/statisical-error/ | 1,669,890,066,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710808.72/warc/CC-MAIN-20221201085558-20221201115558-00745.warc.gz | 500,332,123 | 9,567 | # statisical error
1. The practice this week is about whether the relationship between two variables is significant.
Sydney is an HR rep who sits in on the promotion decisions for the company. She is concerned because it seems that people are getting salary hikes because their managers like them, not because they deserve it. She wants to know if there is a significant relationship between PercentSalaryHike and Years with Current Manager.
What is the correlation between PercentSalaryHike and Years with Current Manager (round to 2 decimal points ex. 1.22)?
2. Is the correlation between PercentSalaryHike and Years with Current Manager significant?
a. Yes, the p-value <.001
b. Yes, the p-value <.01
c. Yes, the p-value <.05
d. No
3. What can Sydney conclude from the data?
a. There is not evidence that the number of years someone is with their manager does not predict their salary hike.
b. Managers do not give salary hikes to employees they like more
c. Managers do give salary hikes to employees they like more
d. There is evidence that the number of years someone is with their manager does not predict their salary hike. | 250 | 1,144 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2022-49 | latest | en | 0.971028 |
https://www.teacherspayteachers.com/Store/Alyssa-Klier | 1,527,497,730,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794872114.89/warc/CC-MAIN-20180528072218-20180528092218-00420.warc.gz | 823,649,285 | 32,857 | # Alyssa Klier
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Balancing equations can be especially tricky for some kiddos, and these worksheets can help your students get the hang of it! In these 7 pages, you will find worksheets where students focus on balancing amounts ranging from up to 10, up to 15,
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Blast off with missing addends in your class using these pages for fun practice and review! This file includes: *Missing addend equations with facts up to 20 *"Space Buddies" missing addend story problems *Directions and student recording sheet
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The activities in this 20-page packet were designed to help students practice Common Core Standard: CCSS.Math.Content.1.NBT.C.5 Given a two-digit number, mentally find 10 more or 10 less than the number, without having to count; explain the
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Brush up on your primary missing addend skills! These fun, seasonal handouts are intended for math center/workshop time or as homework. There are 5 worksheets in all: *Missing addends with sums up to 10 *Missing addends in equations with 3
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** Due to the popularity of the 1st grade homework calendars, I am expanding and creating ones that align with 2nd grade Common Core Standards too! I hope you find these helpful! I will have most of them for the 2014-15 year uploaded by the end of
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Looking for a way to have kids practice their addition and subtraction strategies along with comparing quantities? There are two sets of puppy-themed pages here at varying levels where students are asked to compare the sums of addition or
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We all know that student achievement increases when kiddos are involved with tracking their own progress and taking ownership of their learning. If your district is into Robert Marzano like mine is, you hear that a lot! But how to help
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Just a basic little supplemental activity for your math instruction: 2 pages to go with CCSS.Math.Content.1.OA.D.7 Page 1: Students define and create a symbol for the words TRUE and FALSE Students review equations to determine if the answers are
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This is the weekly reading schedule that I use throughout most of the year in my first grade classroom. After reviewing the previous week's assessments, I group students according to what skill(s) they may need further instruction on before filling
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Get your kiddos up and moving while practicing their math skills! This is a classroom favorite and is also great for substitute teachers. My first graders can never get enough of it! This particular game focuses on missing addend equations with
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Work on Common Core math skills in a fun and festive way during the month of October! This mini packet includes: -Fact Families -Ordering Numbers -Comparing numbers (greater than/less than) -Filling in missing numbers -Missing addends -Decomposing
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Scoot is a fun-filled game that gets your kiddos up and moving around the room while reviewing important skills and concepts! My kids love it so much that the room bubbles over with excitement whenever they hear we're going to play it! It's easy to
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Looking for a fun, easy-to-prep way for your students to review how to mentally add and subtract 10 from one- and two-digit numbers? This mini packet should do the trick! Get kids up and moving around while they practice their skills with these
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I knew I was hooked when I first tried making memory books for my class, and I love what a special keepsake it is for them. But I noticed that some of my previous memory book papers could use a little sprucing up, so I created this year-long set
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In my grade level team, we use these monthly homework calendars not only to encourage students to practice their 1st grade skills at home, but also as a way of communicating with parents about the specific things we are looking for when it comes to
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In my class we have surprise, regular visits from the Desk Fairy who comes to see whose desks are the tidiest. This is the same concept, but with pencils. :) Both color and blackline master versions are included. When the Pencil Fairy (named
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This simple worksheet is a quick way to introduce missing addends or to review them and use as a formative assessment. There are 3 addends (one on each bowling pin) that need to create the total number (designated on the bowling balls). Totals
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This game is just like ZAP but with a weather-themed twist! While practicing their counting skills (from either 1-50, 50-100, 100-130, or the entire range of 1-130), students want to draw as many sunny cards from the pile as they can and determine
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Give a dog a bone! In this 6-page, dog-themed activity packet the following is included to help supplement and support student understanding of: Addition/subtraction facts to 10 and to 20 Comparing quantities in equations using the < > =
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In my grade level team, we use these monthly homework calendars not only to encourage students to practice their 1st grade skills at home, but also as a way of communicating with parents about the specific things we are looking for when it comes to
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I taught 3 wonderfully magical and busy years of Kindergarten before movin' on up to 1st grade! This will be my 5th year in 1st grade, and I truly believe it is the best place to be! I teach in a rural, agricultural-based town about an hour north of Seattle in an area with a high number of students who are English Language Learners and/or come from low-SES families. I wouldn't have it any other way. This community is great, the families are wonderful, and these kiddos are a blessing. Due to limited financial resources in my district and our recent shift to the CCSS, I have begun creating bits to help keep my teaching as current and focused as possible. Although I mostly come to TPT to dig into the treasure trove of fabulous lessons and activities from others, I thought I'd share what I do have so far. Thanks for checking me out!
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MY OWN EDUCATIONAL HISTORY | 1,726 | 6,982 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.859375 | 3 | CC-MAIN-2018-22 | longest | en | 0.93412 |
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## Explore printable Division as Repeated Subtraction worksheets for 6th Grade
Division as Repeated Subtraction worksheets for Grade 6 are an excellent resource for teachers looking to enhance their students' understanding of division concepts. These worksheets provide a variety of problems that focus on the concept of division as repeated subtraction, allowing students to practice and master this essential math skill. By incorporating these worksheets into their lesson plans, teachers can ensure that their Grade 6 students develop a strong foundation in division strategies. Furthermore, these worksheets are designed to be engaging and challenging, ensuring that students remain interested and motivated to learn. With a wide range of problems and scenarios, Division as Repeated Subtraction worksheets for Grade 6 are the perfect tool for teachers to help their students excel in math.
Quizizz is a fantastic platform that offers a plethora of resources for teachers, including Division as Repeated Subtraction worksheets for Grade 6. In addition to these worksheets, Quizizz also provides teachers with a variety of other tools and resources to help them create engaging and interactive lessons. Teachers can easily create quizzes, polls, and other activities that can be used in conjunction with the worksheets to reinforce division strategies and other math concepts. The platform also offers a vast library of pre-made quizzes and activities, covering a wide range of topics and grade levels. By utilizing Quizizz in their classrooms, teachers can ensure that their Grade 6 students receive a well-rounded and comprehensive education in math, division, and other essential subjects. | 600 | 2,521 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.96875 | 3 | CC-MAIN-2024-33 | latest | en | 0.935346 |
https://www.trumpingtonfederation.co.uk/page/?title=Week+Beginning+4th+May&pid=2359 | 1,643,368,794,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320305494.6/warc/CC-MAIN-20220128104113-20220128134113-00094.warc.gz | 1,059,576,812 | 16,656 | # Week Beginning 4th May
Hello Year 6!
We hope you had a good weekend and are rested and ready to get started on the work for this week. Hopefully you’ve all had a chance to listen to the books on the Read it Again competition page and vote for your favourite. If you haven’t yet, do that now! https://www.trumpingtonfederation.co.uk/page/?title=Read+it+Again+Competition&pid=2315
Even though we would have only been in school for four days this week, we have put up five activities for reading, writing and maths. If you decide with your adults that you are going to double up on activities on Thursday so that you are done by Friday then that is fine with us!
There is a word document attached at the bottom of this page which includes the instructions for all of the reading activities for the week.
Activity 1: Finding themes
Activity 2: Bag for Life
Activity 3: Author Hunt
Activity 4: Way to the stars
Activity 5: First News True/False Quiz
Maths activities
For maths this week, we are going to be using the Oak National Academy Lessons. The focus for this week is on missing angles and lengths. Click the link each day and it will take you to the menu of lessons. Scroll down to under the fractions lessons to find the angles lessons. You will need paper and pencil in front of you to work with while you do the online lesson.
Monday: Find missing angles
Tuesday: Compare and classify triangles
Thursday: Find unknown angles in triangles
Friday: Calculate unknown angles in regular polygons
Maths chilli challenges
There are five maths chilli challenge activities this week - see the attached word document at the bottom of the page.
Writing activities
Your writing activities this week have been filmed by teachers in school. There is one video for each day of the week. The instructions for what you need to do are in the video. The pictures are in a power point attachment at the bottom of the page.
Monday
Tuesday
Wednesday
Thursday
Friday
Other subjects
Science
https://museumofzoologyblog.com/2020/04/27/animal-lifecycles/
Scroll down to the ‘Mammals’ section and there are instructions for an investigation into how our proportions change as we grow. All you will need is a tape measure. This links to the work we did on evolution and inheritance.
Take part in Miss Chilcott’s Cambridge Walk Bingo (on the Home Learning Updates page)
Music
Learn the song of the week from the ‘SingUp’ website: https://www.singup.org/singupathome/song-of-the-week
The Royal Scottish National Orchestra can teach you how to play a Samba in your kitchen. Watch the first video on the website. https://www.rsno.org.uk/rsnochallenge/
Art/History
The Fitzwilliam Museum in Cambridge has developed a programme on its website called ‘Look, Think, Do’. You can choose which of the Fitzwilliam artefacts you want to focus on. https://www.fitzmuseum.cam.ac.uk/lookthinkdo/about
Art
Some of the Disney animators have been recording tutorials on how to draw popular Disney characters: https://www.youtube.com/watch?v=Lv5qFYJNTbs&list=PLflf8xbnbx65Z2oa2F9uqKZxJJn-4Nynr
Enter the John Lewis competition for designing a Super-Hero Bear. https://www.trumpingtonfederation.co.uk/blog/?pid=2220&nid=34&storyid=2171
P.S. Happy Birthday Marko! And belated birthday wishes to Callum, Emily, Georgia and Clara! | 789 | 3,324 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2022-05 | longest | en | 0.938432 |
https://in4tips.com/how-do-you-pass-a-signal-through-a-filter-in-matlab/ | 1,656,983,819,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656104506762.79/warc/CC-MAIN-20220704232527-20220705022527-00560.warc.gz | 375,300,116 | 15,083 | # How do you pass a signal through a filter in Matlab?
## How do you pass a signal through a filter in Matlab?
Once you have obtained the B and A vector coefficients you can filter your signal using the function filter. y=filter(x,A,B) x(is your input signal) and y your filtered signal.
## How do you remove noise from a speech signal in Matlab?
Listen to the audio file. %%2) Filter the audio sample data to remove noise from the signal. m1 = length(sample_data); % Original sample length.
## How do I create a filter in Matlab?
Designing the Filter
1. Start the app from the MATLAB® command line.
2. In the Response Type pane, select Bandpass.
3. In the Design Method pane, select IIR, and then select Butterworth from the selection list.
4. For the Filter Order, select Specify order, and then enter 6 .
5. Set the Frequency Specifications as follows:
## What is the disadvantage of moving average filter?
Disadvantages. Inflexible frequency response: nudging a conjugate zero pair results in non-unity coefficients. Poor lowpass filter (frequency domain): slow roll-off and terrible stopband attenuation characteristics.
## Why averaging is low pass filter?
When we say that a signal has high frequency components we mean that the values change rapidly with time. So x had rapid changes in amplitude, while y does not have that much of rapid changes in values. This is the intuition behind why averaging is equivalent to low-pass filtering (disallowing high frequencies).
median filter
Median filter
## Which is the better filter for removing salt and pepper noise?
Therefore, the AM filter can perform better than the standard median filter. The switching median (SM) filter [20] is a popular type of salt-and-pepper noise removing technique in recent years. The SM filter partitions the filtering process into two steps: noise detection and noise restoration.
## How we can remove salt and pepper noise?
It presents itself as sparsely occurring white and black pixels. An effective noise reduction method for this type of noise is a median filter or a morphological filter. For reducing either salt noise or pepper noise, but not both, a contraharmonic mean filter can be effective.
## How do you remove salt noise in Matlab?
Use the Median Filter block to eliminate the black and white speckles in the image. Use the default parameters. The Median Filter block replaces the central value of the 3-by-3 neighborhood with the median value of the neighborhood. This process removes the noise in the image.
## How do you add salt pepper noise in Matlab?
J = imnoise( I ,’salt & pepper’) adds salt and pepper noise, with default noise density 0.05. This affects approximately 5% of pixels. J = imnoise( I ,’salt & pepper’, d ) adds salt and pepper noise, where d is the noise density. This affects approximately d*numel(I) pixels.
## How do you introduce a sound in Matlab?
out = awgn( in , snr ) adds white Gaussian noise to the vector signal in . This syntax assumes that the power of in is 0 dBW. out = awgn( in , snr , signalpower ) accepts an input signal power value in dBW. To have the function measure the power of in before adding noise, specify signalpower as ‘measured’ .
## What does it mean to add noise to an image?
Image noise is random variation of brightness or color information in images, and is usually an aspect of electronic noise. Image noise is an undesirable by-product of image capture that obscures the desired information. | 753 | 3,478 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.65625 | 3 | CC-MAIN-2022-27 | latest | en | 0.88065 |
https://math.stackexchange.com/questions/3279922/confused-about-differentiating-negative-log-likelihood-function | 1,566,565,493,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027318421.65/warc/CC-MAIN-20190823130046-20190823152046-00398.warc.gz | 552,994,303 | 30,037 | # Confused about differentiating negative log likelihood function?
I was wondering if you could provide some clarifications regarding the derivation of the negative log likelihood function.
Let $$\ell := \frac{1}{N}\sum_{n=1}^{N}\left[-log(\sum_{n=1}^{N} \alpha(x)\;\mathcal{N}\left(\mu(x),\;\sigma(x))\right)\right]$$ be our mean negative log likelihood function where:
\begin{align*} x, y &= \mathbb{R}^{N} \\ \alpha(x) &= \frac{e^{f(x_{i})}}{\sum_{j=1}^{N}e^{f(x_{j})}} \\ \mu(x) &= f(x) \\ \sigma(x) &= e^{f(x)} \\ f(x) &= W_{\alpha,\mu,\sigma}x + b_{\alpha,\mu,\sigma}\;\mbox{some affine transformation} \end{align*}
Now the derivatives are accordingly:
\begin{align*} \gamma &= -\frac{1}{\sum_{n=1}^{N}\mathcal{N}(\mu(x),\;\sigma(x))}\alpha(x)\;\mathcal{N}(\mu(x),\;\sigma(x)) \\ \frac{\partial\ell}{\partial\mu} &= \gamma\frac{\mu-y}{N\sigma^{2}} \\ \frac{\partial\ell}{\partial\sigma} &= \gamma(\frac{1}{N} - \frac{(y-\mu)^{2}}{N\sigma^{2}}) \\ \frac{\partial\ell}{\alpha} &= \frac{\alpha - \gamma}{N} \end{align*}
Now my confusion starts with $$\gamma$$, I believe that the numerator should have been $$N\cdot\alpha(x)\cdot(1-\alpha(x))\cdot\mathcal{N}(\mu, \sigma)$$ ?
Am I missing something here?
According to my understanding $$\frac{1}{N}\sum_{n=1}^{N}-log(\phi(x_{n})) = -log(\phi(x_{n}))$$ and $$\sum_{n=1}^{N}\alpha(x)\mathcal{N}(\mu(x),\sigma(x)) = N\cdot\alpha(x)\cdot\mathcal{N}(\mu(x),\sigma(x))$$.
Then the derivative of $$\frac{\partial-log(\phi(x_{n}))}{\partial\phi} = \frac{1}{\phi(x_{n})}\phi\prime(x_{n})$$ if we let $$\phi(x_{n}) = \sum_{n=1}^{N}\alpha(x)\mathcal{N}(\mu(x),\sigma(x))$$.
Did I miss something in the derivation? I can't see clearly where does $$\gamma$$ come from if $$log(f(x)g(x)) = log(f(x)) + log(g(x))$$ then the derivative should be $$\frac{1}{f(x)}f\prime(x) + \frac{1}{g(x)}g\prime(x)$$ | 679 | 1,848 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 12, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.640625 | 4 | CC-MAIN-2019-35 | latest | en | 0.610056 |
https://artofsmart.com.au/physics/year-11-hsc-physics-module-2/ | 1,656,548,971,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103646990.40/warc/CC-MAIN-20220630001553-20220630031553-00159.warc.gz | 145,961,067 | 29,653 | BlogLearnGuide to Year 11 Physics Module 2: Dynamics
# Guide to Year 11 Physics Module 2: Dynamics
Just starting Year 11 Physics Syllabus Module 2: Dynamics?
As you settle in to the sometimes-wacky world that is thinking like a Physicist, our comprehensive guide to Year 11 Physics Module 2: Dynamics will help you become a Physics force to be reckoned with.
The new Module 1 Kinematics made a point of throwing students a bit in the deep-end, to get some heavy-lifting out of the way and important recurring skills nailed down early on.
If you need some help with Year 11 Physics Module 1: Kinematics, check out our guide here!
Breaking through to Module 2, you’ve already laid solid groundwork behind the skills/tools and logic of physics problems solving. Through Module 2, you’ll notice a familiar pattern developing in how to successfully attack problems!
In this article, we’ll cover how to prepare, use class time and smash through Year 11 Physics Module 2: Dynamics on the journey towards your Band 6!
### What is Dynamics all about?
Kinematics describes the states of motion we witness objects in. You developed a stack of tools for this, like vector addition of velocities and displacements.
Dynamics analyses the coming together of forces (push/pulls – also vectors).
Their combined action can work to bring about changes in motion; or conspire to make sure that motion stays unchanged (including constantly no motion, or any other motion status quo).
In other words, dynamics studies the reasons underneath all the types of motion (changing and unchanging) studied in Module 1: Kinematics.
This time, your toolbox is vector addition of forces (married to Newton’s 3 Laws).
You’ve already met acceleration (without its cause). You’ll be adding words like momentum, inertia and energy to this description of unbalanced force response.
### Breakdown of Physics Module 2: Dynamics
There are three main ‘Inquiry Questions’ for Year 11 Physics Module 2: Dynamics:
Inquiry Question 1: How are forces produced between objects and what effects do forces produce?
Inquiry Question 2: Inquiry question: How can the motion of objects be explained and analysed?
Inquiry Question 3: How is the motion of objects in a simple system dependent on the interaction between the objects?
Are you thinking this module sounds like the old module Moving About? It’s similar, but expanded upon.
Syllabus commands like “identify” and “describe” are out; things like “evaluate” and “’analyse” have taken over.
Your vector addition skills, practiced in Kinematics, are going to come out refined, but red-hot from overuse.
The module starts out by laying down Newton’s Laws (Inquiry Question 1). The rest of the module trains you to make skilled application of them to complex situations.
You’ll be analysing force vectors in real-world static and accelerated motion situations (in 2D), guided by Newton and solving for things like rope tensions and road frictions (Inquiry Question 2)
Later, you’ll look at how unbalanced force leads to energy and momentum changes (through velocity change); and the resulting details of collisions in 2D (momentum vectors) that fall out of Newton’s Laws (Inquiry Question 3).
This proper treatment of the vectors and maths of physics extends from Module 1 and gives a stronger foundation for Year 12 and university-level Physics/Maths/Engineering.
## How to get a Band 6 in Physics Module 2: Dynamics
Here’s a breakdown of the knowledge and understanding this module targets for mastery development:
Dynamics begs you to become expert in three areas:
1. Applying Newton’s Laws to Force vector addition in 2D
2. How net force changes energy (with displacement) and momentum (with time)
3. Applying Momentum Conservation (from Newton) to collisions in 2D (vector addition)
### Firstly, identify your misconceptions of Newton’s Laws and mark them for demolition.
Everyone brings aboard a suite of totally natural misconceptions, when they embark on Module 2. It’s probably *the* most misconception-blocked Physics topic.
You’ve had a lifetime of pushing objects around to arrive with assumptions that work well if you want to throw a ball to someone here on Earth.
But how do you know what effect your hand really has on the ball, if gravity and air are always getting in the way changing the results?
Here’s what research finds are the most common Newton’s Laws misconceptions:
Misconception: Sustained motion needs sustained force (implying motion “dissipates”)
Truth: Unbalanced force is needed to alter motion.
Misconception: There is no motion without force
Truth: There is no change in motion without unbalanced force.
Misconception: Equal and/or opposite reactions “cancel each other out”.
Truth: When A pushes B, B pushes back on A just as hard (something can only receive a push by pushing off something else – these are forces on different bodies always)
Misconception: Heavier objects fall faster.
Truth: Everything falls identically by gravity (hard-to-move objects are pulled harder).
Misconception: Inertia keeps things moving but doesn’t make it hard to start things moving.
Truth: Inertia (mass) measures resistance/sensitivity to any change in motion.
Unlearning these to make way for the truth will take some effort!
If you don’t have access to a lot of “frictionless/airless” environments (not many of us do…) to really see the force-machinery under the hood, there’s a tonne of interactive simulators online that let you push things around and watch the motion change!
A great site to experiment with is this one from PhET University of Colorado!
#### PhET Simulator – University of Colorado
Secondly, get used to your new world view being defined by Newton’s Laws.
This step’s a bit harder… You need to try and develop a “feel” or intuition for Newton’s Laws.
The best way to do this is to obsess over Newton’s Laws.
• When you’re walking; when you’re driving; when you’re playing sport – convince yourself shoes and tires are designed to maximally grip the Earth and throw it backwards.
• Convince yourself balls stop rolling because the ground pulls them back.
• Convince yourself plane rides feel identical to sitting motionlessly in a chair on the ground (no net force) except when the plane is changing its motion.
• Convince yourself rockets push off gases (and do so all the way up!).
• Convince yourself the cannon and cannonball are equally pushed apart, but the heavier cannon is less force-sensitive (higher mass).
• Convince yourself swinging a golf club exactly as hard at a bowling ball is less effective.
• Convince yourself gravity pulls heavier things harder to make them fall the same (how hard would you have to swing at the bowling ball to make it fly the same as a golfball?!).
• Convince yourself you need to touch the accelerator to cover up road friction and produce a no-net-force situation to maintain unchanging speed when you drive. Convince yourself rolling balls slow down because they can’t do exactly this!
View the world through the lens of Newton when you’re walking to school or doing anything – convince yourself of the reality of your replaced misconceptions.
##### Here’s your very own intuition checklist!
This checklist outlines the ideas and concepts to keep in mind when you’re thinking about Newton’s Laws and how they operate in reality:
1. Any force exerted on any object is always exerted by another object.
2. Whenever one object exerts a force on a second object, the second object exerts an equal force in the opposite direction on the first. (When something pushes something else, it gets pushed back just as hard.)
3. Action-reaction pairs always act on different bodies. They cannot cancel each other, as only forces acting on the same object can cancel.
4. The resulting accelerations of the objects (under equal force) is controlled by their masses (their inertia).
5. The resulting acceleration is proportional to the amount of unbalanced force acting, and the objects “lightness” (force sensitivity; the force is divided up over less mass)
### Step 2: Use Prescribed Strategy for Problem Solving
Firstly, start practicing Newton’s Law Force diagrams!
This is something you become skilled at with practice. Deliberately burn through as many practice questions as you can get your hands on.
Fall off the bike the exact required number of times that stand between you and that moment where you go “Woah, I’m doing it! I can do this!”
For everyone, in this topic, that number requires some determination!
To make this easier, we’ve outlined exactly how to do it below:
#### Use this one to solve Force situations:
1. Sketch the scenario in full.
2. Re-sketch objects individually (free-body diagram), showing all the forces acting on that object as labelled vectors (whose length and direction are as accurate as possible). Don’t show forces that the chosen object exerts on other objects to put force on itself (the full action-reaction partners)
Label vectors clearly identifying the actor and acted-upon e.g. for Force A exerts on B, write
1. is a vector sum; it’s usually helpful to resolve vectors into components and sum them along (x and y) axes. Choose one axis along the direction of acceleration (or zero acceleration). Pick a direction in this axis to call positive and stick to it.
2. For each object, write a true Net Force/Newton’s Law Statement in the x anddirections separately: i.e. and equal either max, may or zero. These components can reform a resultant ΣF vector later.
#### Use this one to solve Conserved Momentum (1D + 2D collision) situations:
1. Sketch the scenario in full
2. Decompose velocity vectors along axes of interest (x and y i.e. axes with no pre-collision momenta, and all of it)
3. Write a true Momentum Conservation statement along each axis separately. i.e. pfinalx = pinitalx and pfinaly = pinitaly
4. Bring your above x-momentum and y-momentum statements together to solve.
And practice, practice, practice until it starts becoming effortless! (Everyone has to concentrate hard when they first ride a bike – afterwards, it’s more automatic!)
### Step 3: Recast the role of class-time in your life
Because the syllabus is more internationally aligned, there are riches of resources out there already, beyond your prescribed text, that can be used to throw yourself into the content ahead of your class.
This recasts your class time in the role of a revision session with an expert consultant (teacher), ready to answer your pre-formed questions.
## How to Prepare for Year 11 Physics Module 2: Dynamics
### Step 1: Gather your resources
Use online learning to launch yourself at least a week ahead of your class and pre-learn the concepts. Bring already-identified sticking points to your class lessons. Be proactive!
Crunch through as many questions as it takes to feel confident (present challenging ones to your tutor and teacher to make the solving strategies known to you).
### Step 2: Rewrite your Newton’s Laws misconceptions
Make a concerted effort to roll back your earthbound preconceptions, to be replaced with the updated truth.
This is an important skill not only for Physics, but for all sciences and life in general…
Keep Newton’s Laws in the back of your mind outside of class and convince yourself of what forces really do – and that your rewritten conceptions are indeed truth. Develop a familiarity and feel for Newton’s Laws. Notice how they work as you move through everyday life.
### Step 3: Use your notes as a gap-identifying tool
Rewrite your class notes as if you had to explain the concepts to someone with no knowledge of Physics. Translate your notes from textbook-speak to familiar, everyday terms.
If you’re having trouble formulating a way to teach the concept to a non-scientist, you’ve just ID’d an area you don’t understand as deeply as you first thought! Run to your textbook/online to further develop a way to explain the underlying idea in simple terms!
### Step 4: Hone your problem-solving skills
Most of this course is mastered through polishing the unique, unfamiliar problem-solving skills Physics questions require. A whole host of solving strategies are not obvious, and just a matter of having seen them.
You’ll get accustomed to the common ways certain problems (like force diagrams, or collisions) are solved by making the strategies known to yourself (by failing to solve a question then seeking the solution!) and practicing their application.
Never feel incapable because the way was not immediately clear. Worked solutions will often need consulting first, as Physics is a wholly new way of thinking!
## On the hunt for other Year 11 Physics resources?
Check out the other module breakdowns we’ve created for Year 11 Physics below:
## Are you looking for some extra help with Year 11 Physics Module 2: Dynamics?
#### We pride ourselves on our inspirational Year 11 Physics coaches and mentors!
We offer tutoring and mentoring for Years K-12 in a large variety of subjects, with personalised lessons conducted one-on-one in your home or at our state of the art campus in Hornsby!
To find out more and get started with an inspirational tutor and mentor get in touch today!
Give us a ring on 1300 267 888, email us at [email protected] or check us out on Facebook!
Adrian Wendeborn is a qualified Science and Maths teacher with a Physics/Chemistry double-major degree from USYD and a GDipEd from UQ. Adrian has taught in QLD and NSW and has worked with Art of Smart Education as a campus teacher, tutor, resource developer and Head of Faculty.
## 45,861 students have a head start...
Get exclusive study content & advice from our team of experts delivered weekly to your inbox! | 2,925 | 13,759 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.15625 | 4 | CC-MAIN-2022-27 | latest | en | 0.870191 |
https://help.highbond.com/acl/11/topic/com.acl.language.help/lang_ref_functions/r_tan.html | 1,582,853,336,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875146940.95/warc/CC-MAIN-20200228012313-20200228042313-00461.warc.gz | 400,597,315 | 2,459 | # TAN( ) function
Returns the tangent of an angle expressed in radians, with a precision of 15 decimal places.
## Syntax
`TAN(radians)`
## Parameters
Numeric. The measurement of the angle in radians.
Numeric.
## Remarks
The three trigonometric functions in ACL – SIN( ), COS( ), and TAN( ) – support performing the Mantissa Arc Test associated with Benford’s Law.
If your input is in degrees you can use the PI( ) function to convert the input to radians: (degrees * PI( )/180) = radians. If required, you can round or truncate the return value using the DEC( ) function.
## Examples
Example Return value
TAN(0.785398163397448)
0.999999999999999
The return value is the tangent of the specified number of radians. The radians value is equivalent to 45 degrees.
TAN(45 * PI( )/180)
0.999999999999999
The return value is the tangent of 45 degrees.
DEC(TAN(45 * PI( )/180),3)
1.000
The return value is the tangent of 45 degrees, rounded to 3 decimal places.
Related reference
COS( ) function
PI( ) function
SIN( ) function | 284 | 1,040 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.828125 | 3 | CC-MAIN-2020-10 | latest | en | 0.602663 |
https://ch.mathworks.com/matlabcentral/answers/525134-grouping-matrix-data-into-variables-when-they-have-the-same-value-in-the-first-column?s_tid=prof_contriblnk | 1,723,077,510,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640713903.39/warc/CC-MAIN-20240808000606-20240808030606-00587.warc.gz | 123,571,337 | 27,362 | # Grouping matrix data into variables when they have the same value in the first column
5 views (last 30 days)
sgericeb on 13 May 2020
Commented: sgericeb on 13 May 2020
Apologies if this question has been answered somewhere but I have been checking for a while and couldn't find the answer to this specific problem.
I have a matrix of two columns i.e.
{A} {B}
1 22
1 19
1 15
5 12
5 23
5 45
5 1
5 4
6 12
6 12
6 23
6 9
10 2
10 3
10 29
Is there a way I can group the data in column B into variables depending on their value in column A? For example in column B I want 22, 19 and 15 grouped together into a variable, i.e. x, as they all have the same value of 1 in column A. I then want 12, 23, 45, 1 and 4 grouped into a variable, i.e. y, as they all have the same value of 5 in column 1 and so on.
However, the issue I have is that there are around 357 different values of A in Column A and I cannot write them all out manually into the code as it will take forever. Is there a way MATLAB can automatically create the 357 new variables I have shown above automatically by finding all the values in column B which have the same value of A and grouping them? I would then like to obtain the mean and sandard deviation for the data in each of the 357 new data sets.
I hope that is clear enough.
Any help would be greatly appreciated.
Stephen23 on 13 May 2020
KSSV on 13 May 2020
Let A, B be your first and second column.
[C,ia,ib] = unique(A) ;
N = length(C) ;
iwant = cell(N,1) ;
for i = 1:N
iwant{i} = B(ib==i) ;
end
KSSV on 13 May 2020
You see the dimensions of each cell in iwant will e different. You cannot put them into a matrix. If you want to put them into a matrix, you have to pad NaN or zeros to each cell and make a matrix, Is that okay for you?
sgericeb on 13 May 2020
Yes thanks, I undertand the padding. I'm just not sure how to do it for all 357 new column vectors yet to merge them into a new file | 565 | 1,913 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.171875 | 3 | CC-MAIN-2024-33 | latest | en | 0.934423 |
http://support.hfm.io/1.6/api/linear-1.20.7/Linear-Epsilon.html | 1,722,996,410,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640667712.32/warc/CC-MAIN-20240807015121-20240807045121-00535.warc.gz | 25,275,964 | 2,895 | linear-1.20.7: Linear Algebra
Copyright (C) 2012-2015 Edward Kmett BSD-style (see the file LICENSE) Edward Kmett provisional portable Safe Haskell98
Linear.Epsilon
Description
Testing for values "near" zero
Synopsis
# Documentation
class Num a => Epsilon a where #
Provides a fairly subjective test to see if a quantity is near zero.
>>> nearZero (1e-11 :: Double)
False
>>> nearZero (1e-17 :: Double)
True
>>> nearZero (1e-5 :: Float)
False
>>> nearZero (1e-7 :: Float)
True
Minimal complete definition
nearZero
Methods
nearZero :: a -> Bool #
Determine if a quantity is near zero.
Instances
# abs a <= 1e-12 Methods # abs a <= 1e-6 Methods # abs a <= 1e-6 Methods # abs a <= 1e-12 Methods Epsilon (V0 a) # MethodsnearZero :: V0 a -> Bool # Epsilon a => Epsilon (V1 a) # MethodsnearZero :: V1 a -> Bool # Epsilon a => Epsilon (V2 a) # MethodsnearZero :: V2 a -> Bool # Epsilon a => Epsilon (V3 a) # MethodsnearZero :: V3 a -> Bool # Epsilon a => Epsilon (V4 a) # MethodsnearZero :: V4 a -> Bool # Epsilon a => Epsilon (Plucker a) # MethodsnearZero :: Plucker a -> Bool # (RealFloat a, Epsilon a) => Epsilon (Quaternion a) # MethodsnearZero :: Quaternion a -> Bool # Epsilon (f a) => Epsilon (Point f a) # MethodsnearZero :: Point f a -> Bool # (Dim k n, Epsilon a) => Epsilon (V k n a) # MethodsnearZero :: V k n a -> Bool # | 423 | 1,346 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.796875 | 3 | CC-MAIN-2024-33 | latest | en | 0.666444 |
https://kmath.cn/math/detail.aspx?id=6277 | 1,685,257,959,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224643585.23/warc/CC-MAIN-20230528051321-20230528081321-00520.warc.gz | 387,487,950 | 5,076 | \begin{aligned} & A^2=A A=\left(\begin{array}{lll} \lambda & 1 & 0 \\ 0 & \lambda & 1 \\ 0 & 0 & \lambda \end{array}\right)\left(\begin{array}{ccc} \lambda & 1 & 0 \\ 0 & \lambda & 1 \\ 0 & 0 & \lambda \end{array}\right)=\left(\begin{array}{ccc} \lambda^2 & 2 \lambda & 1 \\ 0 & \lambda^2 & 2 \lambda \\ 0 & 0 & \lambda^2 \end{array}\right) \\ & A^3=A^2 A=\left(\begin{array}{ccc} \lambda^2 & 2 \lambda & 1 \\ 0 & \lambda^2 & 2 \lambda \\ 0 & 0 & \lambda^2 \end{array}\right)\left(\begin{array}{ccc} \lambda & 1 & 0 \\ 0 & \lambda & 1 \\ 0 & 0 & \lambda \end{array}\right)=\left(\begin{array}{ccc} \lambda^3 & 3 \lambda^2 & 3 \lambda \\ 0 & \lambda^3 & 3 \lambda^2 \\ 0 & 0 & \lambda^3 \end{array}\right) \\ & A^4=A^3 A=\left(\begin{array}{ccc} \lambda^3 & 3 \lambda^2 & 3 \lambda \\ 0 & \lambda^3 & 3 \lambda^2 \\ 0 & 0 & \lambda^3 \end{array}\right)\left(\begin{array}{ccc} \lambda & 1 & 0 \\ 0 & \lambda & 1 \\ 0 & 0 & \lambda \end{array}\right)=\left(\begin{array}{ccc} \lambda^4 & 4 \lambda^3 & 6 \lambda^2 \\ 0 & \lambda^4 & 4 \lambda^3 \\ 0 & 0 & \lambda^4 \end{array}\right) \\ & \end{aligned}
$$\lambda, \lambda^2, \lambda^3, \lambda^4, 1,2 \lambda, 3 \lambda^2, 4 \lambda^3, 0,1,3 \lambda, 6 \lambda^2 .$$
$$\begin{gathered} \lambda, \lambda^2, \lambda^3, \lambda^4, \cdots, \lambda^n, 1,2 \lambda, 3 \lambda^2, 4 \lambda^3, \cdots, n \lambda^{n-1} \\ 0,1,3 \lambda, 6 \lambda^2, \cdots, \frac{n(n-1)}{2} \lambda^{n-2} \end{gathered}$$
$$A^k=\left(\begin{array}{ccc} \lambda^k & k \lambda^{k-1} & \frac{k(k-1)}{2} \lambda^{k-2} \\ 0 & \lambda^k & k \lambda^{k-1} \\ 0 & 0 & 0 \end{array}\right),$$
$$A^n=\left(\begin{array}{ccc} \lambda^n & n \lambda^{n-1} & \frac{1}{2} n(n-1) \lambda^{n-2} \\ 0 & \lambda^n & n \lambda^{n-1} \\ 0 & 0 & \lambda^n \end{array}\right) .$$
\begin{aligned} A^{n+1}=A^n A & =\left(\begin{array}{ccc} \lambda^n & n \lambda^{n-1} & \frac{n(n-1)}{2} \lambda^{n-2} \\ 0 & \lambda^n & n \lambda^{n-1} \\ 0 & 0 & \lambda^n \end{array}\right)\left(\begin{array}{lll} \lambda & 1 & 0 \\ 0 & \lambda & 1 \\ 0 & 0 & \lambda \end{array}\right) \\ & =\left(\begin{array}{ccc} \lambda^{n+1} & (n+1) \lambda^n & \frac{(n+1) n}{2} \lambda^{n-1} \\ 0 & \lambda^{n+1} & (n+1) \lambda^n \\ 0 & 0 & \lambda^{n+1} \end{array}\right) \end{aligned}
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①此题难易度如何 ②此题推荐度如何 确定 | 1,139 | 2,408 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.34375 | 4 | CC-MAIN-2023-23 | longest | en | 0.198145 |
https://blog.richmond.edu/physicsbunn/2016/01/31/electability-update/ | 1,675,746,670,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764500384.17/warc/CC-MAIN-20230207035749-20230207065749-00423.warc.gz | 155,646,992 | 14,810 | # Electability update
As I mentioned before, a fair amount of conversation about US presidential politics, especially at this time in the election cycle, is speculation about the “electability” of various candidates. If your views are aligned with one party or the other, so that you care more about which party wins than which individual wins, it’s natural to throw your support to the candidate you think is most electable. The problem is that you may not be very good at assessing electability.
I suggested that electability should be thought of as a conditional probability: given that candidate X secures his/her party’s nomination, how likely is the candidate to win the general election? The odds offered by the betting markets give assessments of the probabilities of nomination and of victory in the general election. By Bayes’s theorem, the ratio of the two is the electability.
Here’s an updated version of the table from my last post, giving the candidates’ probabilities:
PartyCandidateNomination ProbabilityElection ProbabilityElectability
DemocratClinton70.54463
DemocratSanders28.519.568
RepublicanBush8.53.541
RepublicanCruz13.55.40
RepublicanRubio32.51546
RepublicanTrump47.5
29.562
As before, these are numbers from PredictIt, which is a betting market where you can go wager real money.
If you use numbers from PredictWise, they look quite different:
PartyCandidateNomination ProbabilityElection ProbabilityElectability
DemocratClinton845363
DemocratSanders16850
RepublicanBush7343
RepublicanCruz8225
RepublicanRubio321341
RepublicanTrump511835
PredictWise aggregates information from various sources, including multiple betting markets as well as polling data. I don’t know which one is better. I do know that if you think PredictIt is wrong about any of these numbers, then you can go there and place a bet. Since PredictWise is an aggregate, there’s no correspondingly obvious way to make money off of it. If you do think the PredictWise numbers are way off, then it’s probably worth looking around at the various betting markets to see if there are bets you should be making: since PredictWise got its values in large part from these markets, there may be.
To me, the most interesting numbers are Trump’s. Many of my lefty friends are salivating over the prospect of his getting the nomination, because they think he’s unelectable. PredictIt disagrees, but PredictWise agrees. I don’t know what to make of that, but it remains true that, if you’re confident Trump is unelectable, you have a chance to make some money over on PredictIt.
My old friend John Stalker, who is an extremely smart guy, made a comment on my previous post that’s worth reading. He raises one technical issue and one broader issue.
The technical point is that whether you can make money off of these bets depends on the bid-ask spread (that is, the difference in prices to buy or sell contracts). That’s quite right. I would add that you should also consider the opportunity cost: if you make these bets, you’re tying up your money until August (for bets on the nomination) or November (for bets on the general election). In deciding whether a bet is worthwhile, you should compare it to whatever investment you would otherwise have made with that money.
John’s broader claim is that “electability” as that term is generally understood in this context means something different from the conditional probabilities I’m calculating:
I suspect that by the term “electability” most people mean the candidate’s chances of success in the general election assuming voters’ current perceptions of them remain unchanged, rather than their chances in a world where those views have changed enough for them to have won the primary.
You should read the rest yourself.
I think that I disagree, at least for the purposes that I’m primarily interested in. As I mentioned, I’m thinking about my friends who hope that Trump gets the nomination because it’ll sweep a Democrat into the White House. I think that they mean (or at least, they should mean) precisely the conditional probability I’ve calculated. I think that they’re claiming that a world in which Trump gets the nomination (with whatever other events or changes go along with that) is a world in which the Democrat wins the Presidency. That’s what my conditional probabilities are about.
But as I said, John’s an extremely smart guy, so maybe he’s right and I’m wrong.
### Ted Bunn
I am chair of the physics department at the University of Richmond. In addition to teaching a variety of undergraduate physics courses, I work on a variety of research projects in cosmology, the study of the origin, structure, and evolution of the Universe. University of Richmond undergraduates are involved in all aspects of this research. If you want to know more about my research, ask me!
## 8 thoughts on “Electability update”
1. The common perception, particularly among the Left, is that Trump is a buffoon. This is true. What is not true is that he is the worst of the bunch. He gets a lot of press because of his antics, but if you compare the positions, he is the leftmost of the Republican candidates. A very low bar, to be sure.
There is an episode of All In The Family (probably the best sitcom ever made) from the 1970s which revolves around the presidential election. The punch line of the entire episode was the Archie entered Ronald Reagan as a write-in vote. The laughter brought the house down. It was a joke. Then.
As such, it wouldn’t surprise me if Trump wins.
2. Timothy Savage says:
I’m with the Stalk on this one. I don’t think the betting markets are as good predictors in this case than more tried-and-true models of voter behavior. Now Trump is a very unconventional candidate, so you could argue that the old models don’t apply to him. But Stalk’s point about the ability to shift position to the general are well-taken. Trump has had very consistent positions on a number of positions long before he decided to enter politics. Moreover, his name recognition is 100% and he’s been in the public eye for decades, so most people already have well-developed views of him that are unlikely to change drastically based on the primary results.
So if Trump can’t appeal to a broader set of voters, his only chance of winning would be to change the makeup of the electorate. His base of voters is whites without a college education, who traditionally do not have high turnout numbers. Fivethirtyeight.com has a very nice tool where you can fiddle with turnout numbers to see how the election swings.
http://projects.fivethirtyeight.com/2016-swing-the-election/
Trump would need a significant increase in both the numbers of non-college educated whites voting AND the percentage of them voting Republican, and/or a significant drop in the number of blacks and Latinos to have a chance to win. The former would require the kind of retail, get-out-the-vote effort that Trump lacks (and which both Obama and Ted Cruz excel at). The latter seems unlikely given the strong emotions that Trump invokes, particularly among Hispanics.
That being said, I am certainly not one of those liberals who hopes for a Trump victory to propel a Democrat to the White House. The most important thing, in my view, is that both parties nominate people who, even if we disagree with their policies, could at least be counted on to not drive the country off a cliff. Trump, with his pre-WWII isolationist views, poses a risk that we cannot afford.
3. There are (at least) two different issues here. One is whether the futures markets are reliable indicators. The other is whether the conditional probabilities are the correct things to calculate to assess electability. John and I differ on the latter: even if the futures markets are reliable measures of some sort of probability, he doesn’t think that what I’m doing with them is the right thing.
On the former question, I’m not one of those people who believes the futures markets are infallible. I think that they’re one particular distillation of conventional wisdom, with all the associated advantages and drawbacks. I’m interested in them primarily because, unlike most polling data, they allow a calculation of those conditional probabilities.
The main thing I want is a rhetorical point I can make against those who hope Trump gets the nomination because they’re sure he’s unelectable. I claim that, whatever the flaws in the futures market, they provide this: to anyone who claims this, I can say, “Wanna bet?” If you’re confident that Trump is unelectable, then you can make money off of that conviction.
4. Timothy Savage says:
While I can’t claim to be as good as math as you, I still question the assumptions behind your conditional probability. The chances of any given candidate winning the presidency after getting nominated is strongly correlated to who their opponent is. The betting markets you’re citing don’t take that into account. Pollsters, on the other hand, do:
http://www.realclearpolitics.com/epolls/2016/president/2016_presidential_race.html
I think a more accurate conditional probability would be to calculate the chances of each individual matchup (Trump v. Clinton, Trump v. Sanders, Rubio v. Sanders, etc.) and then multiply that times each candidates chances of prevailing in that matchup, and aggregate the results. It would be a lot of work, though.
5. The opponent is certainly important.
I remember when Carter became president. I remember someone saying “Reagan would have carried Texas”, meaning that if Reagan, rather then Gerald Ford, had been the Republican candidate, then Carter would not have done as well.
While I can see where people are coming from when they say they would like Trump to be the Republican nominee because he is unelectable, thus ensuring a Democratic win, I don’t think that he is unelectable at all. When they were starting out, Schwarzenegger and Reagan were considered to be jokes by many.
Anyone who thinks that Trump is unelectable is vastly overestimating the US electorate.
6. Tim — You’re quite right that the probability of a candidate winning depends on the opponent, and that one way to calculate that probability would be a weighted average of the candidate beating each opponent:
P(Trump wins) = P(Trump beats Clinton)*P(Clinton is the opponent) + P(Trump beats Sanders)*P(Sanders is the opponent) + ….
But I don’t know of any good way to estimate all of those probabilities. There is polling data for various head-to-head matchups, but those don’t tell you about probabilities of various outcomes.
Fortunately, you don’t need all of those individual probabilities to get the overall probability of victory. To be precise, you can get the opinion of the bettors in the prediction markets about that probability, simply by asking them what odds they would be willing to accept for various bets. If a bettor is willing to bet that Trump will win the election at odds of 2:1, they’re saying that they believe that there’s a 1/3 chance that he’ll win. They’re making a claim about that complicated probability sum you mentioned, but you don’t need to know or think about all of those ingredients to understand the final probability.
Suppose that I say that the Red Sox have a 40% probability of winning next year’s World Series. You could write that probability as a messy sum: it’s
P(Red Sox beat Mets in WS)*P(Red Sox and Mets both make it to WS) + P(Red Sox beat Phillies in WS)*P(Red Sox and Phillies both make it to WS) + …
That’s all true, but I don’t need all that complication in order to talk about the probability that the Sox win the series.
The specific claim I’m making is this: if you believed (as of Monday) that Trump’s conditional probability of winning the election, given that he wins the nomination, is significantly different from 62%, then you could have made money off of that conviction by placing suitable bets at PredictIt at odds that are in your favor.
7. “The main thing I want is a rhetorical point I can make against those who hope Trump gets the nomination because they’re sure he’s unelectable.” That’s what I was afraid you were doing. I believe any attempt to do this using conditional probabilities is doomed.
The conditional probabilities reflect the betting markets’ best guess of future events in the world we really live in, one where primary voters and caucus participants choose their parties’ nominees. When I think about who I hope they nominate I am answering a question about a different world, one where I get to pick their nominees for them. In that calculation I will presumably view electability favourably for the party I hope wins in November and unfavourably for the party I hope loses.
In practice I’m with Tim on this one, and wouldn’t try to game the system by choosing someone I thought would make a terrible president, but that’s not really the issue here. Neither is the issue the prediction markets’ skill, or lack thereof, in assigning probabilities to future events. The issue is that when I’m deciding who I hope the parties nominate I’m imagining a world where I get to decide that, because otherwise the question doesn’t make much sense. That world is very different from a world where the candidates have to persuade millions of other people to vote for them, and only reach the general election if they have succeeded in doing that. It’s the latter world which the prediction markets are attempting to predict.
The standard political science model of elections is what in physics terms we might call a hidden variables theory. Candidates are described by some directly observable variables, like experience and ideological positions, and some hidden variables which represent those aspects of candidate quality which we can’t observe except through their effect on election outcomes. Every election provides additional data points for estimating those variables, via Bayes’ Theorem.
“Electability”, to the extent that it means anything, is a proxy for the candidate quality variables, both the directly measurable ones like experience, and these hidden variables. For most of these candidates we have very few measurements. Clinton has contested a general election for the Senate and a presidential primary. Trump has never run for anything. The information we get from the outcome of this years’ primaries is therefore quite valuable, particularly if that outcome is unexpected.
My estimate of Carson’s electability, for example, will change enormously if he wins the nomination. When the prediction markets evaluate his odds in the general election what they are doing, or at least should be doing, is estimating that probability based on what they know now plus an imagined additional measurement: his presumed Republican primary win. When I say today that I don’t think he’s very electable I’m basing that on our present state of knowledge, where she’s never won an election for anything and doesn’t seem likely to.
In short, I don’t believe one can freely mix the language of hope with that of probability theory and expect all the resulting statements to be meaningful.
8. “When I think about who I hope they nominate I am answering a question about a different world, one where I get to pick their nominees for them.”
Then I agree that the conditional probabilities I calculate are not right for you. But I still think that they’re right for a certain category of people. When lefties I know say that they hope Trump is the nominee because the Democrats will beat him, I don’t think they’re saying, “I wish that the world were a different sort of place, in which Trump wins the nomination.” They’re saying, “On averaging over all the unknown possible ways the world might be (with their associated probabilities), the outcomes in which Trump is the nominee are better for my party than the outcomes in which Trump is not the nominee.”
Of course, I don’t know for sure that that’s what they’re thinking, but I think it is. If they’re thinking the way you suggest, then I agree that you’re right.
To be more specific, for someone who is thinking of voting strategically in the opposing party’s primary (e.g., voting for Trump to help the Democrats), my model is closer to the right one than yours. That person is considering what marginal change they should be trying to effect in the world as it actually is, not imagining a world in which they got to choose the nominee.
Like you, I wouldn’t vote strategically in this way, for at least two reasons:
1. I don’t trust either the prediction markets or my own instincts well enough to be sure I’d do it right.
2. Such an action doesn’t feel ethical to me. Casting a vote for, e.g., Donald Trump is, in my opinion, an unethical act for reasons that are not merely utilitarian. | 3,492 | 16,803 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.6875 | 3 | CC-MAIN-2023-06 | latest | en | 0.92703 |
blog.bedfordteam.com | 1,368,984,165,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368697843948/warc/CC-MAIN-20130516095043-00042-ip-10-60-113-184.ec2.internal.warc.gz | 27,960,975 | 17,108 | # Estate Planning & Cost Segregation
Contrary to popular misconception, the concepts of estate planning and cost segregation are not mutually exclusive. In fact, a cost segregation study can enhance the estate planning process by lowering the tax burden of the property owner. Rather than trying to describe this, we will use an example to explain how they work together.
In the example, we assume that a husband and wife own property jointly. They acquired this property by a purchase for \$10,000,000 10 years ago and have been depreciating this property under MACRS 39-year class. At this point in time, a cost segregation study is performed for adoption in 2011 with the following results.
### Example
5-year property equals 18% or \$1,800,000.15-year property equals 15% or \$1,500,000.
39-year property equals 67% or \$6,700,000.
The basis property is \$5,000,000 for the husband and \$5,000,000 for the wife.
The husband dies in 2013 and a valuation is done on the property, as is required by law. The new valuation has the building at \$12,000,000.
The basis of the husband is stepped up to \$6,000,000 as it passes through the estate to the wife. Therefore, even though the valuation is for \$12,000,000, the depreciable basis is only \$11,000,000. The placed in service date remains the same for the wife’s original assets, but a new placed in service date is used for the portion that passes through the estate.
At this point, a second cost segregation study can be performed as the property has been revalued. The results are shown in the table below on the line for 2013.
Let us further assume that the wife dies in 2017. The valuation placed a new value on the property of \$15,000,000. As the entire property has gone through the estate, the valuation and the depreciable basis are the same and a third cost segregation study is done.
Please note that because the property went through the estate no depreciation recapture was required even though the owners were allowed to step up the basis of the property.
*Years where a cost segregation is performed.
The above discusses just one example of how cost segregation can be used in the estate planning process. Of course, there are other applications, such as the issues associated with partnerships and the implications of §754.
### Bottom Line
The two key benefits of using cost segregation in the estate plan are the acceleration of depreciation and the ability to avoid potential recapture. Both have a significant and positive impact on cash flow.
## About Bedford Strategies and Solutions
Bedford is an independent professional services firm specializing in commercial real estate consulting since 2002.
Posted on September 26, 2011, in Commercial Real Estate, Cost Segregation, Estate Planning, Property Tax and tagged , , , . Bookmark the permalink. Leave a Comment. | 626 | 2,857 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.640625 | 3 | CC-MAIN-2013-20 | latest | en | 0.957276 |
https://doc.cgal.org/5.0/Kernel_23/classCGAL_1_1Ray__2.html | 1,718,474,321,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861605.77/warc/CC-MAIN-20240615155712-20240615185712-00806.warc.gz | 187,072,728 | 5,535 | CGAL 5.0 - 2D and 3D Linear Geometry Kernel
CGAL::Ray_2< Kernel > Class Template Reference
#include <CGAL/Ray_2.h>
## Definition
An object r of the data type Ray_2 is a directed straight ray in the two-dimensional Euclidean plane $$\E^2$$.
It starts in a point called the source of r and goes to infinity.
Is Model Of:
Kernel::Ray_2
## Creation
Ray_2 (const Point_2< Kernel > &p, const Point_2< Kernel > &q)
introduces a ray r with source p and passing through point q.
Ray_2 (const Point_2< Kernel > &p, const Direction_2< Kernel > &d)
introduces a ray r starting at source p with direction d.
Ray_2 (const Point_2< Kernel > &p, const Vector_2< Kernel > &v)
introduces a ray r starting at source p with the direction of v.
Ray_2 (const Point_2< Kernel > &p, const Line_2< Kernel > &l)
introduces a ray r starting at source p with the same direction as l.
## Operations
bool operator== (const Ray_2< Kernel > &h) const
Test for equality: two rays are equal, iff they have the same source and the same direction.
bool operator!= (const Ray_2< Kernel > &h) const
Test for inequality.
Point_2< Kernelsource () const
returns the source of r.
Point_2< Kernelpoint (const Kernel::FT i) const
returns a point on r. More...
Direction_2< Kerneldirection () const
returns the direction of r.
Vector_2< Kernelto_vector () const
returns a vector giving the direction of r.
Line_2< Kernelsupporting_line () const
returns the line supporting r which has the same direction.
Ray_2< Kernelopposite () const
returns the ray with the same source and the opposite direction.
## Predicates
bool is_degenerate () const
ray r is degenerate, if the source and the second defining point fall together (that is if the direction is degenerate).
bool is_horizontal () const
bool is_vertical () const
bool has_on (const Point_2< Kernel > &p) const
A point is on r, iff it is equal to the source of r, or if it is in the interior of r.
bool collinear_has_on (const Point_2< Kernel > &p) const
checks if point p is on r. More...
## Miscellaneous
Ray_2< Kerneltransform (const Aff_transformation_2< Kernel > &t) const
returns the ray obtained by applying t on the source and on the direction of r.
## ◆ collinear_has_on()
template<typename Kernel >
bool CGAL::Ray_2< Kernel >::collinear_has_on ( const Point_2< Kernel > & p ) const
checks if point p is on r.
This function is faster than function has_on() if the precondition checking is disabled.
Precondition
p is on the supporting line of r.
## ◆ point()
template<typename Kernel >
Point_2 CGAL::Ray_2< Kernel >::point ( const Kernel::FT i ) const
returns a point on r.
point(0) is the source, point(i), with i>0, is different from the source.
Precondition
$$i \geq0$$. | 751 | 2,730 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.640625 | 3 | CC-MAIN-2024-26 | latest | en | 0.662026 |
http://lupus.sg/coffee-mate-hjdz/viewtopic.php?id=elasticity-of-demand-formula-calculus-4b9ad2 | 1,618,289,260,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038072082.26/warc/CC-MAIN-20210413031741-20210413061741-00072.warc.gz | 63,969,522 | 16,048 | # elasticity of demand formula calculus
When the price rises, quantity demanded falls for almost any good, but it falls more for some than for others. Gaining proficiency in managerial economics involves a lot of calculations. Income Elasticity of Demand Formula. Calculating the Price Elasticity of Demand. Read on to learn how to calculate the price elasticity of demand with the midpoint method! To do this, the change in demand is divided by the original demand and multiplied by 100. The point elasticity formula is only useful for data points close to each other in value. Section 2.10: Elasticity of Demand. ... How To Calculate Price Elasticity Of Demand. The formula for Elasticity measures how demand reacts to price changes. 4) Calculate the x-Intercept of the Demand Function Next, we can update the primary function to include the actual slope (instead of m). Category of goods based on their own price elasticity of demand. Ultimately, your goal is to determine how you can maximize your profits. We divide the change in quantity by initial quantity to calculate a percentage. By using point elasticity of demand, we can calculate the elasticity of demand using the below formula: One downside of the midpoint method is that the elasticity value loses its importance as both points become more separated. We ignore the negative or positive signs of the elasticity calculation results when classifying goods. Formula to Calculate Income Elasticity of Demand. In this article, we will look at the concept of elasticity of demand … In order to use this equation, we must have quantity alone on the left-hand side, and the right-hand side be some function of the other firm's price. Consider the price-demand equation given by p = 8/5 - 12,500/ x. Income Elasticity of Demand formula calculates the reflection of the consumer behavior or change in demand of the product because of change in the real income of the consumers those who purchase the product. The following equation is used to calculate the income elasticity demand of an object. Calculating Elasticity. Or will revenue increase because demand didn't drop very much? The formula for calculating elasticity is: $\displaystyle\text{Price Elasticity of Demand}=\frac{\text{percent change in quantity}}{\text{percent change in price}}$. Arc elasticity. For our examples of price elasticity of demand, we will use the price elasticity of demand formula. But what about revenue = price $$\times$$ quantity? That allows us to calculate the x-intercept (again, we don’t use the y-intercept because the axes are flipped) of the demand function by plugging in the values of one ordered pair and solving the resulting equation for b. This responsiveness can also be measured with elasticity by the income elasticity of demand. % change in qua n ti t y demanded % change in p r i c e. We can use this equation to calculate the effect of price changes on quantity demanded, and on therevenue received by firms before and after any price change.. For example, if the price of a daily newspaper increases from £1.00 to £1.20p, and the daily sales falls from 500,000 to 250,000, the PED will be: (In the case of the midpoint formula, the average of the two prices and quantities is used.) Exercise: Calculating the Price Elasticity of Demand. 1. If the price rises from $50 to$70, we divide 20/50 = 0.4 = 40%. The price elasticity of demand (which is often shortened to demand elasticity) is defined to be the percentage change in quantity demanded, q, divided by the percentage change in price, p. The formula for the demand elasticity (ǫ) is: ǫ = p q dq dp. We know that demand functions are decreasing, so when the price increases, the quantity demanded goes down. So, below is the formula for the Income Elasticity of Demand. So our elasticity of demand right over here is negative 1. The coefficient of price-elasticity of demand that is obtained at a point on the demand curve is called the point (price-) elasticity of demand, and it is given by the formula (2.1) or (2.2). This is because the formula uses the same base for both cases. Income elasticity of demand indicates whether a product is a normal good or an inferior good.When the quantity demanded of a product increases with … This means the particular prices and quantities don’t matter, and everything is treated as a percent change, as Grove City College accurately states.. When the price increases will revenue go down because the demand dropped so much? For example, if two goods A and B are consumed together i.e. The law of demand states that as the price of the commodity or the product increases, the demand for that product or the commodity will … So the absolute value of the elasticity of demand, right over here, is equal to 1. The advantage of the is Midpoint Method is that one obtains the same elasticity between two price points whether there is a price increase or decrease. Note that elasticity can also be expressed as . In other words, if the price increases by 1%, the demand will decrease by E%. So this right over here. Or it's absolute value is 1. Income elasticity of demand is the ratio of percentage change in quantity of a product demanded to percentage change in the income level of consumer. We can use two methods to calculate the elasticity of demand, point elasticity, and arc elasticity. Figure 2. Income elasticity of demand (e N D) In Topic 3 we also explained how goods can be normal or inferior depending on how a consumer responds to a change in income. Use this calculator to determine the elasticity of your product. All price elasticity of demand have a negative sign, so it’s easiest to think about elasticity in absolute value. This calculator uses the midpoint formula for the elasticity of demand. That is why some economists favor the approach of point elasticity. The formula for the price elasticity itself of demand is as follows: Own price elasticity of demand (OPE) =% Change in quantity demanded of Product X /% Change of price of Product X. Once points become too far apart, the arc elasticity formula is more accurate: . Price elasticity of demand formula is (% Change in Quantity Demanded / % Change in Price). Price Elasticity of Demand Example. Price Elasticity of Demand = 43.85% / 98%. The price elasticity of demand is a way of measuring the effect of changing price on an item, and the resulting total number of sales of the item. Note that the law of demand implies that dq/dp < 0, and so ǫ will be a negative number. How to calculate price elasticity of demand? You cannot calculate the point elastic directly because it produces bias. Widget Inc. decides to reduce the price of its product, Widget 1.0 from $100 to$75. Cross elasticity of demand is the ratio of percentage change in quantity demanded of a product to percentage change in price of a related product.. One of the determinants of demand for a good is the price of its related goods. In economics, elasticity is the measure of how much buyers and sellers respond to changes in market conditions. A good's price elasticity of demand is a measure of how sensitive the quantity demanded of it is to its price. Price elasticity of demand is a very useful concept because it shows how responsive quantity demanded is to a change in price. It is a measure of responsiveness of quantity demanded to changes in consumers income. Using this formula it is easy to show the following results. This Demonstration shows two ways to calculate the price elasticity of demand: the point elasticity formula and the arc elasticity formula. With income elasticity of demand… Point elasticity. Calculate E(p), the elasticity of demand, and use it to find the value of p for which E(p) = 1. 1/5 times negative 5 over 1-- it is negative 1. When the elasticity is less than 1, we say that demand is inelastic. A change in the price of a commodity affects its demand.We can find the elasticity of demand, or the degree of responsiveness of demand by comparing the percentage price changes with the quantities demanded. This price elasticity of demand calculator helps you to determine the price elasticity of demand using the midpoint elasticity formula. Calculating Price Elasticity of Demand. And this is just because 2 over 10 is the same thing as 1/5. The formula for elasticity of demand can be formulated two different ways depending on what is available to you at the time. Under point elasticity, you need a mathematical function (demand curve) to define the relationship between price and quantity demanded. This "midpoint" or "arc" elasticity formula is the version used in most introductory texts. The price elasticity gives the percentage change in quantity demanded when there is a one percent increase in price, holding everything else constant. Economists use the concept of price elasticity of demand to describe how the quantity demanded changes in response to a price change. Arc elasticity of demand (arc PED) is the value of PED over a range of prices, and can be calculated using the standard formula: More formally, we can say that PED is the ratio of the quantity demanded to the percentage change in price. To work out elasticity of demand, it is necessary to first calculate the percentage change in quantity demanded and a percentage change in price. That is the case in our demand equation of Q = 3000 - 4P + 5ln(P'). Thus we differentiate with respect to P' and get: Let’s look at the practical example mentioned earlier about cigarettes. Once you will have calculated its value, you can head straight to the optimum price calculator to deduce what price is the best for your product. Price elasticity of demand is a measurement that determines how demand for goods or services may change in response to … Price Elasticity of Demand = 0.45 Explanation of the Price Elasticity formula. Our equation is as follows: $\frac{\%\Delta Q}{\%\Delta Income}$ In this video, explore a simple way to calculate the price elasticity of demand, how to interpret that calculation, and how price elasticity of demand varies along a demand curve. Where IED is the income elasticity of demand Let’s calculate the elasticity from points B to A and from points G to H, shown in Figure 2, below. The same process is used to work out the percentage change in price. The company predicts that the sales of Widget 1.0 will increase from 10,000 units a month to 20,000 units a month. Formula to calculate elasticity. This is because the formula uses the same base for both cases. Let’s calculate the elasticity between points A and B and between points G and H shown in Figure 1. I ed = FD – ID / IF – II. Calculate the best price of your product based on the price elasticity of demand. The formula for calculating income elasticity of demand is the percent change in quantity demanded divided by the percent change in income. Demanded is to its price the percentage change in price positive signs of the midpoint formula the. Managerial economics involves a lot of calculations of calculations economics involves a lot of calculations of how sensitive the demanded. Shown in Figure 1 a change in quantity demanded falls for almost any,... Product based on the price elasticity of demand is divided by the income elasticity of demand over! This is just because 2 over 10 is the measure of how buyers... Use the price elasticity of demand formula is the measure of how sensitive the quantity demanded %... Price \ ( \times \ ) quantity for our examples of price elasticity of demand we ignore the negative elasticity of demand formula calculus! Each other in value midpoint formula, the change in quantity by initial quantity to calculate the best price your! Is ( % change in quantity demanded / elasticity of demand formula calculus change in quantity demanded to changes in response a... By the percent change in quantity by initial quantity to calculate the income of... 2, below 1, we say that demand functions are decreasing, so when the price elasticity the. A measure of how sensitive the quantity demanded is to a and B are consumed together i.e change quantity! Given by P = 8/5 - 12,500/ x elasticity demand of an object revenue because! 'S price elasticity of demand to describe how the quantity demanded changes in response to a change quantity. Arc elasticity formula is the same process is used to calculate the elasticity demand. Calculation results when classifying goods to H, shown in Figure 2, below is the measure responsiveness. Elasticity by the income elasticity of demand have a negative sign, so when price. = 3000 - 4P + 5ln ( P ' and get: price elasticity of demand, elasticity. Can use two methods to calculate a percentage most introductory texts price of its,. Than 1, we say that demand is inelastic differentiate with respect to P ). Accurate: of how much buyers and sellers respond to changes in consumers income and multiplied 100... ' ) in the case in our demand equation of Q = 3000 - 4P + 5ln ( P ). Close to each other in value in our demand equation of Q = 3000 4P. Because demand did n't drop very much useful for data points close to each other in value determine the of. ( P ' and get: price elasticity of demand decreasing, when... ( \times \ ) quantity demand did n't drop very much and H shown in Figure 2, is... Two prices and quantities is used to calculate the point elastic directly because it shows how quantity. Learn how to calculate the price elasticity of demand = 43.85 % / 98 % why some favor... Demanded when there is a one percent increase in price calculator to determine the elasticity calculation results when classifying.. Lot of calculations in economics, elasticity is the version used in most introductory.... I ed = FD – ID / if – II is why economists... Arc elasticity for the income elasticity of demand for both elasticity of demand formula calculus in absolute value ǫ will be a negative,... More accurate: elasticity formula is the case of the midpoint method learn how to calculate elasticity. Own price elasticity of demand is a very useful concept because it shows how responsive quantity.. The percent change in price = 0.45 Explanation of the elasticity of demand same is! Demand with the midpoint formula for the elasticity from points B to and. Company predicts that the sales of Widget 1.0 from $50 to$ 75 implies that dq/dp <,... Are decreasing, so when the price elasticity of demand can be formulated two different depending... Divide 20/50 = 0.4 = 40 % that dq/dp < 0, and so will. Of quantity demanded falls for almost any good, but it falls more for than! Have a negative number 10 is the same base for both cases in value! If – II % change in price this calculator uses the same thing as 1/5 predicts that the of. Uses the same process is used to calculate a percentage get: price elasticity of.... Calculating income elasticity of demand 1 -- it is negative 1 Widget decides... Is a one percent increase in price absolute value get: price elasticity of,... Negative sign, so it ’ s easiest to think about elasticity in absolute value in. Price and quantity demanded divided by the original demand and multiplied by 100 between price and demanded! What is available to you at the practical example mentioned earlier about cigarettes 20/50! Demanded falls for almost any good, but it falls more for some for... The best price of its product, Widget 1.0 will increase from 10,000 units a month to 20,000 units month! That the law of demand used in most introductory texts our examples of price elasticity demand! It produces bias the demand dropped so much - 4P + 5ln ( P ' and get: elasticity... Price, holding everything else constant quantity to calculate the price elasticity formula is only useful for points... Much buyers and sellers respond to changes in consumers income to calculate the price! How the quantity demanded changes in response to a price change uses the same base for cases... Between points a and from points G to H, shown in Figure 1 %! Law of demand with the midpoint method the two prices and quantities is used. what... Case in our demand equation of Q = 3000 - 4P + 5ln ( P ' and get price. Function ( demand curve ) to define the relationship between price and quantity demanded is to a from! Formula and the arc elasticity formula is more accurate: elasticity calculation when! Is a very useful concept because it shows how responsive quantity demanded of it is easy show. The midpoint formula for the income elasticity demand of an object demanded to changes in response to a change quantity. Ǫ will be a negative sign, so when the price increases will revenue go because. Differentiate with respect to P ' and get: price elasticity of demand, say. Negative sign, so it ’ s calculate the elasticity of demand = 43.85 % / 98.! \ ) quantity increase in price responsive quantity demanded to changes in market conditions price rises, quantity falls... Formula and the arc elasticity formula this midpoint '' or arc '' elasticity formula and the arc formula. Shows how responsive quantity demanded falls for almost any good, but it falls more for some than others... In economics, elasticity is less than 1, we will use the concept of price elasticity demand. Same process is used to calculate a percentage the average of the price increases will go... Demand can be formulated two different ways depending on what is available you., point elasticity formula is the case of the midpoint method economists the... What about revenue = price \ ( \times \ ) quantity in by! S easiest to think about elasticity in absolute value of the midpoint method 50 to $,! Falls for almost any good, but it falls more for some than for others the same thing as.! Market conditions ID / if – II elasticity gives the percentage change in price, holding else! Can also be measured with elasticity by the income elasticity of demand is a very useful concept because it how! We divide the change in price ) introductory texts points a and from points G H! Will decrease by E % change in quantity demanded when there is a measure of of. The original demand and multiplied by 100 the midpoint method demand is a one percent increase in price look. - 4P + 5ln ( P ' ) demand = 0.45 Explanation of the two prices and quantities used. ) to define the relationship between price and quantity demanded easiest to think about elasticity in absolute value.. Define the relationship between price and quantity demanded to changes in market conditions and get price. A good 's price elasticity of demand = 43.85 % / 98.. Points G to H, shown in Figure 1 how the quantity demanded of it is negative.! In managerial economics involves a lot of calculations demand to describe how the elasticity of demand formula calculus demanded / change... Same thing as 1/5 shows two ways to calculate the income elasticity of have! Company predicts that the law of demand very useful concept because it shows how responsive quantity of! In our demand equation of Q = 3000 - 4P + 5ln ( '. Proficiency in managerial economics involves a lot of calculations process is used to calculate the between! – II price )$ 70, we say that demand is inelastic have a negative.... Most introductory texts quantity demanded when there is a one percent increase in.... Learn how to calculate the elasticity between points a and from points B a. Introductory texts, if the price elasticity of your product based on the price,. Demand can be formulated two different ways depending on what is available to you at the practical example mentioned about. A mathematical function ( demand curve ) to define the relationship between price and demanded! Because 2 over 10 is the same base for both cases the price elasticity of demand in... The average of the midpoint method is equal to 1 negative number it falls more for some than others! Quantity by initial quantity to calculate the point elasticity, you need a mathematical (. How you can not calculate the price elasticity of demand can be formulated different!
free vector | 4,156 | 20,009 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.09375 | 4 | CC-MAIN-2021-17 | latest | en | 0.896712 |
http://mathhelpforum.com/discrete-math/165366-hand-shake-theorem.html | 1,529,875,013,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267867055.95/warc/CC-MAIN-20180624195735-20180624215735-00032.warc.gz | 195,873,197 | 11,274 | 1. ## hand shake theorem
Hi,
I was wondering if someone can show me an example similar to this one, so that I can figure this one out for myself?
The problem is:
In a group of 25 people, is it possible for each person to shake hands with exactly 3 other people? Explain
2. In a simple graph the number of odd vertices must be even.
3. If you consider the people as vertices of a graph, and these vertices are connected if the people shake hands. Then what you want to know is if it is possible to draw a graph with 25 vertices each of degree 3, you now use the handshaking lemma, does this help?
4. Originally Posted by hmmmm
If you consider the people as vertices of a graph, and these vertices are connected if the people shake hands. Then what you want to know is if it is possible to draw a graph with 25 vertices each of degree 3, you now use the handshaking lemma, does this help?
So are you saying if I can draw a graph with 25 vertices with 3 edges each, this problem is true if not this problem is false???
5. It is impossible to draw a simple graph of order 25 in which all vertices have degree 3.
6. Originally Posted by Plato
In a simple graph the number of odd vertices must be even.
Please explain this logic. I don't understand it. Ie: if we have 3 vertices, what has to be even??
7. Well yes but I am not saying that you do it I am then saying you should consider the handshaking lemma and draw a conclusion from that if you can or cannot draw the graph
The handshaking lemma: Sum of the degree's of all vertices = 2*number of edges
[Math]\sum\(d_{i}(V)=2*\#E[/tex]
8. It is even easier that that.
In a simple graph each edge has two vertices.
The sum of the degrees of the vertices must be an even number.
If we have 25 odd vertices that sum is 75 which is odd. Impossible.
9. Thanks Guys
10. yeah I was putting that because in the title it says handshaking so I thought it was probably look for the handshaking lemma and that the sum of the degree's of the vertices is even is a trivial consequence of the handshaking lemma
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# shakehand theorem
Click on a term to search for related topics. | 519 | 2,135 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.0625 | 4 | CC-MAIN-2018-26 | latest | en | 0.953252 |
https://byjusexamprep.com/How-to-Calculate-UPSC-Prelims-Total-Score-after-Negative-Marking-i-184fbbf0-3f7d-11e7-a52b-4a0c51f5de35 | 1,652,993,849,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662530066.45/warc/CC-MAIN-20220519204127-20220519234127-00642.warc.gz | 202,353,711 | 75,762 | # How to Calculate UPSC Prelims Total Score after Negative Marking
By Arpit Kumar Jain|Updated : May 23rd, 2017
It is essential that one knows about the marking pattern of UPSC IAS civil service exam. This help them in calculating their probable marks after the exam. As well as, before the exam it helps aspirants in getting an estimate about how many question to attempt to get certain marks.
In prelims, every question is of 2 marks. So, if a student answers a question correctly then 2 marks will be awarded to him. But there is also a penalty on wrong attempt. For every question marked wrong UPSC deduct one-third of the question’s mark. Since every question carries 2 marks, so penalty for every wrong answer will be 2/3 i.e. 0.67 marks.
So, suppose a student attempted 79 questions in the prelims exam. Now if he marks 64 question correctly but 15 question wrong. Then his expected score would be: 64(2) – 15(.67) = 128 – 10 = 118 marks.
Student should understand this method of calculating marks correctly. As this will help students in finding how much question they need to attempt to get a particular score.
For example, if a student aim to get 120 marks in UPSC IAS prelims then he can have several permutations of the question to be attempted to get those marks.:
• If he attempted 90 questions than he can afford to marks around 22 question wrong: 68(2) – 22(.67) = 121.33 marks.
• If he attempted 80 questions than he can afford to marks around 22 question wrong: 65(2) – 15(.67) = 120 marks.
• If he attempted 70 questions than he can afford to marks around 8 question wrong: 62(2) – 8(.67) = 118.67 marks.
It is generally seen that those students who don’t try to estimate, how many question they need to attempt to get a particular score, generally over attempts the number of question which led to large negative marking. | 451 | 1,849 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2022-21 | latest | en | 0.951653 |
https://myassignmenthelp.com/us/vt/cmda3605-mathematical-modeling/dynamical-system.html | 1,720,995,766,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514654.12/warc/CC-MAIN-20240714220017-20240715010017-00322.warc.gz | 359,452,578 | 19,525 | Get Instant Help From 5000+ Experts For
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MATLAB and Discrete-Time Dynamical Systems Homework
## Problem 1: Web network ranking using Google PageRank algorithm
I need the answers generated for this HW in a latex file so I will need the original .tex file and the pdf. It involves the use of MATLAB and code needs to be included in the report. For reference purposes, I have attached a previous HW and HW submission report. A lot of MATLAB files need to be used for this assignment therefore they are attached too.
1. (25 pts) (Modified from Leon.pdf) Consider a Web network of four sites linked together as shown below. If the Google PageRank algorithm is used to rank these pages, what is the transition matrix A. Assume that the Web surfer will follow a link on the current page 80 percent of the time. What will be the Google ranking of these pages?
2. (25 pts) In this problem we will modify our Hertz Car Rental Modeling example from Lecture 7 and model the resulting revised model as a non-homogeneous discrete-time dynamical system. Set-up is as before similar. Hertz has three locations around Detroit: City Airport (CA), Downtown (DT), and Metropolitan Airport (MA). But return patterns are different. It is observed that in a given month, 85% of the cars rented from CA are returned back to CA, 5% to DT, and 5% to MA. As you can see this does not add up to 100%.
This means that 5% of the cars rented from CA are returned to locations outside the larger Detroit area. Similarly, 90% of the cars rented from DT are returned back to DT, 1% to CA, and 5% to MA. And finally 85% of the cars rented from MA are returned back to MA 1% to DT, and another 9% to to CA. However, it was also observed that every month, a certain number of cars rented outside the Detroit area (rented in locations other than these three) are returned to these three Detroit Hertz locations. CA receives 10 such additional cars every month, DT receives 2, and MA receives 50.
(a) Let xca(k), xdt(k), and xma(k) denote the number of cars in CA, DT, and MA in
month t. Define the state-vector as x(k) =xca(k))xdt(k)
xma(k) and model this problem in
the form of nonlinear discrete-time dynamical systems x(k + 1) = Ax(k) + b. What is A and what is b? What do A and b represent?
(b) Assume that initially there were 500 cars in every location. Write a Matlab code to simulate the evolution of the number of cars in these three locations for 5 years. (be careful about the time unit in your model). Your Matlab code should take the initial condition provided in the question. Plot the evolution of the number of cars in every location using the subplot command. Make sure to label the all the axes 2 appropriately. The file lake−simul.m on Canvas is a good guide/help with this coding question.
## Problem 2: Hertz Car Rental modeling
(c) Repeat the previous part assuming initially there were 250 cars in CA, 250 cars in DT, and 1000 cars in MA.
(d) ) Parts (b) and (c) simulated the dynamics using different initial conditions. What do you observe about the steady-state behavior in each case? Is this what you have expected? Prove/show theoretically that this observed behavior is exactly what one should expect. Compute the steady-state values for the number of cars in every location. (You might end up with non-integer numbers; this is OK. Our models are
not exact.)
3. (25 pts) Contributions to Saving/Retirement Accounts: Assume that you have a monthly salary of S(k). For the sake of this question, let us assume that as opposed to annual raises, your employee gives smaller, but monthly raises. Your monthly raise rate is r%. Your employee offers a 401k Retirement Plan and contributes, every month, to your retirement account e% of your monthly salary. And your retirement plan gains, on average, g% per month, following the gains in the stock market. You are also allowed to make pre-taxed monthly contributions to your plan. Assume that you make monthly
contributions of \$m(k). Your initial retirement account has zero balance and your initial salary is \$S0.
(a) Let R(k) denote your monthly Retirement Account balance and let x(k) = R(k)S(k)
R
2 be the state-vector. Construct the discrete dynamical system x(k + 1) = Ax(k) + b(k), x(0) = x0, k = 0, 1, 2, . . . ,(months), where b(k), x0 ∈ R
and A ∈ R
2×2 that models the monthly evolution of R(k) and S(k) in terms of your monthly raise of r%, your employee’s monthly contribution rate e%, the monthly gain g%, and your monthly contribution of \$m(k). Note that this dynamical system has a “forcing term” b(k).
(b) Take S0 = 4, 000, m(k) = 1.005k × 50 (you start with an initial contribution of \$50 and increase it every month by 0.5%), r = 0.5%, e = 2%, and g = 1%. Assume that you work for 360 months (30 years). Write a Matlab script that simulates the evolution of R(k) and S(k) during your employment. How much savings will you have in the end? How much of this was your own contribution?
(c) You have decided that you can indeed start with an initial monthly contribution of \$100 instead, i.e., m(k) = 1.005k × 100. Repeat Part (a). How much savings do you have now after 30 years? You might think that “of course I have more in my savings, because I have contributed more”. You are right. But, that is not the
whole story. Compute the difference in your savings and compute the difference in your own contributions? Are they the same? How do you explain this?
4. (25 pts) Let λ be an eigenvalue of A with the corresponding eigenvector v.
(a) Let α be a scalar. Show that λ − α is an eigenvalue of A − αI with the same eigenvector v. Our eigenvalue proof for I − A in Lecture Notes 10 should help a lot for this question.
(b) Let µ be a scalar such that λ 6= µ. Show that 1 λ − µ is an eigenvalue of (A − µI)−1 with the same eigenvector v.
(c) Show that the eigenvalue λ is given by
λ =
v
TAv
vT v
Therefore, if we are given an eigenvector of a matrix, we can easily find the corresponding eigenvalue by this simple operation. Note that in this formula both v
TAv
and v
T v are scalars. | 1,543 | 6,224 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-30 | latest | en | 0.948076 |
http://steamexperiments.com/experiment/build-a-boat/ | 1,721,909,724,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763858305.84/warc/CC-MAIN-20240725114544-20240725144544-00691.warc.gz | 30,790,391 | 20,890 | Build a Boat WIP
Meta Description
Design your own boat. Investigate how different materials and shapes affect movement. Master the physics of flotation with this hands-on experiment.
Learning Objectives
Observe how the shape of a boat affects its movement and stability.
Awareness that different materials have different buoyancies.
Introduction of some basic terms related to ship structure.
Key Terms
Buoyancy
The ability of an object to float in a given fluid, or rise through it if it has been fully submerged.
Density
A property of a material, expressed as the mass per unit volume.
Hull
The body of a boat or ship. The hull is watertight.
Keel
The bottommost part of a ship’s structure. The hull is built around the keel. The keel runs along the centreline of the hull.
The law of floatation
When an object is freely floating, it displaces an amount of fluid which has the same weight as itself.
Upthrust
The upward force exerted by a fluid on an object which is partially or fully immersed in the fluid. This force is equal to the weight of the fluid displaced by the floating object.
Step 1
Familiarize yourself with the locations of the hull and keel on a boat by following the link.
Step 2
The first part of the experiment investigates the effect of hull shape. Ask an adult to help you cut out different full shapes from the balsa wood, cork sheet, styrofoam and cardboard. Basic shapes such as triangles and squares can be used. You can also use the shapes in the link, which are similar to those found in actual ships
Step 3
Cover the cardboard hull with duct tape to make it watertight.
Step 4
Fill the container (basin, aquarium or pool) with just enough water to carry out the experiment.
Step 5
Gently push each boat through the water and observe the boat’s speed, distance moved and stability. Record your results. The table below is an example results table which you could use.
Material Hull Shape Speed Mobility Stability Styrofoam Deep V High Easy to move Unstable
Step 6
Add a small weight to each of the boats and observe how this affects the motion of each. For a fair test, ensure that the same weight is added to each boat.
Step 7
The second part of the experiment investigates the effect of the keel. Cut two identical hull shapes from the same material.
Step 8
Cut out two different keels for the chosen hull shape. One of the keels should run along the whole length of the hull, while the other should be shorter, covering only a quarter of the length.
Step 9
Use the hot glue gun to attach the keels to the hulls.
Step 10
Test out each of the boats in the water and observe the effect of the keel on each boat’s movement by filling in a similar results table to before.
Step 11
Add the same small weight to the boats and observe how loading affects the motion of each. Again, ensure that the same weight is added to each of the boats.
1. Craft knives are very sharp. Only adults should use these knives. Replace the guard after every use and keep out of reach of young children.
2. Where possible, use recycled materials. For example, use a styrofoam block from a packaged appliance, cardboard from an old shoebox, scrap wood from the DIY store and/or collect rainwater for the experiment.
3. The hot glue gun should be used with the help of an adult and under strict supervision since the hot glue is a burn risk. Younger children should not be allowed to operate the device.
4. Use adequate protection to prevent the work surface from being damaged by the hot glue or the water.
5. To ensure a fair test, all boats should be loaded with the same weight.
6. If the experiment is being carried out by very young children, ensure that the weights used to load the boats do not pose a choking hazard.
Imagine you’re going rowing for a day at sea with your friend. But your friend is a bit anxious and has overpacked. They’ve got a thick, heavy book about sailing, a huge rucksack full of food and a great big sun umbrella. What do you think may happen once all those things are loaded onto the little boat and you both get in? Why does a boat sink when it’s loaded past a certain point?
A boat floats because it displaces water. This means it pushes water out of the way. In turn, the water produces an upward force on the boat, which keeps it afloat. This force is equal to the weight of the water displaced. The more weight placed in the boat, the lower it sinks in the water, the more water it displaces and the greater the force acting on it.
When the weight of the boat and added weight exceeds a certain value, the boat sinks. This is because the density of the boat has become greater than that of the water. What does this mean? When the sunken boat is submerged in the water, the volume of the water displaced by the boat is equal to the volume of the boat itself, right? However, the weight of this displaced water is less than the weight of the boat, meaning that the upward force on the boat is too weak to lift the boat upwards and keep it above the surface.
Why do ships made of steel float, even though steel is denser than water?
The boat is hollow. Average density of boat, air and cargo is less than that of water.
What do sails do?
The wind produces a force on the sail, which moves the ship.
Why do boats have different shapes?
Shape depends on load, required stability, sailing conditions and required speed.
What are the fastest types of boats?
Catamarans provide the best stability and some of the highest speeds.
Why do the boats sink with loading?
Average density of vessel and loading is greater than the density of water.
This experiment investigated three aspects of ship design: material, hull shape and the keel.
The material used to build the ship affects its buoyancy. The water exerts an upward force on the boat which is equal to the weight of the water the boat displaces. This force is called the upthrust force. A material floats in water because its density is less than or equal to the density of the water. All the boats floated, indicating that their overall density was less than that of water (1000 kg/m3). For a boat with the same hull and keel, the less dense the material, the faster it can skim the water, but it may be less stable since it has a lower inertia.
Hull shape affects maneuverability and speed. Round bottom boats maximize the amount of water displaced and provide a smooth ride, making them ideal for cruise ships and sea boats. Boats with Shallow V-shaped hulls move faster than those with round hulls because the shape increases the upward forces exerted by the water on the boat as it moves. While boats with round and deep V-shaped hulls cut through the water, flat bottom boats travel on it. This makes them popular for travelling in shallow waters but they are less comfortable in rough conditions.
The keel can affect speed and stability. A long keel along the length of the boat can give a comfortable ride and increase control over the ship’s motion. However, this slows the boat down. Boats with shorter keels may have less stable structures but have higher speeds and lighter steering.
The choice of material affects the maneuverability of the boat. According to the law of floatation, when an object floats freely in a fluid, the weight of the fluid displaced is equal to the weight of the object. In order for an object to float, its overall density must be less than that of water. Table 1 shows typical densities of the materials used and that of pure water.
Material Density (kg/m3) Water 1000 Balsa Wood 160 Cork 250 Styrofoam 14 Cardboard 700
The density of the material affects the overall mass of the boat. The heavier the boat, the greater it’s inertia and the less speed it will tend to gain compared to a lighter boat of the same design, provided with a similar initial push. A larger inertia, however, could make the boat more stable in choppy waters.
The shape of the hull affects maneuverability. Displacement hulls (round hulls) are only subjected to the upthrust force of the water, which keeps them afloat. V-shaped hulls exploit hydrodynamic lift, which enables them to achieve better speeds. Semi-planar hulls have a shallower V than planing hulls. Semi-planar hulls offer good mileage and a relatively comfortable ride, making them ideal for yachts and trawlers. Planing hulls (include flat bottomed and specially shaped deep v-shaped hulls) are built for power and speed but this can result in ‘slamming’, leading to an uncomfortable journey. They are designed to experience better lift than the shallower hulls previously mentioned. Planing hulls are usually used in small, recreational vessels.
Applications
Boats and ships are used in various industries. Cargo ships transport products across the globe. In fact, 40% of EU products travel by ship before reaching consumers. Navies use a range of different vessels, from aircraft carriers to tugboats. Boats are an indispensable tool to the fishing industry. Ships are also used for leisure and tourism, such as yachts and cruise liners. Sport keel boats are used in competitive sports. The design of the ship depends on the application. Designers must factor, amongst other things, the sailing conditions, loading, speed range and density of the material the boat will be made of.
Research
Shipping generates 4% of the EU’s greenhouse gas emissions. Current research has been focused on producing more efficient ships which would cost less to run and be more environmentally friendly. One study investigated how innovative hull design and propulsion mechanisms could improve efficiency. The study took ideas which worked in theory and tested them using advanced computer simulations. Questions the research explored included: Can these technologies be applied to full-sized ships? By how much can the efficiency be increased? How much would it cost? How cost effective are these technologies? What are the implementation problems? The simulations not only confirmed that the technologies hold promise, but are now helping researchers solve issues with implementing them on actual ships.
• Try adding sails to the boats and observe how they affect speed and stability. A mast could be made using a straw or ice cream sticks. The sails could be made using paper or lightweight material from old clothes. Testing the boats in a draft or on a breezy day may give the best results.
• Try loading each boat to its maximum capacity. To do this, keep adding weights to each of the boats until they just remain afloat before sinking. Use weighing scales to record the maximum load each boat could support. Compare the maximum loads to the material used, hull shape and boat size. What do you observe?
• Add rudders and tillers to the boat. What materials can be used for this purpose? Observe how motion can be altered and controlled using these structural features.
Subjects
Education
Time Required
• ~1 hour
• Preparation: 40 minutes
• Conducting: 30 minutes
• Clean Up: 10 minutes
Cost
Recommended Age
Number of People
Supervision
Materials
Balsa Wood
Cork sheet
Corrugated cardboard
Craft knife
Duct tape
Hot glue gun
Hot glue sticks
Large basin of water/paddling pool/ aquarium
Objects to serve as weights (eg small pebbles, clay, coins)
Paper
Pen
Styrofoam trays
Contributors
First published: September 30, 2017 | 2,450 | 11,359 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.359375 | 3 | CC-MAIN-2024-30 | latest | en | 0.924905 |
https://iitutor.com/vce-mathematics/ | 1,721,387,266,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514902.63/warc/CC-MAIN-20240719105029-20240719135029-00368.warc.gz | 263,602,782 | 27,577 | # Foundation Mathematics Units 1 and 2
Not Enrolled
25 Lessons
## VCE Foundation Mathematics Units 1 and 2 – Space and Shape
Geometry Circles Similarity Pythagoras Theorem
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69 Lessons
## VCE Foundation Mathematics Units 1 and 2 – Patterns and Number
Whole Numbers Fractions Decimals Integers Directed Numbers Percentages Ratios Rates Algebraic Expressions
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40 Lessons
## VCE Foundation Mathematics Units 1 and 2 – Data
Descriptive Statistics Representing Data Exploring Data
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42 Lessons
## VCE Foundation Mathematics Units 1 and 2 – Measurement
4.1 Time4.2 Metric Units4.3 Measurement4.4 Parameters4.5 Area4.6 Surface Area
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# General Mathematics Units 1 and 2
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26 Lessons
## VCE General Mathematics Units 1 and 2 – Algebra and Structure
Algebraic Expressions Basic Equations Linear Equations Simultaneous Equations Application of Linear Equations
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## VCE General Mathematics Units 1 and 2 – Arithmetic and Number
2.1 Directed Numbers2.2 Number Skills2.3 Ratios2.4 Decimals2.5 Percentages2.6 Interest Rates2.7 Spending Money
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Matrices Understanding Networks Number Sequences
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## VCE General Mathematics Units 1 and 2 – Geometry, Measurement and Trigonometry
4.1 Measurement4.2 Perimeter4.3 Area4.4 Surface Area4.5 Volume4.6 Similarity4.7 Pythagoras Theorem4.8 Trigonometry4.9 Non-right Angled Triangles
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## VCE General Mathematics Units 1 and 2 – Graphs of Linear and Non-Linear Relations
5.1 Motions in Straight Lines5.2 Kinematics5.3 Reciprocal Model5.4 Locus5.5 parameters5.6 Functions
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## VCE General Mathematics Units 1 and 2 – Statistics
6.1 Descriptive Statistics6.2 Representing Data6.3 Exploring data6.4 Linear Modelling
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# Mathematical Methods Units 1 and 2
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65 Lessons
## VCE Mathematical Methods Units 1 and 2 – Functions and Graphs
1.1 Coordinate Geometry1.2 Functions1.3 Polynomials1.4 Trigonometry1.5 Radian Measure1.6 Exponential Graphs1.7 Logarithmic Graphs1.8 Exponential Growth and Decay
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## VCE Mathematical Methods Units 1 and 2 – Algebra
2.1 Algebraic Expressions2.2 Parameters2.3 Transformations2.4 Polynomials2.5 Simultaneous Equations2.6 Inverse Functions2.7 Equations2.8 Exponents2.9 Logarithms
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4.1 Probability4.2 Counting Techniques
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Whole Numbers Number Skills Number Sequences Arithmetic Sequences and Series Geometric Sequences and Series Principal of Mathematical Induction Complex Numbers
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32 Lessons
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Set Theory Venn Diagrams Counting Techniques Descriptive Statistics
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Graph Theory
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Geometry Circle Geometry Similarity Logic Pythagoras Theorem Trigonometry Non-Right Angled Triangles Vectors
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13 Lessons
## VCE Specialist Mathematics Units 1 and 2 – Graphs of Linear and Non-Linear Relations
Motions in Straight Lines Graphs Applications of Differentiation
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37 Lessons
## VCE Specialist Mathematics Units 1 and 2 – Statistics
Descriptive Statistics Simulation Representing Data Exploring Data
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# Mathematical Methods Units 3 and 4
Not Enrolled
28 Lessons
## VCE Mathematical Methods Units 3 and 4 – Functions and Graphs
Graphs Exponential Graphs Logarithmic Graphs Trigonometric Graphs Transformations
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30 Lessons
## VCE Mathematical Methods Units 3 and 4 – Algebra
Polynomials Inverse Functions Inverse Trigonometric Functions Functions Trigonometric Graphs Simultaneous Equations
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39 Lessons
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Rates of Change Differentiation Application of Differentiation Integration Application of Integration
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18 Lessons
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Descriptive Statistics Discrete Random Variables Binomial Distribution Statistical Inference
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# Specialist Mathematics Units 3 and 4
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27 Lessons
## VCE Specialist Mathematics Units 3 and 4 – Functions and Graphs
1.1 Rational Functions1.2 Absolute Value Functions1.3 Reciprocal Trigonometric Functions1.4 Trigonometric Identities1.5 Inverse Trigonometric Functions
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2.1 Rational Functions2.2 Complex Numbers
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## VCE Specialist Mathematics Units 3 and 4 – Calculus
3.1 Inverse Trigonometric Functions3.2 Differentiation3.3 Integration3.4 Area by Integration3.5 Volume by Integration3.6 Differential Equations3.7 Kinematics3.8 Related Rates of Change
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4.1 Vectors4.2 Vector Applications
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5.1 Mechanics5.2 Motions in Straight Lines
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6.1 Expectation6.2 Sample Mean6.3 Discrete Random Variables6.4 Continuous Random Variables6.5 Hypothesis Testing
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Our online courses are meticulously crafted to adhere to the Victorian Curriculum and Assessment Authority outlines, ensuring a comprehensive and suitable learning experience. | 2,203 | 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} | 3.015625 | 3 | CC-MAIN-2024-30 | latest | en | 0.670665 |
http://www.subcommittee.com/phpBB3/viewtopic.php?f=21&t=3768 | 1,412,158,975,000,000,000 | text/html | crawl-data/CC-MAIN-2014-41/segments/1412037663417.12/warc/CC-MAIN-20140930004103-00476-ip-10-234-18-248.ec2.internal.warc.gz | 756,413,295 | 11,383 | ## Motor Theory - More Voltage?
R/C Submarine modelers
Okay, so I've forgotten some of the basics and here my confusion:
I have two DC motors of the same physical size. One operates on 6 volts the other on 12 volts. When I look at the current requirements, the 6 volt says it wants 3.5 amps and the 12 volt motor says it only wants .8 amps! Originally I thought if you double the voltage you half the current. This seems to be 1/4th the current or maybe the square root of the power. Can anyone explain?
Thanks, Ed
Just one more wire!
SC# 2268
WB6NSN
Ramius-II
SubCommittee Member
Posts: 394
Joined: Tue Apr 01, 2003 6:35 pm
Location: Lomita California
you are right about half the voltage, double the amps.
because volt x amp = watts (power).
But this goes for the same motor. and the trick
is to force the current through the motor.
Motors can differ, e.g. in wire thickness, number of
magnetic "spokes" on the axle etc.
Those variations cause the motors to have varying
resistance. worse, the magnetics cause the motor
to have varying resistance during revolution.
a stalled motor will be equivalent to the wire it is made of
(i,e, a short) while the motor can have a far higher
resistance while rotating, because of the creation and
destruction of magnetic fields.
Resistance is what limits the amount of current
through the motor for a given voltage.
up the voltage, and more current will flow,
giving more power output.
hope this helps !
Regards,
Ronald van Aalst
--------------------------------------------
Just here to Learn
raalst
SubCommittee Member
Posts: 1319
Joined: Wed Feb 26, 2003 8:46 am
Location: netherlands, the hague
Thanks Ron!
It just had me a bit lost. I was thinking in terms of Power= I2 X R.
Ed
Just one more wire!
SC# 2268
WB6NSN
Ramius-II
SubCommittee Member
Posts: 394
Joined: Tue Apr 01, 2003 6:35 pm
Location: Lomita California
Do trust yourself ! you are quite right
power = V x I
V = I x R
=> power = (I x R) x I
it is just that the R depends on the motortype
and also that large currents tend to lower V
(meaning that relative to the resistance of the
load/motor the resistance of the battery itself
becomes a factor.)
Regards,
Ronald van Aalst
--------------------------------------------
Just here to Learn
raalst
SubCommittee Member
Posts: 1319
Joined: Wed Feb 26, 2003 8:46 am
Location: netherlands, the hague
raalst wrote:you are right about half the voltage, double the amps.
because volt x amp = watts (power).
But this goes for the same motor. and the trick
is to force the current through the motor.
Motors can differ, e.g. in wire thickness, number of
magnetic "spokes" on the axle etc.
Those variations cause the motors to have varying
resistance. worse, the magnetics cause the motor
to have varying resistance during revolution.
a stalled motor will be equivalent to the wire it is made of
(i,e, a short) while the motor can have a far higher
resistance while rotating, because of the creation and
destruction of magnetic fields.
Resistance is what limits the amount of current
through the motor for a given voltage.
up the voltage, and more current will flow,
giving more power output.
hope this helps !
Hi,
Say we have a motor thats supposed to run on 8.4v. The standard battery pack is rated 8.4v - 600mah.
If we change this battery to 8.4v - 3000mah, will the motor's RPM increase and if why?
Thanks
bo@ti
Registered User
Posts: 29
Joined: Sat Jun 14, 2003 8:19 pm
Location: Norway
Hi,
Say we have a motor thats supposed to run on 8.4v. The standard battery pack is rated 8.4v - 600mah.
If we change this battery to 8.4v - 3000mah, will the motor's RPM increase and if why?
Thanks
Hi Bo@ti, No, the RPM should stay the same, unless the new battery had a lower internal resistace (V=IR for the battery as well), and allowed more current through at a given voltage. It the 3000mah battery is the same cell size as the 600mah, then its highly likely that it has a lower internal resisatance.
Anymore thoughts on resettable fuses?
Trafalgar
Thanx, for confirming what I thought.
No news on the fuses nope Maybee just bypass'em? I don't know
bo@ti
Registered User
Posts: 29
Joined: Sat Jun 14, 2003 8:19 pm
Location: Norway
"Say we have a motor thats supposed to run on 8.4v. The standard battery pack is rated 8.4v - 600mah.
If we change this battery to 8.4v - 3000mah, will the motor's RPM increase and if why?
"
With the example above, the only thing that will change is the amount of time between charges. The motor will continue to draw the same as it does on the 600mAh battery.
The current rating of a battery can be compared to a gas tank in your car. If you can go 350 miles on a 20 gallon gas tank (600mAh), you'll go 700 miles on a 40 gallon tank(1200mAh).
Skip Asay
SubTech
The U.S. of A - Land of the Free BECAUSE of the Brave
Skip Asay
Registered User
Posts: 255
Joined: Fri Feb 21, 2003 11:04 pm
Location: Sanford, NC
Can anybody, explain why the amp doesn`t have anything to do with the rpm?
Im on an Norwegian board, and 2 idiots keep saying that they have tested it and they know, by listening, or watching the motor, that amp DOES raise the rpm.
I even filmed the darn thing. But no response
Thanx
bo@ti
Registered User
Posts: 29
Joined: Sat Jun 14, 2003 8:19 pm
Location: Norway
Hi Bo:
The amps (current) does have an effect as the current is part of the overall power of the motor. If you connect your motor to a power supply with current limited output then yes, you can "limit" the power available to the motor. Turns out that flux or gauss is a fuction more of voltage than current. Therefore, the magnetic field of the coils in the motor are greater with a higher voltage that current. Basically, you can not force more current through a motor without raising the voltage. The two guys are probably raising the voltage and watching the amp meter go up at the same time.
Where most people get into trouble is by using a small wire size to their motors thus the wire acts like a resistor. The wire starts to get warm which tells you the power for the motor is being converted to heat before the motor can use it. Direct Current (D.C.) goes through the entire cross section of the wire and there are charts that say how much current a given wire size can handle. For main drive motors, I would not use less than 18 ga. wire and would lean more towards the 12-14 range.
Ed
Just one more wire!
SC# 2268
WB6NSN
Ramius-II
SubCommittee Member
Posts: 394
Joined: Tue Apr 01, 2003 6:35 pm
Location: Lomita California
Hi, Ed and Thanks.
bo@ti
Registered User
Posts: 29
Joined: Sat Jun 14, 2003 8:19 pm
Location: Norway
Welcome Bo:
By the way, I noticed that Engler now has a 12 volt version of the piston tanks! Maybe as a result of the design work I did? The Relay circuit can control either 6 or 12 volt motors and the "electronics" are fed from the receiver which is on 6 volts thus only the motors need to be changed! If the current draw is consistant with the other motors, that is 1/4 the amps, we now have a good extended run time.
Ed
Just one more wire!
SC# 2268
WB6NSN
Ramius-II
SubCommittee Member
Posts: 394
Joined: Tue Apr 01, 2003 6:35 pm
Location: Lomita California
Hi, yes I noticed
But I didden't look on the price tag
bo@ti
Registered User
Posts: 29
Joined: Sat Jun 14, 2003 8:19 pm
Location: Norway
Hey RamiusII and bo@ti, the running time of the typhoon is longer than 4 hours, do you really need more than that ? If you answer yes then you are a much more dedicated bubble head than I am
There are 10 kind of people, those who undertand binary and those who don't !
Sub-Systems
ThierryC
Registered User
Posts: 176
Joined: Fri Feb 21, 2003 3:52 am
Location: Perpignan, France
bo@ti wrote:Can anybody, explain why the amp doesn`t have anything to do with the rpm?
Im on an Norwegian board, and 2 idiots keep saying that they have tested it and they know, by listening, or watching the motor, that amp DOES raise the rpm.
I even filmed the darn thing. But no response
Thanx
If you connect a 8.4v Nicad to a motor it will draw more current than if you hook it up to 9v dry cells (or even 8.4v dry cell if it existed) . Similarly some Nicad's will provide more current than other Nicads.
Trafalgar
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### Which Letters Of The Alphabet Are Graphs of Polynomials?
This light exercise is to get your students thinking about polynomial graph behavior. You could extend this exercise to guess they degrees and signs of the leading coefficients if desired. Simply print out a worksheet of all the letters of the alphabet and ask your students which letters are graphs of polynomials based on the following distinctions
1. Polynomial graphs are continuous. You can draw them without lifting your pen
2. Polynomial graphs have no sharp corners or cusps, they are smooth (see pic below).
(From www.mathisfun.com)
See mathisfun.com for a great tutorial and pics.
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Different teachers have different teaching methods or techniques. Here i must say that your technique is also sound effective which i must try to use it. On the other hand, Some more teaching techniques i got from .http://www.idm-serial-number.com/2017/03/idm-keygen.html | 551 | 2,645 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2018-30 | latest | en | 0.944092 |
ERROR: type should be string, got "https://www.proz.com/kudoz/russian_to_english/science/70378-%FD%F4%F4%E5%EA%F2_%C4%E6%EE%E7%E5%F4%F1%EE%ED%E0.html" | 1,513,165,364,000,000,000 | text/html | crawl-data/CC-MAIN-2017-51/segments/1512948522999.27/warc/CC-MAIN-20171213104259-20171213124259-00397.warc.gz | 776,910,968 | 18,048 | # эффект Джозефсона
## English translation: the Josephson Effect
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GLOSSARY ENTRY (DERIVED FROM QUESTION BELOW)
Russian term or phrase: эффект Джозефсона English translation: the Josephson Effect Entered by:
00:49 Jul 27, 2001
Russian to English translations [PRO]
Science
Russian term or phrase: эффект Джозефсона протекание сверхпроводящего тока через тонкий (около 10 А) слой диэлектрика, разделяющий два сверхпроводника.
the Josephson Effect Explanation:Electromagnetic Metrology DC&LF: Resistance & Voltage Measurements Josephson Effect If a thin insulating barrier between two superconductors (a Josephson junction) is cooled in liquid helium to 4.2 K and exposed to electromagnetic radiation of frequency, f, then the DC voltage across the junction VJ assumes discrete values given by VJ = nfKJ where n is an integer which identifies a constant voltage step in a current-voltage characteristic and KJ is the Josephson constant. The Josephson constant is believed to be equal to the ratio h/2e where h is Planck’s constant and e is the electronic charge. In January 1990 a value of 483597.9 GHz/V was adopted for the Josephson constant in SI units with a relative 1s uncertainty of ± 0.4 ppm. Voltage standards based on arrays of Josephson junctions have become increasingly widespread in recent years. These arrays, which consist of several thousand individual junctions, can now be used to measure voltage standards directly at the 1 V and 10 V level.
Selected response from:
Vidmantas Stilius
Local time: 13:42
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Summary of answers provided
nathe Josephson Effect
Vidmantas Stilius
16 mins
the Josephson Effect
Explanation:
Electromagnetic Metrology
DC&LF: Resistance & Voltage Measurements
Josephson Effect
If a thin insulating barrier between two superconductors (a Josephson junction) is cooled in liquid helium to 4.2 K and exposed to electromagnetic radiation of frequency, f, then the DC voltage across the junction VJ assumes discrete values given by VJ = nfKJ where n is an integer which identifies a constant voltage step in a current-voltage characteristic and KJ is the Josephson constant. The Josephson constant is believed to be equal to the ratio h/2e where h is Planck’s constant and e is the electronic charge. In January 1990 a value of 483597.9 GHz/V was adopted for the Josephson constant in SI units with a relative 1s uncertainty of ± 0.4 ppm.
Voltage standards based on arrays of Josephson junctions have become increasingly widespread in recent years. These arrays, which consist of several thousand individual junctions, can now be used to measure voltage standards directly at the 1 V and 10 V level.
Reference: http://www.npl.co.uk/npl/cem/dclf/voltres/josephson_effect.h...
Vidmantas StiliusLocal time: 13:42Native speaker of: LithuanianPRO pts in pair: 118
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A model of an electric car can travel 400 miles before it needs to be recharged. What kind of quantity is the range of this model of car?
00:43
### Video Transcript
A model of an electric car can travel 400 miles before it needs to be recharged. What kind of quantity is the range of this model of car?
So we have a model car. And he can go some amount of miles before it needs to be charged. The car’s range is 400 miles. And the miles measure distance. The range of a model of car in this scenario is a measure of distance. | 139 | 600 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2021-25 | latest | en | 0.944807 |
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# P299 Physics - Physics 299 Important Constants and Equations Note Some basic simple equations you are supposed to know Here we are providing you
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Physics 299 Important Constants and Equations Note: Some basic simple equations you are supposed to know. Here we are providing you with only some important equations that were introduced in the chapters given below. Constants: g = 9.80 m/s mass of electron = 9.11 x 10 -31 Kg mass of proton = 1.67 x 10 -27 Kg μ o = 4π x10 -7 T.m / A k = 9 x 10 9 N . m 2 / C 2 ε o = 8.85 x 10 -12 C 2 / N.m 2 ; k = 1/ (4π ε o ) e = |1.60x 10 -19 | C; charge on electron or proton h = 6.63 x 10 -34 J.s; Planck’s constant 1 eV = 1.60x 10 -19 J Physics 298 information Roots: x = (-b ± (b 2 – 4 a c)) / (2 a) 1. A x = A cos α; A y = A sin α 2. α = tan -1 (A y / A x ) 3. A = (A x 2 + A y 2 ) Other important equations: 4. x = (v o + v) t/2 5. v = v o + a t 6. x = v o t + ½ a t 2 7. v 2 = v o 2 + 2 a x 8. K = ½ mv 2 9. w = F d cos ϑ 10. U = mgh 11. p = mv 12. v = r ω 14. F c = mv 2 / r; centripetal force Chapter 20 Electric Charge, Force, and Field F = k |q 1 q 2 | / r 2 ; Coulomb’s Law
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## This note was uploaded on 06/03/2010 for the course PHYS 299 taught by Professor Hoston,amahd,bakanowski during the Summer '08 term at University of Louisville.
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https://inches-to-meters.appspot.com/4630-inches-to-meters.html | 1,632,205,230,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780057158.19/warc/CC-MAIN-20210921041059-20210921071059-00036.warc.gz | 363,873,793 | 6,317 | Inches To Meters
# 4630 in to m4630 Inches to Meters
in
=
m
## How to convert 4630 inches to meters?
4630 in * 0.0254 m = 117.602 m 1 in
A common question is How many inch in 4630 meter? And the answer is 182283.464567 in in 4630 m. Likewise the question how many meter in 4630 inch has the answer of 117.602 m in 4630 in.
## How much are 4630 inches in meters?
4630 inches equal 117.602 meters (4630in = 117.602m). Converting 4630 in to m is easy. Simply use our calculator above, or apply the formula to change the length 4630 in to m.
## Convert 4630 in to common lengths
UnitUnit of length
Nanometer1.17602e+11 nm
Micrometer117602000.0 µm
Millimeter117602.0 mm
Centimeter11760.2 cm
Inch4630.0 in
Foot385.833333333 ft
Yard128.611111111 yd
Meter117.602 m
Kilometer0.117602 km
Mile0.0730744949 mi
Nautical mile0.0635 nmi
## What is 4630 inches in m?
To convert 4630 in to m multiply the length in inches by 0.0254. The 4630 in in m formula is [m] = 4630 * 0.0254. Thus, for 4630 inches in meter we get 117.602 m.
## Alternative spelling
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https://digoptionemyuc.netlify.app/rodenburg73982vym/how-to-calculate-gas-oil-ratio-nyx.html | 1,679,558,883,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296945030.59/warc/CC-MAIN-20230323065609-20230323095609-00560.warc.gz | 245,721,841 | 12,351 | ## How to calculate gas oil ratio
GOR (Gas-Oil Ratio) The gas-oil ratio is a measure of the volume of gas produced along with oil from the same well. How it’s important to us Gas-oil ratio is a measurement typically displayed along with other basic well metadata. It is a determining factor in whether the well is classified as a gas or oil well (as its primary hydrocarbon
GAS-OIL RATIO. CALCULATION. MISC. TILROA. Suy11. RAILROAD COMMISSION. OF TEXAS. James E. (Jim) Nugent, Chairman. Mack Wallace 10 May 2013 Gas-oil Ratio Calculation - Free download as PDF File (.pdf), Text File (.txt) or read online for free. Engineering. Gas:oil ratios (GOR) for source rocks with different kerogen types subjected to Volume of hydrocarbon gas (C1–C5) generated and calculated GOR for. 26 Sep 2017 Bubble point pressure and solution gas-oil ratio are closely linked. Several correlations are available to estimate these values. The same gas to oil ratio of less than 1750 cubic feet and API gravity less than 40 degrees). It is the default method required by KDHE for all facilities with flash emissions. Generally, PVT properties are determined either experimentally or calculated Another important factor comprising gas oil ratio (GOR), and mole % of C7+,
## 26 Sep 2017 Bubble point pressure and solution gas-oil ratio are closely linked. Several correlations are available to estimate these values. The same
22 Oct 2019 Q827. How do I determine the gas to oil ratio (GOR) for my oil well operations to decide whether I need to report emissions from completions The GOR of a reservoir fluid is a key parameter in determining the equation of state of a crude oil. Equations of state that are commonly used in petroleum 1 Oct 2014 The calculated pressure drops are presented in graphical format as a function of the oil stock tank volume flow rate and GOR. Variation of thermo begins with the material balance written as: Defining the ratio of the initial gas cap volume to the initial oil volume as: and plugging into the equation yields: Let:
### The gas gravity affects the calculations of gas viscosity, compressibility, compressibility factor, and solution gas-oil-ratio. Compressibility Factor. The gas
The fuel-oil mixture ratio for all Weed Eater products is 40:1. You can obtain this ratio by mixing 3.2 oz. of two cycle air cooled engine oil with one gallon of regular gas. Over a period of time oil will separate from gasoline. To properly insure oil and gas mixture, we recommend thoroughly shaking your oil and gas mixture before refueling your unit. GOR is the ratio of volumetric flow of produced gas to the volumetric flow of crude oil for crude oil and gas mixture sample. For any hydrocarbon mixture produced from an oil production well, the proportion of liquid and vapor phases in the mixture changes with changing temperature and pressure conditions. A properly calibrated injector pump injects just enough oil to provide proper engine lubrication. Now days the most common gas to oil ratio is 50:1 and 40:1. 50:1 is the most common, and it means there are 50 parts of gas to 1 part of oil. The operator must calculate how much oil to add to the fuel to obtain the correct gas to oil ratio. For 5 gallons of gas, with a 32:1 ratio, the amount of oil required is 20 US ounces (156.25 ml), 4 ounces per gallon of gas (31.25 ml per liter). Experts agree that it is better to err toward the side of too much oil, rather than not enough. Typical break in period ratios are about twice the normal; The ratio of produced gas to produced oil, commonly abbreviated GOR. See: produced fluid. 2. n. [Production] The ratio of the volume of gas that comes out of solution to the volume of oil at standard conditions. See: gas/liquid ratio (GLR), gas volume fraction (GVF) For most purposes, the solution GOR at the bubblepoint is the value of interest. At pressures above the bubble point pressure the oil is said to be undersaturated. Below the bubblepoint pressure, the gas begins to come out of solution and form a free gas phase, and the oil is said to be saturated. Easily calculate the appropriate premix for a 2 stroke engine with this gas oil mix ratio calculator. Calculate fuel mixtures 50:1, 40:1 and more.
### GAS-OIL RATIO. CALCULATION. MISC. TILROA. Suy11. RAILROAD COMMISSION. OF TEXAS. James E. (Jim) Nugent, Chairman. Mack Wallace
empirical correlations for calculating the reservoir fluid properties such as bubble point pressure, oil formation volume factor, solution gas oil ratio, gas formation 13 Aug 2009 In order to determine OWR, volume of oil/water/solid in drilling mud comes from a retort analysis. A sample The formulas below demonstrate how to calculate oil water ratio from retort data. a) % oil calculation gas /oil ratio. Gas/oil ratio calculator. Calculate the ratio of oil to gasoline. Liters Gallons (US) Gallons (UK). The volume of gasoline, liters. - +. Mixture Ratio. :1 - +. Calculate. Step 1: Begin by checking you are using the right measurement, as different countries use different units, i.e. Step 2: Mixing should take place in a fuel can, not inside the tank. Step 3: Once you have mixed the fuel, pour it into the tank and you're good to go. GOR is the ratio of volumetric flow of produced gas to the volumetric flow of crude oil for crude oil and gas mixture sample. The gas oil ratio is a measure of the volume of gas produced along with oil from the same well. Gas oil ratio is a measurement typically displayed along with other basic well metadata.
## The ratio of produced gas to produced oil, commonly abbreviated GOR. See: produced fluid. 2. n. [Production] The ratio of the volume of gas that comes out of solution to the volume of oil at standard conditions. See: gas/liquid ratio (GLR), gas volume fraction (GVF)
22 Oct 2019 Q827. How do I determine the gas to oil ratio (GOR) for my oil well operations to decide whether I need to report emissions from completions The GOR of a reservoir fluid is a key parameter in determining the equation of state of a crude oil. Equations of state that are commonly used in petroleum 1 Oct 2014 The calculated pressure drops are presented in graphical format as a function of the oil stock tank volume flow rate and GOR. Variation of thermo begins with the material balance written as: Defining the ratio of the initial gas cap volume to the initial oil volume as: and plugging into the equation yields: Let: 15 Dec 2019 (1) The following formula shall be used in the determination of the allowable of a gas well producing with a gas-oil ratio of 100,000 or more. contact the webmaster. Gas Oil Mixture Ratio Calculator. Enter the volume of Gas (petrol): U.S. Gallons | U.K. Gallons | Metric Liters. Enter the desired Ratio: :1
contact the webmaster. Gas Oil Mixture Ratio Calculator. Enter the volume of Gas (petrol): U.S. Gallons | U.K. Gallons | Metric Liters. Enter the desired Ratio: :1 | 1,608 | 6,909 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.234375 | 3 | CC-MAIN-2023-14 | latest | en | 0.920239 |
https://metanumbers.com/1534473 | 1,643,186,036,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320304928.27/warc/CC-MAIN-20220126071320-20220126101320-00020.warc.gz | 454,007,759 | 7,393 | 1534473 (number)
1,534,473 (one million five hundred thirty-four thousand four hundred seventy-three) is an odd seven-digits composite number following 1534472 and preceding 1534474. In scientific notation, it is written as 1.534473 × 106. The sum of its digits is 27. It has a total of 3 prime factors and 6 positive divisors. There are 1,022,976 positive integers (up to 1534473) that are relatively prime to 1534473.
Basic properties
• Is Prime? No
• Number parity Odd
• Number length 7
• Sum of Digits 27
• Digital Root 9
Name
Short name 1 million 534 thousand 473 one million five hundred thirty-four thousand four hundred seventy-three
Notation
Scientific notation 1.534473 × 106 1.534473 × 106
Prime Factorization of 1534473
Prime Factorization 32 × 170497
Composite number
Distinct Factors Total Factors Radical ω(n) 2 Total number of distinct prime factors Ω(n) 3 Total number of prime factors rad(n) 511491 Product of the distinct prime numbers λ(n) -1 Returns the parity of Ω(n), such that λ(n) = (-1)Ω(n) μ(n) 0 Returns: 1, if n has an even number of prime factors (and is square free) −1, if n has an odd number of prime factors (and is square free) 0, if n has a squared prime factor Λ(n) 0 Returns log(p) if n is a power pk of any prime p (for any k >= 1), else returns 0
The prime factorization of 1,534,473 is 32 × 170497. Since it has a total of 3 prime factors, 1,534,473 is a composite number.
Divisors of 1534473
6 divisors
Even divisors 0 6 4 2
Total Divisors Sum of Divisors Aliquot Sum τ(n) 6 Total number of the positive divisors of n σ(n) 2.21647e+06 Sum of all the positive divisors of n s(n) 682001 Sum of the proper positive divisors of n A(n) 369412 Returns the sum of divisors (σ(n)) divided by the total number of divisors (τ(n)) G(n) 1238.74 Returns the nth root of the product of n divisors H(n) 4.15382 Returns the total number of divisors (τ(n)) divided by the sum of the reciprocal of each divisors
The number 1,534,473 can be divided by 6 positive divisors (out of which 0 are even, and 6 are odd). The sum of these divisors (counting 1,534,473) is 2,216,474, the average is 3,694,12.,333.
Other Arithmetic Functions (n = 1534473)
1 φ(n) n
Euler Totient Carmichael Lambda Prime Pi φ(n) 1022976 Total number of positive integers not greater than n that are coprime to n λ(n) 170496 Smallest positive number such that aλ(n) ≡ 1 (mod n) for all a coprime to n π(n) ≈ 116308 Total number of primes less than or equal to n r2(n) 8 The number of ways n can be represented as the sum of 2 squares
There are 1,022,976 positive integers (less than 1,534,473) that are coprime with 1,534,473. And there are approximately 116,308 prime numbers less than or equal to 1,534,473.
Divisibility of 1534473
m n mod m 2 3 4 5 6 7 8 9 1 0 1 3 3 3 1 0
The number 1,534,473 is divisible by 3 and 9.
• Deficient
• Polite
Base conversion (1534473)
Base System Value
2 Binary 101110110101000001001
3 Ternary 2212221220100
4 Quaternary 11312220021
5 Quinary 343100343
6 Senary 52520013
8 Octal 5665011
10 Decimal 1534473
12 Duodecimal 620009
20 Vigesimal 9bg3d
36 Base36 ww09
Basic calculations (n = 1534473)
Multiplication
n×y
n×2 3068946 4603419 6137892 7672365
Division
n÷y
n÷2 767236 511491 383618 306895
Exponentiation
ny
n2 2354607387729 3613081462070681817 5544175950347985339777441 8507388303058324108284309223593
Nth Root
y√n
2√n 1238.74 115.342 35.1957 17.266
1534473 as geometric shapes
Circle
Diameter 3.06895e+06 9.64138e+06 7.39722e+12
Sphere
Volume 1.51344e+19 2.95889e+13 9.64138e+06
Square
Length = n
Perimeter 6.13789e+06 2.35461e+12 2.17007e+06
Cube
Length = n
Surface area 1.41276e+13 3.61308e+18 2.65779e+06
Equilateral Triangle
Length = n
Perimeter 4.60342e+06 1.01957e+12 1.32889e+06
Triangular Pyramid
Length = n
Surface area 4.0783e+12 4.25806e+17 1.25289e+06
Cryptographic Hash Functions
md5 9a4478f5bef499b95443b98fd38e8439 3a8d2d7712d422c6e1b3642e4d528548697bd771 0ac6ab7e0a9ef8fed0f686bfc232cae5eb161d8145348fdfe9afc13a2dd69f37 43219f4fa5c83febf6dfb52a9848eecff2d494a2d7eddd1d3b13db7fa5af075a0eda2f2e330858f61bbb765005e6ece6b9234b09367ad094c7de8dab1fa5a78a 1a25aae08ebc84547e9867cf3270734a1bf54485 | 1,503 | 4,200 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.59375 | 4 | CC-MAIN-2022-05 | latest | en | 0.824884 |
http://programmersheaven.com/discussion/176046/ReportPost.aspx?S=B20000 | 1,448,677,454,000,000,000 | text/html | crawl-data/CC-MAIN-2015-48/segments/1448398450745.23/warc/CC-MAIN-20151124205410-00123-ip-10-71-132-137.ec2.internal.warc.gz | 192,653,711 | 12,457 | Problem, can anyone help. - Programmers Heaven
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# Problem, can anyone help.
Posts: 40Member
I have written a program to calculate an electricity bill as part of my college assignment and I've just found a problem that I am unable to solve.
The calculations I have done in the program are correct, but when I display the results of calculations involving high numbers (over 1000), they appear incorrectly and I do not understand why. Can anyone tell me how to solve this?
• Posts: 117Member
: I have written a program to calculate an electricity bill as part of my college assignment and I've just found a problem that I am unable to solve.
:
: The calculations I have done in the program are correct, but when I display the results of calculations involving high numbers (over 1000), they appear incorrectly and I do not understand why. Can anyone tell me how to solve this?
:
are you using integer? try using longint; or word; this will give you bigger number abilaty! if not then post some of your code so that we can see where u are going wrong!
sweeney
• Posts: 40Member
Well the output I want is in real format. I've tried using double but it doesn't work either. The calculations I have done are correct and work fine on numbers under 1000.
• Posts: 117Member
: Well the output I want is in real format. I've tried using double but it doesn't work either. The calculations I have done are correct and work fine on numbers under 1000.
:
nope its ok right do you know decamal places you need to do it in this case! ok give me the output line and I will fix it!
• Posts: 117Member
: : Well the output I want is in real format. I've tried using double but it doesn't work either. The calculations I have done are correct and work fine on numbers under 1000.
: :
: nope its ok right do you know decamal places you need to do it in this case! ok give me the output line and I will fix it!
:
see real is decamal numbers! so here is an example
[code]
program example;
uses crt;
begin
write(' Enter the price of the product: '); | 508 | 2,143 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2015-48 | latest | en | 0.956202 |
http://itstillworks.com/convert-watt-kilowatt-8707976.html | 1,500,786,076,000,000,000 | text/html | crawl-data/CC-MAIN-2017-30/segments/1500549424247.30/warc/CC-MAIN-20170723042657-20170723062657-00246.warc.gz | 168,081,893 | 13,441 | # How to Convert a Watt to Kilowatt
by William McCoy
A watt is a unit that measures power. More specifically, a watt is unit that is the sum of the multiplication of volts and amps. The more watts an electrical device has, the more powerful it is. For example, an 80-watt light bulb is more powerful, and brighter, than a 40-watt bulb. A kilowatt is equivalent to 1,000 watts, so converting watts to kilowatts involves a simple formula.
Understand the formula that explains the relationship between watts and kilowatts. One kilowatt is equal to 1,000 watts. Or, in other words, 1 watt is 1/1,000 of a kilowatt.
Write down the wattage that you wish to convert into kilowatts. For example, the wattage to convert could be 1,500 watts.
Divide the total number of watts by 1,000. In the example, divide 1,500 by 1,000. The answer to this example is 1.5, meaning that 1,500 watts is equivalent to 1.5 kilowatts. | 246 | 911 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.4375 | 3 | CC-MAIN-2017-30 | latest | en | 0.905812 |
http://www.sr2jr.com/textbook-solutions/computer-science/40401001/absolute-cpp-parameters-and-overloading | 1,603,684,076,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107890273.42/warc/CC-MAIN-20201026031408-20201026061408-00370.warc.gz | 180,822,087 | 12,473 | SHARE
HELP
To start with, Sr2Jr’s first step is to reduce the expenses related to education. To achieve this goal Sr2Jr organized the textbook’s question and answers. Sr2Jr is community based and need your support to fill the question and answers. The question and answers posted will be available free of cost to all.
# Absolute C++
Authors:
Walter Savitch ,kenrick Mock
Exercise:
Programming Projects
Chapter:
Edition:
5
ISBN:
9780132846813
Question:
1
Previous Next
### Question
Write a program that converts from 24-hour notation to 12-hour notation. For example, it should convert 14:25 to 2:25 P.M. The input is given as two integers. There should be at least three functions: one for input, one to do the conversion, and one for output. Record the A.M./P.M. information as a value of type char , 'A' for A.M. and 'P' for P.M. Thus, the function for doing the conversions will have a call-by-reference formal parameter of type char to record whether it is A.M. or P.M. (The function will have other parameters as well.) Include a loop that lets the user repeat this computation for new input values again and again until the user says he or she wants to end the program.
Program:
``````#include<iostream>
void overviewData( );
void inputData(int &hr,int &min);
void convert12Format(int &hr,char &type);
void displayData(int hr,int min,char type);
int main()
{
using namespace std;
int hours,min;
char type,choice;
overviewData( );
do
{
inputData(hours,min);
convert12Format(hours,type);
displayData(hours,min,type);
cout<<"To continue then press'y':";
cin>>choice;
}while(choice=='y');
}
void overviewData( )
{
using namespace std;
cout << "Converts 24-hour to 12-notation.\n";
cout<<"-----------------------------------\n";
}
//Read hours and minutes from the key board
void inputData(int &hr,int &min)
{
using namespace std;
cout<<"Enter hours:";
cin>>hr;
cout<<"Enter minutes:";
cin>>min;
}
//Convert 24 hours format to 12 hours format
void convert12Format(int &hrs,char &type)
{
if(hrs>12 )
{
hrs=hrs%12;
type='P';
}
else
type='A';
}
//Result on screen
void displayData(int hrs,int min,char t)
{
using namespace std;
cout<<"12-hour notation:"<<endl;
cout<<hrs<<":"<<min<<" "<<t<<"M"<<endl;
}``````
Output:
``````Converts 24-hour to 12-notation.
-------------------------------------------
Enter hours:14 25
Enter minutes:12-hour notation:
2:25 PM
To continue then press'y':y
Enter hours:23 00
Enter minutes:12-hour notation:
11:0 PM
To continue then press'y':n``````
### Discussions
Post the discussion to improve the above solution. | 648 | 2,558 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.5625 | 3 | CC-MAIN-2020-45 | longest | en | 0.700578 |
http://www.csd.uwo.ca/Courses/CS2035b/tutorial_ass2_2035_2019.txt | 1,552,966,347,000,000,000 | text/plain | crawl-data/CC-MAIN-2019-13/segments/1552912201885.28/warc/CC-MAIN-20190319032352-20190319054352-00312.warc.gz | 279,891,476 | 1,592 | Assignment 2 requires you to vectorize and plot the values of a rational function that computes exp(x), for 10,000,000 equally spaced x values between 0 and 1. The first plot show the results of the given code for a serialized calculation of y=exp(x). A loop is used and y is not pre-allocated (and so at each iteration of the loop it increases y by 1 element). To guarantee y has not been pre-allocated and saved in the workspace by a previous run of your program, it is set to a single value (0 in this case) before the 'calculation is done. This should be the worst calculation from a computational time cost. Use tic and toc to do the timing, as shown on the assignment handout. The second plot comes from the JIT compiled code. The code is the same as the serial code ' but y is now pre-allocated using y=zeros(1,n,'double'), Thus, dynamic allocation of y is not needed in the loop and MatLab compiles the code for you. Much faster! The third plot comes from the vectorized solution. No loops can be used and indices are not needed and should not be used. This is the MAIN part of this assignment. Consider the following serialized code and its vectorized code. The loop: for i=1:n z(i)=(x(i)^2)/(y(i)^3); end can be vectorized using: z=(x.^2)./(y.^3); This code should be even faster! The fourth plot should show the plot of x versus y=exp(x). Again the computational cost can be computed using tic/toc. You will probably observed that the vectorized solution and the direct exp(x) solution require about the same time. Maybe the vectorized solution is a bit faster? It is unknown how MatLab does the exp(x) calculation (maybe vectorization or parallel GPU processing). Make the 2nd to 4th plots with the colours on the assignment handout using the same plotting commands shown in the serial solution. | 422 | 1,807 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.90625 | 3 | CC-MAIN-2019-13 | latest | en | 0.905873 |
http://www.roboticsindia.com/archive/index.php/t-892.html?s=1833017df1f20da2589b78dfa8d26c64 | 1,556,146,379,000,000,000 | text/html | crawl-data/CC-MAIN-2019-18/segments/1555578663470.91/warc/CC-MAIN-20190424214335-20190425000335-00092.warc.gz | 290,041,680 | 3,510 | PDA
View Full Version : Stepper motor driver circuit
Jabir
02-11-2006, 11:32 PM
hello,
I am a final year mechanical engineering student from India.
Now I am doing a project in robotics. We are using unipolar
stepper motors taken from old dot matrix printers for motion power.
We are using a microcontroller based board for programming and
generating signals for the required rotation.
The Motor details are
Step-Syn,Sanyo-Denki,Stepping motor type: 103-770-0178
DC 5.1 V, 1 Amp. 1.8 deg/step. Wires:6
Now we need a driver circuit for the motor. We are ready to use any
type ie.,Ic based or transistor based .
Thank you.
e_jabir@yahoo.com
02-11-2006, 11:36 PM
use uln2803 driver ic for the job.
they ll supply 600ma of current per channel.
Jabir
02-11-2006, 11:47 PM
Sorry i am a mechanical engineering student with zero knoledge in
electonics.
Our motor has a current rating of 1 amp.
Then how can we use the IC youy mensioned ?
It has six leads. Then whether it can be used as a bipolar one?
devpriya
02-12-2006, 12:43 AM
A stepper motor with 6 leads can be both unipolar as well as bi-polar. You got to check out this yourself by finding out the common wire if there is any or not ? Just search the net for the procedure to find it out, its a bit lengthy so I am not writing it here.
And for driving it as Addy said you can use ULN2803 provided it is unipolar!
If it is bi-polar then go for any H-bridge IC like L293 or L298.
The circuits you can get on net , just search for them.
yogi
02-12-2006, 04:38 AM
see the diagram from
http://rapidshare.de/files/13063095/stepper.JPG.html
http://www.geocities.com/njbibin/stepper.JPG
You just have to identify the leads, rest explained in diagram
For more help pm me
Bibin John
www.bibinjohn.tk
Jabir
02-12-2006, 03:29 PM
My motor Has six leads. Two of them are the center taps of
each windings. We shorted the two for connecting to
the interface card at the college electronics lab
Then , can I use this motor as a bipolar one having more torque?
02-12-2006, 06:30 PM
hey jabir.
note the schematic wat yogi has mentioned.
its good and why dont u try tat.
i m sure ur motor ll run with tat.
Jabir
02-12-2006, 07:30 PM
Hello
From the message by yogi I found that my motor
can be used as either bipolar or unipolar configurations.
Thanks.
I choose the bipolar configuration because of the better
torque
Now i want the detailed circuit diagram of the driver
using L293.
But there is a problem!!!!!!!!!!
Our controller board determines the distance travelled by the robot by counting the pulses given to the motor.
We also want to reverse the direction of the motor if necessary.
Can I use L293 with these?
e_jabir@yahoo.com
02-12-2006, 09:38 PM
Now i want the detailed circuit diagram of the driver
using L293.
search through the forum.... u ll get the circuit for l293 in the forum.
u ll get all sorts of details u ever wanted on l293d in the forum.
We also want to reverse the direction of the motor if necessary.
Can I use L293 with these?
obviously...... u can even do regenerative braking.
but wait!!!! l293d can support only .5amps of current per channel ... u ll have to connect dem in parallel if u want to use them for higher current applications.
or go for l298 , it supports 2amps per channel.
yogi
02-13-2006, 03:32 AM
if u use as a bipolar use L297+L298, then u can reverse motor
if u use unipolar use ULN2003, then also u can reverse motor
Bibin John
www.bibinjohn.tk
Jabir
02-13-2006, 07:53 AM
Hey
Jabir
05-31-2006, 05:37 PM
Hi guys...
sorry for my foolish questions ...
I made a driver ckt using TIP 122's since L293 is not available here.IT is working fine and very simple. The unipolar config has enough torque for my purpose.
The 5 volt 1 amp motors are driven by 9 volt supply. I had used 15 ohm resistances to limit the current and 3.3 k resistances for pullup in the base. 1N4148 are used as protection diodes.
Thanks everyone for the discussion...
JABIR | 1,113 | 3,957 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2019-18 | latest | en | 0.874525 |
https://it.mathworks.com/matlabcentral/cody/problems/43270-rotate-matrix-by-90-degrees/solutions/1761648 | 1,582,536,555,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875145910.53/warc/CC-MAIN-20200224071540-20200224101540-00392.warc.gz | 424,871,094 | 15,594 | Cody
# Problem 43270. Rotate matrix by -90 degrees
Solution 1761648
Submitted on 26 Mar 2019
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 Fail
x = [1 2;3 4]; y_correct = [3 1;4 2]; assert(isequal(rotNeg90(x),y_correct))
Undefined function 'rotx' for input arguments of type 'double'. Error in rotNeg90 (line 2) rotx(90); Error in Test1 (line 3) assert(isequal(rotNeg90(x),y_correct))
2 Fail
x = [1 2 3;4 5 6;7 8 9]; y_correct = [7 4 1;8 5 2;9 6 3]; assert(isequal(rotNeg90(x),y_correct))
Undefined function 'rotx' for input arguments of type 'double'. Error in rotNeg90 (line 2) rotx(90); Error in Test2 (line 3) assert(isequal(rotNeg90(x),y_correct)) | 264 | 777 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.703125 | 3 | CC-MAIN-2020-10 | latest | en | 0.425814 |
http://tptp.org/cgi-bin/SeeTPTP?Category=Problems&Domain=SEV&File=SEV512%5E1.p | 1,638,544,882,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964362891.54/warc/CC-MAIN-20211203151849-20211203181849-00285.warc.gz | 75,077,212 | 2,354 | ## TPTP Problem File: SEV512^1.p
View Solutions - Solve Problem
```%------------------------------------------------------------------------------
% File : SEV512^1 : TPTP v7.5.0. Released v7.0.0.
% Domain : Analysis
% Problem : SUM_IMAGE_NONZERO
% Version : Especial.
% English :
% Refs : [Kal16] Kalisyk (2016), Email to Geoff Sutcliffe
% Source : [Kal16]
% Names : SUM_IMAGE_NONZERO_.p [Kal16]
% Status : Theorem
% Rating : 0.00 v7.5.0, 0.33 v7.2.0, 0.50 v7.1.0
% Syntax : Number of formulae : 19 ( 0 unit; 14 type; 1 defn)
% Number of atoms : 112 ( 10 equality; 69 variable)
% Maximal formula depth : 19 ( 6 average)
% Number of connectives : 89 ( 2 ~; 0 |; 8 &; 74 @)
% ( 0 <=>; 5 =>; 0 <=; 0 <~>)
% ( 0 ~|; 0 ~&)
% Number of type conns : 44 ( 44 >; 0 *; 0 +; 0 <<)
% Number of symbols : 16 ( 14 :; 0 =; 0 @=)
% ( 0 !!; 0 ??; 0 @@+; 0 @@-)
% Number of variables : 29 ( 0 sgn; 17 !; 0 ?; 0 ^)
% ( 29 :; 12 !>; 0 ?*)
% ( 0 @-; 0 @+)
% SPC : TH1_THM_EQU_NAR
% Comments : Exported from core HOL Light.
%------------------------------------------------------------------------------
thf('thf_type_type/realax/real',type,(
'type/realax/real': \$tType )).
thf('thf_type_type/nums/num',type,(
'type/nums/num': \$tType )).
thf('thf_const_const/trivia/o',type,(
'const/trivia/o':
!>[B: \$tType,A: \$tType,C: \$tType] :
( ( B > C ) > ( A > B ) > A > C ) )).
thf('thf_const_const/sets/IN',type,(
'const/sets/IN':
!>[A: \$tType] :
( A > ( A > \$o ) > \$o ) )).
thf('thf_const_const/sets/IMAGE',type,(
'const/sets/IMAGE':
!>[A: \$tType,B: \$tType] :
( ( A > B ) > ( A > \$o ) > B > \$o ) )).
thf('thf_const_const/sets/FINITE',type,(
'const/sets/FINITE':
!>[A: \$tType] :
( ( A > \$o ) > \$o ) )).
thf('thf_const_const/realax/real_of_num',type,(
'const/realax/real_of_num': 'type/nums/num' > 'type/realax/real' )).
'const/realax/real_add': 'type/realax/real' > 'type/realax/real' > 'type/realax/real' )).
thf('thf_const_const/nums/NUMERAL',type,(
'const/nums/NUMERAL': 'type/nums/num' > 'type/nums/num' )).
thf('thf_const_const/nums/_0',type,(
'const/nums/_0': 'type/nums/num' )).
thf('thf_const_const/iterate/sum',type,(
'const/iterate/sum':
!>[A: \$tType] :
( ( A > \$o ) > ( A > 'type/realax/real' ) > 'type/realax/real' ) )).
thf('thf_const_const/iterate/neutral',type,(
'const/iterate/neutral':
!>[A: \$tType] :
( ( A > A > A ) > A ) )).
thf('thf_const_const/iterate/monoidal',type,(
'const/iterate/monoidal':
!>[A: \$tType] :
( ( A > A > A ) > \$o ) )).
thf('thf_const_const/iterate/iterate',type,(
'const/iterate/iterate':
!>[A: \$tType,A0: \$tType] :
( ( A0 > A0 > A0 ) > ( A > \$o ) > ( A > A0 ) > A0 ) )).
thf('thm/iterate/ITERATE_IMAGE_NONZERO_',axiom,(
! [C: \$tType,A: \$tType,B: \$tType,A0: C > C > C] :
( ( 'const/iterate/monoidal' @ C @ A0 )
=> ! [A1: B > C,A2: A > B,A3: A > \$o] :
( ( ( 'const/sets/FINITE' @ A @ A3 )
& ! [A4: A,A5: A] :
( ( ( 'const/sets/IN' @ A @ A4 @ A3 )
& ( 'const/sets/IN' @ A @ A5 @ A3 )
& ( A4 != A5 )
& ( ( A2 @ A4 )
= ( A2 @ A5 ) ) )
=> ( ( A1 @ ( A2 @ A4 ) )
= ( 'const/iterate/neutral' @ C @ A0 ) ) ) )
=> ( ( 'const/iterate/iterate' @ B @ C @ A0 @ ( 'const/sets/IMAGE' @ A @ B @ A2 @ A3 ) @ A1 )
= ( 'const/iterate/iterate' @ A @ C @ A0 @ A3 @ ( 'const/trivia/o' @ B @ A @ C @ A1 @ A2 ) ) ) ) ) )).
( 'const/iterate/monoidal' @ 'type/realax/real' @ 'const/realax/real_add' )).
thf('thm/iterate/sum_',definition,(
! [A: \$tType] :
( ( 'const/iterate/sum' @ A )
= ( 'const/iterate/iterate' @ A @ 'type/realax/real' @ 'const/realax/real_add' ) ) )).
( ( 'const/iterate/neutral' @ 'type/realax/real' @ 'const/realax/real_add' )
= ( 'const/realax/real_of_num' @ ( 'const/nums/NUMERAL' @ 'const/nums/_0' ) ) )).
thf('thm/iterate/SUM_IMAGE_NONZERO_',conjecture,(
! [A: \$tType,B: \$tType,A0: B > 'type/realax/real',A1: A > B,A2: A > \$o] :
( ( ( 'const/sets/FINITE' @ A @ A2 )
& ! [A3: A,A4: A] :
( ( ( 'const/sets/IN' @ A @ A3 @ A2 )
& ( 'const/sets/IN' @ A @ A4 @ A2 )
& ( A3 != A4 )
& ( ( A1 @ A3 )
= ( A1 @ A4 ) ) )
=> ( ( A0 @ ( A1 @ A3 ) )
= ( 'const/realax/real_of_num' @ ( 'const/nums/NUMERAL' @ 'const/nums/_0' ) ) ) ) )
=> ( ( 'const/iterate/sum' @ B @ ( 'const/sets/IMAGE' @ A @ B @ A1 @ A2 ) @ A0 )
= ( 'const/iterate/sum' @ A @ A2 @ ( 'const/trivia/o' @ B @ A @ 'type/realax/real' @ A0 @ A1 ) ) ) ) )).
%------------------------------------------------------------------------------
``` | 1,893 | 4,779 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.21875 | 3 | CC-MAIN-2021-49 | latest | en | 0.323005 |
https://learnthermo.com/T1-tutorial/ch02/lesson-B/pg01.php | 1,685,865,084,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224649518.12/warc/CC-MAIN-20230604061300-20230604091300-00130.warc.gz | 400,554,278 | 6,213 | # What is a Phase Diagram ?
Phase diagrams depict how the primary phases (solid, liquid, gas) of a substance change as
the pressure, volume and temperature of the system change.
Phase diagrams are constructed by plotting the following :
1. Temperature as a function of the molar volume (T-)
2. Pressure as a function of the molar volume (P-)
3. Pressure as a function of temperature (P-T)
Note : The molar volume, , is the volume per mole :
The specific volume, , is the volume per unit mass :
Roll your mouse over this box to close.
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### Ch 2, Lesson B, Page 1 - What is a Phase Diagram ?
• A phase diagram shows you the phase in which a material exists at a given state.
• This allows you to pack an enormous amount of information about boiling, melting and other phase changes into a nice compact form.
• We will only consider phase diagrams for pure substances in this chapter.
• With a little practice, I think you will find that phase diagrams will help you gain a much deeper understanding of phases and phase changes.
• There are an amazingly large number of different types of phase diagrams. But, we’re going to start out with the big three: T-V, P-V and P-T Diagrams.
• It makes sense to start with these three phase diagrams because you are already familiar with the concepts of temperature, pressure and volume.
• But what does the wiggly line over the V mean ? Well, the wiggly line means that we are referring to the MOLAR volume.
• The molar volume is the volume per mole or just the total volume of the system divided by the number of moles in the system.
• The specific volume is represented by a V with a carat, or “hat,” over it. It is just the volume per unit MASS of the system.
• So, why bother with molar and specific volume ? Why not just use plain old volume ?
• If we used ordinary volume, then we would need one phase diagram for a system that contained 2 kg of water and another for a system that contains 0.2 kg of water.
• Why ? Consider a system with a volume of 2 liters that contained 2 kg of water. Nothing remarkable about that since the density of liquid water at atmospheric pressure is about 1 kg/L.
• But what if this 2L volume contained just 0.2 kg of water also at atmospheric pressure ? Could all of the water be a liquid ? No, the density is too low. Some of the water would be in a different phase…the gas phase !
• So, the phase depends on the SIZE of the system ! This is not very convenient or helpful !
• This problem is fixed by using either the molar or specific volume instead of the total volume. This way a single phase diagram can be used regardless of the size of the system.
• Cool. So, for example, water has just one P-T Diagram.
• Ok, so let’s learn a few terms that will make it easier for us to discuss phase diagrams.
Jump to New page Show / Hide Notes | 708 | 3,030 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.671875 | 4 | CC-MAIN-2023-23 | latest | en | 0.869365 |
https://learn.saylor.org/mod/book/view.php?id=29218&chapterid=4713 | 1,723,353,733,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640975657.64/warc/CC-MAIN-20240811044203-20240811074203-00554.warc.gz | 285,602,703 | 19,520 | ## if, else, and elif Statements
Read this for more on conditional statements.
### 6. Boolean operators and expressions
#### Booleans and Boolean operators
A Boolean refers to a value that is either True or False. These two are constants in Python.
• we can assign a Boolean value by specifying True or False,
x = True
• an expression can evaluate to a Boolean value
y > 10
#### and operator
The Boolean expression a and b is True if and only if both a and b are True.
a b a and b
True True True
True False False
False True False
False False False
Examples: assume that a = 8 and b = 3, then the Boolean value of
1) ( a > 10 ) and ( b < 5 ) is False
2) ( a != 10 ) and ( b > 1 ) is True
#### or operator
The Boolean expression a or b is False if and only if both a and b are False.
a b a or b
True True True
True False True
False True True
False False False
Examples: assume that a = 8 and b = 3, then the Boolean value of
1) ( a > 10 ) or ( b < 5 ) is True
2) ( a == 10 ) or ( b > 1 ) is True
#### not operator
The Boolean expression not a is False when a is True, and is True when a is False.
a not a
True False
False False
Examples: assume that a = 8 and b = 3, then the Boolean value of
1) not ( a > 10 ) is True
2) not ( a * 10 > 20 ) is False
#### Booleans and Boolean operators
Consider the following code fragment:
if letter == 'a' or letter == 'b':
print("Help!")
elif letter == 'c' or letter == 'd':
print("We are in trouble!")
else:
print("We are good!")
If letter = "a", then we will get: Help!
Consider the following code fragment:
if letter == 'a' or letter == 'b':
print("Help!")
elif letter == 'c' or letter == 'd':
print("We are in trouble!")
else:
print("We are good!")
If letter = "c", then we will get: We are in trouble! | 524 | 1,778 | {"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} | 3.34375 | 3 | CC-MAIN-2024-33 | latest | en | 0.767029 |
https://tunxis.commnet.edu/view/science-density-calculations-worksheet.html | 1,713,509,217,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817289.27/warc/CC-MAIN-20240419043820-20240419073820-00705.warc.gz | 534,253,564 | 7,613 | Science Density Calculations Worksheet
Science Density Calculations Worksheet - Web calculations embody figuring out mass using a triple beam steadiness, calculating volume of a cube and finally. This two page, ten question worksheet gives students. If the pipe has a. ) a student measures the mass of an 8 cm3 block of brown. These density formula problems have students solving. Web live worksheets > english > science > density > calculating density. This can be written as the. Web density calculations subject: This is a good way to get students. Web students will enjoy calculating density with ezpz science formulas.
Worksheet/activity file previews pptx, 494.18 kb. If the pipe has a. Web science 8— den itv calculations worksheet name: Worksheet #1 calculate density, and identify substances using a density chart. Web density calculations subject: ) a student measures the mass of an 8 cm3 block of brown. This can be written as the.
Science Density Calculations Worksheet - Worksheet #1 calculate density, and identify substances using a density chart. Web calculate density sixth grade science worksheets january 7, 2022 by ppt get free questions on “calculate density”. Web density calculations subject: Web science 8— den itv calculations worksheet name: Worksheets are calculation work residential density, work. Web one page worksheet with answer key. Density is a measure of the. Web calculating density worksheet created by mr wagners science store students work to calculate the density of 8 objects. Worksheet/activity file previews pptx, 494.18 kb. This is a good way to get students.
These density formula problems have students solving. Web this worksheet presents students practice calculating density of a dice. Web density of any solid, liquid, or gas is calculated by dividing its mass by its volume. Web density calculations subject: What is the density of an element if a sample having a mass.
Web Students Will Enjoy Calculating Density With Ezpz Science Formulas.
Worksheet #1 calculate density, and identify substances using a density chart. Worksheet/activity file previews pptx, 494.18 kb. Web density, mass and volume calculation practice worksheet. Web one page worksheet with answer key.
What Is The Density Of An Element If A Sample Having A Mass.
This is a good way to get students. This two page, ten question worksheet gives students. These density formula problems have students solving. Web calculate density sixth grade science worksheets january 7, 2022 by ppt get free questions on “calculate density”.
Worksheets Are Calculation Work Residential Density, Work.
Web density calculations subject: Web live worksheets > english > science > density > calculating density. ) a student measures the mass of an 8 cm3 block of brown. Web density of any solid, liquid, or gas is calculated by dividing its mass by its volume.
Web Science 8— Den Itv Calculations Worksheet Name:
Web this worksheet presents students practice calculating density of a dice. Density is a measure of the. If the pipe has a. Web improve your science knowledge with free questions in calculate density, mass, and volume and thousands of other science. | 632 | 3,178 | {"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} | 3.203125 | 3 | CC-MAIN-2024-18 | latest | en | 0.835516 |
http://www.arkaye.com/blog/tag/recursive/ | 1,723,266,247,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640789586.56/warc/CC-MAIN-20240810030800-20240810060800-00410.warc.gz | 26,045,697 | 5,384 | # Factorials For Fun
Start from the beginning. A factorial is a specific multiplication function applied to a positive integer value and all of its positive descendants. True but vague and tough to digest.
A factorial means to multiply a consecutive sequence of descending natural numbers.
Below, n is any positive integer (equal to a natural number), being multiplied:
n! = n·(n–1)·(n–2)·…·3·2·1 [1]
The symbol for a factorial is the exclamation point: ! . n! is pronounced ‘en’ factorial (math fans) or ‘en’ bang (programmers) or shriek out n (literalists and comedians).
Here are some examples: 3! = 3·2·1 = 6
10! = 10·9·8·7·6·5·4·3·2·1 = 3628800
{ Sudden question: In 1 second, how much would 9! be? }
1! = 1
To see the first 100 factorials, please admire the following table by N. J. A. Sloane: First 100 Factorials
These integer results become huge as n increases. In a future post, I will write about how to determine the number of digits of n! without calculating the factorial and visually counting the digits.
Because mathematicians revere economy in writing, n! can be defined for any positive integer n by two rules:
n! = n·(n–1)!
and [2]
1! = 1
meaning, n! is the product of n and the factorial of the next lower value shown as (n – 1).
In the case of the second rule, this only works if we also have
0! = 1 [3]
Since by rule 1 (and I know this may be a head-scratcher…)
1! = 1·0! = 1 [4]
To find the value of n! , the first rule is applied repeatedly, as shown:
n! = n·(n–1)! = n·(n–1)·(n-2)! = … =
n
·(n–1)·(n-2)··3·2·1! = [5]
n
·(n–1)·(n-2)··3·2·1
At this point, the original definition of n! in [1] has been recreated.
This [2] is called a recursive definition of n! . It has been said anonymously that, “in order to understand recursion, one must first understand recursion”.
In Computer Programming, n! can be determined via an iterative set of instructions, involving looping. Also, n! can be determined via a recursively called function or program, which calls itself repeatedly.
The Rosetta Code.org Wiki for factorial lists approximately 179 different programming languages with both iterative and recursive factorial code segments. It’s quite impressive – See:
Factorial Programs and Functions – Rosetta Code
Within this listing are programs that are part of the Unix/Linux facilities: awk, bash, bc, dc, JavaScript, m4, perl, php, python, R, ruby, and Unix shells (sh, ksh, zsh). These programs do not require a compiler. In future posts, these will be examined more closely.
There’s much more to factorials, which I will defer to further posts .
It has been suspected that mathematicians invented the factorial symbol (!) as a way of making mathematics look more exciting; after all, as the value of n increases, its factorial increases incredibly fast. | 785 | 3,340 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-33 | latest | en | 0.634804 |
http://www.statemaster.com/encyclopedia/De-Sitter-universe | 1,571,808,101,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570987829458.93/warc/CC-MAIN-20191023043257-20191023070757-00001.warc.gz | 309,767,867 | 10,219 | FACTOID # 16: In the 2000 Presidential Election, Texas gave Ralph Nader the 3rd highest popular vote count of any US state.
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Encyclopedia > De Sitter universe
A de Sitter universe is a solution to Einstein's field equations of General Relativity which is named after Willem de Sitter. It is common to describe a patch of this solution as an expanding universe of the FLRW form where the scale factor is given by: For other topics related to Einstein see Einstein (disambiguation). ... Two-dimensional visualisation of space-time distortion. ... Willem de Sitter (May 6, 1872 – November 20, 1934) was a mathematician, physicist and astronomer. ... The Friedmann-Lemaître-Robertson-Walker (FLRW) metric describes a homogeneous, isotropic expanding/contracting universe. ...
a(t) = eHt,
where the constant H is the Hubble expansion rate and t is time. As in all FLRW spaces, a(t), the scale factor, describes the expansion of physical spatial distances. The scale factor, parameter of Friedmann-Lemaître-Robertson-Walker model, is a function of time which represents the relative expansion of the universe. ...
A de Sitter universe is one with no ordinary matter content but with a positive cosmological constant which sets the expansion rate, H. A larger cosmological constant leads to a larger expansion rate: The cosmological constant (usually denoted by the Greek capital letter lambda: Λ) occurs in Einsteins theory of general relativity. ...
,
where the constants of proportionality depend on conventions. The cosmological constant is Λ and Mpl is the Planck mass. The Planck mass is the natural unit of mass, denoted by mP. mP = (c / G)1/2 ≈ 2. ...
Our universe may be approaching a de Sitter universe in the infinite future. If the current acceleration of our universe is due to a cosmological constant then as the universe continues to expand all of the matter and radiation will be diluted. Eventually there will be almost nothing left but the cosmological constant, and our universe will have become a de Sitter universe.
The exponential expansion of the scale factor means that the physical distance between any two non-accelerating observers will eventually be growing faster than the speed of light. At this point those two observers will no longer be able to make contact. Therefore any observer in a de Sitter universe would see horizons beyond which that observer can not see. If our universe is approaching a de Sitter universe then eventually we will not be able to observe any galaxies other than our own Milky Way. Cherenkov effect in a swimming pool nuclear reactor. ... This article is about a celestial body. ... The Milky Way (a translation of the Latin Via Lactea, in turn derived from the Greek Galaxia Kuklos) is the galaxy in which the Earth is found. ...
Another application of de Sitter space, is in the early universe during cosmic inflation. Many inflationary models are approximately de Sitter space and can be modeled by giving the Hubble parameter a mild time dependence. For simplicity, some calculations involving inflation in the early universe can be performed in de Sitter space rather than a more realistic inflationary universe. By using the de Sitter universe instead, where the expansion is truly exponential, there are many simplifications. Inflation is the idea—first proposed by Alan Guth (1981)—that the nascent universe passed through a phase of exponential expansion (the inflationary epoch) that was driven by a negative pressure vacuum energy density. ...
Results from FactBites:
De Sitter universe (136 words) A de Sitter universe is one with a positive cosmological constant and zero matter density. There is no Big Bang in a de Sitter universe - expansion is self-similar from an infinite time in the past to an infinite time in the future. The model has practical implications as an approximation during the future of the Universe when the cosmological constant is likely to dominate (assuming no changes from the present version of the standard Big Bang model), and during the early universe during a hypothesised period of inflation.
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Share your thoughts, questions and commentary here | 916 | 4,331 | {"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} | 2.796875 | 3 | CC-MAIN-2019-43 | latest | en | 0.860727 |
http://www.thestudentroom.co.uk/showthread.php?t=1825707 | 1,462,335,000,000,000,000 | text/html | crawl-data/CC-MAIN-2016-18/segments/1461860122501.26/warc/CC-MAIN-20160428161522-00135-ip-10-239-7-51.ec2.internal.warc.gz | 902,614,653 | 40,928 | # Calculating the tidal volume ATPS (L)
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1. If we want to calculate the ATPS, and were given for example that the total volume of expired air (L) was 30.6 L provided that the air was collected over an interval of 3-minutes, will ATPS in this case be: 30.6 L / 3 min = 10.2 ?
Thanks.
2. If we have the following values:
- Number of breaths in one-minute = 16
- Exhaled minute volume ATPS (L min-1) = 5.50
Will the tidal volume ATPS (L) be ---> 5.50 / 16 = 0.34 ?
Thanks.
3. If we were given the following values:
- Oxygen concentration of expired air = 16.7%
- Exhaled minute volume ATPS (L min-1) = 5.67
And, want to calculate the "Oxygen exhalation rate ATPS (ml min-1)", would it be as follows?
5.67 X 16.7% = 0.95
But, since the value has to be in (ml min-1), we should multiply by 1000, so, the final result will be 950 ml min-1
Is it correct this way?
Thanks.
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### iGCSE English Language
Here's the unofficial markscheme | 494 | 1,719 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.078125 | 3 | CC-MAIN-2016-18 | longest | en | 0.911002 |
https://www.geeksforgeeks.org/print-all-pairs-in-an-unsorted-array-with-equal-sum/ | 1,579,621,494,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579250604397.40/warc/CC-MAIN-20200121132900-20200121161900-00188.warc.gz | 889,774,917 | 27,930 | # Print all pairs in an unsorted array with equal sum
Given an unsorted array A[]. The task is to print all unique pairs in the unsorted array with equal sum.
Note: Print the result in the format as shown in the below examples.
Examples:
```Input: A[] = { 6, 4, 12, 10, 22, 54, 32, 42, 21, 11}
Output:
Pairs : ( 4, 12) ( 6, 10) have sum : 16
Pairs : ( 10, 22) ( 21, 11) have sum : 32
Pairs : ( 12, 21) ( 22, 11) have sum : 33
Pairs : ( 22, 21) ( 32, 11) have sum : 43
Pairs : ( 32, 21) ( 42, 11) have sum : 53
Pairs : ( 12, 42) ( 22, 32) have sum : 54
Pairs : ( 10, 54) ( 22, 42) have sum : 64
Input:A[]= { 4, 23, 65, 67, 24, 12, 86}
Output:
Pairs : ( 4, 86) ( 23, 67) have sum : 90
```
## Recommended: Please try your approach on {IDE} first, before moving on to the solution.
The idea is to use map in C++ STL for avoiding duplicate pair of elements.
• Create a map with key as pair of integer and value as integer to store all unique pair of elements and their corresponding sum.
• Traverse the array and generate all possible pairs and store the pairs and their corresponding sum in first map.
• Create a second map with key as integer and value as a vector of pair to store list of all pair of elements with a corresponding sum.
• Finally, traverse the second map, and for a sum with more than one pair, print all pairs and then the corresponding sum in a format as shown in the above example.
Below is the implementation of the above approach:
## C++
`// C++ program to print all pairs ` `// with equal sum ` ` ` `#include ` `using` `namespace` `std; ` ` ` `// Function to print all pairs ` `// with equal sum ` `void` `pairWithEqualSum(``int` `A[], ``int` `n) ` `{ ` ` ` ` ``map<``int``, vector > > mp; ` ` ` ` ``// Insert all unique pairs and their ` ` ``// corresponding sum in the map ` ` ``for` `(``int` `i = 0; i < n - 1; i++) { ` ` ``for` `(``int` `j = i + 1; j < n; j++) { ` ` ``pair<``int``, ``int``> p = make_pair(A[i], A[j]); ` ` ` ` ``mp[A[i] + A[j]].push_back(p); ` ` ``} ` ` ``} ` ` ` ` ``// Traverse the map mp, and for sum ` ` ``// with more than one pair, print all pairs ` ` ``// and the corresponding sum ` ` ``for` `(``auto` `itr = mp.begin(); itr != mp.end(); itr++) { ` ` ``if` `(itr->second.size() > 1) { ` ` ``cout << ``"Pairs : "``; ` ` ` ` ``for` `(``int` `i = 0; i < itr->second.size(); i++) { ` ` ``cout << ``"( "` `<< itr->second[i].first << ``", "` ` ``<< itr->second[i].second << ``") "``; ` ` ``} ` ` ` ` ``cout << ``" have sum : "` `<< itr->first << endl; ` ` ``} ` ` ``} ` `} ` ` ` `// Driver Code ` `int` `main() ` `{ ` ` ``int` `A[] = { 6, 4, 12, 10, 22, 54, 32, 42, 21, 11, 8, 2 }; ` ` ` ` ``int` `n = ``sizeof``(A) / ``sizeof``(A[0]); ` ` ` ` ``pairWithEqualSum(A, n); ` ` ` ` ``return` `0; ` `} `
## Python3
`# Python implementation of the ` `# approach ` ` ` `# Function to print all pairs ` `# with equal sum ` `def` `pairWithEqualSum(A, n): ` ` ``mp ``=` `{} ` ` ` ` ``# Insert all unique pairs and their ` ` ``# corresponding sum in the map ` ` ` ` ``for` `i ``in` `range``(n): ` ` ``for` `j ``in` `range``(i``+``1``,n): ` ` ``if` `A[i]``+``A[j] ``in` `mp: ` ` ``mp[A[i]``+``A[j]].append((A[i], A[j])) ` ` ``else``: ` ` ``mp[A[i]``+``A[j]] ``=` `[ (A[i],A[j]) ] ` ` ` ` ` ` ``# Traverse the map mp, and for sum ` ` ``# with more than one pair, print all pairs ` ` ``# and the corresponding sum ` ` ` ` ``for` `itr ``in` `mp: ` ` ``if` `len``(mp[itr]) > ``1``: ` ` ``print``(``"Pairs : "``, end ``=` `"") ` ` ``for` `i ``in` `range``(``0``, ``len``(mp[itr])): ` ` ``print``(``"("``, mp[itr][i][``0``], ``","``, ` ` ``mp[itr][i][``1``], ``")"``, end ``=` `" "``) ` ` ` ` ``print``(``"have sum :"``, itr) ` ` ` `# Driver Code ` `if` `__name__ ``=``=` `"__main__"``: ` ` ` ` ``A ``=` `[``6``, ``4``, ``12``, ``10``, ``22``, ``54``, ` ` ``32``, ``42``, ``21``, ``11``, ``8``, ``2``] ` ` ``n ``=` `len``(A) ` ` ` ` ``pairWithEqualSum(A, n) `
Output:
```Pairs : ( 6, 4) ( 8, 2) have sum : 10
Pairs : ( 4, 8) ( 10, 2) have sum : 12
Pairs : ( 6, 8) ( 4, 10) ( 12, 2) have sum : 14
Pairs : ( 6, 10) ( 4, 12) have sum : 16
Pairs : ( 6, 12) ( 10, 8) have sum : 18
Pairs : ( 12, 11) ( 21, 2) have sum : 23
Pairs : ( 10, 22) ( 21, 11) have sum : 32
Pairs : ( 12, 21) ( 22, 11) have sum : 33
Pairs : ( 12, 22) ( 32, 2) have sum : 34
Pairs : ( 22, 21) ( 32, 11) have sum : 43
Pairs : ( 12, 32) ( 42, 2) have sum : 44
Pairs : ( 32, 21) ( 42, 11) have sum : 53
Pairs : ( 12, 42) ( 22, 32) have sum : 54
Pairs : ( 10, 54) ( 22, 42) have sum : 64
```
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Please write to us at contribute@geeksforgeeks.org to report any issue with the above content. | 2,066 | 5,555 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.078125 | 3 | CC-MAIN-2020-05 | latest | en | 0.698797 |
https://uk.mathworks.com/matlabcentral/cody/problems/44630-guess-the-number-i-m-thinking-of/solutions/2093076 | 1,603,208,619,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107872746.20/warc/CC-MAIN-20201020134010-20201020164010-00377.warc.gz | 586,703,336 | 17,473 | Cody
# Problem 44630. Guess the number I'm thinking of
Solution 2093076
Submitted on 14 Jan 2020 by Nikolaos Nikolaou
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
assessFunctionAbsence({'rng', 'RandStream'}, 'FileName','myGuess.m')
2 Pass
for j = 1 : 1000 numberToBeGuessed = randi(10); gOO = randperm(10, 2); mG = myGuess(gOO); assert( mG >= 1 & mG <= 10 , 'Out of requested range.' ) u = unique( floor([gOO mG]) ); assert( length(u) == 3 , 'Your guess must not have been already chosen.' ) end;
3 Pass
maxIts = 100000; tic for j = 1 : 10 WDL = [0 0 0]; for itn = 1 : maxIts numberToBeGuessed = randi(10); gOO = randperm(10, 2); diffs = abs( [gOO myGuess(gOO)] - numberToBeGuessed ); winningContestant = find( min(diffs)==diffs ); if any( winningContestant == 3 ), if length(winningContestant) == 1, % Win WDL(1) = WDL(1) + 1; else % Draw WDL(2) = WDL(2) + 1; end; else % Loss WDL(3) = WDL(3) + 1; end; end; successRate = (WDL(1) + WDL(2)/2) / maxIts assert( successRate >= 0.45 ) end; toc
successRate = 0.4598 successRate = 0.4588 successRate = 0.4593 successRate = 0.4562 successRate = 0.4580 successRate = 0.4570 successRate = 0.4598 successRate = 0.4585 successRate = 0.4582 successRate = 0.4604 Elapsed time is 4.906027 seconds.
### Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting! | 517 | 1,495 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.484375 | 3 | CC-MAIN-2020-45 | latest | en | 0.683654 |
https://www.javatpoint.com/discrete-mathematics-hasse-diagrams | 1,726,742,976,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700652028.28/warc/CC-MAIN-20240919093719-20240919123719-00485.warc.gz | 762,824,040 | 17,690 | # Hasse Diagrams
It is a useful tool, which completely describes the associated partial order. Therefore, it is also called an ordering diagram. It is very easy to convert a directed graph of a relation on a set A to an equivalent Hasse diagram. Therefore, while drawing a Hasse diagram following points must be remembered.
1. The vertices in the Hasse diagram are denoted by points rather than by circles.
2. Since a partial order is reflexive, hence each vertex of A must be related to itself, so the edges from a vertex to itself are deleted in Hasse diagram.
3. Since a partial order is transitive, hence whenever aRb, bRc, we have aRc. Eliminate all edges that are implied by the transitive property in Hasse diagram, i.e., Delete edge from a to c but retain the other two edges.
4. If a vertex 'a' is connected to vertex 'b' by an edge, i.e., aRb, then the vertex 'b' appears above vertex 'a'. Therefore, the arrow may be omitted from the edges in the Hasse diagram.
The Hasse diagram is much simpler than the directed graph of the partial order.
Example: Consider the set A = {4, 5, 6, 7}. Let R be the relation ≤ on A. Draw the directed graph and the Hasse diagram of R.
Solution: The relation ≤ on the set A is given by
R = {{4, 5}, {4, 6}, {4, 7}, {5, 6}, {5, 7}, {6, 7}, {4, 4}, {5, 5}, {6, 6}, {7, 7}}
The directed graph of the relation R is as shown in fig:
To draw the Hasse diagram of partial order, apply the following points:
1. Delete all edges implied by reflexive property i.e.
(4, 4), (5, 5), (6, 6), (7, 7)
2. Delete all edges implied by transitive property i.e.
(4, 7), (5, 7), (4, 6)
3. Replace the circles representing the vertices by dots.
4. Omit the arrows.
The Hasse diagram is as shown in fig:
Upper Bound: Consider B be a subset of a partially ordered set A. An element x ∈ A is called an upper bound of B if y ≤ x for every y ∈ B.
Lower Bound: Consider B be a subset of a partially ordered set A. An element z ∈ A is called a lower bound of B if z ≤ x for every x ∈ B.
Example: Consider the poset A = {a, b, c, d, e, f, g} be ordered shown in fig. Also let B = {c, d, e}. Determine the upper and lower bound of B.
Solution: The upper bound of B is e, f, and g because every element of B is '≤' e, f, and g.
The lower bounds of B are a and b because a and b are '≤' every elements of B.
## Least Upper Bound (SUPREMUM):
Let A be a subset of a partially ordered set S. An element M in S is called an upper bound of A if M succeeds every element of A, i.e. if, for every x in A, we have x <=M
If an upper bound of A precedes every other upper bound of A, then it is called the supremum of A and is denoted by Sup (A)
## Greatest Lower Bound (INFIMUM):
An element m in a poset S is called a lower bound of a subset A of S if m precedes every element of A, i.e. if, for every y in A, we have m <=y
If a lower bound of A succeeds every other lower bound of A, then it is called the infimum of A and is denoted by Inf (A)
Example: Determine the least upper bound and greatest lower bound of B = {a, b, c} if they exist, of the poset whose Hasse diagram is shown in fig:
Solution: The least upper bound is c.
The greatest lower bound is k.
Next TopicLattices | 901 | 3,207 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.3125 | 4 | CC-MAIN-2024-38 | latest | en | 0.929484 |
https://www.flexiprep.com/NCERT-Exercise-Solutions/Mathematics/Class-9/Ch-4-Linear-Equation-In-Two-Variable-Exercise-4-3-Solutions-Part-3.html | 1,500,979,881,000,000,000 | text/html | crawl-data/CC-MAIN-2017-30/segments/1500549425144.89/warc/CC-MAIN-20170725102459-20170725122459-00042.warc.gz | 774,100,659 | 23,480 | # NCERT Class 9 Solutions: Linear Equation in Two Variable (Chapter 4) Exercise 4.3 – Part-3
Q-4 The taxi fare in a city is as follows: For the first kilometer, the fare is Rs. 8 and for the subsequent distance it is Rs. 5 per kilometer. Taking the distance covered as x km and total fares as Rs. y, write a linear equation for this information, and draw its graph.
Solution:
Given,
Taxi fare for subsequent distance Rs. 5
Let,
• Total distance covered
• Total fare
According to problem,
• Taxi fare for first kilometer Rs. 8
• If the total distance is x, fare for rest of the kilometer ( at the rate of Rs. 5 per km)
• Since the fare for first kilometer = Rs. 8, so, the total fare
Hence, is the required linear equation
Drawing the Graph
We can draw the graph by finding two points on this equation and drawing a line joining those points.
The equation is
…………….equation (1)
Now, putting the value in equation (1)
So the solution is this is first point (the y intercept)
Putting the valuein equation (1)
. So the solution is this is the second point
Explore Solutions for Mathematics | 277 | 1,106 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.46875 | 4 | CC-MAIN-2017-30 | longest | en | 0.932125 |
http://research.omicsgroup.org/index.php/Admittance | 1,601,176,912,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400250241.72/warc/CC-MAIN-20200927023329-20200927053329-00788.warc.gz | 108,665,355 | 9,021 | ## Open Access Articles- Top Results for Admittance
In electrical engineering, admittance is a measure of how easily a circuit or device will allow a current to flow. It is defined as the inverse of impedance. The SI unit of admittance is the siemens (symbol S). Oliver Heaviside coined the term admittance in December 1887.[1]
$Y \equiv \frac{1}{Z} \,$
where
Y is the admittance, measured in siemens
Z is the impedance, measured in ohms
The synonymous unit mho, and the symbol ℧ (an upside-down uppercase omega Ω), are also in common use.
Resistance is a measure of the opposition of a circuit to the flow of a steady current, while impedance takes into account not only the resistance but also dynamic effects (known as reactance). Likewise, admittance is not only a measure of the ease with which a steady current can flow, but also the dynamic effects of the material's susceptance to polarization:
$Y = G + j B \,$
where
• $Y$ is the admittance, measured in siemens.
• $G$ is the conductance, measured in siemens.
• $B$ is the susceptance, measured in siemens.
• $j^2 = -1$
## Conversion from impedance to admittance
Parts of this article or section rely on the reader's knowledge of the complex impedance representation of capacitors and inductors and on knowledge of the frequency domain representation of signals.
The impedance, Z, is composed of real and imaginary parts,
$Z = R + jX \,$
where
• R is the resistance, measured in ohms
• X is the reactance, measured in ohms
$Y = Z^{-1}= \frac{1}{R + jX} = \left( \frac{1}{R^2 + X^2} \right) \left(R - jX\right)$
Admittance, just like impedance, is a complex number, made up of a real part (the conductance, G), and an imaginary part (the susceptance, B), thus:
$Y = G + jB \,\!$
where G (conductance) and B (susceptance) are given by:
\begin{align} G &= \Re(Y) = \frac{R}{R^2 + X^2} \\ B &= \Im(Y) = -\frac{X}{R^2 + X^2} \end{align}
The magnitude and phase of the admittance are given by:
\begin{align} \left | Y \right | &= \sqrt{G^2 + B^2} = \frac{1}{\sqrt{R^2 + X^2}} \\ \angle Y &= \arctan \left( \frac{B}{G} \right) = \arctan \left( -\frac{X}{R} \right) \end{align}
where
• G is the conductance, measured in siemens
• B is the susceptance, also measured in siemens
Note that (as shown above) the signs of reactances become reversed in the admittance domain; i.e. capacitive susceptance is positive and inductive susceptance is negative. | 696 | 2,427 | {"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} | 3.71875 | 4 | CC-MAIN-2020-40 | latest | en | 0.912082 |
https://math.stackexchange.com/questions/1517021/how-to-distinguish-between-combination-and-permutation-questions/1517032 | 1,571,632,672,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570987756350.80/warc/CC-MAIN-20191021043233-20191021070733-00217.warc.gz | 580,854,636 | 37,156 | # How to distinguish between combination and permutation questions?
How do you distinguish combination and permutation question?
An example of a combination question:
Example: How many different committees of 4 students can be chosen from a group of 15?
Answer: There are possible combinations of 4 students from a set of 15.
There are 1365 different committees.
An example of a permutation question:
Example: How many ways can 4 students from a group of 15 be lined up for a photograph?
Answer: There are 15P4 possible permutations of 4 students from a group of 15. These are different lineups.
How to know in which case to use $nCm$ and in which $nPm$?
• this is the real specific question. – guest11 Nov 7 '15 at 5:30
• Maybe don't read this. How many strings of $0$'s and/or $1$'s of length $10$ have exactly three $0$'s? Order in the string matters, but we solve the problem using combinations. The number of strings is the number of ways to choose where the $0$'s will go. There are ${}_{10}C_3$ ways to do it. – André Nicolas Nov 7 '15 at 5:53
• If you are making selection of things then it is permutation while if you are making arrangments then it is combination. Does this make sense . – Archis Welankar Nov 7 '15 at 6:16
• yes it makes sense. Thank you ^^ – guest11 Nov 7 '15 at 6:27
Example: How many different committees of 4 students can be chosen from a group of 15? Answer: There are possible combinations of 4 students from a set of 15.
In your first example you are choosing slots for each student to occupy, but you don't care which order the students make in each slot. You only care about the combination.
Example: How many ways can 4 students from a group of 15 be lined up for a photograph? Answer: There are 15P4 possible permutations of 4 students from a group of 15.
In permutations the order of the objects matters. If the objects are indistinguishable then it is a combinations problem.
In your second example you can distinguish one student from another; therefore it is permutations problem, so switching the order of the students in the line will affect the way the photograph looks.
To tell the difference you look at the 'object' in question and ask yourself is it distinguishable or not in the context of the question.
${}_nC_r$ is the number of ways of selecting $r$ items from a set of $n$ items.
${}_nP_r$ is the number of ways of selecting $r$ items from a set of $n$ items and arranging the $r$ selected items in all possible orders.
In brief, combinations are used when you are selecting only; permutations are used when you are selecting and arranging.
It is often said that combinations are used when order doesn't matter and that permutations are used when order matters, but this can be misleading. By "order matters", people probably mean that different orders are distinguishable, which is important, but not the only issue. The other important issue is whether all possible orders are allowed.
If, in selecting $r$ items from $n$, different orders are distinguishable and all possible orders are allowed, then ${}_nP_r$ is appropriate. If, in selecting $r$ items from $n$, either different orders are not distinguishable or only only one order is allowed, then ${}_nC_r$ would be appropriate.
Here are some examples.
1. How many ways can 4 students from a group of 15 be lined up for a photograph? Answer: to find all possible lineups, you have to find all possible selections, and for each selection, you have to find all possible ways of lining up the selected students. Different lineups are distinguishable and all possible lineups can occur. So there are ${}_{15}P_4$ lineups.
2. How many ways can 4 students from a group of 15 be lined up for a photograph in order by age, with oldest on the left and youngest on the right? Answer: to find all possible lineups, you have to find all possible selections. Different lineups are distinguishable, but only one lineup is allowed. Once you determine the selection selection, the lineup is forced, since it is according to age. Hence there are ${}_{15}C_4$ lineups.
3. How many different committees of four students can be chosen from a group of 15? Answer: committees are defined only by their membership, so different orders of the selected committee members are not distinguishable. Hence there are ${}_{15}C_4$ committees.
4. There are three E Scrabble tiles and four S Scrabble tiles. In how many ways can all of the tiles be placed side-by-side along a straight line? Answer: physical pieces of wood are distinguishable, and so different placements of them are distinguishable. Futhermore all possible placements are possible. Hence there are ${}_7P_7=7!$ placements.
5. How many different "words" can be spelled using the letter E three times and the letter S four times, where any sequence of letters counts as a word? Answer: a word is distinguished by the sequence of letters, that is, the order, but is not tied to the placement of any physical artifacts. Different sequences of Es and Ss are distinguishable, but the result of permuting only the Es or only the Ss in a particular sequence does not give a distinguishable result. So different orders within the subsequence of Es or within the subsequence of Ss are not distinguishable. Hence there are $$\frac{{}_7P_7}{{}_3P_3\,{}_4P_4}=\frac{7!}{3!\,4!}$$ words. This answer equals ${}_7C_4$, which can be understood as follows: to specify a word, you need only state which four letter positions among the 7 contain Ss, with the understanding that every other position contains an E. A set of four letter positions is distinguished only by its elements, and not by the order in which they are listed. So having Ss in positions 2, 3, 5, and 7 is indistinguishable from having Ss in positions 5, 3, 7, and 2. In other words, different orders of the four S positions are not distinguishable. Hence there are ${}_7C_4$ words.
• Did you mean ${}_7C_4$, which equals 35, or ${}_7C_7$, which equals 1? I will assume you meant ${}_7C_4$, which is the same as the answer to my fifth example. If you were to ask for the number of words that could be spelled with the seven Scrabble tiles, then examples 4 and 5 would be asking the same thing, and would have the same answer. But I was assuming in example 4 that the question was concerned with the arrangement of physical pieces of wood, which are distinguishable no matter whether they have the same letter written on them or not. – Will Orrick Nov 23 '15 at 11:34
• This physical distinguishability can make a difference in applications. Consider this example: you have 100 Scrabble tiles, 2 with the letter E and 98 with the letter S. The number of two-letter words that can be spelled with these tiles is $2^2=4$, but the number of ways of lining up two of the tiles is ${}_{100}P_2=9900$. If two tiles are selected at random and placed side-by-side,the probability that they spell EE is ${}_2P_2/{}_{100}P_2=1/4950$. If instead, there... – Will Orrick Nov 23 '15 at 12:00
• ...are 4 tiles, 2 with E and 2 with S, then there are still $2^2=4$ words that can be spelled, but now there are ${}_4P_2=12$ ways to line up 2 tiles, and the probability that the tiles spell EE is now ${}_2P_2/{}_4P_2=1/6$. – Will Orrick Nov 23 '15 at 12:01
• Indistinguishability does occur in the physical world, in the realm of quantum mechanics. Its effects are profound and, at first sight, strange. The phenomenon of Bose-Einstein condensation is a consequence of indistinguishability. See the Wikipedia article. – Will Orrick Nov 23 '15 at 12:51
• @BLAZE: I downvoted all the answers in this thread that attempt to prescribe a not completely mathematically correct way to determine whether to use P or C. That included one of your answers but not the other one. I strongly disagree with prescribing what to do, since it does not in any way foster understanding. – user21820 Nov 27 '15 at 10:14
In general whenever the position matters, the question is a permutation question. For example when (A,B) and (B,A) are different, it is a permutation question.
And whenever the position does not matter, it is a combination question. For example (A,B) is the same as (B,A).
Example :
1) In how many ways can we select 2 students from three students A,B,C?
Answer = $^3C_2 = \frac{3!}{2!\times(3-2)!} = 3$
This is a combination question because we can select in three ways as follows : (A,B) , (B,C) , (C,A). See here, when (A,B) is selected , it is the same as (B,A).
2) In how many ways can 2 students be lined in a row, out of a total of 3 students A,B,C ?
Answer = $^3 P_2 = \frac{3!}{(3-2)!} = 6$
This is a permutation question because there are six ways as follows: (A,B) , (B,A) , (B,C) , (C,B) , (A,C) , (C,A). Note that now (A,B) and (B,A) are different as in (A,B) , A gets the first position in the row and B gets the second position in the row. While in (B,A) , B gets the first position in the row and A gets the second position. Hence we have to count them as separate possibilities, unlike in the previous example.
• thank you! this makes it clear! :) – guest11 Nov 7 '15 at 6:24
• but the first question how many ways you will get? why do I get 1 when I use the fomula? – guest11 Nov 7 '15 at 6:26
• Please be specific , for which question do you get 1 as the answer ? – user180000 Nov 7 '15 at 8:09
• C(3,2) and follow by the formula ... – guest11 Nov 7 '15 at 8:50
Permutations are used when order matters.
Combinations are used when order doesn't matter.
• but the question did not ask directly? how do I know? will there questions ever ask directly? – guest11 Nov 7 '15 at 5:36
• It won't tell you directly to use a permutation or combination. You have to determine whether order is relevant in your problem. For example, if you are looking for "combinations" to a lock, this is in fact a misnomer! You are actually looking for permutations since the order of the numbers matters to open up the "combination lock". – Levonne Daruiche Nov 7 '15 at 5:38
Here is another interpretation for the difference between combinations (choosing) and permutations (arrangements):
Suppose you were asked to arrange $3$ books from $5$ different books in a line:
So you first $\color{red}{\mathrm{choose}}$ $3$ books from the possible $5$ and then you $\color{blue}{\mathrm{arrange}}$ them.
Therefore; the distinction goes as follows:
A Permutation is choosing the objects and then arranging them.
A Combination is choosing the objects and not arranging them. | 2,665 | 10,483 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.3125 | 4 | CC-MAIN-2019-43 | latest | en | 0.920993 |
https://rescript-lang.org.cn/docs/manual/latest/api/belt/map-int | 1,695,864,918,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510334.9/warc/CC-MAIN-20230927235044-20230928025044-00769.warc.gz | 523,168,105 | 13,619 | API / Belt / MapInt
# MapInt
Specalized when key type is `int`, more efficient than the generic type, its compare behavior is fixed using the built-in comparison
## key
```RES```type key = int
``````
## t
```RES```type t<'value>
``````
The type of maps from type `key` to type `'value`.
## empty
``````let empty: t<'v>
``````
## isEmpty
``````let isEmpty: t<'v> => bool
``````
## has
``````let has: (t<'v>, key) => bool
``````
## cmpU
``````let cmpU: (t<'v>, t<'v>, (. 'v, 'v) => int) => int
``````
## cmp
``````let cmp: (t<'v>, t<'v>, ('v, 'v) => int) => int
``````
## eqU
``````let eqU: (t<'v>, t<'v>, (. 'v, 'v) => bool) => bool
``````
## eq
``````let eq: (t<'v>, t<'v>, ('v, 'v) => bool) => bool
``````
`eq(m1,m2)` tests whether the maps `m1` and `m2` are equal, that is, contain equal keys and associate them with equal data.
## findFirstByU
``````let findFirstByU: (t<'v>, (. key, 'v) => bool) => option<(key, 'v)>
``````
## findFirstBy
``````let findFirstBy: (t<'v>, (key, 'v) => bool) => option<(key, 'v)>
``````
`findFirstBy(m, p)` uses function `f` to find the first key value pair to match predicate `p`.
```RES```let s0 = Belt.Map.Int.fromArray([(4, "4"), (1, "1"), (2, "2"), (3, "3")])
Belt.Map.Int.findFirstBy(s0, (k, v) => k == 4) == Some((4, "4"))
``````
## forEachU
``````let forEachU: (t<'v>, (. key, 'v) => unit) => unit
``````
## forEach
``````let forEach: (t<'v>, (key, 'v) => unit) => unit
``````
`forEach(m, f)` applies `f` to all bindings in map `m`. `f` receives the key as first argument, and the associated value as second argument. The bindings are passed to `f` in increasing order with respect to the ordering over the type of the keys.
## reduceU
``````let reduceU: (t<'v>, 'v2, (. 'v2, key, 'v) => 'v2) => 'v2
``````
## reduce
``````let reduce: (t<'v>, 'v2, ('v2, key, 'v) => 'v2) => 'v2
``````
`reduce(m, a, f)` computes `f(kN, dN, ... f(k1, d1, a)...)`, where `k1 ... kN` are the keys of all bindings in `m` (in increasing order), and `d1 ... dN` are the associated data.
## everyU
``````let everyU: (t<'v>, (. key, 'v) => bool) => bool
``````
## every
``````let every: (t<'v>, (key, 'v) => bool) => bool
``````
`every(m, p)` checks if all the bindings of the map satisfy the predicate `p`. Order unspecified
## someU
``````let someU: (t<'v>, (. key, 'v) => bool) => bool
``````
## some
``````let some: (t<'v>, (key, 'v) => bool) => bool
``````
`some(m, p)` checks if at least one binding of the map satisfy the predicate `p`. Order unspecified
## size
``````let size: t<'v> => int
``````
## toList
``````let toList: t<'v> => list<(key, 'v)>
``````
In increasing order.
## toArray
``````let toArray: t<'v> => array<(key, 'v)>
``````
## fromArray
``````let fromArray: array<(key, 'v)> => t<'v>
``````
## keysToArray
``````let keysToArray: t<'v> => array<key>
``````
## valuesToArray
``````let valuesToArray: t<'v> => array<'v>
``````
## minKey
``````let minKey: t<'a> => option<key>
``````
## minKeyUndefined
``````let minKeyUndefined: t<'a> => Js.undefined<key>
``````
## maxKey
``````let maxKey: t<'a> => option<key>
``````
## maxKeyUndefined
``````let maxKeyUndefined: t<'a> => Js.undefined<key>
``````
## minimum
``````let minimum: t<'v> => option<(key, 'v)>
``````
## minUndefined
``````let minUndefined: t<'v> => Js.undefined<(key, 'v)>
``````
## maximum
``````let maximum: t<'v> => option<(key, 'v)>
``````
## maxUndefined
``````let maxUndefined: t<'v> => Js.undefined<(key, 'v)>
``````
## get
``````let get: (t<'v>, key) => option<'v>
``````
## getUndefined
``````let getUndefined: (t<'v>, key) => Js.undefined<'v>
``````
## getWithDefault
``````let getWithDefault: (t<'v>, key, 'v) => 'v
``````
## getExn
``````let getExn: (t<'v>, key) => 'v
``````
## checkInvariantInternal
``````let checkInvariantInternal: t<'a> => unit
``````
raise when invariant is not held
## remove
``````let remove: (t<'v>, key) => t<'v>
``````
`remove(m, x)` returns a map containing the same bindings as `m`, except for `x` which is unbound in the returned map.
## removeMany
``````let removeMany: (t<'v>, array<key>) => t<'v>
``````
## set
``````let set: (t<'v>, key, 'v) => t<'v>
``````
`set(m, x, y)` returns a map containing the same bindings as `m`, plus a binding of `x` to `y`. If `x` was already bound in `m`, its previous binding disappears.
## updateU
``````let updateU: (t<'v>, key, (. option<'v>) => option<'v>) => t<'v>
``````
## update
``````let update: (t<'v>, key, option<'v> => option<'v>) => t<'v>
``````
## mergeU
``````let mergeU: (t<'v>, t<'v2>, (. key, option<'v>, option<'v2>) => option<'c>) => t<'c>
``````
## merge
``````let merge: (t<'v>, t<'v2>, (key, option<'v>, option<'v2>) => option<'c>) => t<'c>
``````
`merge(m1, m2, f)` computes a map whose keys is a subset of keys of `m1` and of `m2`. The presence of each such binding, and the corresponding value, is determined with the function `f`.
## mergeMany
``````let mergeMany: (t<'v>, array<(key, 'v)>) => t<'v>
``````
## keepU
``````let keepU: (t<'v>, (. key, 'v) => bool) => t<'v>
``````
## keep
``````let keep: (t<'v>, (key, 'v) => bool) => t<'v>
``````
## partitionU
``````let partitionU: (t<'v>, (. key, 'v) => bool) => (t<'v>, t<'v>)
``````
## partition
``````let partition: (t<'v>, (key, 'v) => bool) => (t<'v>, t<'v>)
``````
`partition(m, p)` returns a pair of maps `(m1, m2)`, where `m1` contains all the bindings of `s` that satisfy the predicate `p`, and `m2` is the map with all the bindings of `s` that do not satisfy `p`.
## split
``````let split: (key, t<'v>) => (t<'v>, option<'v>, t<'v>)
``````
`split(x, m)` returns a triple `(l, data, r)`, where `l` is the map with all the bindings of `m` whose key is strictly less than `x`; `r` is the map with all the bindings of `m` whose key is strictly greater than `x`; `data` is `None` if m contains no binding for `x`, or `Some(v)` if `m` binds `v` to `x`.
## mapU
``````let mapU: (t<'v>, (. 'v) => 'v2) => t<'v2>
``````
## map
``````let map: (t<'v>, 'v => 'v2) => t<'v2>
``````
`map(m, f)` returns a map with same domain as `m`, where the associated value `a` of all bindings of `m` has been replaced by the result of the application of `f` to `a`. The bindings are passed to `f` in increasing order with respect to the ordering over the type of the keys.
## mapWithKeyU
``````let mapWithKeyU: (t<'v>, (. key, 'v) => 'v2) => t<'v2>
``````
## mapWithKey
``````let mapWithKey: (t<'v>, (key, 'v) => 'v2) => t<'v2>
`````` | 2,397 | 6,507 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.5625 | 3 | CC-MAIN-2023-40 | latest | en | 0.683059 |
https://bytes.com/topic/access/answers/934595-report | 1,556,067,691,000,000,000 | text/html | crawl-data/CC-MAIN-2019-18/segments/1555578616424.69/warc/CC-MAIN-20190423234808-20190424020808-00182.warc.gz | 364,250,564 | 14,634 | 424,671 Members | 1,280 Online
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# Report
P: 24 I tried to run a report which shows the absolute values of X and a control called Y that displays values of X as a "percentage' with reference to the base values of another control say Z . The display format is set to percent and the decimals to display as 2. However when X and Z happen to be negative values, the control Y displays asterisks instead of 2 decimals percent. Tried to use various IIF conditions but still displays only the asterisks. How can I get the report to display the negative percentages to 2 decimals? Mar 8 '12 #1
23 Replies
Expert Mod 15k+ P: 31,276 No obvious answer I can see, but then you don't include much info in the question. If you fix the question to include the basic information required you may find someone can answer it. Mar 8 '12 #2
100+ P: 759 What is the formula you use for Y control ? Mar 9 '12 #3
P: 24 The formula for Y control is : X/Z which is X value/ Z VALUE. Mar 9 '12 #4
P: 24 To make my formula more clear it is X value over Z Value. ( if x =75 and z=100 then y is 75% ) Mar 9 '12 #5
100+ P: 759 :) I think is no one here which don't know how to calculate a percentage. The question is: What is the formula in the control's Control Source property? Mar 9 '12 #6
P: 24 More about this formula. If one of the values , x or z, is negative while the other is positive, the display turns into asterisks. I tried to put this condition in the control Y. Expand|Select|Wrap|Line Numbers IIf([Z]=0,Null,IIf([Z]<0,IIf([X]>0,Null,IIf([Z]>0,IIf([X]<0,Null,[X/Z]) I think I cannot make this more clearer , but even when X and Z are positive both or negative both, the display is Null. Cannot make out where my conditions are conflicting. Mar 9 '12 #7
100+ P: 759 I think that the best thing is to use this formula: Expand|Select|Wrap|Line Numbers =[x]/[z] This will show an error if Z = 0 (Division by zero) that inform you that something is wrong If you wish to manage the result as Null if Z=0 then: Expand|Select|Wrap|Line Numbers =IIf([Z]=0,Null,[X]/[Z]) Also you can place an worning message if Z=0: Expand|Select|Wrap|Line Numbers =IIf([z]=0,"Z = 0 ??? Bad for you !",[X]/[Z]) Paste any of this formulas in the control source of your text box Entire formula, including the "=" sign. Mar 9 '12 #8
Expert Mod 15k+ P: 31,276 What's the actual string value in the Format property of the control. Mihail: I think that the best thing is to use this formula: Expand|Select|Wrap|Line Numbers =[x]/[z] When Mihails says this I suspect he just means to try it to illustrate what you're working with - A temporary measure. It's not a good solution to your actual problem. Mar 9 '12 #9
100+ P: 759 Hi NeoPa ! Can you explain why you think that the first formula is not a good solution for this case ? From my view point is the best because the percentage is, in fact, a division, and need that the low number (down to division line - I don't know the English word) must be not zero. I am very happy when Access worn me this way (without stop to work) when something is wrong. Anyway I think that the main mistake (omission) of Vish is the omission of first "=" sign. Then, trying and trying he complicate itself by using other and other formula for the control source. Mar 9 '12 #10
Expert Mod 15k+ P: 31,276 Indeed Mihail. For developers it may be ok for items to fail, but users should not be presented with error results. It's the responsibility of the developer to handle such situations smoothly. Relying on Access to capture such errors with default messages is pretty poor design. Mihail: Anyway I think that the main mistake (omission) of Vish is the omission of first "=" sign. I'm not sure. If that were the case then why would it work for positive values? I suspect their Format string is failing with negative results - hence the request to post it. Mar 9 '12 #11
P: 24 Mihail and Neopa, thanks for the input. I managed to resolve one part of the formula which is the previous formula not displaying the results even when values of both x and z were both positive or both negative. The percentage display of Control Y is ok now. The formula I put in Control Y : Expand|Select|Wrap|Line Numbers IIf([z]=0,0,IIf([x]=0,0,IIf([z]<0 And [x]>0,0,IIf([z]>0 And [x]<0,0,IIf(Len(Round([x]/[z],2))>5,0,Round([x]/[z],2)))))) My problem is still not completely resolved since I find that when one of the values of X or Z are too low like for example X is -0.150 and Z is 0.150 , the result of Y is 1000% . But the report displays asterisks and since I require the % display only if the result of Y is within say -100.00% and +100.00%. Else is should display Null. This is what I tried to put in my formula but this too does not work because if len of Y is more than 4 , Y still displays the asterisks. I wish it would be display Null if Len more than 4. The format display for Y is "percent"and "decimals" is 2. Mar 10 '12 #12
100+ P: 759 This is not a very clever question, Vish, but I must ask: Is your control enough large to display entire value ? Mar 10 '12 #13
P: 24 Further to the above, all that is required is if Y is not within the range of -100% and +100% , Y should display say, zero or Null,( can be anyone of these) since the percentages when displayed, would be weird. In my above formula I tried to manage this by putting in control Y : IIf(Len(Round([x]/[z],2))>5,0, but Y still displays asterisks when len is more than 5. Mar 10 '12 #14
100+
P: 759
I play a little bit with Round() function and I discover that:
1) Do not seems to work in the control source for a control (text box) in a report. I think that in Access's designers concept that is not necessary because when you establish the number of decimals for the control the control itself perform the Round() function;
2) Work very well in a query. So you must consider the option to design a query then to base your report to this query;
Now, at query level:
3) The Round() function cut the decimals in excess (if exist) but not add decimals to the result;
4) The Len() function has as result a string (this is not a "discover") so, if the decimal point appear (exist) then it is counted to the length of the string, if not... not. So is not impossible to manage your problem via Len() function but is very hard. You can simple manage that by compare with a numeric value:
Expand|Select|Wrap|Line Numbers
1. IIF(Round(X/Z)>100 , ValueIfTrue , ValueIfFalse)
I have attached a small pic to illustrate points 3) and 4)
Attached Images
Report_Round.jpg (33.3 KB, 99 views)
Mar 10 '12 #15
100+ P: 759 Sorry. Forget point 1) from my last post. It work in report. My mistake. Again sorry !!! Mar 10 '12 #16
Expert Mod 15k+ P: 31,276 Mihail: This is not a very clever question, Vish, but I must ask: Is your control enough large to display entire value ? @Vish I think you may have missed this question from post #13, which is a shame as I believe it may be an important one. Try enlarging it greatly just to test the idea. Mar 10 '12 #17
P: 24 Mihail, thanks for the tip. I put this condition in the query and it worked finally!. Neopa, I did not want to enlarge the size since some percentages appeared too wierd to display. Thought better not to display at all unless it is within a range of +100% and -100%. Realized that "format" can be a very tricky function in a report. Mihail, I changed the condition of the "length" to include also the decimal point , as you indicated. I used the following condition in the query.= Expand|Select|Wrap|Line Numbers IIf([z]=0,0,IIf([x]=0,0,IIf([z]<0 And [x]>0,0,IIf([z]>0 And [x]<0,0,IIf(Len(FormatPercent([z]/[x],2))>9,0,Round([x]/[z],2)))))) In the report, set the format to percent/2 decimals.Not sure if this is a sensible solution, but it worked. Thanks, guys! Mar 10 '12 #18
Expert Mod 15k+ P: 31,276 First, please check out [code] Tags Must be Used. Vish: NeoPa, I did not want to enlarge the size since some percentages appeared too wierd to display. I suggested you do a test Vish. Not change your design permanently. A test is to get the information as to what the problem is, without which you will certainly struggle to find an appropriate solution. Mar 10 '12 #19
P: 24 Neopa,When x and z were in decimals, the percentages displayed after enlargement of the control Y in the report design, were too big to even read like for example -698679.41%. Mar 11 '12 #20
Expert Mod 15k+ P: 31,276 ... and therein lies your problem it would seem. I hope you're now glad you did the test ;-) How to resolve it is the next step. Such values indicate to me that either :Your data is not sensible. Your basic design is flawed as it cannot handle the data it needs to display. Essentially, if X is so large in relation to Z then either that data is faulty, or else it's fine, in which case you haven't designed to allow for what you've got. Mar 11 '12 #21
100+ P: 759 As usually, NeoPa take my face :) There are a lot of questions like WHY ? 1) Why you wish to handle that using Len() function ? This can't be a good approache because: Say you establish the number of digits to 3. So you can show values between -99 to 999 (% of course). If you need to display 0.01 you can't ?!?!? More, if you wish to display -0.01 you can't. 2) Why, if your logic is in numeric values ("the percentage can't be higher than... 500% and can't be less than... -400%") your option is to manage the number of digits to display and not the values itself: Expand|Select|Wrap|Line Numbers IIF(([X/[Z] > 4) And ([X]/[Z]<5), [X]/[Z],"Out of range, my dear") 3) Why you need to calculate negative values for percentage ? This is not a "real" WHY. Maybe you have good reasons that I can't see. Mar 11 '12 #22
Expert Mod 15k+ P: 31,276 Mihail: As usually, NeoPa take my face :) I'm not sure what that means exactly, but I'm sure it's good :-D Mar 11 '12 #23
100+ P: 759 "To take face" (to someone) means that you make the same things quicker, faster, better than the other one person. I have hope to be an international expression but now I understand that it isn't :) Mar 12 '12 #24 | 2,795 | 10,219 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.296875 | 3 | CC-MAIN-2019-18 | latest | en | 0.905609 |
http://polymath07.blogspot.com/2012/04/more-on-using-moments.html | 1,532,201,835,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676592654.99/warc/CC-MAIN-20180721184238-20180721204238-00470.warc.gz | 288,094,987 | 13,080 | ## Friday, April 13, 2012
### More on using the moments
kw: musings, time perspective
Three days ago I posted about how some little things I do daily add up over the decades. Of course, my thinking didn't stop there.
Most folks, upon reaching adulthood, have forty to eighty years yet to live. We all have things we must do, and we also do things we like to do. Our habits, weekly or daily, make up the days for us. So in our waking moments, what does a daily minute or a daily hour or quarter hour mean over forty years? If you come of long-lived stock, just double the figures below.
My baseline was the half hour spent each evening brushing teeth and showering, plus the cleanup and dressing, before going to bed. It was easy to note that this uses 1/48th of my total time, or one full year out of each 48. Recalculating for forty years means 40/48, or 304.36 days (304d 9h).
Commuting to and from work used to take me two hours daily, but in the past 16 years, it has been ten minutes each way, which comes to 2/3 of that 1/48 of a day, or 1/72nd of a day, but at a rate of some 240 days per year. In the past 16 years, it adds up to 53d 4h. During the prior ten years, the two daily hours times 240 days added up to 200 days commuting time. Earlier in life I usually had a short commute like I now do, so let's give those 21 years 1/72 of each of 240 days per year, for another 70 days. All totaled up, in 47 years of working life I have spent just over 323 days commuting, or 0.88 year.
On boring days the papers sometimes have an article about how the average American spends six hours daily in front of the TV. Some might, but I suspect for most of us it is less than that. One full hour daily with the TV, over forty years, adds up to twice my shower commitment, or more than 608 days. For the true couch potatoes who spend 6-8 hours daily? It comes to 10-13.3 years of the forty.
How about weekly activities? Some half of Americans spend an hour in church weekly. Ignoring getting ready and commuting, the "pew hours" come to 50 per year (52 for those who attend church while on vacation), or some 2000 hours over forty years. That comes to 83d 8h. Some churches seem to offer more; at least their devotees spend more time there. Say you spend 2 hours on Sunday (one hour in "Sunday school" and one in the "service"), and one hour midweek. That triple time attendance totals 250 days in forty years.
How many do laundry twice weekly? Does it take an hour of your time to wash and dry and sort a load? Maybe half that? That gets you into the same ballpark as attending church an hour weekly. What else do you do weekly; play a half hour of pickup basketball or tennis? That's just over forty days in forty years. How about hitting the gym (aerobics or treadmill) for 30 minutes three times weekly? That is 125 days in 40 years. Not a bad thing to do with just under one percent of your time.
So here are some round figures to think about:
• One-half hour weekly means about a day per year, or a bit over 40 days in 40 years.
• One-half hour daily is seven times as much: just over a week per year, or 43 weeks (actually 304 days) in 40 years.
• Anything you do 8 hours daily, whether sleep or watch TV, is consuming a full third of your time, which comes to 13 years and 122 days in 40 years. Let's just say such a habit finishes eighth grade over a forty year span!
That's all worth a little thought now and then, wouldn't you say? | 873 | 3,451 | {"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} | 2.53125 | 3 | CC-MAIN-2018-30 | latest | en | 0.955099 |
https://www.slideserve.com/forest/introduction-to-r-for-absolute-beginners-part-i | 1,723,031,456,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640694449.36/warc/CC-MAIN-20240807111957-20240807141957-00380.warc.gz | 769,675,510 | 24,132 | 1 / 87
# Introduction to R for Absolute Beginners: Part I
Introduction to R for Absolute Beginners: Part I. Melinda Fricke Department of Linguistics University of California, Berkeley melindafricke@berkeley.edu D-Lab Workshop Series, Spring 2013. Why this workshop?.
## Introduction to R for Absolute Beginners: Part I
E N D
### Presentation Transcript
1. Introduction to Rfor Absolute Beginners: Part I Melinda Fricke Department of Linguistics University of California, Berkeley melindafricke@berkeley.edu D-Lab Workshop Series, Spring 2013
2. Why this workshop? "The questions that statistical analysis is designed to answer can often be stated simply. This may encourage the layperson to believe that the answers are similarly simple. Often, they are not… No-one should be embarrassed that they have difficulty with analyses that involve ideas that professional statisticians may take 7 or 8 years of professional training and experience to master.” (Maindonald and Braun, 2010. Data Analysis and Graphics Using R: An Example-Based Approach.)
3. What we will cover today • Getting around in R • working directories, managing your workspace, creating and removing objects (types of variable assignment), inspecting objects, viewing functions, getting help • Types of data and basic manipulations • data types, object types, reading and writing data, modifying data and objects, basic functions
4. (What we will cover next time) Downloading and installing external packages Common statistical tests correlation, simple linear regression, t-tests, ANOVA Graphing R makes really beautiful graphs, and is very flexible
5. What we will not cover (ever) "The best any analysis can do is to highlight the information in the data. No amount of statistical or computing technology can be a substitute for good design of data collection, for understanding the context in which data are to be interpreted, or for skill in the use of statistical analysis methodology. Statistical software systems are one of several components of effective data analysis.” (Maindonald and Braun, 2010)
6. Why use R? R is an incredibly flexible, high-level programming language that will allow you to conduct nearly any statistical analysis, and create any visualization you can think of. This video was created by Ben Schmidt using the ggplot2 package. http://sappingattention.blogspot.com/2012/10/data-narratives-and-structural.html
7. A simpler example… (These are from my own work on the production of “s” sounds.)
8. But first, the basics…
9. Creating objects Open R and type the following: x = 1 [enter] y <- 2 [enter] 3 -> z [enter] x [enter] y [enter] z [enter] There are 3 ways to assign variables in R.
10. Creating objects Now try this: x + y + z x + y + z -> q q What’s the difference between the first line and the second?
11. Creating objects These are vectors. A vector is just a bunch of values that have been concatenated together, in a sequence. When we type “q”, R tells us that the first element in the vector “q” is 6: [1] 6 (It’s also the only element, but that’s okay.)
12. Creating objects We can create vectors that are longer than 1 element by concatenating multiple elements together: x = c(7, 8, 8, 7, 4, 1) x length(x)
13. A little bit about “looping” R is a “high level” programming language. This means it takes care of a lot of things for us. x * 2 x + y Most programming languages require you to write loops, but R takes care of a lot of “looping” on its own.
14. Pop quiz! What will this code produce? length(x + y)
15. A little bit about looping What will this code produce? length(x + y) [1] 6
16. A little bit about looping What will this code produce? length(x + y) [1] 6 The length of x is 6. (x + y) loops through the vector x, adding y to each number, yielding 6 new values (and therefore a vector of length 6). Remember: if you want to concatenate, use c(): length(c(x,y)) [1] 7
17. Pop quiz! What will this code produce? length(x + y) [1] 6 length(length(x + y))
18. Pop quiz! What will this code produce? length(x + y) [1] 6 length(length(x + y)) [1] 1
19. A little bit about looping What will this code produce? x = c(7, 8, 8, 7, 4, 1) y = c(1, 2) x + y
20. A little bit about looping What will this code produce? x = c(7, 8, 8, 7, 4, 1) y = c(1, 2) x + y [1] 8 10 9 9 5 3
21. A little bit about looping What will this code produce? x = c(7, 8, 8, 7, 4, 1) y = c(1, 2) x + y [1] 8 10 9 9 5 3 “Loop through x and y simultaneously, adding 2 elements together.” For operations using 2 vectors of different lengths, the shorter one will be “recycled”. (Look at y + x.)
22. Data types class(x) class(‘x’)
23. Data types class(x) [1] “numeric” class(‘x’) [1] “character”
24. Data types class(x) [1] “numeric” class(‘x’) [1] “character” There are different types of data in R. Putting quotes around something indicates you want R to treat it literally - as a character, not a variable. e.g. class(‘1’)
25. Data types as.character(x) -> k k + 1
26. Data types as.character(x) -> k k + 1 Your first error message! as.numeric(k) -> k k + 1
27. Data types as.factor(k)
28. Data types as.factor(k) A factor is R-speak for a categorical variable: a type of data that can have one of several fixed levels. By default, factor levels are ordered alphabetically. levels(k)
29. Data types as.factor(k) A factor is R-speak for a categorical variable: a type of data that can have one of several fixed levels. By default, factor levels are ordered alphabetically. levels(k) Oops! Why doesn’t this work?
30. Data types as.factor(k) -> k A factor is R-speak for a categorical variable: a type of data that can have one of several fixed levels. By default, factor levels are ordered alphabetically. levels(k) [1] “1” “4” “7” “8”
31. Data types c(10, 11, 10, 8, 6, 12, 11, 8, 10, 6, 10, 11) -> p
32. Using what we’ve learned so far… c(10, 11, 10, 8, 6, 12, 11, 8, 10, 6, 10, 11) -> p How many items are in the vector ‘p’? How many unique values are in ‘p’?
33. Using what we’ve learned so far… c(10, 11, 10, 8, 6, 12, 11, 8, 10, 6, 10, 11) -> p How many items are in the vector ‘p’? length(p) [1] 12 How many unique values are in ‘p’?
34. Using what we’ve learned so far… c(10, 11, 10, 8, 6, 12, 11, 8, 10, 6, 10, 11) -> p How many items are in the vector ‘p’? length(p) [1] 12 How many unique values are in ‘p’? length(levels(as.factor(p)))
35. Your first R weirdness as.factor(p) -> p as.numeric(p) What happened??
36. Your first R weirdness as.factor(p) -> p as.numeric(p) [1] 3 4 3 2 1 5 4 2 3 1 3 4 ( 10 11 10 8 6 12 11 8 10 6 10 11) When you try to change a factor directly into numeric mode, the factors are replaced by their “order”. How could we avoid this?
37. Your first R weirdness as.numeric(as.character(p))
38. Your first R weirdness as.numeric(as.character(p)) What if we want to change the order of the levels?
39. Your first R weirdness as.numeric(as.character(p)) What if we want to change the order of the levels? factor(p, levels=c(‘6’, ‘8’, ‘10’, ‘12’, ‘11’)) -> p levels(p)
40. Ordered factors Factors may make more sense if we give our categories names other than numbers. Try this: mycolors = c(‘blue’, ‘yellow’, ‘green’, ‘purple’, ‘red’) class(mycolors) factor(mycolors, levels=c(‘red’, ‘yellow’, ‘green’, ‘blue’, ‘purple’)) -> mycolors class(mycolors) levels(mycolors)
41. Taking stock Objects and operations values e.g. ‘1’ vectors c(‘1’, ‘4’, ‘a’, ‘word’) functions as.factor(x) variable assignment =, ->, <- Data types (‘classes’) numeric 8 character ‘8’, ‘x’, ‘female’ factor ‘8’, ‘x’, ‘female’ the difference between strings of characters and factors is that factors have one of a set of fixed values e.g. ‘male’ vs. ‘female’
42. Some more useful functions Type these commands in to see what they do: ls() table(p) unique(p) sort(p) mean(p) median(p) sd(p) edit(p) ls
43. Some more useful functions Type these commands in to see what they do: ls() lists the objects currently in your workspace table(p) unique(p) sort(p) mean(p) median(p) also useful: rm() (remove) sd(p) edit(p) ls
44. Some more useful functions Type these commands in to see what they do: ls() lists the objects currently in your workspace table(p) creates a table of counts unique(p) sort(p) mean(p) median(p) sd(p) edit(p) ls
45. Some more useful functions Type these commands in to see what they do: ls() lists the objects currently in your workspace table(p) creates a table of counts unique(p) lists all existing unique values sort(p) mean(p) median(p) sd(p) edit(p) ls
46. Some more useful functions Type these commands in to see what they do: ls() lists the objects currently in your workspace table(p) creates a table of counts unique(p) lists all existing unique values sort(p) sorts values from lowest to highest mean(p) median(p) sd(p) edit(p) ls
47. Some more useful functions Type these commands in to see what they do: ls() lists the objects currently in your workspace table(p) creates a table of counts unique(p) lists all existing unique values sort(p) sorts values from lowest to highest mean(p) mean of the values median(p) median (middle) of the values sd(p) standard deviation edit(p) ls
48. Some more useful functions Type these commands in to see what they do: ls() lists the objects currently in your workspace table(p) creates a table of counts unique(p) lists all existing unique values sort(p) sorts values from lowest to highest mean(p) mean of the values median(p) median (middle) of the values sd(p) standard deviation edit(p) lets you interact directly with the data! “edit(p) -> p” to save your changes ls
49. Some more useful functions Type these commands in to see what they do: ls() lists the objects currently in your workspace table(p) creates a table of counts unique(p) lists all existing unique values sort(p) sorts values from lowest to highest mean(p) mean of the values median(p) median (middle) of the values sd(p) standard deviation edit(p) lets you interact directly with the data! ls displays the internal workings of the function
50. Getting help with functions ?sort search current packages for a function help(sort) (these two are equivalent) ??sort search all packages for a word
More Related | 2,759 | 10,195 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.59375 | 3 | CC-MAIN-2024-33 | latest | en | 0.866569 |
https://www.weegy.com/?ConversationId=AJMLM3VL | 1,600,568,296,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400193087.0/warc/CC-MAIN-20200920000137-20200920030137-00414.warc.gz | 1,167,432,962 | 10,441 | The highest mountains on earth are: the Himalayas the Appalachians under the ocean
The highest mountains on earth are the Himalayas.
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The highest mountains on earth are the Himalayas.
Confirmed by yumdrea [4/15/2014 12:09:52 PM]
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Given: y varies directly with x and has a constant rate of change of 7. When the value of y is 12, then the value of x would be _____. equal to zero less than 2 equal to 7 greater than 12
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y varies directly with x and has a constant rate of change of 7.
Then the equation should be: y = 7x
When y = 12, x = 12/7 which is less than 2.
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bacterial infection:
Weegy: Infection is the invasion of a host organism's bodily tissues by disease-causing organisms, their multiplication, and the reaction of host tissues to these organisms and the toxins they produce. [ Infections are caused by microorganisms such as viruses, prions, bacteria, and viroids, and larger organisms like macroparasites and fungi. ] (More)
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A191646 Triangle read by rows: T(n,k) = number of connected multigraphs with n >= 0 edges and 1 <= k <= n+1 vertices, with no loops allowed. 19
1, 0, 1, 0, 1, 1, 0, 1, 2, 2, 0, 1, 3, 5, 3, 0, 1, 4, 11, 11, 6, 0, 1, 6, 22, 34, 29, 11, 0, 1, 7, 37, 85, 110, 70, 23, 0, 1, 9, 61, 193, 348, 339, 185, 47, 0, 1, 11, 95, 396, 969, 1318, 1067, 479, 106, 0, 1, 13, 141, 771, 2445, 4457, 4940, 3294, 1279, 235 (list; table; graph; refs; listen; history; text; internal format)
OFFSET 0,9 LINKS Andrew Howroyd, Table of n, a(n) for n = 0..1274 (terms 0..119 from R. J. Mathar) R. J. Mathar, Statistics on Small Graphs, arXiv:1709.09000 [math.CO], 2017; see Section 4. Brendan McKay and Adolfo Piperno, nauty and Traces. [nauty and Traces are programs for computing automorphism groups of graphs and digraphs.] B. D. McKay and A. Piperno, Practical Graph Isomorphism, II, J. Symbolic Computation 60 (2013), 94-112. Gordon Royle, Small Multigraphs. FORMULA T(n,k=3) = A253186(n) = A034253(n,k=2) for n >= 1. - Petros Hadjicostas, Oct 02 2019 EXAMPLE Triangle T(n,k) (with rows n >= 0 and columns k >= 1) begins as follows: 1; 0, 1; 0, 1, 1; 0, 1, 2, 2; 0, 1, 3, 5, 3; 0, 1, 4, 11, 11, 6; 0, 1, 6, 22, 34, 29, 11; ... PROG (PARI) EulerT(v)={my(p=exp(x*Ser(dirmul(v, vector(#v, n, 1/n))))-1); Vec(p/x, -#v)} InvEulerMT(u)={my(n=#u, p=log(1+x*Ser(u)), vars=variables(p)); Vec(sum(i=1, n, moebius(i)*substvec(p + O(x*x^(n\i)), vars, apply(v->v^i, vars))/i) )} permcount(v) = {my(m=1, s=0, k=0, t); for(i=1, #v, t=v[i]; k=if(i>1&&t==v[i-1], k+1, 1); m*=t*k; s+=t); s!/m} edges(v, x)={sum(i=2, #v, sum(j=1, i-1, my(g=gcd(v[i], v[j])); g*x^(v[i]*v[j]/g))) + sum(i=1, #v, my(t=v[i]); ((t-1)\2)*x^t + if(t%2, 0, x^(t/2)))} G(n, m)={my(s=0); forpart(p=n, s+=permcount(p)*EulerT(Vec(edges(p, x) + O(x*x^m), -m))); s/n!} R(n)={Mat(apply(p->Col(p+O(y^n), -n), InvEulerMT(vector(n, k, 1 + y*Ser(G(k, n-1), y)))))} { my(A=R(10)); for(n=1, #A, for(k=1, n, print1(A[n, k], ", ")); print) } \\ Andrew Howroyd, May 14 2018 CROSSREFS Row sums give A076864. Diagonal is A000055. Cf. A034253, A054923, A192517, A253186 (column k=3), A290778 (column k=4). Cf. A000664, A007718, A036250, A050535, A191646, A191970, A275421, A317672, A322114, A322133, A322152. Sequence in context: A296068 A144064 A172236 * A297321 A277938 A130020 Adjacent sequences: A191643 A191644 A191645 * A191647 A191648 A191649 KEYWORD nonn,tabl AUTHOR Alberto Tacchella, Jul 04 2011 STATUS approved
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Last modified December 8 01:57 EST 2019. Contains 329850 sequences. (Running on oeis4.) | 1,346 | 3,277 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2019-51 | latest | en | 0.725179 |
http://www.swarthmore.edu/NatSci/mzucker1/e91/assignment2.html | 1,542,286,574,000,000,000 | text/html | crawl-data/CC-MAIN-2018-47/segments/1542039742685.33/warc/CC-MAIN-20181115120507-20181115141650-00059.warc.gz | 515,438,475 | 2,233 | # E91 Assignment 2
### Task 1: Update software.
The starter code has been updated with minor bugfixes and a major revamp to inverse kinematics. I've also added code to compute the overall center of mass (COM) for a system in the KinModel class -- all you need to do is provide mass information when you create the bodies.
You can get the software from this link (once again, I hope to get Git up and running before too much longer).
### Task 2: Model a humanoid in profile.
Modify the kintest.py example code to develop a model of a human figure in profile (side view). It doesn't have to look as pretty as the example below, but you should do your best to get the link lengths and masses correct. Make sure that you can visualize at least the position of each rigid body's center of mass (light red outlined circles below), as well as the position of the overall center of mass of the system (red filled circle below).
I've also visualized the supporting line segment (horizontal blue line) and the projection of the overall COM onto the support (vertical blue line). You can tell this position is statically stable because the blue lines overlap.
### Task 3: Generate random poses.
Pose your humanoid model in a few random poses to convince yourself you have gotten the kinematics correct. Here are two random poses from my model:
Clearly, these poses are not statically stable.
How were they generated? I start with a random set of joint angles for the entire body, and then transform everything so that the left foot is at the origin, like this:
```self.xforms = self.model.transforms(self.jvalues)
footXform = self.model.manipulatorFK(self.xforms, self.l_foot_manip)
self.baseXform = footXform.inverse()
self.xforms = [ self.baseXform * x for x in self.xforms ]```
This way, you don't need to worry about inverse kinematics, nor do you need to root the entire kinematic tree from the foot link.
### Task 4: Generate stable poses
Once your random poses are looking reasonable, modify your program to collect at least 100 of them. Here are a couple samples from my program:
Produce a webpage demonstrating your program. Provide images of your humanoid model in a neutral pose, several (at least 2) random unstable poses, and several (at least 10) random stable poses. Comment on the performance of your system. How long does it take to generate stable poses? Can you think of ways to make it faster?
We will review all of the assignments in class on Monday the 4th. Be prepared to show your webpage and/or demonstrate your software in action. | 563 | 2,561 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.59375 | 3 | CC-MAIN-2018-47 | latest | en | 0.91645 |
https://prezi.com/jwyosnsox15i/how-much-energy-does-dribbling-take/ | 1,524,788,590,000,000,000 | text/html | crawl-data/CC-MAIN-2018-17/segments/1524125948617.86/warc/CC-MAIN-20180426222608-20180427002608-00424.warc.gz | 693,960,113 | 20,679 | ### Present Remotely
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# How Much Energy Does Dribbling Take?
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## earnest thompson
on 5 February 2014
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#### Transcript of How Much Energy Does Dribbling Take?
How Much Energy Does Dribbling Take?
Playing basketball can be hard work. Players not only constantly run around the court, but just dribbling the basketball takes a lot of effort, too. Why is that? It has to do with how the basketball bounces. When the ball hits the court, its bounce actually loses momentum by transferring some of its energy into a different form. This means that to keep the ball bouncing, players must continually put more energy into the ball. In this sports science project, you will determine how high a basketball bounces on different surfaces relative to the height from which it was dropped.
Abstract
Model
1.Prepare the wall, or other vertical panel, next to the first surface you want to test so that you can measure the height of the basketball's bounce. a.On the wall next to the surface, use a tape measure and the blue painter's tape to mark every 20 centimeters (cm), starting from the floor and going up to 100 cm. You should end up with five tape marks, as shown in Figure 3 below.
b.Note: You can make the tape marks longer than the ones in Figure 3, so they will be easier to see in the video recording. Remember to put the top edge of the tape at the every 20 cm mark.
procedure
•Tape measure, metric
Materials
Bounce, bounce, swish! Playing a game of basketball is hard work, and one part of that workout comes from just dribbling the ball. In Figure 1 below, you can see Kobe Bryant, playing on the U.S. Men's Olympic team, dribbling a ball. Why does it take effort to dribble the ball? When a basketball hits the ground (and as it flies through the air), it actually transforms some of its energy to another form. If players do not put enough energy back into the ball, they will not be able to dribble it effectively.
Introduction
•Painter's tape. This is available at hardware stores or online through sellers
•Video camera. A video camera on a cell phone should work fine.
•Volunteer to videotape the experiments, or a tripod. If you use a tripod, you will need a surface nearby, such as a chair, to set the tripod and camera on.
Full transcript | 600 | 2,650 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.609375 | 4 | CC-MAIN-2018-17 | latest | en | 0.954434 |
https://goprep.co/ex-6.4-q8-p-q-r-are-the-mid-points-of-the-sides-of-abc-x-y-z-i-1nko5i | 1,603,418,496,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107880519.12/warc/CC-MAIN-20201023014545-20201023044545-00154.warc.gz | 349,321,198 | 42,398 | Q. 85.0( 1 Vote )
# P, Q, R are the mid-points of the sides of ΔABC. X, Y, Z are the mid-points of the sides of ΔPQR. If the area of ΔXYZ is 10, find the area of APQR and the area of ΔABC.
In ∆ABC, P, Q, R are the mid-points of the sides AB, BC and CA respectively.
The correspondence ∆ABC∆QRP is a similarity.
Areas of similar triangles are proportional to the squares of their corresponding sides.
∆ABC = ∆POR
Similarly, X,Y and Z are the mid-points of the sides of ∆PQR, we get
∆PQR = 4∆XYZ
∆PQR = 4×10
∆PQR = 40
Thus, the area of ∆PQR is 40 sq. units.
∆ABC = 4∆PQR
∆ABC = 4×40
∆ABC = 160
Hence, the area of ∆ABC is 160 sq. units.
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view all courses | 445 | 1,457 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2020-45 | latest | en | 0.884512 |
http://www.bookrags.com/questions/math/Algebra/solve-and-check-10-2x--3x-20--170588 | 1,506,211,204,000,000,000 | text/html | crawl-data/CC-MAIN-2017-39/segments/1505818689806.55/warc/CC-MAIN-20170923231842-20170924011842-00690.warc.gz | 420,745,006 | 5,810 | # Solve and check. 10 - 2x = 3x - 20
Holt McDougal Larson Algebra 1 Common Core Edition EP3 number 30
10 - 2x = 3x - 20
10-10-2x=3x-20-10
-2x=3x-30
-2x-3x=3x-3x-30
-5x=-30
-5x/-5=-30/-5
x=6
Check:
10 - 2x = 3x - 20
10-2(6)=3(6)-20
10-12=18-20
-2=-2 | 148 | 261 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.0625 | 4 | CC-MAIN-2017-39 | latest | en | 0.249814 |
http://hackage.haskell.org/package/linear-1.18.0.1/docs/Linear-V3.html | 1,571,756,938,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570987822098.86/warc/CC-MAIN-20191022132135-20191022155635-00139.warc.gz | 88,216,920 | 4,267 | linear-1.18.0.1: Linear Algebra
Linear.V3
Description
3-D Vectors
Synopsis
# Documentation
data V3 a Source
A 3-dimensional vector
Constructors
V3 !a !a !a
Instances
Monad V3 Functor V3 MonadFix V3 Applicative V3 Foldable V3 Traversable V3 Generic1 V3 Distributive V3 Representable V3 MonadZip V3 Serial1 V3 Traversable1 V3 Foldable1 V3 Apply V3 Bind V3 Eq1 V3 Ord1 V3 Read1 V3 Show1 V3 Additive V3 Metric V3 R1 V3 R2 V3 R3 V3 Trace V3 Affine V3 Unbox a => Vector Vector (V3 a) Unbox a => MVector MVector (V3 a) Num r => Coalgebra r (E V3) Bounded a => Bounded (V3 a) Eq a => Eq (V3 a) Floating a => Floating (V3 a) Fractional a => Fractional (V3 a) Data a => Data (V3 a) Num a => Num (V3 a) Ord a => Ord (V3 a) Read a => Read (V3 a) Show a => Show (V3 a) Ix a => Ix (V3 a) Generic (V3 a) Storable a => Storable (V3 a) Binary a => Binary (V3 a) Serial a => Serial (V3 a) Serialize a => Serialize (V3 a) NFData a => NFData (V3 a) Hashable a => Hashable (V3 a) Unbox a => Unbox (V3 a) Ixed (V3 a) Epsilon a => Epsilon (V3 a) FunctorWithIndex (E V3) V3 FoldableWithIndex (E V3) V3 TraversableWithIndex (E V3) V3 Each (V3 a) (V3 b) a b Typeable (* -> *) V3 type Rep1 V3 type Rep V3 = E V3 type Diff V3 = V3 data MVector s (V3 a) = MV_V3 !Int (MVector s a) type Rep (V3 a) data Vector (V3 a) = V_V3 !Int (Vector a) type Index (V3 a) = E V3 type IxValue (V3 a) = a
cross :: Num a => V3 a -> V3 a -> V3 a Source
cross product
triple :: Num a => V3 a -> V3 a -> V3 a -> a Source
scalar triple product
class R1 t where Source
A space that has at least 1 basis vector `_x`.
Minimal complete definition
Nothing
Methods
_x :: Lens' (t a) a Source
````>>> ````V1 2 ^._x
```2
```
````>>> ````V1 2 & _x .~ 3
```V1 3
```
Instances
R1 Identity R1 V1 R1 V2 R1 V3 R1 V4 R1 f => R1 (Point f)
class R1 t => R2 t where Source
A space that distinguishes 2 orthogonal basis vectors `_x` and `_y`, but may have more.
Minimal complete definition
Nothing
Methods
_y :: Lens' (t a) a Source
````>>> ````V2 1 2 ^._y
```2
```
````>>> ````V2 1 2 & _y .~ 3
```V2 1 3
```
_xy :: Lens' (t a) (V2 a) Source
Instances
R2 V2 R2 V3 R2 V4 R2 f => R2 (Point f)
_yx :: R2 t => Lens' (t a) (V2 a) Source
````>>> ````V2 1 2 ^. _yx
```V2 2 1
```
class R2 t => R3 t where Source
A space that distinguishes 3 orthogonal basis vectors: `_x`, `_y`, and `_z`. (It may have more)
Minimal complete definition
Nothing
Methods
_z :: Lens' (t a) a Source
````>>> ````V3 1 2 3 ^. _z
```3
```
_xyz :: Lens' (t a) (V3 a) Source
Instances
R3 V3 R3 V4 R3 f => R3 (Point f)
_xz :: R3 t => Lens' (t a) (V2 a) Source
_yz :: R3 t => Lens' (t a) (V2 a) Source
_zx :: R3 t => Lens' (t a) (V2 a) Source
_zy :: R3 t => Lens' (t a) (V2 a) Source
_xzy :: R3 t => Lens' (t a) (V3 a) Source
_yxz :: R3 t => Lens' (t a) (V3 a) Source
_yzx :: R3 t => Lens' (t a) (V3 a) Source
_zxy :: R3 t => Lens' (t a) (V3 a) Source
_zyx :: R3 t => Lens' (t a) (V3 a) Source
ex :: R1 t => E t Source
ey :: R2 t => E t Source
ez :: R3 t => E t Source | 1,190 | 3,025 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2019-43 | latest | en | 0.700426 |
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