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http://www.reddit.com/r/compsci/comments/1geq45/code_that_writes_tests_and_optimizes_other_code_ai/ | 1,386,427,727,000,000,000 | text/html | crawl-data/CC-MAIN-2013-48/segments/1386163054610/warc/CC-MAIN-20131204131734-00063-ip-10-33-133-15.ec2.internal.warc.gz | 511,723,738 | 19,573 | [–] 24 points25 points ago
One cutting-edge (and super-cool) technique is program synthesis: Synthesis.
This method can take a lot of time to achieve nontrivial results, however. But it's pretty amazing.
Otherwise, the field of autotuning generally tries to do what you describe (which returns an optimally-performing kernel on any given machine). FFTW was one of the pioneers in doing this, and further frameworks were developed for certain mathematical kernels (ATLAS, SPIRAL, OSKI.. MAGMA for GPU's). These generally tune to minimize bandwidth usage by tuning for machine-specific memory hierarchies, often by running thousands of microbenchmarks.
[–] 5 points6 points ago
On a similar theme, but using genetic programming: evolving gzip for CUDA.
[–][S] 0 points1 point ago
Awesome, this is what I was looking for.
[–] 2 points3 points ago
Regarding the matrices thing, can you give a better description of what you're trying to achieve?
[–][S] 0 points1 point ago
Let say, as a toy example, you want automatically created code that will sort. With this framework, you would give it a randomly arranged vector and a sorted one and it would eventually return code that sorts vectors. Then, you could have it select among possible solutions that use the fewest resources (i.e. cpu, memory).
[–] 1 point2 points ago
Don't you mean that you'll provide a lot of samples?
[–][S] 0 points1 point ago
Right, many vector pair examples or 2 very large vectors. It would be interesting to see how little data you could give something like this, and get back a recognizable/known to work algorithm.
[–] 0 points1 point ago
Maybe instead of thinking of input/output samples, you could imagine providing a function that would measure the correctness of the result. For some things that means trying out known input/output pairs, but for other things, that could mean doing a simple algorithm to verify the results. For example, you can verify the output array is sorted without knowing how to sort. You could even measure "how sorted" it is.
[–] 0 points1 point ago
It seems to me like this might be something a quantum computer could handle. See for example the tutorials for the DWave system regarding image recognition. You could have the quantum computer learn the algorithm based on the training data you give it, and then spit the algorithm out in some way that would be suitable for the average computer.
[–] 0 points1 point ago
Artificial Neural Networks are trained this way, FYI.
[–][S] 0 points1 point ago
Yea, it would be tough to get code-generation to work as well as the deep-net stuff is in computer vision, but if it did, it might be possible to understand what exactly (some of) the code is doing, re-use useful modules, or improve something a silly computer missed.
[–] 0 points1 point ago
This was quite a big field in the 80s as part of the research effort into 5th generation languages. The Programmer's Apprentice was one project back then that you might find interesting.
[–] 2 points3 points ago
You probably want something closer to machine learning / regression analysis. It's possible with Google's Prediction API or a variety of (more difficult) prediction modeling tools out there.
If it's up to a framework to understand converting [1, 2] to [2, 4], how can it assume you're multiplying? You could be assigning numbers randomly. You could be mandating that anything other than 1 or 2 would return 0. It could be 2, 2*2, 2*2*2...
[–][S] 0 points1 point ago
There would be infinite solutions to [1, 2] to [2, 4], but if it optimized toward a solution that uses the least amount of code, M2=M1*2; might be the winner. Depending on the size of the input, the relationships between them and the optimization constraints, sometimes there will be many, one, or no solution(s). Accuracy will vary by situation. That's cool because it can be dealt with by someone who can either improve the input, the parameters, or go in a make a few tweaks in the generated code.
[–] -1 points0 points ago
If human checkers are going to intervene to make sure that your program wrote the right algorithm, why not have them write the algorithm themselves? Computer-written algorithms are good for things like statistics or uncertain relationships. Writing the logic for an algorithm lets you catch mistakes like if(salary < 0){ do something } which a computer wouldn't expect or understand as incorrect answers.
[–][S] 0 points1 point ago
The better-for-a-human algorithms are to test that code generation is working as expected. The cool stuff happens when new algorithms are created that humans are bad at making.
[–] 2 points3 points ago
This sounds like genetic programming to me.
[–] -1 points0 points ago
I feel like this shouldn't be possible, i.e. the halting problem
[–] 4 points5 points ago
Why would the halting problem show up here? We aren't asking if a program halts, only that it generates a program which outputs the correct answer. We don't necessarily need to run said program to demonstrate that -- we can skirt it in one of two ways. Either by construction -- ie, never allowing the program to self-reference / loop (essentially enumerating every case), this will approximate a 'correct' (potentially non-halting) algorithm, but can be proved to halt by construction.
Alternately, we could operate by training a finite approximation algorithm, eg a neural net, or some appropriate markov chain, to approximate the answer as well.
If you want to restrict us to producing source code in some 'real' language, then we can adopt a genetic approach and use a simple timeout mechanism.
In short -- the halting problem isn't an issue unless you start asking if the program halts. Asking if a program halts in general is hard, but by either re-framing the problem in terms of known-to-halt methods, or as a construction in a non-turing-complete language, or using a timeout mechanism, you can skirt the issue pretty easily.
[–] 0 points1 point ago
I admit I was too vague. When I said "i.e. the halting problem", I meant corollaries of the halting problem in computability theory. One such corollary says that given two programs A and B, it is impossible to write a program that decides if A and B produce the same output for any input. This means in the original question, if the ai algorithm came up with a program, it wouldn't be able to know for sure its program is completely correct. However after rereading the thread, this problem seems heuristic by nature, and its definitely possible to make a good heuristic algorithm.
[–] 0 points1 point ago
That's talking about two arbitrary programs, but if, again, program B is constructed by from program A using only transformations that are known to conserve semantics, then we're solving for a very specific case, and we do know that programs A and B produce the same output for any input. Isn't that, generally speaking, the entirety of compiler design?
[–] 0 points1 point ago
Program A is the solution, and B is the constructed one, but we don't know A at the start, that's the one we're trying to find. So we can't construct B from A using transformations.
[–] 1 point2 points ago
I think the halting problem would be an issue here but not an entirely unworkable one. You could randomly alter the instructions and then time-out if it takes too long, not actually caring if it halts or not.
[–] 1 point2 points ago
We clearly cannot be solving the halting problem, yet we are able to write code. We just don't have the techniques, intelligence, etc, yet, in order to write a computer program that does that sort of thing, though we are getting pretty damn good.
[–] 0 points1 point ago
well it's not doing the first step very well if it needs the rest of the steps .^
[–] 0 points1 point ago
quickcheck is a marvelous little automatic testing framework. http://en.wikipedia.org/wiki/QuickCheck
It generates test cases automatically and takes your logical assertions about how the function should work and verifies it. | 1,827 | 8,040 | {"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} | 2.515625 | 3 | CC-MAIN-2013-48 | latest | en | 0.939955 |
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Message board for the users of flat assembler.
Index > Main > String to number and vice versa ??
Author
Thread
saigon
Joined: 29 May 2006
Posts: 62
saigon
Hello!
I have been looking thourgh these forums, but didn't find a solution to this.
I need to convert a string (which has only numbers) to a number. And then again to convert a number into a string.
Here's an example of what I am trying to approach in pseudo-code:
Code:
```MyString db "12345"
MyNumber equ 0
ChangeStringIntoNumber MyString, MyNumber ; MyNumber now holds 12345 (as integer)
INC MyNumber ; Increment the MyNumber (now it's 12346)
ChangeNumberIntoString MyNumber, MyString ; Convert the number back to string (MyString is now "12346") ```
Please note that this is only an example of thousands.
Thanks! I appreciate any kind of help!
EDIT: I just forgot to mention that this shouldn't be OS or library dependent. Thanks!
12 Jun 2006, 17:59
rugxulo
Joined: 09 Aug 2005
Posts: 2341
Location: Usono (aka, USA)
rugxulo
unsigned decimal to ASCII
I don't know of any ASCII to decimal routines offhand (besides my own, and it's 16-bit and probably not good enough for what you want).
13 Jun 2006, 00:35
zhak
Joined: 12 Apr 2005
Posts: 501
Location: Belarus
zhak
You should check the latest MASM distributive, for sure. There are a lot of examples on how to convert numbers (bin,dec,hex) to ASCII and vice versa.
13 Jun 2006, 01:37
saigon
Joined: 29 May 2006
Posts: 62
saigon
@rugxulo: Please go ahead and post your code, I am sure it will help me.
@zhak: I'll look into the MASM package to see if it helps.
Thanks!
13 Jun 2006, 08:40
Quantum
Joined: 24 Jun 2005
Posts: 122
Quantum
It's too simple converting an ASCII string to dec. The main idea:
Code:
```int s2dec(char* s){
int dec = 0;
while(*s){
if(*s >= '0' && *s <= '9')
dec = dec * 10 + *s - '0';
s++;
}
return dec;
}
```
Just convert to asm and optimize.
13 Jun 2006, 14:58
saigon
Joined: 29 May 2006
Posts: 62
saigon
I need some help converting that to FASM, any idea where I could start? Thanks!
13 Jun 2006, 15:10
UCM
Joined: 25 Feb 2005
Posts: 285
Location: Canada
UCM
Here is my (crappy) translation of what Quantum just wrote:
Code:
```s2dec:
;In registers:
;edx=Zero-terminated ASCII string
;Return value in eax, will not notice if overflow
xor eax,eax
xor ecx,ecx ;Make sure high 3 bytes of ecx are always zero.
.loop:
mov cl, [edx]
cmp ecx,'9'
ja .next ;Value out of range.
sub ecx,'0'
jb .next
;Note: the previous sub is a trick. Since cmp is the same as sub,
;but it does not affect registers, and I would subtract '0' anyway,
;I use sub instead.
lea eax,[eax*5] ;Multiply eax by 5.
shl eax,2 ;Multiply eax by 2. These two instructions are the same as
;an "imul eax,10" but faster.
add eax,ecx ;dec = dec * 10 + *s (*s = ecx)
.next:
inc edx ;s++
cmp [edx],byte 0 ;if end
jne .loop ;if not, continue
ret
```
EDIT: changed jb to ja, whoops
_________________
This calls for... Ultra CRUNCHY Man!
Ta da!! *crunch*
Last edited by UCM on 13 Jun 2006, 23:19; edited 1 time in total
13 Jun 2006, 21:29
Reverend
Joined: 24 Aug 2004
Posts: 408
Location: Poland
Reverend
http://board.flatassembler.net/topic.php?t=4377&start=53
It is code for fasmlib and there you have procs for converting numbers to strings and vice versa. Includes floating point numbers.
13 Jun 2006, 22:00
LocoDelAssembly
Your code has a bug
Joined: 06 May 2005
Posts: 4624
Location: Argentina
LocoDelAssembly
Code:
```cmp ecx,'9'
jb .next
```
It should be
Code:
```cmp ecx,'9'
ja .next ```
Code:
`shl eax,2 ;Multiply eax by 2. `
Should be
Code:
`shl eax,1 ;Multiply eax by 2. `
[/edit]
13 Jun 2006, 23:17
rugxulo
Joined: 09 Aug 2005
Posts: 2341
Location: Usono (aka, USA)
rugxulo
Okay, well, it's in the archive below (but, seriously, don't expect anything much, I'm no crazy smart coder like most here, just a hobbyist!). It shouldn't be too hard to find better code, though, since this kind of thing is needed by most everybody.
<EDIT> slightly updated the MAKE.BAT, plus included a smaller-by-two-bytes .COM </EDIT>
saigon wrote:
@rugxulo: Please go ahead and post your code, I am sure it will help me.
@zhak: I'll look into the MASM package to see if it helps.
Thanks!
Description: BASE 1.9.7 -- number base converter for DOS public domain rugxulo AT bellsouth DOT net Download Filename: base197.7z Filesize: 15.42 KB Downloaded: 825 Time(s)
Last edited by rugxulo on 13 Oct 2006, 14:52; edited 1 time in total
14 Jun 2006, 00:29
saigon
Joined: 29 May 2006
Posts: 62
saigon
Thanks rugxulo! This will help me a lot! (And, by the way, I am a hobbyist too )
16 Jun 2006, 06:55
Borsuc
Joined: 29 Dec 2005
Posts: 2465
Location: Bucharest, Romania
Borsuc
Here's my optimized 32-bit version (works in base10, unsigned, and converts to ASCII):
Code:
```; edx points at string
xor ecx, ecx
mov cl, BYTE PTR [edx]
xor eax, eax
xor cl, '0' ;Translates '0'..'9' to 0..9 (nasty trick I'd say )
cmp cl, 10
jae end
convert_loop:
lea eax, DWORD PTR [eax+4*eax] ; eax*=5
inc edx
lea eax, DWORD PTR [ecx+2*eax] ; eax=eax*2+ecx
mov cl, BYTE PTR [edx]
xor cl, '0'
cmp cl, 10
jb convert_loop
end:
ret ```
The arrangement of the instructions might be bad, I know.. if someone can correct it (and also inform me of it, so I can learn from that) and make the CPU to perform this better in the pipeline, feel free to do so
26 Jun 2006, 16:03
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Website powered by rwasa. | 1,897 | 6,310 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2022-33 | latest | en | 0.856272 |
https://stemgeeks.net/puzzle/@quantumdeveloper/3rq4d1-math-contest-24-2-sbi | 1,643,370,661,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320305494.6/warc/CC-MAIN-20220128104113-20220128134113-00267.warc.gz | 594,412,180 | 105,319 | # Math Contest #25 [2 SBI]
in #puzzle2 years ago (edited)
Here you can keep your brain fit by solving math related problems and also earn SBI or sometimes other rewards by doing so.
The problems usually contain a mathematical equation that in my opinion is fun to solve or has an interesting solution.
I will also only choose problems that can be solved without additional tools(at least not if you can calculate basic stuff in your head), so don't grab your calculator, you won't need it.
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# Rules
#### You have 4 days to solve it.
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# Problem
Given two vectors (a, 5a, 2a/x) and (3a, 4a, x).
For what values of a and x are the two vectors orthogonal to each other?
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Sort:
Anyway in order to be orthogonal the scalar product must be zero.
The scalar product is 23a^2 + 2a , so "a" has to be -2/23 and x is free. But x has to be different from 0, otherwise the first vector is not defined. Another solution is a = 0 and x different from 0 but we get a null vector for the first one and it is not so much interesting.
Bye
I am stumped... but I will go with 0
The vector (0, 0, 0) cannot be orthogonal to another vector because it has no direction.
in reality (0,0,0) is orthogonal to every vector, since the scalar product between him and any vectors is always 0
;)
You are right. It depends how you define it.
The definition I use is that two vectors are orthogonal if they form a right angle.
Ok that's fine.
This concept is not at all trivial.
The physicists say that a vector is an object that has 3 features: length, direction and verse.
So if (0,0,0) is a vector must have one direction!!
But rightly as you say it has no direction.
On the other side the mathematicians define a vector as an object that "live" in a space with a certain number of dimensions: in this case 3.
So (0,0,0) is a vector respect to this point of view.
Moreover it's true that if two vector form a right angle, their scalar product is 0, and this is true not by definition but it is a fact. You can check it by using whatever method you prefer...arctg,Pythagorean theorem,...
So I think that the null vector is an extension of the physical vector concept, like I'm sure you know it happens for the elements of an Hilbert space on quantum physics, where the L2 complex functions are the vectors and the scalar product is an integral!!
That's really cool!! | 973 | 3,220 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.265625 | 3 | CC-MAIN-2022-05 | latest | en | 0.828132 |
http://math.stackexchange.com/questions/37868/number-of-distinct-nets-of-dual-polyhedra | 1,469,612,647,000,000,000 | text/html | crawl-data/CC-MAIN-2016-30/segments/1469257826759.85/warc/CC-MAIN-20160723071026-00144-ip-10-185-27-174.ec2.internal.warc.gz | 168,053,816 | 17,185 | # Number of distinct nets of dual polyhedra
There are 11 non-congruent nets of a cube as well as 11 distinct nets of an octahedron. Both a dodecahedron and an icosahedron have 43380 distinct nets.
Is it true that any pair of dual convex polyhedra always have the same number of distinct nets (let's forget about self-intersections for simplicity)? I believe, there should be some simple proof.
-
I am not sure how to define a net in general, but if we disregard self-intersection, a net should correspond to a choice of edges that are cut, so that the faces are still connected, but there are no circles (which would prohibit the flattening of the net).
So if I regard the polyhedron as a graph, a net is a choice of edges=dual edges to remove, so that the dual graph becomes a tree. So I think that you are looking at spanning trees of graphs and their duals.
And the complement of the edges of a spanning trees viewed as dual edges form a spanning tree of the dual graph, so their number is equal.
(In particular, if you regard the corresponding edges of the cube and the octahedron, each net of the cube corresponds to a net of the octahedron where one has cut open exactly those edges that have not been cut in the cube.)
-
Look at the case of two regular tetrahedra pasted together along an equilateral triangle face (bi-pyramid) vs. a triangular prism, with squares and equilateral triangles, and do the cuts for each of these convex polyhedra along edges to get their "nets."
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https://brainly.com/question/174433 | 1,485,035,952,000,000,000 | text/html | crawl-data/CC-MAIN-2017-04/segments/1484560281226.52/warc/CC-MAIN-20170116095121-00020-ip-10-171-10-70.ec2.internal.warc.gz | 791,339,721 | 8,988 | 2014-11-04T19:34:46-05:00
### This Is a Certified Answer
Certified answers contain reliable, trustworthy information vouched for by a hand-picked team of experts. Brainly has millions of high quality answers, all of them carefully moderated by our most trusted community members, but certified answers are the finest of the finest.
3/10 > 1/5
First make both denominators the same.
3/10 =3 /10
1/5 =2 /10
3/10 > 2/10
Hope that helps you. :-)
Thank you it does
You're very welcome.
2014-11-04T20:26:11-05:00
3/10>1/5
1/5 is equal to 2/10
So 3/10> 2/10 | 184 | 563 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2017-04 | latest | en | 0.909775 |
http://www.terrylove.com/forums/printthread.php?t=28780&pp=15&page=3 | 1,398,078,828,000,000,000 | text/html | crawl-data/CC-MAIN-2014-15/segments/1397609539705.42/warc/CC-MAIN-20140416005219-00049-ip-10-147-4-33.ec2.internal.warc.gz | 682,417,883 | 5,653 | # Troubleshooting two 3-way switch setups
Show 40 post(s) from this thread on one page
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• 11-23-2010, 01:27 PM
Thatguy
Jim Port
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[So much for freedom of speech! :D I guess this comes under freedom of listening or not. . .]
I found it is much easier to predict the results given the schematic instead of the other way around, but I can't resist a good puzzle.
Quote:
Originally Posted by mgabbard
#1 down, #2 down - lights are on
#1 up, #2 down - lights are on
Taking only the case shown above, if you step through on a piece of paper all 3 ways of hooking up the 3 terminals of a working #1 switch to 3 wires and with switch #1 in both positions [6 diagrams in all] I don't see how this outcome is possible unless you have #22 wiring with 3 travelers.
An "impossible outcome" means that we are assuming something that is not true.
What's good about these is that we will probably learn something new. Plus, nobody else wants these budget-busters. :D
To find out what it is I'd say check #1 switch with an ohmmeter. If it checks good something has changed inside the walls and that's a separate troubleshooting procedure involving a voltmeter and incand. bulbs.
There are many different wiring diagrams for this, but I can only find two schematics. In any case it's a source in series with a load in series with an SPST switch made up of two 3-way and zero or more 4-way switches.
• 11-23-2010, 04:18 PM
Jim Port
How would something have changed within the walls? All the wiring should be in accessible junction boxes. If the box was buried there would be no way to make a change to it.
The example posted is the way that 3 ways work. There is nothing to troubleshoot.
• 07-21-2011, 04:59 AM
roncarter
Quote:
Originally Posted by Thatguy
This assumes two travelers between the switches.
Assign each switch a number.
With switch #1 & switch #2 down, light is on or off?
With #1 up & #2 down, light is on or off?
With #1 down and #2 up, light is on or off?
With #1 & #2 up, light is on or off?
In principle, with the answers filled in, the problem(s) can be pinpointed.
Test cases welcome. . .
1 dn 2 dn = off
1 up 2 dn = off
1 dn 2 up = on
1 up 2 up = on
I replaced porceline sockets (working properly) with can lights and found that I MUST have made an error, and now I seem to be missing the nuetral for 4 of the new lights. If I hook up a jumper wire from the nuetral at sw#1(where feed comes from) the 4 lights work but trips the breaker in one of the variations of switch configurations as follows:
1 dn 2 dn = off
1 up 2 dn = trip
1 dn 2 up = on
1 up 2 up = trip
• 07-21-2011, 06:05 AM
hj
Switch 3 is probably, but not necessarily, the miswired one, and there are several ways it could be miswired. One problem is that you would have to test switches 1 and 2 under ALL possible configurations to ensure that 3 is actually the misfit.
• 07-21-2011, 02:56 PM
roncarter
3 way switch troubles
I now have all 3 switches working so they can turn the lights all off and on, BUT... the bulbs (all same wattage) are dim. and if I remove any bulb, the rest of the lamps go off.
• 07-21-2011, 04:26 PM
DonL
Ron,
That will save you electricity, being wired in series.
Nice Job.
Have a Great evening.
DonL
• 07-21-2011, 07:10 PM
roncarter
Thanks DonL, but they aren't wired in series. At least not in what might be considered the "normal" series circuit.
It seems that one or more of them are 'hung' on travelers, rather than the neutral and hots. I have 4 boxes to go check to see where I hooked up wrong! All of the boxes are attached to the original ceiling joists 14" above the new 5/8FC ceiling, which obviously makes it very difficult to get to for troubleshooting purposes. I posted here to pick ya'lls brains about where I might start looking to possibly save some aggravation.
• 08-19-2011, 03:41 PM
Sinestro
Assign each switch a number.
With switch #1 & switch #2 down, light is on or off?
With #1 up & #2 down, light is on or off?
With #1 down and #2 up, light is on or off?
With #1 & #2 up, light is on or off?
off
on
on
on
The weird thing is, if the downstairs switch is on, then the upstairs switch can't turn off. Both on and off keep the light on.
This one has been baffling a friend and I for weeks. Any help would be appreciated.
• 08-19-2011, 08:11 PM
ActionDave
It would be my bet that the upstairs switch is not functional due to miswiring, bad switch or it just plain never worked.
• 08-21-2011, 06:46 AM
Sinestro
Quote:
Originally Posted by ActionDave
It would be my bet that the upstairs switch is not functional due to miswiring, bad switch or it just plain never worked.
The downstairs switch does the same thing though. Turn on the upstairs switch and you can't turn it off downstairs.
This all worked until we took down a wall and one of the other light switches
• 08-21-2011, 07:01 AM
LLigetfa
My very first electrical wiring was with two 3-way switches. I think I was around 10 or 11 at the time. My father couldn't figure it out and hired an electrician to help. Unfortunately the electrician couldn't figure it out either and while they were gone to buy more fuses, I took up the challenge.
It was so simple that I could not believe two grown men could not figure it out. I just followed the schematic on the side of the box the switch came in. Also, not having any more fuses, I just screwed a lighbulb into the fuse panel to test. Since it was in series, it was dim but had I got the wiring wrong, rather than blow a fuse, the bulb would be bright.
When my father and the electrician returned with the fuses, they thought I was pretty bright too.
• 08-22-2011, 09:06 PM
ActionDave
Quote:
Originally Posted by Sinestro
The downstairs switch does the same thing though. Turn on the upstairs switch and you can't turn it off downstairs.
This all worked until we took down a wall and one of the other light switches
Alright then. You have a miswired switch. Nothing more than a WAG but I bet the power came to the light first and after the remodel work a wire was crossed.
• 08-23-2011, 07:50 AM
Sinestro
Quote:
Originally Posted by ActionDave
Alright then. You have a miswired switch. Nothing more than a WAG but I bet the power came to the light first and after the remodel work a wire was crossed.
Apologies as I am a bit of a Luddite, but what is a WAG?
Also, I have read on here that you guys use a "light bulb test". What does that entail?
• 08-23-2011, 08:41 AM
jwelectric
Quote:
Originally Posted by Sinestro
Also, I have read on here that you guys use a "light bulb test". What does that entail?
Anyone who answers this question will have their post deleted.
Light bulb testers are death in a socket and unsafe advice will be deleted.
• 08-23-2011, 08:59 AM
nukeman
WAG = Wild A** Guess
Show 40 post(s) from this thread on one page
Page 3 of 5 First 12345 Last | 1,877 | 6,979 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2014-15 | latest | en | 0.932112 |
https://www.mrexcel.com/board/threads/challenge-balance-of-numbers-over-months.1088318/ | 1,603,508,600,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107881640.29/warc/CC-MAIN-20201024022853-20201024052853-00628.warc.gz | 822,458,330 | 19,573 | # Challenge! balance of numbers over months!
#### dannyok90
##### Board Regular
Hi All,
I have a challenge for you.
Im trying to come up with a way to balance hours over a period of time:
30 hours over 6 months for example
now, easily thats 5 hours a month.
but if i used a scale of 1-10:
10 meaning the majority of hours were going to be used at the end of the six months
1 meaning the majority of hours were going to be used at the start of the six months
if i set the balance to 1 (majority of hours to be used at the start)
how could i get excel to divvy the rest of the hours up going down in scale until the final (6th) month?
I've been at this for hours and cant think of anything, i've come close but no cigar!
any help?
Thanks, Dan
### Excel Facts
Square and cube roots
The =SQRT(25) is a square root. For a cube root, use =125^(1/3). For a fourth root, use =625^(1/4).
#### MickG
##### MrExcel MVP
Perhaps:-An Arithmetical Progression (adding common difference)
#### jim may
##### Well-known Member
FWIW....
Excel 2010
ABCDEFG
1
2One Method of Allocation is known as "SUM OF THE DIGITS" which takes the Sum of the Periods and
3for whatever period you want to use, say 6 (Months, Years,, whatever) allows you to apportion as:
46/21, 5/21, 4/21, 3/21, 2/21 and finally 1/21
5
6
7Total Hours to Allocate30
8Over Number of Periods6IncreasingDecreasing
9Sum of Digits - where x = 6(x²+x)/221
10Periods (below)16
1111.4285714298.571428571
1222.8571428577.142857143
1334.2857142865.714285714
1445.7142857144.285714286
1557.1428571432.857142857
1668.5714285711.428571429
17
18Sum >>>30.0000030.00000
19
</tbody>
Sheet1
Worksheet Formulas
CellFormula
C9=((C8*C8)+C8)/2
F11=(\$C\$7*(\$E11/\$C\$9))
G11=(\$C\$7*(OFFSET(\$E\$11,16-ROW(),0)/\$C\$9))
F18=SUM(F11:F16)
G18=SUM(G11:G16)
</tbody>
<tbody>
</tbody>
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219 | 637 | 1,929 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.71875 | 4 | CC-MAIN-2020-45 | latest | en | 0.822649 |
https://www.yourdissertation.co.uk/samples/momentum-of-an-object/ | 1,642,915,748,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320304134.13/warc/CC-MAIN-20220123045449-20220123075449-00687.warc.gz | 1,120,865,530 | 7,939 | Jul 22, 2017
# momentum of an object
This paper concentrates on the primary theme of momentum of an object in which you have to explain and evaluate its intricate aspects in detail. In addition to this, this paper has been reviewed and purchased by most of the students hence; it has been rated 4.8 points on the scale of 5 points. Besides, the price of this paper starts from £ 40. For more details and full access to the paper, please refer to the site.
# Discussion: momentum of an object
INSTRUCTIONS:
Please read the discussions below, and tell what you really like about each one. Make sure you understand the physics involved. Please say positive things and keep them separate.
I would not recommend anyone short of superman, to try and stop a car that is moving the same speed as a mosquito with their bare hand. Momentum (p) is the product of the mass (m) and velocity (v) or an object. Without going into detail, it is easy to see that the mass of a mosquito is an awful lot smaller than that of a car. Therefore just based off this formula the momentum of a 100kg car is much bigger than a 2.5 mg mosquito. However, say an individual is convinced that they are superman, and need to be talked out of attempting to stop the car; there are multiple equations and theorems that are useful in doing so. The momentum of the mosquito is ( 2.5 x 10^-6 kg) (3m/s) =.0000075 or 7.5x 10^-6 kg*m/s. This is clearly an extremely small momentum, and because of this is would be easy to stop a mosquito while it is in flight. The momentum of a small car is (1000kg) (3 m/s)=3000 kg*m/s. This is clearly a lot bigger of a number which would make it harder to stop it. If you take this and solve for momentum and kinetic energy, you would use the formula k=1/2(m)(v)^2=1/2(m)(p/m)^2. So 1/2(2.5x10-6) (7.5x10-6/2.5x10-6)^2 =1.125x10-27 for the mosquito. The small car in regards to the same formula, is 1/2 (1000)(3000/1000)^2=4500. As we look again at the differences in kinetic energy that it takes to move these two, we`ll see that it takes a very miniscule amount of force to create movement for the mosquito, and a very large amount of force to do the same for the car. This can be thought of in respect to its opposite reaction; it would take a large amount of force to cause the car to stop, and a small to cause the mosquito.
Impulse states that; when a force acts on an object, the objects change in momentum depends on the force and on the time interval which it acts. In regards to our discussion; the force omitted by the person must be enough to slow the car or mosquito from 3m/s to 0m/s. Well use the formula, J=m(Vfx-Vix); we do not have to worry about the y-axis because there is none in this problem. Let`s plus our numbers in J=1000kg(0m/s-3m/s)= -3000J. The force that it would take to stop the car is a force that isn`t possible for the average person. In comparison with the mosquito; again let`s plug in our numbers. J=(2.5x10^-6kg)(0m/s-3m/s)=-7.5x10^-6. To avoid confusion the negative sign states the direction of the force
Momentum of an object is the product of its mass and velocity. Momentum of an object tends to be in the same direction of velocity. The average mass is about 2.5 mg so about 2.5x10-6 kg. The average velocity of a mosquito is around 0.67 m/s. This means that the average momentum for a mosquito is 1.8x10-6 kg(m)/s. The average mass of a car is around 2000kg. If this car was moving at 0.67 m/s its momentum would be 1340 kg(m)/s. So even though the car is moving at the same velocity of the car, its greater mass means that it has a higher momentum. There would not be any way for a person to stop this car with their hand.
CONTENT:
Discussion: momentum of an objectStudent:Professor:Course title:Date:Discussion: momentum of an objectDiscussion 1: This first discussion is very enlightening as it clearly contrasts the momentum of a car and that of a mosquito. The momentum of a small vehicle is (1000kg) (3 m/s)=3000 kg*m/s, while that of a mosquito is just (2.5 x 10^-6 kg)Â (3m/s) =.0000075 or 7.5x 10^-6 kg*m/s. Given that the momentum of a mosquito is considerably smaller relative to that of a car, it is very easy to stop a mosquito while it is in flight, and it would be harder to ...
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### bentley4
Hi everyone!
1. Is a linear equation the same as a polynomial of first(or 0th) degree?
2. The book 'Mathematics for physicists and engineers' by springer(publisher) states that an example of a linear (DE) equation is 5.dy(x)/dx = x.y(x). Yet I read somewhere else that f(x,y)=a.x.y+b is not linear.(polynomial of 2nd degree).
Is one them wrong? Or are both right because in the first example y is a depent variable while in the second example y is an independant variable?
2. Feb 29, 2012
### Vorde
I think you have a language mix up here. The first equation is a linear differential equation, which means something different than linear equation in the basic algebraic sense.
3. Feb 29, 2012
### Fredrik
Staff Emeritus
If you denote the operator that takes a differentiable function f to its derivative f' by D, then your differential equation can be written as Df(x)=xf(x), or equivalently, as (D-Q)f=0, where I have defined a new operator Q by Qf(x)=xf(x). The equation (D-Q)f=0 is said to be linear because the operator D-Q is linear.
The function f defined by f(x,y)=axy+b is clearly not linear.
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# Minimum Quant and Verbal for Total Score > 600
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Minimum Quant and Verbal for Total Score > 600 [#permalink]
### Show Tags
01 Mar 2013, 20:54
Hi, I'm wondering if I want to achieve total GMAT score above 600 (it doesn't have to be greater than 650):
1. What would be the combinations of the safe minimum Quant and Verbal correct answers?
2. What would be the combinations of the safe minimum Quant and Verbal scores?
Thanks so much for the information!
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Re: Minimum Quant and Verbal for Total Score > 600 [#permalink]
### Show Tags
01 Mar 2013, 23:52
stewartlife wrote:
Hi, I'm wondering if I want to achieve total GMAT score above 600 (it doesn't have to be greater than 650):
1. What would be the combinations of the safe minimum Quant and Verbal correct answers?
2. What would be the combinations of the safe minimum Quant and Verbal scores?
Thanks so much for the information!
Check the score matrix below:
gmat-scores-83890.html
what is so special about 600-650 ?
_________________
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Re: Minimum Quant and Verbal for Total Score > 600 [#permalink]
### Show Tags
01 Mar 2013, 23:54
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Expert's post
Hi there,
There are many possible combinations of quant and verbal scores that will yield a 600 or higher. If you are fairly balanced between quant and verbal, you might aim for a 40 or higher in quant and a 33 or higher in verbal. Although that verbal score seems much lower, it actually represents a higher percentile score than the 40 in quant, in part because there is more competition on the quant side. Note that these are on the low end--to be safe, you would want to aim somewhat higher. Naturally, if you are much stronger in one subject, you can underperform on the other, although some schools may consider your subscores in addition to your overall test score.
As to the number correct, because the GMAT is a computer adaptive test, it's not possible to predict a score based on the number correct/incorrect. Most test-takers will miss about 40% of the questions. Naturally, it's great (but practically impossible) to get all the questions right, and it's not so great (and also fairly difficult) to miss them all. However, two people with the same number of missed problems might have drastically different scores, depending on *which* problems they missed. You get rewarded more for getting hard problems right, and penalized more for missing easy problems. You also get *no* credit for answering experimental questions correctly, and you won't know which these are. You will also receive an additional penalty for timing out at the end and leaving questions unanswered.
So what's the best response to all of this? See how close you are to your score goal, and then work on improving your content knowledge, problem-solving process, and time management. Don't worry so much about the number of problems you are getting right, especially since this is something you can't directly control. Just focus on getting more problems correct without overinvesting time and effort on any one problem. If a problem is too tough to do in 2 minutes, let it go!
I hope this helps. Let me know if I can shed any more light on your specific situation. Good luck!
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Re: Minimum Quant and Verbal for Total Score > 600 [#permalink]
### Show Tags
02 Mar 2013, 03:33
DmitryFarber, thanks so much for the detailed reply.
ConnectTheDots, thanks for the link. My target score for my business school application is >600, and hopefully >650. I know it's not high compared to many people here, but it's high enough for me.
Re: Minimum Quant and Verbal for Total Score > 600 [#permalink] 02 Mar 2013, 03:33
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# Minimum Quant and Verbal for Total Score > 600
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Powered by phpBB © phpBB Group and phpBB SEO Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®. | 1,537 | 6,136 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5625 | 4 | CC-MAIN-2016-50 | latest | en | 0.852732 |
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How do you estimate the variation for reset ICs, including the detection voltage temperature? The rate of change per 1C is indicated by the temperature coefficient [VDET/ΔT][ppm/℃]. The detection voltage at ambient temperature Ta is VDET(Ta)=VDET(25℃)×(1+((Ta-25℃)×Tc/1000000)). In order to approximate the variation including the temperature, with a variation of ±1% and temperature coefficient Tc=±360ppm/℃ at ambient temperature Ta the detection voltage VDET can be estimated as VDET(max)=VDET×1.01+VDET×1.01×|Ta-25|×360/1000000VDET(min)=VDET×0.99-VDET×0.99×|Ta-25|×360/1000000. Refer to the datasheet regarding the variation that takes into account the detection voltage of the BD52xxG-2C/BD53xxG-2C series along with the BD71L4L-1 overvoltage detection IC. | 233 | 768 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.90625 | 3 | CC-MAIN-2024-30 | latest | en | 0.829012 |
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A158570 A007814((2n-1)!!+1) 1
1, 2, 4, 1, 1, 2, 5, 1, 1, 2, 4, 1, 1, 2, 5, 1, 1, 2, 4, 1, 1, 2, 5, 1, 1, 2, 4, 1, 1, 2, 5, 1, 1, 2, 4, 1, 1, 2, 5, 1, 1, 2, 4, 1, 1, 2, 5, 1, 1, 2, 4, 1, 1, 2, 5, 1, 1, 2, 4, 1, 1, 2, 5, 1, 1, 2, 4, 1, 1, 2, 5, 1, 1, 2, 4, 1, 1, 2, 5, 1, 1, 2, 4, 1, 1, 2, 5, 1, 1, 2, 4, 1, 1, 2, 5, 1, 1, 2, 4, 1, 1, 2, 5, 1, 1 (list; graph; refs; listen; history; text; internal format)
OFFSET 1,2 COMMENTS It is a periodic sequence of period 8. Also the decimal expansion of the constant 124112510/99999999. [From R. J. Mathar, Apr 02 2009] Terms of the simple continued fraction of 399/[5*sqrt(5595)-99]. [From Paolo P. Lava, Aug 05 2009] LINKS FORMULA a(n)=1, if n==0,1,4,5 mod 8; a(n)=2, if n==2,6 mod 8; a(n)=4, if n==3 mod 8; a(n)=5, if n==7 mod 8. G.f.: -x*(1+2*x+4*x^2+x^3+x^4+2*x^5+5*x^6+x^7)/((x-1)*(1+x)*(x^2+1)*(x^4+1)) [From R. J. Mathar, Apr 02 2009] a(n)=(1/224)*{17*(n mod 8)+129*[(n+1) mod 8]-67*[(n+2) mod 8]-11*[(n+3) mod 8]+17*[(n+4) mod 8]+101*[(n+5) mod 8]-39*[(n+6) mod 8]-11*[(n+7) mod 8]}, with n>=0 [From Paolo P. Lava, Mar 30 2009] MATHEMATICA Table[Which[Mod[n, 8]==7, 5, Mod[n, 8]==3, 4, MemberQ[{2, 6}, Mod[n, 8]], 2, True, 1], {n, 120}] (* or *) PadRight[{}, 120, {1, 2, 4, 1, 1, 2, 5, 1}] (* Harvey P. Dale, May 05 2020 *) CROSSREFS Sequence in context: A336434 A007738 A186520 * A295224 A074749 A194524 Adjacent sequences: A158567 A158568 A158569 * A158571 A158572 A158573 KEYWORD nonn AUTHOR Vladimir Shevelev, Mar 21 2009 STATUS approved
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Last modified May 14 07:13 EDT 2021. Contains 343879 sequences. (Running on oeis4.) | 979 | 2,006 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.53125 | 4 | CC-MAIN-2021-21 | longest | en | 0.411723 |
https://math.stackexchange.com/questions/3054558/when-we-refer-to-topology-on-a-metric-space-s-do-we-mean-the-topology-genera | 1,579,423,980,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579250594333.5/warc/CC-MAIN-20200119064802-20200119092802-00000.warc.gz | 545,816,842 | 30,800 | # When we refer to topology on a metric space $S$ , do we mean the topology generated by open ball?
I've just learned a throrem which states that : A metric space has the structure of a topological space in which the open sets are unions of balls .
But the theorem only told me there "exist" one topology with respect to the metric.When we refer to topology on a metric space $$S$$ , do we mean the topology generated by open ball ?
Yes, a metric space $$(S, d)$$ induces a topology $$\mathcal{T}$$ which is generated by the basis $$\mathcal{B} = \{B(x, r) \ | \ x \in S \ \text{ and } \ r > 0\}$$
When authors refer to the topology on this metric space $$(S, d)$$, they usually mean the topology $$\mathcal{T}$$ above, which you can think of as the topology generated by open balls.
Yes. So you can say that a set $$U$$ is open iff for each $$x\in U$$, there exists an open ball $$B(x,r)$$ centered at $$x$$ with $$B(x,r)\subset U$$. | 257 | 937 | {"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": 11, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.125 | 3 | CC-MAIN-2020-05 | latest | en | 0.92302 |
http://mathforum.org/library/drmath/view/62009.html | 1,498,164,240,000,000,000 | text/html | crawl-data/CC-MAIN-2017-26/segments/1498128319902.52/warc/CC-MAIN-20170622201826-20170622221826-00418.warc.gz | 259,870,047 | 3,759 | Associated Topics || Dr. Math Home || Search Dr. Math
### Shortcut for Comparing Fractions
```Date: 01/14/2003 at 07:50:56
From: Joe
Subject: Comparing fractions
We are going over comparing fractions in class and my teacher showed a
little shortcut. He used the example of 2/3 and 3/5. He said you can
cross multiply and get 9/10. The numerator (the product of the
interior terms, 3 and 3) always corresponds to the second of the two
fractions, and the denominator (the product of the exterior terms, 2
and 5) corresponds to the first fraction.
He showed us many examples and I have tried many myself and this
always works. 9 is less than 10, so 3/5 is less than 2/3. My question
is WHY?
```
```
Date: 01/14/2003 at 11:20:56
From: Doctor Ian
Subject: Re: Comparing fractions
Hi Joe,
Suppose I have two quantities, and I want to know which is larger.
One way to do that is to divide one by the other, and see if I end up
with something less than, greater to, or equal to 1.
This is trivial when you do it with integers:
7 / 5 > 1 so 7 > 5
3 / 4 < 1 so 3 < 4
But it also works with fractions. Suppose I use '?' to represent '=',
'>', or '<'. That is, I'm going to use it as a 'variable' for these
relational operators. Then if I want to compare two fractions, a/b
and c/d, I can do this:
a/b
----- ? 1
c/d
Of course, to divide by a fraction, I invert and multiply, so this is
the same as
a d
- * - ? 1
b c
If I get something greater than 1, a/b must be larger. If I get
something less than 1, a/b must be smaller. If I get 1, the two
fractions must be equal.
Let's try it with your example:
2/3
----- ? 1
3/5
2 5
- * - ? 1
3 3
10
-- > 1
9
So 'cross-multiplying' is just what you have to do to divide the first
fraction by the second. Comparing the result to 1 tells you whether
the numerator (first fraction) or denominator (second fraction) is
larger.
Does that make sense?
You can get a better feel for what is going on if you use letters
instead of numbers. For example, suppose my two fractions are a/b and
(a+1)/b. I know that the second one has to be larger, right? Let's
see what happens when we divide:
a/b
--------- ? 1
(a+1)/b
a b
- * ----- ? 1
b (a+1)
a
----- < 1
(a+1)
Let's try comparing two equivalent fractions. If we have a fraction
a/b, we can make an equivalent fraction by multiplying by k/k, where k
is anything other than zero:
a/b
--------- ? 1
(ka)/(kb)
a kb
- * ---- ? 1
b ka
kab
--- = 1
kab
For fun, you might consider trying to see which is larger: a/b, or
(a+k)/(b+k). That is, when you increment both the numerator and
denominator of a fraction by the same amount, is the resulting
fraction larger, or smaller?
or anything else.
- Doctor Ian, The Math Forum
http://mathforum.org/dr.math/
```
Associated Topics:
Middle School Algebra
Middle School Fractions
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Math Forum Home || Math Library || Quick Reference || Math Forum Search | 935 | 3,220 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.4375 | 4 | CC-MAIN-2017-26 | longest | en | 0.919803 |
http://stackoverflow.com/questions/21662041/how-to-find-the-intersection-of-two-nfa | 1,444,202,156,000,000,000 | text/html | crawl-data/CC-MAIN-2015-40/segments/1443736682773.27/warc/CC-MAIN-20151001215802-00015-ip-10-137-6-227.ec2.internal.warc.gz | 304,695,898 | 19,543 | # How to find the intersection of two NFA
In DFA we can do the intersection of two automata by doing the cross product of the states of the two automata and accepting those states that are accepting in both the initial automata. Union is performed similarly. How ever although i can do union in NFA easily using epsilon transition how do i do their intersection?
-
You can use the cross-product construction on NFAs just as you would DFAs. The only changes are how you'd handle ε-transitions. Specifically, for each state (qi, rj) in the cross-product automaton, you add an ε-transition from that state to each pair of states (qk, rj) where there's an ε-transition in the first machine from qi to qk and to each pair of states (qi, rk) where there's an ε-transition in the second machine from rj to rk.
Alternatively, you can always convert the NFAs into DFAs and then compute the cross product of those DFAs.
Hope this helps!
-
so there is no easy method like just adding epsilon as in the union case? – Aditya Nambiar Feb 9 '14 at 17:47
@AdityaNambiar- Not to the best of my knowledge. – templatetypedef Feb 9 '14 at 17:48
there is a huge mistake in the templatetypedef 's answer .
the product automaton of L1 and L2 which are NFAs :
new states Q = product of the states of L1 and L2 .
now the transition function :
a is a symbol in the union of both automatons alphabets
delta( (q_1,q_2) , a) = delta_L1(q_1 , a) X delta_L2(q_2 , a)
which means you should multiply the set that is the resualt of delta_L1(q_1 , a) with the set that results from delta_L2(q_1 , a).
the problem in the templatetypedef's answer is that the product result (qk ,rk) is not mentioned .
-
An alternative to constructing the product automaton is allowing more complicated acceptance criteria. Ordinarily, an NFA accepts an input string when it has reached any one of a set of accepting final states. That can be extended to boolean combinations of states. Specifically, you construct the automaton for the intersection like you do for the union, but consider the resulting automaton to accept an input string only when it is in (what corresponds to) accepting final states in both automata.
-
We can also use De Morgan's Laws: A intersection B = (A' U B')'
Taking the union of the compliments of the two NFA's is comparatively simpler, especially if you are used to the epsilon method of union.
-
Hmm... how do you take the negation of an NFA, other than by turning it into a DFA first? – Stefan May 6 at 21:18 | 627 | 2,511 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.03125 | 3 | CC-MAIN-2015-40 | latest | en | 0.907946 |
https://www.teacherspayteachers.com/Product/Fraction-Operations-and-Integer-Concepts-Unit-2572628 | 1,493,428,788,000,000,000 | text/html | crawl-data/CC-MAIN-2017-17/segments/1492917123102.83/warc/CC-MAIN-20170423031203-00611-ip-10-145-167-34.ec2.internal.warc.gz | 975,641,468 | 26,191 | Total:
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# Fraction Operations and Integer Concepts Unit
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61.98 MB | 92 Unit Pages + 52 Answer Key Pages pages
### PRODUCT DESCRIPTION
Fraction Operations and Integer Concepts Unit for 6th Grade Math
6th Grade Math Curriculum Unit 2
This unit includes five multi-day lessons that cover a review of 5th grade skills and some new 6th grade skills :
* Equivalent Fractions and Simplest Form
* Adding and Subtracting Mixed Numbers
* Multiplying Fractions and Mixed Numbers
* Dividing Fractions and Mixed Numbers by Fractions
* Understanding Integers
* Absolute Value
* Plotting in a Coordinate Plane
Included in this resource:
•Weekly warm up recording sheets
•Weekly exit ticket sheets
•Blank lesson plans
•Unit vocabulary sheet
•Unit pre-assessment
•Unit post assessment
•Warm ups
•Fold and Flip Notes
•Practice assignments (for homework or classwork)
Download the preview to see exactly what is included with each lesson as well as sample pages!
This unit is part of my 6th Grade Math Curriculum. If you'd like to purchase the curriculum, please click here! I will not be able to offer bundle discounts for teachers who have purchased individual units.
Activities are not included in this resource. My 6th Grade Curriculum is offered as an add on to my 6th Grade Math Curriculum Resources Mega Bundle. There will be no overlap between this and the Curriculum Resources Mega Bundle.
Unit 1 – Operations with Whole Numbers and Decimals {A 5th Grade Review Unit}
Unit 2 – Fraction Operations and Integer Concepts
Unit 3 – Ratios, Rates, Percents and Proportions
Unit 4 – Expressions
Unit 5 – Equations and Inequalities
Unit 6 – Geometry
Unit 7 – Data Displays
Total Pages
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\$20.00 | 516 | 2,045 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2017-17 | longest | en | 0.899129 |
http://tutor-usa.com/topic/worksheets/equations?page=1 | 1,560,683,309,000,000,000 | text/html | crawl-data/CC-MAIN-2019-26/segments/1560627998100.52/warc/CC-MAIN-20190616102719-20190616124719-00228.warc.gz | 183,060,239 | 9,057 | # Free Equations Worksheets
This free algebra worksheet contains 14 problems on solving equations involving fractions and mixed numbers. Students must use addition, subtraction, and multiplication of fractions and mixed numbers...
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Worksheet (Pre-Algebra) | 357 | 1,788 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2019-26 | latest | en | 0.908202 |
https://acemyhomeworkwriters.com/choose-a-local-company-in-the-industry-of-your-liking-from-the-following-list-restaurant-hotel-hospital-internet-cable-construction-grocery-store/ | 1,639,000,094,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964363598.57/warc/CC-MAIN-20211208205849-20211208235849-00077.warc.gz | 165,521,016 | 11,880 | # Choose a local company in the industry of your liking from the following list Restaurant/ hotel, hospital, internet/cable, construction, grocery store,
Learning Objective: State Mandated Outcomes numbered 3,5,7,8 and 10.
3.Compute descriptive statistics, construct graphs for data analysis, and interpret outcomes.
5.Calculate expected values to evaluate multiple outcomes of a decision.
7.Construct and interpret confidence intervals for means and proportions.
8.Formulate, perform, and interpret hypotheses tests (one and two population parameters).
10.Use statistical software to graph, compute, and analyze statistical data
Choose a local company in the industry of your liking from the following list:
Restaurant/ hotel,
hospital,
internet/cable,
construction,
grocery store,
insurance,
automobile
banking/fianace/loan
You should be done with the following four First in Part A:
Step by Step Instructions:
1.Make sure, you do have a data access from the chosen company in the above industry for two different time periods or two different locations in the same time period.
2.Suppose you work on quality control in that company. Your job is to do monthly checks on the quality of service by collecting data on key areas. Your company is watching sales by collecting data on location, and three variables.
3.Select the three variables of your choice (based on data availability).
4.You are to prepare a Managerial Report answering the following (show the work for each of the following at the end of managerial report)
1. Use appropriate descriptive statistics to summarize each of the three variables. (Covered in Chapter-2)
2. Use appropriate descriptive statistics to summarize each of the three variables for the comparison year/quarter/location (dependent on what you chose in #1 – location or time). (Covered in Chapter-2)
3. Compare your summary results. Discuss any specific statistical results in visuals that would help the reader to understand your point. (Covered in Chapter-2)
Example
Make sure, you do have a data access from that company. Suppose you work on quality control in that company. Your job is to do monthly checks on the quality of service by collecting data on key areas. Your company is watching sales by collecting data on location, and three variables – list price, sale price, and number of days it takes to sell each residential single-family housing unit. Each single-family housing unit is classified as “Collin” if it is located within Collin county or “Non-Collin” if it is located near (within 50 miles), but not in Collin county.
Part-A:
You are to prepare a Managerial Report answering the following (show the work for each of the following at the end of managerial report)
1. Use appropriate descriptive statistics to summarize each of the three variables for the Collin housing units. (Covered in Chapter-2)
2. Use appropriate descriptive statistics to summarize each of the three variables for the Non-Collin housing units. (Covered in Chapter-2)
3. Compare your summary results. Discuss any specific statistical results that would help a real estate agent understand the DFW housing market. (Covered in Chapter-2)
Part-B :
4. Develop a 95% confidence interval estimate of the population mean sales price and population mean number of days to sell for Collin housing units. Interpret your results.
5. Develop a 95% confidence interval estimate of the population mean sales price and population mean number of days to sell for Non-Collin housing units. Interpret your results.
Assume the branch manager requested estimates of the mean selling price of Collin housing units with a margin of error of \$40,000 and the mean selling price of Non-Collin housing units with a margin of error of \$15,000. Using 95% confidence, how large should the sample sizes be?
7. Your company just signed contracts for two new listings: a Collin housing unit with a list price of \$589,000 and a Non-Collin housing unit with a list price of \$285,000. What is your estimate of the final selling price and number of days required to sell each of these units?
Part-B
Excel Exercise
1. Develop a 95% confidence interval estimate of the population mean sales price and population mean number of days to sell for Collin housing units using excel. Attach the excel file to your report. Also enlist the steps explaining how you used excel.
2. Develop a 95% confidence interval estimate of the population mean sales price and population mean number of days to sell for Non-Collin housing units using excel. Attach the excel file to your report. Also enlist the steps explaining how you used excel.
(Hint: Review the guidelines for Interval Estimation Using Excel)
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https://worlddatabaseofhappiness.eur.nl/distributional-findings/burger-veenhoven-2016-study-bf-2007-13679/ | 1,721,043,431,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514696.4/warc/CC-MAIN-20240715102030-20240715132030-00454.warc.gz | 548,259,197 | 11,701 | # Distributional findings
## Study Burger & Veenhoven (2016): study BF 2007
Public
15+ aged, general public, Burkina Faso, 2007
Survey name
INT-GallupWorldPoll2015
Sample
Probability multi-stage random
Respondents
N = 1000
Non Response
Assessment
Interview: face-to-face
## Happiness measure(s)
Full text:
Self report on 10 questions:
A Did you feel well-rested yesterday?
B Were you treated with respect yesterday?
C Did you smile and laugh a lot yesterday
D Did you learn or do something interesting yesterday?
Did you experience the following feelings during A LOT of the day yesterday?
E How about enjoyment?
F How about physical pain?
G How about worry?
I How about stress?
J How about anger?
Rated:
0: no
1: yes
Computation:
- Average % positive affect = (A+B+C+D+E)/5
- Average % negative affect = (F+G+H+I+J)/5
- Affect Balance = Average % positive affect - Average % negative affect
Transformation to range 0-10:
(Score[range-100-100] + 100)/200*10
These questions do not measure hedonic level of affect of individual persons adequately, since yesterday's mood does not always correspond with typical average mood. Yet these questions can be used for measuring hedonic level in aggregates, such as nations, since individual variations balance out in big samples.
Transformation to 0-10 scale: (percent abs)*0.05 + 5
Classification:
A-AB-yd-mq-v-2-mb
Observed distribution
Summary Statistics
On original range -100 - 100 On range 0-10
Mean:
33.00 6.65
SD:
- -
Full text:
Self report on single question:
Here is ladder representing the 'ladder of life'. Let's suppose the top of the ladder represents the best possible life for you; and the bottom, the worst possible life for you. On which step of the ladder do you feel you personally stand at the present time?
10 best possible
9
8
7
6
5
4
3
2
1
0 worst possible life
This question was followed (not preceded) by items on life 5 years ago and 5 years from now.
Classification:
C-BW-c-sq-l-11-c
Observed distribution
Summary Statistics
On original range 0 - 10 On range 0-10
Mean:
4.00 4.00
SD:
- - | 568 | 2,082 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-30 | latest | en | 0.861206 |
http://www.cordovaalumni.com/simple-manufacturing-process-flow-chart/sales-process-flowchart-template-lucidchart-19/ | 1,566,282,494,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027315222.56/warc/CC-MAIN-20190820045314-20190820071314-00168.warc.gz | 238,941,432 | 13,585 | # Simple Manufacturing Process Flow Chart Sales Flowchart Template Lucidchart
By Sophie Dulhunty at November 25 2018 21:28:11
In mathematics, method of solving a problem by repeatedly using a simpler computational method. A basic example is the process of long division in arithmetic. The term algorithm is now applied to many kinds of problem solving that employ a mechanical sequence of steps, as in setting up a computer program. The sequence may be displayed in the form of a flowchart in order to make it easier to follow. As with algorithms used in arithmetic, algorithms for computers can range from simple to highly complex.
After deciding on the points you want to make in your upcoming presentation, you need to figure out how to support those points. For example, if your point is that your company has the largest market share in the industry, quote the research (hopefully done by a third party) that says so. This applies to both business presentations and educational presentations. The support you provide for your message is essential for an effective presentation.
## Gallery of Simple Manufacturing Process Flow Chart
Rice Huskers: These huskers remove the husk (outer covering) from the paddy rice during the processing. Paddy Separators: It makes the brown rice more efficient. Plano-Shifters: This makes the rice more uniform and give rice proper size and grading with a high speed. Color Sorters: These color sorters give a proper color to the rice and define its shade. The basic structure and the process followed in the rice milling industries and rice milling plants include: 1. Quantity of Rice (In Abundance) ; 2. Pre - Cleaning ; 3. Steaming ; 4. Drying ; 5. Packaging ; 6. Grading and Sorting ; 7. Polishing ; 8. Removal of Husk
This method seems to be the inefficient method as the chances or the profits to rise at the higher level often degrades. The marketing costs and the net returns tend to decrease that prevents the consumers from earning accountable outputs in time. Further, the latest and modern rice milling units make use of more efficient technology that hep them yield a large net return and the calculated costs become better for the large quantity of paddy. Roughly, we can say that capacity utilization using the modern methods is approximately 70% in comparison to the conventional methods used (approx. 45%). Rice is the staple food and is one of the man source among all the food grains. | 511 | 2,446 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.578125 | 3 | CC-MAIN-2019-35 | latest | en | 0.91509 |
https://www.antiessays.com/free-essays/Healthy-Spring-Water-Company-710863.html | 1,653,385,249,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662570051.62/warc/CC-MAIN-20220524075341-20220524105341-00313.warc.gz | 719,362,509 | 10,629 | # Healthy Spring Water Company
347 Words2 Pages
HEALTHY SPRING WATER COMPANY DEFINING THE PRICE-VOLUME TRADEOFF FOR A 20% PRICE INCREASE The Healthy Spring Water Company sells bottled water for offices and homes. The price of the water is \$20 per 10 gallon bottle and the company currently sells 2000 bottles per day. Following is the company's income and costs on a daily basis. Sales revenue \$40,000 Incremental Variable cost \$16,000 Nonincremental Fixed cost \$20,000 The company is enjoying stable demand with its current pricing, but management is looking for ways to increase profitability. One suggestion is that the company reposition its water as a premium product, justifying a higher price. If successful, the company believes that it could charge 20% more for its water than it does now. 1. What is the maximum sales loss (in % and units) that Healthy Spring could tolerate before a 20% price increase would fail to make a positive contribution to its profitability? The maximum sales loss would be 25% before they can tolerate before a 20% price increase, which amount to 1,500 bottles per day instead of 2,000. 2. In order for Healthy Spring to reposition itself as a premium water, management believes that it will have to upgrade the packaging of its product. The company will deliver the water in glass rather than plastic bottles and the bottles will be "safety sealed" to insure their cleanliness until the covering is removed in the customer's home. These changes will add \$1.00 per bottle to the variable cost of sales. Calculate the new break even given the increase in variable costs. After charging \$1.00 more per bottle, the company would have to sell 1,600 bottles per day which would be 20% sales loss. 3. To reposition its water as a premium product, Healthy Spring will require an increase in its advertising and promotion budget of \$900 daily. What is | 410 | 1,888 | {"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} | 2.515625 | 3 | CC-MAIN-2022-21 | latest | en | 0.952678 |
https://nrich.maths.org/public/topic.php?code=97&cl=3&cldcmpid=5642 | 1,585,623,902,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585370499280.44/warc/CC-MAIN-20200331003537-20200331033537-00193.warc.gz | 574,785,122 | 10,003 | # Resources tagged with: 2D shapes and their properties
Filter by: Content type:
Age range:
Challenge level:
### There are 65 results
Broad Topics > Angles, Polygons, and Geometrical Proof > 2D shapes and their properties
### Roaming Rhombus
##### Age 14 to 16 Challenge Level:
We have four rods of equal lengths hinged at their endpoints to form a rhombus ABCD. Keeping AB fixed we allow CD to take all possible positions in the plane. What is the locus (or path) of the point. . . .
### What Shape?
##### Age 7 to 14 Challenge Level:
This task develops spatial reasoning skills. By framing and asking questions a member of the team has to find out what mathematical object they have chosen.
### Circles, Circles Everywhere
##### Age 7 to 14
This article for pupils gives some examples of how circles have featured in people's lives for centuries.
### LOGO Challenge 10 - Circles
##### Age 11 to 16 Challenge Level:
In LOGO circles can be described in terms of polygons with an infinite (in this case large number) of sides - investigate this definition further.
### Arclets Explained
##### Age 11 to 16
This article gives an wonderful insight into students working on the Arclets problem that first appeared in the Sept 2002 edition of the NRICH website.
### LOGO Challenge 12 - Concentric Circles
##### Age 11 to 16 Challenge Level:
Can you reproduce the design comprising a series of concentric circles? Test your understanding of the realtionship betwwn the circumference and diameter of a circle.
### LOGO Challenge 11 - More on Circles
##### Age 11 to 16 Challenge Level:
Thinking of circles as polygons with an infinite number of sides - but how does this help us with our understanding of the circumference of circle as pi x d? This challenge investigates. . . .
### What's Inside/outside/under the Box?
##### Age 7 to 14
This article describes investigations that offer opportunities for children to think differently, and pose their own questions, about shapes.
### LOGO Challenge 6 - Triangles and Stars
##### Age 11 to 16 Challenge Level:
Recreating the designs in this challenge requires you to break a problem down into manageable chunks and use the relationships between triangles and hexagons. An exercise in detail and elegance.
### Not So Little X
##### Age 11 to 14 Challenge Level:
Two circles are enclosed by a rectangle 12 units by x units. The distance between the centres of the two circles is x/3 units. How big is x?
### Squaring the Circle
##### Age 11 to 14 Challenge Level:
Bluey-green, white and transparent squares with a few odd bits of shapes around the perimeter. But, how many squares are there of each type in the complete circle? Study the picture and make. . . .
### What Shape for Two
##### Age 7 to 14 Challenge Level:
'What Shape?' activity for adult and child. Can you ask good questions so you can work out which shape your partner has chosen?
### Circumspection
##### Age 14 to 16 Challenge Level:
M is any point on the line AB. Squares of side length AM and MB are constructed and their circumcircles intersect at P (and M). Prove that the lines AD and BE produced pass through P.
##### Age 14 to 16 Challenge Level:
Given a square ABCD of sides 10 cm, and using the corners as centres, construct four quadrants with radius 10 cm each inside the square. The four arcs intersect at P, Q, R and S. Find the. . . .
### Floored
##### Age 11 to 14 Challenge Level:
A floor is covered by a tessellation of equilateral triangles, each having three equal arcs inside it. What proportion of the area of the tessellation is shaded?
### Semi-detached
##### Age 14 to 16 Challenge Level:
A square of area 40 square cms is inscribed in a semicircle. Find the area of the square that could be inscribed in a circle of the same radius.
### Pentagonal
##### Age 14 to 16 Challenge Level:
Can you prove that the sum of the distances of any point inside a square from its sides is always equal (half the perimeter)? Can you prove it to be true for a rectangle or a hexagon?
### First Forward Into Logo 4: Circles
##### Age 7 to 16 Challenge Level:
Learn how to draw circles using Logo. Wait a minute! Are they really circles? If not what are they?
### Lawnmower
##### Age 14 to 16 Challenge Level:
A kite shaped lawn consists of an equilateral triangle ABC of side 130 feet and an isosceles triangle BCD in which BD and CD are of length 169 feet. A gardener has a motor mower which cuts strips of. . . .
##### Age 14 to 16 Challenge Level:
Investigate the properties of quadrilaterals which can be drawn with a circle just touching each side and another circle just touching each vertex.
### Pi, a Very Special Number
##### Age 7 to 14
Read all about the number pi and the mathematicians who have tried to find out its value as accurately as possible.
### Efficient Packing
##### Age 14 to 16 Challenge Level:
How efficiently can you pack together disks?
### Square Areas
##### Age 11 to 14 Challenge Level:
Can you work out the area of the inner square and give an explanation of how you did it?
### Circle Packing
##### Age 14 to 16 Challenge Level:
Equal circles can be arranged so that each circle touches four or six others. What percentage of the plane is covered by circles in each packing pattern? ...
### LOGO Challenge - Circles as Animals
##### Age 11 to 16 Challenge Level:
See if you can anticipate successive 'generations' of the two animals shown here.
### Salinon
##### Age 14 to 16 Challenge Level:
This shape comprises four semi-circles. What is the relationship between the area of the shaded region and the area of the circle on AB as diameter?
### Hex
##### Age 11 to 14 Challenge Level:
Explain how the thirteen pieces making up the regular hexagon shown in the diagram can be re-assembled to form three smaller regular hexagons congruent to each other.
### Curvy Areas
##### Age 14 to 16 Challenge Level:
Have a go at creating these images based on circles. What do you notice about the areas of the different sections?
##### Age 14 to 16 Challenge Level:
The sides of a triangle are 25, 39 and 40 units of length. Find the diameter of the circumscribed circle.
### From One Shape to Another
##### Age 7 to 14
Read about David Hilbert who proved that any polygon could be cut up into a certain number of pieces that could be put back together to form any other polygon of equal area.
### Opposite Vertices
##### Age 11 to 14 Challenge Level:
Can you recreate squares and rhombuses if you are only given a side or a diagonal?
### Like a Circle in a Spiral
##### Age 7 to 16 Challenge Level:
A cheap and simple toy with lots of mathematics. Can you interpret the images that are produced? Can you predict the pattern that will be produced using different wheels?
### Blue and White
##### Age 11 to 14 Challenge Level:
Identical squares of side one unit contain some circles shaded blue. In which of the four examples is the shaded area greatest?
### Crescents and Triangles
##### Age 14 to 16 Challenge Level:
Can you find a relationship between the area of the crescents and the area of the triangle?
### Three Four Five
##### Age 14 to 16 Challenge Level:
Two semi-circles (each of radius 1/2) touch each other, and a semi-circle of radius 1 touches both of them. Find the radius of the circle which touches all three semi-circles.
### LOGO Challenge 2 - Diamonds Are Forever
##### Age 7 to 16 Challenge Level:
The challenge is to produce elegant solutions. Elegance here implies simplicity. The focus is on rhombi, in particular those formed by jointing two equilateral triangles along an edge.
### Tricircle
##### Age 14 to 16 Challenge Level:
The centre of the larger circle is at the midpoint of one side of an equilateral triangle and the circle touches the other two sides of the triangle. A smaller circle touches the larger circle and. . . .
### LOGO Challenge 8 - Rhombi
##### Age 7 to 16 Challenge Level:
Explore patterns based on a rhombus. How can you enlarge the pattern - or explode it?
### Some(?) of the Parts
##### Age 14 to 16 Challenge Level:
A circle touches the lines OA, OB and AB where OA and OB are perpendicular. Show that the diameter of the circle is equal to the perimeter of the triangle
### Square Pegs
##### Age 11 to 14 Challenge Level:
Which is a better fit, a square peg in a round hole or a round peg in a square hole?
### Holly
##### Age 14 to 16 Challenge Level:
The ten arcs forming the edges of the "holly leaf" are all arcs of circles of radius 1 cm. Find the length of the perimeter of the holly leaf and the area of its surface.
### Tied Up
##### Age 14 to 16 Short Challenge Level:
How much of the field can the animals graze?
### Dividing the Field
##### Age 14 to 16 Challenge Level:
A farmer has a field which is the shape of a trapezium as illustrated below. To increase his profits he wishes to grow two different crops. To do this he would like to divide the field into two. . . .
### Track Design
##### Age 14 to 16 Challenge Level:
Where should runners start the 200m race so that they have all run the same distance by the finish?
### Towering Trapeziums
##### Age 14 to 16 Challenge Level:
Can you find the areas of the trapezia in this sequence?
### Lying and Cheating
##### Age 11 to 14 Challenge Level:
Follow the instructions and you can take a rectangle, cut it into 4 pieces, discard two small triangles, put together the remaining two pieces and end up with a rectangle the same size. Try it!
### Trapezium Four
##### Age 14 to 16 Challenge Level:
The diagonals of a trapezium divide it into four parts. Can you create a trapezium where three of those parts are equal in area?
### From All Corners
##### Age 14 to 16 Challenge Level:
Straight lines are drawn from each corner of a square to the mid points of the opposite sides. Express the area of the octagon that is formed at the centre as a fraction of the area of the square.
### Fitting In
##### Age 14 to 16 Challenge Level:
The largest square which fits into a circle is ABCD and EFGH is a square with G and H on the line CD and E and F on the circumference of the circle. Show that AB = 5EF. Similarly the largest. . . .
### The Medieval Octagon
##### Age 14 to 16 Challenge Level:
Medieval stonemasons used a method to construct octagons using ruler and compasses... Is the octagon regular? Proof please. | 2,430 | 10,425 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2020-16 | latest | en | 0.881963 |
http://sage.knou.ac.kr/home/pub/380/ | 1,660,337,636,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882571758.42/warc/CC-MAIN-20220812200804-20220812230804-00022.warc.gz | 42,384,889 | 2,370 | 1장-과제-김진서 1번.
1602 days 전, kjseo123 작성
var('x') f(x)= (4*x^3+x-5)/(10*x^2+12*x+7) plot(f(x), (x, -10, 10), ymax=5, ymin=-5)
var('x') f(x)= (4*x^3+x-5)/(10*x^2+12*x+7) plot(f(x), (x, -10, 10), ymax=5, ymin=-5) limit(f(x), x=+oo)
+Infinity +Infinity | 131 | 250 | {"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} | 3.375 | 3 | CC-MAIN-2022-33 | latest | en | 0.101753 |
http://mathcentral.uregina.ca/QQ/database/QQ.09.09/h/kamarah1.html | 1,675,865,890,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764500813.58/warc/CC-MAIN-20230208123621-20230208153621-00772.warc.gz | 24,292,157 | 3,642 | SEARCH HOME
Math Central Quandaries & Queries
Question from Kamarah, a student: If a rooster is worth 5 coins and a hen is worth 3 coins and chicks 3 together are worth 1 coin, how many roosters, hen and chicks totaling 100 can be bought with 100 coins? notes: you must buy at least one of each type of fowl no fractional fowl chicks are purchased in multiple of 3 find 3 solutions
Hi Kamarah,
I would try what some educators call guess-and-check but it's much more than guessing, it's intelligent trial and error.
I started thinking about roosters since they are the most expensive. You could spend all 100 coins on roosters but you would only get 20 creatures and you need 100. If you bought 19 roosters it would leave you with 5 coins and the most you could buy with 5 coins would be 15 chick giving 19 + 15 = 34 creatures. again not nearly enough, so lets drop down on the number of roosters to 15 (75 coins) and be more organized by using a table. (I am ignoring the fact that I need hens also but hopefully I can fix that later.)
roosters hens chicks total
number 15 15
coins spent 75 75
This leaves 25 coins and I could spend them all on 75 chicks giving
roosters hens chicks total
number 15 75 90
coins spent 75 25 100
which is still too few creatures.
roosters hens chicks total
number 14 14
coins spent 70 70
This leaves 30 coins and spending them all on chicks gives
roosters hens chicks total
number 14 90 104
coins spent 70 30 100
This is too many creatures so I could spend 3 coins on hens giving
roosters hens chicks total
number 14 1 81 96
coins spent 70 3 27 100
Almost there!
What about 13 rosters? 12 roosters? 11? 10?
Penny
Math Central is supported by the University of Regina and The Pacific Institute for the Mathematical Sciences. | 489 | 1,796 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2023-06 | latest | en | 0.939153 |
http://hyperphysics.phy-astr.gsu.edu/Hbase/electric/hsehld.html | 1,369,250,762,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368702414478/warc/CC-MAIN-20130516110654-00009-ip-10-60-113-184.ec2.internal.warc.gz | 126,664,964 | 2,640 | # Household Wiring
The standard U.S. household wiring design has two 120 volt "hot" wires and a neutral which is at ground potential. The two 120 volt wires are obtained by grounding the centertap of the transformer supplying the house so that when one hot wire is swinging positive with respect to ground, the other is swinging negative. This versatile design allows the use of either hot wire to supply the standard 120 volt household circuits. For higher power applications like clothes dryers, electric ranges, air conditioners, etc. , both hot wires can be used to produce a 240 volt circuit.
Polarized receptacles Breakers Ground wire
Electrical Code Variations
What happens to the electric charge in household circuits?
Index
Practical circuit concepts
HyperPhysics*****Electricity and magnetism R Nave
Go Back
# Polarized Receptacles
The high voltage (about 120 volts effective, 60 Hz AC) is supplied to the smaller prong of the standard polarized U.S. receptacle. It is commonly called the "hot wire". If an appliance is plugged into the receptacle, then electric current will flow through the appliance and then back to the wider prong, the neutral. The neutral wire carries the current back to the electrical panel and from there to the earth (ground). The ground wire is not a part of the electrical circuit, but is desirable for prevention of electric shock.
The two receptacles in a common "duplex" receptacle receive power from the same circuit leading from the main electrical supply panel. They are wired in parallel so that two appliances which are plugged into the receptacle receive the same voltage, but can draw different amounts of electric current. Parallel wiring is the standard for 120 volt circuits in the entire house, making possible the independent use of all appliances, supplied by the same voltage.
Why is one prong wider? Example circuit
Index
Practical circuit concepts
HyperPhysics*****Electricity and magnetism R Nave
Go Back
# Polarized Receptacles
The wider prong on the polarized plug will permit it to be plugged in only with the correct polarity. The narrower prong is the "hot" lead and the switch to the appliance is placed in that lead, gauranteeing that no voltage will reach the appliance when it is switched off. A non-polarized plug may have the switch in the neutral leg and thus be a shock hazard even when it is switched off.
Index
Practical circuit concepts
HyperPhysics*****Electricity and magnetism R Nave
Go Back
# The Electrical Code and Variations
U.S. electrical wiring is governed by a general electrical code. For example, current code dictates three-prong polarized receptacles and dictates the use of ground fault interrupters in locations where an electrical appliance may be dropped in water.
No attempt to summarize the code will be made here, but a few examples may be instructive. One recent variation which is in force in some locations is the requirement that the neutral tie block and ground wire tie block be separate. The neutral tie block is grounded at the center tap of the transformer which supplies the house, and the ground tie block is tied directly to ground via a ground stake or other grounding mechanism. This is different from the circuit panel illustration which reflects the long-standing practice of having a common neutral and ground tie block which was grounded at the house as well as at the transformer.
Index
Practical circuit concepts
HyperPhysics*****Electricity and magnetism R Nave
Go Back | 721 | 3,517 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2013-20 | latest | en | 0.873456 |
https://community.tes.com/threads/x-bar.384/page-3 | 1,582,754,703,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875146562.94/warc/CC-MAIN-20200226211749-20200227001749-00362.warc.gz | 324,410,064 | 18,770 | # x bar
Discussion in 'Mathematics' started by Colleen_Young, Apr 18, 2005.
1. ### wibble73
Today in itself appears to have generated some serious threads though...
Another lemsip needed here - why do colds come along when things are busiest?
2. ### Colleen_YoungOccasional commenter
Poor Wibble
You are right - lots of serious threads.
3. ### wibble73
Nothing that some lemsip and a good night's sleep won't put right!
If you are feeling like you do some Maths, please tell me I'm not going mad:
A tank takes 5 hours to fill, how long will it take if the flow rate is increased by 50%
(Either I'm going mad or the kids have all got it wrong...)
3hrs 20mins - yes?
4. ### mrs bombastic
we could keep this thread for all the light hearted stuff and leave all the 'heavy' stuff elsewhere? Can you imagine how long it would be?
5. ### mrs bombastic
oop!- sorry to hear you're not well wibblewobble- get well soon. Maybe a days rest from TES...?
6. ### dydx
Agreed Wibs...5 divided by 1.5
7. ### wibble73
So, because they ain't showing their working, how did they get 3hrs 45??
8. ### dydx
Probably all copying from each other Wibs....see what the brightest one in class has done...then her/his mate...
9. ### wibble73
Ooooo, that's cynical...
Found one with working:
First part of questions says teh flow rate increases by 100% - This gives an answer of 2hr 30,
So an increase of 50% is 2hr 30 + 1hr 15, intriguing concept...
10. ### dydx
No ....it's not cynical...that's what they do...you know 'What did you get for Q2' I've seen them do it in the playground. Your'e too trusting Wibs...get several of them who got it wrong to stand up infront of you to explain to the rest of the class how they did it. You'll soon rumble them...
11. ### Colleen_YoungOccasional commenter
mrs b, isn't this a place where we can discuss whatever we want including Maths? Someimes we might have a Mathematical quickie we want to run by colleages in x bar.
12. ### wibble73
I'm with you there Casy.
Was beginning to doubt my solutions for a moment - and definitely not something I would post a thread about normally.
Anayway, hope I haven't given anyone the lurgy - I'll take my mug with me just to be on the safe side! Bed calls...
Take care, W
13. ### mrs bombastic
true- being tee-total I guess I'm not that used to being in a bar. Discuss, discuss, discuss away.
14. ### bethylou
Apology to Wibble for only just replying - no Stats stuff must have been someone else.
I was doing some Stats but I hadn't mentioned that on here and now I am doing Shape instead. That and moving averages were sprung on me last minute.
I can't wait to get a chance to teach some algebra but I bet the kids will ruin it when I do!! (sorry had a bad day today....)
15. ### bbibblerNew commenter
We have all been naughty number crunchers, I have only one thing to say to the tes web staff
2p2q/p
16. ### wibble73
Tcho ho!!!
Don't worry the fun's not going away - just need to keep it to single thread.
People do have a point, we wouldn't want to drive them away, they have much to offer in the way of resources.
17. ### Colleen_YoungOccasional commenter
where did my last post go? apologies if this appears twice.
naughty number crunchers - I like that bbibbler.
mrs b this is no ordinary bar - it's the mathematician's pad - a place to relax and chat.
A fine range of tea, coffee and soft drinks is always available.
18. ### mrs bombastic
let's keep the x bar!
19. ### bethylou
Drinks are on me I have just had the best lesson of my teaching career yet.
I swear that they didn't even breathe unless I had given them permission and they were hanging on my every word (year 8 set 2).
Thanks to everyone on here for advice on motivation - it has definitely made a difference.
That and positive letters home which I had advice on in the behaviour forum. This forum is such good value (infinitely I guess it being free!!).
Red wine anyone?
20. ### mrs bombastic
congrats! Did you try the 'talk'? | 1,039 | 4,017 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.5625 | 3 | CC-MAIN-2020-10 | latest | en | 0.939333 |
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Continuous and Discrete Time Signals and Systems (Mandal & Asif) solutions - chap06
# Continuous and Discrete Time Signals and Systems (Mandal & Asif) solutions - chap06
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Chapter 6: Laplace Transform Problem 6.1 (a) + + = + = = ±² ±³ ´ ±² ±³ ´ II t s I t s st t st t st dt e dt e dt e t u e dt e t u e dt e t x s X 0 ) 4 ( 0 ) 5 ( 4 5 ) ( ) ( ) ( ) ( . Integral I reduces to [ ] ( 5) ( 5) 1 0 0 1 1 0 1 provided Re{( 5)} 0 : Re{ } 5 ( 5) ( 5) 5 s t s t e I e dt s ROC R s s s s + + = = = = + > > − + + + , while integral II reduces to [ ] 0 0 (4 ) (4 ) 1 1 1 1 0 provided Re{(4 )} 0 : Re{ } 4 (4 ) (4 ) 4 s t s t e II e dt s ROC R s s s s −∞ −∞ = = = = > < . The Laplace transform is therefore given by 1 2 1 1 9 ( ) : :( 5 Re{ } 4) 5 4 ( 5)( 4) X s I II with ROC R R R or R s s s s s = + = = = < < + + . (b) + + = + = = = ±² ±³ ´ ±² ±³ ´ II t s I t s st t st t st t st dt e dt e dt e e dt e e dt e e dt e t x s X 0 ) 3 ( 0 ) 3 ( 0 3 0 3 3 ) ( ) ( . Integral I reduces to [ ] 0 0 (3 ) (3 ) 1 1 1 1 0 Re{(3 )} 0 : Re{ } 3 (3 ) (3 ) 3 s t s t e I e dt provided s ROC R s s s s −∞ −∞ = = = = > < , while integral II reduces to [ ] ( 3) ( 3) 1 0 0 1 1 0 1 Re{( 3)} 0 : Re{ } 3 ( 3) ( 3) 3 s t s t e II e dt provided s ROC R s s s s + + = = = = + > > − + + + The Laplace transform is therefore given by 1 2 2 1 1 6 ( ) : :( 3 Re{ } 3) 3 3 9 X s I II with ROC R R R or R s s s s = + = = = < < + .
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2 Chapter 6 (c) ( ) 0 2 2 10 10 1 2 ( ) ( ) cos(10 ) ( ) st st j t j t st j X s x t e dt t t u t e dt t e e e dt −∞ −∞ −∞ = = = ( ) ( ) [ ] 3 3 3 0 0 2 ( 10) 2 ( 10) 1 1 2 2 0 0.5 ( 10) 2 2 ( 10) 0 0.5 ( 10) 2 2 ( 10) 0.5 0.5 ( 10) ( 10) 2( 10) 2 ( 10) 2( 10) 2 s 10 2 0 s j t s j t j j j s j t s j j s j t s j j j s j t e dt t e dt e s j t s j t e s j t s j t j + −∞ −∞ −∞ + + −∞ = = + + + + + + ≠ ± = [ ] 3 3 3 ( 10) 1 1 ( 10) ( 10) ( 1 2 0 Re {s j10}<0 ROC: Re{s}<0 s j s j s j s j j j + + + ± = = 3 3 2 2 3 3 2 3 2 3 0) ( 10) 6 10 2000 60 2000 ( 10) ( 10) ( 100) ( 100) ROC: Re{s}<0 s j s j j s s j s j s s j + + + + = = (d) + + == = = ± ± ± ² ± ± ± ³ ´ ± ± ± ² ± ± ± ³ ´ II t s I t s st t st dt t e dt t e dt e t e dt e t x s X 0 ) 3 ( 0 ) 3 ( 3 ) 5 cos( ) 5 cos( ) 5 cos( ) ( ) ( . Integral I reduces to [ ] 0 0 (3 ) (3 ) (3 ) 2 2 1 2 2 2 2 1 cos(5 ) (3 ) cos(5 ) 5 sin( ) (3 ) 5 1 ( 3) (3 0) (0 0) Re{(3 )} 0 : Re{ } 3 (3 ) 5 ( 3) 5 s t s t s t I e t dt s e t e bt s s s provided s ROC R s s s −∞ −∞ = = + + = + + = > < + + while integral II reduces to [ ] [ ] 3 } Re{ : 0 )} 3 Re{( 5 ) 3 ( ) 3 ( ) 0 ) 3 ( ( ) 0 0 ( 5 ) 3 ( 1 ) 5 sin( 5 ) 5 cos( ) 3 ( 5 ) 3 ( 1 ) 5 cos( 1 2 2 2 2 0 ) 3 ( ) 3 ( 2 2 0 ) 3 ( > > + + + + = + + + + + = + + + + = = + + + s R ROC s provided s s s s t e t e s s dt t e II t s t s t s The Laplace transform is therefore given by 1 2 2 2 2 2 3 3 ( ) : :( 3 Re{ } 3) ( 3) 5 ( 3) 5 s s X s I II with ROC R R R or R s s s + = + = = < < + + + .
Solutions 3 (e) 7 ( 7) 0 ( ) ( ) cos(9 ) ( ) cos(9 ) st t st s t X s x t e dt e t u t e dt e t dt −∞ −∞ = = = [ ] ( 7) ( 7) 2 2 0 2 2 2 2 1 ( 7) cos(9 ) 9 sin(9 ) provided Re{( 7)} 0 ( 7) 9 1 (0 0) ( ( 7) 0) ( 7) 9 ( 7) ( 7) 9 s t s t s e t e t s s s s s s = + >
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{[ snackBarMessage ]} | 1,770 | 3,560 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.671875 | 4 | CC-MAIN-2018-17 | latest | en | 0.640102 |
https://flashman.neocities.org/MD/knowls/subsection.CCD.MVT.knowl | 1,713,091,593,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296816879.25/warc/CC-MAIN-20240414095752-20240414125752-00124.warc.gz | 243,242,714 | 170,551 | SubSection CCD.MVT The Mean Value Theorem
Theorem CCD.MVT. The Mean Value Theorem:
If $f$ is a continuous function on the interval $[a,b]$ and a differentiable function on the interval $(a,b)$, then there is a number $c \in (a,b)$ where $$f'(c) = \frac{f(b) - f(a)}{b-a}$$ or $$f(b) = f(a) + f'(c) (b-a)$$ .
The Mean Value Theorem and 1st Derivative Analysis:
The analysis of the first derivative, traditionally treated using graphs, is visualized in this section with mapping diagrams. Extremes and critical numbers and values are connected to mapping diagrams.
CCD.MVT.FDA : First Derivative Analysis. Visualizing the derivative for an interval with the "derivative vector" in a mapping diagram supports first derivative analysis for monotonic function behavior.
Mapping Diagram Examples: $1:x^2; 2: 3x^2-2x^3 ; 3: 2x^3-3x^2; 4: \sin(\frac {\pi}2 x)$
This is a Java Applet created using GeoGebra from www.geogebra.org - it looks like you don't have Java installed, please go to www.java.com
Graphs of functions and mapping diagrams visualize first derivative analysis.
The Mean Value Theorem and 2nd Derivative Analysis:
CCD.MVT.SDA : Second Derivative Analysis. The analysis of the second derivative, traditionally treated using graphs, is visualized in this section with mapping diagrams. Using acceleration to interpret the second derivative connects the second derivative analysis to the (rate of) change of the derivative.
If $f''(x) \gt 0$ for an interval then $f'(x)$ is increasing for that interval and $f(x)$ is accelerating for that interval.
This is a Java Applet created using GeoGebra from www.geogebra.org - it looks like you don't have Java installed, please go to www.java.com
Notice how the points on the graph are paired with the arrows on the mapping diagram.
CCD.MVT.SFDA : First and Second Derivative Analysis. Visualizing the derivative for an interval with the "derivative vector" in a mapping diagram supports first derivative analysis for extremes, critical numbers and values, and the first and second derivative tests for extremes.
Mapping Diagram and Graphs for First and Second Derivative Analysis Examples
This is a Java Applet created using GeoGebra from www.geogebra.org - it looks like you don't have Java installed, please go to www.java.com | 560 | 2,285 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.90625 | 4 | CC-MAIN-2024-18 | latest | en | 0.828024 |
http://www.dcode.fr/euclidean-division | 1,511,197,904,000,000,000 | text/html | crawl-data/CC-MAIN-2017-47/segments/1510934806086.13/warc/CC-MAIN-20171120164823-20171120184823-00634.warc.gz | 380,266,717 | 5,934 | Search for a tool
Euclidean Division
Tool to make an euclidean division automatically. Euclidean division is an arithmetical operation which associates to two numbers: the dividend and the divisor, two others numbers resulting from the division operation : the quotient and the remainder.
Results
Euclidean Division -
Tag(s) : Mathematics, Arithmetics
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# Euclidean Division
## Euclidean Division A/B
Tool to make an euclidean division automatically. Euclidean division is an arithmetical operation which associates to two numbers: the dividend and the divisor, two others numbers resulting from the division operation : the quotient and the remainder.
### How to calculate the quotient of the euclidean division?
Quotient q is the integer part of the division a/b where a is the dividend and b the divisor.
Example: Consider the division $$43/21 = 2.047619...$$, the quotient equals $$2$$ (the integer part).
### How to calculate the remainder of the euclidean division?
Rest is the result of $$r = a - q \times b$$
Example: Consider the division $$43/21$$, the quotient equals $$2$$ and the remainder equals $$43 - 21 \times 2 = 1$$, indeed $$43 = 2 \times 21 + 1$$.
### How to make the remainder a positive value?
To get a positive remainder, subtract $$1$$ to the quotient and add it to the negative remainder.
Example: If $$a = 15, b = 4$$, then ou can have $$q = 4, r = -1$$ (negative) because $$b \times q + r = 4 \times 4 - 1 = 15 = a$$.
Example: Another possibility is to take $$q = ( 4 - 1 ) = 3$$ and then have $$r = 3$$ (positive) and $$b \times q + r = 4 \times 3 + 3 = 15 = a$$
### Why the name Euclidean Division?
The name comes from Euclid, a mathematician.
### What are the limits of this software?
This software is not limited, it can calculated with any number, including big numbers with arbitrary precision | 563 | 2,260 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 2, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.46875 | 4 | CC-MAIN-2017-47 | latest | en | 0.874052 |
https://skipperkongen.dk/category/fun-and-trivia/page/2/ | 1,716,085,670,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971057631.79/warc/CC-MAIN-20240519005014-20240519035014-00840.warc.gz | 471,622,257 | 20,668 | # Fun and trivia
## When to be most careful about catching the flu?
Continuing on my blogification of Peter Norvigs excellent talk, the question is, when to watch out for the flu, e.g. if you live in Denmark? 1) Go to www.google.com/trends/ 2) Type in the word “influenza” 3) Select your geographical region (Denmark in my case) 4) See data up to year 2008, to avoid the graph …
## How long is a year?
How to find out how long a year is on Earth by only analyzing text? This approach is lifted straight from an excellent and very inspiring talk by Peter Norvig. 1) Go to www.google.com/trends/ 2) Type in the word “Icecream” 3) Measure the distance between the peaks (turns out that the average is exactly the …
## A gallery of big graphs
I like graphs. There is a gallery of big ones over at nd.edu
## Playing with GraphViz and MathGenealogy data
Math in Genealogy is a great project (donate online). Sven Köhler from Potsdam, Germany has written a python script for visualizing the database, which I’m going to try. First step is to clone the git repo: \$ git clone git@github.com:tzwenn/MathGenealogy.git\$ git clone git@github.com:tzwenn/MathGenealogy.git His instructions are quite simple: \$ ./genealogy.py –search 38586 # 30 seconds …
## Gregory Palamas 1363
This stuff blows my mind. Gregory Palamas (1296–135), spelled Γρηγόριος Παλαμάς in greek, was a monk on Mount Athos, a place I’ve visited with my father two times. It is a beautiful peninsula in northern Greece, scattered with old monasteries. Furthermore, only men have been allowed on the peninsula for around a thousand years. Palamas …
## DIKU APL Group IRC channel
Come join the DIKU APL channel on IRC:
## A new side to Bill Gates
I don’t know why, but I really liked to read this post by Bill Gates. In an odd way it was heart warming :-) Three Things I’ve Learned From Warren Buffett Maybe I should start reading Bill Gates blog as well, just for more of that feel good vibe I got from reading the blog …
## NSA back door
This is geek humor at it’s best, and is one of the funniest reactions to PRISM I have seen. Here is a screenshot from the github project flask-nsa The project description simply goes: “Give the NSA free backdoor access to your Flask app”. <3
## Surviving a startup with small children in the house
The question: Is it possible to be an entrepreneur while having small children and a wife? Here is what a bunch of entrepreneurs say about that: Entrepreneur Verdict Jason RobertsPasadena He says yes. Has three kids aged 6, 4 and 2 and several startups behind him or in the works, including AppIgnite. Jason CalacanisLos Angeles …
## Evolving database algorithms through human experiments
Here is something fun to do on a sunny day. The idea is the following: A group of people collectively designing an algorithm by playing a game. | 695 | 2,851 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-22 | latest | en | 0.873673 |
https://practicaldev-herokuapp-com.global.ssl.fastly.net/sreisner/heres-why-mapping-a-constructed-array-in-javascript-doesnt-work-55di | 1,624,150,696,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623487653461.74/warc/CC-MAIN-20210619233720-20210620023720-00368.warc.gz | 410,298,482 | 43,513 | ## DEV Community is a community of 639,914 amazing developers
We're a place where coders share, stay up-to-date and grow their careers.
# Here's Why Mapping a Constructed Array in JavaScript Doesn't Work
Shawn Reisner Originally published at itnext.io on ・4 min read
### Scenario
For the sake of demonstration, suppose you need to generate an array of numbers from 0 to 99. How might you do this? Here’s one option:
``````const arr = [];
for (let i = 0; i < 100; i++) {
arr[i] = i;
}
``````
If you’re like me, seeing a traditional for-loop in JavaScript makes you slightly uncomfortable. In fact, I haven’t written a traditional for-loop in years thanks to higher-order functions such as forEach, map, filter, and reduce. Declarative functional programming for the win!
Maybe you haven’t yet drunk the functional programming Kool-aid and you’re thinking the above solution seems perfectly fine. It technically is, but after you’ve had a taste of the magic of higher-order functions, you’re probably thinking, “There must be a better way.
My first reaction to this problem was, “I know! I’ll create an empty array of length 100 and map the indices to each element!” JavaScript allows you create an empty array of length n with the Array constructor, like this:
``````const arr = Array(100);
``````
Perfect, right? I have an array of length 100, so now I just need to map the index to each element.
``````const arr = Array(100).map((_, i) => i);
console.log(arr[0] === undefined); // true
``````
What the h*ck!? The first element of the array should be 0, but it’s actually undefined!
### Explanation
There’s an important technical distinction I need to make here in order to explain why this happened. Internally, JavaScript arrays are objects, with numbers as keys. For example:
``````['a', 'b', 'c']
``````
is essentially equivalent to this object:
``````{
0: 'a',
1: 'b',
2: 'c',
length: 3
}
``````
When you access the element at index 0 in an array, you’re really just accessing an object property whose key is 0. This is important because when you think of arrays as objects in conjunction with the way these higher-order functions are implemented (don’t worry, I did that part for you), the cause of our problem makes perfect sense.
When you create a new array with the Array constructor, it creates a new array object with its length property set to the value you passed in, but otherwise the object is a vacuum. There are no index keys in the object representation of the array whatsoever.
``````{
//no index keys!
length: 100
}
``````
You get undefined when you try to access the array value at index 0, but it’s not that the value undefined is stored at index 0, it’s that the default behavior in JavaScript is to return undefined if you try to access the value of an object for a key that does not exist.
It just so happens that the higher-order functions map, reduce, filter, and forEach iterate over the index keys of the array object from 0 to length, but the callback is only executed if the key exists on the object. This explains why our callback is never called and nothing happens when we call the map function on the array — there are no index keys!
### Solution
As you now know, what we need is an array whose internal object representation contains a key for each number from 0 to length. The best way to do this is to spread the array out into an empty array.
``````const arr = [...Array(100)].map((_, i) => i);
console.log(arr[0]);
// 0
``````
Spreading the array into an empty array results in an array that’s filled with undefined at each index.
``````{
0: undefined,
1: undefined,
2: undefined,
...
99: undefined,
length: 100
}
``````
This is because the spread operator is simpler than the map function. It simply loops through the array (or any iterable, really) from 0 to length and creates a new index key in the enclosing array with the value returned from the spreading array at the current index. Since JavaScript returns undefined from our spreading array at each of its indices (remember, it does this by default because it doesn’t have the index key for that value), we end up with a new array that’s actually filled with index keys, and therefore map-able (and reduce-able, filter-able, and forEach-able).
### Conclusion
We discovered some of the implications of arrays being internally represented as objects in Javascript, and learned the best way to create arrays of arbitrary length filled with whatever values you need.
Happy coding! 👍
If you liked this post, follow me! Or at least throw me a clap or two. You can spare one measly clap, right?
Medium : @shawn.webdev
dev.to : @sreisner
Instagram : @shawn.webdev
## Discussion (16)
Elarcis • Edited
``````Array.from({ length: 100 }, (_, i) => i);
``````
Florian Reuschel
Whaaat.
I'm honestly baffled. Used `Array.from()` so often but never knew about the mapping argument. Well, today I learned. Thank you!
Jon Bristow • Edited
Why not just `[...Array(5).keys()];`?
For history’s sake, it’s equivalent to:
``````Array.apply(null, Array(5)).map(function(_, i) {
return i;
});
``````
Thorsten Hirsch • Edited
Shawn's point was not to find a solution that does not use a for loop. Shawn's point was to demonstrate why the code that one would expect to work does not:
``````Array(100).map((_, i) => i)
``````
It looks simple, but does not work in Javascript, due to this array-is-an-object thingy. That's the point.
I'm not sure if I would go with Shawn's spread operator solution or if I'd just go back to a simple for loop. But I'm pretty sure that Array.apply(null, ...) and Elarcis' Array.from(...) are not solutions that are easy to understand (no offence), they both seem to be workarounds.
Florian Reuschel • Edited
To be honest, with regard to the article, I find it quite weird that `Array(5).keys()` does work. 🤔
Jon Bristow
It doesn’t!
The `...` fires first and produces `Array.apply(null, Array(5)).keys()`
Then that is expanded to `Array(undefined, undefined, undefined, undefined, undefined).keys()`.
Since the Array was created by a list of items, keys returns `[0,1,2,3,4]`
JavaScript needs a range operator or a Texas range notation. Just so I don’t need to hide the splat under a function to increase readability.
Florian Reuschel
Makes sense, thanks. I struggle with splat operator execution order from time to time.
Jon Bristow
In trying to rewrite the above code I ended up having to experiment a bit to figure out what was happening when... it’s certainly far from obvious.
Juan Martin Gimenez
I think my question is out of the scope of the topic, but:
When we do:
`Array(100);`
If there are no index keys in the object representation of the array (the object is a vacuum), why we can spread it?? 🤯
`const arr = [...Array(100)]`
Thank you for this interesting post!
Salute from 🇦🇷!
Shawn Reisner • Edited
Great question. The reason we can spread an array with no index keys is because spread's implementation doesn't require that any index keys exist, unlike map. It blindly loops from 0 to 99, asks the array for array[index], and places whatever's returned in the new array.
Franco Traversaro
I know this isn't a constructive comment, but... How the heck a map on a deconstructed array with an anonymous function can be more appealing then a for loop? Really, I don't want to start a flame, I just don't understand.
Shawn Reisner
The for loop is an appealing construct. It's simple. You understand exactly what's happening algorithmically without much of a second thought. The appeal of the higher order map function is that it abstracts away the algorithmic details so that you're telling the code what to, not how to do it. In my experience, code written this way is easier to understand, debug, and maintain, and generally produces fewer bugs. This becomes more true when we need to chain the effects of these higher order functions, but even in cases like this we still benefit from the fact that there's almost nowhere for a bug to hide.
Franco Traversaro
I don't agree... You know, you've just written an article about this approach not working with a naive approach 😁
That was obviously a joke, but still for me it's very hard to understand what that code do. And I feel a bit of horror thinking about the CPU cycles wasted with so many calls... | 2,002 | 8,342 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.6875 | 3 | CC-MAIN-2021-25 | latest | en | 0.880723 |
https://www.mathhomeworkanswers.org/84/what-is-6-4-y-6-2-5?show=35056 | 1,568,961,448,000,000,000 | text/html | crawl-data/CC-MAIN-2019-39/segments/1568514573832.23/warc/CC-MAIN-20190920050858-20190920072858-00457.warc.gz | 942,467,428 | 26,068 | Can you help me solve this problem. I am having trouble
If I understand the problem correctly you need to convert the fraction into a decimal or the decimal into a fraction. Let's convert 6 and 2 fifths into a decimal by dividing 2 by 5
We get
6.4 - y = 6.4
Then get y alone by subtracting 6.4 from both sides.
-y = 0
Then you multiply both sides by (- ) and you get y = 0 because anything times 0 = 0
by Level 4 User (9.5k points)
What is 2/3=3R?
what is the percentage of 24.5?
by
There are 4 classes that are saling pies for \$0.50.First class sales 3 2/6.Second class sales 2 5/6.third class sales 4 1/6 and fouth class sales 1 1/6.How do i know how many money did they collected in total?
by
1x1 2x2+16
Round 173.494
by
Round it to WHAT?
Rounding it to the nearest hundreds gives you 200.
Rounding it to the nearest tens gives you 170.
Rounding it to nearest ones gives you 173.
Rounding it to nearest tenths gives you 173.5
Rounding it to the nearest hundredths gives you 173.49
Wirte the fraction as a decimal 599/10?
by
how to simpfie fractions like 9 5/8 - 1/3
by
i dont know that
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what is 3/(2+I) simplify
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what is the fraction of 0.7
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how would i round this problem to two decimals 875/90
by
what is 0.03971 correct to 3 decimal places
by
6.4-y=62/5
6.4-y=12.4
y =6.4-12.4
ans is
y = 6
by
HOW TO PUT 128/64 IN LOWEST FORM OF FRACTION
by
293,95+176,66+1150+163=
by
write an equivalent decimal for 6.0250.
by
thrice a number x yields fifteen
by
what is the zero in the linear function 5/6x-25
by
21.025 in fractional notation
by
what is 5/6 of 567
by
6.4-y=6 2/5
6.4-y=6.4
-y=0
y=0
by
find common multiple for 5 and 6
by
6.4-y=62/5
6.4-y=6.4
-y=0
y=0
by
Solve :-
6.4 - y = 62 /5
By simple cross multiplication method :-
6.4 x 5 - y = 62
32 - y = 62
- y = 62 - 32
- y = 30
by Level 3 User (3.9k points)
TO SOLVE FOR y ,you transpose 6.4 to the right to make it -6.4.that is
-y = 6 2/5 -6.4 that is
-y = 6.4 -6.4
so -y = 0 you divide -1 both side to make the value y into positive so that is
-y/-1 = 0/-1
y= 0 it is only zero because theres no number that has negative zero>..................................thats all
by
6.4 - y = 6 2
6.4 - 6 2/6 = y
y = 6.4 - 6 2/5
at this point you can convert decimals to fractions or fractions to decimals
6.4 = 6 4/10 = 6 2/5
so:
y = 6 2/5 - 6 2/5 = 0, SO: Y = 0
You try converting fractions to decimals, and solve the problem.
by
they are equivalent
by
12+15 ans 25
by
is very nice
by
6.4 - y = 6 2/5
We need to convert both the numbers to the same unit, either both of them be fractions, or both of them be decimals. You can solve it either way, i've solved it by converting both the numbers to decimal.
6.4 - y = 6 (2/5) 2
6.4 - y = 6 4/10
6.4 - y = 6.4
y = 6.4 - 6.4
y = 0
by
First yo do multiplication. Get a calculator and do 62/5=? 12.4
Then you try to get the y alone so you minus the 6.4. So your work should look like this:
y = 12.4-6.4
Now do 12.4-6.4=? 6
y = 6
by
i dont really know
by
jjjxjjuuidfzchjuhcjxhcjzxhcjkchkxhckjfhdnxchukygfjkgn
by
6.4 - y = 6 2/5
6.4 = y + 6 2/5
6.4 = y + 6.4
6.4 - 6.4 = y
y=0
by
6.4-y=6 2/5 so 6 2/5=6.4 so 6.4-y=6.4 that makes y=0
by
Solve :-
6.4 - y = 62 / 5
By simple cross multiplication method :-
32.0 - 5 y = 62
-5 y = 62 + 32
- 5 y = 94
-y = 94 / 5
- y = 18.8
by Level 3 User (3.9k points)
6.4 - 6 2/5 = y * step1;
6.4 - 6.4 = y * step2;
0 = y * step3;
y = 0 * step 4;
by
~friendy nieborhood spiderman
by
What are the solutions of ? Write the solutions as either the union or the intersection of two sets.
A.
B.
C.
D.
by
round to the nearest 2 decimal places, 9.42/3.47=
by
Estimate the fraction:
A B C D
0---------•---•---------•---•---------1
by
i dont kno how to do this
by
can u explain it better?
by
y = 2/5 or .6
by
como puedo convertir un decimal a octal??
by
XE09153285B42
by
Q: 6.4 - y = 62/5 -----------> (1)
y = ?
Soln:
Multiply both sides of eq (1) by 5, we get:
5(6.4) - 5y = 62
=> 32 - 5y = 62
=> -5y = 62 - 32
=> -5y = 30 --------------------->(2)
now, by dividing both sides of eq (2) by -5, we get:
y = -6 {ANS}
by
6.4-y=62/5
6.4=62/5+y
6.4-62/5=62/5+y-62/5
6.4-62/5=y
0=y
by
the answer of this qs is 18.8
by Level 1 User (700 points)
change the fraction 4/13 to a decimal round off to the nearest tousandth
by
6.4=6.4 +y
y=6.4-6.4=0
by
6.4-y=62/5
6.4-y=12.4
-y=12.4-6.4
-y=6.0
y=-6
by
actually i need some math for my class plus the answers
by
by
6.4 - y=6 2/5
6.4 - 6 2/5 or 6.4(mixed number converted to decimal) = y
0 = y or y = 0
by
round each decimal too the nerest tenth 13.13
by
6.4-y=6 2/5 =24-y=32/5 => - y=32/5-24 = -y=32/5-110/5 = -y=-78/5 So -y=78/5
by
Ans: We know 6 2/5 = 6.4
Then, 6.4 - y = 6.4
or, 6.4 - 6.4 = y
Therefore y=0
by
027.010, 270.1, 027.9, 027.089, 027.15, 271.89, 027.001, 027.011
by
what is the fractional notation for the ratio for 8.2 to 10
by
What is 5 control 3 under binomial distribution?
by
6.4-y=6 2/5=
-y=6.4-6.4
-y=0
by
6.4 - Y = 6 2/5
6.4 = 6 2/5 +Y
6.4 / 6 2/5 =Y
6.4 / 32/5 = Y
6.4 /6.4 =Y
1 = Y
by
87 5/9
by
6.4-y = 6 2/5
y = 6,4 or 6 2/5 / 6 2/5
y = 1
by
the dewey is toleleotjfurhc cvur6ehxs js7d kfj ;vgvffffffhtsadfghjkl ncxxxxxfc,
by
64-10y=124=y =-6
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fraction to decimal 3 1/2
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gfytdthduyycvbdhtdtfth
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How do you compare decimals?
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13 - 2
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First,change 6 2/5 to decimal and we get 6.2. S0... 6.4 -6.2 ___ .2 or 2/5
by
y = -(12/5 + 6.4)
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I need help with standard grid for ged test
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(w+6)4
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10101101
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Is 60% greater than or less than 2/3
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by Level 2 User (1.8k points)
what is 7/25 with a denominator of 100
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The population of the Arab people, almost 12 5. Write this number as a standard
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1/3(6b+9)=
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I am not really sure. I know that there is a good website called mathcalculators.com but maybe that can help you. Sorry though!
by
Which is the least to the greatest decimal 0.2,0.06,0.21,0.25?
by
6(x+10)+ 3x -5
by
not sure
by
6.4 - y = 6 2/5
6.4-y=6.25
-y=6.25-6.4
-y=-.15
y=.15 ans.
Afsispeaks
by
14/24
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nombre:nehemias castellon panozo
direccion:av. mariscal santa cruz
hobby:deporte
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I am having trouble in Roundings Fractions And Mixed Numbers
I would like to show me how to do this question 31/4.
by Level 1 User (140 points)
(13x + 23y) + (45x + 47y)
by
y,0.
by
6.4=6 2/5
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by Level 9 User (42.1k points)
what is 1/8 of 1 1/2
by | 2,832 | 6,755 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2019-39 | latest | en | 0.926421 |
http://www.self.gutenberg.org/article/WHEBN0002732031/Laning%20and%20Zierler%20system | 1,606,415,428,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141188899.42/warc/CC-MAIN-20201126171830-20201126201830-00641.warc.gz | 145,254,886 | 18,303 | #jsDisabledContent { display:none; } My Account | Register | Help
# Laning and Zierler system
Article Id: WHEBN0002732031
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Title: Laning and Zierler system Author: World Heritage Encyclopedia Language: English Subject: Collection: Compilers Publisher: World Heritage Encyclopedia Publication Date:
### Laning and Zierler system
The Laning and Zierler system (sometimes called "George" by its users) was one of the first operating algebraic compilers, that is, a system capable of accepting mathematical formulae in algebraic notation and producing equivalent machine code (the term compiler had not yet been invented and the system was referred to as "an interpretive program"). It was implemented in 1954 for the MIT WHIRLWIND by J. Halcombe Laning and Neal Zierler. It is preceded by the UNIVAC A-2, IBM Speedcoding and a number of systems that were proposed but never implemented.
## Contents
• Description 1
• Sample Program 2
• Applications 3
• Influence on FORTRAN 4
• Notes 5
• References 6
## Description
The system accepted formulae in a more or less algebraic notation. It respected the standard rules for operator precedence, allowed nested parentheses, and used superscripts to indicate exponents. It was among the first programming systems to allow symbolic variable names and allocate storage automatically.
The system also automated the following tasks: floating point computation, linkage to subroutines for the basic functions of analysis (sine, etc.) and printing, and arrays and indexing.
The system accepted input on punched tape produced by a Friden Flexowriter. The character set in use at the Whirlwind installation included "upper-case" (superscript) digits and a hyphen, which were used to indicate array indices, function codes, and (integer) exponents. Like other programming notations of its time, the system accepted only single-letter variable names and multiplication was indicated by juxtaposition of operands. A raised dot was available to indicate multiplication explicitly (the character was created by filing off the lower half of a colon!) The system also included support for solution of linear differential equations via the Runge-Kutta method.
The system was described in an 18-page typewritten manual written for people familiar with mathematics but perhaps unfamiliar with computers. It contains almost nothing in the way of an introduction to computer hardware.
## Sample Program
The following example, taken from page 11 of the system's manual, evaluates \cos x for x = 0, 0.1, ..., 1 using the Taylor series expansion. The implementation is not terribly efficient, and the system already includes \cos x in its subroutine library, but the example serves to give a flavor of the system's syntax. Note that division in the system is evaluated after multiplication:
1 x = 0,
z = 1 - x2/2 + x4/2·3·4 - x6/2·3·4·5·6
+ x8/2·3·4·5·6·7·8 - x10/2·3·4·5·6·7·8·9·10,
PRINT x, z.
e = x - 1.05,
CP 1,
STOP
## Applications
Few applications were written for the system. One documented application, authored by Laning and Zierler themselves, involved a problem in aeronautics. The problem required seven systems of differential equations to express, and had been given to the Whirlwind because it was too large for MIT's Differential Analyzer to handle. The authors, exploiting the Runge-Kutta feature of their programming system, produced a 97-statement program in two and half hours. The program ran successfully the first time.
## Influence on FORTRAN
Some sources have said that the Laning and Zierler system was the inspiration for FORTRAN. John Backus himself admitted to having contributed to this misconception:
The effect of the Laning and Zierler system on the development of FORTRAN is a question which has been muddled by many misstatements on my part. For many years I believed that we had gotten the idea for using algebraic notation in FORTRAN from seeing a demonstration of the Laning and Zierler system at MIT. (Backus )
After reviewing documentation from the time, Backus learned that the FORTRAN project was "well underway" when he and his team got a chance to see Laning and Zierler's work:
[W]e were already considering algebraic input considerably more sophisticated than that of Laning and Zierler's system when we first heard of their pioneering work... [I]t is difficult to know what, if any, new ideas we got from seeing the demonstration of their system. (Backus, op cit)
## Notes
1. ^ J. W. Backus, The history of FORTRAN I, II and III. Proceedings First ACM SIGPLAN conference on History of programming languages
## References
• Backus, J. W. The history of FORTRAN I, II and III. Proceedings First ACM SIGPLAN conference on History of programming languages (Available on line).
• Laning, J.H. and N. Zierler. A Program For Translation of Mathematical Equations for Whirlwind I. Engineering Memorandum E-364, Instrumentation Laboratory, Massachusetts Institute of Technology. (Available on line).
• Sammet, Jean E., "Programming Languages: History and Fundamentals" Prentice-Hall, 1969
• "The Early Development of Programming Languages" in A History of Computing in the Twentieth Century, New York, Academic Press, 1980. ISBN 0-12-491650-3 | 1,203 | 5,251 | {"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} | 2.53125 | 3 | CC-MAIN-2020-50 | latest | en | 0.90832 |
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# JEE Main 2019 Paper - 2 with Solution (After Noon, 8 April)
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JEE Main 2019 Paper - 2 with Solution (After Noon, 8 April) - Question 1
### A damped harmonic oscillator has a frequency of 5 oscillations per second. The amplitude drops to half its value for every 10 oscillations. The time it will take to dropto 1/1000 of the original amplitude is close to:
Detailed Solution for JEE Main 2019 Paper - 2 with Solution (After Noon, 8 April) - Question 1
JEE Main 2019 Paper - 2 with Solution (After Noon, 8 April) - Question 2
### A cell of internal resistance r drives current through an external resistance R. Thepower delivered by the cell to the external resistance will be maximum when:
Detailed Solution for JEE Main 2019 Paper - 2 with Solution (After Noon, 8 April) - Question 2
Power maximum when R=r.
JEE Main 2019 Paper - 2 with Solution (After Noon, 8 April) - Question 3
### A uniform rectangular thin sheet ABCD of mass M has length a and breadth b, as shown in the figure. If the shaded portion HBGO is cut–off, the coordinates of the centre of mass of the remaining portion will be
Detailed Solution for JEE Main 2019 Paper - 2 with Solution (After Noon, 8 April) - Question 3
JEE Main 2019 Paper - 2 with Solution (After Noon, 8 April) - Question 4
In a line of sight radio communication, a distance of about 50 km is kept betweenthe transmitting and receiving antennas. If the height of the receiving antenna is 70m, then the minimum height of the transmitting antenna should be:(Radius of the Earth = 6.4*106 m)
Detailed Solution for JEE Main 2019 Paper - 2 with Solution (After Noon, 8 April) - Question 4
JEE Main 2019 Paper - 2 with Solution (After Noon, 8 April) - Question 5
In a simple pendulum experiment for determination of acceleration due to gravity(g), time taken for 20 oscillations is measured by using a watch of 1 second leastcount. The mean value of time taken comes out to be 30 s. The length of pendulum ismeasured by using a meter scale of least count 1 mm and the value obtained is55.0 cm. The percentage error in the determination of g is close to:
Detailed Solution for JEE Main 2019 Paper - 2 with Solution (After Noon, 8 April) - Question 5
JEE Main 2019 Paper - 2 with Solution (After Noon, 8 April) - Question 6
A parallel plate capacitor has 1μF capacitance. One of its two plates is given +2μCcharge and the other plate, +4μC charge. The potential difference developed acrossthe capacitor is:
Detailed Solution for JEE Main 2019 Paper - 2 with Solution (After Noon, 8 April) - Question 6
JEE Main 2019 Paper - 2 with Solution (After Noon, 8 April) - Question 7
Two magnetic dipoles X and Y are placed at a separation d, with their axes perpendicular to each ot.her. The dipole moment 0f Y is twice that of X. A particle of charge q is passing through their mid–point P, at angle θ = 45° with the
horizontal line, as shown in figure. What would be the magnitude of force on the particleat that instant ? (d is much larger than the dimensions of the dipole)
Detailed Solution for JEE Main 2019 Paper - 2 with Solution (After Noon, 8 April) - Question 7
JEE Main 2019 Paper - 2 with Solution (After Noon, 8 April) - Question 8
A solid sphere and solid cylinder of identical radii approach an incline with the same linear velocity (see figure). Both roll without slipping all throughout. The two climb maximum heights hsph and hcyl on the incline. The ratio
is given by:
Detailed Solution for JEE Main 2019 Paper - 2 with Solution (After Noon, 8 April) - Question 8
JEE Main 2019 Paper - 2 with Solution (After Noon, 8 April) - Question 9
A nucleus A, with a finite de–broglie wavelength λA, undergoes spontaneous fission into two nuclei B and C of equal mass. B flies in the same direction as that of A, while C flies in the opposite direction with a velocity equal to half of that of B. The de–Broglie wavelengths λB and λC of B and C are respectively :
Detailed Solution for JEE Main 2019 Paper - 2 with Solution (After Noon, 8 April) - Question 9
JEE Main 2019 Paper - 2 with Solution (After Noon, 8 April) - Question 10
The temperature, at which the root mean square velocity of hydrogen molecules equals their escape velocity from the earth, is closest to:
[Boltzmann Constant kB =1.38 X 10–23 J/K Avogadro Number NA = 6.02 X 1026 /kg Radius of Earth : 6.4 X 106 m
Gravitational acceleration on Earth = 10 ms–2]
Detailed Solution for JEE Main 2019 Paper - 2 with Solution (After Noon, 8 April) - Question 10
(But according to given data in the question no answer, avogadro number is given wrong )
JEE Main 2019 Paper - 2 with Solution (After Noon, 8 April) - Question 11
A circuit connected to an ac source of emf e=e0 sin(100t) with t in seconds, gives aphase difference of π/4 between the emf e and current i. Which of the followingcircuits will exhibit this?
Detailed Solution for JEE Main 2019 Paper - 2 with Solution (After Noon, 8 April) - Question 11
JEE Main 2019 Paper - 2 with Solution (After Noon, 8 April) - Question 12
Young’s moduli of two wires A and B are in the ratio 7:4. Wire A is 2m long and hasradius R. Wire B is 1.5 m long and has radius 2 mm. If the two wires stretch by thesame length for a given load, then the value of R is close to :
Detailed Solution for JEE Main 2019 Paper - 2 with Solution (After Noon, 8 April) - Question 12
JEE Main 2019 Paper - 2 with Solution (After Noon, 8 April) - Question 13
A body of mass m1 moving with an unknown velocity of , undergoes a collinearcollision with a body of mass m2 moving with a velocity . After collision, m1 andm2 move with velocities of and , respectively. If m2=0.5 m1 and v3 =0.5 v1 ,then v1 is:
Detailed Solution for JEE Main 2019 Paper - 2 with Solution (After Noon, 8 April) - Question 13
JEE Main 2019 Paper - 2 with Solution (After Noon, 8 April) - Question 14
A particle starts from origin O from rest and moves with a uniform acceleration along the positive x– axis. Identify the figure that incorrectly represent the motion qualitatively. (a = acceleration, = velocity, x = displacement, t = time)
Detailed Solution for JEE Main 2019 Paper - 2 with Solution (After Noon, 8 April) - Question 14
JEE Main 2019 Paper - 2 with Solution (After Noon, 8 April) - Question 15
A positive point charge is released from rest at a distance ro from a positive line charge with uniform density. The speed of the point charge, as a function of instantaneous distance r from line charge, is proportional to:
Detailed Solution for JEE Main 2019 Paper - 2 with Solution (After Noon, 8 April) - Question 15
JEE Main 2019 Paper - 2 with Solution (After Noon, 8 April) - Question 16
The electric field in a region is given by , where E is in NC–1 and x is in metres. The values of constants are A=20 SI unit and B=10 SI unit. If the potential at x=1 isV1 and that at x= –5 is V2, then V1 –V2is:
Detailed Solution for JEE Main 2019 Paper - 2 with Solution (After Noon, 8 April) - Question 16
JEE Main 2019 Paper - 2 with Solution (After Noon, 8 April) - Question 17
The ratio of mass densities of nuclei of 40Ca and 16O is close to:
Detailed Solution for JEE Main 2019 Paper - 2 with Solution (After Noon, 8 April) - Question 17
Radius of the nucleus is given by R= ROA1/3 So, Density constant
JEE Main 2019 Paper - 2 with Solution (After Noon, 8 April) - Question 18
Detailed Solution for JEE Main 2019 Paper - 2 with Solution (After Noon, 8 April) - Question 18
JEE Main 2019 Paper - 2 with Solution (After Noon, 8 April) - Question 19
Two very long, straight, and insulated wires are kept at 90° angle from each other in xy–plane as shown in the figure.
These wires carry currents of equal magnitude I, whose directions are shown in the figure. The net magnetic field at point P will be:
Detailed Solution for JEE Main 2019 Paper - 2 with Solution (After Noon, 8 April) - Question 19
JEE Main 2019 Paper - 2 with Solution (After Noon, 8 April) - Question 20
A common emitter amplifier circuit, built using an npn transistor, is shown in the figure. Its dc current gain is 250, RC =1KΩ and VCC = 10V. What is the minimum base current for VCE to reach saturation?
Detailed Solution for JEE Main 2019 Paper - 2 with Solution (After Noon, 8 April) - Question 20
JEE Main 2019 Paper - 2 with Solution (After Noon, 8 April) - Question 21
If Surface tension (S), Moment of Inertia (I) and Planck's constant (h), were to betaken as the fundamental units, the dimensional formula for Linear momentumwould be:
Detailed Solution for JEE Main 2019 Paper - 2 with Solution (After Noon, 8 April) - Question 21
JEE Main 2019 Paper - 2 with Solution (After Noon, 8 April) - Question 22
An electric dipole is formed by two equal and opposite charges q with separation d.The charges have same mass m. It is kept in a uniform electric field E. If it is slightly rotated from its equilibrium orientation, then its angular frequency ω is:
Detailed Solution for JEE Main 2019 Paper - 2 with Solution (After Noon, 8 April) - Question 22
JEE Main 2019 Paper - 2 with Solution (After Noon, 8 April) - Question 23
A rocket has to be launched from earth in such a way that it never returns. If E is the minimum energy delivered by the rocket launcher, what should be the mjnimum energy that the launcher should have if the same rocket is to be launched from the surface of the moon ? Assume that the density of the earth and the moon are equal and that the earth's volume is 64 times the volume of the moon.
Detailed Solution for JEE Main 2019 Paper - 2 with Solution (After Noon, 8 April) - Question 23
JEE Main 2019 Paper - 2 with Solution (After Noon, 8 April) - Question 24
A convex lens (of focal length 20 cm) and a concave mirror, having their principalaxes along the same lines, are kept 80 cm apart from each other. The concavemirror is to the right of the convex lens. When an object is kept at a distance of 30cm to the left of the convex lens, its image remains at the same position even if theconcave mirror is removed. The maximum distance of the object for which thisconcave mirror, by itself would produce a virtual image would be
Detailed Solution for JEE Main 2019 Paper - 2 with Solution (After Noon, 8 April) - Question 24
Focal length of concave mirror is 10cm
JEE Main 2019 Paper - 2 with Solution (After Noon, 8 April) - Question 25
In the circuit shown, a four–wire potentiometer is made of a 400 cm long wire,which extends between A and B. The resistance per unit length of the potentiometer wire is r= 0.01 Ω/cm If an ideal voltmeter is connected as shown with jockey J at 50 cm from end A, the expected reading of the voltmeter will be:
Detailed Solution for JEE Main 2019 Paper - 2 with Solution (After Noon, 8 April) - Question 25
Total resistance of potentiometer wire is 4Ω
Total potential drop across wire is 4/6*3V =2V
JEE Main 2019 Paper - 2 with Solution (After Noon, 8 April) - Question 26
A rectangular solid box of length 0.3 m is held horizontally, with one of its sides on the edge of a platform of height 5 m. When released, it slips off the table in a very short time τ =0.01 s, remaining essentially horizontal. The angle by which it would rotate when it hits the ground will be (in radians) close to
Detailed Solution for JEE Main 2019 Paper - 2 with Solution (After Noon, 8 April) - Question 26
JEE Main 2019 Paper - 2 with Solution (After Noon, 8 April) - Question 27
Calculate the limit of resolution of a telescope objective having a diameter of 200cm,if it has to detect Light of wavelmgth 500 nm coming from a star.
Detailed Solution for JEE Main 2019 Paper - 2 with Solution (After Noon, 8 April) - Question 27
JEE Main 2019 Paper - 2 with Solution (After Noon, 8 April) - Question 28
In the figure shown, what is the current (in Ampere) drawn from the battery? You are given:
Detailed Solution for JEE Main 2019 Paper - 2 with Solution (After Noon, 8 April) - Question 28
Reff = 160/3
JEE Main 2019 Paper - 2 with Solution (After Noon, 8 April) - Question 29
The magnetic field of an electromagnetic wave is given by:
The associated electric field will be:
Detailed Solution for JEE Main 2019 Paper - 2 with Solution (After Noon, 8 April) - Question 29
JEE Main 2019 Paper - 2 with Solution (After Noon, 8 April) - Question 30
The given diagram shows four processes i.e., isochoric, isobaric, isothermal and adiabatic. The correct assignment of the processes, in the same order is given by:
Detailed Solution for JEE Main 2019 Paper - 2 with Solution (After Noon, 8 April) - Question 30
PVn = constant
If n = 0, isobaric
n = 1 isothermal
n = ∞ isochoric
JEE Main 2019 Paper - 2 with Solution (After Noon, 8 April) - Question 31
The maximum prescribed concentration of copper in drinkmg water is:
JEE Main 2019 Paper - 2 with Solution (After Noon, 8 April) - Question 32
The major product obtained in the following reaction is
Detailed Solution for JEE Main 2019 Paper - 2 with Solution (After Noon, 8 April) - Question 32
JEE Main 2019 Paper - 2 with Solution (After Noon, 8 April) - Question 33
Fructose and glucose can be distinguished by :
Detailed Solution for JEE Main 2019 Paper - 2 with Solution (After Noon, 8 April) - Question 33
Barfoed's test: Positive for Fructose and glucose.
Seliwanoff’s test: Positive for Fructose and negative for glucose.
Benedict’s test: Positive for Fructose and glucose.
Fehling’s test: Positive for Fructose and glucose.
JEE Main 2019 Paper - 2 with Solution (After Noon, 8 April) - Question 34
The structure of Nylon–6 is:
Detailed Solution for JEE Main 2019 Paper - 2 with Solution (After Noon, 8 April) - Question 34
JEE Main 2019 Paper - 2 with Solution (After Noon, 8 April) - Question 35
The covalent alkaline earth metal halide (X = Cl, Br, I) is :
Detailed Solution for JEE Main 2019 Paper - 2 with Solution (After Noon, 8 April) - Question 35
Beryllium halides are covalent due to more polarizing power of small Be2+ ion with more charge density.
JEE Main 2019 Paper - 2 with Solution (After Noon, 8 April) - Question 36
0.27 g of a long chain fatty acid was dissolved in 100 cm3 of hexane. 10 mL of this solution was added dropwise to the surface of water in a round watch glass. Hexane evaporates and a monolayer is formed. The distance from edge to centre of the watch glass is 10 cm. What is the height of the monolayer ? [Density of fatty acid = 0.9 g cm-3 ; p = 3 ]
Detailed Solution for JEE Main 2019 Paper - 2 with Solution (After Noon, 8 April) - Question 36
JEE Main 2019 Paper - 2 with Solution (After Noon, 8 April) - Question 37
For the following reactions, equilibrium constans are given:
The equilibrium constant for the reaction,
Detailed Solution for JEE Main 2019 Paper - 2 with Solution (After Noon, 8 April) - Question 37
JEE Main 2019 Paper - 2 with Solution (After Noon, 8 April) - Question 38
The correct statement about ICl5and ICl-4 is:
Detailed Solution for JEE Main 2019 Paper - 2 with Solution (After Noon, 8 April) - Question 38
ICl5 have 4 B.P and 1 L.P. It is square pyramidal. ICl-4have 4 B.P. and 2 L.P. The two L.P occupy opposite corners of octahedral. So it square planar.
JEE Main 2019 Paper - 2 with Solution (After Noon, 8 April) - Question 39
The major product of the following reaction is:
Detailed Solution for JEE Main 2019 Paper - 2 with Solution (After Noon, 8 April) - Question 39
JEE Main 2019 Paper - 2 with Solution (After Noon, 8 April) - Question 40
The compound that inhibits the growth of tumors is:
Detailed Solution for JEE Main 2019 Paper - 2 with Solution (After Noon, 8 April) - Question 40
cis–platini is used as anti–tumer agent.
JEE Main 2019 Paper - 2 with Solution (After Noon, 8 April) - Question 41
For a reaction scheme , if the rate of formation of B is set to be zero then the concentration of B is given by:
Detailed Solution for JEE Main 2019 Paper - 2 with Solution (After Noon, 8 April) - Question 41
JEE Main 2019 Paper - 2 with Solution (After Noon, 8 April) - Question 42
Among the following molecules/ ions,
Which one is diamagnetic and has the shortest bond length?
Detailed Solution for JEE Main 2019 Paper - 2 with Solution (After Noon, 8 April) - Question 42
Bond order
is diamagnetic since all electrons are paired and have shortest bond length due to triple bond.
JEE Main 2019 Paper - 2 with Solution (After Noon, 8 April) - Question 43
The major product of the following reaction is:
Detailed Solution for JEE Main 2019 Paper - 2 with Solution (After Noon, 8 April) - Question 43
JEE Main 2019 Paper - 2 with Solution (After Noon, 8 April) - Question 44
Calculate the standard cell potential (in V) of the cell in which following reaction takes place:
Fe2+ (aq) + Ag+ (aq) → Fe3+ (aq) + Ag (s)
Given that
Detailed Solution for JEE Main 2019 Paper - 2 with Solution (After Noon, 8 April) - Question 44
JEE Main 2019 Paper - 2 with Solution (After Noon, 8 April) - Question 45
The major product in the following reaction is:
Detailed Solution for JEE Main 2019 Paper - 2 with Solution (After Noon, 8 April) - Question 45
JEE Main 2019 Paper - 2 with Solution (After Noon, 8 April) - Question 46
The major product obtained in the following reaction is:
Detailed Solution for JEE Main 2019 Paper - 2 with Solution (After Noon, 8 April) - Question 46
Intramolecular aldol condensation
JEE Main 2019 Paper - 2 with Solution (After Noon, 8 April) - Question 47
Which one of the following alkenes when treated with HCl yields majority an anti Maxkovnikov product?
Detailed Solution for JEE Main 2019 Paper - 2 with Solution (After Noon, 8 April) - Question 47
JEE Main 2019 Paper - 2 with Solution (After Noon, 8 April) - Question 48
The calculated spin–only magnetic moments (BM) of the anionic and cationic species of [Fe(H2O)6]2 and [Fe(CN)6], respectively, are :
Detailed Solution for JEE Main 2019 Paper - 2 with Solution (After Noon, 8 April) - Question 48
The compound containing [Fe(H2O)6]2 cation and [Fe(CN)6] anion must be [Fe(H2O)6]2 [Fe(CN)6]. It contains 2[Fe(H2O)6]2+ cation and [Fe(CN)6]4– anion in which Iron is present as Fe2+ ion. In [Fe(H2O)6]2+ since H2O is weak ligand there will be 4 unpaired electrons with 4.9 BM while in [Fe(CN)6]4– all the electrons are paired as the CN is strong ligand with zero unpaired electrons and zero magnetic moment.
JEE Main 2019 Paper - 2 with Solution (After Noon, 8 April) - Question 49
The IUPAC symbol for the element with atomic number 119 would be:
Detailed Solution for JEE Main 2019 Paper - 2 with Solution (After Noon, 8 April) - Question 49
For 1 the abreviation is U (from unium) and for 9 is e (from enneium). So the IUPAC symbol is uue.
JEE Main 2019 Paper - 2 with Solution (After Noon, 8 April) - Question 50
The percentage composition of carbon by mole in methane is :
Detailed Solution for JEE Main 2019 Paper - 2 with Solution (After Noon, 8 April) - Question 50
% of C by mole = 1/5 x 100 = 20%
JEE Main 2019 Paper - 2 with Solution (After Noon, 8 April) - Question 51
The Mond process is used for the:
Detailed Solution for JEE Main 2019 Paper - 2 with Solution (After Noon, 8 April) - Question 51
Nickel is purified by Mond’s process using the formation of unstable volatile compound. Ni(CO)4.
JEE Main 2019 Paper - 2 with Solution (After Noon, 8 April) - Question 52
The statement that is INCORRECT about the interstitial compounds is :
Detailed Solution for JEE Main 2019 Paper - 2 with Solution (After Noon, 8 April) - Question 52
They are relatively non reactive
JEE Main 2019 Paper - 2 with Solution (After Noon, 8 April) - Question 53
Polysubstitution is a major drawback in
Detailed Solution for JEE Main 2019 Paper - 2 with Solution (After Noon, 8 April) - Question 53
Polyalkylation – Products of Friedel–Crafts are even more reactive than starting material. Alkyl groups produced in Friedel–Crafts Alkylation are electron–donating substituents meaning that the products are more susceptible to electrophilic attack than reactant.
JEE Main 2019 Paper - 2 with Solution (After Noon, 8 April) - Question 54
For the solution of the gases w, x, y and z in water at 298 K, the Henrys law constants (KH) are 0.5, 2, 35 and 40 kbar, respectively. The correct plot for the given data is :
Detailed Solution for JEE Main 2019 Paper - 2 with Solution (After Noon, 8 April) - Question 54
KH for w, x, y, z are 0.5, 2, 35 & 40.
P = KH X
P = KH (1 - Xwater)
P = KH - KHXwater (y = c - mx)
KH values increases in the order z > y > x > w.
JEE Main 2019 Paper - 2 with Solution (After Noon, 8 April) - Question 55
Consider the bcc unit cells of the solids 1 and 2 with the position of atoms as shown below. The radius of atom B is twice that of atom A. The unit cell edge length is 50% more in solid 2 than in 1. What is the approximate packing efficiency in solid 2?
Detailed Solution for JEE Main 2019 Paper - 2 with Solution (After Noon, 8 April) - Question 55
√3a = 2r + 4r = 6r
a = 2√3r
Packing fraction =
i.e. 90% is the packed in the solid B
JEE Main 2019 Paper - 2 with Solution (After Noon, 8 April) - Question 56
5 moles of an ideal gas at 100 K are allowed to undergo reversible compression till its temperature becomes 200 K. If CV = 28 JK-1 mol–1, calcuIate Δ U and Δ PV for this process. ( R = 8.0 JK -1 mol-1 )
Detailed Solution for JEE Main 2019 Paper - 2 with Solution (After Noon, 8 April) - Question 56
ΔU = 28x5x100 = 14kJ
Δ(pV) = p2v2 - p1v2 = nR (T- T1)
= 5x8.314x100 ≈ 4KJ
JEE Main 2019 Paper - 2 with Solution (After Noon, 8 April) - Question 57
The ion that has sp3d2 hybridization for the central atom, is:
Detailed Solution for JEE Main 2019 Paper - 2 with Solution (After Noon, 8 April) - Question 57
ICl4- have 4 B.P. and 2 L.P (steric number 6). So central atom is involved in sp3d2 hybridization.
JEE Main 2019 Paper - 2 with Solution (After Noon, 8 April) - Question 58
If p is the momentum of the fastest electron ejected from a metal surface after the irradiation of light having wavelength λ , then for 1.5 p momentum of the photoelectron, the wavelength of the light should be: (Assume kinetic energy of ejected photoelectron to be very high in comparison to work function) :
Detailed Solution for JEE Main 2019 Paper - 2 with Solution (After Noon, 8 April) - Question 58
W0 is too small in comparision to K.E.
JEE Main 2019 Paper - 2 with Solution (After Noon, 8 April) - Question 59
Which of the following compounds will show the maximum ‘enol’ content ?
Detailed Solution for JEE Main 2019 Paper - 2 with Solution (After Noon, 8 April) - Question 59
CH3COCH2COOC2H5 %enol = 0.00025
CH3COCH2CONH2 % enol = <7.5
CH3COCH2COCH3 %enol = 72
CH3COCH3 %enol = 0.00025
JEE Main 2019 Paper - 2 with Solution (After Noon, 8 April) - Question 60
The strength of 11.2 volume solution of H2O2 is:
[Given that molar mass of H=1 g mol–1 and O=16 g mol–1]
Detailed Solution for JEE Main 2019 Paper - 2 with Solution (After Noon, 8 April) - Question 60
JEE Main 2019 Paper - 2 with Solution (After Noon, 8 April) - Question 61
The tangent and the normal lines at the point ( √3,1) to the circle x2 + y2 = 4 and the x-axis form a triangle. The area of this triangle (in square units) is:
Detailed Solution for JEE Main 2019 Paper - 2 with Solution (After Noon, 8 April) - Question 61
JEE Main 2019 Paper - 2 with Solution (After Noon, 8 April) - Question 62
Let and ,for some real x.Then is possible if :
Detailed Solution for JEE Main 2019 Paper - 2 with Solution (After Noon, 8 April) - Question 62
JEE Main 2019 Paper - 2 with Solution (After Noon, 8 April) - Question 63
If three distinct numbers a, b, c are in G.P. and the equations ax2 + 2bx +c =0 and dx2 + 2ex + f = 0 have a common root, then which one of the following statements is correct ?
Detailed Solution for JEE Main 2019 Paper - 2 with Solution (After Noon, 8 April) - Question 63
JEE Main 2019 Paper - 2 with Solution (After Noon, 8 April) - Question 64
The sum is equal to
Detailed Solution for JEE Main 2019 Paper - 2 with Solution (After Noon, 8 April) - Question 64
JEE Main 2019 Paper - 2 with Solution (After Noon, 8 April) - Question 65
If the system of linear equations
x – 2y + kz =1
2x + y + z = 2
3x – y – kz =3
has a solution (x, y, z), z ≠ 0, then (x, y) lies on the straight line whose equation is:
Detailed Solution for JEE Main 2019 Paper - 2 with Solution (After Noon, 8 April) - Question 65
From the given eqs, (1) + (3) → 4x – 3y – 4 = 0
JEE Main 2019 Paper - 2 with Solution (After Noon, 8 April) - Question 66
The tangent to the parabola y2=4x at the point where it intersects the circle x2 + y2 = 5 in the first quadrant, passes through the point :
Detailed Solution for JEE Main 2019 Paper - 2 with Solution (After Noon, 8 April) - Question 66
y2 = 4x,x2 + y2 = 5 ⇒ x2 + 4x -5= 0
x =1 (x ≠-5) ⇒ y =2 Q (y>0)
Eq of tangent is 2y = 2(x+1) ⇒ x – y + 1 = 0
JEE Main 2019 Paper - 2 with Solution (After Noon, 8 April) - Question 67
If where C is a constant of integration, then the function f(x) is equal to :
Detailed Solution for JEE Main 2019 Paper - 2 with Solution (After Noon, 8 April) - Question 67
JEE Main 2019 Paper - 2 with Solution (After Noon, 8 April) - Question 68
Suppose that the points (h, k), (1, 2) and (–3, 4) lie on the line L1. If a line L2 passing through the points (h, k) and (4, 3) is perpendicular to L1, then k/h equals:
Detailed Solution for JEE Main 2019 Paper - 2 with Solution (After Noon, 8 April) - Question 68
eq of L1 is x + 2y = 5 ⇒
JEE Main 2019 Paper - 2 with Solution (After Noon, 8 April) - Question 69
Let the numbers 2, b, c be in an A.P. and If det(A) ∈ [2,16],then c lies in the interval:
Detailed Solution for JEE Main 2019 Paper - 2 with Solution (After Noon, 8 April) - Question 69
JEE Main 2019 Paper - 2 with Solution (After Noon, 8 April) - Question 70
The minimum number of times one has to toss a fair coin so that the probability of observing at least one head is at least 90% is:
Detailed Solution for JEE Main 2019 Paper - 2 with Solution (After Noon, 8 April) - Question 70
JEE Main 2019 Paper - 2 with Solution (After Noon, 8 April) - Question 71
Two vertical poles of heights, 20 m and 80 m stand apart on a horizontal plane. The height (in meters) of the point of intersection of the Lines joining the top of each pole to the foot of the other, from this horizontal plane is :
Detailed Solution for JEE Main 2019 Paper - 2 with Solution (After Noon, 8 April) - Question 71
JEE Main 2019 Paper - 2 with Solution (After Noon, 8 April) - Question 72
Let , where g is a non–zero even function. If f(x+5)=g(x), then equals:
Detailed Solution for JEE Main 2019 Paper - 2 with Solution (After Noon, 8 April) - Question 72
g(x) = f '(x) & g (- x) = g (x) = f (x + 5) = f (5 - x) and also f(x) is odd.
JEE Main 2019 Paper - 2 with Solution (After Noon, 8 April) - Question 73
Let be a diiferentiable function Satisfying f '(3) + f '(2) = 0 . Then is equal to
Detailed Solution for JEE Main 2019 Paper - 2 with Solution (After Noon, 8 April) - Question 73
JEE Main 2019 Paper - 2 with Solution (After Noon, 8 April) - Question 74
Let be defined as
where [t] denotes the greatest integer less than or equal to t. Then, f is discontinuous at:
Detailed Solution for JEE Main 2019 Paper - 2 with Solution (After Noon, 8 April) - Question 74
JEE Main 2019 Paper - 2 with Solution (After Noon, 8 April) - Question 75
let and A(α) is area of the region S (α) . If for a λ, 0 < λ < 4, A(λ) : A(4) = 2 : 5 , then λ equals:
Detailed Solution for JEE Main 2019 Paper - 2 with Solution (After Noon, 8 April) - Question 75
JEE Main 2019 Paper - 2 with Solution (After Noon, 8 April) - Question 76
The number of integral values of m for which the equation (1 + m2) x2 - 2 (1 + 3m) x + (1 + 8m) = 0 has no real root is:
Detailed Solution for JEE Main 2019 Paper - 2 with Solution (After Noon, 8 April) - Question 76
D < 0 ⇒ 8m3 -8m2 + 2m > 0 ⇒ 2m (2m-1)2 > 0 ⇒ m > 0
JEE Main 2019 Paper - 2 with Solution (After Noon, 8 April) - Question 77
If the eccmtricity of the standard hyperbola passing through the point (4, 6) is 2, then the equation of the tangent to the hyperbola at (4, 6) is :
Detailed Solution for JEE Main 2019 Paper - 2 with Solution (After Noon, 8 April) - Question 77
Eq. of hyperbola is it passes through (4,6) ⇒ a2=4.
Eq. of tangent is 2x–y=2.
JEE Main 2019 Paper - 2 with Solution (After Noon, 8 April) - Question 78
If , then is equal to:
Detailed Solution for JEE Main 2019 Paper - 2 with Solution (After Noon, 8 April) - Question 78
JEE Main 2019 Paper - 2 with Solution (After Noon, 8 April) - Question 79
The height of a right circular cylinder of maximum volume inscribed in a sphere of radius 3 is :
Detailed Solution for JEE Main 2019 Paper - 2 with Solution (After Noon, 8 April) - Question 79
JEE Main 2019 Paper - 2 with Solution (After Noon, 8 April) - Question 80
The number of four–digit numbers strictly greater than 4321 that can be formed using the digits 0, 1, 2, 3, 4, 5 (repetition of digits is allowed) is :
Detailed Solution for JEE Main 2019 Paper - 2 with Solution (After Noon, 8 April) - Question 80
Total no. = 310.
JEE Main 2019 Paper - 2 with Solution (After Noon, 8 April) - Question 81
Let be written as f(x) =f1(x) +f2(x), where f1(x) is an even function and f2(x) is an odd function. Then f1(x + y) +f1(x – y) equals:
Detailed Solution for JEE Main 2019 Paper - 2 with Solution (After Noon, 8 April) - Question 81
JEE Main 2019 Paper - 2 with Solution (After Noon, 8 April) - Question 82
The vector equation of the plane through the line of intersection of the planes x+y+ z= 1 and 2x+3y+4z=5 which is perpendicular to the plane x – y+ z =0 is :
Detailed Solution for JEE Main 2019 Paper - 2 with Solution (After Noon, 8 April) - Question 82
Line of intersection of the planes x+y+z=1, 2x+3y+4z=5 is
Eq. of required plane is
JEE Main 2019 Paper - 2 with Solution (After Noon, 8 April) - Question 83
Which one of the following statements is not a tautology ?
Detailed Solution for JEE Main 2019 Paper - 2 with Solution (After Noon, 8 April) - Question 83
JEE Main 2019 Paper - 2 with Solution (After Noon, 8 April) - Question 84
Given that the slope of the tangent to a curve y = y ( x) at any point (x,y) is 2y/xIf the curve passes through the centre of the circle x2 + y2 – 2x – 2y = 0, then its equation is :
Detailed Solution for JEE Main 2019 Paper - 2 with Solution (After Noon, 8 April) - Question 84
JEE Main 2019 Paper - 2 with Solution (After Noon, 8 April) - Question 85
If a point R(4, y, z) lies on the line segment joining the points P (2,-3, 4) and Q(8, 0, 10), then the distance of R from the origin is:
Detailed Solution for JEE Main 2019 Paper - 2 with Solution (After Noon, 8 April) - Question 85
Here, P, Q, R are collinear ⇒
JEE Main 2019 Paper - 2 with Solution (After Noon, 8 April) - Question 86
In an ellipse, with centre at the origin, if the difference of the lengths of major axis and minor axis is 10 and one of the foci is at (0, 5 √3) , then the length of its latus rectum is:
Detailed Solution for JEE Main 2019 Paper - 2 with Solution (After Noon, 8 April) - Question 86
Here, given ellipse is vertical ellipse & b – a = 5, b2 – a2 = 75
JEE Main 2019 Paper - 2 with Solution (After Noon, 8 April) - Question 87
If the fourth term in the binomial expansion of is equal to 200,and x >1, then the value of x is:
Detailed Solution for JEE Main 2019 Paper - 2 with Solution (After Noon, 8 April) - Question 87
JEE Main 2019 Paper - 2 with Solution (After Noon, 8 April) - Question 88
If the lengths of the sides of a triangle are in A.P. and the greatest angle is double the smallest, then a ratio of lengths of the sides of this triangle is:
Detailed Solution for JEE Main 2019 Paper - 2 with Solution (After Noon, 8 April) - Question 88
JEE Main 2019 Paper - 2 with Solution (After Noon, 8 April) - Question 89
If f(1) = 1, f '(1) = 3, then the derivative of f(f(f ( x))) + (f ( x))2 of x=1 is
Detailed Solution for JEE Main 2019 Paper - 2 with Solution (After Noon, 8 April) - Question 89
JEE Main 2019 Paper - 2 with Solution (After Noon, 8 April) - Question 90
A student scores the following marks in five tests : 45, 54, 41, 57, 43. His score is not known for the sixth test. If the mean score is 48 in the six tests, then the standard deviation of the marks in six tests is :
Detailed Solution for JEE Main 2019 Paper - 2 with Solution (After Noon, 8 April) - Question 90
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how to prove the area of the trapezium is 1/2(a+b)h
1. 👍 0
2. 👎 0
3. 👁 137
1. Draw a diagonal, and calculate the area of the two resulting triangles:
(1/2)ah + (1/2)bh
=(1/2)(a+b)h
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asked by Jack rider on April 10, 2018
5. Math
A large pizza at Tony's Pizzeria is a circle with a 14-inch diameter. Its box is a rectangular prism that is 1 14 8 inches long, 1 14 8 inches wide, and 3 1 4 inches tall. Your job is to design a crazy new shape for a large pizza.
asked by Naomi on March 30, 2017
6. geometry
pqrs is a parallelogram. pr and sq are its diagonals prove area triangle sor =area triangle pqo
asked by DIPI on December 26, 2017 | 841 | 2,485 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.0625 | 4 | CC-MAIN-2020-34 | latest | en | 0.925588 |
https://www.energyvanguard.com/blog/a-humidifier-is-a-bandaid-the-problem-is-infiltration/ | 1,716,795,110,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971059037.23/warc/CC-MAIN-20240527052359-20240527082359-00271.warc.gz | 659,607,142 | 26,540 | # A Humidifier Is a Bandaid — The Problem Is Infiltration
Does your heating system have a humidifier attached to it? If so, it’s most likely there to treat a symptom while leaving the underlying cause of that symptom alone. The symptom is dry air in winter. Unless you’re in an extremely dry or cold climate, the cause is not dry air. The real cause is lack of air-sealing.
First, do you know if your home has a humidifier? If you look at the furnace/air handler and see a smaller attachment that has connections for water, electricity, and drainage, that’s most likely a humidifier. The photo here shows what one model looks like. What they do is introduce water vapor into your duct system so that the heated air in wintertime carries it into your house.
Ideally, you want the relative humidity (RH) in your home to be in the range of about 30% to 50%. For a lot of us, the summertime problem is keeping the RH low enough, but it’s the opposite in wintertime, even in humid climates. The air in homes often gets too dry. You’ll notice it because you get the static electricity shocks every once in a while, or you have to keep putting lotion on your dry skin.
### Why is your home’s humidity too low?
The company trying to sell you that humidifier says your home’s humidity is too low because there’s not enough water vapor in the indoor air. Yeah, yeah, that’s true, but it’s like saying you’re naked because you don’t have any clothes on. You’re just repeating the same thing in different words. What we really want to know is why you don’t have any clothes on…uh, I mean why the humidity in your home is so low.
The answer to that question is: Cold air is dry air. Understand that and you’ll know what I mean when I say a humidifier is a bandaid, not a real solution. The chart below, called a psychrometric chart, has the explanation.
I won’t give a full lesson on psychrometrics here, but let’s focus on two points on the chart, point A and point B. Point A corresponds to outdoor air when it’s moderately cold, 32° F. Let’s say it’s also raining and the relative humidity is 100%. Among other things, the psychrometric chart shows how temperature and relative humidity change. For our purposes today, here’s what you need to know about the chart:
• When the temperature changes, you’re moving left (lower temperature) or right (higher temperature).
• Relative humidity is shown as the curves, which are all bunched up together at the lower left. They spread out and move upward as the temperature increases.
When outside air at 32° F, 100% RH leaks into the house, it gets warmed up. Your heating system has to do extra work and you pay extra money to heat up that outside air, but we’re interested in what happens to the humidity right now. As that air heats up, it moves to the right. If you keep your thermostat at 70° F, the conditions of that outside air that leaked in eventually land at point B.
If you look carefully, you’ll se that we crossed 8 RH curves to get to point B, each covering a range of 10% RH. That means we dropped from 100% RH when the air was still outside to 20% RH after it leaked in and warmed up. Wow! That’s a big drop in relative humidity.
Of course, the 20% RH applies only to the volume of air that leaked in, not all the air in your home. The 20% RH air mixes with whatever you have in the house already, and the final number will be the RH of the combined air masses. The key to your low humidity, though, is that the more infiltration you have, the lower your indoor air’s relative humidity will be.
That brings us back to where we started, so we can see now that when the relative humidity in your home is too low, a humidifier is treating the symptom. It’s a bandaid, not a real solution. If you really want to solve the problem, seal the air leaks. It’s a lot less expensive, too, because there’s no operating cost for air-sealing.
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#### This Post Has 13 Comments
1. D Gough says:
Allison,
Allison,
Dry homes are not always leaky homes. There are some cases where the homeowner’s activities just don’t produce enough humidity to stay in the “zone”. I have seen it happen about 4-5 times in the last 20 years.
I’ve seen substantially airtight homes with airtight ducts, with dry crawl space or basements and/or the homeowner(s) apparently didn’t cook much, perhaps didn’t breathe much, had very few house plants, no wood storage indoors, no pets, little to no mopping and maybe one shower a week whether they needed it our not. Plus, there were bath exhaust fans that really worked. The winter RH stayed consistently below 30%. Stuff would shrink and the homeowner complained about static electricity yada yada… A steam humidifier is the only thing that would add sufficient moisture to solve the problem.
Plus, with unconfessed sin in the home, I also can see supply air only ventilation being contributory to dry interiors.
2. Allison Bailes says:
Dale S.:
Dale S.: There’s just no telling what code officials are thinking sometimes, but overall, I think codes are improving. There’s going to be a session on this topic at this year’s RESNET conference called Do Building Codes Really Incorporate Building Science?
D Gough: Yes, I’ll admit that infiltration isn’t going to be the culprit in every case. I mentioned extremely dry or cold climates as possible exceptions in the article, but yes, tight homes could have problems, too, if they don’t generate much moisture inside. Overventilation can cause problems as well, and I have firsthand experience with that one in the green home that I built
Darrel T.: You could go back to the house (if it still exists) and do a blower door test. ;~)
3. John Poole says:
Darrel’s comment reminded of
Darrel’s comment reminded of those decorative cast iron “humidifiers” that you sometimes see advertised for use with woodburning stoves.
You fill them with water and place them on your stove top, and they humidify the surrounding air. I have no direct experience with this, but I’d imagine homes that rely heavily on wood or pellet stoves, especially cookstoves, would have pretty severe dry air problems in the winter.
4. Mesa AC Repairman says:
Out in Arizona we have a big
Out in Arizona we have a big problem with houses getting too dry all year round. While we dont have to use heaters very often they are still used a few month of year and during that time there is a lot of dry air problems that come about. But it is like that most of the year so we really have to dig deep to get to the root of the problems homeowners have with dry air.
5. John Proctor says:
One other problem with
One other problem with humidifiers is that many of them bypass air from the supply side to the return side creating a bypass loop that detracts from both cooling and heating efficiency. Most of the bypass humidifiers have a manual damper to shut off the recirculation in the summer, but the homeowner usually doesn’t do anything with it.
6. Mary Beth says:
While we’re on the subject of
While we’re on the subject of humidification, any thoughts about hanging wet clothes in the house to dry — and how bad this action might be for adding moisture to the home from the inside-out. The dryer-hater in me hates to waste the propane. We inside-dwellers don’t mind having a little extra humidity in the dry winter. Usually I dry things half-way, then hang them up to dry the rest of the way. Please tell me how bad this is!?
7. Allison Bailes says:
John P:
John P: That depends on how big the stove is and also how efficient. I once had a woodstove with a catalytic converter, and once we got the fire going and closed the damper so that the exhaust went through the converter, it drew very little air.
Mesa AC: Indeed you’re right. I mentioned that my remarks didn’t apply to extremely dry climates because you may be too dry anyway. Of course, good air-sealing is still important because then the humidity you add is more likely to stay in the home.
John P.: Great point. I’ve seen those bypass humidifiers here in the Southeast, and you’re right – homeowners won’t close a damper because they don’t even know it’s there. And most certainly won’t go down into the crawl space twice a year to open and close it if they do know.
Mary Beth: What you’re doing should be fine in winter, when the air in your home is probably dryer anyway. It could cause problems in the summer, though, unless you live in a dry climate like Colorado or Arizona. I’d suggest buying a digital thermo-hygrometer (which measures temperature and relative humidity) and keep it near your drying rack. If the RH gets up towards 60% or above, you’ll want to do something differently.
8. Matthew Cooper says:
If cooking and cleaning
If cooking and cleaning activities include the proper use of exhaust fans and infiltration is controlled, a properly sized heating system will still reduce humidity below levels considered comfortable in low-humidity winter climates in my experience. Properly controlled steam humidifiers such as those manufactured by Norco can effectively work and in many cases are necessary. For new construction homes if the humidity isn’t properly maintained and there are hardwood floors, the flooring manufacturer/installer will not warranty the flooring.
We also have designed and installed many systems where a homeowner needs elevated humidity for either health issues or expensive art collections. Controls are a key component along with effective ADPI for proper air mixing. Devices such as cold-snap indicators on windows tied into DDC systems can overcome the inherent challenges posed by the introduction of moisture into the properly controlled environment.
9. Jaap Weel says:
This is very interesting to
This is very interesting to know. I always wondered how heating a house could somehow cause the humidity to go down and where the heck the water ended up. Now I understand a bit better.
For now, I’m living in an apartment where I cannot do much about air leakage. The landlord clearly has bigger fish to fry, for now, than sealing the building; there’s some real deferred maintenance the prior owner has pushed off onto the current, and it looks like they’re investing considerable money just getting the building back into its original 1960s condition, let alone upgrading it to 21st-century standards.
Besides, us tenants are paying the heating bill, and the housing market in my area is heating up, so they’re having no trouble at the moment filling every vacancy at rapidly increasing prices.
So for now, I’ll keep running my little drugstore humidifiers. But when (hopefully soon) I have a home of my own, I’ll consider improved air sealing first.
10. Amy A says:
As a homeowner, and
As a homeowner, and researcher I like to review information based on studies. This blog spoke to the science part of my interest. Great Job!
Here is my question: I get the point to SEAL, but what I do not know is who is considered a professional in sealing?
11. Allison Bailes says:
Amy A: I
Amy A: I wish there were an easy answer to your question. For existing homes, the best answer is to see if you can find a company with BPI certified professionals or a RESNET Energy Smart contractor. There are plenty of companies that do good air-sealing work without having those, however, but that makes it harder for the homeowners who don’t know the details to know if you’re being hoodwinked or not.
Also, I’d suggest getting a full energy audit or home performance assessment. Here’s an article I wrote about that: How to Choose a Company to Do a Home Energy Audit.
12. Dirk Muir says:
I am heating an 80year old
I am heating an 80year old home near Thunder Bay, NW Ontario (Canada) with an indoor woodfurnace. Night time temperatures are frequently in the -20 to -40’F range. Not only do we have static sparks and dry skin & lips, but I frequently wake with my mouth and throat completely dry. My doctor has suggested that a persistent cough and throat irritation is due to low humidity in our house. This is despite running 20-30 litres a day through two humidifiers. Through your postings, I now understand that the woodfurnace is driving infiltration of air, that once warmed in the house, has very low relative humidity that the humidifiers can’t compensate for. Other than sealing the house (a major refit), is there anything else one can do to buffer humidity levels? What about feeding exterior air direct to the furnace to reduce the infiltration gradient?
13. John Proctor says:
@Dirk
@Dirk
Anything you can do to reduce air infiltration will help. Supplying combustion air straight to the furnace is a help as well as any amount of air sealing (generally starting high in the house. | 2,839 | 12,889 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.703125 | 3 | CC-MAIN-2024-22 | latest | en | 0.922767 |
https://cs.nyu.edu/pipermail/fom/2012-March/016384.html | 1,566,554,281,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027318243.40/warc/CC-MAIN-20190823083811-20190823105811-00287.warc.gz | 413,905,170 | 2,726 | # [FOM] Equivalence of Downward Löwenheim-Skolem theorem and the Axiom of Dependent Choice
Christian Espindola christian.espindola at gmail.com
Fri Mar 30 11:58:14 EDT 2012
```Dear all,
It is known that the Axiom of Dependent Choice (DC) is enough to derive in
ZF the following form of the (downward) Löwenheim-Skolem theorem (LS):
Every model M of a first order theory T with countable signature has an
elementary submodel N which is at most countable. But the standard
reference (Consequences of the axiom of choice by Howard and Rubin) does
not contain any mention of the reverse implication (neither the book nor
the website).
I believe that DC and LS are equivalent over ZF, according to the following
argument:
That DC implies LS is standard. Perhaps the shortest proof adapts Henkin's
construction in the following way: Let T be a first order theory with
countable signature with a model M. Use Dependent Choice to extend M to a
model M' of the Henkinized theory T' (this is necessary because the
interpretation of the constants added in each level depends on that of the
constants of previous levels). Finally, an elementary submodel N which is
at most countable can be constructed by taking as its underlying universe X
the interpretation in M' of all constants of T' and restricting to X the
interpretation in M of the symbols of T (the restriction of functions is
well defined because T' is a Henkin theory). Note that this amounts to
repeat Henkin's proof of the completeness theorem except that we use Th(M')
(the theory of the model M') as the maximal consistent extension of T'.
That LS implies DC could be seen in this way: Let S be a set with a binary
relation R such that for every x in S, the set of y's in S such that xRy is
nonempty. Consider the theory T over the language {R} that contains a
binary relation symbol, and whose only non logical axiom is \$\forall x
\exists y R(x, y)\$. Then S is a model of T with the obvious interpretation
of R. By LS, it has a submodel N whose underlying set is in bijection with
either some finite ordinal or the natural numbers. Hence, a sequence x_n
such that x_nRx_{n+1} can be defined by recursion taking at each step the
minimum element (according to the bijection) of the nonempty set of
elements of N which are related to x_n.
My question: is this equivalence known (if it's correct?). If so, where can
I find a reference?
Many thanks,
Christian
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https://www.coursehero.com/file/5994257/852HW1/ | 1,498,203,313,000,000,000 | text/html | crawl-data/CC-MAIN-2017-26/segments/1498128320023.23/warc/CC-MAIN-20170623063716-20170623083716-00661.warc.gz | 853,710,482 | 46,478 | # 852HW1 - and S y for spin-1/2 e Show explicitly that S 2 =...
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PHYS852 Quantum Mechanics II, Spring 2009 HOMEWORK ASSIGNMENT 1 1. [25 points] This problem shows you how to derive the matrix representations of spin operators from Frst principles. a.) ±or spin 1/2, use the eigenvalue equation S z | m s a = p m s | m s a to Fnd the components of the two-by-two matrix representation of S z in the basis of its own eigenstates. b.) ±rom the deFnition S ± = S x + iS y , write S x and S y in terms of S + and S - . c.) The relation J ± | j, m j a = p r j ( j + 1) m j ( m j ± 1) | j, m j ± 1 a is valid for any angular momentum operators. Apply this to the spin-1/2 case to Fnd the matrix elements of S + and S - in the basis of eigenstates of S z . d.) ±rom your previous answers, derive the matrix representations of S x
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Unformatted text preview: and S y for spin-1/2. e.) Show explicitly that S 2 = v S · v S = p 2 s ( s + 1). 2. [15 points] Consider an electron whose position is held Fxed, so that it can be described by a simple two-component spinor. The initial state of the particle is spin-up ( m s = 1 / 2) with respect to the z-axis. At time t = 0 a uniform magnetic Feld is applied along the y-axis. What is the state-vector of this system at any arbitrary time t > 0. 3. [15 points] Cohen-Tannoudji problem 9.1, page 990 4. [15 points] Cohen-Tonnoudji problem 9.2, page 990 5. [20 points] Cohen-Tannoudji problem 9.3, page 991 1...
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## This note was uploaded on 10/25/2010 for the course PHYSICS PHYS 851 taught by Professor Michaelmoore during the Fall '08 term at Michigan State University.
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# In the figure given below, PQ and RS are two mirrors placed parallel to each other. An incident ray AB strikes the mirror PQ at B, the reflected ray moves along the path BC and strikes the mirror RS at C and again reflects back along CD. Prove that AB ll CD.
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Hint: Here, in this question, we can see that PQ and RS are the two mirrors that are placed parallel to each other. So, these two mirrors will act like the two parallel lines. And, as the mirrors PQ and RS are parallel to each other, therefore, the ray BC is acting like a transversal. So, we shall solve this question according to the result that we get when we perform any operation with the measures of the angles.
So, let us now know about the different angles formed when a transversal cuts two parallel lines. Here are some of them:
1. Corresponding Angles
2. Vertically Opposite Angles
3. Linear Pair
4. Alternate Interior Angles
5. Alternate Exterior Angles
We will then construct two normals to PQ and RS at B and C respectively. As the normals will be incident on two parallel lines, they will also be parallel to each other. We will then use the properties of the different angles formed when two parallel lines are cut by a transversal and try to prove the given condition.
As mentioned in the question, PQ||RS. Thus, BC acts as a transversal to two parallel lines.
Now, let us construct a ‘m’ normal to PQ at B and a normal ‘n’ to RS at C.
We know that PQ||RS. Since the normals we drew are drawn to two different parallel lines, the normals will also be parallel to each other. Thus, m||n.
Now, to make it convenient, we will mark the angles at x,y,${{i}_{1}},{{i}_{2}},{{r}_{1}}$ and ${{r}_{2}}$ as shown in the figure.
Thus, our figure will look as follows:
Now, if we consider the lines PQ and RS, we can see that BC is the transversal.
Here, we can see that $\angle x$ and $\angle y$ are alternate interior angles. We also know that alternate interior angles are always equal.
Thus, $\angle x=\angle y$
Now, ‘m’ and ‘n’ are normals at PQ and RS respectively. Now, if we look at the normal ‘m’, we can see that $\angle x$ and $\angle {{r}_{1}}$ sum to ${{90}^{\circ }}$.
Thus, $\angle x+\angle {{r}_{1}}={{90}^{\circ }}$ …..(i)
Similarly, if we look at the normal ‘n’, we can see that:
$\angle y+\angle {{i}_{2}}={{90}^{\circ }}$ …..(ii)
From the equations (i) and (ii), we can see that:
\begin{align} & \angle x+\angle {{r}_{1}}={{90}^{\circ }} \\ & \Rightarrow \angle x={{90}^{\circ }}-\angle {{r}_{1}} \\ & \angle y+\angle {{i}_{2}}={{90}^{\circ }} \\ & \Rightarrow \angle y={{90}^{\circ }}-\angle {{i}_{2}} \\ \end{align}
Now, we have already established that $\angle x=\angle y$,
Thus, by putting in the values of $\angle x$ and $\angle y$, we get:
\begin{align} & \angle x=\angle y \\ & \Rightarrow {{90}^{\circ }}-\angle {{r}_{1}}={{90}^{\circ }}-\angle {{i}_{2}} \\ & \Rightarrow \angle {{r}_{1}}=\angle {{i}_{2}} \\ \end{align}
Also, we know that the angle of incidence is equal to the angle of reflection.
Hence,
\begin{align} & \angle {{i}_{1}}=\angle {{r}_{1}} \\ & \angle {{i}_{2}}=\angle {{r}_{2}} \\ \end{align}
Since, $\angle {{r}_{1}}=\angle {{i}_{2}}$, we can say that:
$\angle {{i}_{1}}=\angle {{r}_{1}}=\angle {{i}_{2}}=\angle {{r}_{2}}$
Since, these 4 angles are equal, the sum of any two of them will be equal to the sum of the remaining two angles.
And therefore,
$\angle {{i}_{1}}+\angle {{r}_{1}}=\angle {{i}_{2}}+\angle {{r}_{2}}$
Now, if we look at lines AB and CD, we can see that BC is a line cutting them and the sum of $\angle {{i}_{1}}$ and $\angle {{r}_{1}}$ and the sum of $\angle {{i}_{2}}$ and $\angle {{r}_{2}}$ are alternate interior angles.
Hence, AB ll CD as the alternate interior angles are equal ($\angle {{i}_{1}}+\angle {{r}_{1}}=\angle {{i}_{2}}+\angle {{r}_{2}}$).
Hence, proved.
Note: Let us now know about the different angles formed when a transversal cuts two parallel lines.
LINEAR PAIR: They are two adjacent angles which sum up to ${{180}^{\circ }}$. Here, linear pair is made by the following angles:
i. $\angle 1\text{ and }\angle 2$
ii. $\angle 4\text{ and }\angle 3$
iii. $\angle 5\text{ and }\angle 6$
iv. $\angle 7\text{ and }\angle 8$
v. $\angle 1\text{ and }\angle \text{4}$
vi. $\angle 2\text{ and }\angle 3$
vii. $\angle 5\text{ and }\angle 8$
viii. $\angle 6\text{ and }\angle 7$
VERTICALLY OPPOSITE ANGLES: they are the pair of opposite angles made when two lines intersect each other. They’re always equal. Here, pair of vertically opposite angles is formed by:
i. $\angle 1\text{ and }\angle 3$
ii. $\angle 2\text{ and }\angle 4$
iii. $\angle 5\text{ and }\angle 7$
iv. $\angle 6\text{ and }\angle 8$
CORRESPONDING ANGLES: angles in the corresponding positions on the two parallel lines with respect to the transversal are always equal. Here, pairs of corresponding angles are:
i. $\angle 1\text{ and }\angle 5$
ii. $\angle 2\text{ and }\angle 6$
iii. $\angle 4\text{ and }\angle 8$
iv. $\angle 3\text{ and }\angle 7$
ALTERNATE INTERIOR ANGLES: angles in between the parallel lines and opposite to one another, i.e. the alternate interior angles are always equal. Here, pairs of alternate interior angles are:
i. $\angle 4\text{ and }\angle 6$
ii. $\angle 5\text{ and }\angle 3$
ALTERNATE EXTERIOR ANGLES: angles outside the parallel lines and opposite to one another, i.e. the alternate exterior angles are always equal. Here, the pairs of alternate exterior angles are:
i. $\angle 1\text{ and }\angle 7$
ii. $\angle 8\text{ and }\angle 2$
CO – INTERIOR ANGLES: angles in the interior of the parallel lines on the same side of the transversal always sum up to ${{180}^{\circ }}$. Here, the pairs of co-interior angles are:
i. $\angle 4\text{ and }\angle 5$
ii. $\angle 3\text{ and }\angle 6$
CO – EXTERIOR ANGLES: angles on the exterior of the parallel lines on the same side of the transversal always up to ${{180}^{\circ }}$. Here, the pairs of co-exterior angles are:
i. $\angle 1\text{ and }\angle 8$
ii. $\angle 2\text{ and }\angle 7$
It is very important to solve this question very carefully keeping all these different types of pairs of angles in mind as if there is any mistake; the answer can come out to be wrong. | 1,875 | 6,344 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 6, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.6875 | 5 | CC-MAIN-2024-26 | latest | en | 0.933034 |
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TypeLabelDescription
Statement
Theoremfta1lem 24901* Lemma for fta1 24902. (Contributed by Mario Carneiro, 26-Jul-2014.)
𝑅 = (𝐹 “ {0}) & (𝜑𝐷 ∈ ℕ0) & (𝜑𝐹 ∈ ((Poly‘ℂ) ∖ {0𝑝})) & (𝜑 → (deg‘𝐹) = (𝐷 + 1)) & (𝜑𝐴 ∈ (𝐹 “ {0})) & (𝜑 → ∀𝑔 ∈ ((Poly‘ℂ) ∖ {0𝑝})((deg‘𝑔) = 𝐷 → ((𝑔 “ {0}) ∈ Fin ∧ (♯‘(𝑔 “ {0})) ≤ (deg‘𝑔)))) (𝜑 → (𝑅 ∈ Fin ∧ (♯‘𝑅) ≤ (deg‘𝐹)))
Theoremfta1 24902 The easy direction of the Fundamental Theorem of Algebra: A nonzero polynomial has at most deg(𝐹) roots. (Contributed by Mario Carneiro, 26-Jul-2014.)
𝑅 = (𝐹 “ {0}) ((𝐹 ∈ (Poly‘𝑆) ∧ 𝐹 ≠ 0𝑝) → (𝑅 ∈ Fin ∧ (♯‘𝑅) ≤ (deg‘𝐹)))
Theoremquotcan 24903 Exact division with a multiple. (Contributed by Mario Carneiro, 26-Jul-2014.)
𝐻 = (𝐹f · 𝐺) ((𝐹 ∈ (Poly‘𝑆) ∧ 𝐺 ∈ (Poly‘𝑆) ∧ 𝐺 ≠ 0𝑝) → (𝐻 quot 𝐺) = 𝐹)
Theoremvieta1lem1 24904* Lemma for vieta1 24906. (Contributed by Mario Carneiro, 28-Jul-2014.)
𝐴 = (coeff‘𝐹) & 𝑁 = (deg‘𝐹) & 𝑅 = (𝐹 “ {0}) & (𝜑𝐹 ∈ (Poly‘𝑆)) & (𝜑 → (♯‘𝑅) = 𝑁) & (𝜑𝐷 ∈ ℕ) & (𝜑 → (𝐷 + 1) = 𝑁) & (𝜑 → ∀𝑓 ∈ (Poly‘ℂ)((𝐷 = (deg‘𝑓) ∧ (♯‘(𝑓 “ {0})) = (deg‘𝑓)) → Σ𝑥 ∈ (𝑓 “ {0})𝑥 = -(((coeff‘𝑓)‘((deg‘𝑓) − 1)) / ((coeff‘𝑓)‘(deg‘𝑓))))) & 𝑄 = (𝐹 quot (Xpf − (ℂ × {𝑧}))) ((𝜑𝑧𝑅) → (𝑄 ∈ (Poly‘ℂ) ∧ 𝐷 = (deg‘𝑄)))
Theoremvieta1lem2 24905* Lemma for vieta1 24906: inductive step. Let 𝑧 be a root of 𝐹. Then 𝐹 = (Xp𝑧) · 𝑄 for some 𝑄 by the factor theorem, and 𝑄 is a degree- 𝐷 polynomial, so by the induction hypothesis Σ𝑥 ∈ (𝑄 “ 0)𝑥 = -(coeff‘𝑄)‘(𝐷 − 1) / (coeff‘𝑄)‘𝐷, so Σ𝑥𝑅𝑥 = 𝑧 − (coeff‘𝑄)‘ (𝐷 − 1) / (coeff‘𝑄)‘𝐷. Now the coefficients of 𝐹 are 𝐴‘(𝐷 + 1) = (coeff‘𝑄)‘𝐷 and 𝐴𝐷 = Σ𝑘 ∈ (0...𝐷)(coeff‘Xp𝑧)‘𝑘 · (coeff‘𝑄) ‘(𝐷𝑘), which works out to -𝑧 · (coeff‘𝑄)‘𝐷 + (coeff‘𝑄)‘(𝐷 − 1), so putting it all together we have Σ𝑥𝑅𝑥 = -𝐴𝐷 / 𝐴‘(𝐷 + 1) as we wanted to show. (Contributed by Mario Carneiro, 28-Jul-2014.)
𝐴 = (coeff‘𝐹) & 𝑁 = (deg‘𝐹) & 𝑅 = (𝐹 “ {0}) & (𝜑𝐹 ∈ (Poly‘𝑆)) & (𝜑 → (♯‘𝑅) = 𝑁) & (𝜑𝐷 ∈ ℕ) & (𝜑 → (𝐷 + 1) = 𝑁) & (𝜑 → ∀𝑓 ∈ (Poly‘ℂ)((𝐷 = (deg‘𝑓) ∧ (♯‘(𝑓 “ {0})) = (deg‘𝑓)) → Σ𝑥 ∈ (𝑓 “ {0})𝑥 = -(((coeff‘𝑓)‘((deg‘𝑓) − 1)) / ((coeff‘𝑓)‘(deg‘𝑓))))) & 𝑄 = (𝐹 quot (Xpf − (ℂ × {𝑧}))) (𝜑 → Σ𝑥𝑅 𝑥 = -((𝐴‘(𝑁 − 1)) / (𝐴𝑁)))
Theoremvieta1 24906* The first-order Vieta's formula (see http://en.wikipedia.org/wiki/Vieta%27s_formulas). If a polynomial of degree 𝑁 has 𝑁 distinct roots, then the sum over these roots can be calculated as -𝐴(𝑁 − 1) / 𝐴(𝑁). (If the roots are not distinct, then this formula is still true but must double-count some of the roots according to their multiplicities.) (Contributed by Mario Carneiro, 28-Jul-2014.)
𝐴 = (coeff‘𝐹) & 𝑁 = (deg‘𝐹) & 𝑅 = (𝐹 “ {0}) & (𝜑𝐹 ∈ (Poly‘𝑆)) & (𝜑 → (♯‘𝑅) = 𝑁) & (𝜑𝑁 ∈ ℕ) (𝜑 → Σ𝑥𝑅 𝑥 = -((𝐴‘(𝑁 − 1)) / (𝐴𝑁)))
Theoremplyexmo 24907* An infinite set of values can be extended to a polynomial in at most one way. (Contributed by Stefan O'Rear, 14-Nov-2014.)
((𝐷 ⊆ ℂ ∧ ¬ 𝐷 ∈ Fin) → ∃*𝑝(𝑝 ∈ (Poly‘𝑆) ∧ (𝑝𝐷) = 𝐹))
14.1.5 Algebraic numbers
Syntaxcaa 24908 Extend class notation to include the set of algebraic numbers.
class 𝔸
Definitiondf-aa 24909 Define the set of algebraic numbers. An algebraic number is a root of a nonzero polynomial over the integers. Here we construct it as the union of all kernels (preimages of {0}) of all polynomials in (Poly‘ℤ), except the zero polynomial 0𝑝. (Contributed by Mario Carneiro, 22-Jul-2014.)
𝔸 = 𝑓 ∈ ((Poly‘ℤ) ∖ {0𝑝})(𝑓 “ {0})
Theoremelaa 24910* Elementhood in the set of algebraic numbers. (Contributed by Mario Carneiro, 22-Jul-2014.)
(𝐴 ∈ 𝔸 ↔ (𝐴 ∈ ℂ ∧ ∃𝑓 ∈ ((Poly‘ℤ) ∖ {0𝑝})(𝑓𝐴) = 0))
Theoremaacn 24911 An algebraic number is a complex number. (Contributed by Mario Carneiro, 23-Jul-2014.)
(𝐴 ∈ 𝔸 → 𝐴 ∈ ℂ)
Theoremaasscn 24912 The algebraic numbers are a subset of the complex numbers. (Contributed by Mario Carneiro, 23-Jul-2014.)
𝔸 ⊆ ℂ
Theoremelqaalem1 24913* Lemma for elqaa 24916. The function 𝑁 represents the denominators of the rational coefficients 𝐵. By multiplying them all together to make 𝑅, we get a number big enough to clear all the denominators and make 𝑅 · 𝐹 an integer polynomial. (Contributed by Mario Carneiro, 23-Jul-2014.) (Revised by AV, 3-Oct-2020.)
(𝜑𝐴 ∈ ℂ) & (𝜑𝐹 ∈ ((Poly‘ℚ) ∖ {0𝑝})) & (𝜑 → (𝐹𝐴) = 0) & 𝐵 = (coeff‘𝐹) & 𝑁 = (𝑘 ∈ ℕ0 ↦ inf({𝑛 ∈ ℕ ∣ ((𝐵𝑘) · 𝑛) ∈ ℤ}, ℝ, < )) & 𝑅 = (seq0( · , 𝑁)‘(deg‘𝐹)) ((𝜑𝐾 ∈ ℕ0) → ((𝑁𝐾) ∈ ℕ ∧ ((𝐵𝐾) · (𝑁𝐾)) ∈ ℤ))
Theoremelqaalem2 24914* Lemma for elqaa 24916. (Contributed by Mario Carneiro, 23-Jul-2014.) (Revised by AV, 3-Oct-2020.)
(𝜑𝐴 ∈ ℂ) & (𝜑𝐹 ∈ ((Poly‘ℚ) ∖ {0𝑝})) & (𝜑 → (𝐹𝐴) = 0) & 𝐵 = (coeff‘𝐹) & 𝑁 = (𝑘 ∈ ℕ0 ↦ inf({𝑛 ∈ ℕ ∣ ((𝐵𝑘) · 𝑛) ∈ ℤ}, ℝ, < )) & 𝑅 = (seq0( · , 𝑁)‘(deg‘𝐹)) & 𝑃 = (𝑥 ∈ V, 𝑦 ∈ V ↦ ((𝑥 · 𝑦) mod (𝑁𝐾))) ((𝜑𝐾 ∈ (0...(deg‘𝐹))) → (𝑅 mod (𝑁𝐾)) = 0)
Theoremelqaalem3 24915* Lemma for elqaa 24916. (Contributed by Mario Carneiro, 23-Jul-2014.) (Revised by AV, 3-Oct-2020.)
(𝜑𝐴 ∈ ℂ) & (𝜑𝐹 ∈ ((Poly‘ℚ) ∖ {0𝑝})) & (𝜑 → (𝐹𝐴) = 0) & 𝐵 = (coeff‘𝐹) & 𝑁 = (𝑘 ∈ ℕ0 ↦ inf({𝑛 ∈ ℕ ∣ ((𝐵𝑘) · 𝑛) ∈ ℤ}, ℝ, < )) & 𝑅 = (seq0( · , 𝑁)‘(deg‘𝐹)) (𝜑𝐴 ∈ 𝔸)
Theoremelqaa 24916* The set of numbers generated by the roots of polynomials in the rational numbers is the same as the set of algebraic numbers, which by elaa 24910 are defined only in terms of polynomials over the integers. (Contributed by Mario Carneiro, 23-Jul-2014.) (Proof shortened by AV, 3-Oct-2020.)
(𝐴 ∈ 𝔸 ↔ (𝐴 ∈ ℂ ∧ ∃𝑓 ∈ ((Poly‘ℚ) ∖ {0𝑝})(𝑓𝐴) = 0))
Theoremqaa 24917 Every rational number is algebraic. (Contributed by Mario Carneiro, 23-Jul-2014.)
(𝐴 ∈ ℚ → 𝐴 ∈ 𝔸)
Theoremqssaa 24918 The rational numbers are contained in the algebraic numbers. (Contributed by Mario Carneiro, 23-Jul-2014.)
ℚ ⊆ 𝔸
Theoremiaa 24919 The imaginary unit is algebraic. (Contributed by Mario Carneiro, 23-Jul-2014.)
i ∈ 𝔸
Theoremaareccl 24920 The reciprocal of an algebraic number is algebraic. (Contributed by Mario Carneiro, 24-Jul-2014.)
((𝐴 ∈ 𝔸 ∧ 𝐴 ≠ 0) → (1 / 𝐴) ∈ 𝔸)
Theoremaacjcl 24921 The conjugate of an algebraic number is algebraic. (Contributed by Mario Carneiro, 24-Jul-2014.)
(𝐴 ∈ 𝔸 → (∗‘𝐴) ∈ 𝔸)
Theoremaannenlem1 24922* Lemma for aannen 24925. (Contributed by Stefan O'Rear, 16-Nov-2014.)
𝐻 = (𝑎 ∈ ℕ0 ↦ {𝑏 ∈ ℂ ∣ ∃𝑐 ∈ {𝑑 ∈ (Poly‘ℤ) ∣ (𝑑 ≠ 0𝑝 ∧ (deg‘𝑑) ≤ 𝑎 ∧ ∀𝑒 ∈ ℕ0 (abs‘((coeff‘𝑑)‘𝑒)) ≤ 𝑎)} (𝑐𝑏) = 0}) (𝐴 ∈ ℕ0 → (𝐻𝐴) ∈ Fin)
Theoremaannenlem2 24923* Lemma for aannen 24925. (Contributed by Stefan O'Rear, 16-Nov-2014.)
𝐻 = (𝑎 ∈ ℕ0 ↦ {𝑏 ∈ ℂ ∣ ∃𝑐 ∈ {𝑑 ∈ (Poly‘ℤ) ∣ (𝑑 ≠ 0𝑝 ∧ (deg‘𝑑) ≤ 𝑎 ∧ ∀𝑒 ∈ ℕ0 (abs‘((coeff‘𝑑)‘𝑒)) ≤ 𝑎)} (𝑐𝑏) = 0}) 𝔸 = ran 𝐻
Theoremaannenlem3 24924* The algebraic numbers are countable. (Contributed by Stefan O'Rear, 16-Nov-2014.)
𝐻 = (𝑎 ∈ ℕ0 ↦ {𝑏 ∈ ℂ ∣ ∃𝑐 ∈ {𝑑 ∈ (Poly‘ℤ) ∣ (𝑑 ≠ 0𝑝 ∧ (deg‘𝑑) ≤ 𝑎 ∧ ∀𝑒 ∈ ℕ0 (abs‘((coeff‘𝑑)‘𝑒)) ≤ 𝑎)} (𝑐𝑏) = 0}) 𝔸 ≈ ℕ
Theoremaannen 24925 The algebraic numbers are countable. (Contributed by Stefan O'Rear, 16-Nov-2014.)
𝔸 ≈ ℕ
14.1.6 Liouville's approximation theorem
Theoremaalioulem1 24926 Lemma for aaliou 24932. An integer polynomial cannot inflate the denominator of a rational by more than its degree. (Contributed by Stefan O'Rear, 12-Nov-2014.)
(𝜑𝐹 ∈ (Poly‘ℤ)) & (𝜑𝑋 ∈ ℤ) & (𝜑𝑌 ∈ ℕ) (𝜑 → ((𝐹‘(𝑋 / 𝑌)) · (𝑌↑(deg‘𝐹))) ∈ ℤ)
Theoremaalioulem2 24927* Lemma for aaliou 24932. (Contributed by Stefan O'Rear, 15-Nov-2014.) (Proof shortened by AV, 28-Sep-2020.)
𝑁 = (deg‘𝐹) & (𝜑𝐹 ∈ (Poly‘ℤ)) & (𝜑𝑁 ∈ ℕ) & (𝜑𝐴 ∈ ℝ) (𝜑 → ∃𝑥 ∈ ℝ+𝑝 ∈ ℤ ∀𝑞 ∈ ℕ ((𝐹‘(𝑝 / 𝑞)) = 0 → (𝐴 = (𝑝 / 𝑞) ∨ (𝑥 / (𝑞𝑁)) ≤ (abs‘(𝐴 − (𝑝 / 𝑞))))))
Theoremaalioulem3 24928* Lemma for aaliou 24932. (Contributed by Stefan O'Rear, 15-Nov-2014.)
𝑁 = (deg‘𝐹) & (𝜑𝐹 ∈ (Poly‘ℤ)) & (𝜑𝑁 ∈ ℕ) & (𝜑𝐴 ∈ ℝ) & (𝜑 → (𝐹𝐴) = 0) (𝜑 → ∃𝑥 ∈ ℝ+𝑟 ∈ ℝ ((abs‘(𝐴𝑟)) ≤ 1 → (𝑥 · (abs‘(𝐹𝑟))) ≤ (abs‘(𝐴𝑟))))
Theoremaalioulem4 24929* Lemma for aaliou 24932. (Contributed by Stefan O'Rear, 16-Nov-2014.)
𝑁 = (deg‘𝐹) & (𝜑𝐹 ∈ (Poly‘ℤ)) & (𝜑𝑁 ∈ ℕ) & (𝜑𝐴 ∈ ℝ) & (𝜑 → (𝐹𝐴) = 0) (𝜑 → ∃𝑥 ∈ ℝ+𝑝 ∈ ℤ ∀𝑞 ∈ ℕ (((𝐹‘(𝑝 / 𝑞)) ≠ 0 ∧ (abs‘(𝐴 − (𝑝 / 𝑞))) ≤ 1) → (𝐴 = (𝑝 / 𝑞) ∨ (𝑥 / (𝑞𝑁)) ≤ (abs‘(𝐴 − (𝑝 / 𝑞))))))
Theoremaalioulem5 24930* Lemma for aaliou 24932. (Contributed by Stefan O'Rear, 16-Nov-2014.)
𝑁 = (deg‘𝐹) & (𝜑𝐹 ∈ (Poly‘ℤ)) & (𝜑𝑁 ∈ ℕ) & (𝜑𝐴 ∈ ℝ) & (𝜑 → (𝐹𝐴) = 0) (𝜑 → ∃𝑥 ∈ ℝ+𝑝 ∈ ℤ ∀𝑞 ∈ ℕ ((𝐹‘(𝑝 / 𝑞)) ≠ 0 → (𝐴 = (𝑝 / 𝑞) ∨ (𝑥 / (𝑞𝑁)) ≤ (abs‘(𝐴 − (𝑝 / 𝑞))))))
Theoremaalioulem6 24931* Lemma for aaliou 24932. (Contributed by Stefan O'Rear, 16-Nov-2014.)
𝑁 = (deg‘𝐹) & (𝜑𝐹 ∈ (Poly‘ℤ)) & (𝜑𝑁 ∈ ℕ) & (𝜑𝐴 ∈ ℝ) & (𝜑 → (𝐹𝐴) = 0) (𝜑 → ∃𝑥 ∈ ℝ+𝑝 ∈ ℤ ∀𝑞 ∈ ℕ (𝐴 = (𝑝 / 𝑞) ∨ (𝑥 / (𝑞𝑁)) ≤ (abs‘(𝐴 − (𝑝 / 𝑞)))))
Theoremaaliou 24932* Liouville's theorem on diophantine approximation: Any algebraic number, being a root of a polynomial 𝐹 in integer coefficients, is not approximable beyond order 𝑁 = deg(𝐹) by rational numbers. In this form, it also applies to rational numbers themselves, which are not well approximable by other rational numbers. This is Metamath 100 proof #18. (Contributed by Stefan O'Rear, 16-Nov-2014.)
𝑁 = (deg‘𝐹) & (𝜑𝐹 ∈ (Poly‘ℤ)) & (𝜑𝑁 ∈ ℕ) & (𝜑𝐴 ∈ ℝ) & (𝜑 → (𝐹𝐴) = 0) (𝜑 → ∃𝑥 ∈ ℝ+𝑝 ∈ ℤ ∀𝑞 ∈ ℕ (𝐴 = (𝑝 / 𝑞) ∨ (𝑥 / (𝑞𝑁)) < (abs‘(𝐴 − (𝑝 / 𝑞)))))
Theoremgeolim3 24933* Geometric series convergence with arbitrary shift, radix, and multiplicative constant. (Contributed by Stefan O'Rear, 16-Nov-2014.)
(𝜑𝐴 ∈ ℤ) & (𝜑𝐵 ∈ ℂ) & (𝜑 → (abs‘𝐵) < 1) & (𝜑𝐶 ∈ ℂ) & 𝐹 = (𝑘 ∈ (ℤ𝐴) ↦ (𝐶 · (𝐵↑(𝑘𝐴)))) (𝜑 → seq𝐴( + , 𝐹) ⇝ (𝐶 / (1 − 𝐵)))
Theoremaaliou2 24934* Liouville's approximation theorem for algebraic numbers per se. (Contributed by Stefan O'Rear, 16-Nov-2014.)
(𝐴 ∈ (𝔸 ∩ ℝ) → ∃𝑘 ∈ ℕ ∃𝑥 ∈ ℝ+𝑝 ∈ ℤ ∀𝑞 ∈ ℕ (𝐴 = (𝑝 / 𝑞) ∨ (𝑥 / (𝑞𝑘)) < (abs‘(𝐴 − (𝑝 / 𝑞)))))
Theoremaaliou2b 24935* Liouville's approximation theorem extended to complex 𝐴. (Contributed by Stefan O'Rear, 20-Nov-2014.)
(𝐴 ∈ 𝔸 → ∃𝑘 ∈ ℕ ∃𝑥 ∈ ℝ+𝑝 ∈ ℤ ∀𝑞 ∈ ℕ (𝐴 = (𝑝 / 𝑞) ∨ (𝑥 / (𝑞𝑘)) < (abs‘(𝐴 − (𝑝 / 𝑞)))))
Theoremaaliou3lem1 24936* Lemma for aaliou3 24945. (Contributed by Stefan O'Rear, 16-Nov-2014.)
𝐺 = (𝑐 ∈ (ℤ𝐴) ↦ ((2↑-(!‘𝐴)) · ((1 / 2)↑(𝑐𝐴)))) ((𝐴 ∈ ℕ ∧ 𝐵 ∈ (ℤ𝐴)) → (𝐺𝐵) ∈ ℝ)
Theoremaaliou3lem2 24937* Lemma for aaliou3 24945. (Contributed by Stefan O'Rear, 16-Nov-2014.)
𝐺 = (𝑐 ∈ (ℤ𝐴) ↦ ((2↑-(!‘𝐴)) · ((1 / 2)↑(𝑐𝐴)))) & 𝐹 = (𝑎 ∈ ℕ ↦ (2↑-(!‘𝑎))) ((𝐴 ∈ ℕ ∧ 𝐵 ∈ (ℤ𝐴)) → (𝐹𝐵) ∈ (0(,](𝐺𝐵)))
Theoremaaliou3lem3 24938* Lemma for aaliou3 24945. (Contributed by Stefan O'Rear, 16-Nov-2014.)
𝐺 = (𝑐 ∈ (ℤ𝐴) ↦ ((2↑-(!‘𝐴)) · ((1 / 2)↑(𝑐𝐴)))) & 𝐹 = (𝑎 ∈ ℕ ↦ (2↑-(!‘𝑎))) (𝐴 ∈ ℕ → (seq𝐴( + , 𝐹) ∈ dom ⇝ ∧ Σ𝑏 ∈ (ℤ𝐴)(𝐹𝑏) ∈ ℝ+ ∧ Σ𝑏 ∈ (ℤ𝐴)(𝐹𝑏) ≤ (2 · (2↑-(!‘𝐴)))))
Theoremaaliou3lem8 24939* Lemma for aaliou3 24945. (Contributed by Stefan O'Rear, 20-Nov-2014.)
((𝐴 ∈ ℕ ∧ 𝐵 ∈ ℝ+) → ∃𝑥 ∈ ℕ (2 · (2↑-(!‘(𝑥 + 1)))) ≤ (𝐵 / ((2↑(!‘𝑥))↑𝐴)))
Theoremaaliou3lem4 24940* Lemma for aaliou3 24945. (Contributed by Stefan O'Rear, 16-Nov-2014.)
𝐹 = (𝑎 ∈ ℕ ↦ (2↑-(!‘𝑎))) & 𝐿 = Σ𝑏 ∈ ℕ (𝐹𝑏) & 𝐻 = (𝑐 ∈ ℕ ↦ Σ𝑏 ∈ (1...𝑐)(𝐹𝑏)) 𝐿 ∈ ℝ
Theoremaaliou3lem5 24941* Lemma for aaliou3 24945. (Contributed by Stefan O'Rear, 16-Nov-2014.)
𝐹 = (𝑎 ∈ ℕ ↦ (2↑-(!‘𝑎))) & 𝐿 = Σ𝑏 ∈ ℕ (𝐹𝑏) & 𝐻 = (𝑐 ∈ ℕ ↦ Σ𝑏 ∈ (1...𝑐)(𝐹𝑏)) (𝐴 ∈ ℕ → (𝐻𝐴) ∈ ℝ)
Theoremaaliou3lem6 24942* Lemma for aaliou3 24945. (Contributed by Stefan O'Rear, 16-Nov-2014.)
𝐹 = (𝑎 ∈ ℕ ↦ (2↑-(!‘𝑎))) & 𝐿 = Σ𝑏 ∈ ℕ (𝐹𝑏) & 𝐻 = (𝑐 ∈ ℕ ↦ Σ𝑏 ∈ (1...𝑐)(𝐹𝑏)) (𝐴 ∈ ℕ → ((𝐻𝐴) · (2↑(!‘𝐴))) ∈ ℤ)
Theoremaaliou3lem7 24943* Lemma for aaliou3 24945. (Contributed by Stefan O'Rear, 16-Nov-2014.)
𝐹 = (𝑎 ∈ ℕ ↦ (2↑-(!‘𝑎))) & 𝐿 = Σ𝑏 ∈ ℕ (𝐹𝑏) & 𝐻 = (𝑐 ∈ ℕ ↦ Σ𝑏 ∈ (1...𝑐)(𝐹𝑏)) (𝐴 ∈ ℕ → ((𝐻𝐴) ≠ 𝐿 ∧ (abs‘(𝐿 − (𝐻𝐴))) ≤ (2 · (2↑-(!‘(𝐴 + 1))))))
Theoremaaliou3lem9 24944* Example of a "Liouville number", a very simple definable transcendental real. (Contributed by Stefan O'Rear, 20-Nov-2014.)
𝐹 = (𝑎 ∈ ℕ ↦ (2↑-(!‘𝑎))) & 𝐿 = Σ𝑏 ∈ ℕ (𝐹𝑏) & 𝐻 = (𝑐 ∈ ℕ ↦ Σ𝑏 ∈ (1...𝑐)(𝐹𝑏)) ¬ 𝐿 ∈ 𝔸
Theoremaaliou3 24945 Example of a "Liouville number", a very simple definable transcendental real. (Contributed by Stefan O'Rear, 23-Nov-2014.)
Σ𝑘 ∈ ℕ (2↑-(!‘𝑘)) ∉ 𝔸
14.2 Sequences and series
14.2.1 Taylor polynomials and Taylor's theorem
Syntaxctayl 24946 Taylor polynomial of a function.
class Tayl
Syntaxcana 24947 The class of analytic functions.
class Ana
Definitiondf-tayl 24948* Define the Taylor polynomial or Taylor series of a function. TODO-AV: 𝑛 ∈ (ℕ0 ∪ {+∞}) should be replaced by 𝑛 ∈ ℕ0*. (Contributed by Mario Carneiro, 30-Dec-2016.)
Tayl = (𝑠 ∈ {ℝ, ℂ}, 𝑓 ∈ (ℂ ↑pm 𝑠) ↦ (𝑛 ∈ (ℕ0 ∪ {+∞}), 𝑎 𝑘 ∈ ((0[,]𝑛) ∩ ℤ)dom ((𝑠 D𝑛 𝑓)‘𝑘) ↦ 𝑥 ∈ ℂ ({𝑥} × (ℂfld tsums (𝑘 ∈ ((0[,]𝑛) ∩ ℤ) ↦ (((((𝑠 D𝑛 𝑓)‘𝑘)‘𝑎) / (!‘𝑘)) · ((𝑥𝑎)↑𝑘)))))))
Definitiondf-ana 24949* Define the set of analytic functions, which are functions such that the Taylor series of the function at each point converges to the function in some neighborhood of the point. (Contributed by Mario Carneiro, 31-Dec-2016.)
Ana = (𝑠 ∈ {ℝ, ℂ} ↦ {𝑓 ∈ (ℂ ↑pm 𝑠) ∣ ∀𝑥 ∈ dom 𝑓 𝑥 ∈ ((int‘((TopOpen‘ℂfld) ↾t 𝑠))‘dom (𝑓 ∩ (+∞(𝑠 Tayl 𝑓)𝑥)))})
Theoremtaylfvallem1 24950* Lemma for taylfval 24952. (Contributed by Mario Carneiro, 30-Dec-2016.)
(𝜑𝑆 ∈ {ℝ, ℂ}) & (𝜑𝐹:𝐴⟶ℂ) & (𝜑𝐴𝑆) & (𝜑 → (𝑁 ∈ ℕ0𝑁 = +∞)) & ((𝜑𝑘 ∈ ((0[,]𝑁) ∩ ℤ)) → 𝐵 ∈ dom ((𝑆 D𝑛 𝐹)‘𝑘)) (((𝜑𝑋 ∈ ℂ) ∧ 𝑘 ∈ ((0[,]𝑁) ∩ ℤ)) → (((((𝑆 D𝑛 𝐹)‘𝑘)‘𝐵) / (!‘𝑘)) · ((𝑋𝐵)↑𝑘)) ∈ ℂ)
Theoremtaylfvallem 24951* Lemma for taylfval 24952. (Contributed by Mario Carneiro, 30-Dec-2016.)
(𝜑𝑆 ∈ {ℝ, ℂ}) & (𝜑𝐹:𝐴⟶ℂ) & (𝜑𝐴𝑆) & (𝜑 → (𝑁 ∈ ℕ0𝑁 = +∞)) & ((𝜑𝑘 ∈ ((0[,]𝑁) ∩ ℤ)) → 𝐵 ∈ dom ((𝑆 D𝑛 𝐹)‘𝑘)) ((𝜑𝑋 ∈ ℂ) → (ℂfld tsums (𝑘 ∈ ((0[,]𝑁) ∩ ℤ) ↦ (((((𝑆 D𝑛 𝐹)‘𝑘)‘𝐵) / (!‘𝑘)) · ((𝑋𝐵)↑𝑘)))) ⊆ ℂ)
Theoremtaylfval 24952* Define the Taylor polynomial of a function. The constant Tayl is a function of five arguments: 𝑆 is the base set with respect to evaluate the derivatives (generally or ), 𝐹 is the function we are approximating, at point 𝐵, to order 𝑁. The result is a polynomial function of 𝑥.
This "extended" version of taylpfval 24958 additionally handles the case 𝑁 = +∞, in which case this is not a polynomial but an infinite series, the Taylor series of the function. (Contributed by Mario Carneiro, 30-Dec-2016.)
(𝜑𝑆 ∈ {ℝ, ℂ}) & (𝜑𝐹:𝐴⟶ℂ) & (𝜑𝐴𝑆) & (𝜑 → (𝑁 ∈ ℕ0𝑁 = +∞)) & ((𝜑𝑘 ∈ ((0[,]𝑁) ∩ ℤ)) → 𝐵 ∈ dom ((𝑆 D𝑛 𝐹)‘𝑘)) & 𝑇 = (𝑁(𝑆 Tayl 𝐹)𝐵) (𝜑𝑇 = 𝑥 ∈ ℂ ({𝑥} × (ℂfld tsums (𝑘 ∈ ((0[,]𝑁) ∩ ℤ) ↦ (((((𝑆 D𝑛 𝐹)‘𝑘)‘𝐵) / (!‘𝑘)) · ((𝑥𝐵)↑𝑘))))))
Theoremeltayl 24953* Value of the Taylor series as a relation (elementhood in the domain here expresses that the series is convergent). (Contributed by Mario Carneiro, 30-Dec-2016.)
(𝜑𝑆 ∈ {ℝ, ℂ}) & (𝜑𝐹:𝐴⟶ℂ) & (𝜑𝐴𝑆) & (𝜑 → (𝑁 ∈ ℕ0𝑁 = +∞)) & ((𝜑𝑘 ∈ ((0[,]𝑁) ∩ ℤ)) → 𝐵 ∈ dom ((𝑆 D𝑛 𝐹)‘𝑘)) & 𝑇 = (𝑁(𝑆 Tayl 𝐹)𝐵) (𝜑 → (𝑋𝑇𝑌 ↔ (𝑋 ∈ ℂ ∧ 𝑌 ∈ (ℂfld tsums (𝑘 ∈ ((0[,]𝑁) ∩ ℤ) ↦ (((((𝑆 D𝑛 𝐹)‘𝑘)‘𝐵) / (!‘𝑘)) · ((𝑋𝐵)↑𝑘)))))))
Theoremtaylf 24954* The Taylor series defines a function on a subset of the complex numbers. (Contributed by Mario Carneiro, 30-Dec-2016.)
(𝜑𝑆 ∈ {ℝ, ℂ}) & (𝜑𝐹:𝐴⟶ℂ) & (𝜑𝐴𝑆) & (𝜑 → (𝑁 ∈ ℕ0𝑁 = +∞)) & ((𝜑𝑘 ∈ ((0[,]𝑁) ∩ ℤ)) → 𝐵 ∈ dom ((𝑆 D𝑛 𝐹)‘𝑘)) & 𝑇 = (𝑁(𝑆 Tayl 𝐹)𝐵) (𝜑𝑇:dom 𝑇⟶ℂ)
Theoremtayl0 24955* The Taylor series is always defined at the basepoint, with value equal to the value of the function. (Contributed by Mario Carneiro, 30-Dec-2016.)
(𝜑𝑆 ∈ {ℝ, ℂ}) & (𝜑𝐹:𝐴⟶ℂ) & (𝜑𝐴𝑆) & (𝜑 → (𝑁 ∈ ℕ0𝑁 = +∞)) & ((𝜑𝑘 ∈ ((0[,]𝑁) ∩ ℤ)) → 𝐵 ∈ dom ((𝑆 D𝑛 𝐹)‘𝑘)) & 𝑇 = (𝑁(𝑆 Tayl 𝐹)𝐵) (𝜑 → (𝐵 ∈ dom 𝑇 ∧ (𝑇𝐵) = (𝐹𝐵)))
Theoremtaylplem1 24956* Lemma for taylpfval 24958 and similar theorems. (Contributed by Mario Carneiro, 31-Dec-2016.)
(𝜑𝑆 ∈ {ℝ, ℂ}) & (𝜑𝐹:𝐴⟶ℂ) & (𝜑𝐴𝑆) & (𝜑𝑁 ∈ ℕ0) & (𝜑𝐵 ∈ dom ((𝑆 D𝑛 𝐹)‘𝑁)) ((𝜑𝑘 ∈ ((0[,]𝑁) ∩ ℤ)) → 𝐵 ∈ dom ((𝑆 D𝑛 𝐹)‘𝑘))
Theoremtaylplem2 24957* Lemma for taylpfval 24958 and similar theorems. (Contributed by Mario Carneiro, 31-Dec-2016.)
(𝜑𝑆 ∈ {ℝ, ℂ}) & (𝜑𝐹:𝐴⟶ℂ) & (𝜑𝐴𝑆) & (𝜑𝑁 ∈ ℕ0) & (𝜑𝐵 ∈ dom ((𝑆 D𝑛 𝐹)‘𝑁)) (((𝜑𝑋 ∈ ℂ) ∧ 𝑘 ∈ (0...𝑁)) → (((((𝑆 D𝑛 𝐹)‘𝑘)‘𝐵) / (!‘𝑘)) · ((𝑋𝐵)↑𝑘)) ∈ ℂ)
Theoremtaylpfval 24958* Define the Taylor polynomial of a function. The constant Tayl is a function of five arguments: 𝑆 is the base set with respect to evaluate the derivatives (generally or ), 𝐹 is the function we are approximating, at point 𝐵, to order 𝑁. The result is a polynomial function of 𝑥. (Contributed by Mario Carneiro, 31-Dec-2016.)
(𝜑𝑆 ∈ {ℝ, ℂ}) & (𝜑𝐹:𝐴⟶ℂ) & (𝜑𝐴𝑆) & (𝜑𝑁 ∈ ℕ0) & (𝜑𝐵 ∈ dom ((𝑆 D𝑛 𝐹)‘𝑁)) & 𝑇 = (𝑁(𝑆 Tayl 𝐹)𝐵) (𝜑𝑇 = (𝑥 ∈ ℂ ↦ Σ𝑘 ∈ (0...𝑁)(((((𝑆 D𝑛 𝐹)‘𝑘)‘𝐵) / (!‘𝑘)) · ((𝑥𝐵)↑𝑘))))
Theoremtaylpf 24959 The Taylor polynomial is a function on the complex numbers (even if the base set of the original function is the reals). (Contributed by Mario Carneiro, 31-Dec-2016.)
(𝜑𝑆 ∈ {ℝ, ℂ}) & (𝜑𝐹:𝐴⟶ℂ) & (𝜑𝐴𝑆) & (𝜑𝑁 ∈ ℕ0) & (𝜑𝐵 ∈ dom ((𝑆 D𝑛 𝐹)‘𝑁)) & 𝑇 = (𝑁(𝑆 Tayl 𝐹)𝐵) (𝜑𝑇:ℂ⟶ℂ)
Theoremtaylpval 24960* Value of the Taylor polynomial. (Contributed by Mario Carneiro, 31-Dec-2016.)
(𝜑𝑆 ∈ {ℝ, ℂ}) & (𝜑𝐹:𝐴⟶ℂ) & (𝜑𝐴𝑆) & (𝜑𝑁 ∈ ℕ0) & (𝜑𝐵 ∈ dom ((𝑆 D𝑛 𝐹)‘𝑁)) & 𝑇 = (𝑁(𝑆 Tayl 𝐹)𝐵) & (𝜑𝑋 ∈ ℂ) (𝜑 → (𝑇𝑋) = Σ𝑘 ∈ (0...𝑁)(((((𝑆 D𝑛 𝐹)‘𝑘)‘𝐵) / (!‘𝑘)) · ((𝑋𝐵)↑𝑘)))
Theoremtaylply2 24961* The Taylor polynomial is a polynomial of degree (at most) 𝑁. This version of taylply 24962 shows that the coefficients of 𝑇 are in a subring of the complex numbers. (Contributed by Mario Carneiro, 1-Jan-2017.)
(𝜑𝑆 ∈ {ℝ, ℂ}) & (𝜑𝐹:𝐴⟶ℂ) & (𝜑𝐴𝑆) & (𝜑𝑁 ∈ ℕ0) & (𝜑𝐵 ∈ dom ((𝑆 D𝑛 𝐹)‘𝑁)) & 𝑇 = (𝑁(𝑆 Tayl 𝐹)𝐵) & (𝜑𝐷 ∈ (SubRing‘ℂfld)) & (𝜑𝐵𝐷) & ((𝜑𝑘 ∈ (0...𝑁)) → ((((𝑆 D𝑛 𝐹)‘𝑘)‘𝐵) / (!‘𝑘)) ∈ 𝐷) (𝜑 → (𝑇 ∈ (Poly‘𝐷) ∧ (deg‘𝑇) ≤ 𝑁))
Theoremtaylply 24962 The Taylor polynomial is a polynomial of degree (at most) 𝑁. (Contributed by Mario Carneiro, 31-Dec-2016.)
(𝜑𝑆 ∈ {ℝ, ℂ}) & (𝜑𝐹:𝐴⟶ℂ) & (𝜑𝐴𝑆) & (𝜑𝑁 ∈ ℕ0) & (𝜑𝐵 ∈ dom ((𝑆 D𝑛 𝐹)‘𝑁)) & 𝑇 = (𝑁(𝑆 Tayl 𝐹)𝐵) (𝜑 → (𝑇 ∈ (Poly‘ℂ) ∧ (deg‘𝑇) ≤ 𝑁))
Theoremdvtaylp 24963 The derivative of the Taylor polynomial is the Taylor polynomial of the derivative of the function. (Contributed by Mario Carneiro, 31-Dec-2016.)
(𝜑𝑆 ∈ {ℝ, ℂ}) & (𝜑𝐹:𝐴⟶ℂ) & (𝜑𝐴𝑆) & (𝜑𝑁 ∈ ℕ0) & (𝜑𝐵 ∈ dom ((𝑆 D𝑛 𝐹)‘(𝑁 + 1))) (𝜑 → (ℂ D ((𝑁 + 1)(𝑆 Tayl 𝐹)𝐵)) = (𝑁(𝑆 Tayl (𝑆 D 𝐹))𝐵))
Theoremdvntaylp 24964 The 𝑀-th derivative of the Taylor polynomial is the Taylor polynomial of the 𝑀-th derivative of the function. (Contributed by Mario Carneiro, 1-Jan-2017.)
(𝜑𝑆 ∈ {ℝ, ℂ}) & (𝜑𝐹:𝐴⟶ℂ) & (𝜑𝐴𝑆) & (𝜑𝑀 ∈ ℕ0) & (𝜑𝑁 ∈ ℕ0) & (𝜑𝐵 ∈ dom ((𝑆 D𝑛 𝐹)‘(𝑁 + 𝑀))) (𝜑 → ((ℂ D𝑛 ((𝑁 + 𝑀)(𝑆 Tayl 𝐹)𝐵))‘𝑀) = (𝑁(𝑆 Tayl ((𝑆 D𝑛 𝐹)‘𝑀))𝐵))
Theoremdvntaylp0 24965 The first 𝑁 derivatives of the Taylor polynomial at 𝐵 match the derivatives of the function from which it is derived. (Contributed by Mario Carneiro, 1-Jan-2017.)
(𝜑𝑆 ∈ {ℝ, ℂ}) & (𝜑𝐹:𝐴⟶ℂ) & (𝜑𝐴𝑆) & (𝜑𝑀 ∈ (0...𝑁)) & (𝜑𝐵 ∈ dom ((𝑆 D𝑛 𝐹)‘𝑁)) & 𝑇 = (𝑁(𝑆 Tayl 𝐹)𝐵) (𝜑 → (((ℂ D𝑛 𝑇)‘𝑀)‘𝐵) = (((𝑆 D𝑛 𝐹)‘𝑀)‘𝐵))
Theoremtaylthlem1 24966* Lemma for taylth 24968. This is the main part of Taylor's theorem, except for the induction step, which is supposed to be proven using L'Hôpital's rule. However, since our proof of L'Hôpital assumes that 𝑆 = ℝ, we can only do this part generically, and for taylth 24968 itself we must restrict to . (Contributed by Mario Carneiro, 1-Jan-2017.)
(𝜑𝑆 ∈ {ℝ, ℂ}) & (𝜑𝐹:𝐴⟶ℂ) & (𝜑𝐴𝑆) & (𝜑 → dom ((𝑆 D𝑛 𝐹)‘𝑁) = 𝐴) & (𝜑𝑁 ∈ ℕ) & (𝜑𝐵𝐴) & 𝑇 = (𝑁(𝑆 Tayl 𝐹)𝐵) & 𝑅 = (𝑥 ∈ (𝐴 ∖ {𝐵}) ↦ (((𝐹𝑥) − (𝑇𝑥)) / ((𝑥𝐵)↑𝑁))) & ((𝜑 ∧ (𝑛 ∈ (1..^𝑁) ∧ 0 ∈ ((𝑦 ∈ (𝐴 ∖ {𝐵}) ↦ (((((𝑆 D𝑛 𝐹)‘(𝑁𝑛))‘𝑦) − (((ℂ D𝑛 𝑇)‘(𝑁𝑛))‘𝑦)) / ((𝑦𝐵)↑𝑛))) lim 𝐵))) → 0 ∈ ((𝑥 ∈ (𝐴 ∖ {𝐵}) ↦ (((((𝑆 D𝑛 𝐹)‘(𝑁 − (𝑛 + 1)))‘𝑥) − (((ℂ D𝑛 𝑇)‘(𝑁 − (𝑛 + 1)))‘𝑥)) / ((𝑥𝐵)↑(𝑛 + 1)))) lim 𝐵)) (𝜑 → 0 ∈ (𝑅 lim 𝐵))
Theoremtaylthlem2 24967* Lemma for taylth 24968. (Contributed by Mario Carneiro, 1-Jan-2017.)
(𝜑𝐹:𝐴⟶ℝ) & (𝜑𝐴 ⊆ ℝ) & (𝜑 → dom ((ℝ D𝑛 𝐹)‘𝑁) = 𝐴) & (𝜑𝑁 ∈ ℕ) & (𝜑𝐵𝐴) & 𝑇 = (𝑁(ℝ Tayl 𝐹)𝐵) & (𝜑𝑀 ∈ (1..^𝑁)) & (𝜑 → 0 ∈ ((𝑥 ∈ (𝐴 ∖ {𝐵}) ↦ (((((ℝ D𝑛 𝐹)‘(𝑁𝑀))‘𝑥) − (((ℂ D𝑛 𝑇)‘(𝑁𝑀))‘𝑥)) / ((𝑥𝐵)↑𝑀))) lim 𝐵)) (𝜑 → 0 ∈ ((𝑥 ∈ (𝐴 ∖ {𝐵}) ↦ (((((ℝ D𝑛 𝐹)‘(𝑁 − (𝑀 + 1)))‘𝑥) − (((ℂ D𝑛 𝑇)‘(𝑁 − (𝑀 + 1)))‘𝑥)) / ((𝑥𝐵)↑(𝑀 + 1)))) lim 𝐵))
Theoremtaylth 24968* Taylor's theorem. The Taylor polynomial of a 𝑁-times differentiable function is such that the error term goes to zero faster than (𝑥𝐵)↑𝑁. This is Metamath 100 proof #35. (Contributed by Mario Carneiro, 1-Jan-2017.)
(𝜑𝐹:𝐴⟶ℝ) & (𝜑𝐴 ⊆ ℝ) & (𝜑 → dom ((ℝ D𝑛 𝐹)‘𝑁) = 𝐴) & (𝜑𝑁 ∈ ℕ) & (𝜑𝐵𝐴) & 𝑇 = (𝑁(ℝ Tayl 𝐹)𝐵) & 𝑅 = (𝑥 ∈ (𝐴 ∖ {𝐵}) ↦ (((𝐹𝑥) − (𝑇𝑥)) / ((𝑥𝐵)↑𝑁))) (𝜑 → 0 ∈ (𝑅 lim 𝐵))
14.2.2 Uniform convergence
Syntaxculm 24969 Extend class notation to include the uniform convergence predicate.
class 𝑢
Definitiondf-ulm 24970* Define the uniform convergence of a sequence of functions. Here 𝐹(⇝𝑢𝑆)𝐺 if 𝐹 is a sequence of functions 𝐹(𝑛), 𝑛 ∈ ℕ defined on 𝑆 and 𝐺 is a function on 𝑆, and for every 0 < 𝑥 there is a 𝑗 such that the functions 𝐹(𝑘) for 𝑗𝑘 are all uniformly within 𝑥 of 𝐺 on the domain 𝑆. Compare with df-clim 14843. (Contributed by Mario Carneiro, 26-Feb-2015.)
𝑢 = (𝑠 ∈ V ↦ {⟨𝑓, 𝑦⟩ ∣ ∃𝑛 ∈ ℤ (𝑓:(ℤ𝑛)⟶(ℂ ↑m 𝑠) ∧ 𝑦:𝑠⟶ℂ ∧ ∀𝑥 ∈ ℝ+𝑗 ∈ (ℤ𝑛)∀𝑘 ∈ (ℤ𝑗)∀𝑧𝑠 (abs‘(((𝑓𝑘)‘𝑧) − (𝑦𝑧))) < 𝑥)})
Theoremulmrel 24971 The uniform limit relation is a relation. (Contributed by Mario Carneiro, 26-Feb-2015.)
Rel (⇝𝑢𝑆)
Theoremulmscl 24972 Closure of the base set in a uniform limit. (Contributed by Mario Carneiro, 26-Feb-2015.)
(𝐹(⇝𝑢𝑆)𝐺𝑆 ∈ V)
Theoremulmval 24973* Express the predicate: The sequence of functions 𝐹 converges uniformly to 𝐺 on 𝑆. (Contributed by Mario Carneiro, 26-Feb-2015.)
(𝑆𝑉 → (𝐹(⇝𝑢𝑆)𝐺 ↔ ∃𝑛 ∈ ℤ (𝐹:(ℤ𝑛)⟶(ℂ ↑m 𝑆) ∧ 𝐺:𝑆⟶ℂ ∧ ∀𝑥 ∈ ℝ+𝑗 ∈ (ℤ𝑛)∀𝑘 ∈ (ℤ𝑗)∀𝑧𝑆 (abs‘(((𝐹𝑘)‘𝑧) − (𝐺𝑧))) < 𝑥)))
Theoremulmcl 24974 Closure of a uniform limit of functions. (Contributed by Mario Carneiro, 26-Feb-2015.)
(𝐹(⇝𝑢𝑆)𝐺𝐺:𝑆⟶ℂ)
Theoremulmf 24975* Closure of a uniform limit of functions. (Contributed by Mario Carneiro, 26-Feb-2015.)
(𝐹(⇝𝑢𝑆)𝐺 → ∃𝑛 ∈ ℤ 𝐹:(ℤ𝑛)⟶(ℂ ↑m 𝑆))
Theoremulmpm 24976 Closure of a uniform limit of functions. (Contributed by Mario Carneiro, 26-Feb-2015.)
(𝐹(⇝𝑢𝑆)𝐺𝐹 ∈ ((ℂ ↑m 𝑆) ↑pm ℤ))
Theoremulmf2 24977 Closure of a uniform limit of functions. (Contributed by Mario Carneiro, 18-Mar-2015.)
((𝐹 Fn 𝑍𝐹(⇝𝑢𝑆)𝐺) → 𝐹:𝑍⟶(ℂ ↑m 𝑆))
Theoremulm2 24978* Simplify ulmval 24973 when 𝐹 and 𝐺 are known to be functions. (Contributed by Mario Carneiro, 26-Feb-2015.)
𝑍 = (ℤ𝑀) & (𝜑𝑀 ∈ ℤ) & (𝜑𝐹:𝑍⟶(ℂ ↑m 𝑆)) & ((𝜑 ∧ (𝑘𝑍𝑧𝑆)) → ((𝐹𝑘)‘𝑧) = 𝐵) & ((𝜑𝑧𝑆) → (𝐺𝑧) = 𝐴) & (𝜑𝐺:𝑆⟶ℂ) & (𝜑𝑆𝑉) (𝜑 → (𝐹(⇝𝑢𝑆)𝐺 ↔ ∀𝑥 ∈ ℝ+𝑗𝑍𝑘 ∈ (ℤ𝑗)∀𝑧𝑆 (abs‘(𝐵𝐴)) < 𝑥))
Theoremulmi 24979* The uniform limit property. (Contributed by Mario Carneiro, 27-Feb-2015.)
𝑍 = (ℤ𝑀) & (𝜑𝑀 ∈ ℤ) & (𝜑𝐹:𝑍⟶(ℂ ↑m 𝑆)) & ((𝜑 ∧ (𝑘𝑍𝑧𝑆)) → ((𝐹𝑘)‘𝑧) = 𝐵) & ((𝜑𝑧𝑆) → (𝐺𝑧) = 𝐴) & (𝜑𝐹(⇝𝑢𝑆)𝐺) & (𝜑𝐶 ∈ ℝ+) (𝜑 → ∃𝑗𝑍𝑘 ∈ (ℤ𝑗)∀𝑧𝑆 (abs‘(𝐵𝐴)) < 𝐶)
Theoremulmclm 24980* A uniform limit of functions converges pointwise. (Contributed by Mario Carneiro, 27-Feb-2015.)
𝑍 = (ℤ𝑀) & (𝜑𝑀 ∈ ℤ) & (𝜑𝐹:𝑍⟶(ℂ ↑m 𝑆)) & (𝜑𝐴𝑆) & (𝜑𝐻𝑊) & ((𝜑𝑘𝑍) → ((𝐹𝑘)‘𝐴) = (𝐻𝑘)) & (𝜑𝐹(⇝𝑢𝑆)𝐺) (𝜑𝐻 ⇝ (𝐺𝐴))
Theoremulmres 24981 A sequence of functions converges iff the tail of the sequence converges (for any finite cutoff). (Contributed by Mario Carneiro, 24-Mar-2015.)
𝑍 = (ℤ𝑀) & 𝑊 = (ℤ𝑁) & (𝜑𝑁𝑍) & (𝜑𝐹:𝑍⟶(ℂ ↑m 𝑆)) (𝜑 → (𝐹(⇝𝑢𝑆)𝐺 ↔ (𝐹𝑊)(⇝𝑢𝑆)𝐺))
Theoremulmshftlem 24982* Lemma for ulmshft 24983. (Contributed by Mario Carneiro, 24-Mar-2015.)
𝑍 = (ℤ𝑀) & 𝑊 = (ℤ‘(𝑀 + 𝐾)) & (𝜑𝑀 ∈ ℤ) & (𝜑𝐾 ∈ ℤ) & (𝜑𝐹:𝑍⟶(ℂ ↑m 𝑆)) & (𝜑𝐻 = (𝑛𝑊 ↦ (𝐹‘(𝑛𝐾)))) (𝜑 → (𝐹(⇝𝑢𝑆)𝐺𝐻(⇝𝑢𝑆)𝐺))
Theoremulmshft 24983* A sequence of functions converges iff the shifted sequence converges. (Contributed by Mario Carneiro, 24-Mar-2015.)
𝑍 = (ℤ𝑀) & 𝑊 = (ℤ‘(𝑀 + 𝐾)) & (𝜑𝑀 ∈ ℤ) & (𝜑𝐾 ∈ ℤ) & (𝜑𝐹:𝑍⟶(ℂ ↑m 𝑆)) & (𝜑𝐻 = (𝑛𝑊 ↦ (𝐹‘(𝑛𝐾)))) (𝜑 → (𝐹(⇝𝑢𝑆)𝐺𝐻(⇝𝑢𝑆)𝐺))
Theoremulm0 24984 Every function converges uniformly on the empty set. (Contributed by Mario Carneiro, 3-Mar-2015.)
𝑍 = (ℤ𝑀) & (𝜑𝑀 ∈ ℤ) & (𝜑𝐹:𝑍⟶(ℂ ↑m 𝑆)) & (𝜑𝐺:𝑆⟶ℂ) ((𝜑𝑆 = ∅) → 𝐹(⇝𝑢𝑆)𝐺)
Theoremulmuni 24985 A sequence of functions uniformly converges to at most one limit. (Contributed by Mario Carneiro, 5-Jul-2017.)
((𝐹(⇝𝑢𝑆)𝐺𝐹(⇝𝑢𝑆)𝐻) → 𝐺 = 𝐻)
Theoremulmdm 24986 Two ways to express that a function has a limit. (The expression ((⇝𝑢𝑆)‘𝐹) is sometimes useful as a shorthand for "the unique limit of the function 𝐹"). (Contributed by Mario Carneiro, 5-Jul-2017.)
(𝐹 ∈ dom (⇝𝑢𝑆) ↔ 𝐹(⇝𝑢𝑆)((⇝𝑢𝑆)‘𝐹))
Theoremulmcaulem 24987* Lemma for ulmcau 24988 and ulmcau2 24989: show the equivalence of the four- and five-quantifier forms of the Cauchy convergence condition. Compare cau3 14713. (Contributed by Mario Carneiro, 1-Mar-2015.)
𝑍 = (ℤ𝑀) & (𝜑𝑀 ∈ ℤ) & (𝜑𝑆𝑉) & (𝜑𝐹:𝑍⟶(ℂ ↑m 𝑆)) (𝜑 → (∀𝑥 ∈ ℝ+𝑗𝑍𝑘 ∈ (ℤ𝑗)∀𝑧𝑆 (abs‘(((𝐹𝑘)‘𝑧) − ((𝐹𝑗)‘𝑧))) < 𝑥 ↔ ∀𝑥 ∈ ℝ+𝑗𝑍𝑘 ∈ (ℤ𝑗)∀𝑚 ∈ (ℤ𝑘)∀𝑧𝑆 (abs‘(((𝐹𝑘)‘𝑧) − ((𝐹𝑚)‘𝑧))) < 𝑥))
Theoremulmcau 24988* A sequence of functions converges uniformly iff it is uniformly Cauchy, which is to say that for every 0 < 𝑥 there is a 𝑗 such that for all 𝑗𝑘 the functions 𝐹(𝑘) and 𝐹(𝑗) are uniformly within 𝑥 of each other on 𝑆. This is the four-quantifier version, see ulmcau2 24989 for the more conventional five-quantifier version. (Contributed by Mario Carneiro, 1-Mar-2015.)
𝑍 = (ℤ𝑀) & (𝜑𝑀 ∈ ℤ) & (𝜑𝑆𝑉) & (𝜑𝐹:𝑍⟶(ℂ ↑m 𝑆)) (𝜑 → (𝐹 ∈ dom (⇝𝑢𝑆) ↔ ∀𝑥 ∈ ℝ+𝑗𝑍𝑘 ∈ (ℤ𝑗)∀𝑧𝑆 (abs‘(((𝐹𝑘)‘𝑧) − ((𝐹𝑗)‘𝑧))) < 𝑥))
Theoremulmcau2 24989* A sequence of functions converges uniformly iff it is uniformly Cauchy, which is to say that for every 0 < 𝑥 there is a 𝑗 such that for all 𝑗𝑘, 𝑚 the functions 𝐹(𝑘) and 𝐹(𝑚) are uniformly within 𝑥 of each other on 𝑆. (Contributed by Mario Carneiro, 1-Mar-2015.)
𝑍 = (ℤ𝑀) & (𝜑𝑀 ∈ ℤ) & (𝜑𝑆𝑉) & (𝜑𝐹:𝑍⟶(ℂ ↑m 𝑆)) (𝜑 → (𝐹 ∈ dom (⇝𝑢𝑆) ↔ ∀𝑥 ∈ ℝ+𝑗𝑍𝑘 ∈ (ℤ𝑗)∀𝑚 ∈ (ℤ𝑘)∀𝑧𝑆 (abs‘(((𝐹𝑘)‘𝑧) − ((𝐹𝑚)‘𝑧))) < 𝑥))
Theoremulmss 24990* A uniform limit of functions is still a uniform limit if restricted to a subset. (Contributed by Mario Carneiro, 3-Mar-2015.)
𝑍 = (ℤ𝑀) & (𝜑𝑇𝑆) & ((𝜑𝑥𝑍) → 𝐴𝑊) & (𝜑 → (𝑥𝑍𝐴)(⇝𝑢𝑆)𝐺) (𝜑 → (𝑥𝑍 ↦ (𝐴𝑇))(⇝𝑢𝑇)(𝐺𝑇))
Theoremulmbdd 24991* A uniform limit of bounded functions is bounded. (Contributed by Mario Carneiro, 27-Feb-2015.)
𝑍 = (ℤ𝑀) & (𝜑𝑀 ∈ ℤ) & (𝜑𝐹:𝑍⟶(ℂ ↑m 𝑆)) & ((𝜑𝑘𝑍) → ∃𝑥 ∈ ℝ ∀𝑧𝑆 (abs‘((𝐹𝑘)‘𝑧)) ≤ 𝑥) & (𝜑𝐹(⇝𝑢𝑆)𝐺) (𝜑 → ∃𝑥 ∈ ℝ ∀𝑧𝑆 (abs‘(𝐺𝑧)) ≤ 𝑥)
Theoremulmcn 24992 A uniform limit of continuous functions is continuous. (Contributed by Mario Carneiro, 27-Feb-2015.)
𝑍 = (ℤ𝑀) & (𝜑𝑀 ∈ ℤ) & (𝜑𝐹:𝑍⟶(𝑆cn→ℂ)) & (𝜑𝐹(⇝𝑢𝑆)𝐺) (𝜑𝐺 ∈ (𝑆cn→ℂ))
Theoremulmdvlem1 24993* Lemma for ulmdv 24996. (Contributed by Mario Carneiro, 3-Mar-2015.)
𝑍 = (ℤ𝑀) & (𝜑𝑆 ∈ {ℝ, ℂ}) & (𝜑𝑀 ∈ ℤ) & (𝜑𝐹:𝑍⟶(ℂ ↑m 𝑋)) & (𝜑𝐺:𝑋⟶ℂ) & ((𝜑𝑧𝑋) → (𝑘𝑍 ↦ ((𝐹𝑘)‘𝑧)) ⇝ (𝐺𝑧)) & (𝜑 → (𝑘𝑍 ↦ (𝑆 D (𝐹𝑘)))(⇝𝑢𝑋)𝐻) & ((𝜑𝜓) → 𝐶𝑋) & ((𝜑𝜓) → 𝑅 ∈ ℝ+) & ((𝜑𝜓) → 𝑈 ∈ ℝ+) & ((𝜑𝜓) → 𝑊 ∈ ℝ+) & ((𝜑𝜓) → 𝑈 < 𝑊) & ((𝜑𝜓) → (𝐶(ball‘((abs ∘ − ) ↾ (𝑆 × 𝑆)))𝑈) ⊆ 𝑋) & ((𝜑𝜓) → (abs‘(𝑌𝐶)) < 𝑈) & ((𝜑𝜓) → 𝑁𝑍) & ((𝜑𝜓) → ∀𝑚 ∈ (ℤ𝑁)∀𝑥𝑋 (abs‘(((𝑆 D (𝐹𝑁))‘𝑥) − ((𝑆 D (𝐹𝑚))‘𝑥))) < ((𝑅 / 2) / 2)) & ((𝜑𝜓) → (abs‘(((𝑆 D (𝐹𝑁))‘𝐶) − (𝐻𝐶))) < (𝑅 / 2)) & ((𝜑𝜓) → 𝑌𝑋) & ((𝜑𝜓) → 𝑌𝐶) & ((𝜑𝜓) → ((abs‘(𝑌𝐶)) < 𝑊 → (abs‘(((((𝐹𝑁)‘𝑌) − ((𝐹𝑁)‘𝐶)) / (𝑌𝐶)) − ((𝑆 D (𝐹𝑁))‘𝐶))) < ((𝑅 / 2) / 2))) ((𝜑𝜓) → (abs‘((((𝐺𝑌) − (𝐺𝐶)) / (𝑌𝐶)) − (𝐻𝐶))) < 𝑅)
Theoremulmdvlem2 24994* Lemma for ulmdv 24996. (Contributed by Mario Carneiro, 8-May-2015.)
𝑍 = (ℤ𝑀) & (𝜑𝑆 ∈ {ℝ, ℂ}) & (𝜑𝑀 ∈ ℤ) & (𝜑𝐹:𝑍⟶(ℂ ↑m 𝑋)) & (𝜑𝐺:𝑋⟶ℂ) & ((𝜑𝑧𝑋) → (𝑘𝑍 ↦ ((𝐹𝑘)‘𝑧)) ⇝ (𝐺𝑧)) & (𝜑 → (𝑘𝑍 ↦ (𝑆 D (𝐹𝑘)))(⇝𝑢𝑋)𝐻) ((𝜑𝑘𝑍) → dom (𝑆 D (𝐹𝑘)) = 𝑋)
Theoremulmdvlem3 24995* Lemma for ulmdv 24996. (Contributed by Mario Carneiro, 8-May-2015.) (Proof shortened by Mario Carneiro, 28-Dec-2016.)
𝑍 = (ℤ𝑀) & (𝜑𝑆 ∈ {ℝ, ℂ}) & (𝜑𝑀 ∈ ℤ) & (𝜑𝐹:𝑍⟶(ℂ ↑m 𝑋)) & (𝜑𝐺:𝑋⟶ℂ) & ((𝜑𝑧𝑋) → (𝑘𝑍 ↦ ((𝐹𝑘)‘𝑧)) ⇝ (𝐺𝑧)) & (𝜑 → (𝑘𝑍 ↦ (𝑆 D (𝐹𝑘)))(⇝𝑢𝑋)𝐻) ((𝜑𝑧𝑋) → 𝑧(𝑆 D 𝐺)(𝐻𝑧))
Theoremulmdv 24996* If 𝐹 is a sequence of differentiable functions on 𝑋 which converge pointwise to 𝐺, and the derivatives of 𝐹(𝑛) converge uniformly to 𝐻, then 𝐺 is differentiable with derivative 𝐻. (Contributed by Mario Carneiro, 27-Feb-2015.)
𝑍 = (ℤ𝑀) & (𝜑𝑆 ∈ {ℝ, ℂ}) & (𝜑𝑀 ∈ ℤ) & (𝜑𝐹:𝑍⟶(ℂ ↑m 𝑋)) & (𝜑𝐺:𝑋⟶ℂ) & ((𝜑𝑧𝑋) → (𝑘𝑍 ↦ ((𝐹𝑘)‘𝑧)) ⇝ (𝐺𝑧)) & (𝜑 → (𝑘𝑍 ↦ (𝑆 D (𝐹𝑘)))(⇝𝑢𝑋)𝐻) (𝜑 → (𝑆 D 𝐺) = 𝐻)
Theoremmtest 24997* The Weierstrass M-test. If 𝐹 is a sequence of functions which are uniformly bounded by the convergent sequence 𝑀(𝑘), then the series generated by the sequence 𝐹 converges uniformly. (Contributed by Mario Carneiro, 3-Mar-2015.)
𝑍 = (ℤ𝑁) & (𝜑𝑁 ∈ ℤ) & (𝜑𝑆𝑉) & (𝜑𝐹:𝑍⟶(ℂ ↑m 𝑆)) & (𝜑𝑀𝑊) & ((𝜑𝑘𝑍) → (𝑀𝑘) ∈ ℝ) & ((𝜑 ∧ (𝑘𝑍𝑧𝑆)) → (abs‘((𝐹𝑘)‘𝑧)) ≤ (𝑀𝑘)) & (𝜑 → seq𝑁( + , 𝑀) ∈ dom ⇝ ) (𝜑 → seq𝑁( ∘f + , 𝐹) ∈ dom (⇝𝑢𝑆))
Theoremmtestbdd 24998* Given the hypotheses of the Weierstrass M-test, the convergent function of the sequence is uniformly bounded. (Contributed by Mario Carneiro, 9-Jul-2017.)
𝑍 = (ℤ𝑁) & (𝜑𝑁 ∈ ℤ) & (𝜑𝑆𝑉) & (𝜑𝐹:𝑍⟶(ℂ ↑m 𝑆)) & (𝜑𝑀𝑊) & ((𝜑𝑘𝑍) → (𝑀𝑘) ∈ ℝ) & ((𝜑 ∧ (𝑘𝑍𝑧𝑆)) → (abs‘((𝐹𝑘)‘𝑧)) ≤ (𝑀𝑘)) & (𝜑 → seq𝑁( + , 𝑀) ∈ dom ⇝ ) & (𝜑 → seq𝑁( ∘f + , 𝐹)(⇝𝑢𝑆)𝑇) (𝜑 → ∃𝑥 ∈ ℝ ∀𝑧𝑆 (abs‘(𝑇𝑧)) ≤ 𝑥)
Theoremmbfulm 24999 A uniform limit of measurable functions is measurable. (This is just a corollary of the fact that a pointwise limit of measurable functions is measurable, see mbflim 24270.) (Contributed by Mario Carneiro, 18-Mar-2015.)
𝑍 = (ℤ𝑀) & (𝜑𝑀 ∈ ℤ) & (𝜑𝐹:𝑍⟶MblFn) & (𝜑𝐹(⇝𝑢𝑆)𝐺) (𝜑𝐺 ∈ MblFn)
Theoremiblulm 25000 A uniform limit of integrable functions is integrable. (Contributed by Mario Carneiro, 3-Mar-2015.)
𝑍 = (ℤ𝑀) & (𝜑𝑀 ∈ ℤ) & (𝜑𝐹:𝑍⟶𝐿1) & (𝜑𝐹(⇝𝑢𝑆)𝐺) & (𝜑 → (vol‘𝑆) ∈ ℝ) (𝜑𝐺 ∈ 𝐿1)
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Copyright terms: Public domain < Previous Next > | 22,510 | 37,625 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.515625 | 4 | CC-MAIN-2024-22 | latest | en | 0.420175 |
https://aakashsrv1.meritnation.com/ask-answer/question/whaat-is-the-meaning-of-congruent-trapezium/perimeter-and-area/3194362 | 1,675,231,166,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764499911.86/warc/CC-MAIN-20230201045500-20230201075500-00634.warc.gz | 98,361,386 | 8,040 | # whaat is the meaning of congruent trapezium ?????
A congruent trapezium may be defined as the one having all its sides and angles equal to the corresponding sides and angles of another trapezium.
For example, in the figure given below, PQRS is a congruent trapezium corresponding to trapezium ABCD in which
AB = PQ, BC = QR, CD = RS and AD = PS.
Also, ∠A = ∠P, ∠B = ∠Q, ∠C = ∠R and ∠D =∠S
• 1
What are you looking for? | 140 | 426 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.671875 | 3 | CC-MAIN-2023-06 | latest | en | 0.920163 |
http://www.mathworks.com/help/phased/ref/phased.custommicrophoneelement.step.html?requestedDomain=www.mathworks.com&nocookie=true | 1,516,417,052,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084888878.44/warc/CC-MAIN-20180120023744-20180120043744-00005.warc.gz | 496,153,163 | 15,006 | # Documentation
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# step
System object: phased.CustomMicrophoneElement
Package: phased
Output response of microphone
## Syntax
```RESP = step(H,FREQ,ANG) ```
## Description
### Note
Starting in R2016b, instead of using the `step` method to perform the operation defined by the System object™, you can call the object with arguments, as if it were a function. For example, `y = step(obj,x)` and `y = obj(x)` perform equivalent operations.
`RESP = step(H,FREQ,ANG)` returns the microphone’s magnitude response, `RESP`, at frequencies specified in `FREQ` and directions specified in `ANG`.
### Note
The object performs an initialization the first time the `step` method is executed. This initialization locks nontunable properties (MATLAB) and input specifications, such as dimensions, complexity, and data type of the input data. If you change a nontunable property or an input specification, the System object issues an error. To change nontunable properties or inputs, you must first call the `release` method to unlock the object.
## Input Arguments
`H` Microphone object. `FREQ` Frequencies in hertz. `FREQ` is a row vector of length L. `ANG` Directions in degrees. `ANG` can be either a 2-by-M matrix or a row vector of length M. If `ANG` is a 2-by-M matrix, each column of the matrix specifies the direction in the form [azimuth; elevation]. The azimuth angle must be between –180 and 180 degrees, inclusive. The elevation angle must be between –90 and 90 degrees, inclusive. If `ANG` is a row vector of length M, each element specifies a direction’s azimuth angle. In this case, the corresponding elevation angle is assumed to be 0.
## Output Arguments
`RESP` Response of microphone. `RESP` is an M-by-L matrix that contains the responses of the microphone element at the M angles specified in `ANG` and the L frequencies specified in `FREQ`.
## Examples
expand all
Construct a custom cardioid microphone with an operating frequency of 500 Hz. Find the microphone response in the directions: (0,0) degrees azimuth and elevation and (40,50) degrees azimuth and elevation.
```sCustMic = phased.CustomMicrophoneElement; sCustMic.PolarPatternFrequencies = [500 1000]; sCustMic.PolarPattern = mag2db([... 0.5+0.5*cosd(sCustMic.PolarPatternAngles);... 0.6+0.4*cosd(sCustMic.PolarPatternAngles)]); fc = 700; ang = [0 0; 40 50]'; resp = step(sCustMic,fc,ang)```
```resp = 1.0000 0.7424 ```
## Algorithms
The total response of a custom microphone element is a combination of its frequency response and spatial response. `phased.CustomMicrophoneElement` calculates both responses using nearest neighbor interpolation and then multiplies them to form the total response. When the `PolarPatternFrequencies` property value is nonscalar, the object specifies multiple polar patterns. In this case, the interpolation uses the polar pattern that is measured closest to the specified frequency.
## See Also
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#### Hybrid Beamforming for Massive MIMO Phased Array Systems
Download the white paper | 796 | 3,369 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2018-05 | latest | en | 0.793157 |
https://math.stackexchange.com/questions/1839221/is-the-least-squares-solution-unique | 1,623,799,260,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623487621627.41/warc/CC-MAIN-20210615211046-20210616001046-00543.warc.gz | 352,373,332 | 42,783 | # Is the least-squares solution unique?
I am looking for a line closest to $(-5, -2)$, $(-2, 0)$, $(-1, 0)$, $(2, 3)$, $(5, 4)$ using the least square solution. So I set the line as $$ax+by+c=0$$ let $a=1$ (where $a$ is not $0$ obviously) and got
$$\begin{pmatrix} -2 & 1 \\ 0 & 1 \\ 0 & 1 \\ 3 & 1 \\ 4 & 1\\ \end{pmatrix} \begin{pmatrix} b \\ c \\ \end{pmatrix}= \begin{pmatrix} 5 \\ 2 \\ 1 \\ -2 \\ -5\\ \end{pmatrix}$$
Then I solved it by multiplying $A^T$on the both sides.
But the $(b, c)$ I got here was different from that of the usual solution using $$y=ax+b$$ Although the two lines are almost identical (which implies I am not that wrong), they are still different. What's the matter?
• Why did you assume that $a = 1$? Why did you reuse the same symbols $a,b$ to denote something else? – Rodrigo de Azevedo Jun 25 '16 at 13:22
• What was the solution $(b, c)$ that you got and what was the $y=mx+y_0$ solution? – Noble Mushtak Jun 25 '16 at 13:26
• @RodrigodeAzevedo since the ratio of a, b, and c is important, a=1 doesn't matter unless a=0. I am sorry for the confusion of the reuse. The second a, b denotes the coefficients of another line. They are different. – Mike Park Jun 25 '16 at 13:31
• Actually, I'm also getting two different solutions by these different methods and I don't know why. I got: $$(b,c)=\left(-\frac{37}{24}, \frac{209}{120}\right)$$ Is that what you got? – Noble Mushtak Jun 25 '16 at 13:32
• @NobleMushtak the fisrt line was 120x-185y+209=0, and the second was 65x-98y+111=0 – Mike Park Jun 25 '16 at 13:33
## 3 Answers
I think the reason we get different solutions here is because we're measuring different squares.
In the first equation, we want to minimize the distance from the vector on the right side made up of x-coordinates to the range of the matrix on the left side made up of y-coordinates and constants.
In the second equation, our vector has y-coordinates and we want to minimize the distance to the range of the matrix made up of x-coordinate and constants.
Yes, we're approximating the same data set with the same purpose in mind, so we get similar points, but we're working with different vectors and matrices in order to do that, and that gives us different least squares metrics to approximate this line with, meaning that we're going to get different answers because we're using different metrics, but similar answers because we're still approximating the same data set.
• From a statistical perspective, I think the first method assumes the measurements of $x$ are noisy, while the second assumes the measurements of $y$ are noisy. – Jakob Hansen Jun 25 '16 at 14:08
• So it's like minimizing horizontal vs vertical distances. And I think those will result in different lines in most cases probably. – jdods Jun 25 '16 at 14:08
• Then if I set constant term on the right side, would it give a best answer because it both considers x and y distances? I am just wondering. – Mike Park Jun 26 '16 at 3:08
• I just calculated the sum of square distances. The third method (constant on the right side) gave the best answer, then the first one (x on the right) and then the second one (y on the right). – Mike Park Jun 26 '16 at 3:28
• @MikePark How did you put the constant on the right side? Did you just say $c=1$ and then had a vector of ones on the right side? If that's what you did, I think that is a good idea. – Noble Mushtak Jun 26 '16 at 13:39
By inspection, no the least squares is not unique. The nullspace $\mathcal{N}\left( \mathbf{A}^{*} \right)$ is not trivial. The column space has dimension $m=5$, and we only have 2 linearly independent vectors.
To continue the problem, a guess was made about the notation and true problem. Start with the $m=5$ data points $\left\{ x_{k}, y_{k} \right\}_{k=1}^{m}$ and the trial function. $$y(x) = a_{0} + a_{1} x$$ The linear system is \begin{align} \mathbf{A} x &= y \\ % \left[ \begin{array}{rr} 1 & -5 \\ 1 & -2 \\ 1 & -1 \\ 1 & 2 \\ 1 & 5 \\ \end{array} \right] % \left[ \begin{array}{c} a_{0} \\ a_{1} \end{array} \right] % &= % \left[ \begin{array}{r} -2 \\ 0 \\ 0 \\ 3 \\ 4 \\ \end{array} \right] % \end{align} The general least squares problem is defined as $$x_{LS} = \left\{ x\in\mathbb{C}^{n} \colon \lVert \mathbf{A} x_{LS} - b \rVert_{2}^{2} \text{ is minimized} \right\}$$ which has the general solution $$x_{LS} = \mathbf{A}^{\dagger} b + \left( \mathbf{I}_{n} - \mathbf{A}^{\dagger}\mathbf{A} \right) z, \quad z\in\mathbb{C}^{n}$$ which is in general an affine space shown with the dashed red line below.
Pose the normal equations \begin{align} \mathbf{A}^{*} \mathbf{A} x &= \mathbf{A}^{*} b \\ \left[ \begin{array}{cc} \mathbf{1} \cdot \mathbf{1} & \mathbf{1} \cdot x \\ x \cdot \mathbf{1} & x \cdot x \end{array} \right] % \left[ \begin{array}{c} a_{0} \\ a_{1} \end{array} \right] % &= % \left[ \begin{array}{c} \mathbf{1} \cdot y \\ x \cdot y \end{array} \right] \\[3pt] % \left[ \begin{array}{rr} 5 & -1 \\ -1 & 59 \end{array} \right] % \left[ \begin{array}{c} a_{0} \\ a_{1} \end{array} \right] % &= % \left[ \begin{array}{r} 1 \\ -59 \end{array} \right]. % % \end{align} % The solution is \begin{align} x_{LS} % &= \left( \mathbf{A}^{*}\mathbf{A} \right)^{-1} \mathbf{A}^{*} y \\[3pt] % &= % \frac{1}{294} \left[ \begin{array}{rr} 59 & 1 \\ 1 & 5 \\ \end{array} \right] % \left[ \begin{array}{r} 1 \\ -59 \end{array} \right] \\[3pt] % &= % \frac{1}{294} \left[ \begin{array}{r} 331 \\ 185 \end{array} \right] % \end{align} As shown in Difference between orthogonal projection and least squares solution, the normal equations solution in the pseudoinverse solution. but we need the rest of the minimizers on the dashed line.
Knowing that $$\mathcal{N}\left( \mathbf{A}^{*} \right) = \text{span } \left\{ \ % \left[ \begin{array}{r} 7 \\ -10 \\ 0 \\ 0 \\ 3 \end{array} \right], \ % \left[ \begin{array}{r} 4 \\ -7 \\ 0 \\ 3 \\ 0 \end{array} \right],\ % \left[ \begin{array}{r} 1 \\ -4 \\ 3 \\ 0 \\ 0 \end{array} \right]\ \right\} %$$ the full least squares solution is $$x_{LS} = % \frac{1}{294} \left[ \begin{array}{r} 331 \\ 185 \end{array} \right] % + \alpha % \left[ \begin{array}{r} 7 \\ -10 \\ 0 \\ 0 \\ 3 \end{array} \right] % + \beta % \left[ \begin{array}{r} 4 \\ -7 \\ 0 \\ 3 \\ 0 \end{array} \right] % + \gamma % \left[ \begin{array}{r} 1 \\ -4 \\ 3 \\ 0 \\ 0 \end{array} \right]$$ where $\alpha$, $\beta$, and $\gamma$ are arbitrary complex constants.
Imagine you have two data points $$(2,4)$$ and $$(2,6)$$. Now, if we put the slope formula into matrix form, we will have $$\begin{pmatrix}2 & 1 \\2 & 1 \\\end{pmatrix}$$ as our $$X$$ matrix. First column is for the input $$X$$ and second for the slope "$$b$$". We set "$$b$$" as one because we want to give it freedom to move (otherwise, $$0$$ would kill the "$$b$$" out of the formula, $$mx = y$$). Then, we have the vector $$(M ,B)$$ for our unknown coefficients. These two matrices multiplied will give the output vector $$(4 ,6)$$.
Now, try to think of how you would graph through these two data points. The problem is that there are infinite solutions depending on what you set the bias "$$b$$" to be equal to. Why do we have infinite least square solutions? Well, The column space is not full rank and we have a dependency!
This is the most intuitive way to think in my opinion. Once you understand in 2d, you just "rely" it will hold on to $$n$$ dimensions (and there is no reason to not hold). | 2,445 | 7,416 | {"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": 14, "wp-katex-eq": 0, "align": 3, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.0625 | 4 | CC-MAIN-2021-25 | latest | en | 0.907991 |
https://www.idealhometuition.com/2014/04/motion-in-straight-line-ncert-solutions_2402.html | 1,618,788,923,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038862159.64/warc/CC-MAIN-20210418224306-20210419014306-00372.warc.gz | 895,845,485 | 11,110 | ## Pages
### Motion in a straight line NCERT Solutions Class 11 Physics - Solved Exercise Question 3.16
Question 3.16
Look at the graphs (a) to (d) (Fig. 3.20) carefully and state, with reasons, which of these cannot possibly represent one-dimensional motion of a particle.
Solution:
(a) The given x-t graph, shown in (a), does not represent one-dimensional motion of the particle. This is because a particle cannot have two positions at the same instant of time.
(b) The given v-t graph, shown in (b), does not represent one-dimensional motion of the particle. This is because a particle can never have two values of velocity at the same instant of time.
(c) The given v-t graph, shown in (c), does not represent one-dimensional motion of the particle. This is because speed being a scalar quantity cannot be negative.
(d) The given v-t graph, shown in (d), does not represent one-dimensional motion of the particle. This is because the total path length travelled by the particle cannot decrease with time. | 229 | 1,016 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.828125 | 3 | CC-MAIN-2021-17 | latest | en | 0.914301 |
http://docplayer.net/21803520-Chapter-22-outside-a-uniformly-charged-sphere-the-field-looks-like-that-of-a-point-charge-at-the-center-of-the-sphere.html | 1,619,126,366,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618039604430.92/warc/CC-MAIN-20210422191215-20210422221215-00290.warc.gz | 27,657,159 | 27,225 | # Chapter 22. Outside a uniformly charged sphere, the field looks like that of a point charge at the center of the sphere.
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1 Chapte.3 What is the magnitude of a point chage whose electic field 5 cm away has the magnitude of.n/c. E E C.5 An atom of plutonium-39 has a nuclea adius of 6.64 fm and atomic numbe Z94. Assuming that the positive chage is distibuted unifomly within the nucleus, what ae the magnitude and diection of the electic field at the suface of the nucleus due to the positive chage. Outside a unifomly chaged sphee, the field looks like that of a point chage at the cente of the sphee. E C ( m) N / C!.8 In Fig. -31, paticle 1 of chage 1 5 and paticle of chage + ae fixed to an x-axis (a) As a multiple of distance L, at what coodinate on the axis is the net electic field of the paticles zeo? (a) Sketch the net electic field lines. E1 E 1 (x,)
2 E 1 E 1 x E E 1 5 x (x L) 5 x (x L) 5 x (x L) x 1 (5 + 1)L 3.11 Figue -34 shows two chaged paticles on an x axis: C at x 3.m and C at x +3.m. What ae the (a) magnitude and (b) diection (elative to the positive diection of the x axis) of the net electic field poduced at point P at y +4.m x (., 4.) (-3.,.) (+3.,.) We can do this poblem most easily by using the full vecto fom fo the electic field due to a point chage. ( 3.) i ˆ + (4. ) ˆ j m ˆ + + 3ˆ i + 4 ˆ j 5 ( (3))ˆ i + (4 ) ˆ j ˆ (3) + 4 5m 3ˆ i + 4 ˆ j 5
3 E E + + E C 4 π ε (5m) 3 ˆ 5 i + 4 ˆ j C 4 π ε (5m) 6 i ˆ + ˆ j N / C ˆ i C 4 π ε (5m) 3 ˆ i + 4 ˆ j 5.13 In Fig. -36, thee paticles ae fixed in place and have chages 1 +e and 3 +e. Distance a 6.µm. What ae the (a)magnitude and (b) diection of the net electic field at point P due to the paticles. 1 E E 3 E 1 3 The fields due to chages 1 and cancel exactly. This leaves us only to calculate the field due to 3. 1 a + a a E N / C at 45 a /.19. Find the magnitude and diection of the electic field at point P due to the electic dipole. You may assume that >>d
4 (,d/ ) + (, ) - (,-d/ ) We begin by witing the exact field. ( ) i ˆ + ( d ) ˆ j ( ) i ˆ + ( ( d )) ˆ j + ( d ) + ( d ) E E + + E ˆ + ˆ ˆ + + i ˆ d ˆ j + ( d ) ˆ i ˆ + d ˆ j + ( d ) We can substitute in. In the last step, since d/ is vey small we ignoe this tem. emembe that the diection of p is in the +j diection in this poblem (fom - to + chage).
5 E E + + E ˆ + ˆ ( + ( d ) ) i ˆ d ˆ j + ( d ) ( + ( d ) ) i ˆ + d ˆ j + ( d ) (- indicates downwad) d ( + ( d ˆ j ) ) 3/ p 3 (1 + ( d ˆ j ) ) 3/ p 3.4 In Fig -44, a thin glass od foms a semicicle of adius 5.cm. Chage is unifomly distibuted along the od, with pc in the uppe half and 4.5 pc in the l lowe half. What ae the (a) magnitude and (b) diection elative to the positive diection of the x axis of the electic field E a P, the cente of the cicle. Because of the symmety in the poblem, we can see that the net field will point downwad. We also can see that the contibution fom the bottom uate cicle is eual to the contibution fom the top uate cicle. Because of this, we only need to compute the downwad component due to the top uate cicle and multiply by. d d E We begin by defining a chage pe unit length π /
6 We now find the component of inteest and integate... d de cosθ de d d dθ The last facto of is because thee ae two uate ings. π / π / d dθ cosθ π / π 8ε tot 4ε cosθ dθ.5 In Fig -45, two cuved plastic ods, one of chage + and one of chage -, fo a cicle of adius in the XY plane.. Te x axis passes though thei connection points and the chage is distibuted unifomly on both ods. What ae the magnitude and diection of the electic field E poduces at P, the cente of the cicle. Because of the symmety in the poblem, we can see that the net field will point downwad. We also can see that the contibution fom the bottom half cicle is eual to the contibution fom the top half cicle. Because of this, we only need to compute the downwad component due to the top half cicle and multiply by. d θ θ de We begin by defining a chage pe unit length
7 π We now find the component of inteest and integate... The last facto of is because thee ae two half ings. d de cosθ de d d dθ π / π / π / π / d dθ cosθ πε tot π / π / πε cosθ dθ.6
8 i de dez i z { Because of the symmety, we only need to compute the field along the axis (called z). de z de cosθ de d + z d 1 dϕ cosθ z + z E z E z π π π de z 1 dϕ ( + z ) 1 z dϕ ( + z ) 3/ 1 z ε ( + z ) 3/ z + z Now that we know the field, we take the deivative with espect to z and set it eual to zeo to find the location of the maximum field. m E z 1z f ( + z ) 3/ de z dz m 1( - z ) f ( + z ) 5/ z 1
9 -3. In Fig. -51, positive chage 7.81pC is spead unifomly along a thin nonconducting od of length L 14.5cm. What ae the (a) magnitude and (b) diection (elative to the x axis of the electic field poduced at a distance 6.cm fom the od along its pependicula bisecto. (,d) -L / L / Fist we wite d fo a little length of chage We then wite the,, ˆ fo the chage d. d dx ( x) + (d ) ( x) i ˆ + (d ) ˆ j ˆ x i ˆ + d ˆ j x + d We can then compute the x and y components of the Electic Field. You may need dx x (a + x ) 3/ a a + x
10 E E x d ˆ d dx d (x + d ) 3/ dx x i ˆ + d ˆ j (x + d ) x + d d L / d + L / 4 L / d + L / 4 d L d + L / 4 πε d L 4d + L dx d (x + d ) 3/ d x d + x L / L / Now we can compute with numbes C C / m.145m d.6m C / m πε.6m.145m 4(.6m) + (.145m) 1.43N / C.33 In Fig -5, a semi infinite non conducting od (that is, infinite in one diection only), has unifom linea chage density. Show that the electic field Ep at point P makes an angle of 45 degees with the od and this esult is independent of the the distance. x d L d E (,- )
11 ( x) + ( ) ( x) i ˆ + ( ) ˆ j ˆ x i ˆ ˆ j x + We can then compute the x and y components of the Electic Field. You may need dx (a + x ) 3/ x a a + x We can compute each integal and then take the limit as L goes to infinity. We find that in this limit, the components ae identical fo all. This means that the atio of these components is 1 and the angle is 45 degees fo all. E E x d ˆ dx (x + ) lim :L E x 1 dx (x + ) 1 + L 1 dx (x + ) 3/ L + L L + L lim L x i ˆ ˆ j x + x x + x dx (x + ) 3/ dx (x + ) 3/ 1 + x x + x L L.34 This poblem is woked out in detail in section 3-7 of the book. We also woked this poblem in detail in class. Please eview the deivation thee.
12 σ C / m.5 1 m z 1 1 m E z σ ε (1 z z + ) N / C.39 An electon is eleased fom est in a unifom electic field of magnitude. # 1 4 N/C. Calculate the acceleation of the electon (ignoe gavitation). F ma E 1.6 # 1 a m E -19 C 9.1 # 1-31 \$. # 14 N/C 3.5 # 1 kg 15 m/s.47 In Millikan s expeiment, an oil dop of adius 1.64 µm and density ρ.851g / cm 3 is suspended in chambe C (Fig. -14) when a downwad electic field E N / C is applied. Find the chage on the dop in tems of e F E If the doplet is suspended, we know that the net foce is zeo. This means that the weight is balanced by an upwad electic foce. We wite this balance mg F E mg E mg mg E We can wite the mass in tems of the volume of the dop.
13 m ρ 4 3 π g cm π ( cm) g kg mg E kg 9.8m / s C N / C 5e
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Spiotechnics! Septembe 7, 2011 Amanda Zeingue, Michael Spannuth and Amanda Zeingue Dieential Geomety Poject 1 The Beginning The geneal consensus of ou goup began with one thought: Spiogaphs ae awesome.
### The Binomial Distribution
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### HW6 Solutions Notice numbers may change randomly in your assignments and you may have to recalculate solutions for your specific case.
HW6 Solutions Notice numbers may change randomly in your assignments and you may have to recalculate solutions for your specific case. Tipler 22.P.053 The figure below shows a portion of an infinitely
### Open Economies. Chapter 32. A Macroeconomic Theory of the Open Economy. Basic Assumptions of a Macroeconomic Model of an Open Economy
Chapte 32. A Macoeconomic Theoy of the Open Economy Open Economies An open economy is one that inteacts feely with othe economies aound the wold. slide 0 slide 1 Key Macoeconomic Vaiables in an Open Economy
### Electric Potential. otherwise to move the object from initial point i to final point f
PHY2061 Enched Physcs 2 Lectue Notes Electc Potental Electc Potental Dsclame: These lectue notes ae not meant to eplace the couse textbook. The content may be ncomplete. Some topcs may be unclea. These
### Worked Examples. v max =?
Exaple iction + Unifo Cicula Motion Cicula Hill A ca i diing oe a ei-cicula hill of adiu. What i the fatet the ca can die oe the top of the hill without it tie lifting off of the gound? ax? (1) Copehend
### Problems of the 2 nd and 9 th International Physics Olympiads (Budapest, Hungary, 1968 and 1976)
Poblems of the nd and 9 th Intenational Physics Olympiads (Budapest Hungay 968 and 976) Péte Vankó Institute of Physics Budapest Univesity of Technology and Economics Budapest Hungay Abstact Afte a shot | 10,151 | 33,555 | {"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} | 3.34375 | 3 | CC-MAIN-2021-17 | latest | en | 0.857227 |
https://overnightwriters.com/2022/04/28/what-is-a-salt-bridge-and-how-does-it-work-are-there-other-devices-used-in-galvanic-cells-instead-of-a-salt-bridge/ | 1,679,452,550,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296943749.68/warc/CC-MAIN-20230322020215-20230322050215-00088.warc.gz | 517,575,680 | 20,552 | # What is a salt bridge and how does it work? Are there other devices used in galvanic cells instead of a salt bridge?
A salt bridge is an inverted U-tube that contains an electrolyte and connects the two half-cells in a galvanic cell.
The salt bridge maintains electrical neutrality in the internal circuit.
See
Consider the galvanic cell shown below.
If there were no salt bridge, an excess of positive charge (##”Zn”^(2+)## ions) would build up in the left hand cell. This would oppose more ##”Zn”^(2+)## ions going into solution.
The ##”Cu”^(2+)## ions plating out on the copper would lead to an excess of negative charge (##”SO”_4^(2-)## ions) in solution. This would oppose the loss of more ##”Cu”^(2+)## ions.
The flow of electrons through the wire would soon come to a halt
.
The salt bridge keeps the contents of each half-cell separate and allows the flow of ions to maintain a balance of charge between them.
As positive charge builds up in the left hand cell, ##”SO”_4^(2-)## ions can flow in from the bridge and ##”Zn”^(2+)## ions can flow into the bridge.
As negative charge builds up in the right hand cell, ##”K”^+## ions can flow in from the bridge and ##”SO”_4^(2-)## ions can flow into the bridge.
Cations are flowing in one direction, and anions are flowing in the opposite direction. This countercurrent flow of ions is the internal circuit.
The process continues until the system reaches equilibrium.
Instead of a salt bridge, you can use a porous barrier of clay or fritted glass.
Some labs use beakers joined by a fritted glass disk.
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# M02 Q2
Author Message
Manager
Joined: 27 Dec 2009
Posts: 164
Kudos [?]: 133 [0], given: 3
### Show Tags
30 Apr 2010, 02:19
If x and y are positive integers, is x^16 -y^8 +345(y^2) divisible by 15?
1. x is a multiple of 25, and y is a multiple of 20
2. y = x^2
It is obvious that statement 2 is sufficient.
From Statement 1 , we can also conclude that it is not divisable by 3 and therefore not divisible by 15 .
OA :
[Reveal] Spoiler:
B
What am I missing ?
Kudos [?]: 133 [0], given: 3
CIO
Joined: 02 Oct 2007
Posts: 1216
Kudos [?]: 989 [0], given: 334
### Show Tags
30 Apr 2010, 02:24
Hi,
This question has already been discussed here:
m02-02-not-convinced-with-method-used-to-explaing-the-ans-76875.html
If you can't find the discussion you need in the thread above, please use Forum Search or Custom Google Search available at the top of every page.
Please do not create duplicate topics. Thank you for cooperation .
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Kudos [?]: 989 [0], given: 334
Manager
Joined: 27 Dec 2009
Posts: 164
Kudos [?]: 133 [0], given: 3
### Show Tags
30 Apr 2010, 03:24
dzyubam wrote:
Hi,
This question has already been discussed here:
m02-02-not-convinced-with-method-used-to-explaing-the-ans-76875.html
If you can't find the discussion you need in the thread above, please use Forum Search or Custom Google Search available at the top of every page.
Please do not create duplicate topics. Thank you for cooperation .
Hi,
Thanks for pointing out the correct thread. Do the search always - this time it was with no luck in finding the right thread
I should have tried more.Apology .
Thanks once more.
Kudos [?]: 133 [0], given: 3
Re: M02 Q2 [#permalink] 30 Apr 2010, 03:24
Display posts from previous: Sort by
# M02 Q2
Moderator: Bunuel
Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®. | 833 | 3,029 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.421875 | 3 | CC-MAIN-2017-51 | latest | en | 0.880394 |
https://askworksheet.com/subtracting-decimals-worksheet/ | 1,723,160,011,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640741453.47/warc/CC-MAIN-20240808222119-20240809012119-00062.warc.gz | 77,041,933 | 26,965 | # Subtracting Decimals Worksheet
Add and subtract decimals up to 3 digits. This decimals worksheet will produce addition and subtraction problems.
Subtracting Decimals Horizontal Format Mixed Decimals Adding Decimals Subtracting Decimals
### This worksheet contains a mixture of addition and subtraction problems with decimal numbers.
Subtracting decimals worksheet. Our grade 5 addition and subtraction of decimals worksheets provide practice exercises in adding and subtracting numbers with up to 3 decimal digits. Add and subtract decimals these grade 6 decimals worksheets provide practice in adding and subtracting decimals of varying lengths a skill for which pencil and paper practice is critical to attain mastery. Worksheets math grade 5 decimals addition subtraction.
Quick links to download preview the below listed worksheets. This assemblage of 350 printable worksheets is designed to provide ample practice in subtracting decimals the skills included here provide the 4th grade 5th grade and 6th grade children practice in learning decimal subtraction in both formats horizontal and vertical rows and columns and to subtract decimals using tools like number lines grids and base ten block models and much more. 21 decimal subtraction worksheets are given in this section with numbers varying from 1 digit to 5 digits and decimal places ranging between 1 and 3.
1 digit 1 decimal place 1 digit 2 decimal place 1 digit 3 decimal place 2 digit 2 decimal place 3 digit 2 decimal place 4 digit 3 decimal place 5 digit 3 decimal place. The hundredths subtraction problems are well suited for practice with subtracting monetary amounts the final set of worksheets includes a mixture of problems with different place valued arguments. You may select up to 25 problems per worksheet.
These worksheets introduce various types of subtraction problems with decimals including tenths hundredths and thousandths problems. Just redefine the blocks so the big block is a one the flat is a tenth the rod is a hundredth and the little cube is a thousandth. Model and subtract decimals using base ten blocks so students can see how decimals really work.
Three levels of difficulty with 5 worksheets each. To check the skills acquired attempt this set of printable worksheets with mixed decimal numbers. Download the set 15 worksheets.
Sample grade 6 decimal subtraction worksheet. Subtracting decimals worksheets base ten blocks can be used for decimal subtraction. Multiplication worksheets with decimals.
It may be configured for 1 2 and 3 digits on the right and up to 4 digits on the left of the decimal. Decimals worksheets addition and subtraction worksheets with decimals vertical format. Subtraction worksheets with decimals these decimals worksheets may be configured for 1 2 and 3 digits on the right of the decimal and up to 4 digits on the left of the decimal subtraction problems.
Subtracting decimals column subtraction mixed review. Subtract the decimals in vertical format column and find the difference. Requires students to re write each horizontal problem in vertical notation.
You may select up to 25 subtraction problems for these decimals worksheets.
Adding And Subtracting Hundredths A Decimals Worksheet Decimals Worksheets Decimals Subtraction
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Adding And Subtracting Decimals With Up To Two Places Before And After The Decimal A Decimals Worksh Subtracting Decimals Decimals Worksheets Adding Decimals | 968 | 5,099 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.40625 | 3 | CC-MAIN-2024-33 | latest | en | 0.840028 |
https://fastresearchers.com/project-management-mathematics-homework-help/ | 1,660,445,543,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882571993.68/warc/CC-MAIN-20220814022847-20220814052847-00609.warc.gz | 266,195,050 | 18,641 | # Project management | Mathematics homework help
1) The Reynolds Construction Company has received a contract to build a water purification plant; managers at Reynolds have identified the following tasks to assist them with the construction of the remote control building. In addition to identifying the tasks needed for this project, they estimated the duration and direct costs associated with completing each task at its “Normal” point (the duration with the minimum direct costs) and the duration and direct costs associated with the “Crash” point (that is, if the task is maximally compressed). The managers assume that the marginal compression costs are constant (for example, the marginal compression cost for task B would be (\$26,000 – \$14,000)/(6 – 4 weeks) = \$6000/wk).
Normal Crash
Task Description Predecessor Duration(wks) Cost(\$) Duration(wks) Cost(\$
A Procure materials Start 3 5,000 2 10,000
B Prepare site Start 6 14,000 4 26,000
C Dpt approval Start 2 2,500 1 5,000
D Prebabricate bldg. A 5 10,000 3 18,000
E Obtain city approvals C 2 8,000 2 8,000
F Install Lines A 7 11,000 5 17,500
G Erect bldg. B,D,E 4 10,000 2 24,000
Totals 61,000 108,000
If all tasks are performed at their “normal” times as indicated in the table above, the project is estimated to cost Reynolds a total of \$61,000 in direct costs and result in an expected project makespan of 12 weeks. However, managers at Reynolds want to explore the time-cost trade-offs associated with compressing the project and have come to you for assistance.
Specifically, Reynolds’ managers have asked you to do the following. Starting with all tasks performed at their normal duration (such that the project makespan is 12 weeks), reduce the makespan of the project by one week at a time. For each iteration (e.g., 11 weeks, 10 weeks, etc), find the task(s) that should be compressed to minimize the marginal increase in direct costs at each iteration. Continue compressing the project until at least one critical path exists with all tasks set at their crash times (that is, no further project compression is possible). Construct a graph showing the total direct cost (on the Y axis) versus the project makespan (on the X axis). Clearly indicate the minimum possible project makespan.
2) The managers at Reynolds Construction Company now realize that their calculations in problem #1 do not include overhead/indirect costs that they estimate at \$5100/week (this cost includes the rental fee for security fencing, an overhead construction crane needed for the duration of this project, and your services as manager of this construction project). They also have signed a contract that calls for a penalty cost of \$3200 per week for each week that the makespan exceeds ten weeks. Given these costs, formulate and solve a linear programming model that finds the baseline schedule that minimizes the total costs of this project (where total costs are the sum of direct, indirect/overhead, and penalty costs).
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http://slideplayer.com/slide/4393700/ | 1,638,209,358,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964358786.67/warc/CC-MAIN-20211129164711-20211129194711-00170.warc.gz | 76,316,451 | 24,019 | Presentation on theme: "CHAPTER 1 Economic Models: Trade-offs and Trade"— Presentation transcript:
Models in Economics A model is a simplified representation of a real situation that is used to better understand real-life situations. Create a real but simplified economy Ex.: Cigarettes in World War II prison camps Simulate an economy on a computer Ex.: Tax models, money models The “other things equal” assumption means that all other relevant factors remain unchanged. Note to the instructor: Mentioning “Ceteris Paribus”, the Latin word for “other things equal” might be interesting for some students.
Trade-offs: The Production Possibility Frontier (PPF)
The production possibility frontier (PPF) illustrates the trade-offs facing an economy that produces only two goods. It shows the maximum quantity of one good that can be produced for any given production of the other. The PPF improves our understanding of trade-offs by considering a simplified economy that produces only two goods by showing this trade-off graphically.
Tom’s Trade-offs: The Production Possibility Frontier
Increasing Opportunity Cost
Economic Growth Production is initially at point A (20 fish and 25 coconuts), it can move to point E (25 fish and 30 coconuts). Economic growth results in an outward shift of the PPF because production possibilities are expanded. The economy can now produce more of everything.
Tom and Hank’s Opportunity Costs of Fish and Coconuts
Tom’s Opportunity Cost Hank’s Opportunity Cost One fish 3/4 coconut 2 coconuts One coconut 4/3 fish 1/2 fish
Specialize and Trade Both castaways are better off when they each specialize in what they are good at and trade. It’s a good idea for Tom to catch the fish for both of them, because his opportunity cost of a fish in terms of coconuts not gathered is only 3⁄4 of a coconut, versus 2 coconuts for Hank. Correspondingly, it’s a good idea for Hank to gather coconuts for the both of them.
Transactions: The Circular-Flow Diagram
Trade takes the form of barter when people directly exchange goods or services that they have for goods or services that they want. The circular-flow diagram is a model that represents the transactions in an economy by flows around a circle.
The Circular-Flow Diagram
Circular-Flow of Economic Activities
A household is a person or a group of people that share their income. A firm is an organization that produces goods and services for sale. Firms sell goods and services that they produce to households in markets for goods and services. Firms buy the resources they need to produce—factors of production—in factor markets.
Growth in the U.S. Economy from 1962…
…to 1988
Using Models Positive economics is the branch of economic analysis that describes the way the economy actually works. Normative economics makes prescriptions about the way the economy should work. A forecast is a simple prediction of the future.
Using Models Economists can determine correct answers for positive questions, but typically not for normative questions, which involve value judgments. The exceptions are when policies designed to achieve a certain prescription can be clearly ranked in terms of efficiency. It is important to understand that economists don’t use complex models to show “how clever they are,” but rather because they are “not clever enough” to analyze the real world as it is.
When and Why Economists Disagree
There are two main reasons economists disagree: They may disagree about which simplifications to make in a model. They may disagree about values. | 730 | 3,555 | {"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} | 2.828125 | 3 | CC-MAIN-2021-49 | latest | en | 0.893171 |
https://acmore.cc/problem/LOCAL/7406 | 1,718,436,405,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861584.65/warc/CC-MAIN-20240615062230-20240615092230-00175.warc.gz | 63,554,797 | 8,181 | # 7406：Percolation
###### 题目描述
SSPUer is studying a system called percolation. We model a percolation system using an n-by-n grid of sites. Each site is either open or blocked. A full site is an open site that can be connected to an open site in the top row via a chain of neighboring (left, right, up, down) open sites. We say the system percolates if there is a full site in the bottom row. In other words, a system percolates if we fill all open sites connected to the top row and that process fills at least one open site on the bottom row.
(For the insulating/metallic materials example, the open sites correspond to metallic materials, so that a system that percolates has a metallic path from top to bottom, with full sites conducting. )
Now, giving you the size of the system n, the number of open site m and the position (i,j) of each open site. The position of the top-left grid is (1,1) and the position of the bottom-right grid is (n,n). Your task is to calculate the sum of full sites and tell whether the system percolates or not.
###### 输入解释
For each test case, the first line contains two integers n, m, representing the size of the system and the number of open site.
There are m lines following, each contains two integer i, j, representing the position (i,j) is an open site. (1 <= i <= j <= n)
n<=500
###### 输出解释
For each test case, output a single line containing an integer and a string, representing the sum of full sites and whether the system percolates, if the system percolates output “YES”, if not output “NO”.
###### 输入样例
2 2
1 1
2 2
###### 输出样例
1 NO
###### 提示
The sample is shown on the picture below, which has 1 full site and the system does not percolate. (position (2,2) is open but not full)
时间上限 内存上限 1000 MS 128 MB | 461 | 1,756 | {"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} | 2.765625 | 3 | CC-MAIN-2024-26 | latest | en | 0.879481 |
https://cs.stackexchange.com/questions/18067/task-dependency-combinatorics | 1,582,739,933,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875146414.42/warc/CC-MAIN-20200226150200-20200226180200-00509.warc.gz | 327,512,058 | 32,406 | here is a competitive programming question:
[Task, Dependency(s)] : [1, NA], [2,1], [3,2], [4,1], [5,4], [6, 3 and 5], [7,6], [8,7], [9,6], [10, 8 and 9].
I inferred the following:
• Any sequence will always start with 1, since it has no dependencies.
• 6 will always be in the 6th position of any sequence.
• 10 will always be at the last position.
Then, by trial and error, and listing all possibilities for the two separate parts of the sequence (1 _ _ _ _ 6 and 6 _ _ _ 10), I got 6x3 = 18 possibilities. However, for a larger set of data, these deductions would not be easy. What is the way to calculate this logically and find the number of possibilities, and how can this be integrated into a program?
(I have tried to represent the question as clearly as possible, but you can visit this link to view the question (Q. No 4): http://www.iarcs.org.in/inoi/2013/zio2013/zio2013-qpaper.pdf)
I am a high school student preparing for a programming competition, and I haven't taken many courses on algorithm design, so this might be a trivial question - please excuse me!
• Seems like you need to count the number of linear (or total) orders on the set $\{1,\dots,N\}$ that contain a given partial order on $\{1,\dots,N\}$. See mathoverflow.net/q/45875/7252 for discussion. – András Salamon Nov 16 '13 at 13:49
It is a good question, certainly not trivial.
This problem is essentially counting the number of topological orderings. A topological ordering/sort is basically a valid ordering of completing the tasks in a dependency graph.
According to the link that @AndrásSalamon put into his comment, it is #P-complete: this means it is very hard to actually compute for a large number of tasks and dependencies.
They are expecting you to write a program that does it for just the small number of tasks that they give you. You probably need to do brute-force; i.e. try all possible orders, check if it is valid, and count them.
Alternatively, some smart variation of brute force, such as enumerating all possible topological orderings, and counting them.
Frequently this is the type of question asked by a competition:
• An impossibly difficult question if extended to large input size,
• Therefore, some sort of brute-force must be used,
• But a smart variation of brute-force,
• That can depend on the specific characteristics and limitations of the input they give you; if you don't take advantage of these limitations, and/or if you don't figure out the variation they want you to do, then your program will over-run its time limits.
You can also take a look at On Computing the Number of Topological Orderings of a Directed Acyclic Graph. To run this, you would make a dependency graph of the input tasks, then count the topological orderings using the algorithm(s) in this paper. I am not sure if this would finish in time on the input data you have; but one way to know would be to try it. | 713 | 2,911 | {"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} | 3.546875 | 4 | CC-MAIN-2020-10 | latest | en | 0.935495 |
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# Eye movement occurs more rapidly during dreams than when
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Eye movement occurs more rapidly during dreams than when [#permalink]
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16 Sep 2010, 19:03
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Eye movement occurs more rapidly during dreams than when waking.
a) more rapidly during dreams than when waking
b) when dreaming more rapidly than waking hours
c) more rapidly during dreaming than waking
d) more rapidly during dreams than during the period of time when a person is awake
e) more rapidly when dreaming than when waking
OA will follow...
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16 Sep 2010, 19:32
I'm stuck between D and E.
d) more rapidly during dreams than during the period of time when a person is awake
Well formed sentence. But we are talking about eye movement when waking up not when a person is entirely awake.
e) more rapidly when dreaming than when waking
This seems to be the best answer. Although it can be better written...more rapidly when dreaming than when a person is waking up.
I go for E.
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16 Sep 2010, 20:15
i think it is E. When is used correctly and it is parallel on 2nd opart
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17 Sep 2010, 02:59
Though D is wordy but it looks better to me.
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17 Sep 2010, 13:46
OA IS D
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17 Sep 2010, 14:05
i would think it would be E... but just noticed OA is D
close call. the only reason i can think of for D is that waking cannot be just used as a parallel with dreaming since it is usually used as a verb and for a short time "while awake" is what is usual for the waking hours.... anyone else can explain how the lengthy D is right?
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17 Sep 2010, 16:30
B incorrectly associates ‘more rapidly’ with dreaming rather than with eye movement.a,c and e incorrectly refer to the process of waking rather than the period of time of being awake.Hence d is correct.
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18 Sep 2010, 00:21
i was also confused between d and e,
i believe there is nothing called as ' i am waking
it has to be ' i am awake'.
d seems to be the only right answer, grammatically.
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18 Sep 2010, 00:45
weird question. I felt C is correct? whats wrong in that?
What is the source?
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18 Sep 2010, 05:17
Source:Manhattan OG verbal Companion
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19 Sep 2010, 11:38
I picked D because 'waking' and 'when waking' is awkward in C and E.
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Re: Eye movement [#permalink] 19 Sep 2010, 11:38
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# Eye movement occurs more rapidly during dreams than when
Moderators: GMATNinjaTwo, GMATNinja
Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®. | 1,648 | 5,597 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2017-51 | latest | en | 0.908359 |
http://theoretical-physics-digest.wikia.com/wiki/Supernova_Remnants | 1,534,553,338,000,000,000 | text/html | crawl-data/CC-MAIN-2018-34/segments/1534221213247.0/warc/CC-MAIN-20180818001437-20180818021437-00654.warc.gz | 408,123,617 | 47,816 | ## FANDOM
148 Pages
Supernova remnants is a collective name for changes in the environment of the progenitor due to the explosion. It is usually divided into three phases. The first is the ejecta dominated stage. At this stage the ejecta are in ballistic motion (constant velocity). It ends when the swept up mass is the same order of magnitude as the ejecta mass. For a supernova energy $E_e = 10^{51} erg$, ejecta mass of $M_e = 1 M_{\odot}$ and an environment where the density is uniform and equal to $n \approx 1 \frac{1}{cm^3}$ , the end of the ejecta dominated stage is at a radius of about 3.4 parsec. Since the velocity is uniform and can be approximated as $v_e \approx \sqrt{\frac{E_e}{M_e} }$, this phase lasts for about 500 years.
Next comes the Sedov Taylor stage. At this stage the interior of the shock wave evolves adiabatically. This stage ends when the energy loss through radiation becomes important. The cooling function is approximately constant in a wide range of temperatures and we will assume that it is equal to $\Lambda \approx 10^{-22} erg \cdot cm^3 / s$[1]. The density of the supernova is assumed to remain the same (because the explosion neither generates nor destroys particles). Equating the cooling time to the age of the supernova yields
$t \approx \frac{E_e}{\Lambda n^2 V} \approx \frac{E_e}{\Lambda n^2 R^3}$
The relation between the radius and the time is given by
$R \approx \left( \frac{E_e}{\rho} \right)^{1/5} t^{2/5}$
Where $mu$ is the atomic mass and $\rho = \mu n$.Combining these two equations and solving for the time yields
$t \approx \frac{E_e^{2/11} \mu^{3/11}}{n^{7/11} \Lambda^{5/11} } \approx 200\cdot 10^3 years$
This corresponds to a radius of about 40 parsec. A more detailed calculation[2] suggests that the end of the Sedov Taylor phase occurs an order of magnitude earlier.
The final stage is the radiation dominated stage. At this stage the remnants radiate the excess energy and merge with the surrounding ISM. In principle, it should take the same time as the duration of the ST phase. However, at low temperatures the cooling function is smaller, so it might take longer.
## References Edit
1. Sutherland, Ralph S & Dopita, M. A., Cooling Functions for low - density astrophysical plasmas, ApjS 88, 253 (1993)
2. J. M. Blonding et al, Transition to the radiative phase in supernova remnants, ApJ 500:342-354 (1998) | 638 | 2,392 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.140625 | 3 | CC-MAIN-2018-34 | longest | en | 0.86399 |
http://physicsinventions.com/question-about-electric-field/ | 1,563,361,235,000,000,000 | text/html | crawl-data/CC-MAIN-2019-30/segments/1563195525136.58/warc/CC-MAIN-20190717101524-20190717123524-00383.warc.gz | 123,462,464 | 9,736 | # Question about Electric field.
1. The problem statement, all variables and given/known data
A point charge $q_{1}=15.00\mu C$ is held fixed in space. From a horizontal distance of $3.00cm$ , a small sphere with mass $4.00*10^{-3}kg$ and charge $q_{2}=+2.00\mu$C is fired toward the fixed charge with an initial speed of $38.0m/s$ . Gravity can be neglected.
What is the acceleration of the sphere at the instant when its speed is $20.0m/s$ ?
Express your answer with the appropriate units.
2. Relevant equations
$a = \frac{k*q_{1}*q_{2}}{m*r^{2}}$
3. The attempt at a solution
$a = \frac{9*10^{9}*15*10^{-6}*2*10^{-6}}{4*10^{-3}*r^{2}}$
$a = 67.5 * \frac{1}{r^{2}}$
By integration,
$v = -67.5 * \frac{1}{r} + C$
Initial conditions: $v = 38 m/s, r = 0.03m$
$38 = -67.5 * \frac{1}{0.03} + C$
$C = 2288$
$v = -67.5 * \frac{1}{r} + 2288$
When $v = 20m/s$,
$20 = -67.5 * \frac{1}{r} + 2288$
$r = 0.029761904$
$a = 67.5 * \frac{1}{0.029761904^{2}}$
$a ≈ 76200 m/s^{2}$
Since both charges are +ve,
$a = -76200 m/s^{2}$
However, the solution is not correct.
May anyone pointing out the errors? Thank you very much.
http://ift.tt/1epvqUa | 429 | 1,152 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.890625 | 4 | CC-MAIN-2019-30 | latest | en | 0.794665 |
http://intelgo.ru/water-saturation/ | 1,670,241,603,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446711016.32/warc/CC-MAIN-20221205100449-20221205130449-00780.warc.gz | 24,302,631 | 7,784 | Water saturation
The fraction of water in a given pore space. Unless otherwise state water saturation is the fraction of formation water in the undisturbed zone. The saturation is known as the total water saturation if the pore space is the total porosity, but is known as effective . Water saturation determination.
Complexities arise because there are a number of independent approaches that can be . Techniques for calculating. Application of each water. We calculate water saturation from the effective porosity and the resistivity log. Hydrocarbon saturation is (one) minus the water saturation. Empirically, it has been found that P has a normal distribution for intervals which are 1per cent saturated . This page is a highly abbreviated version of Chapter on this website.
For more about eater resistivity, water saturation , alternate methods and examples, go to the Main Index Page. The fourth step in a log analysis is to determine the water resistivity since most methods for computing water . However, some authors have used them interchangeably and mean to be the same. Could someone please explain the difference . What is the Archie equation?
Archie developed his famous equation to calculate, from well log parameters, the water saturation (Sw) of the uninvaded zone in a formation next to a borehole. Sw = water saturation of the uninvaded zone . Sw interpretations should be made accounting for h, r3 and pore throat size distribution in the reservoir. The Archie equation is the most widely used method of determining Sw. It can be calculated from the effective porosity and the resistivity of the log.
It is the determination of petrochemical calculations used to quantify the hydrocarbon saturation. Moreover, it can be described . Saturation can be expressed as volume per volume units. This paper presents the method used to calibrate the log data to the OBC water saturations and reviews the methods used to derive resistivity and formation factor values. It also presents field measurements of as-received core-plug resistivity and water – saturation values from two oil-base-mud (OBM) -cored wells. Etissue increases with increasing turgor pressure, which has been verified for a growing tissue.
It is also known to depend upon the size of the tissue. Hitherto, many scientists attempted to estimate accurately water saturation from well-logging data which has a continuous record without losing information. Therefore, various model were introduced to relate reservoir properties . Formation porosity and water saturation play important role in evaluating potential oil reservoirs and for drafting development plans for new oil fields. Dielectric measurements of moist soils are used to determine water saturations. The degree of satu- ration of water is determined using a mixing law for the composite structure (soil, water, and air) which is verified experimentally.
Online calculator, figures and tables showing water saturation (vapor) pressure at temperatures ranging from to 3°C and from to 700°F – in Imperial and SI Units. Перевод контекст water saturation c английский на русский от Reverso Context: The emission rate depends on a complex array of factors like soil structure, pH, temperature, type of crop, water saturation and nitrogen fertilizer. | 652 | 3,294 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.03125 | 3 | CC-MAIN-2022-49 | latest | en | 0.921044 |
https://daltyboy11.github.io/evm-puzzles/ | 1,713,813,593,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296818337.62/warc/CC-MAIN-20240422175900-20240422205900-00363.warc.gz | 163,349,790 | 4,025 | # EVM Puzzles
The EVM puzzles test your understanding of EVM bytecode. I went down the rabbit hole to understand contract bytecode and here I present to you my solutions. A helpful source for understanding how opcodes interact with the stack can be found here.
# Puzzle 1
The lines
CALLVALUE
JUMP
To jump to the destination we need CALLVALUE = 0x08.
# Puzzle 2
We want to choose a call value and code size such that CODESIZE - CALLVALUE == 0x06. The codesize is 10 bytes so we need the call value to be 4.
# Puzzle 3
To jump to line 4 the call data size must be 4 bytes. Any 4 byte payload will do, like 0x00000000.
# Puzzle 4
For this challenge we want the XOR of two arguments to jump to the desired line. It will be helpful to look at CALLVALUE and CODESIZE, and the line’s binary representations.
The jump destination is line 0A, or 1010 (base 2). The code size is 12 bytes, or 1100 (base 2). We want to choose a value x such that x ^ 1100 == 1010. The answer is 6.
# Puzzle 5
For this challenge I found it helpful to visualize the stack after each instruction.
CALLVALUE => stack = [CALLVALUE]
DUP1 => stack = [CALLVALUE, CALLVALUE]
MUL => stack = [CALLVALUE**2]
PUSH2 0100 => stack = [0x0100, CALLVALUE**2]
EQ => ?
What do we have on the stack after EQ? Well, that depends on CALLVALUE. Recall the EQ opcode consumes two items from the stack and produces one. If the two items are equal it will push 0x01 onto the stack, otherwise it will push 0x00 onto the stack. In our case we want it to push 0x01 so we must set CALLVALUE = sqrt(0x0100) = 16 (base 10)
# Puzzle 6
The CALLDATALOAD opcode takes element x off the stack and reads the 32 byte word in calldata starting at offset x. In this challenge we’re reading from offset 0x00. We want the first 32 bytes of call data to be the jump destination: 0x000000000000000000000000000000000000000000000000000000000000000a
# Puzzle 7
This challenge uses the calldata to deploy a contract with the CREATE opcode. We essentially need to write creation + runtime contract bytecode where the deployment bytecode is only a single byte in length.
## Contract creation bytecode
A quick primer on contract creation bytecode: the bytecode loads the runtime bytecode into memory and returns it. We could write something like this
#################
Creation bytecode
#################
00 PUSH1 0x01 (0x6001)
02 PUSH1 0x0c (0x600c)
04 PUSH1 0x00 (0x6000)
06 CODECOPY (0x39)
07 PUSH1 0x01 (0x6001)
09 PUSH1 0x00 (0x6000)
0b RETURN (0xF3)
################
Runtime bytecode
################
0c INVALID (0xFE)
This code copies 1 byte of code to memory starting at line 0c and returns that byte. If we sent this bytecode with a contract creation transaction then it would create a contract whose bytecode is just the single byte 0xFE. This is exactly what we need to solve the puzzle.
As a bytecode string the answer is: 0x6001601160003960016000F3FE.
# Puzzle 8
This is another creation bytecode puzzle. The code calls our newly created contract and jumps only if the CALL opcode returns 0. Let’s look at our opcode reference to figure out how to make CALL return 0: if the call is successful it returns 1, otherwise it returns 0. Therefore our newly created contract should revert. All we have to do is change the runtime bytecode from our previous solutions to be REVERT.
#################
Creation bytecode
#################
00 PUSH1 0x01 (0x6001)
02 PUSH1 0x0c (0x600c)
04 PUSH1 0x00 (0x6000)
06 CODECOPY (0x39)
07 PUSH1 0x01 (0x6001)
09 PUSH1 0x00 (0x6000)
0b RETURN (0xF3)
################
Runtime bytecode
################
0c REVERT (0xFD)
As a bytecode string the answer is: 0x6001600c60003960016000F3FD.
# Puzzle 9
For the first jump we need 0x03 < CALLDATASIZE and for the second jump we need CALLVALUE * CALLDATASIZE = 0x08. To satisfy the constraints we’ll choose a 4 byte long calldata, e.g. 0x00000000 and a value of 2.
# Puzzle 10
For the first jump we need CODESIZE > CALLVALUE. For the second jump we need CALLDATASIZE % 3 == 0 and CALLVALUE + 0x0a = 0x1b. A call value of 15 and calldata 0x000000 will satisfy the constraints.
Written on April 30, 2022 | 1,176 | 4,239 | {"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} | 2.90625 | 3 | CC-MAIN-2024-18 | latest | en | 0.816457 |
https://math.stackexchange.com/questions/1709285/if-a-third-ball-is-drawn-from-urn-u-2-then-what-is-the-probability-that-it-is | 1,621,135,876,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243991659.54/warc/CC-MAIN-20210516013713-20210516043713-00436.warc.gz | 405,058,745 | 38,012 | # If a third ball is drawn from urn $U_2$ then what is the probability that it is white?
An urn $U_1$ has $3$ white balls and $5$ black balls. $4$ balls were drawn from this urn and put into another empty urn $U_2$. Now $2$ balls were drawn from the urn $U_2$ and were found to be white. If a third ball is drawn from urn $U_2$ then what is the probability that it is white?
• Welcome to MSE. Please include your thoughts and efforts (work in progress) in this and future posts. You are more likely to receive positive/constructive feedback that way. Formatting your post helps too. Formatting tips here. – Em. Mar 22 '16 at 20:45
Urn 2 is irrelevant. We can imagine we are looking at the balls as they come out of Urn 1. If the first two are white, there is one white left, so the probability the third is white is $1/6$.
After moving $4$ balls to urn $2$, the probability distribution for the number of white balls is $\{ 0 : \frac{5}{70}, 1: \frac{30}{70}, 2: \frac{30}{70}, 3: \frac{5}{70}\}$. (For example, we can choose 2 white balls in $\binom32 \binom52 = 30$ ways.)
So the chances of pulling $2$ white balls out of urn $2$ are $$\frac{5}{70}\frac34\frac23 + \frac{30}{70}\frac12\frac13= \frac5{140}+\frac{10}{140} = \frac{30}{280}$$ And the chances of pulling $2$ white balls out of urn $2$ and then pulling a third white ball out are $$\frac{5}{70}\frac34\frac23\frac12 = \frac5{280}$$ So the probability of that second event given that the first event happened is $$\frac{5}{30} = \frac16$$ | 457 | 1,504 | {"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} | 4.625 | 5 | CC-MAIN-2021-21 | latest | en | 0.940663 |
https://kidsworksheetfun.com/special-right-triangles-worksheet-30-60-90-answers/ | 1,708,697,171,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474412.46/warc/CC-MAIN-20240223121413-20240223151413-00253.warc.gz | 333,722,591 | 25,371 | # Special Right Triangles Worksheet 30-60-90 Answers
1 a 2 2 b 45 2 4 x y 45 3 x y 3 2 2 45 4 x y 3 2 45 5 6 x y 45 6 2 6 y x 45 7 16 x y 60 8 u v 2 30 1. The most frequently studied right triangles the special right triangles are the 30 60 90 triangles followed by the 45 45 90 triangles.
Special Right Triangles Formulas 30 60 90 And 45 45 90 Special Right Triangles Examples Pictures And Interac Math Methods Studying Math Mathematics Education
Special right triangles worksheet 30-60-90 answers. Displaying top 8 worksheets found for 30 60 90 answer key. Saperstein established and sent on the road in 1927. Use the answers to reveal the name of the team that abraham m.
Leave your answers as radicals in simplest form. The perimeter of a square is 24 cm. Math worksheets 30 60 90 triangles a 30 60 90 triangle is a special type of right triangle.
Special right triangles 45 45 90 30 60 90 created aug. An equilateral triangle has a side len th of 0 inches. Special right triangles date period find the missing side lengths.
Special right triangles use the 30 60 90 and 45 45 90 triangle relationships to solve for the missing sides. 2 2019 by user linda gregory i use this activity to have my students discover the relationships between the sides on 45 45 90 and 30 60 90 triangles. There are two special right triangles that will continually appear throughout your study of mathematics.
Day 3 special right triangles 30 60 90 warm up use the information marked on the figure to find the value of x. What is special about 30 60 90 triangles is that the sides of the 30 60 90 triangle always have the same ratio. Then find the requested measure.
1 date period v j2o0c1x5w ukvuvt at isgomftt wpahrgex rlpleck q l aul ln zr isgqhotksv vroexswesrwvke d 1 find the missing side lengths. Long leg 14 short leg 20 600 12 00 600 600 600 i2 sketch the figure that is described. Some of the worksheets for this concept are work 4 special 30 60 90 triangles find the missing side leave your answers as infinite geometry answer keys to special right triangles bar graph of ticket sales 30 60 90 triangle practice 65 75 80 90 100 40 60 70 90 30 60 90 right triangles and algebra examples.
1 12 m n 30 2 72 ba 30 3 x y 5. Although all right triangles have special features trigonometric functions and the pythagorean theorem. Worksheet by kuta software llc geometry 30 60 90 triangle practice name id.
Special right triangles hypotenuse 2n hypotenuse 2 short leg long leg leg find the value of x and y in each triangle. Therefore if we are given one side we are able to easily find the other sides using the ratio of 1 2 3. Express all answers in simplest radical form.
Find the length of the diagonal of the square. The 30º 60º 90º triangle and the 45º 45º 90º triangle the special nature of these triangles is their ability to yield exact answers instead of decimal approximations when dealing with trigonometric functions.
Right Triangle Speical Right Triangles Interactive Notebook Pages Teaching Geometry Studying Math Right Triangle
Special Right Triangles Interactive Notebook Page 30 60 90 Trigonometry Worksheets Teaching Geometry Right Triangle
Right Triangles Special 30 60 90 Riddle Practice Worksheet Practices Worksheets Right Triangle Geometry Worksheets
Right Triangles Special Right Tris Notes Practice Task Cards Riddle Bundle Right Triangle Special Right Triangle Math Notes
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Day 1 Hw Special Right Triangles 45 45 90 30 60 90 Special Right Triangle Right Triangle Problem Solving
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Special Right Triangles 30 60 90 And The 45 45 90 Special Right Triangle Right Triangle Sentence Structure | 1,215 | 5,077 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.625 | 4 | CC-MAIN-2024-10 | latest | en | 0.816988 |
https://iphone.apkpure.com/strength-of-material/com.twominds.strengthofmaterials | 1,571,847,670,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570987834649.58/warc/CC-MAIN-20191023150047-20191023173547-00501.warc.gz | 532,754,999 | 12,691 | Strength of material
# Strength of material
Free
0
0 Ratings
Release Date
2018-05-14
Size
22.9 MB
### Screenshots for iPhone
Strength of material Description
The app is a complete free handbook of Strength of Materials with diagrams and graphs.
* It is part of Mechanical Engineering which brings important topics, notes, news & blog on the subject.
* Set Reminder for Particular topic to read that Topic at your own choice time.
* It has Quiz for General Test to Grow your IQ Level.
* Through the search functionality within the app one can review any subject related concept instantly.
*Download the App as quick reference guide & ebook on this Engineering mathematics subject.
It covers 80 topics of Maths in detail. These 80 topics are divided in 5 chapters.
It provides quick revision and reference to the topics like a detailed flash card. Each topic is complete with diagrams, equations and other forms of graphical representations for easy understanding.
The application serves to both engineering students and professionals.
Some of the topics Covered in this application are:
1. Tensile Stress
2. Compressive Stress
3. Shear Stress
4. Volumetric Stress
5. Volumetric Strain
6. Shear Strain
7. Compressive Strain
8. Tensile Strain
9. Stress strain Diagram
10. Thermal Stress
11. Poisson Ratio
12. Three Modulus
13. Temperature Stress in Composite Bar
14. Strain Energy
15. Modulus of Ressilince and Toughness
17. Strain Energy in Impact Load
18. Hook's law
19. Stress, strain and change in length
20. Change in length when one both ends are free
21. Change in length when one both ends are fixed
22. Composite bar in tension or compression
23. Principal Stress and Principal Plane
24. Maximum Shear Stress
25. Theories of Elastic Failure
26. Maximum Principal Stress Theory
27. Maximum Shear Stress Theory
28. Maximum Principal Strain Theory
29. Total Strain Energy per Unit Volume Theory
30. Maximum Shear Strain Energy per Unit Volume Theory
31. Mohr’s Rupture Theory for Brittle Materials
32. Mohr’s Circle
33. Introduction to Bending Moment and Shearing Force
34. Shearing Force, and Bending Moment in a Straight Beam
35. Sign Conventions for Bending Moments and Shearing Forces
36. Bending Of Beams
37. Procedure for Drawing Shear Force and Bending Moment Diagram
38. SFD & BMD of Cantilever Carrying Load at Its One End
39. SFD and BMD of Simply Supported Beam Subjected To a Central Load
40. SFD and BMD of A Cantilever Beam Subjected To U.D.L
41. Simply Supported Beam Subjected To U.D.L
42. Simply Supported Beam Carrying UDL & End Couples
43. Points of Inflection
44. Theory of Bending: Assumption and General Theory
45. Elastic Flexure Formula
46. Beams of Composite Cross Section
47. Flexural Buckling Of a Pin-Ended Strut
48. Rankine-Gordon Formula
49. Comparison of the Rankine-Gordon and Euler Formulae
50. Effective Lengths of Struts
51. Struts and Columns with One End Fixed and the Other Free
52. Thin Cylinder
53. Members Subjected to Axisymmetric Load
54. ANALYSIS: Pressurized thin walled cylinder
55. Longitudinal Stress: Pressurized thin walled cylinder
56. Change in Dimensions: Pressurized thin walled cylinder
57. Volumetric Strain or Change in the Internal Volume
58. Cylindrical Vessel with Hemispherical Ends
59. Thin rotating ring or cylinder
60. Stresses in thick cylinders
61. Stresses in thick cylinders
62. Representation of radial and circumferential strain
63. A thick cylinder with both external and internal pressure
64. The stress distribution within the cylinder wall
65. Methods of increasing the elastic strength of a thick cylinder by pre-stressing
SOM also called also called mechanics of materials is part of materials science & mechanical engineering education courses and technology degree programs of various universities.
Strength of material 2 Update
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Developer Apps | 990 | 4,123 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2019-43 | latest | en | 0.796688 |
https://www.crazy-numbers.com/en/9169 | 1,566,704,763,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027322170.99/warc/CC-MAIN-20190825021120-20190825043120-00396.warc.gz | 773,055,563 | 4,132 | Discover a lot of information on the number 9169: properties, mathematical operations, how to write it, symbolism, numerology, representations and many other interesting things!
## Mathematical properties of 9169
Is 9169 a prime number? No
Is 9169 a perfect number? No
Number of divisors 4
List of dividers 1, 53, 173, 9169
Sum of divisors 9396
## How to write / spell 9169 in letters?
In letters, the number 9169 is written as: Nine thousand hundred and sixty-nine. And in other languages? how does it spell?
9169 in other languages
Write 9169 in english Nine thousand hundred and sixty-nine
Write 9169 in french Neuf mille cent soixante-neuf
Write 9169 in spanish Nueve mil ciento sesenta y nueve
Write 9169 in portuguese Nove mil cento sessenta e nove
## Decomposition of the number 9169
The number 9169 is composed of:
2 iterations of the number 9 : The number 9 (nine) represents humanity, altruism. It symbolizes generosity, idealism and humanitarian vocations.... Find out more about the number 9
1 iteration of the number 1 : The number 1 (one) represents the uniqueness, the unique, a starting point, a beginning.... Find out more about the number 1
1 iteration of the number 6 : The number 6 (six) is the symbol of harmony. It represents balance, understanding, happiness.... Find out more about the number 6
Other ways to write 9169
In letter Nine thousand hundred and sixty-nine
In roman numeral MMMMMMMMMCLXIX
In binary 10001111010001
In octal 21721
In US dollars USD 9,169.00 (\$)
In euros 9 169,00 EUR (€)
Some related numbers
Previous number 9168
Next number 9170
Next prime number 9173
## Mathematical operations
Operations and solutions
9169*2 = 18338 The double of 9169 is 18338
9169*3 = 27507 The triple of 9169 is 27507
9169/2 = 4584.5 The half of 9169 is 4584.500000
9169/3 = 3056.3333333333 The third of 9169 is 3056.333333
91692 = 84070561 The square of 9169 is 84070561.000000
91693 = 770842973809 The cube of 9169 is 770842973809.000000
√9169 = 95.754895436213 The square root of 9169 is 95.754895
log(9169) = 9.1235835080499 The natural (Neperian) logarithm of 9169 is 9.123584
log10(9169) = 3.9623219727296 The decimal logarithm (base 10) of 9169 is 3.962322
sin(9169) = 0.96591518532113 The sine of 9169 is 0.965915
cos(9169) = -0.25885875447054 The cosine of 9169 is -0.258859
tan(9169) = -3.7314371974662 The tangent of 9169 is -3.731437 | 754 | 2,381 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.75 | 4 | CC-MAIN-2019-35 | latest | en | 0.729933 |
https://cdcvs.fnal.gov/redmine/projects/qss/wiki/N02_2015_07_16 | 1,603,385,451,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107879673.14/warc/CC-MAIN-20201022141106-20201022171106-00243.warc.gz | 260,523,712 | 5,094 | # Example N02, results updated to July 16, 2015¶
## Parameters¶
### Physical System Parameters and Constant Values¶
B = 1.0 tesla = 1E-3 (MeV/eplus)*(ns/mm*mm)
K = 10.9181415106 MeV
c = 299.792458
M = 0.510998910 MeV
m = M/c^2
gamma = 22.3662720421293741
v_0 = 0.999*c
### Geant4 Parameters¶
trackMaxSteps = 200000
trackMaxLen = 10 m
stepMax = 0.1 mm
DeltaOne = 1.0e-5 mm
DeltaInt = 1.0e-6 mm
EpsilonMin = 1.0e-3
EpsilonMax = 1.0e-3
MinStep = 1.0e-2 mm
DeltaChord = 0.25 mm
DeltaQrel = 1E-6
DeltaQmin = 1E-9
tf = ?
### Analytic Baseline¶
r = (M/c^2) * gamma * v_0 / B
T = 2*pi / (B/((M/c^2) * gamma))
w = 2*pi / T
x0 = 0
y0 = -r
x(t) = x0 + r * sin(w*t)
y(t) = y0 + r * cos(w*t)
### Geant4 Baseline Parameters¶
No baseline was generated in Geant 4 up to date.
trackMaxSteps =
trackMaxLen =
stepMax =
DeltaOne =
DeltaInt =
EpsilonMin =
EpsilonMax =
MinStep =
DeltaChord =
### Baseline Power Devs / QSS Parameters¶
DeltaQrel = 1E-9
DeltaQmin = 1E-12
tf = ?
x,y,z vs time
x,y,z vs length
xyz 3D line plot
x,y,z vs time
x,y,z vs length
xyz 3D line plot
x,y,z vs time
x,y,z vs length
xyz 3D line plot
x,y,z vs time
x,y,z vs length
xyz 3D line plot
x,y,z vs time
x,y,z vs length
xyz 3D line plot
## Analyses:¶
### Metrics¶
#### Power Devs / QSS¶
NumIntegrationSteps (per integrator and total)
SimWallClockTime (with all logging/outputs disabled)
#### Geant4¶
NumInvocations to {RightHandSide,OneGoodStep,ChordFinder,ComputeStep}
SimWallClockTime (with all logging/outputs disabled)
### Error¶
#### Power Devs / QSS¶
##### Using Analytic Baseline¶
X_Error, Y_Error, Z_Error, Position_Error vs time
X_Error, Y_Error, Z_Error, Position_Error vs length
X_Error, Y_Error, Z_Error, Position_Error histogram
##### Using Power Devs / QSS Baseline¶
X_Error, Y_Error, Z_Error, Position_Error vs time
X_Error, Y_Error, Z_Error, Position_Error vs length
X_Error, Y_Error, Z_Error, Position_Error histogram
##### Using Geant4 Baseline¶
X_Error, Y_Error, Z_Error, Position_Error vs time
X_Error, Y_Error, Z_Error, Position_Error vs length
X_Error, Y_Error, Z_Error, Position_Error histogram
#### Geant4¶
##### Using Analytic Baseline¶
X_Error, Y_Error, Z_Error, Position_Error vs time
X_Error, Y_Error, Z_Error, Position_Error vs length
X_Error, Y_Error, Z_Error, Position_Error histogram
##### Using Power Devs / QSS Baseline¶
X_Error, Y_Error, Z_Error, Position_Error vs time
X_Error, Y_Error, Z_Error, Position_Error vs length
X_Error, Y_Error, Z_Error, Position_Error histogram
##### Using Geant4 Baseline¶
X_Error, Y_Error, Z_Error, Position_Error vs time
X_Error, Y_Error, Z_Error, Position_Error vs length
X_Error, Y_Error, Z_Error, Position_Error histogram | 914 | 2,694 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.671875 | 3 | CC-MAIN-2020-45 | longest | en | 0.522171 |
http://portraitofacreative.com/books/appl-of-differential-algebra-to-single-particle-dynamics-in-storage-rings | 1,553,440,585,000,000,000 | text/html | crawl-data/CC-MAIN-2019-13/segments/1552912203462.50/warc/CC-MAIN-20190324145706-20190324171706-00013.warc.gz | 165,622,494 | 8,801 | Similar algebra books
Topics in Computational Algebra
The most goal of those lectures is first to in short survey the basic con nection among the illustration idea of the symmetric crew Sn and the speculation of symmetric services and moment to teach how combinatorial equipment that come up obviously within the conception of symmetric capabilities result in effective algorithms to specific quite a few prod ucts of representations of Sn when it comes to sums of irreducible representations.
Additional resources for Appl of Differential Algebra to Single-Particle Dynamics in Storage Rings
Example text
F/+ 1, given by Eq. 26), are obtained once li(z-") is obtained. 2 Eigenfunctions of the Linear Normal Form Since (:r-T_-l(z-')) in Eq. 26)is a Lie operator, it follows that we can obtain Fi(_ if we can decompose fi(_ and hi(J) into polynomials of the eigenfunctions of T_-I(_ or T_(_. For simplicity without losing generality, we shall consider a transverse map. Then T_-I(z -') -- e:fi'Y: _. e:p=J=+l_Ju:, 1 where J_ - ½(x 2 + p_) and Jy - ½(y2 + p_). It is clear that _± - :_(x 4- ipx) and _± = 7_(y 1 -4-ipp) are eigenfunctions of the Lie operators • Jx " and • Jy ", respectively, as we have •j.
Tl cos(#0 + 6#a ++ . 12) ' and - 4 det(/Mi) hi = (ai - di)2 + (bi + ci) 2 = - -[Tr(iMi)12 L jsm/_0 . A canonical transformation can then be made to obtain,+1M(6) asfollows: ,+_g(6) = (I- 6'A_)_M(6)(I + _'A,) = (I- 6_A,)(R___(6) + 6_M_)(I + 6_A_) + 6i+1(I -- 6iAi)iMi+l(I +... t,)_ + `5i/_i)] ' and Tr(iMi) _ui --" ai + di = 2 sin/_0 2 sin/_0 • The above process is then iterated until we obtain the nth-order symplectic generation matrix 2,,(,5) -- I + 6nAn and then make the (n + 1)th canonical transformation to obtain (I- `snAn)nM(6)(I + 6nAn) = Rh(6) + a(6n+l) Tracing back the n + 1 canonical M(6) transformations, = Ao(I + 6Aa ) .
Density = m-hp(z-") = p (m-n_) k Currently, the nth-turn ,th-tllrn the required h_,_rn results are to be reported. 48 50 order and the phase-space prr_tqlp I"nn(_X"J _rfff 1_ rg_........ _d D_f area _ ; 1 _d 5 Dispersed " Betatron Motion As discussed in Chapter 2, there is always energy spread around the nominal energy in the particle beam. The energy spread causes closed-orbit spread. These dispersed closed orbits with respect to the reference orbit are functions of the longitudinal position, s, and of the energy deviation, 6 = AE/Eo (or the off-momentum, 6 =/Xp/po); that is, • = where Xc is a vector representing the transverse phase-space dispersed closed orbit, and its transpose is given by = coordinates of the yo,p ,o). | 794 | 2,585 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2019-13 | longest | en | 0.799049 |
http://slideplayer.com/slide/4217285/ | 1,519,567,405,000,000,000 | text/html | crawl-data/CC-MAIN-2018-09/segments/1518891816462.95/warc/CC-MAIN-20180225130337-20180225150337-00733.warc.gz | 313,729,932 | 18,798 | # Kinetic Molecular Theory 1.pertaining to motion. 2. caused by motion. 3. characterized by movement: Running and dancing are kinetic activities. ki ⋅ net.
## Presentation on theme: "Kinetic Molecular Theory 1.pertaining to motion. 2. caused by motion. 3. characterized by movement: Running and dancing are kinetic activities. ki ⋅ net."— Presentation transcript:
Kinetic Molecular Theory 1.pertaining to motion. 2. caused by motion. 3. characterized by movement: Running and dancing are kinetic activities. ki ⋅ net ⋅ ic Origin: 1850–55; < Gk kīnētikós moving, equiv. to kīnē- (verbid s. of kīneîn to move) + -tikos Source: Websters Dictionary
CA Standards
The Nature of Gases Gases expand to fill their containers Gases are fluid – they flow Gases have low density 1/1000 the density of the equivalent liquid or solid Gases are compressible Gases effuse and diffuse Gases expand to fill their containers Gases are fluid – they flow Gases have low density 1/1000 the density of the equivalent liquid or solid Gases are compressible Gases effuse and diffuse
Kinetic Molecular Theory Particles of matter are ALWAYS in motion Volume of individual particles is zero. Collisions of particles with container walls cause the pressure exerted by gas. Particles exert no forces on each other. Average kinetic energy is proportional to Kelvin temperature of a gas.
Kinetic Energy of Gas Particles At the same conditions of temperature, all gases have the same average kinetic energy. m = mass v = velocity small moleculesFASTER At the same temperature, small molecules move FASTER than large molecules
Diffusion describes the mixing of gases. The rate of diffusion is the rate of gas mixing. Diffusion is the result of random movement of gas molecules The rate of diffusion increases with temperature Small molecules diffuse faster than large molecules Diffusion
Graham’s Law of Diffusion M 1 M 1 = Molar Mass of gas 1 M 2 M 2 = Molar Mass of gas 2
Purification of Uranium-235 Using Gaseous Diffusion
Download ppt "Kinetic Molecular Theory 1.pertaining to motion. 2. caused by motion. 3. characterized by movement: Running and dancing are kinetic activities. ki ⋅ net."
Similar presentations | 590 | 2,239 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.09375 | 3 | CC-MAIN-2018-09 | latest | en | 0.85916 |
http://slideshowes.com/doc/213322/solving-linear-programming-problems-using-excel | 1,493,459,339,000,000,000 | text/html | crawl-data/CC-MAIN-2017-17/segments/1492917123484.45/warc/CC-MAIN-20170423031203-00135-ip-10-145-167-34.ec2.internal.warc.gz | 362,098,473 | 12,817 | ```Solving Linear Programming
Problems Using Excel
Ken S. Li
Southeastern Louisiana University
Linear Programming
The Problem
An optimization model is a linear program if it has
continuous variables, a single objective function, and
all constraints are linear equalities or inequalities
History
Linear programming was conceptually developed before
World War II by the outstanding Russian mathematician
A.N. Kolmogorov.
Who and When
Major contributors:
Kantorovich
Stigler
Dantzig
Karmarkar
Gonzaga
Wright
Terlaky
Todd
1945
1947
1984
1992
1996
1995
1991
Standard Formulation
Mathematical Formulation
minimize cx
subject to
Ax = b
x >= 0
Standard Formulation
where x is the vector of variables to be solved for, A
is a matrix of known coefficients,
and c and b are vectors of known coefficients.
The expression "cx" is called the objective
function,
and the equations "Ax=b" are called the
constraints.
Standard Formulation
Usually A has more columns than rows, and Ax=b
is therefore quite likely to be under-determined,
leaving great latitude in the choice of x with
which to minimize cx. The word "Programming"
is used here in the sense of "planning"; the
necessary relationship to computer programming
was incidental to the choice of name. Hence the
phrase "LP program" to refer to a piece of
software is not a redundancy, although I tend to
use the term "code" instead of "program" to avoid
the possible ambiguity.
Applications
Applications
Linear and integer programming have proved
valuable for modeling many and diverse types of
problems in planning, routing, scheduling,
assignment, and design. Industries that make use
of LP and its extensions include transportation,
energy, telecommunications, and manufacturing
of many kinds.
Applications
Specific Applications
1. Development of a production schedule that will satisfy
future demands for a firm’s production and at the
same time minimize total production and inventory
costs
2. Selection of the product mix in a factory to make best
use of machine-hours and labor-hours available while
maximizing the firm’s products
Application - Contd
3. Determination of grades of petroleum products to
yield the maximum profit
4. Selection of different blends of raw materials to feed
mills to produce finished feed combinations at
minimum cost
5. Determination of a distribution system that will
minimize total shipping cost from several warehouses
to various market locations
Computational Method
Simplex Methods
Simplex methods, introduced by Dantzig about 50
years ago, visit "basic" solutions computed by
fixing enough of the variables at their bounds to
reduce the constraints Ax = b to a square system,
which can be solved for unique values of the
remaining variables. Basic solutions represent
extreme boundary points of the feasible region
defined by Ax = b, x >= 0, and the simplex
method can be viewed as moving from one such
point to another along the edges of the boundary.
Computational Method
Interior Point Methods
Barrier or interior-point methods, by contrast,
visit points within the interior of the feasible
region. These methods derive from techniques for
nonlinear programming that were developed and
popularized in the 1960s by Fiacco and
McCormick, but their application to linear
programming dates back only to Karmarkar's
innovative analysis in 1984.
Interior Point Method
Step 1: Choose any feasible interior point solution,
x(0) 0 and set solution index t=0.
Step 2: If any component of x ( t ) is 0, or if recent
steps have made no significant change in the
solution value, stop. Current point is either
optimal or very nearly so.
Step 3: Construct the next move direction
x
(t 1)
Xt Pt c
(t )
Interior Point Method -contd
Where
x1( t )
0
Xt
0
0
x2( t )
0
0
0
(t )
xn ,
T 1
t
Pt I A (At A ) At , ct Xt c.
T
t
Interior Point Method - contd
Step 4: If there is no limit on feasible moves in the
direction x (t 1) (all components are
nonnegative), stop ; the given model is
unbounded. Otherwise, construct the step size
1
x
( t 1)
1
t
X
.
Interior Point Method - contd
Step 4: compute the new solution
x (t 1) x (t ) x (t 1)
Then let t t 1, and return to Step 2.
A Simple Example
The Marriott Tub Company manufactures two lines of
bathtubs, called Model A and model B. Every tub
requires a certain amount of steel and zinc; the company
has available a total of 25,000 pounds of steel and 6,000
pounds of zinc. Each model A bathtub requires a total of
125 pounds of steel and 20 pounds of zinc, and each
yields a profit of \$90. Each model B bathtub can be sold
for a profit of \$70; it in turn requires 100 pounds of steel
and 30 pounds of zinc. Find the best production mix of
the bathtubs.
The Formulation
Maximize
Subject to
P 90 x 70 y
125 x 100 y 25000
20 x 30 y 6000
x, y 0
Where x and y are the numbers of model A
and model B bathtubs that the company
will make, respectively.
Solving by Interior Point
Method
x1
x2
x3
x4
Obj
Initial
1st
2nd
3 rd
100
100
2500
1000
16000
129.4 86.4
178.8
818.8
17698
152.8 58.7
3.3
1184
17861
157
53.6
5.7
1250
17882
4 th
5 th
6 th
7 th
8 th
189
13.60
8.3
1810
17962
199.6 0.4
6.5
1995
17992
199.7 0.3
0.1
1994
17994
199.9 0.1
0.1
1999
17998
200
0
2000
18000
0
Solving by Excel
/linearprog.xls
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Ques:- Three different containers contain different quantities of mixture milk and water, whose measurements are 403 Kg, 434 Kg and 465 Kg. What biggest measure must be there to measure all the different quantities exactly?
A. 1 Kg
B. 7 Kg
C. 14 Kg
D. 31 Kg
31
cos
434-403=31
465-434=31
Ques:- A milk vendor has 2 cans of milk. The first contains 25% water and the rest milk. The second contains 50% water. How much milk should he mix from each of the containers so as to get 12 litres of milk such that the ratio of water to milk is 3 : 5?
A. 4 litres, 8 litres
B. 5 litres, 7 litres
C. 6 litres, 6 litres
D. 7 litres, 5 litres
Ans- 6,6 litres
Let x Litres from can1 (where, water = x/4 litres and milk = 3x/4 litres)
Now
12-x liters from can 2 ( where, water = 6-x/2, milk = 6-x/2)
Now given
water/milk = 3/5
0.25x+6-0.5x / 0.75 +6-0.5x = 3/5
6-0.25x/6+0.25 = 3/5
Solve it you get
x = 6 litres from can1
Then 12-x= 12-6=6 litres from can2
Ques:- if a car starts from A towards B with some velocity due to some problem in the engine afetr travelling 30 km .if the car goes 4/5 th of its of actual velociteythe car reaches B 45 min later to the actual time.if the car engine fails after travelling 45km the car reaches the destination B 36min late to the actual time ,wat is the initial velocity of car and what is the distance between A and B on km.
20&130
Ques:- Why do you wish to join a new company??
Ques:- A cube of side 4cm is painted with colors red,blue,green in such a way that opposite sides are painted in the same colour . this cube is now cut into 64 cubes of equal size .then 1) how many have atleast two sides painted in different colours.
Ques:- A fathers age is reverse of sons age. One year back fathers age was twice of sons age. What is the fathers current age?
73.
one year back father = 72 and son = 36.
Ques:- Why our company hire you?
Ques:- The fraction EVE/DID = 0,TALKTALKTALKTALK… is a normal fraction that can also be written as a recurring decimal. Which fraction is this (equal letters are equal ciphers)?
Ques:- Can you work under pressure?Give an example.
Ques:- How would you describe yourself as a person?
Ques:- Whom would you pick as a leader icon and why?
Ques:- Speak for 2 minutes on some topic
Ques:- Arrange these 13 Alphabets into 3 Love Words?
Ques:- Back ground
Ques:- Evaluate x from the the equation : 6x – 27 + 3x = 4 + 9 – x
4
Ques:- A rectangular grass field is 75 m * 55 m, it has a path of 2.5 m wide all round it on the outside. Find the area of the path and the cost of constructing it at Rs.2 per sq m?
Ques:- A is a working partner and B is a sleeping partner in the business. A puts in Rs.15000 and B Rs.25000, A receives 10% of the profit for managing the business the rest being divided in proportion of their capitals. Out of a total profit of Rs.9600, money received by A is?
Explanation:
15:25 => 3:5
9600*10/100 = 960
9600 – 960 = 8640
8640*3/8 = 3240 + 960
= 4200
Ques:- A box contains 3 blue marbles, 4 red, 6 green marbles and 2 yellow marbles. If two marbles are drawn at random, what is the probability that at least one is green?
Ques:- If you were given 10 lakhs what business you would like to open and why?
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The Teacher-Author indicated this resource includes assets from Google Workspace (eg. docs, slides, etc.).
#### Also included in
1. Google Paperless Practice - 4th Grade Fractions Bundle {Supports Common Core NF standards} Engage your students in practicing key fraction skills with these interactive DIGITAL resources that work with Google Slides™. No more copies to be made, no more printer ink, and no more lost papers! Plus,
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### Description
Engage your students with this interactive DIGITAL resource that works with Google Slides™. No more copies to be made, no more printer ink, and no more lost papers!
With this 24-slide digital resource, your students will practice recognizing and generating equivalent fractions. Students will love interacting with the movable pieces and typing their responses on these slides!
This resource was created to support 4th grade Common Core standard 4.NF.1: Explain why a fraction a/b is equivalent to a fraction (n × a)/(n × b) by using visual fraction models, with attention to how the number and size of the parts differ even though the two fractions themselves are the same size. Use this principle to recognize and generate equivalent fractions.
• Instructions for opening, sharing, and using this Google Slides™ file
• 24 interactive slides for your students to complete
If you have a Google classroom, this activity is sure to make practicing these key skills way more fun than if they were done with paper and pencil.
IMPORTANT: This is a digital resource, so please only purchase this resource if you have the capabilities in your classroom to use it (computers, laptops, or tablets, Internet access, and a Google account).
If you have any issues with this resource, please feel free to email me directly at brittney@games4gains.com. I'll be happy to try to help you!
Think ahead and save \$\$ ★ by buying this digital resource as part of this discounted
4TH GRADE FRACTIONS DIGITAL PRACTICE BUNDLE, which includes the following resources:
→ 4th Grade Equivalent Fractions Digital Practice {4.NF.1}
→ 4th Grade Comparing Fractions Digital Practice {4.NF.2}
→ 4th Grade Decomposing Fractions Digital Practice {4.NF.3B}
→ 4th Grade Subtracting Fractions Digital Practice {4.NF.3}
→ 4th Grade Subtracting Mixed Numbers Digital Practice {4.NF.3C}
→ 4th Grade Multiplying Fractions Digital Practice {4.NF.4}
→ 4th Grade Adding Fractions with Denominators of 10 and 100 Digital Practice {4.NF.5}
→ 4th Grade Comparing Decimals {4.NF.7}
Want to see all of the digital resources that we have available?
~~~~~~~~~
Your students may also enjoy these 4th grade fraction activities from Games 4 Gains:
4th Grade Fractions Games Pack - BEST SELLER
4th Grade Fractions Monster Mystery Pictures Pack - BEST SELLER!
Equivalent Fractions 'Clip and Flip' Cards
Equivalent Fractions Board Game
Equivalent Fractions Bingo
Mixed Numbers and Improper Fractions Bump Games
Adding and Subtracting Fractions Bump Games Bundle
~~~~~~~~~
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This purchase is for single classroom use only. Sharing this resource with multiple teachers, an entire school, or an entire school system is strictly forbidden. Multiple licenses are available at a discount.
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http://studentsmerit.com/paper-detail/?paper_id=69489 | 1,480,926,699,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698541556.70/warc/CC-MAIN-20161202170901-00293-ip-10-31-129-80.ec2.internal.warc.gz | 255,099,934 | 7,021 | #### Details of this Paper
##### Write a program that reads a power of 10
Description
solution
Question
You MUST use a switch statement in your solution to this problem.;A million is 106 and a billion is 109. Write a program that reads a power of 10 (6, 9;12, etc.) and displays how big the number is (million, billion, etc.). Display an;appropriate message for the input value that has no corresponding word. The table;below shows the correspondence between the power and the word for that power.;*See the table below for help.;Power of 10 Number;6 Million;9 Billion;12 Trillion;15 Quadrillion;18 Quintillion;21 Sextillion;24 Septillion;27 Octillion;30 Nonillion;100 Googol
Paper#69489 | Written in 18-Jul-2015
Price : \$22 | 187 | 731 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2016-50 | longest | en | 0.770349 |
https://ohbug.com/uva/10660/ | 1,726,561,863,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651750.27/warc/CC-MAIN-20240917072424-20240917102424-00702.warc.gz | 391,191,219 | 2,110 | # Citizen attention offices
The Mayor of a city wants to improve the attentions of the local government to the citizens. For that, five offices to attend to consults will be open in the city. The Mayor thinks that the attentions would be better if the offices are close to where the people live. To decide where to open the offices, the city has been divided in 25 areas, in a square: 01234 56789 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 and the quantity of people living in each square is counted, thus assigning to each area the number (an integer number) of thousands of people living there. For example: 00231 52220 11230 00215 21204 The number of people in each area is less or equal to 10 million. For each area, the minimum distance to the five offices is obtained as follows: the distance from one area to an office is obtained as the number of movements to go from the area to the area where the office will be installed (the movements are in horizontal and vertical; for example, from (1,1) to (2,3) three movements are necessary). And we want to minimize the sum of the minimum distances from all the areas, but having into account the quantity of people living in each area. Thus, for each area the distance to an office is multiplied by the value associated to the area. Input The input begins with a line where the number of test cases (t) is indicated. The data for each test case appear in sucessive lines. In each test case the first line contains the number of areas in the city with non null population (n) and in each one of the n following lines three numbers appear, the first and the second indicate the row and the column of the area (they are numbered from 0 to 4) and the third the number of thousand of people living in the area. Output The output consists of a line for each problem, with five numbers with a space between each two numbers, and no space after the last number. The numbers in the solutions are written in increasing order. If there is more than one optimum solution, you must output the solution with lowest first value. If the first value coincides, you must output the solution with lowest second value, and so on.
2/2 Sample Input 4 1 221 4 001 441 041 401 5 001 111 221 331 441 7 422 331 243 211 134 122 101 Sample Output 0 1 2 3 12 0 1 4 20 24 0 6 12 18 24 5 7 8 14 22 | 586 | 2,323 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.453125 | 3 | CC-MAIN-2024-38 | latest | en | 0.945576 |
https://slideplayer.com/slide/9351637/ | 1,685,324,008,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224644574.15/warc/CC-MAIN-20230529010218-20230529040218-00707.warc.gz | 598,839,094 | 16,697 | # Objective: Multiply three-digit numbers by single-digit numbers using the grid method © Hamilton Trust Stepping Up Term 2 Week 4 Day 3.
## Presentation on theme: "Objective: Multiply three-digit numbers by single-digit numbers using the grid method © Hamilton Trust Stepping Up Term 2 Week 4 Day 3."— Presentation transcript:
Objective: Multiply three-digit numbers by single-digit numbers using the grid method © Hamilton Trust Stepping Up Term 2 Week 4 Day 3
Talk me through the steps to solve: 4800 + 240 + 72 = 5112
© Hamilton Trust Stepping Up Term 2 Week 4 Day 3 In pairs, can you work out the missing numbers? 2000 + 360 + 12 = 2372
© Hamilton Trust Stepping Up Term 2 Week 4 Day 3 Choose one number from the red set and one from blue. Multiply the two numbers using the grid method. Write the product. Repeat three times or more. 308 542 416 295 38753875
Download ppt "Objective: Multiply three-digit numbers by single-digit numbers using the grid method © Hamilton Trust Stepping Up Term 2 Week 4 Day 3."
Similar presentations | 258 | 1,044 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2023-23 | latest | en | 0.751813 |
https://www.physicsforums.com/threads/prove-that-a-triangle-with-lattice-points-cannot-be-equilateral.1050888/ | 1,719,344,478,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198866218.13/warc/CC-MAIN-20240625171218-20240625201218-00304.warc.gz | 821,262,556 | 16,114 | # Prove that a triangle with lattice points cannot be equilateral
• I
• JoeAllen
In summary, the conversation discusses the assumption of three points for a triangle with coordinates (a, c), (c, d), and (b, e), where a, b, c, d, and e are all integers. The conversation also mentions using the distance formula to set the distances between each point equal, but this leads to a contradiction and shows that the points do not form a triangle. The conversation also introduces the concept of a 2D lattice, where one of the lattice points can be set as (0,0) without losing generality. However, when using the formula for lattice points, it leads to a contradiction.
JoeAllen
I assumed three points for a triangle P1 = (a, c), P2 = (c, d), P3 = (b, e)
and of course:
a, b, c, d, e∈Z
Using the distance formula between each of the points and setting them equal:
\sqrt { (b - a)^2 + (e - d)^2 } = \sqrt { (c - a)^2 + (d - d)^2 } = \sqrt { (b - c)^2 + (e - d)^2 }(e+d)2 = (c-a)2 - (b-a)2
(e+d)2 = (c-a)2 - (b-c)2
c2 - 2ac - b2 +2ab = -2ac + a2 - b2 + 2bc
c2 + 2ab = a2 + 2bc
c(c - 2b) = a(a - 2b)
Thus, for this to be true, a = c. But in this example, the distance between a and c would be 0. Thus, not a triangle and certainly not an equilateral triangle.
Where did I go wrong here? I'm bored waiting for Calculus II in the Fall and I'm going through Courant's Differential and Integral Calculus on my free time until then (Fall term probably starting in August/September, so I'm not worried if it takes a few months to get comfortable with Courant - Calculus I has been a breeze since I already knew most of the content before taking it).
Last edited:
As for 2D lattice we can make one of the lattice points is (0,0) without losing generality.
Say other points are ##(n_1,n_2),(m_1,m_2)##
$$n_1^2+n_2^2=A$$
$$m_1^2+m_2^2=A$$
$$(n_1-m_1)^2+(n_2-m_2)^2=A$$
where A is square of the side length. You will find contradiction in this set of formla.
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JEE Main 2023 Maths Question Paper with Solutions (10 April Evening Shift)
JEE Main 2023| 10 April Evening Maths Question Paper- Answer key with Detailed Solutions
JEE NTA Board, the conducting body of JEE Main for 10 April Evening JEE Main 2023 Question Paper, conducted the exam in 13 regional languages.
As Candidates who are yet to give the other JEE Main 2023 shifts and sessions, it becomes essential to get the gist of the overall difficulty level of the Questions asked in the Maths question paper and the analysis from experts and answer keys curated by the master teachers.
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How to Download Question Paper of Maths JEE Main 2023 (April 10 Evening)?
Candidates can download the memory-based questions with solutions of JEE Main 2023 question paper PDF. Just click on the links to download the free pdf of the answers. By downloading the question paper solutions, candidates can analyse the difficulty level of the exam, how much score candidates will get, what could be the cut-off, and what the candidate's rank could be.
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Maths 10 April Evening JEE Main 2023 Question Paper Analysis
The focus of the Maths JEE Main 2023 April 10 shift 2 exams was on Algebra and Coordinate Geometry, with Straight Lines, Parabolas, Circles, Definite Integrals, Limits, Differential Equations, Area, Matrices, Binomial Theorem, Probability, Statistics, Determinants, Vectors, 3D Geometry, Complex numbers, and Progression Series being among the specific topics covered. The exam consisted of multiple-choice questions and numerical-based questions that necessitated extensive calculations. The overall difficulty of the exam was moderate.
Algebra and Coordinate Geometry were given significant importance in the exam, encompassing various mathematical concepts, including calculus, geometry, probability, and statistics. The numerical-based questions were particularly challenging, necessitating lengthy calculations to obtain the correct solution. The exam's objective was to assess a student's overall mathematical ability and capacity to apply mathematical concepts in real-world scenarios.
FAQs on JEE Main 2023 Maths Question Paper with Solutions (10 April Evening Shift)
1. What were the most challenging topics in the Maths section of the JEE Main 2023 10 April Evening shift paper?
Calculus, Complex Numbers, and Probability were among the most difficult topics in the Maths section of the JEE Main 2023 Evening shift paper held on 10th April. These subjects posed a challenge for numerous students due to their complexity and the extensive calculations necessary to resolve the issues.
2. Were there any repeated questions from previous years in the Maths section of the JEE Main 2023 10 April Evening shift paper?
Some questions in the Maths section of the JEE Main 2023 Evening shift paper held on 10th April were identical to the questions from previous years. According to some students, a few questions related to Coordinate Geometry and Binomials seemed familiar from the previous year's papers, but they still needed to be solved. | 830 | 4,064 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.234375 | 3 | CC-MAIN-2023-23 | latest | en | 0.920933 |
http://www.cplusplus.com/forum/beginner/111423/ | 1,503,135,254,000,000,000 | text/html | crawl-data/CC-MAIN-2017-34/segments/1502886105334.20/warc/CC-MAIN-20170819085604-20170819105604-00265.warc.gz | 528,219,518 | 4,031 | Can anyone give me some suggestions on how I could edit the below code to output to the user which quadrant they are in (Northeast, Southeast, Southwest, Northwest) If they are on an axis, simply state that they are (North, South, East, or West). Example: If the user moves so that coordinates are (1,3), they are in the Northeast quadrant.
Thank you
I want to use for example a if (x > 0 && y > 0) then quadrant = "Northeast" but I cannot figure out the syntax to make that work in my below code.
``1234567891011121314151617181920212223242526272829303132333435363738394041424344454647484950515253545556`` `````` #include using namespace std; int main() { bool compass = true; int x = 0; int y = 0; cout << "enter direction: N (north), E (east), S (south), W (west) or Q to pause: (" << x << "," << y << ")"; char direction; cin >> direction; do{ if(direction == 'E'||direction == 'e'){ x += 1; cout << "enter direction: N (north), E (east), S (south), W (west) or Q to pause: (" << x << "," << y << ")"; cin >> direction; } else if(direction == 'S'||direction == 's'){ y -= 1; cout << "enter direction: N (north), E (east), S (south), W (west) or Q to pause:(" << x << "," << y << ")"; cin >> direction; } else if(direction == 'N'||direction == 'n'){ y += 1; cout << "enter direction: N (north), E (east), S (south), W (west) or Q to pause:(" << x << "," << y << ")"; cin >> direction; } else if(direction == 'W'||direction == 'w'){ x -= 1; cout << "enter direction: N (north), E (east), S (south), W (west) or Q to pause:(" << x << "," << y << ")"; cin >> direction; } else if (direction == 'q' || direction == 'Q') system ("pause"); }while(compass == true); return 0; }``````
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Topic archived. No new replies allowed. | 517 | 1,737 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2017-34 | latest | en | 0.90295 |
http://www.readbag.com/rutherglen-science-mq-au-wchen-lnicafolder-ica10 | 1,590,500,924,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347390758.21/warc/CC-MAIN-20200526112939-20200526142939-00434.warc.gz | 204,985,279 | 9,381 | `INTRODUCTION TO COMPLEX ANALYSISW W L CHENcW W L Chen, 1986, 2008.This chapter originates from material used by the author at Imperial College, University of London, between 1981 and 1990. It is available free to all individuals, on the understanding that it is not to be used for financial gain, and may be downloaded and/or photocopied, with or without permission from the author. However, this document may not be kept on any information storage and retrieval system without permission from the author, unless such system is not accessible to any individuals other than its owners.Chapter 10RESIDUE THEORY10.1. Cauchy's Residue Theorem If we extend Cauchy's integral theorem to functions having isolated singularities, then the integral is in general not equal to zero. Instead, each singularity contributes a term called the residue. Our principal aim in this section is to show that this residue depends only on the coefficient of (z -z0 )-1 in the Laurent expansion of the function near the singularity z0 , since all the other powers of z - z0 has single valued integrals and so integrate to zero. Definition. By a simple closed contour or Jordan contour, we mean a contour : [A, B] C such that (t1 ) = (t2 ) whenever t1 = t2 , with the one exception (A) = (B). THEOREM 10A. Suppose that a function f is analytic in a simply connected domain D, except for an isolated singularity at z0 , and that-1f1 (z) =n=-an (z - z0 )nis the principal part of f at z0 . Suppose further that C is a Jordan contour in D followed in the positive (anticlockwise) direction and not passing through z0 . Then 1 2iChapter 10 : Residue Theoryf1 (z) dz =Ca-1 0if z0 lies inside C, if z0 lies outside C.page 1 of 11 Introduction to Complex AnalysisxxxxxcW W L Chen, 1986, 2008Proof. Suppose first of all that z0 is outside C. Then z0 is in the exterior domain of C which also contains the point at . It follows that z0 can be joined to the point at by a simple polygonal curve L, as shown in the picture below. D Lz0CThe Jordan contour C is clearly contained in the simply connected domain obtained when L is deleted from the complex plane. In fact, it is contained in a simply connected domain which is a subset of D \ L, as shown by the shaded part in the picture above. Clearly f is analytic in this simply connected domain, so it xxxxx from Theorem 9B that follows f (z) dz = 0.CSuppose next that z0 is inside C. Then there exists r > 0 such that the closed disc {z : |z - z0 | r} is inside C. Let denote the boundary of this disc, followed in the positive (anticlockwise) direction. Dz0 CWe now draw a horizontal line through the point z0 . Following this line to the left from z0 , it first intersects and then C (for the first time). Draw a line segment joining these two intersection points. Similarly, following this line to the right from z0 , it first intersects and then C (for the first time). Again draw a line segment joining these two intersection points. Note that these two line segments are inside C and outside . We now divide C into two parts by cutting it at the two intersection points mentioned. It can be shown that one part of this, together with the part of above the horizontal line and the two line segments, gives rise to a simple closed contour C + followed in the positive directionChapter 10 : Residue Theory page 2 of 11 Introduction to Complex AnalysiscW W L Chen, 1986, 2008xxxxx and which can be shown to lie in a simply connected domain lying in D but not containing z0 . Clearly f is analytic in this simply connected domain, so that f (z) dz = 0,C+in view of Theorem 9B. C+z0C-Similarly, the other part of C, together with the part of below the horizontal line and the two line segments, gives rise to a simple closed contour C - followed in the positive direction and which again can be shown to lie in a simply connected domain lying in D but not containing z0 . Clearly f is analytic in this simply connected domain, so that f (z) dz = 0.C-It is easily seen that f (z) dz -C f (z) dz =C+f (z) dz +C-f (z) dz,so that f (z) dz =C f (z) dz.By Theorem 8E, we have f (z) dz = 2ia-1 .It follows that f (z) dz = 2ia-1 .CFinally, note that f2 (z) = f (z) - f1 (z) is analytic in D, so that f2 (z) dz = 0,Cwhence f (z) dz =C Cf1 (z) dz.The result follows.Chapter 10 : Residue Theory page 3 of 11 Introduction to Complex AnalysiscW W L Chen, 1986, 2008Definition. The value a-1 in Theorem 10A is called the residue of the function f at z0 , and denoted by res(f, z0 ). We are now in a position to state and prove a simple version of Cauchy's residue theorem. THEOREM 10B. Suppose that the function f is analytic in a simply connected domain D, except for isolated singularities at z1 , . . . , zk . Suppose further that C is a Jordan contour in D followed in the positive (anticlockwise) direction and not passing through z1 , . . . , zk . Then 1 2ikf (z) dz =C zj j=1 inside Cres(f, zj ).Proof. For every j = 1, . . . , k, let fj (z) denote the principal part of f (z) at zj . By Theorem 8D, fj is analytic in C except at zj . It follows that the functionkg(z) = f (z) -j=1fj (z)is analytic in D, so that g(z) dz = 0Cby Theorem 9B, and sokf (z) dz =C j=1 Cfj (z) dz.The result now follows from Theorem 10A.10.2. Finding the Residue In order to use Theorem 10B to evaluate the integral f (z) dz,Cwe need a technique to evaluate the residues at the isolated singularities. Suppose that f (z) has a removable singularity at z0 . Then f (z) has a Taylor series expansion which is valid in a neighbourhood of z0 . The residue is clearly 0. Suppose that f (z) has a simple pole at z0 . Then we can write f (z) = a-1 + g(z), z - z0where g(z) is analytic at z0 , so that (z - z0 )g(z) 0 as z z0 . It follows that the residue is given by a-1 = lim (z - z0 )f (z).zz0 Chapter 10 : Residue Theory page 4 of 11 Introduction to Complex AnalysiscW W L Chen, 1986, 2008Suppose that f (z) has a pole of order m at z0 . Then we can write f (z) = a-m a-m+1 a-1 + + ... + + g(z), (z - z0 )m (z - z0 )m-1 z - z0where g(z) is analytic at z0 , so that (z - z0 )m f (z) = a-m + a-m+1 (z - z0 ) + . . . + a-1 (z - z0 )m-1 + (z - z0 )m g(z) is analytic at z0 . Differentiating m - 1 times gives dm-1 dm-1 ((z - z0 )m f (z)) = a-1 (m - 1)! + m-1 ((z - z0 )m g(z)). dz m-1 dz Since g(z) is analytic at z0 , we have lim dm-1 ((z - z0 )m g(z)) = 0. dz m-1zz0It follows that the residue is given by a-1 = 1 dm-1 lim ((z - z0 )m f (z)). (m - 1)! zz0 dz m-1Definition. A function is said to be meromorphic in a domain D if it is analytic in D except for poles. Example 10.2.1. The function f (z) = has simple poles at z = ±i/2, with residues xxxxx and res f, - i 2 = lim z+ i 2 f (z) = e2iz e =- . 4i z-i/2 4(z - i/2) lim res f, i 2 = lim z- i 2 f (z) = lim e2iz e-1 = 4i zi/2 4(z + i/2) e2iz 1 + 4z 2zi/2z-i/2i/2 1 -i/2 CIt follows from Cauchy's residue theorem that if C = {z : |z| = 1} is the circle with centre 0 and radius 1, followed in the positive (anticlockwise) direction, then e2iz dz = 2i 1 + 4z 2 e-1 e - 4i 4i = 2 1 -e . epage 5 of 11C Chapter 10 : Residue Theory Introduction to Complex AnalysiscW W L Chen, 1986, 2008Example 10.2.2. The function f (z) = has a pole of order 4 at z = 0, with residue res(f, 0) = 1 1 1 d3 d3 lim 3 (z 4 f (z)) = lim 3 ez = . 3! z0 dz 3! z0 dz 6 ez z4It follows from Cauchy's residue theorem that if C is any Jordan contour with 0 inside and followed in the positive (anticlockwise) direction, then ez dz = 2i z4 1 6 = i . 3CExample 10.2.3. Suppose that a function f is analytic in a simply connected domain D, and that z0 D. Suppose further that C is a Jordan contour in D, followed in the positive (anticlockwise) direction and with z0 inside. If f (z0 ) = 0, then the function F (z) = has a simple pole at z0 , with residuezz0f (z) z - z0lim (z - z0 )F (z) = f (z0 ).Applying Cauchy's residue theorem, we obtain Cauchy's integral formula 1 2i f (z) dz = f (z0 ). z - z0CIf f (z0 ) = 0, then F (z) has a removable singularity at z0 . The same result follows instead from Cauchy's integral theorem.10.3. Principle of the Argument In this section, we shall show that the residue theorem, when applied suitably, can be used to find the number of zeros of an analytic function, as well as the number of zeros minus the number of poles of a meromorphic function. The main idea underpinning our discussion can be summarized by the following two results. THEOREM 10C. Suppose that a function f is analytic in a neighbourhood of z0 . Suppose further that f has a zero of order m at z0 . Then the function f /f is analytic in a punctured neighbourhood of z0 , with a simple pole at z0 with residue m. THEOREM 10D. Suppose that a function f is analytic in a punctured neighbourhood of z0 . Suppose further that f has a pole of order m at z0 . Then the function f /f is analytic in a punctured neighbourhood of z0 , with a simple pole at z0 with residue -m.Chapter 10 : Residue Theory page 6 of 11 Introduction to Complex AnalysiscW W L Chen, 1986, 2008Proof of Theorem 10C. We can write f (z) = (z-z0 )m g(z), where g(z) is analytic in a neighbourhood of z0 and g(z0 ) = 0. Then f (z) m(z - z0 )m-1 g(z) + (z - z0 )m g (z) m g (z) = = + . f (z) (z - z0 )m g(z) z - z0 g(z) Since g(z) is analytic in a neighbourhood of z0 and g(z0 ) = 0, the function g (z)/g(z) is analytic in a neighbourhood of z0 . The result follows. Proof of Theorem 10D. We can write f (z) = (z - z0 )-m g(z), where g(z) is analytic in a neighbourhood of z0 and g(z0 ) = 0. Then -m(z - z0 )-m-1 g(z) + (z - z0 )-m g (z) -m f (z) g (z) = = + . f (z) (z - z0 )-m g(z) z - z0 g(z) Since g(z) is analytic in a neighbourhood of z0 and g(z0 ) = 0, the function g (z)/g(z) is analytic in a neighbourhood of z0 . The result follows. The main result in this section is the Principle of the argument, as stated below. THEOREM 10E. Suppose that a function f is meromorphic in a simply connected domain D. Suppose further that C is a Jordan curve in D, followed in the positive (anticlockwise) direction, and that f has no zeros or poles on C. If N denotes the number of zeros of f in the interior of C, counted with multiplicities, and if P denotes the number of poles of f in the interior of C, counted with multiplicities, then 1 2i f (z) dz = N - P. f (z)CProof. Note that by Theorems 10C and 10D, the poles of the function f /f are precisely at the zeros and poles of f . Furthermore, a zero of f of order m gives rise to a residue m for f /f , so that the residues of f /f arising from the zeros of f are equal to the number of zeros of f counted with multiplicities, and this number is N . On the other hand, a pole of f of order m gives rise to a residue -m for f /f , so that the residues of f /f arising from the poles of f are equal to minus the number of poles of f counted with multiplicities, and this number is P . It follows that the sum of the residues is equal to N - P . The result now follows from Theorem 10B applied to the function f /f . Remarks. (1) Note that 1 2i f (z) 1 1 1 dz = var(log f (z), C) = var(i arg f (z), C) = var(arg f (z), C). f (z) 2i 2i 2CIt follows that the conclusion of Theorem 10E can be expressed in the form N -P = 1 var(arg f (z), C), 2in terms of the variation of the argument of f (z) along the Jordan curve C. (2) Note also that 1 2iChapter 10 : Residue TheoryCf (z) 1 dz = f (z) 2if (C)dw = n(f (C), 0). wpage 7 of 11 Introduction to Complex AnalysiscW W L Chen, 1986, 2008(3) Theorem 10E can be generalized in the following way. Suppose that a function f is meromorphic in a simply connected domain D, and that all its zeros and poles in D are simple. Suppose further that C is a Jordan curve in D, followed in the positive (anticlockwise) direction, and that f has no zeros or poles on C. If a1 , . . . , aN denote the zeros of f in the interior of C, and if b1 , . . . , bP denote the poles of f in the interior of C, then 1 2i f (z) g(z) dz = f (z)N Pg(aj ) -j=1 k=1g(bk )(1)Cfor every function g analytic in D. To see this, simply note that any simple zero or simple pole z0 of f , where g(z0 ) = 0, gives rise to a simple pole of (f /f )g with residue lim (z - z0 ) f (z) f (z) g(z) = g(z0 ) lim (z - z0 ) = zz0 f (z) f (z) g(z0 ) -g(z0 ) if z0 is a simple zero of f , if z0 is a simple pole of f ;zz0on the other hand, if g(z0 ) = 0, then (f /f )g has a removable singularity at z0 . In fact, (1) remains valid if the zeros and poles of f are of higher order, provided that all zeros and poles are counted with multiplicities. Note also that the choice g(z) = 1 in D gives Theorem 10E again. A particular useful choice of f is given by the entire function f (z) = sin z, with simple zeros at every n Z. Since cos z f (z) = = cot z, f (z) sin z it follows from (1) that 1 2i g(z) cot z dz =C n inside Cg(n)(2)for every function g analytic in D. This may be used to obtain a variety of infinite series expansions. xxxxx See Chapter 16. Example 10.3.1. To find the number of zeros of the function f (z) = z 4 + z 3 - 2z 2 + 2z + 4 in the first quadrant of the complex plane, we use the Jordan curve C = C1 C2 C3 , where C1 = [0, R] is the straight line segment along the real axis from 0 to R, C2 is the circular path : [0, /2] C, given by (t) = Reit , and C3 = [iR, 0] is the straight line segment along the imaginary axis from iR to 0. Here R is taken to be a large positive real number.C2 C3C1 On C1 , we have z = x > 0, so that f (z) = f (x) = x4 + x3 - 2x2 + 2x + 4 Rx4 + x3 + 4 2x + 4if 0 x 1 if x 1is clearly positive, so that var(arg f (z), C1 ) = 0. Next, note that f (z) = z 4 1 +Chapter 10 : Residue Theoryz 3 - 2z 2 + 2z + 4 z4.page 8 of 11 Introduction to Complex AnalysiscW W L Chen, 1986, 2008On C2 , we have |z| = R, so that z 3 - 2z 2 + 2z + 4 R3 + 2R2 + 2R + 4 2R3 2 < 4 = z4 R4 R R whenever R > 8, say. It follows that on C2 when R is large enough, we have f (z) = R4 e4it (1 + w), where |w| < 2/R, so that var(arg f (z), C2 ) = 2 + 1 , where 1 0 as R . Finally, on C3 , we have z = iy, where y > 0, so that f (z) = f (iy) = (y 4 + 2y 2 + 4) + i(2y - y 3 ) = (y 2 + 1)2 + 3 + i(2y - y 3 ). Note that Ref (iy) > 0, so that f (iy) is in the first or fourth quadrant of the complex plane. In fact, when R > 0 is large, f (iR) is much nearer the real axis than the imaginary axis, while f (0) = 4 is on the positive real axis. It follows that var(arg f (z), C3 ) = 2 , where 2 0 as R . We now conclude that var(arg f (z), C) = 2 + 1 + 2 , where 1 , 2 0 as R . On the other hand, C is a closed contour, so that var(arg f (z), C) must be an integer multiple of 2. It follows that var(arg f (z), C) = 2. Note now that the function f has no poles in the first quadrant. It follows from the Argument principle that f has exactly one zero inside the contour C for all large R. Hence f has exactly one zero in the first quadrant of the complex plane. To find the number of zeros in a region, the following result provides an opportunity to either bypass the Argument principle or at least enable one to apply the Argument principle to a simpler function. Needless to say, the proof is based on an application of the Argument principle. ´ THEOREM 10F. (ROUCHE'S THEOREM) Suppose that functions f and g are analytic in a simply connected domain D, and that C is a Jordan contour in D. Suppose further that |f (z)| > |g(z)| on C. Then f and f + g have the same number of zeros inside C. We shall give two proofs of this result. The first is the one given in most texts. First Proof of Theorem 10F. Consider the function F (z) = f (z) + g(z) . f (z)The condition |f (z)| > |g(z)| on C ensures that both f and f + g have no zeros on C. On the other hand, note that |F (z) - 1| = g(z) <1 f (z) (3)for every z C. By Remark (2) after Theorem 10E, we have 1 2i F (z) 1 dz = F (z) 2i dw = n(F (C), 0). wCF (C)In view of (3), the closed contour F (C) is contained in the open disc {w : |w - 1| < 1} with centre 1 and radius 1. This disc does not contain the point 0, so that n(F (C), 0) = 0. Hence 1 2i F (z) dz = 0. F (z)CIt follows from the Argument principle that the function F has the same number of zeros and poles inside C. Note now that the poles of F are precisely the zeros of f , and the zeros of F are precisely the zeros of f + g.Chapter 10 : Residue Theory page 9 of 11 Introduction to Complex AnalysiscW W L Chen, 1986, 2008Second Proof of Theorem 10F. For every [0, 1], let N ( ) = 1 2i f (z) + g (z) dz. f (z) + g(z)CThe condition |f (z)| > |g(z)| on C ensures that |f (z) + g(z)| |f (z)| - |g(z)| |f (z)| - |g(z)| > 0 on C, so that f + g does not have any zeros (or poles) on C. In fact, there is a positive lower bound for |f (z) + g(z)| on C independent of . It follows easily from this that N ( ) is continuous in [0, 1]. By the Argument principle, N ( ) is an integer for every [0, 1]. Hence N ( ) must be constant in [0, 1]. In particular, we must have N (0) = N (1). Clearly, N (0) is the number of zeros of f inside C, and N (1) is the number of zeros of f + g inside C. Example 10.3.2. To determine the number of solutions of ez = 2z + 1 with |z| < 1, we write f (z) = -2z and g(z) = ez - 1,so that f (z) + g(z) = ez - 2z - 1. We therefore need to find the number of zeros of f + g inside the unit circle C = {z : |z| = 1}. Clearly, f has precisely one zero, at z = 0, inside C. On the other hand, note that ez - 1 = If z C, then |ezt | et , and so |g(z)| = |ez - 1| 1 0[0,z]e d =1 0ezt z dt.|ezt z| dt 1 0et dt = e - 1.Since |f (z)| = 2 whenever z C, it follows that |f (z)| > |g(z)| on C. By Rouch´'s theorem, f + g has e precisely one zero inside C.Chapter 10 : Residue Theorypage 10 of 11 Introduction to Complex AnalysiscW W L Chen, 1986, 2008Problems for Chapter 10 1. a) Write down the Taylor series for ew about the origin w = 0. 2 b) Using the substitution w = 1/z 2 in (a), find the Laurent series for the function e1/z about the origin z = 0. 2 c) Find the residue of the function e1/z at the origin z = 0. 2 d) What type of singularity does the function e1/z have at the origin z = 0? 2. Suppose that f (z) = g(z)/h(z), where the functions g(z) and h(z) are analytic at z0 . Suppose further that g(z0 ) = 0 and h(z) has a simple zero at z0 . Use l'Hopital's rule to show that res(f, z0 ) = lim g(z) g(z0 ) = . h (z) h (z0 )zz03. For each of the functions f (z) given below, find all the singularities in C, find the residues at these singularities, and evaluate the integrals f (z) dzCandCf (z) dz,where C and C are circular paths centred at the origin z = 0, of radius 1/2 and 2 respectively, followed in the positive (anticlockwise) direction: z z3 + 2 1 b) f (z) = 4 c) f (z) = 4 a) f (z) = z(z - 1) z +1 (z - 1)(z + 1) 4. Suppose that C is a circular path centred at the origin z = 0, of radius 1, followed in the positive (anticlockwise) direction. Show each of the following: ez ez a) dz = i; b) dz = i. 2 3 C 4z + 1 C z 5. Find the number of zeros of f (z) = z 4 + z 3 + 5z 2 + 2z + 4 in the first quadrant of the complex plane. Find also the number of zeros of the function in the fourth quadrant. 6. Consider the equation 2z 5 + 8z - 1 = 0. a) Writing f (z) = 2z 5 and g(z) = 8z - 1, use Rouch´'s theorem to show that all the roots of this e equation lie in the open disc {z : |z| < 2}. b) Writing f (z) = 8z - 1 and g(z) = 2z 5 , use Rouch´'s theorem to show that this equation has e exactly one root in the open disc {z : |z| < 1}. c) How many roots does this equation have in the open annulus {z : 1 < |z| < 2}? Justify your assertion. 7. Show that the equation z 6 + 4z 2 = 1 has exactly two roots in the open disc {z : |z| < 1}. [Hint: Use Rouch´'s theorem. You will need to make a good choice for f (z) and g(z). Do not give e up if your first guess does not work.]Chapter 10 : Residue Theorypage 11 of 11 `
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CB638-FM.tex | 6,554 | 20,043 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2020-24 | latest | en | 0.905241 |
https://www.school-for-champions.com/algebra.htm | 1,685,765,723,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224649105.40/warc/CC-MAIN-20230603032950-20230603062950-00723.warc.gz | 1,093,575,857 | 4,175 | # Understanding Algebra
by Ron Kurtus (updated 16 March 2023)
Algebra is a branch of mathematics that uses letters of the alphabet to represent objects, numbers or even groups of numbers or expressions. In many cases, algebra uses shorthand mathematical notation to simplify complex or very large terms. A major goal in algebra is to solve equations.
I've found that knowing about Algebra has been very useful. Understanding algebraic methods is important in science, computing and other math courses. Algebra is used in so many fields that the subject can even help in your career.
The purpose of these lessons is to help you gain understanding the basics of Algebra, such that you do well in school and become a champion in the subject.
## Basics
What is Algebra?
Purpose of Algebra is to Solve Equations
## Terminolgy
Terminology Used in Algebra
Variables and Constants in Algebra
Algebraic Operations and Symbols
Algebraic Expressions
## Properties
Product of Two Negative Numbers is Positive
Properties or Laws of Addition and Multiplication
Algebra of Objects
Imaginary Numbers
## Working with expressions
Combining Like Terms in an Expression
Multiplying Binomial Expressions
## Exponents and roots
Exponents
Exponent Rules
Multiplication with Exponents
Division with Exponents
Exponents to the Power of 10
Square Roots
Newton's Square Root Approximation
## Solving equations
Linear Equations
Solving a Linear Equation with One Variable
Completing the Square Method
## Resources
Algebra Resources
Arithmetic
Author's Credentials
## Survey results
See results of survey questions
## My books
School for Champions
## Subjects in website
### Let's make the world a better place
Be the best that you can be.
Use your knowledge and skills to help others succeed.
Don't be wasteful; protect our environment.
### Live Your Life as a Champion:
Seek knowledge and gain skills
Do excellent work
Be valuable to others
Have utmost character
#### Be a Champion!
The School for Champions helps you become the type of person who can be called a Champion. | 435 | 2,098 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.984375 | 3 | CC-MAIN-2023-23 | latest | en | 0.904743 |
http://telliott99.blogspot.com/2010/01/flatten-list-in-python.html | 1,531,957,761,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676590362.13/warc/CC-MAIN-20180718232717-20180719012717-00237.warc.gz | 350,029,931 | 13,859 | ## Friday, January 29, 2010
### Flatten a list in Python
I asked another question on Stack Overflow, this time about "flattening" a list of lists. If the input is
`L = [[[1, 2, 3], [4, 5]], 6]`
the output should be
`[1, 2, 3, 4, 5, 6]`
The (simple) solution that I liked the best is a generator (i.e. it returns an iterator). I changed the variable names slightly:
`import collectionsIt = collections.Iterabledef flatten(L): for e in L: if isinstance(e, It) and not isinstance(e, basestring): for sub in flatten(e): yield sub else: yield e`
In the interpreter:
`>>> L = [[[1, 2, 3], [4, 5]], 6]>>> flatten(L)>>> list(flatten(L))[1, 2, 3, 4, 5, 6]`
As you can see, this works, and it also works if one of the elements is a string or unicode object (that's the reason for checking isinstance(e, basestring).
`>>> list(flatten([[['abc',2,3],[4,5]],6]))['abc', 2, 3, 4, 5, 6]`
Recursion will fail if the "depth" is too great. SO user unutbu came up with another version that doesn't use recursion, which Alex Martelli simplified:
`def genflat(L, ltypes=collections.Sequence): L = list(L) while L: while L and isinstance(L[0], ltypes): L[0:1] = L[0] if L: yield L.pop(0)`
There's a bit of trickery here. Suppose we're just starting with
`L = [[[1, 2, 3], [4, 5]], 6]`
The first element, `L[0]`, is `[[1, 2, 3], [4, 5]]`
so we go into the second while and we repeatedly do
`L[0:1] = L[0]`
`>>> L = [[[1, 2, 3], [4, 5]], 6]>>> L[0][[1, 2, 3], [4, 5]]>>> L[:1][[[1, 2, 3], [4, 5]]]>>> L[1:][6]`
We continue until `L[0]` is not an iterable…
`>>> L[0:1] = L[0]>>> L[[1, 2, 3], [4, 5], 6]>>> L[0:1] = L[0]>>> L[1, 2, 3, [4, 5], 6]`
Assigning `L[0:1]` to be = `L[0]` removes the nesting!
Note that this will choke on a string element (we need to guard against that as we did in the first version.
And a little matter of style. For short-lived variables I prefer simple names: i (an index), N (a numerical constant), s (a string), L (a list), D (a dictionary). For many people, capitalized words (even words of a single letter) should be reserved for class objects. This instinct is so strong in them that the people posting on SO followed my usage of a single letter "l" for the list variable, but made it lowercase. Of course, this is also a bad idea, because it can be confused with the digit "one", but they probably figured that was my problem. | 795 | 2,366 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3 | 3 | CC-MAIN-2018-30 | latest | en | 0.82685 |
https://appliedmath.arizona.edu/event-categories/analysis-seminar | 1,623,895,182,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623487626465.55/warc/CC-MAIN-20210617011001-20210617041001-00096.warc.gz | 109,576,092 | 12,287 | # Analysis Seminar
Boundary Integral Formulations in Linear Thermo-Elasto Dynamics
In this theoretical work, we study boundary integral formulations for an interior/exterior initial boundary value problem arising from the thermo-elasto-dynamic equations in a homogeneous and isotropic domain. The time dependence is handled through a passage to the Laplace domain. In the Laplace domain, combined single- and double-layer potential boundary integral operators are introduced and proven to be coercive . Based on the Laplace domain estimates, it is possible to prove the existence, uniqueness and regularity properties of solutions in the time domain, as well as regularity requirements for problem data. This analysis may serve as the mathematical foundation for discretization schemes based on the combined use of the boundary element method and convolution quadrature.
Place: Zoom: https://arizona.zoom.us/j/99410014231 Password: “arizona” (all lower case)
## When
12:30 p.m. May 4, 2021
## Where
Online
Some tumor growth models and connections between them
This talk concerns PDEs modeling tumor growth. We show that a novel free boundary problem arises via the stiff-pressure limit of a certain model. We take a viscosity solutions approach; however, since the system lacks maximum principle, there are interesting challenges to overcome. We also discuss connections between these problems and other PDEs arising in tumor growth modeling. This is joint work with Inwon Kim.
Zoom: https://arizona.zoom.us/j/99410014231 Password: “arizona” (all lower case)
## When
12:30 p.m. April 27, 2021
## Where
Online
Low degree spline approximation of surfaces and functions
Splines (piecewise polynomial or, more generally, rational functions) are commonly used for numerical approximation of (arbitrary) functions in a variety of contexts. For example, the celebrated Bezier/NURBS curves and surfaces are ubiquitous in geometric design and computer graphics. Some of the most widely used frameworks involve tensor products of 1d cubic polynomials which lead to degree 9 polynomials in 3d. This high degree leads to excessive amounts of data and long evaluation times, which makes the utility of such approximations in real-time environments quite limited, even with the state of the art hardware. One of the ways to deal with this issue is to develop methods for low degree approximation of geometric data. There are, however, some geometric and topological obstructions to this. In my talk I will review the history of this subject, highlight some specific problems and ideas of how to address them.
Zoom: https://arizona.zoom.us/j/99410014231 Password: “arizona” (all lower case)
## When
12:30 p.m. April 20, 2021
## Where
Online
The Boltzmann equation near and far from equilibrium
If an ideal gas is in thermodynamic equilibrium, the distribution of particle velocities is given explicitly by the Maxwellian (a.k.a. Maxwell-Boltzmann) distribution. If the gas is out of equilibrium, modeling the dynamics is naturally much more difficult. The evolution equation satisfied by the particle density, known as the Boltzmann equation, lacks a suitable well-posedness theory despite its long history and widespread use in statistical physics. So far, global solutions have only been constructed for initial data that is sufficiently close to an equilibrium state. In this talk, I will describe progress from the last few years on the "large-data" or "far-from-equilibrium" regime, including joint work with C. Henderson and A. Tarfulea on local existence, instantaneous filling of vacuum regions, and continuation criteria.
Zoom: https://arizona.zoom.us/j/99410014231 Passworappappd: “arizona” (all lower case)
## When
12:30 p.m. April 13, 2021
## Where
Online
Geometry and Dynamics of Phase Transitions in Periodic Media
In this talk we discuss recent progress on homogenization of the Allen-Cahn equation in a periodic medium to Brakke's formulation of anisotropic mean curvature flow. The starting point of our work is the recent Gamma-convergence result of Cristoferi-Fonseca-Hagerty-Popovici, and we combine it with the Sandier-Serfaty scheme for the convergence of gradient flows. Along the way, our analysis also provides resolution to a long-standing open problem in geometry: what do distance functions to planes look like, in periodic Riemannian metrics on \$R^n\$ that are conformal to the Euclidean one? The corresponding question about distance functions to points, has a long history that goes back to Gromov and Burago. This represents joint work with Irene Fonseca (CMU), Rustum Choksi and Jessica Lin (McGill).
Zoom: https://arizona.zoom.us/j/99410014231 Password: “arizona” (all lower case)
## When
12:30 p.m. April 6, 2021
## Where
Online
Numerical approximation of statistical solutions of hyperbolic systems of conservation laws
Statistical solutions are time-parameterized probability measures on spaces of integrable functions, which have been proposed recently as a framework for global solutions for multi-dimensional hyperbolic systems of conservation laws. We present a numerical algorithm to approximate statistical solutions of conservation laws and show that under the assumption of ‘weak statistical scaling’, which is inspired by Kolmogorov’s 1941 turbulence theory, the approximations converge in an appropriate topology to statistical solutions. We will show numerical experiments which indicate that the assumption might hold true.
Zoom: https://arizona.zoom.us/j/99410014231 Password: “arizona” (all lower case)
## When
12:30 p.m. March 30, 2021
## Where
Online
On the inviscid problem for the Navier-Stokes equations
The question of whether the solution of the Navier-Stokes equation converges to the solution of the Euler equation as the viscosity vanishes is one of the fundamental problems in fluid dynamics. In the talk, we will review current results on this problem. We will also present a recent result, joint with Vlad Vicol and Fei Wang, which shows that the inviscid limit holds for the initial data that is analytic only close to the boundary of the domain, and has finite Sobolev regularity in the interior.
Zoom: https://arizona.zoom.us/j/99410014231 Password: “arizona” (all lower case)
## When
12:30 p.m. March 23, 2021
## Where
Online
The effect of surface tension on thin elastic rods
A recent experiment by Mora et al. showed that surface tension can cause a thin elastic rod to spontaneously bend when immersed in a fluid [PRL 111, 114301]. This experiment is remarkable in that it defies common intuition: the typical examples of large deformations caused by surface tension involve objects floating on the surface of a fluid, and the deformations are driven by a difference in surface tension between the two sides of the object. We use nonlinear elasticity and dimension reduction to derive an equation for the energy of a thin rod with surface tension, correctly predicting which rods bent when immersed in fluid and which rods did not.
Zoom: https://arizona.zoom.us/j/99410014231 Password: “arizona” (all lower case)
## When
12:30 p.m. March 16, 2021
## Where
Online
The impact of Gårding inequalities on uniqueness
In this talk we discuss the question of uniqueness of minimizers of integral quasiconvex functionals. It is known that, in this setting, no uniqueness is to be expected in general. However, the situation is different if we impose suitable smallness conditions on the Dirichlet boundary datum. Employing strategies typically arising in regularity theory, the quasiconvexity condition allows us to establish a Gårding inequality that, in turn, leads to our uniqueness result and other interesting observations regarding uniqueness of minimizers. The talk is based on a joint work with Jan Kristensen.
Zoom: https://arizona.zoom.us/j/99410014231 Password: “arizona” (all lower case)
## When
12:30 p.m. March 2, 2021
## Where
Zoom
Nonexistence of subcritical solitary waves
We consider waves on the surface of an incompressible fluid, in particular localized solitary waves which travel at a constant speed. It is a long-standing conjecture that such waves must be "supercritical", traveling faster than infinitesimal periodic waves. While there are physical grounds for expecting subcritical solitary waves to be extremely rare, it seems impossible to turn these ideas into a rigorous proof. I will outline a surprisingly simple proof of this conjecture, which hinges instead on the properties of an auxiliary function related to momentum conservation. This is joint work with Vladimir Kozlov and Evgeniy Lokharu.
Zoom: https://arizona.zoom.us/j/99410014231 Password: “arizona” (all lower case)
## When
12:30 p.m. Feb. 23, 2021
Online | 1,948 | 8,888 | {"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} | 2.5625 | 3 | CC-MAIN-2021-25 | latest | en | 0.877621 |
https://www.physicsforums.com/threads/matlab-code-problem-with-differential-equations.897726/ | 1,603,803,605,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107894175.55/warc/CC-MAIN-20201027111346-20201027141346-00058.warc.gz | 873,529,255 | 16,191 | # Matlab code problem with differential equations
## Homework Statement
For a following differential equation
d^2y/dx^2-4y=(e^x)/x
Find the solution using numerical methods
## Homework Equations
d^2y/dx^2-4y=(e^x)/x
## The Attempt at a Solution
Matlab:
%num
dx=0.01;
x=1:dx:3;
l=zeros(1,length(x));
m=zeros(1,length(x));
l(1)=1;
m(1)=0.25;
for ii=2:length(x)
l(ii) = l(ii-1)+m(ii-1)*dx;
m(ii) = m(ii-1) -(1/4) * (l(ii-1)) * dx +(exp(2*x*(ii-1)/(x(ii-1))))*dx;<----- this line keeps throwing an error
end
plot(x,l);
any assistance would be greatly appreciated
Last edited by a moderator:
Related Engineering and Comp Sci Homework Help News on Phys.org
jedishrfu
Mentor
What error are you getting?
If its during execution then perhaps a print statement inside your for loop printing out the results of L(ii) and M(ii) and some of your expressions will give you some insight into what going on.
You should be able to use the computer to help you find the error in your code by tracing its operations.
Delta2
Homework Helper
Gold Member
The way the code is it adds at every step ##e^{\frac{2x}{x}}dx## while I believe you want to add ##\frac{e^{2x}}{x}dx##. Check carefully the parentheses in the (exp(2*x)...) term at that line.
Also what it seems to be a typo it says exp(2*x*(ii-1)... while I believe it should be exp(2*x(ii-1))...
Also not sure since I am rusty on numerical methods and matlab code but this looks like the Euler Method for solving ODEs correct?
Last edited:
jedishrfu
Mentor
Good catch that x*(ii-1) is clearly wrong as the program needs x(ii-1) | 471 | 1,578 | {"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} | 3.484375 | 3 | CC-MAIN-2020-45 | latest | en | 0.884466 |
https://bornholm-sommerhus.info/relationship-between-and/relationship-between-vo2-max-and-heart-rate-formula.php | 1,571,556,091,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570986703625.46/warc/CC-MAIN-20191020053545-20191020081045-00281.warc.gz | 388,159,173 | 9,212 | # Relationship between vo2 max and heart rate formula
Literature shows numerous studies modelling the relationship between steady Figure 1. Model of HR-derived VO2-estimation. HR(max) = (maximal) heart rate. The regression analysis of the quadratic equation also indicated that there was also a significant relationship between the VO2 max and HR. VO2 Max is the most accurate measure of cardio-respiratory fitness. It measures the maximum amount of oxygen that the body can use in one minute per.
To continue improving, you need to keep stressing these systems in a healthy way. By designing a workout plan that targets specific heart rate zones, you can improve your endurance, heart rate and speed. Heart Rate Zone Basics Training within specific heart rate zones can help you reach your fitness goals.
### Relationship Between Heart Rate & VO2 Max | Healthy Living
The first zone, known as the energy-efficient zone, is 60 to 70 percent of your maximum heart rate and develops basic endurance. The anaerobic zone is 80 to 90 percent of your maximum heart rate and helps your develop your lactic acid system by improving your anaerobic threshold.
This is the point when you are working out so hard that your muscles start to burn, eventually forcing you to slow down. Finally there is the red line zone, which is 90 to percent of your maximum heart rate. Training in this zone helps you build speed.
When you are working out, your muscles are pulling oxygen from your bloodstream. By improving this rate, you can work out at a higher intensity or duration. In a meta-analysis of more than 50 studies, published in "Sports Medicine" inthe authors concluded that workouts performed at 90 to percent of your VO2 max pace bring about the largest improvements in VO2 max performance. This was defined as the slowest pace that you reach your maximum oxygen consumption.
### Heart Rate and Percent VO2max Conversion Calculator
For most people, this falls somewhere between your one-mile race pace and two-mile race pace, or how quickly you would run a mile in a race. The study is aimed to investigate the relationship between the aerobic capacity and the RPE based on the measurement of heat rate HR of workers from the Metal Industries of Isfahan.
The subjects were male workers from metal components manufacturers in Isfahan selected by using random sampling based on statistic method.
The subjects were examined by using ergometer in accordance with A strand 6 minutes cycle test protocol. Furthermore, the subjects were asked to rate their status based on the Borg rating scale at the end of each minute. Additionally, their heat rates were monitored and recorded automatically at the end of each minutes.
The regression analysis of the quadratic equation also indicated that there was also a significant relationship between the VO2 max and HR. To assess the effectiveness of an ergonomic intervention program, the special tools are needed to gather information on the fitness. By assessing the physical and physiological characteristics of human beings, it is possible to assign him to a task on the basis of his physiological tolerance limits.
Currently, the aerobic capacity has been as the maximal capacity to perform the work. Today, scientists believe that the ability to perform physical exercise should be determined using the aerobic capacity.
## What’s the Relationship Between VO2max and Heart Rate?
Direct methods and indirect methods. The direct methods include using the treadmill, ergometer, and step tests, while the indirect methods include charts and formulas of Astrand and physiological e.
It is worth noting that the direct methods are more accurate but more expensive which need training technicians for setting up and using equipments while time consuming; on the contrary, the indirect methods are useful and effective for assessing aerobic capacity in the industries as to have no such limitations.
In order to measure the level of subjective physical workload, the RPE scale proposed by Borg is very useful,[ 6 ] which is a rating scale as to rating scales are important supplemental measures of psycho - physical performance and work capacity.
Furthermore, this scale also has several applications in the field of medicine and ergonomics. Therefore, to prevent early deterioration of workforce resulting in low efficiency, it is needed to determine the fitness between the individual and the work. However, yet there has been no comprehensive study done on the physical exercise capacity of Iranian workers and, as a result, there is no detailed information in this area;[ 91011 ] perhaps because of the high cost of technician training to operate the special equipments of direct methods such as ergometer, which isn't cost-effective and affordable for the industry. | 917 | 4,813 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2019-43 | latest | en | 0.936595 |
https://quaoar.su/blog/page/to-the-automatic-milling-feature-recognition | 1,719,272,795,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198865490.6/warc/CC-MAIN-20240624214047-20240625004047-00162.warc.gz | 424,193,856 | 12,032 | Manifold Geometry // Многообразная Геометрия
# Open challenges in automatic feature recognition
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«Науку двигают наукоёмкие задачи промышленности». Жорес Алфёров
## Preamble
In this post, we will look at some advanced feature recognition problems. This digest is far from complete, and I'm sure that some of you could add a few more challenges to the list. Still, this text is here to provide insight into problems we have been working on for several years, so take it as a little retrospective. None of the challenges listed below are just "theoretical": huge amounts of energy and time have been invested in resolving these or similar issues in the past by our team. Although this kind of information is pretty hard to comprehend without specific experience, I feel a need to spread the word and maybe increase interest in this exciting topic among our readers.
## Problem 1: how to recognize lathed+milled parts?
As we saw earlier, an MTS (Max Turned State) is a stock shape resulting from the lathing process. Now, let's suppose that we want to add some milling features to the lathed MTS piece and, in this way, finalize the part. This is where some interesting questions about automatic feature recognition arise.
You may want to read a somewhat related discussion on our forum: "How can I extract the turn profile from a solid model?".
Question number one is whether we can automatically determine the production type (turning and/or milling) from the input CAD part. How can we know if a part needs to be lathed in the first place? It appears that its rotational symmetries play a role, but a complete solution is not as simple as it may appear.
Lathed part.
Question number two is whether we can identify and measure how exactly material is to be removed? Answering this question would allow for proper selection of tools.
Knowing how much milling is required is important for preliminary cost estimation. Here, we might also want to distinguish between roughing and surface finishing. Roughing is related to how much of the volume we subtract from the stock shape. Surface finishing is a more delicate process that is concerned with the total surface area traced by a tool. For example, SEER-3D software by Galorath recognizes different types of surface finish operations. And their reference guide is a nice introduction to the topic of CNC cost estimation as such.
Lathed part with a milled feature.
## Problem 2: how about neural networks?
A commonly acknowledged problem in feature recognition is the difficulty of processing interacting features with deterministic methods. There is actually another trendy research branch that is related to artificial intelligence in feature recognition, but I never saw it in action and frankly doubt it's even worth mentioning. Having said that, it would be completely unfair to leave aside the remarkable research on AI-driven feature recognition conducted by Autodesk: their BRepNet "topological message passing system." So far, it looks quite promising as it delves into the complexity of topological data structures and does not try to oversimplify shape description with point clouds or meshes.
The open question, though, is how smoothly would it work compared to traditional rule-based recognition? I was surprised to read in "On-demand manufacturing platforms are taking the sector by storm" a statement that "on-demand manufacturing platforms leverage artificial intelligence (AI) to recognize geometric features of the requested part." Although published in a respectable-looking flyer composed by seemingly serious people, this statement is nothing but a false generalization. To the best of my knowledge, all successful feature recognition approaches working on CAD geometry in MaaS are based on deterministic rules. Still, machine learning can be applied when it comes to quotations. In the latter case, AI is employed to guess similar prices for similar parts while still knowing nothing about features or design intent. That's the whole thing about approximation: recovering missing data from already-known data chunks.
## Problem 3: how to detect allowed milling axes?
Searching for a deterministic feature recognition method should normally be guided by pragmatic sense. What are we actually looking for? Do we really need to comprehend that a given group of faces is a specific variation of a pocket or a slot, possibly interacting with other pockets and slots? Or would it be enough to know that some faces are side-milled while some others are end-milled?
A sample part with its side-milling, end-milling, and impossible (red) milling axes recognized automatically.
As it appears, recognition of milling axes for the face in question is a good step toward extracting somewhat more complete information about the whole machining process. If we know that a group of neighboring faces is side-milled, then we can assume that there's a cutting trajectory (a closed or an open contour) to finish these faces all along.
It is interesting that recognition of milling axes should employ a bit of a manufacturability rationale. For example, the presence of a concave sharp edge in the model would narrow down the milling possibilities around this feature. In some situations, we can cancel many candidate axes by doing a local topology analysis. One of the useful heuristics here is the analysis of vertex convexity, which was recently introduced in Analysis Situs. To check if a vertex is concave, convex, or smooth in a planar domain, we check its neighborhood: the idea inspired by a good old paper from Rochester [Requicha & Voelcker. (1984). Boolean Operations in Solid Modeling: Boundary Evaluation and Merging Algorithms].
Convexity of a vertex is decided based on its neighborhood analysis.
The local analysis is not sufficient, though. We must ensure that all the detected milling axes are not locked globally by any distant features of the part. This is where accessibility tests based on ray casting come into play. Such ray casting is a big topic in itself, so let's leave it for now and revisit it in the future series related to DFM.
A side-milling axis is prohibited because it does not pass accessibility test.
## Problem 4: how to handle interacting features?
Once the milling axes are known, the corresponding AAG attributes are attached to the B-rep faces, and this way we grow the volume of knowledge about a CAD part under revision. The information about milling axes can now be used for deriving a somewhat upper-level knowledge about the part. In particular, the following features can be extracted based on the milling axes:
1. End-milled faces (bottom faces).
2. Side-milled walls contouring the given bottom face.
3. Slots and grooves.
4. Steps.
5. Engravings.
6. ... and others.
Different applications would come with different demands on the extracted properties. Below is an example of an open "step" feature that allows for being machined in two perpendicular directions. The same axes can be used to group individual faces into a somewhat meaningful feature.
Step feature with some extracted props.
## Problem 5: how to identify setups?
Knowing the approach axes for the detected features opens the door to finding setups. A "setup" is a manual (hence time-consuming) procedure of reclamping a part in a machine, e.g., flipping it to access the otherwise hidden features. A good machining simulation will not only derive the number of required setups, but would also check their reliability by assessing the clamping contact zones.
Possible milling ops (setups) for a part.
Finding setups is a challenging optimization problem which employs several criteria. One way to formulate it is as follows. Given a universum of feature variants F, find the shortest possible sequence of operations O to fabricate all the features without repetitions. Let S be the setup finder function that maps the feature universum F to the list of milling operations O:
Setup finder function S.
The order of operations is critical as the order defines the number of ops: depending on the order of operations, a part might require different number of setups as the same feature face generally allows for being fabricated from different directions. Now, if we agree that the goal is to minimize the number of ops, we have to parameterize the function S in a way that any argument's change affects the number of list elements. Such parameterization is not given for granted though. If an argument's variation does not affect the order of ops, then we just waste iterations and reevaluate S as the same sequence of ops over and over again.
The number of possible setup sequences is m! (factorial) where m is the initial number of distinct milling axes. Although direct permutations of those axes might end up in a sensible order, they would not account for the geometric significance of features and will force the setup finder to test pretty hopeless configurations. Therefore, in order to solve the optimization problem, we have to vary the order of ops more or less forcibly while keeping an eye on the tested order, so that the most significant ops take precedence over the less significant ones.
## Problem 6: how to assess rough-milling volumes?
Volume decomposition aims at detailing how raw material is subtracted from a stock shape to produce a part. Decomposition algorithms search for features in the "delta volume" which is the Boolean difference between a stock and a part. Direct decomposition techniques are cumbersome as they require a number of Boolean operations to be performed one after another. Although there are many papers presenting such decomposition techniques with OpenCascade as a kernel, it will be safe to claim that OpenCascade's Booleans can hardly manage realistic parts and only work for simple prismatic cases. Intensive running of Booleans for decomposing parts of even moderate complexity is far from being an efficient (even viable) solution.
Feature-based delta volume decomposition of a prismatic part.
Still, volume decomposition with OpenCascade remains possible. Amazingly enough, the key to efficient and accurate volume decomposition is the same feature recognition we've been talking about the whole article.
Another example of feature-based delta volume decomposition of a prismatic part.
The trick is to use the recognized finished surfaces and milling axes to approximate the corresponding negative volumes. Another simplification that should be fair for MaaS (Manufacturing as a Service) platforms is to avoid computing explicit volumes for secondary features, such as fillets and chamfers. The topology of fillet/chamfer chains is often so tricky that any attempt to evaluate the explicit boundaries of the corresponding negative volumes might be prohibitively time-consuming and error-prone.
We can use the prismaticity property of a feature to approximate its negative volume.
Another direction of research towards the approximation of negative volumes is avoiding Boolean operations on B-rep and switching to a somewhat ad-hoc data structure where Booleans are not that difficult. This can be a voxelization or, potentially better, dexels or even meshes. Boolean operations on meshes are, of course, not that easy to implement, but they promise much higher performance and reliability compared to precise boundary evaluation.
To conclude, if I were to formulate an open challenge in negative volume classification, I would propose to search for a combination of B-rep feature recognition plus mesh-based Boolean operations. In this paradigm, we can use the luxury of analytic geometry to extract feature props. Then, to assess negative volumes for rough milling, we employ meshes, and this way we escape the well-known robustness and efficiency problems of B-rep Booleans (although sacrificing accuracy).
## Problem 7: how to assess turned volumes?
Turned volumes, like milled volumes, must be broken down into specific machining features, such as bores, steps, and grooves. However, unlike milling, the volume breakdown of a turned object may be done in 2D due to its intrinsic rotational symmetry.
Lathed part with multiple milling features.
Max turned state (MTS) of the part above (constructed automatically as a precise B-rep model).
Swept profile face (white) and negative volumes: steps (red), grooves (orange), central bore (yellow).
Knowing a part's bounding bar and the part itself allows us to break down the negative volume into a set of features using appropriate tessellation of the 2D domain. Instead of computationally expensive Booleans, we can use something more appealing: 2D ray casting. Depending on the arrangement of turning features, each vertex of a turned profile shoots a ray in either the horizontal or vertical direction, hitting a part or a stock. The obtained cells can reveal a great deal about tooling, part complexity, and machining time.
## Problem 8: how to conduct manufacturability checks?
CNC feature recognition goes hand in hand with manufacturability analysis. There are two important points about the AFR-DFM relationship:
• To accurately identify features, we have to employ manufacturability analysis elements. This is what we do, for example, to choose milling axis candidates for each CAD face.
• Without DFM, feature recognition is incomplete. However, accurate DFM tests require significantly more computing power than recognition. As a result, comprehensive DFM checks are better arranged as post-processing logic for AFR.
Thickness check in manufacturability analysis reveals potentially too-thin zones.
Among the DFM inspections, we should test feature accessibility, wall thickness, feature clearance, hole depths, and so on. We will definitely return to this intriguing DFM topic in the future series to walk through some of the computational complexities involved. Meanwhile, you might want to read our NUMGRID 2022 conference paper, which delves deeper into the DFM mechanics.
## Conclusions
Fast, reliable, and accurate feature recognition is only part of the equation. Many more components are required to create a piece of CAM software. For an automatic quotation system, the missing puzzle pieces would include machining time estimation logic and pricing criteria. None of these aspects were addressed above. Material removal simulation is an essential tool for realistic CAPP, and it certainly deserves a whole book to explore all of the challenges surrounding it in depth. We barely touched on tooling and said very little about DFM, despite the fact that manufacturability analysis is of great importance.
After many years of research and development in AFR, CAM, and other relevant subjects, I have to admit that I hardly know enough to lecture you here. Our part of the deal is to create geometric support for engineering technology, while engineering is, of course, way more engaged than playing with geometry. Still, making machining CAD more affordable and seeing it work in the industry feels good enough to justify the existence of this blog and Analysis Situs as software.
Want to discuss this? Jump in to our forum. | 2,967 | 15,197 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.65625 | 3 | CC-MAIN-2024-26 | latest | en | 0.931729 |
http://thestargarden.co.uk/19th-Century-wave-theories.html | 1,701,328,205,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100172.28/warc/CC-MAIN-20231130062948-20231130092948-00534.warc.gz | 50,039,806 | 11,764 | How We Came to Know the Cosmos: Light & Matter
# Chapter 5. 19th Century Wave Theories
## 5.1 Young’s double-slit experiments
The British natural philosopher Thomas Young provided strong evidence for Christiaan Huygens’ wave theory of light (discussed in Chapter 3) in 1803 when he published the results of his double-slit experiments.[1]
Young repeated earlier experiments with diffraction (discussed in Chapter 2) but passed the light through more than one slit. Young stated,
• If light is composed of particles, then they should each pass through a single slit, eventually creating two bright patterns on the other side.
• If light is composed of waves, then the slits will cause diffraction, and the light should produce a predictable interference pattern, just like water waves do.
Figure 5.1Image credit The double-slit experiment with particles.
Figure 5.2Image credit The double-slit experiment with waves.
Young showed that light behaves like a wave and creates an interference pattern, which is a consequence of the superposition principle.
Figure 5.3Image credit Young’s sketch showing the results of his double-slit experiment, 1803.
### 5.1.1 The superposition principle
The superposition principle shows that when two waves meet, a new wave is created that has an amplitude equal to the sum of the amplitudes of the two waves it is composed of. This means that if two waves are emitted in the same phase, then they create a wave that is twice their former amplitude, and waves that are out of phase by 180° will become flat.
Figure 5.4Image credit The superposition of waves in (left) and out (right) of phase.
Once Young knew that light is made of waves, he was able to estimate the wavelengths of individual colours using data from Isaac Newton.
## 5.2 Polarisation
In 1816, the French engineer Augustin-Jean Fresnel proposed that light waves have a transverse as well as longitudinal component, and Young made the same discovery independently, the following year.[2] Young and Fresnel went on to explain Newton’s results in terms of their wave theory and, by 1821, they were able to show that light waves are entirely transverse. This allowed them to explain the behaviour Huygens had documented in calcite crystals[3] (discussed in Chapter 3).
Calcite crystals act as polarisers, a term coined by the French mathematician Étienne-Louis Malus in about 1811, and this means that they only allow the ray to propagate in one plane, either horizontally or vertically.[4]
The horizontal component of a group of waves cannot travel through a vertically aligned polariser, and the vertical component cannot travel through a horizontally aligned polariser. When two crystals are placed next to each other, the light will only be able to travel through both if they are aligned the same way.
Figure 5.5Image credit An electromagnetic wave.
Figure 5.6Image credit Vertical polarisation.
People were able to study electricity in the laboratory for the first time in the early 19th century. The Italian natural philosopher Alessandro Volta created the first electric cell in 1800 by placing paper that had been soaked in salt-water between pieces of zinc and copper. This invoked a voltage, a difference in electrical energy between two points. By connecting many cells together, Volta was able to create a battery, known as a voltaic pile.[5]
Figure 5.7Image credit A diagram of a voltaic pile.
Figure 5.8Image credit A voltaic pile.
In 1831, the British natural philosopher Michael Faraday discovered that if a magnet is moved across a copper wire, then this also creates a current. This is known as Faraday’s law of induction and it was soon put to use in the invention of the electric motor.[6]
Faraday discovered the Faraday effect in 1845. This shows that a magnetic field can cause a ray of polarised light to rotate, a horizontal ray will become vertical and a vertical ray will become horizontal. This means that electricity, magnetism, and light must all be connected.[7]
In 1864, the British natural philosopher James Clerk Maxwell combined Faraday’s law of induction with three other equations:[8] the German mathematician Carl Friedrich Gauss’ two laws concerning electric and magnetic fields[9], and the French natural philosopher Andre-Marie Ampere’s law relating magnetic fields to electric current.[10]
Maxwell used these equations to develop an electromagnetic wave equation. The velocity of this wave was calculated to be the same as the speed of light, and so Maxwell concluded that light is a form of electromagnetic radiation.
Maxwell proposed that light is a transverse wave composed of oscillating electric and magnetic fields. Maxwell’s equations predicted that light could have an infinite number of wavelengths, suggesting that light must exist at energies well beyond the visible spectrum.
### 5.3.1 The electromagnetic spectrum
The British astronomer William Herschel had discovered infrared light in 1800. Herschel measured the temperature of different colours using prisms and found that the temperature was highest just beyond the colour red.[11] The German natural philosopher Johann Wilhelm Ritter predicted the existence of ultraviolet light in 1801 when he found that it reacts with silver chloride.[12]
The electromagnetic spectrum was completed by the end of the 19th century. The German physicist Heinrich Hertz demonstrated the existence of radio waves and microwaves by 1888,[13] the German physicist Wilhelm Röntgen discovered X-rays in 1895,[14] and the French chemist Paul Villard discovered gamma-rays in 1900.[15]
By 1900, the light bulb, the tram, and the telephone were invented and electrical transmission lines were developed that would soon allow the public to access electricity in their own homes.[16]
Figure 5.9Image credit ‘First electric street lights in Berlin’, 1884 by Carl Saltzmann.
Figure 5.10Image credit The electromagnetic spectrum, where blue light has a shorter wavelength than red light.
## 5.4 The aether
Maxwell’s electromagnetic theory, like all previous theories of light, relied on the idea that space is filled with a substance known as the aether. The aether was thought to be a medium, like air or water, that allows light waves to travel through space. As the Earth moves through space, it moves through the aether, and the light of the Sun can be dragged forwards in its direction of travel, like sound in the wind.
Galileo’s relativity (discussed in Book I) showed that speeds are additive. This means that if you walk at speed v across the deck of a ship that is moving at speed u, then someone standing on the shore will measure your speed to be v + u. When light from the Sun is travelling in the same direction as the aether, we should measure its speed to be c + a, where a is the velocity of the aether. When light travels in the opposite direction to the aether, we should measure its speed to be c - a.
Figure 5.11Image credit The aether.
In 1887, the American physicists Albert Michelson and Edward Morley devised an experiment that was precise enough to measure the difference in the speed of light as the Earth moves around the Sun.[17] To almost everyone’s surprise, they found that the speed of light moves at the same rate in all directions. This meant that there was no aether, and so no explanation for how light can travel through space. This was explained in the 20th Century, with the discovery of quantum mechanics (discussed in Chapter 8). | 1,608 | 7,489 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.78125 | 3 | CC-MAIN-2023-50 | latest | en | 0.936346 |
https://bestessaywritingservices.org/bus-461-week-4-assignment/ | 1,627,638,444,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046153966.52/warc/CC-MAIN-20210730091645-20210730121645-00435.warc.gz | 144,796,233 | 13,459 | # Bus 461 week 4 assignment
## Multiple Regression
For this assignment, you are required to complete Problem 64 (p. 593) in Chapter 12 of your textbook. Use the appropriate Excel file templates from www.cengagebrain.com. Once complete, post your Excel Document in Waypoint. Show all work.
23-64) Let Yt be the sales during month t (in thousands of dollars) for a photography studio, and let Pt be the price charged for portraits during month t. The data are in the file Week 4 Assignment Chapter 12 Problem 64. Use regression to fit the following model to these data:
Yt = a + b1Yt−1 + b2Pt + et
This equation indicates that last month’s sales and the current month’s price are explanatory variables. The last term, et, is an error term.
1. If the price of a portrait during month 21 is \$10, what would you predict for sales in month 21?
2. Does there appear to be a problem with autocorrelation of the residual? Explain your answer. | 233 | 938 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.9375 | 3 | CC-MAIN-2021-31 | longest | en | 0.859986 |
https://www.studypool.com/questions/150839/assess-your-understanding-of-the-main-statistical-concepts-covered-in-this-cour | 1,544,882,063,000,000,000 | text/html | crawl-data/CC-MAIN-2018-51/segments/1544376826856.91/warc/CC-MAIN-20181215131038-20181215153038-00574.warc.gz | 1,049,933,754 | 24,344 | # assess your understanding of the main statistical concepts covered in this cour
Anonymous
Question description
The purpose of the Exam is to assess your understanding of the main statistical concepts covered in this course. Part I includes three essay questions. All of your responses should be included in a single Word document for submission.
Please include the following general headings for each section of the written exam within your Word document:
Part I: Essay Questions
1. Essay 1
2. Essay 2
3. Essay 3
Part I: Essay Questions
There are three essay questions in this section. You must answer all three questions. The length of each essay should be one to two double-spaced pages (excluding title and reference pages). Use 12-point font and format your paper with regular 1-inch margins. Do not include the essay prompt in your document. It will not count toward the length requirement for your essays.
Essay 1
A group of researchers conducted an experiment to determine which vaccine is more effective for preventing getting the flu. They tested two different types of vaccines: a shot and a nasal spray. To test the effectiveness, 1000 participants were randomly selected with 500 people getting the shot and 500 the nasal spray. Of the 500 people were treated with the shot, 80 developed the flu and 420 did not. Of the people who were treated with the nasal spray, 120 people developed the flu and 380 did not. The level of significance was set at .05. The proportion of people who were treated with the shot who developed the flu = .16, and the proportion of the people who were treated with the nasal spray was .24. The calculated p value = .0008.
For this essay, describe the statistical approaches (e.g., identify the hypotheses and research methods) used in this excerpt from a research study. Interpret the statistical results and examine the limitations of the statistical methods. Finally, evaluate the research study as a whole and apply what you have learned about hypothesis testing and inferential statistics by discussing how you might conduct a follow-up study.
•Describe the research question for this experiment. ◦What were the null and alternative hypotheses?
◦Were the results of this test statistically significant?
◦If so, why were they significant?
•Would the researchers reject or fail to reject the null hypothesis?
•Do the results provide sufficient evidence to support the alternative hypothesis?
•What are some possible limitations to this study?
•Describe the difference between practical and statistical significance.
Essay 2
A researcher has investigated the relationship between IQ and grade point average (GPA) and found the correlation to be .75.
For this essay, critique the results and interpretation of a correlational study.
•Evaluate the correlational result and identify the strength of the correlation.
•Examine the assumptions and limitations of the possible connection between the researcher’s chosen variables.
• Identify and describe other statistical tests that could be used to study this relationship.
•How strong is this correlation? ◦Is this a positive or negative correlation?
◦What does this correlation mean?
•Does this correlation imply that individuals with high Intelligence Quotients (IQ) have high Grade Point Averages (GPA)?
• Does this correlation provide evidence that high IQ causes GPA to go higher? ◦What other variables might be influencing this relationship?
•What is the connection between correlation and causation?
•What are some of the factors that affect the size of this correlation?
•Is correlation a good test for predicting GPA? ◦ If not, what statistical tests should a researcher use, and why?
Essay 3
A researcher has recorded the reaction times of 20 individuals on a memory assessment. The following table indicates the individual times:
2.2 4.7 7.3 4.1
9.5 15.2 4.3 9.5
2.7 3.1 9.2 2.9
8.2 7.6 3.5 2.5
9.3 4.8 8.5 8.1
In this essay, demonstrate your ability to organize data into meaningful sets, calculate basic descriptive statistics, interpret the results, and evaluate the effects of outliers and changes in the variables. You may use Excel, one of the many free online descriptive statistics calculators, or calculate the values by hand and/or with a calculator.
Next, separate the data into two groups of 10; one group will be the lower reaction times, and the second group will be the higher reaction times. Then, address the following points in your essay response:
•Calculate the sum, mean, mode, median, standard deviation, range, skew, and kurtosis for each group.
•How do the two groups differ?
• Are there any outliers in either data group?
•What effect does an outlier have on a sample?
Lastly, double each sample by repeating the same 10 data points in each group. You will have a total of 20 data points for each group. After completing this, address the following in your essay response:
•Calculate the following for the new data groups: sum, mean, mode, median, standard deviation, range, skew, and kurtosis.
•Did any of the values change?
•How does sample size affect those values?
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2325 Tutors | 1,290 | 5,890 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.6875 | 4 | CC-MAIN-2018-51 | latest | en | 0.948694 |
https://jjthetutor.com/physics-1-final-exam-study-guide-review-multiple-choice-practice-problems/ | 1,591,369,148,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590348502097.77/warc/CC-MAIN-20200605143036-20200605173036-00162.warc.gz | 379,960,769 | 16,212 | # Physics 1 Final Exam Study Guide Review – Multiple Choice Practice Problems
This physics video tutorial is for high school and college students studying for their physics midterm exam or the physics final exam. This study guide review tutorial contains 50 multiple choice practice problems.
Algebra Final Exam Review:
https://www.youtube.com/watch?v=RSqhRY74_rA
Calculus Final Exam Review:
https://www.youtube.com/watch?v=ZTx5KJEC-ks&t=5529s
General Chemistry 1 Review:
https://www.youtube.com/watch?v=5yw1YH7YA7c
Organic Chemistry 1 Final Review:
https://www.youtube.com/watch?v=YQIlAfjR2f4
Physics 2 Final Exam Review:
https://www.youtube.com/watch?v=negS7MyRlgI
You can access the full video here:
https://www.patreon.com/MathScienceTutor
New Physics Video Playlist:
https://www.youtube.com/playlist?list=PL0o_zxa4K1BU6wPPLDsoTj1_wEf0LSNeR
Here is a list of topics:
1. One Dimension Kinematic Problems
2. Average Speed and Average Velocity Problems
3. Displacement, Velocity and Acceleration Problems
4. Projectile Motion Physics Problems
5. Magnitude and Direction Angle of Vectors
6. Newton’s Second Law of Motion
7. Inclined Plane Physics Problems & Free Body Diagrams
8. Static Friction and Kinetic Friction
9. Net Force Problems
10. Tension Force and Normal Force Physics Problems
11. Maximum Height and Range of a Projectile
12. Launch Angle of a Projectile
13. Work Energy Principle
14. How To Calculate The Work Required To Lift an Object
15. Kinetic Energy and Potential Energy
16. Newton’s Third Law of Motion
17. Pulley Problems and Acceleration
18. Circular Motion and Tension In a Rope
19. Car Rounding a Curve – Static Friction and Maximum Speed
20. Centripetal Force and Circular Motion
21. Gravitational Force
22. How To Calculate The Speed of a Satellite
23. Conservation of Energy Problems
24. Elastic Potential Energy and Final Speed
25. Average Power, Force, and Velocity
26. Cost of Electricity and Kilowatt hours
27. Velocity Time Graphs, Slope, and Acceleration
28. Inelastic Collisions and Conservation of Momentum
29. Torque, Rotational Inertia, and Angular Acceleration
30. Conservation of Angular Momentum
31. Seesaw Problem With Torques & Mechanical Advantage
32. Rotational Kinematic Problems
33. Angular Speed and Linear Speed Problems
34. Net Force and Momentum
35. How To Calculate The Work Done Given a Force Displacement Graph
36. Work Done By a Force | 598 | 2,399 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.890625 | 3 | CC-MAIN-2020-24 | latest | en | 0.642275 |
https://codeauri.com/blogs/flowchart-pros-cons-examples-guidelines/ | 1,721,101,383,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514726.17/warc/CC-MAIN-20240716015512-20240716045512-00784.warc.gz | 149,213,741 | 22,490 | # Flowchart( Introductions, Pros, Cons, Guidelines,Examples)
## What is a Flowchart?
A flowchart is generally defined as the graphical representation of any algorithm with the use of standard symbols.
Well, a flowchart is also a pictorial/ diagrammatic representation of an algorithm using various shapes like boxes.
It Facilitates visual communications between business executives and programmers with ease.
Well, flowcharts are helpful while explaining the program to others which has to be documented for complex programs.
You may also like: Difference Between Compiler, Interpreter, and Assembler
## Pros of Flowchart:
### Good Communications:
• Flowcharts are regarded as a suitable way of communicating to make understandable representations between programmers and business people.
• So, What flowchart does is; provide the descriptions of algorithms, the ideas, and the logic to any other users, students, etc.
### Comprehensive Analysis:
• Flowchart presents the overview of entire problems and the algorithm steps with solutions
• So, in this way, the problems can be analyzed much more efficiently.
### Effective Documentation:
• The flowchart provides the proper documentation to execute in an algorithm
• It helps the programmers who will take part in re-updating the features in future
## Cons of Flowchart:
### Difficult Logic:
When the features and the ideas to be integrated into the program gets heavy and expand, the difficulty of logic to understanding the flowchart also gets difficult
### Complex in Modification:
If the program needs changes and modification then, it is more likely to redraw the flowchart completely from the beginning itself.
## Guidelines to follow in Flowchart
• Flowcharts should be started from the top of the page and flow down and to the right.
Only standard flowcharting symbols should be used.
• There should be a start and stop on every flowchart.
The flowchart should be clear, neat, and easy to follow. There should be no ambiguity in understanding the flowchart.
• The direction of the flow line of a procedure or system should be from left to right or top to bottom.
• Only one flow line should emerge from a process symbol.
• Only one flow line should enter a decision symbol, but two or three flow lines, one for each possible answer, can leave the decision symbol.
• Only one flow line is used in conjunction with a terminal (Start and Stop) symbol.
• The contents of each symbol should be written legibly. English-like language should be used in flow charts, not specific programming languages.
• If the flowchart becomes complex, connector symbols should be used to reduce the number of flow lines. The intersection of flow lines should be avoided to make the flow chart a more effective and better way of communication.
## Examples of Flowchart
Source: A textbook of C-Programming
Codeauri is Code Learning Hub and Community for every Coder to learn Coding by navigating Structurally from Basic Programming to Front-End Development, Back-End Development to Database, and many more.
## How C Programming is executed (With Compiling Flowchart)?
You may like this: C Programming(Rich History of C’s Development, Advantages) Execution of C-Programming Creating Source Code: Computer instructions are composed within a text editor, delineating specific computational…
## C Programming(Rich History of C’s Development, Advantages)
Back in 1822 Since the dawn of Charles Babbage’s ingenious difference engine back in 1822, computers have been in need of a way to comprehend and execute specific…
## Difference Between Compiler, Interpreter, and Assembler-Codeauri
The Difference Between Compiler, Interpreter, and Assembler is given Below: You may also like: Flowchart( Introductions, Pros, Cons, Guidelines,Examples) Compiler The compiler helps in translating the high-level program…
## Programming Languages( Types, Pros and Cons)-Codeauri
What are the types or levels of Programming Languages? The types or levels of programming languages are divided into two types: Types Of Programming Languages: 1.Machine-level Language :1st…
Your Journey into Code Begins Now: Discover the Wonders of Basic Programming
X | 837 | 4,198 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.8125 | 3 | CC-MAIN-2024-30 | latest | en | 0.898517 |
http://clay6.com/qa/12973/a-coin-is-tossed-6-times-in-how-many-ways-we-can-get-the-3-head-in-the-5-to | 1,480,905,148,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698541518.17/warc/CC-MAIN-20161202170901-00495-ip-10-31-129-80.ec2.internal.warc.gz | 56,359,360 | 26,988 | Browse Questions
# A coin is tossed 6 times In how many ways we can get the $3^{rd}$ head in the $5^{th}$ toss?
$5^{th}$ toss is head(H).
First 4 tosses can be 2 heads (H) and 2 tails(T). and
$6^{th}$ toss can be head or tail (2 ways)
This can be got in $\large\frac{4!}{2!.2!}$$\times 2$ ways.
$=12\: ways$ | 113 | 309 | {"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} | 3.859375 | 4 | CC-MAIN-2016-50 | longest | en | 0.783859 |
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# Complex Numbers And Quadratic Equations
## Complex Numbers and Quadratic Equations PDF Notes, Important Questions and Formulas
DEFINITION
Complex numbers are defined as expressions of the form a + ib where a, b ∊ R & . It is denoted by z i.e. z=a + ib. ‘a’ is called as real part of z (Re z) and ‘b’ is called as imaginary part of z (Im z).
EVERY COMPLEX NUMBER CAN BE REGARDED AS
Purely real Purely imaginary Imaginary If b = 0 If a = 0 If b ≠ 0
ALGEBRA OF COMPLEX NUMBER
The algebraic operations on complex numbers are similar to those on real numbers treating ‘I’ as a polynomial. In equalities in complex numbers are not defined. There is no validity if we say that complex number is positive or negative.
e.g. z > 0, 4 + 2i < 2+4i are meaningless.
However in real numbers if a2 + b2=0 then a = 0 =b
but in complex numbers,
z12+z22=0 does not imply z1=z2=0
Equality In Complex Number
Two complex numbers z1= a1 + ib1 & z2 =a2 +ib2 are equal if and only if their real & imaginary parts coincide.
CONJUGATE COMPLEX
If z=a + ib then its conjugate complex is obtained by changing the sign of its imaginary & is denoted by
GENERAL POLYNOMIAL
A function f defined by f(x) = anxn+an-1xn-1+……..+a1x+a0, where a0, a1, a2….. an ∊ R is called n degree polynomial while coefficient (an ≠0, n ∊ W) is real .If a0 , a2 ….an ∊ C, then it is called complex coefficient polynomial.
A polynomial of degree two in one variable f(x) = y = ax2 + bx + C, where a ≠ 0 & a, b, c ∊ R
a → leading coefficient, c → absolute term/ constant term
If a = 0 then y = bx + c → linear polynomial b ≠ 0
If a = 0, c = 0 then y = bx → odd linear polynomial
1. The solution of the quadratic equation, ax2+ bx + c = 0 is given by
The expression b2-4ac = D is called the discriminant of the quadratic equation.
2. If α & β are the roots of the quadratic equation
Ax2 + bx + c =0 then,
1. α + β= -b/a
2. αβ =c/a
NATURE OF ROOTS
1. Consider the quadratic equation ax2 + bx + c = 0 where a, b, c ∊ R, a ≠ 0 then;
1. D > 0 ⇔ roots are real & distinct (unequal)
2. D = 0 ⇔ roots are real & coincident (equal)
3. D < 0 ⇔ roots are imaginary
4. If p +iq is one root of a quadratic equation, then the other must be the conjugate p – iq & vice versa. ( p, q ∊ R &
2. Consider the quadratic equation ax2+ bx + c = 0 where a, b, c ∊ Q & a≠0 then;
1. If D > 0 & is a perfect square, then roots are rational & unequal.
2. If is one root in this case, (where p is rational & is a surd) then the other root must be the conjugate of it i.e. & vice versa.
GRAPH OF QUADRATIC EXPRESSION
1. The graph between x, y is always a parabola.
2. If a > 0 then the shape of the parabola is concave upwards & if a < 0 then the shape of the parabola is concave downwards.
3. The co-ordinate of vertex are
4. The parabola intersect the y-axis at point (0,c)
5. The x-co-ordinate of point of intersection of parabola with x-axis are the real roots of the quadratic equation f(x) = 0. Hence the parabola may or may not intersect the x-axis at real points.
RELATION BETWEEN ROOTS & COEFICIENTS
A quadratic equation whose roots are α & β is
(x – α)(x - β)=0
i.e. x2-(α + β)X +αβ=0
i.e. x2-(sum of roots)x + product of roots=0
## Ask Experts for IIT-JEE
Queries asked on Sunday and after 7 pm from Monday to Saturday will be answered after 12 pm the next working day.
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https://www.shortlist.com/entertainment/sport/football-chip-messi-cantona-science-art/362180 | 1,537,462,570,000,000,000 | text/html | crawl-data/CC-MAIN-2018-39/segments/1537267156524.34/warc/CC-MAIN-20180920155933-20180920180333-00134.warc.gz | 866,556,626 | 31,828 | Sport
# An in-depth analysis of the beauty of the chip in football
Posted by
Alex Christian
Published
An in-depth analysis of the most graceful sight in sport, by the scientists, scholars and sportsmen who know it best
## The maths genius
### Hugo Castillo Sanchez, chemical engineer
“The greatest players are math machines: they do instant computations to know the ideal conditions to strike a ball. Take the chip. One-on-one, the keeper rapidly approaching, the shooting angle is restricted to lifting the ball over his head into the net. The perfect calculation and the ball travels with a U-shaped, parabolic trajectory that’s impossible to catch.
“For the perfect chip, the ball has to be hit with a velocity less than 20 metres per second. Then, the gravitational force of the ball must dominate over the drag force [air resistance] and the lift force [the ball’s spin]. Finally, the air resistance must be higher than the effect of the rotating ball. Say the attacker is 12 metres from goal. The keeper rushes out, cutting his distance to the ball to six metres. The attacker has to kick the ball with a velocity under 20m/s and a max height greater than the keeper’s – around two metres. However, as the keeper can also jump to stop the shot, the attacker’s strike has to reach a vertical distance of around three metres. Then, the attacker will need to calculate the velocity and angle required. The initial velocity would need to be 10.84 metres per second with a launching angle of 450 for the ball to follow a parabolic path over the keeper and into the net, leaving 1.56 seconds for defenders to block the goal-bound chip.
“That’s the calculation – it takes ages to work out. What’s amazing is that world-class footballers, the likes of Lionel Messi, have it seemingly programmed in them to calculate in an instant, while also knowing the aerodynamic conditions required to score a perfect chip nearly every time.”
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## The keeper
### David James, former Liverpool and England number one
“Back in the day, it was considered unsportsmanlike to chip the keeper. But the reality is, if you’re in the position to be chipped, you should be chipped. The embarrassing part is that you’re just not ready for it – in the split-second it takes you to process the shot, it’s already too late.
“I got chipped by Wayne Rooney for Portsmouth against Manchester United and it was so, so frustrating. Milliseconds after it left his foot, I was looking at the ball, thinking, “This could go in,” rather than running back to save it. I just didn’t see it coming. Perfect chips go in, and that one was inch-perfect. But when you save one, you can’t help but think that they’re cheeky for even trying to attempt it.”
LG, partners of the England men’s team, and David James surprised residents of Wembley Grove, Birmingham, with free LG TVs and sound bars so they can live the game this summer. Watch the video.
## The philosopher
### Stephen Mumford, metaphysicist and author
“The delicacy of the chip plays into the thought of football as an art – finesse over power. The chip is one of the thrills we seek in football. We’re so used to shots generated with speed and strength that when we witness a perfectly executed chip, time appears to slow down.
“It’s all about the placement, the arc, the trajectory – the ball rising and coming down at the exact right moment, inch-perfect. I was there when Eric Cantona did it against Sheffield United in 1995. The keeper nearly reached it, the crossbar nearly deflected it. It was perfect accuracy, executed by a player who believed football was capable of higher artistic values.”
## The player
### Jermaine Jenas, former Spurs and England midfielder
“The chip is technically very difficult. There are so many elements at play. You have to get your foot right underneath the ball, and delicately enough that it immediately goes up and down again. The chip isn’t always arrogance. If you’re squeezed to the side of the box, you’ll know the keeper will be by his near post – chipping it might be your best option to score. The keeper won’t agree, though.
“In training, if you tried to chip them and they managed to get back and catch it, they’d take the ball and volley it as far away as possible, and then tell you to do one. But that’s what makes it one of the most satisfying shots – it’s one of the hardest to pull off.”
## The psychologist
### Dr Victor Thompson, clinical sports psychologist
“A chip involves more personality than skill. All top-flight footballers have the ability, but not necessarily the confidence. Players know a failed chip will mean more flak; psychologically, they’ll be more inclined to perform to the norm and attempt a more ordinary shot on goal. You’re more likely to see a chip when a team is up by four goals, when they can afford to play more freely.
“Confident players are always more likely to attempt the chip – they can block out the worry and focus on the opportunity. When it goes in, it’s a huge psychological boost for the attacking side. Defenders and keepers, meanwhile, will be humiliated. You hate what you’ve just witnessed, yet you can’t fail to admire it at the same time.”
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# Code No: X0222/R07
Set No. 1
II B.Tech I Semester Supplementary Examinations, May 2010 FLUID MECHANICS AND HYDRAULIC MACHINERY (Electrical & Electronic Engineering) Time: 3 hours Max Marks: 80 Answer any FIVE Questions All Questions carry equal marks
1. (a) State Newton’s equation of viscosity and give examples of its application. (b) An oil of viscosity 5 poise is used for lubrication between a shaft and sleeve. The diameter of shaft is 0.5 m and it rotates at 200 rpm. Calculate the horse power lost in the oil for a sleeve length of 100 mm. The thickness of the oil film is 1.0 mm. [8+8] 2. (a) What do you mean by turbulence ? (b) What do you mean by one, two, three dimensional flow? (c) Derive the continuity equation for one dimensional fluid flow. 3. (a) Derive an equation for discharge of a venturimeter. (b) Explain why Cd of a venturimeter is more than that of orifice meter. (c) A horizontal venturimeter with inlet and throat diameters 300mm and mm respectively is used to measure the flow of water. The pressure intensity at inlet is 130KN/m2 . While the vaccum pressure head at the throat is 350mm of mercury. Assuming that 3% of head is lost in between inlet and throat, find Cd of venturimeter and rate of flow. [6+2+8] 4. (a) Derive an expression for the force exerted by a jet striking the curved plate at one end tangentially when the plate is symetrical. (b) A jet of water 120 mm in diameter and moving with a velocity of 25m/sec strikes normally on a flat plate. Determine the power developed and the efficiency of the system when i. the plate is stationary ii. the plate is moving with a velocity of 8m/sec in the direction of the jet. And iii. the plate is moving with a velocity of 8m/sec towards the jet. [7+9] 5. (a) Write what you understand by hydroelectric power station? What are its types ? Discuss the type where water is recycled. (b) What power can be developed at a hydropower plant site given the following details.Drainage area = 600 Km2 Uniform rainfall over drainage area = 200 cm. Water loss in run off = 15 cm Expected net head at the plant = 50m. Efficiency of turbine = 98% [8+8] 1 of 2 Please purchase PDF Split-Merge on www.verypdf.com to remove this watermark. [2+6+8]
Code No: X0222/R07
Set No. 1
6. (a) What are the functions of a draft tube in a reaction turbine ? Derive the equation for its efficiency. (b) A mixed flow reaction turbine has a draft tube of inlet diameter 1.5 m and is set 2 m above tail race level. A vacuum gauge connected to the draft tube indicates a reading of 5.2 m of water. Find discharge of the turbine. Assume the draft tube efficiency as 85%. [8+8] 7. (a) What is governing of turbines ? How can you govern a reaction turbine ? Explain the governing mechanism for reaction turbine with a neat sketch. (b) A turbine operates under a head of 45m at 280 rpm. The discharge is 10m3 / sec and efficiency is 87 %. Find i. Power developed ii. Specific speed. [10+6] 8. (a) Discuss i. ii. iii. iv. v. Manometric efficiency Mechanical efficiency Volumetric efficiency Overall efficiency and NPSH of a centrifugal pump.
(b) Explain how the number of pumps can be decided to be connected in series to deliver the water to a required height at required rate. [10+6]
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Code No: X0222/R07
Set No. 2
II B.Tech I Semester Supplementary Examinations, May 2010 FLUID MECHANICS AND HYDRAULIC MACHINERY (Electrical & Electronic Engineering) Time: 3 hours Max Marks: 80 Answer any FIVE Questions All Questions carry equal marks
1. (a) State Newton’s equation of viscosity and give examples of its application. (b) An oil of viscosity 5 poise is used for lubrication between a shaft and sleeve. The diameter of shaft is 0.5 m and it rotates at 200 rpm. Calculate the horse power lost in the oil for a sleeve length of 100 mm. The thickness of the oil film is 1.0 mm. [8+8] 2. (a) Distinguish between: i. Steady and Un-steady flow. ii. Uniform and Non-uniform flow. iii. Compressible and incompressible flow. (b) A 400 m long pipe has a slope of 1 in 100 and tapers from 1.2 m dia at high end to 0.6 m dia at the low end. The discharge is 100 lit/sec. If the pressure at high end is 100 kN/m2 , find the pressure at the low end. Neglect friction. [8+8] 3. (a) Derive the Darcy-weisbach equation for friction head loss in a pipe. (b) Water is flowing through a horizontal pipe line 1500m long and 200mm in diameter. Pressures at the two ends of the pope line are respectively 12Kpa and 2Kpa. If f = 0.015, determine the discharge through the pipe in litres per minute. Consider only frictional loss. [8+8] 4. (a) Derive an expression for force exerted by a jet on stationary inclined flat plate. (b) A nozzle of 50mm diameter delivers a stream of water at 20m/sec perpendicular to a plate that moves away from the Jet at 5m/sec.find i. the force on the plate ii. work done and iii. the efficiency of jet.
[7+9]
5. (a) What is mass curve? Give the step wise procedure of constructing the same? What are its features and applications (b) Define gross head and net head associated with working of turbines. What losses are to be considered in finding the net head from gross head ? List the efficiencies of turbines. [8+8] 6. (a) What are single jet and multi jet impulse turbines ? Draw neat sketches and explain. 1 of 2 Please purchase PDF Split-Merge on www.verypdf.com to remove this watermark.
Code No: X0222/R07
Set No. 2
(b) Show from fundamentals that hydraulic efficiency of a Pelton wheel is given [6+10] by η= 2[V1 −u][1 + KCosφ]u/V2 with usual notation. 1 7. (a) What do you mean by selection of turbines ? What are scale ratios ? Explain with neat sketches when geometric similarity is achieved between model and prototype? (b) A Pelton wheel turbine develops a power of 1000 Kw under a head of 75m. If the head becomes 25m, what will be the power developed by the turbine. [10+6] 8. (a) What do you understand by pumps in series and pumps in parallel ? How do you decide whether the pumps shall be used in series or parallel. (b) A single acting reciprocating pump running at 60 rpm delivers 0.53 m3 of water per minute. The diameter of the piston is 200 mm and stroke length 300 mm. The suction and delivery heads are 4m and 12m respectively. Determine i. ii. iii. iv. Theoretical discharge Coefficient of discharge Percentage slip of the pump and Power required to run the pump.
[8+8]
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Code No: X0222/R07
Set No. 3
II B.Tech I Semester Supplementary Examinations, May 2010 FLUID MECHANICS AND HYDRAULIC MACHINERY (Electrical & Electronic Engineering) Time: 3 hours Max Marks: 80 Answer any FIVE Questions All Questions carry equal marks
1. (a) Enunciate Newton’s law of Viscosity. Explain the importance of viscosity in fluid motion. What is the effect of temperature on viscosity of water and that of air? (b) A glass tube 20 mm in diameter contains a mercury column with water above the mercury. The surface tension of mercury in contact witht water is 0.36 N/m. Datermine the capillary depression of the mercury. Take θ= 1300 .[8+8] 2. (a) Define the equation of continuity. Obtain an expression for continuity equation for a one- dimensional flow (b) A pipe of diameter 40 cm carries water at a velocity of 25 m/s. The pressures at the point A and B are given as 29.4 N/cm2 and 22.56N/cm2 respectively while the datum head at A and B are 28 m and 30 m. Find the loss of head between A and B. [8+8] 3. (a) Derive an expression for head loss due to sudden enlargement of a pipe (b) An oil of specific gravity 0.85 and viscosity 5CP flows through a pipe of diameter 400mm at the rate of 50 lit/sec. Find the head lost in friction in this pipe of length 1000Km. Assume that f = 0.079/RN where RN is Reynolds Number. [8+8] 4. (a) Derive an expression for the force exerted by a jet striking the curved plate at one end tangentially when the plate is symetrical. (b) A jet of water 120 mm in diameter and moving with a velocity of 25m/sec strikes normally on a flat plate. Determine the power developed and the efficiency of the system when i. the plate is stationary ii. the plate is moving with a velocity of 8m/sec in the direction of the jet. And iii. the plate is moving with a velocity of 8m/sec towards the jet. [7+9] 5. (a) What are the types of hydroelectric power stations? Write about each of them. (b) At the site where hydroelectric plant is proposed the following details are available. Calculate the power that can be developed. Available head = 28 m Catchment area = 420 Km2 Yearly rainfall = 140 cm Rainfall utilization rate = 68% Penstock efficiency = 94% 1 of 2 Please purchase PDF Split-Merge on www.verypdf.com to remove this watermark.
Code No: X0222/R07 Turbine efficiency = 80% Generator efficiency = 84%
Set No. 3
[8+8]
6. (a) What is draft tube and what are its functions in a reaction turbine. (b) Derive the equation for work done in Pelton wheel turbine from fundamentals. [6+10] 7. (a) How is geometric similarity obtained between model and prototype ? Explain with sketches and give examples. (b) A Francis turbine works under the following conditions. Estimate its performance when the head drops down to 45 m. The head = 60m, Speed = 1200 rpm, Discharge = 12 m3 / sec, Efficiency = 88 % [8+8] 8. (a) What are the components of a reciprocating pump ? Explain its working with a neat sketch. What is the use of providing an air vessel ? (b) The diameter of piston of a reciprocating pump is 15 cm and stroke length is 30 cm. The suction pipe diameter, length and suction head are 5 cm, 5m and 2.5 m respectively. The cavitation occurs at 2.0 m of water (abs) and atmospheric pressure is 10 m of water (abs).What is the maximum speed of the pump without cavitation ? [8+8]
2 of 2 Please purchase PDF Split-Merge on www.verypdf.com to remove this watermark.
Code No: X0222/R07
Set No. 4
II B.Tech I Semester Supplementary Examinations, May 2010 FLUID MECHANICS AND HYDRAULIC MACHINERY (Electrical & Electronic Engineering) Time: 3 hours Max Marks: 80 Answer any FIVE Questions All Questions carry equal marks
1. (a) Enunciate Newton’s law of Viscosity. Explain the importance of viscosity in fluid motion. What is the effect of temperature on viscosity of water and that of air? (b) A glass tube 20 mm in diameter contains a mercury column with water above the mercury. The surface tension of mercury in contact witht water is 0.36 N/m. Datermine the capillary depression of the mercury. Take θ= 1300 .[8+8] 2. (a) What is continuity equation? Derive the equation. (b) Derive Euler?s equation of motion along a stream line. State assumptions made in the derivation. [8+8] 3. (a) Describe the principle and working of orifice meter with the help of a neat sketch. (b) In a calibration test of an orifice meter of with orifice diameter 4cm is inserted in a pipe of 10cm diameter the mercury differential U-guage connected to the meter gives a reading of 38cm when 7.5 lit/sec of water flows through the meter. Compute the coefficient of discharge. [8+8] 4. (a) Derive an expression for the force exerted by a jet striking the curved plate at one end tangentially when the plate is symetrical. (b) A jet of water 120 mm in diameter and moving with a velocity of 25m/sec strikes normally on a flat plate. Determine the power developed and the efficiency of the system when i. the plate is stationary ii. the plate is moving with a velocity of 8m/sec in the direction of the jet. And iii. the plate is moving with a velocity of 8m/sec towards the jet. [7+9] 5. (a) What are the types of hydroelectric power stations? Write about each of them. (b) At the site where hydroelectric plant is proposed the following details are available. Calculate the power that can be developed. Available head = 28 m Catchment area = 420 Km2 Yearly rainfall = 140 cm Rainfall utilization rate = 68% Penstock efficiency = 94% Turbine efficiency = 80% Generator efficiency = 84% [8+8] 1 of 2 Please purchase PDF Split-Merge on www.verypdf.com to remove this watermark.
Code No: X0222/R07
Set No. 4
6. (a) What is an axial flow turbines ? What are its parts ? Explain with sketches. (b) Derive the equation for work done on the runner of an axial flow turbine from fundamentals. [6+10] 7. (a) What do you mean by selection of turbines ? What are scale ratios ? Explain with neat sketches when geometric similarity is achieved between model and prototype? (b) A Pelton wheel turbine develops a power of 1000 Kw under a head of 75m. If the head becomes 25m, what will be the power developed by the turbine. [10+6] 8. (a) Draw a neat sketch of an indicator diagram for a single acting reciprocating pump and explain its features. (b) A reciprocating pump has a 20 cm diameter cylinder and a stroke of 120 cm. The pump delivers 5000 litres / minute of water at a static head of 80m. Friction losses in suction and delivery pipe are 2m and 18m respectively. The slip of pump is 2% and efficiency is 90%. Determine the speed of the pump and power required neglecting velocity head in delivery pipe [8+8]
2 of 2 Please purchase PDF Split-Merge on www.verypdf.com to remove this watermark. | 3,729 | 13,194 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.390625 | 3 | CC-MAIN-2019-09 | latest | en | 0.909007 |
https://www.proprofs.com/discuss/q/138693/uraniums-atomic-number-and-has-neutrons-how-many-protons-doe | 1,575,703,337,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575540496492.8/warc/CC-MAIN-20191207055244-20191207083244-00546.warc.gz | 847,022,215 | 60,466 | If uraniums atomic number is is 92 and IT has 10 neutrons, how many - ProProfs Discuss
If Uraniums atomic number is is 92 and it has 10 neutrons, how many protons does it have?
A. 92
B. 86
C. 82
D. 102
This question is part of Atoms--> Protons, Neutrons, Electrons
smokeandbullet
Smokeandbullet
Yes,but isn't the Protrons and Electrons suppose to minus each other and get the Neutrons or just my school is nut.
The answer is 92 because the atomic number reveals information about the protons and electrons in the element. Do not get this confused with the atomic mass. Neutrons in an atom are found by subtracting the number of protons to the atomic mass in the element. (Which is 82, and is also scientifically impossible.)
John Smith | 197 | 743 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.0625 | 3 | CC-MAIN-2019-51 | latest | en | 0.903866 |
http://www.gaoshenghan.me/2015_01_01_archive.html | 1,527,355,109,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794867841.63/warc/CC-MAIN-20180526170654-20180526190654-00178.warc.gz | 390,339,590 | 11,922 | ## Monday, January 12, 2015
### Need a summer internship!
In recent month, I've sent 15 or 20 resumes to different IT companies, but what I have now is only deny. I have to say that a student just learned one semester computer science is pretty hard to get an internship. But I don't just stay at home or library in my 3 months summer vocation or just get an unpaid internship that even impossible to cover foods and traffic cost. \$1000 is enough for me.
The most important is to work in industry and know what the feel is when working in a company.
Written with StackEdit.
## Problem
Implement an iterator over a binary search tree (BST). Your iterator will be initialized with the root node of a BST.
Calling next() will return the next smallest number in the BST.
Note: next() and hasNext() should run in average O(1) time and uses O(h) memory, where h is the height of the tree.
For a binary search tree, kept returning the next smallest element is inorder traversal. Wikipedia has some great picture to show how to traversal in "preorder", "inorder" and "postorder". If you are not familiar with this, just go to Wikipedia.
The LeetCode always encourage us use loop other than recursion to solve tree traversal, because recursion may cost more extra space time for program stack and easy to understand.
When I see the problem, my first idea is push all the elements in the tree to a stack. But this cost O(n) memory. So a idea result is pop one from stack and push one to stack. Most method didn't do this precisely, just an amortized O(1) time and O(h) space. If the tree is pretty unbalanced, it is as same as push all the elements in stack. Here is code: private Stack nodes = new Stack<>();
```public BSTIterator(TreeNode root) {
pushToStack(root);
}
/** @return whether we have a next smallest number */
public boolean hasNext() {
return !nodes.isEmpty();
}
/** @return the next smallest number */
public int next() {
TreeNode tmp = nodes.pop();
pushToStack(tmp.right);
return tmp.val;
}
private void pushToStack(TreeNode node) {
while (node != null) {
nodes.push(node);
node = node.left;
}
}
```
The idea is keep going left until no more elements in left, than to right element and keep going left again. Redo above until no elements in the stack. The problems need to return elements inorder is the same idea. Binary Tree Inorder Tracersal could be solve use the same code structure:
```public List<Integer> inorderTraversal(TreeNode root) {
Stack<TreeNode> nodes = new Stack<>();
ArrayList<Integer> result = new ArrayList<>();
if (root == null)
return result;
pushToStack(root, nodes);
while (!nodes.isEmpty()) {
TreeNode tmp = nodes.pop();
pushToStack(tmp.right, nodes); | 607 | 2,704 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.375 | 3 | CC-MAIN-2018-22 | latest | en | 0.877895 |
https://eduhawks.com/tag/liquid/ | 1,624,250,453,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623488262046.80/warc/CC-MAIN-20210621025359-20210621055359-00354.warc.gz | 215,159,815 | 28,014 | Categories
## In science class, Savannah measures the temperature of a liquid to be 34 celsius. Her teacher wants her to the temp to degrees fahrenheit.What is the temp of Savannahs liquid to the nearest degree fahrenheit?
In science class, Savannah measures the temperature of a liquid to be 34 celsius. Her teacher wants her to the temp to degrees fahrenheit.What is the temp of Savannahs liquid to the nearest degree fahrenheit?
Categories
## Select all that apply. Which of the following statements are true? A solid can diffuse into a liquid, but a solid cannot diffuse into another solid. A solid can diffuse into both a liquid and another solid. A liquid can diffuse into another liquid. A gas can diffuse into another gas.
Select all that apply. Which of the following statements are true? A solid can diffuse into a liquid, but a solid cannot diffuse into another solid. A solid can diffuse into both a liquid and another solid. A liquid can diffuse into another liquid. A gas can diffuse into another gas.
Categories
## What should i write on a essay about glue and liquid corn starch
What should i write on a essay about glue and liquid corn starch
Categories
## HELP PLEASE! 3 SCIENCE QUESTIONS! I'm stuck on my last 3. #1 Compare and contrast chemical change and physical change. #2 Compare and contrast a gas and a plasma. #3 What state of matter has the electrons separated from the nuclei of the atoms? Gas Solid Liquid Plasma
HELP PLEASE! 3 SCIENCE QUESTIONS! I’m stuck on my last 3. #1 Compare and contrast chemical change and physical change. #2 Compare and contrast a gas and a plasma. #3 What state of matter has the electrons separated from the nuclei of the atoms? Gas Solid Liquid Plasma
Categories
## In the story “To Build a Fire,” what information did the man ignore? Even though the creeks freeze solid, liquid water exists near springs. Always make camp and build a fire while there is still daylight. Always take along survival gear and wear protective clothing. No one must travel alone in the Klondike after fifty degrees below zero.
No one must travel alone in the Klondike after fifty degrees below zero. I would say this is the main lesson to learn about this Jack London story which shows that nature can be cruel and unforgiving and at 50 degrees below zero it is nothing to fool with. If there are two people, one can help the other and warn say of snow which may fall down and douse the fire and also both can have matches to start a fire and they can discuss the best course of action in any situation which presents itself such as trusting the ice not to break over a creek for example.
Categories
## The dimensions of a conical funnel are shown below: Nisha closes the nozzle of the funnel and fills it completely with a liquid. She then opens the nozzle. If the liquid drops at the rate of 15 cubic inches per minute, how long will it take for all the liquid to pass through the nozzle? ( use pi= 3.14)
The dimensions of a conical funnel are shown below: Nisha closes the nozzle of the funnel and fills it completely with a liquid. She then opens the nozzle. If the liquid drops at the rate of 15 cubic inches per minute, how long will it take for all the liquid to pass through the nozzle? ( use pi= 3.14)
Categories
## Why is water an excellent solvent for most ionic compounds and polar covalent molecules but not for non-polar compounds? A) Water is a polar molecule. B) Water has a charge like an ion. C) Water is a deionizing substance. D) Water molecules have an unusually high kinetic energy in the liquid state.
A)Water is a polar molecule. When talking about solubility, the rule that “like dissolves like” is usually correct. This means that polar substances are soluble in other polar substances. This occurs because the opposite partially charged ends of different polar molecules are attracted to each other. Similarly, the partially charged ends of polar molecules will be even more attracted to the charged ends of an ionic compound’s molecule. This is why,for example, table salt/ NaCl, which is an ionic compound, dissolves in water.
Categories
## The volume of 100 drops of a liquid is 0.01 fluid ounce. What is the volume of 10 drops?
The volume of 100 drops of a liquid is 0.01 fluid ounce. What is the volume of 10 drops?
Categories
## Which of the following is an example of a scientific question that is not testable? Why is the sky blue? How moist should soil be for a tomato seed to sprout? Is the strength of steel related to the amount of carbon in it? Does the rate of cooling of a liquid depend upon its initial temperature?
Which of the following is an example of a scientific question that is not testable? Why is the sky blue? How moist should soil be for a tomato seed to sprout? Is the strength of steel related to the amount of carbon in it? Does the rate of cooling of a liquid depend upon its initial temperature?
Categories
## Which energy changes are associated with a liquid freezing? Energy is released as potential energy increases. Energy is absorbed as potential energy increases. Energy is released as potential energy decreases. Energy is absorbed as potential energy decreases.
Which energy changes are associated with a liquid freezing? Energy is released as potential energy increases. Energy is absorbed as potential energy increases. Energy is released as potential energy decreases. Energy is absorbed as potential energy decreases.
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## A recipe for candy requires that the liquid ingredients be stirred constantly until the liquid reaches a temperature of 140°C. What type of spoon – wood or metal – will prevent burns to a person’s hand while stirring the hot liquid? Explain your choice.
Sample Response: A wooden spoon will prevent the person’s hand from burning because wood is an insulator. An insulator slows the transfer of thermal energy so that very little thermal energy from the hot liquid will transfer to the person’s hand.
include:
-A wooden spoon will prevent the person’s hand from burning.
-Wood is an insulator.-An insulator slows the transfer of thermal energy.-The wooden spoon will transfer very little thermal energy to the person’s hand.
you’re welcome lol
Categories
## In 1987 the first substance to act as a superconductor at a temperature above that of liquid nitrogen (77 k) was discovered. the approximate formula of this substance is yba2cu3o7. calculate the percent composition by mass of this material.
The molar mass of the elements are:
Yttrium (Y) = 88.91 g / mol
Barium (Ba) = 137.33 g / mol
Copper (Cu) = 63.55 g / mol
Oxygen (O) = 16 g / mol
Calculating for the molar mass of YBa2Cu3O7:
YBa2Cu3O7 = 88.91 + 2(137.33) + 3(63.55) + 7(16) = 666.22
g / mol
The masses of the elements in 1 mole of YBa2Cu3O7 are:
Yttrium (Y) = 88.91 g / mol (1 mole) = 88.91 g
Barium (Ba) = 137.33 g / mol (2 mole) = 274.66 g
Copper (Cu) = 63.55 g / mol (3 mole) = 190.65 g
Oxygen (O) = 16 g / mol (7 mole) = 112 g
We now calculate for the percent composition by mass of
each element:
% Y
= 88.91 g / 666.22 g = 13.35%
% Ba
= 274.66 g / 666.22 g = 41.23%
%Cu
= 190.65 g / 666.22 g = 28.62%
%O =
112 g / 666.22 g = 16.81%
Categories
## An action movie production team needs glass spheres to hold a green liquid that looks like an explosive l. if all of the available 3,392’92 cubic inches of the liquid is to be poured into 30 glass spheres l, what should the diameter of each sphere be. assume that each sphere is filled to the top
We need 30 glass spheres whose TOTAL volume = 3,392.92 cubic inches.
Therefore each of the 30 spheres must have a volume of 3,392.92 / 30 =
113.0973333333
cubic inches
Sphere Volume = 4/3 • π • r³
Sphere Volume / (4/3 • π)
4.1887902048
26.9999994758
diameter = 6 inches
Categories
## The volume of 1000 drops of a liquid is 1 fluid ounce. What is the volume of 100 drops?
Solution:
The Given Rectangle having dimensions
Length = 80 cm
Let the Six squares which has been cut from this rectangle have side of length a cm.
Area of each square = (Side)²= a²
Area of 6 Identical Squares = 6 × a²= 6 a²
If four squares are cut from four corners and two along Length,
then , Length of Box = (80 – 3 a)cm, Breadth of Box = (50 – 2 a)cm, Height = a cm
Volume of Box =V = Length × Breadth × Height
V = (80 – 3 a)× (50 – 2 a)× a
V = 4000 a – 310 a² + 6 a³
For Maximum Volume
V’= 0 , where V’ = Derivative of V with respect to a.
V’= 4000 – 620 a + 18 a²
V’ =0
18 a² – 620 a + 4000= 0
9 a² – 310 a + 2000=0
using determinant method
cm
V”=1 8 a – 310 = -ve
which shows , when a = 8.6 cm , volume is maximum.
So, V = 4000×8.6 – 310×(8.6)²+6×(8.6)³=15288.736 cm³
OR
If four squares are cut from four corners and two along Breadth,
then , Length of Box = (80 – 2 a)cm, Breadth of Box = (50 – 3 a)cm, Height = a cm
Volume of Box =V = Length × Breadth × Height
V = (80 – 2 a)× (50 – 3 a)× a
V = 4000 a – 340 a² + 6 a³
For Maximum Volume
V’= 0 , where V’ = Derivative of V with respect to a.
V’= 4000 – 680 a + 18 a²
V’ =0
18 a² – 680 a + 4000= 0
9 a² – 340 a + 2000=0
Using Determinant method
cm
V”=18 a -340= -ve value when a = 7.6 cm, shows volume is maximum when a = 7.6 cm
V= 4000×7.6 -340 × (7.6)² +6× (7.6)³=13395.456 cubic cm
Categories
## As a substance changes phase from a gas to liquid to solid: A- The spacing between molecules increases. B- It's density generally increases. C- The spacing between molecules is not affected. D- It undergoes chemical change.
As a substance changes phase from a gas to liquid to solid: A- The spacing between molecules increases. B- It’s density generally increases. C- The spacing between molecules is not affected. D- It undergoes chemical change.
Categories
## What is the effect of temperature on the solubility of a solid in a liquid
What is the effect of temperature on the solubility of a solid in a liquid
Categories
## We can change a gas to liquid by the temperature and the pressure. NextReset
We can change a gas to liquid by the temperature and the pressure. NextReset
Categories
## Which term best describes liquid behavior under pressure?
Which term best describes liquid behavior under pressure?
Categories
## At which point could the substance shown exist as a gas or a solid but not a liquid?
Explanation:
Organic compounds are defined as the chemical compounds where one or more carbon atoms are covalently bonded to other atoms of hydrogen, oxygen or nitrogen.
They occur naturally in nature and the main sources of these organic compounds are petroleum, coal and natural gases.
Petroleum are the naturally occurring substances which contain organic compounds in the form of gases, liquid or semi-solid. Thus, they are a major source of organic compound.
Coal is also a form of fossil fuel which are the major source of organic compound.
Natural gas contains hydrocarbons and also is the major source of organic compounds.
Thus, the correct answer is all of these.
Categories
## 55 kg of liquefied natural gas (lng) are stored in a rigid, sealed 0.17 m3 vessel. in this problem, model lng as 100% methane. due to a failure in the cooling/insulation system, the temperature increases to 200 k, which is above the critical temperature; thus, the natural gas will no longer be in the liquid phase.
55 kg of liquefied natural gas (lng) are stored in a rigid, sealed 0.17 m3 vessel. in this problem, model lng as 100% methane. due to a failure in the cooling/insulation system, the temperature increases to 200 k, which is above the critical temperature; thus, the natural gas will no longer be in the liquid phase.
Categories
## According to Le Chatelier’s principle, a change in pressure affects the chemical equilibrium of the reaction system under what condition? The reactants and the products are in the solid phase. The reactants and the products are in the liquid phase. Some of the reactants or the products are in the gaseous phase. Some of the reactants or the products are solid, and some are liquid.
This problem is to apply Roult’s Law.
Roult’s Law states that the vapor pressure, p, of a solution of a non-volatile solute is equal to the vapor pressure of the pure solvent, Po solv, times the mole fraction of the solvent, Xsolv
p = Xsolv * Po sol
X solv = number of moles of solvent / number of moles of solution
The solvent is water and the solute (not volatile) is glycerin.
Number of moles = mass in grams / molar mass
mass of water = 132 ml * 1 g/ml = 132 g
molar mass of water = 18 g/mol
=> number of moles of water = 132 g / 18 g/mol = 7.33333 mol
mass of glycerin = 27.2 g
molar mass of glycerin:, C3H8O3: 3 * 12 g/mol + 8 * 1 g/mol + 3*16 g/mol = 92 g/mol
number of moles of glycerin = 27.2g / 92 g/mol = 0.29565
total number of moles = 7.33333 moles + 0.29565 moles = 7.62898 moles
=> X solv = 7.33333 / 7.62898 = 0.96125
=> p = 0.96125 * 31.8 torr ≈ 30.57 torr ≈ 30.6 torr.
Categories
## According to Le Chatelier’s principle, a change in pressure affects the chemical equilibrium of the reaction system under what condition? The reactants and the products are in the solid phase. The reactants and the products are in the liquid phase. Some of the reactants or the products are in the gaseous phase. Some of the reactants or the products are solid, and some are liquid.
This problem is to apply Roult’s Law.
Roult’s Law states that the vapor pressure, p, of a solution of a non-volatile solute is equal to the vapor pressure of the pure solvent, Po solv, times the mole fraction of the solvent, Xsolv
p = Xsolv * Po sol
X solv = number of moles of solvent / number of moles of solution
The solvent is water and the solute (not volatile) is glycerin.
Number of moles = mass in grams / molar mass
mass of water = 132 ml * 1 g/ml = 132 g
molar mass of water = 18 g/mol
=> number of moles of water = 132 g / 18 g/mol = 7.33333 mol
mass of glycerin = 27.2 g
molar mass of glycerin:, C3H8O3: 3 * 12 g/mol + 8 * 1 g/mol + 3*16 g/mol = 92 g/mol
number of moles of glycerin = 27.2g / 92 g/mol = 0.29565
total number of moles = 7.33333 moles + 0.29565 moles = 7.62898 moles
=> X solv = 7.33333 / 7.62898 = 0.96125
=> p = 0.96125 * 31.8 torr ≈ 30.57 torr ≈ 30.6 torr.
Categories
## Tim mixed 2 parts vinegar and 5 parts water to make a cleaning liquid. How many parts of vinegar did Tim mix with each part of water?
The triangle ABC is similar to triangle LMP. The order here is very important. The letters correspond to one another
A corresponds to L (first letters of each sequence)
B corresponds to M (second letters of each sequence)
C corresponds to P (third letters of each sequence)
In a similar fashion, the segments also correspond to one another.
AB corresponds to LM (first two letters of each sequence)
AC corresponds to LP (first and last letters of each sequence)
BC corresponds to MP (last two letters of each sequence)
————————————
AB corresponds to LM. AB is 4 units long. LM is 2 units long. So AB is twice as long as LM. This ratio (of 2:1) will be applied to every paired corresponding value.
Also, the right angle is at angle M for triangle LMP. The right angle will be at angle B for triangle ABC (since B corresponds to M). The answer will have an x coordinate of 7. So the answer is either choice B or choice C.
If we move 4 units down from point B, we land on (7,-10). That isn’t listed as an answer choice. Let’s try moving 4 units up from point B. We land on (7,-2). This is an answer choice
So the final answer is choice C) (7,-2)
Categories
## Instructions:Select the correct answer. Why does a solid change to liquid when heat is added? A-The spacing between particles decreases. B-Particles lose energy. C-The spacing between particles increases. D-The temperature decreases.
There are two systems in the human body that are responsible for the coordination between the functions of different systems to achieve the unity of the living organism’s body. These two systems are the nervous system and the endocrine system. The action of the nervous system is fast and takes a short time, while the action of the endocrine system is slow and takes a long time.
The functional unit of the nervous system is the nerve cell or the neuron. The neuron consists of a cell body and the axon. The cell body starts with the dendrites that receive the messages or the impulses from other neurons or from different sense organs or receptors. These impulses are then transmitted through the cell body. The cell body contains a nucleus and different organelles which help the nerve cell to perform its functions. The nerve impulse is then transmitted to the axon.
The axon is an extension from the cell body. There are some cells called Schwan cells that secrete a myelin sheath to insulate the axon from the surrounding medium. The insulated axons have more ability to conduct the impulses than non-insulated axons. The axon ends with the terminal arborizations. The terminal arborizations of a nerve cell connect to the dendrites of the next cell or to the afferent organ. The gaps between the dendrites and the terminal arborizations are called the synapses.
The nerve impulse is an electrochemical phenomenon i.e. an electrical phenomenon with a chemical nature. The membrane of the axon acts as a barrier between an outside positively charged medium and an inside negatively charged medium. This makes a potential difference of -70mV. This state is called the resting potential. When the membrane is subjected to a stimulus, the positive charges enter to inside and the negative charges exit to the outside. The potential difference now becomes +40mV. This state is called the depolarization state. The point of stimulation acts as a new stimulus for the next point and so on. The membrane sooner gains its permeability again and the positive charges return to the outside and the negative charges to inside. This state is called repolarization.
The nerve impulse reaches the synapse. There are some neurotransmitters that are excited by the nerve impulse coming and carry the message across the membrane. Some receptors receive theses neurotransmitters on the dendrites of the next neuron. These receptors act as a stimulus for the new cell. | 4,657 | 18,262 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.59375 | 3 | CC-MAIN-2021-25 | latest | en | 0.892614 |
https://www.pw.live/chapter-applications-of-trigonometry/angle-of-depression | 1,670,014,145,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710916.40/warc/CC-MAIN-20221202183117-20221202213117-00097.warc.gz | 1,008,525,306 | 16,873 | # Angle Of Depression
## Applications Of Trigonometry of Class 10
When the object is at a lower level than the observer’s eyes, he has to look downwards to have a view of the object. In that case, the angle which the line of sight makes with the horizontal through the observer’s eye is known as the angle of depression (Figure).
question 1. Find the angle of elevation of the Sun's altitude when the height of shadow of a vertical pole is equal to its height.
Solution: Let the height of the pole AB = x m.
Then length of shadow OB of the pole AB = x m.
[ height of shadow = height of pole]
Let the angle of elevation be θ, i.e. ∠AOB = θ
In right ΔOBA, we have
⇒ tan θ = 1 ⇒ θ = 45º
Hence, the angle of elevation of the Sun's altitude is 45º.
question 2. In figure, what are the angle of depression of the top and bottom of h m tall building from the top of multistoried building?
Solution: From P draw a line PQ || BD or AC, then the angles of depression of the top and bottom of a tall building h m from the top P of the multistoried building are 30º and 45º, respectively.
question 3. From a point 20 m away from the foot of a tower, the angle of elevation of the top of the tower is 30º. Find the height of the tower.
Solution: Let O be a point 20 m away from the tower AB = h m. In right ΔOBA, we have
Thus, the height of the tower is 20/√3 m.
question 4. In figure, what are the angles of depression of the top and bottom of a pole from the top of a tower h m high?
Solution: From O draw a line OQ parallel to AM or BL, then the angles of depression of the top and bottom of a pole from the top O of a tower h m are 45º and 60º respectively.
question 5. A man is standing on the deck of a ship, which is 8 m above water level. He observes the angle of elevations of the top of a hill as 60o and the angle of depression of the base of the hill as 30o. Calculate the distance of the hill from the ship and the height of the hill.
Solution: Let x be distance of hill from man and h + 8 be height of hill which is required.
ΔABC is right triangle ∠ACB.
In right triangle BCD,
Height of hill = h + 8 = = 32m.
Distance of ship from hill = x = 8√3 m. | 579 | 2,174 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2022-49 | latest | en | 0.91347 |
https://cs.stackexchange.com/questions/99130/which-calculating-running-time-are-correct | 1,716,242,668,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058313.71/warc/CC-MAIN-20240520204005-20240520234005-00327.warc.gz | 166,621,795 | 39,792 | # Which calculating running time are correct?
Let $$C$$ be an array with length $$n$$ (assume the elements are cities with some properties). We have some properties sorted by importance. For example, 1.area, 2.population size, .... We use the following algorithm to find the best city according to sorted properties:
We choose the first property and examine each city. Then we delete half of the cities. (assume we can do this by $$O(n)$$ time.) Now we apply the second properties and again delete half of the remain cities. We continue this way until one city remains.
Can we say the running time is equal to $$O(n)$$ according to the recursive function $$T(n)= T(n/2)+ O(n)$$, and the master theorem or the running time is equal to $$n \log(n)$$ because we use binary search and in each iteration, we apply an operator in linear time?
which is correct and why another one is incorrect?
• Where are you using binary search? Oct 26, 2018 at 20:13
The $$T(n)= 2 T(n/2)+O(n)$$ is $$\mathcal{O}(n \log(n))$$
Your recursion $$T(n)=T(n/2)+O(n)$$ is $$\mathcal{O}(n)$$ by The 3rd case of The Master Theorem.
• Does the recursion function $T(n)= T(n/2)+O(n)$ correct for the mentioned algorithm? Oct 26, 2018 at 10:50
• Yes, it is correct. put $cn$ instead of $\mathcal{O}(n)$ and see by recursion tree, actually vine(backbone). Oct 26, 2018 at 10:53 | 372 | 1,351 | {"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": 10, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.03125 | 4 | CC-MAIN-2024-22 | latest | en | 0.907087 |
https://topnotchteaching.com/math/printable-math-games/ | 1,713,038,477,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296816832.57/warc/CC-MAIN-20240413180040-20240413210040-00686.warc.gz | 561,315,163 | 27,470 | # Who Else Wants Some Free & Easy Printable Math Games?
Are you ready to play some fun math games? My students love playing printable math games. They are quick and easy to prepare and just what you need to help kids…
Are you ready to play some fun math games? My students love playing printable math games. They are quick and easy to prepare and just what you need to help kids practice math skills while also having fun.
Here you’ll find some free math games ready to print and then play.
## Free printable math games
Estimating game
This is a game for 2 players that gets your students practising the skill of estimation. You only need a calculator, 2 x 10 sided dice and a copy of the playing board.
Transformation puzzles
Transformation puzzles help your students develop the idea that things can move around by reflecting (flipping), translating (sliding) and rotating (turning). This helps kids to realize that by doing these movements it doesn’t change the size or shape of the thing.
Included are 5 different puzzles that include various irregular figures: frog, toy, present, fish and tree. When your students have created their transformation puzzles they swap with a partner and try to work out the sequence of movement. They then compare with the original to see how they are the same and how they are different.
2D shape dominoes
Here I share with you a 2D shape domino game that will help your students to learn the names of the shapes. The dominoes have a mixture of the various 2D shape pictures as well as properties of 2D shapes.
You could use the dominoes in the usual way, but you could also cut apart each of the pieces to make a matching game. Just print them on thick card and cut apart each piece separately and spread them out face down in the middle. Players turn over 2 cards at a time to try and find 2 that match. If they match they keep them. The player with the most cards at the end is the winner.
## More educational math games and activities
If you liked these games, then you’ll love my Bumper Book of Fun Math Games & Activities.
In this eBook you’ll get over 130 pages of math activities and ideas that are suitable for students in grades 2 – 4.
The games and activities cover a range of math areas, including number, mental math, space, measurement and chance and data.
I know as a busy teacher you’re pulled in so many directions, so I made this eBook as an easy resource that you can use to add to your math teaching.
## 37 Best Math Board Games & Other Math Games To Develop Math Skills
Math games are an essential part of your tool kit to help kids develop math skills. Here are math board games & more you can incorporate into your class.
## 5 Fun Calculator Games And Quick Math Games For Your Math Classroom
Let’s take a look at 5 quick math games including calculator games to liven up math lessons and give students more practice to build and review skills. | 633 | 2,926 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.984375 | 3 | CC-MAIN-2024-18 | latest | en | 0.95347 |
http://nrich.maths.org/6331/clue | 1,503,121,199,000,000,000 | text/html | crawl-data/CC-MAIN-2017-34/segments/1502886105304.35/warc/CC-MAIN-20170819051034-20170819071034-00218.warc.gz | 307,192,484 | 4,762 | ### Rotating Triangle
What happens to the perimeter of triangle ABC as the two smaller circles change size and roll around inside the bigger circle?
### Doodles
Draw a 'doodle' - a closed intersecting curve drawn without taking pencil from paper. What can you prove about the intersections?
### Russian Cubes
I want some cubes painted with three blue faces and three red faces. How many different cubes can be painted like that?
# Iffy Logic
##### Stage: 4 and 5 Challenge Level:
Some cards can make multiple answers. See if you can first use the cards which have an obvious partner to start to eliminate some of the possibilities. | 135 | 639 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2017-34 | latest | en | 0.943279 |
https://casinonowttt.com/data/questions/1001-do-any-casinos-have-solitaire/ | 1,669,524,080,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710192.90/warc/CC-MAIN-20221127041342-20221127071342-00619.warc.gz | 195,093,235 | 35,550 | Do any casinos have Solitaire?
Do any casinos have Solitaire? We ran the answer to a similar question five years ago and the answer then, as now, is no casino that we know of in Vegas or anywhere else for that matter deals Solitaire in any of its myriad forms.
How much does it cost to play Solitaire in a casino? Vegas-style Solitaire involves, as must be expected, a fee to play and a payoff scheme. In our preliminary analysis, the fee to play will be \$52; each card that is in the “suit stacks” at the end of the game pays \$5.
Is Solitaire used for gambling? Most gambling games, even the ones you think have multiple participants, are really solitaire games.
Is it possible to win Solitaire Vegas? If you are playing standard or Vegas solitaire, you can only win by putting every card into the 4 piles on the top right based on the suit and order. Keep in mind that you will not be able to win every game—there could be a combination of cards that are buried and keep you from moving cards into the piles at the top.
Do any casinos have Solitaire? – Powiązane Pytania
What are the odds of winning at Vegas Solitaire?
Interestingly enough, the Pyramid, one of the simplest and most straightforward games of Solitaire has the lowest chance of winning, with the odds oscillating between 0.5 and 5.5%.
What is the trick to winning Solitaire?
Top Strategies to Help Increase Your Winning Chances at Solitaire
1. Learn the Rules.
2. Target Larger Stacks First.
3. Evenly Distribute Tableau Piles.
4. Move Quickly If You’re Playing Timed Solitaire.
5. Think about Color When Filling Spaces.
6. Handle Your Face Down Cards First.
7. Try Creating Stacks of Similar Suits.
What is Vegas Solitaire scoring?
Each new game starts off at 0 points. Each card moved to the foundation receives 10 points. Each card that is moved to a tableau column receives 5 points.
What percentage of Klondike Solitaire games are winnable?
The probability of winning Thoughtful Klondike (with draw three rules) has been calculated as being approximately 82%, more precisely as having a confidence interval of 81.956% ± 0.096%.
How many times can you go through the deck in Solitaire?
You can turn them at once to the waste through three passes via the deck. You can turn a card at a time to the waste with 3 passes through the deck. Turn one card at a time with a single pass through the deck. Turn a single card at a time to the waste, without a limit on passes through the deck.
What is the difference between Klondike and Solitaire?
The main difference between Solitaire and Klondike is in the number of cards and tableau used in each game. Klondike is played using one standard set of cards that are 52 cards while Solitaire is played using two standard sets of cards that are 104 cards.
What is the highest score you can get on Klondike Solitaire?
The game is won when all cards are face up in the tableau and foundations, with none remaining in the deck. Since the maximum end of game score is 740 [755] points (assuming you took no time to win the game!), the largest conceivable bonus is 7400 [7550] points. Total scores of over 6000 can actually be achieved.
It’s Easy to Learn and Win
First and foremost, Solitaire is addictive because it’s easy to learn the ropes. Anyone can play a few games and get the general hang of it, which enables them to score a few lucky wins and feel that they have the chance to be good at it more skillfully.
Why do old people play Solitaire?
Staple characteristics of solitaire, like the game’s quick play-time and relatively easy-to-achieve rewards, lend it to being a coping mechanism for those dealing with stress or mental illness, Mark Griffiths, a behavioral addiction professor at the Nottingham Trent University, told Business Insider in an email.
Is Solitaire a waste of time?
Despite being around for many centuries and having many known benefits, Solitaire isn’t completely harmless. In fact, many would argue “it is a total waste of time”, “a distraction that leads to non-productivity”, and “a childish past-time for grown adults”. Others claim it can even cause addiction to gambling!
Why is Solitaire so relaxing?
Solitaire is also a great game for calming the mind because it puts you into a light meditative state. This is especially true for those who tend to worry or suffer from anxiety frequently. Solitaire gives the mind something to focus on, particularly in times of low action when the opportunity to fret is high.
Does playing Solitaire sharpen your mind?
The game will help reduce tension in your body and mind hence one will more often than not, have a great day and quality sleep at night. What’s more, solitaire is said to improve a person’s decision-making by helping them to enter into a calm meditative state.
Is Solitaire for lonely people?
Solitaire, though, is a very lonely game. And there are other games you can play alone. In fact, until recently most platform games were largely single-player.
Who invented Solitaire?
The game of Solitaire was invented by a French aristocrat during his imprisonment in the Bastille, or at least that is what some versions of the game’s origin story claim, which would date it back to the first half of the 17th century, a time when King Louis XIV used the fortress to imprison nobles who were not related
What is solitaire actually called?
solitaire, also called patience or cabale, family of card games played by one person. Solitaire was originally called (in various spellings) either patience, as it still is in England, Poland, and Germany, or cabale, as it still is in Scandinavian countries.
What is the oldest card game?
Karnöffel is a trick-taking card game which probably came from the upper-German language area in Europe in the first quarter of the 15th century. It first appeared listed in a municipal ordinance of Nördlingen, Bavaria, in 1426 among the games that could be lawfully played at the annual city fête.
What is the fastest game of solitaire ever played?
The World Record for FASTEST Solitaire game is 5 SECONDS! 5. MOBILITYWARE SOLITAIRE SUITE HAS BEEN DOWNLOADED OVER 200 MILLION TIMES ON ANDROID AND iOS.
Is Solitaire a skill or luck?
Some Solitaire games are purely skill-based and require patience, for example, FreeCell. It is good to understand different rules regarding diverse card games. However, when you play Solitaire, your end goal should be to enjoy and have fun, and if it suits you, have a mix of luck and skill to ace it.
How long does it take the average person to finish a game of Solitaire?
HowLongToBeat on Twitter: “It takes on average 5 minutes to beat Solitaire (25 hours for completionists).
What’s the average time to complete a Solitaire game?
It’s noted that the average Solitaire time takes around 11 minutes to complete. Now that you have an estimate on how long Solitaire takes to play, you shouldn’t hesitate to immerse yourself in the Solitaire Social game. Try to complete at least one game in just a few minutes!
Is it possible to win every game of Solitaire?
Note that solitaire is a general term that describes many different card games each having unique rules and winning chances. Although different games may have different winning probabilities, there isn’t an absolute value. Therefore, there’s no way of ensuring that 100% of solitaire games can be won.
How do people finish Solitaire so quickly?
These tips will maximize your chances of winning the game.
1. Expose Larger Stacks First.
2. Don’t Empty a Spot Without a King!
3. Always Keep Color in Mind when Filling a Space.
4. Turn Up the First Deck Card First.
5. Don’t Always Build Ace Stacks.
6. Don’t Move Cards for No Reason.
7. Play the Ace or Two.
What percentage of Solitaire games are won?
Overall, almost 80% of solitaire games are winnable, but players do not win 80% of games played. That is because at least one bad move results in the game being un-winnable. If one lets the cards from the final pile be moved back to the table in order to create more moves, then the odds do go up between 82% and 92%.
What’s the fewest moves in Solitaire?
For the least number of moves needed to round up solitaire, two situations must be considered. First, you need to know there are two types of deals, 1-deal that needs 76 moves and 3-deal that requires 60 moves.
Can Solitaire always be solved?
Solitaire is a game that precedes its computer version, and that means that all the cards are truly shuffled, without the computer peeking in to verify the game is solvable.
Scroll to Top | 1,926 | 8,555 | {"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} | 2.796875 | 3 | CC-MAIN-2022-49 | latest | en | 0.937186 |
https://www.fmaths.com/interesting-about-math/question-how-to-survive-math.html | 1,624,427,638,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623488534413.81/warc/CC-MAIN-20210623042426-20210623072426-00422.warc.gz | 699,435,504 | 12,334 | # Question: How To Survive Math?
## How do you get really good at math?
How to Get Better at Math (While Spending Less Time Studying)
1. Tip #1: Break Down Complex Problems Into Simpler Ones.
2. Tip # 2: Use Simple Numbers.
3. Tip #3: Review the Underlying Concepts.
4. Tip #4: Get Step-by-Step Instructions from an Online Tool.
5. Tip #5: Don’t Rush Your Homework.
6. Learning Math Can Be Satisfying.
## How can I pass math easily?
Here are some tips to help you pass our high school maths easily.
1. Create a Distraction Free Study Environment. Mathematics is a subject that requires more concentration than any other.
2. Master the Key Concepts.
4. Apply Maths to Real World Problems.
5. Practice, Practice and Practice even more.
## How do you get 100 in maths?
Here’s how I studied.
1. Self-study NCERT completely.
2. Solve all the exercises.
3. Solve all the examples.
4. If not understood well, repeat this.
5. Come up with questions and take guidance from your teachers.
6. Solve sample question papers.
7. Last 20 days of practice – most crucial.
You might be interested: Readers ask: What Is The Upside Down U In Math?
## How do I get through a math class?
Follow these tips and you’ll slowly but surely get through that dreaded math class.
1. Form a study group with friends.
3. Make chapter recaps immediately after finishing your practice problems.
4. Take detailed notes during class.
5. Reward yourself for paying attention.
6. Turn to the internet for help.
## Why is math so hard?
Math is a very abstract subject. For students, learning usually happens best when they can relate it to real life. As math becomes more advanced and challenging, that can be difficult to do. As a result, many students find themselves needing to work harder and practice longer to understand more abstract math concepts.
## Are you born good at math?
You might have been right, at least according to a new study by Johns Hopkins University psychologists that suggests that math ability is linked to your inborn “number sense.”
## How do you not fail in math?
The Day of the Test / Test Time
1. Make sure you’re all ready to go.
2. Have something to eat.
3. Review your material, but don’t try to cram in weeks’ worth of math during the five minutes before the test!
4. Follow all test directions carefully.
5. Pencil in any memorized formulas or equations first.
6. If you get stuck, skip it.
## How can I cheat in exams?
How to use cheat exams methods and tricks
1. Don’t bring out secret notes right after the beginning of your exams;
2. Avoid using an eraser because it’s impractical and obvious;
4. Add tiny notes to clothes, such as sweater sleeves or baseball hats;
You might be interested: What Is Math]?
## How can weak students improve in maths?
While there are no hard and fast rules, there are methods that enable weak students to excel in mathematics:
1. Instilling Positivity and Confidence.
2. Scheduling Practice.
3. Tools to Help with Memory.
4. Ask Questions to Test Understanding.
5. Ensure Strong Fundamentals.
6. Focusing on Weaker Topics.
## Is class 10 maths easy?
The level of the questions of Class 10 Maths Basic examination is easy, where all the questions were asked from NCERT itself. Students can easily attempt all the questions within the time frame without facing any difficulty.
## Has anyone got 100% in 10th boards?
Harini, Ritish Agarwal, Devansh, Aryan Bhatt, Siddhant Chandra, Jasraj Singh – just a few names from the list of students who have scored 99% and above in CBSE 10th Result 2020. Shirija is a student of Little Flower School, New Delhi. She has scored a perfect 100 in Hindi, English, Maths and Science.
## Is 10th board exam easy?
It is not easy, but can be done, if you really want to do it. Have patience and work hard. In this article, we will tell you how to prepare for class 10 board exams, score good marks and become a topper.
## How do I become smarter in math?
1. Learn Smarter. Just as people are either left- or right-handed, they also have dominant brain hemispheres.
2. Study Smarter. Because math is a learned skill that requires practice, you may need to spend more time on homework and studying than you do in other subjects.
3. Practice Smarter.
4. Think Smarter.
## How can I be a genius in maths?
Learn the Logic and Process Involved in Solving a Problem Instead, try to understand the concepts behind math. If you see how and why an equation works, you’ll be able to remember it more easily in a pinch. Math theory may seem complicated, but with a little hard work you can begin to figure it out.
You might be interested: FAQ: What Is The Meaning Of Point In Math?
## How can I get high grades without studying?
Go to class. One of the easiest ways to earn good grades without studying much is to show up to your classes and listen to what your teacher has to say. That means not just showing up but being attentive as well. In addition, many professors will make attendance and participation a part of your grade. | 1,159 | 5,023 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.296875 | 3 | CC-MAIN-2021-25 | latest | en | 0.924396 |
thesocietal.in | 1,652,870,246,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662521883.7/warc/CC-MAIN-20220518083841-20220518113841-00179.warc.gz | 656,176,766 | 19,226 | # Pareto Principle – 80/20 Rule
The Pareto Principle states that 80% of consequences come from 20% of the causes, asserting an unequal relationship between inputs and outputs. It is named after the great economist Vilfredo Pareto. This principle is a general reminder that the connection between inputs and outputs isn’t balanced. The Pareto Principle is additionally referred to as the Pareto Rule or the 80/20 Rule.
APPLICATION
The Pareto Principle is often applied in a wide range of areas like manufacturing, management, and human resources. For instance, the efforts of 20% of a corporation’s staff could generate 80% of the firm’s profits. The Pareto Principle can be applied in those businesses that are client-service based. It has been adopted by many coaching and customer relationship management (CRM) software programs.
It can also be applied on a personalised level. Time management is the most ordinary use for the Pareto Principle, as most of the people tend to thinly spread out their time instead of focusing on the most important tasks. In terms of personalised time management, 80% of your work-related output could come from only 20% of your time at work.
CORE PRINCIPLE
At its core, the 80-20 rule is about identifying an entity’s best assets and using them efficiently to create maximum value. For example, a student should try to identify which parts of a textbook will create the most advantage for an upcoming exam and focus on those first. This does not suggest, however, that the student should ignore the other parts of the textbook.
Here are a few examples of the Pareto principle in action:
v20 % of a given employee’s time yields 80 % of their output.
v20 % of software bugs cause 80 % of the software’s failures.
v20 % of a company’s investments produce 80 % of its investment profits.
PARETO CHART
A Pareto chart is highly helpful for prioritising problems so that you can determine which issues have the greatest effect on the outcome of a given situation. It also enables you to take appropriate actions to resolve the most important issues concerning your business.
The looks of a Pareto chart is quite straight forward with vertical bars plotted in decreasing order in relevance to their relative frequencies. The higher the frequency, the greater the effect it will have on your business – so, it is crystal clear where you need to focus your attention on to improve your business. The Pareto chart is an excellent way to pinpoint the issues influencing your sales, productivity and overall success of the business.
THE UPSIDE OF THE 80/20 RULE
When applied to your life and work, the 80/20 Rule can help you separate “the vital few from the trivial many.”The 80/20 Rule is like a form of judo for life and work. By finding precisely the right area to apply pressure, you can get more results with less effort.
vINCREASED PROFITABILITY
The Pareto Principle suggests that a small percentage of your total amount of customers produce the majority of revenue. Similarly, only some of your products and/or services attract the most sales. As such, analysing the leads and current customers you have can help you determine which ones have produced the most revenue in the past so that you can focus on pursuing only the most valuable leads.
vINCREASED PRODUCTIVITY
The Pareto Principle teaches employees to not waste time on the immaterial goals that will not contribute to the long term goal and to focus their resources on the ‘vital few’ that will produce the majority of the results. Similarly, the Pareto Principle can be used to recognise the causes of unproductiveness in the workplace.
THE DOWNSIDE OF THE 80/20 RULE
While the 80/20 split is true for Pareto’s observation, that doesn’t necessarily mean that it is always true. The Pareto Principle is merely an observation and not necessarily a law.
vQUANTITATIVE DATA BASED
Pareto analysis bases its conclusions on quantitative data. But most decisions, particularly in regard to human and social activities, require weighing not only statistics but more qualitative issues to reach a decision.
vFOCUSES ON THE PAST
Traditional Pareto Analysis uses past data. By the time historical data, is being used, things might have changed and the results obtained may not be compatible with the current situation. The sole dependence on past information in Pareto analysis can be deceptive. Small-businesses owners may find that the past data does not precisely represent the company’s current situation.
vMISAPPLICATION
Many people simply don’t understand it. They, therefore, misuse and abuse the rule. Others have a little understanding of the rule but overzealously apply it beyond its brief. In the end, these users generalise an otherwise powerful analytical tool for getting high level “feel for things” thus giving it a bad name.
Categories: Writers' Desk | 1,012 | 4,852 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.03125 | 3 | CC-MAIN-2022-21 | latest | en | 0.943962 |
http://www.jiskha.com/display.cgi?id=1326424961 | 1,498,424,563,000,000,000 | text/html | crawl-data/CC-MAIN-2017-26/segments/1498128320582.2/warc/CC-MAIN-20170625203122-20170625223122-00396.warc.gz | 585,821,838 | 4,265 | # algebra
posted by .
point-slope form linear equation
y-3 = 3(x+1)
what is the equation in standard form of a perpendicular line that passes through (5,1). i think you take 1/3 as slope, reciprocal, I am confused about how to proceed. I know you put variables on one side and constants on other side in standard form.
what is the x-intercept of the perpendicular line?
thanks
• algebra -
I assume the line is perpendicular to
the given line.
y-3 = 3(x+1).
y-3 = 3x+3,
-3x+y = 6,
3x-y = -6.
m1 = -A/B = -3/-1 = 3.
m2 = -1/3 = Negative reciprocal.
P(5,1).
y = mx + b,
y = (-1/3)*5 + b = 1,
b = 1 + 5/3 = 2 2/3 = 8/3.
y = (-1/3)x+8/3,
Convert to STD form:
x/3 + y = 8/3,
X + 3Y = 8.
• algebra -
ADDITION: Let Y = 0 to find X-int.
X + 3Y = 8,
X + 3*0 = 8,
X = 8 = x-int.
• algebra -
thank you so much for your help Henry, I really appreciate it. Ann
I understand now where I went wrong. first mistake i made was using positive 1/3 and then it was downhill from there. | 344 | 977 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.859375 | 4 | CC-MAIN-2017-26 | latest | en | 0.93998 |
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