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http://forum.enjoysudoku.com/eleven-s-variable-replacement-method-and-its-complexity-t39277.html | 1,679,893,171,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296946637.95/warc/CC-MAIN-20230327025922-20230327055922-00667.warc.gz | 17,945,968 | 12,438 | Eleven's variable replacement method and its complexity
Advanced methods and approaches for solving Sudoku puzzles
Eleven's variable replacement method and its complexity
.
Eleven's variable replacement method and its complexity
AFAIK, the method to be discussed here was first introduced by eleven in this thread: http://forum.enjoysudoku.com/an-alternative-way-to-solve-the-hard-17-clue-t30022.html#p200126, at a time when I wasn't very present on the forum.
I became aware of it only here: http://forum.enjoysudoku.com/endor-fins-1-2-ser-8-3-8-9-t38805.html#p301646
Since then, I've been busy with extending CSP-Rules with new features (mainly related to allowing some control over the number of steps and with the Pandiag variant of Latin Squares) and I haven't had too much time to think about it. But this question of complexity remained in the background of my thoughts and it was revived by a new application of it here: http://forum.enjoysudoku.com/post308008.html#p308008
Generally speaking the method consists of replacing sets of undecided values in some house by supposedly decided variables x, y, ..., the real value of which is unknown.
The method is smart and mathematically valid; validity is not the point of discussion here. My interest is about its logical complexity.
Let me introduce my questioning by recalling Mith's example (in the second mentioned thread):
Code: Select all
`Resolution state after Singles:+--------------+----------------+-----------------+! 9 8 6 ! 5 3 124 ! 1247 124 1247 !! 7 24 3 ! 124 9 6 ! 5 124 8 !! 24 5 1 ! 24 7 8 ! 9 6 3 !+--------------+----------------+-----------------+! 12 126 4 ! 9 8 5 ! 1267 3 127 !! 123 1236 9 ! 124 1246 7 ! 8 5 124 !! 8 7 5 ! 3 1246 124 ! 1246 124 9 !+--------------+----------------+-----------------+! 134 134 8 ! 7 5 124 ! 124 9 6 !! 5 14 7 ! 6 124 9 ! 3 8 124 !! 6 9 2 ! 8 14 3 ! 14 7 5 !+--------------+----------------+-----------------+`
mith wrote:The key is to note that, after singles, the digits 124 only appear once each, while almost all of the other digits are placed. Whatever the solution, the digits 124 will appear in some order in other houses. Let's pick column 4, for example, and we'll relabel r235c4 as ABC. We don't know how the digits 124 map to ABC, but let's look at a grid with ABC in place of all the 124s:
Code: Select all
`+-----------------+-------------+------------------+! 9 8 6 ! 5 3 C ! ABC7 ABC ABC7 !! 7 BC 3 ! A 9 6 ! 5 BC 8 !! AC 5 AC ! B 7 8 ! 9 6 3 !+-----------------+-------------+------------------+! ABC ABC6 ABC ! 9 8 5 ! ABC67 3 ABC7 !! AB3 AB36 9 ! C AB6 7 ! 8 5 AB !! 8 7 5 ! 3 AB6 AB ! ABC6 ABC 9 !+-----------------+-------------+------------------+! AB3C AB3C 8 ! 7 5 ABC ! ABC 9 6 !! 5 ABC 7 ! 6 ABC 9 ! 3 8 ABC !! 6 9 ABC ! 8 ABC 3 ! ABC 7 5 !+-----------------+-------------+------------------+`
mith wrote:But this grid just solves with singles. [...]And now we can match this to the original puzzle to find C=1, A=4, B=2
denis_berthier wrote:From a theoretical point of view, there is some interesting (pseudo-)mystery here:
- you start by deleting some information (stating that B=1or2or4 instead of B=2or4 in r3c4), which should make the puzzle harder;
- but then you find a solution with Singles only, whereas the original puzzle required whips[4].
I think this example is enough motivation for the complexity pseudo-mystery.
Last edited by denis_berthier on Wed Aug 11, 2021 8:48 am, edited 4 times in total.
denis_berthier
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Re: Eleven's variable replacement method and its complexity
.
Using eleven's method, all the (Naked, Hidden and Super-Hidden) Subsets become Singles.
A good way to tackle a problem is to start with its most obvious cases.
Take a Naked Pair in a row and rename x and y the respective contents of the two rc-cells. The eliminations that would have been done by the Naked Pair can now be done by Singles.
Obviously, this also works for Naked Triplets and Naked Quads, with resp. 3 and 4 variables.
By super-symmetry, this also works for Hidden Subsets and Super-Hidden ones (Fish). Another way to say this is, the same method can be applied to cells in the rn, cn or bn spaces.
Now, what does this mean in terms of the complexity of eleven's method?
First, the method is not a resolution rule (in the strict sense I've given to this expression); it is a general method that can be applied in very different situations. If a complexity has to be ascribed to something, it cannot be to the method itself, but to each of its specific applications.
So, what is the complexity in the Subsets cases? Remember that the complexity of a resolution rule is the complexity of the condition pattern of the rule; the application part (assert a value or delete a candidate) is generally obvious.
In order to apply the method, the Subset pattern has to be found; that's the only part where complexity appears in the method. Miracle: in all the Subset cases, the complexity of the method is the same as that of the Subset.
Last edited by denis_berthier on Wed Aug 11, 2021 8:48 am, edited 2 times in total.
denis_berthier
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Re: Eleven's variable replacement method and its complexity
.
Complexity in the Mith example
We have to explain why a puzzle in W4 becomes solvable by Singles.
mith wrote:The key is to note that, after singles, the digits 124 only appear once each, while almost all of the other digits are placed.
If that was really the key, it would mean that one has to check all the occurrences of 3 digits as decided values, i.e. all the 81 rc-cells.
The real application of the method is described in the next sentence:
mith wrote:Let's pick column 4, for example, and we'll relabel r235c4 as ABC
Column 4 has 5 cells containing (at least) one of 1 2 4. This means a complexity of 5 - which may be unnecessary but anyway explains the reduction from W4 to W0. I haven't tried all the other possibilities (all the other rows, columns, blocks and all those in the rn, cn and bn spaces), but there may be a simpler choice than column 4.
On the other hand, as applying the method results in a solution with Singles, it may be the case that complexity 5 is really required.
Anyway, the apparent mystery of complexity reduction is explained.
denis_berthier
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Re: Eleven's variable replacement method and its complexity
.
A reader of PBCS and silent reader of this forum wrote:It is not clear to me if you're making some general hypothesis or claim.
You can consider I'm making a very general observation, not limited to the technique of variables replacement. As it's totally obvious, it doesn't deserve to be called a claim.
Let R be any of the ratings I've introduced, i.e. BC, Z, tW, W, gW, B, gB... plus the corresponding ratings with Subsets added, plus those with Finned-Fish added. (Starting from W, most of the time, all are equal). Remember that we are talking of ratings of the hardest step in the resolution path - and the number of steps is not taken into account.
Each of these ratings defines an intrinsic property of a puzzle. This is true not only for Sudoku but for any binary Constraint Satisfaction Problem or any CSP that can be re-expressed in such a form, including of course all those that I have coded in CSP-Rules.
Take R to be any of these ratings. For any puzzle P, R(P) is well defined (possibly infinite). Here is the obvious observation I mentioned above:
Measured in terms of R, the complexity of any transformation applied to P that would result in a puzzle P' with an R rating R(P') strictly smaller than R(P) cannot be smaller than R(P).
The Subsets example in my second post is very clear.
Remarks:
In some cases, expressing the complexity of the transformation in terms of R may not be obvious.
In some cases, using the transformation can lead to a shorter resolution path, but that is off topic, because the number of steps of a resolution path is not an intrinsic property of a puzzle.
denis_berthier
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Re: Eleven's variable replacement method and its complexity
Column 4 has 5 cells containing (at least) one of 1 2 4.
I must be misunderstanding something here. Where are you getting 5 cells from?
mith
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Re: Eleven's variable replacement method and its complexity
mith wrote:
Column 4 has 5 cells containing (at least) one of 1 2 4.
I must be misunderstanding something here. Where are you getting 5 cells from?
No, you're right. I must have confused two puzzles when I wrote this. For a correct explanation of this case, see the discussion page 2 of this thread: http://forum.enjoysudoku.com/endor-fins-1-2-ser-8-3-8-9-t38805.html#p301646
denis_berthier
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Re: Eleven's variable replacement method and its complexity
.
7 months since the last post in this thread. I've been busy with other things and it had dropped off my mind, but some post in the Tridagon thread reminded me of this.
I now have a somewhat better view of the whole replacement process.
Let P be a puzzle and T a resolution theory. Suppose that, at some resolution state RS in the resolution path of P within T, the conditions for applying eleven's variable replacement method apply to 3 cells (it could be 4 or more, say n, we'll get similar results).
What the method does is indeed apply a clever form of T&E(BRT, 2) - more generally T&E(BRT, n-1) - to RS.
I say clever, because it factors branches of T&E, precisely those branches that don't depend on the specific values of x, y, z.... But it is fundamentally T&E(BRT, n-1).
This view explains both:
- why it can't be expressed as a resolution rule;
- why it can (apparently) reduce the T rating of a puzzle; what's measured after applying the method is no longer the T-rating, but a rating based on T and T&R(BRT, n-1).
And it corresponds to what's done in the proof of the method.
denis_berthier
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Re: Eleven's variable replacement method and its complexity
Maybe this is a good place to give another example for applying the method.
This is one of the new 11.8's by hendrik_monard:
Code: Select all
`98.76.5..7.54.98...46......69.8.7....57946.8.8.45..7..4.8...93......46.........2.`
After placing 9r5c4 (TH) we get this grid (ER 9.8):
Code: Select all
` *----------------------------------------------------------------------* | 9 8 123 | 7 6 123 | 5 14 1234 | | 7 123 5 | 4 123 9 | 8 16 1236 | | 123 4 6 | 123 58 58 | 123 79 79 | |-----------------------+----------------------+-----------------------| | 6 9 123 | 8 123 7 | 1234 145 12345 | | 123 5 7 | 9 4 6 | 123 8 123 | | 8 123 4 | 5 123 123 | 7 69 69 | |-----------------------+----------------------+-----------------------| | 4 1267 8 | 126 57 125 | 9 3 157 | | 1235 1237 1239 | 123 5789 4 | 6 157 1578 | | 135 1367 139 | 136 5789 1358 | 14 2 14578 | *----------------------------------------------------------------------*`
We replace 123 by xyz and (wlog) set r2,4,6 in c5 to x,y,z.
Code: Select all
` *----------------------------------------------------------------------* | 9 8 xyz | 7 6 xyz | 5 14 1234 | | 7 xyz 5 | 4 x 9 | 8 16 1236 | | xyz 4 6 | xyz 58 58 | xyz 79 79 | |-----------------------+----------------------+-----------------------| | 6 9 xyz | 8 y 7 | xyz4 145 12345 | | xyz 5 7 | 9 4 6 | xyz 8 xyz | | 8 xyz 4 | 5 z xyz | 7 69 69 | |-----------------------+----------------------+-----------------------|`
With singles we get
Code: Select all
` *----------------------------------------------------------------------* | 9 8 xy | 7 6 yz | 5 14 1234 | | 7 z 5 | 4 x 9 | 8 16 1236 | | xy 4 6 | yz 58 58 | *xz-y 79 79 | |-----------------------+----------------------+-----------------------| | 6 9 xz | 8 y 7 | 4-xyz 145 12345 | | xz 5 7 | 9 4 6 | *xyz 8 *xyz | | 8 y 4 | 5 z x | 7 69 69 | |-----------------------+----------------------+-----------------------|`
Now y cannot be in r3c7 (would be forced to both r1c36).
x in r3c7 forces x in r5c1 (through r3c1,r14c3) and yz in r3c79.
z in r3c7 forces z in r5c1 (through r3c41) and xy in r3c79.
=> -xyzr4c7
[Edit: i made a mistake here, so i need another step:]
This gives singles 1r9c7, 4r4c7,r9c9,r1c8 and we get
Code: Select all
` *-----------------------------------------------------------------* | 9 8 xy | 7 6 yz | 5 4 xyz | | 7 z 5 | 4 x 9 | 8 16 6-y | | xy 4 6 | 12 58 58 | 23 79 79 | |-----------------------+---------------------+-------------------| | 6 9 xz | 8 y 7 | 4 15 xz5 | | xz 5 7 | 9 4 6 | 23 8 xyz | | 8 y 4 | 5 z x | 7 69 69 | |-----------------------+---------------------+-------------------|`
Here it can be easily seen, that x in r1c9 forces z in r5c1 and vice versa, i.e. y in r5c9, so y must be in r15c9 => 6r2c9
Solves with skyscraper.
eleven
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Re: Eleven's variable replacement method and its complexity
.
A better example in relation with this thread is how I applied the method here:
http://forum.enjoysudoku.com/post316286.html?hilit=replacement#p316286
As you can see, when interpreted in terms of T&E, the 3rd and 4th steps are a version of T&E(2): you try to plug in 2 numbers (the 3rd is a consequence of them) and the combinations that are contradictory with the givens are eliminated.
Granted, my initial presentation with isomorphisms is smarter, as is your presentation with x,y,z variables, but remember, I'm talking of an interpretation in a different framework.
For my general philosophy on this point of interpretation, see here http://forum.enjoysudoku.com/the-t-e-depth-of-a-resolution-rule-t39888-5.html.
denis_berthier
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Re: Eleven's variable replacement method and its complexity
Yes, that's how you would program the method:
In the candidates grid replace all occurances of 1,2,3 (including the givens) by 123, then choose a unit with 123's only, place single digits 1,2,3 there and feed the solver with this sukaku.
Try it for all 123 only units and choose the one with the simplest solution (if you can find one).
In the solution then replace all 123's according to the givens to get a solution to the original puzzle.
However no one would do that manually.
eleven
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Re: Eleven's variable replacement method and its complexity
eleven wrote:Yes, that's how you would program the method:
[..]
However no one would do that manually.
OK. But the interpretation as T&E is not about manual solving. It's about explaining (apparent) reduction of complexity.
denis_berthier
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Re: Eleven's variable replacement method and its complexity
eleven wrote:Maybe this is a good place to give another example for applying the method.
This is one of the new 11.8's by hendrik_monard:
Code: Select all
`98.76.5..7.54.98...46......69.8.7....57946.8.8.45..7..4.8...93......46.........2.`
After placing 9r5c4 (TH) we get this grid (ER 9.8):
Hidden Text: Show
Code: Select all
` *----------------------------------------------------------------------* | 9 8 123 | 7 6 123 | 5 14 1234 | | 7 123 5 | 4 123 9 | 8 16 1236 | | 123 4 6 | 123 58 58 | 123 79 79 | |-----------------------+----------------------+-----------------------| | 6 9 123 | 8 123 7 | 1234 145 12345 | | 123 5 7 | 9 4 6 | 123 8 123 | | 8 123 4 | 5 123 123 | 7 69 69 | |-----------------------+----------------------+-----------------------| | 4 1267 8 | 126 57 125 | 9 3 157 | | 1235 1237 1239 | 123 5789 4 | 6 157 1578 | | 135 1367 139 | 136 5789 1358 | 14 2 14578 | *----------------------------------------------------------------------*`
We replace 123 by xyz and (wlog) set r2,4,6 in c5 to x,y,z.
Code: Select all
` *----------------------------------------------------------------------* | 9 8 xyz | 7 6 xyz | 5 14 1234 | | 7 xyz 5 | 4 x 9 | 8 16 1236 | | xyz 4 6 | xyz 58 58 | xyz 79 79 | |-----------------------+----------------------+-----------------------| | 6 9 xyz | 8 y 7 | xyz4 145 12345 | | xyz 5 7 | 9 4 6 | xyz 8 xyz | | 8 xyz 4 | 5 z xyz | 7 69 69 | |-----------------------+----------------------+-----------------------|`
With singles we get
Code: Select all
` *----------------------------------------------------------------------* | 9 8 xy | 7 6 yz | 5 14 1234 | | 7 z 5 | 4 x 9 | 8 16 1236 | | xy 4 6 | yz 58 58 | *xz-y 79 79 | |-----------------------+----------------------+-----------------------| | 6 9 xz | 8 y 7 | 4-xyz 145 12345 | | xz 5 7 | 9 4 6 | *xyz 8 *xyz | | 8 y 4 | 5 z x | 7 69 69 | |-----------------------+----------------------+-----------------------|`
Now y cannot be in r3c7 (would be forced to both r1c36).
x in r3c7 forces x in r5c1 (through r3c1,r14c3) and yz in r3c79.
z in r3c7 forces z in r5c1 (through r3c41) and xy in r3c79.
=> -xyzr4c7
[Edit: i made a mistake here, so i need another step:]
This gives singles 1r9c7, 4r4c7,r9c9,r1c8 and we get
Code: Select all
` *-----------------------------------------------------------------* | 9 8 xy | 7 6 yz | 5 4 xyz | | 7 z 5 | 4 x 9 | 8 16 6-y | | xy 4 6 | 12 58 58 | 23 79 79 | |-----------------------+---------------------+-------------------| | 6 9 xz | 8 y 7 | 4 15 xz5 | | xz 5 7 | 9 4 6 | 23 8 xyz | | 8 y 4 | 5 z x | 7 69 69 | |-----------------------+---------------------+-------------------|`
Here it can be easily seen, that x in r1c9 forces z in r5c1 and vice versa, i.e. y in r5c9, so y must be in r15c9 => 6r2c9
Solves with skyscraper.
Code: Select all
` *--------------------------------------------------------------------* | 9 8 *123 | 7 6 *123 | 5 14 1234 | | 7 *123 5 | 4 *123 9 | 8 16 1236 | |*123 4 6 |*123 58 58 | 123 79 79 | |----------------------+----------------------+----------------------| | 6 9 *123 | 8 #123 7 |A1234 145 12345 | A => impossible pattern |*123 5 7 | 9 4 6 |#123 8 123 | => r4c7=4 | 8 *123 4 | 5 #123 #123 | 7 69 69 | |----------------------+----------------------+----------------------| | 4 1267 8 | 126 57 125 | 9 3 157 | | 1235 1237 1239 | 123 5789 4 | 6 157 1578 | | 135 1367 139 | 136 5789 1358 | 14 2 14578 | *--------------------------------------------------------------------* *-----------------------------------------------------------* | 9 8 *123 | 7 6 *123 | 5 4 #123 | | 7 *123 5 | 4 *123 9 | 8 16 #1236 | |*123 4 6 |*123 58 58 | 23 79 79 | |-------------------+-------------------+-------------------| | 6 9 *123 | 8 #123 7 | 4 15 1235 | |*123 5 7 | 9 4 6 | 23 8 A123 | A => impossible pattern | 8 *123 4 | 5 #123 #123 | 7 69 69 | => r2c9=6 |-------------------+-------------------+-------------------| | 4 126 8 | 126 57 12 | 9 3 57 | | 1235 1237 1239 | 123 579 4 | 6 57 8 | | 35 367 39 | 36 5789 58 | 1 2 4 | *-----------------------------------------------------------*`
It seems that, when you can use “replacement method” then have "impossible pattern" side by side. I'm not sure...
totuan
totuan
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Re: Eleven's variable replacement method and its complexity
totuan wrote:It seems that, when you can use “replacement method” then have "impossible pattern" side by side. I'm not sure...
You have an example where this is true.
But the method is extremely general. You can apply it to a Naked Triplet.
But the converse is true: when you have an impossible k-digit pattern (with appropriate conditions on the places of cells), you can apply k-replacement.
In any case, applying replacement is just that: applying it. It doesn't guarantee to lead to a solution.
denis_berthier
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Joined: 19 June 2007
Location: Paris | 7,293 | 22,721 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.953125 | 3 | CC-MAIN-2023-14 | latest | en | 0.758855 |
http://mail.scipy.org/pipermail/numpy-discussion/2009-January/039497.html | 1,406,836,077,000,000,000 | text/html | crawl-data/CC-MAIN-2014-23/segments/1406510273663.2/warc/CC-MAIN-20140728011753-00110-ip-10-146-231-18.ec2.internal.warc.gz | 174,667,875 | 2,314 | # [Numpy-discussion] Re : [Newbie] Fast plotting
Jean-Baptiste Rudant boogaloojb@yahoo...
Tue Jan 6 07:38:27 CST 2009
```Hello,
I'm not an expert. Something exists in matplotlib, but it's not very efficient.
import matplotlib.mlab
import numpy
N = 1000
X = numpy.random.randint(0, 10, N)
Y = numpy.random.random(N)
recXY = numpy.rec.fromarrays((X, Y), names='x, y')
summary = matplotlib..mlab.rec_groupby(recXY, ('x',), (('y', numpy.mean, 'y_avg'),))
Jean-Baptiste Rudant
________________________________
De : Franck Pommereau <pommereau@univ-paris12.fr>
À : Discussion of Numerical Python <numpy-discussion@scipy.org>
Envoyé le : Mardi, 6 Janvier 2009, 10h35mn 01s
Objet : [Numpy-discussion] [Newbie] Fast plotting
Hi all, and happy new year!
I'm new to NumPy and searching a way to compute from a set of points
(x,y) the mean value of y values associated to each distinct x value.
Each point corresponds to a measure in a benchmark (x = parameter, y =
computation time) and I'd like to plot the graph of mean computation
time wrt parameter values. (I know how to plot, but not how to compute
mean values.)
My points are stored as two arrays X, Y (same size).
In pure Python, I'd do as follows:
s = {} # sum of y values for each distinct x (as keys)
n = {} # number of summed values (same keys)
for x, y in zip(X, Y) :
s[x] = s.get(x, 0.0) + y
n[x] = n.get(x, 0) + 1
new_x = array(list(sorted(s)))
new_y = array([s[x]/n[x] for x in sorted(s)])
Unfortunately, this code is much too slow because my arrays have
millions of elements. But I'm pretty sure that NumPy offers a way to
handle this more elegantly and much faster.
As a bonus, I'd be happy if the solution would allow me to compute also
standard deviation, min, max, etc.
Thanks in advance for any help!
Franck
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``` | 605 | 2,132 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.328125 | 3 | CC-MAIN-2014-23 | longest | en | 0.694553 |
https://www.myassignmenthelp.net/question/beginners-c++-class | 1,669,451,995,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446706285.92/warc/CC-MAIN-20221126080725-20221126110725-00344.warc.gz | 968,656,423 | 13,215 | + 61-7-5641-0117
Beginners C++ Class
C++ Programming Course Help, C++ Programming Help, C++ programming Questions
This is for a beginners c++ class
For this assignment, you’ll be creating a Bubble Sort algorithm. This algorithm is capable of taking a list of numbers and placing them in order, however it is also known as one of the least efficient means of doing so.
For your assignment, you’ll need to not only sort the given input (and display it in its final sorted form) but you’ll also need to keep track of how many comparisons, and reassignments happen. So comparing two numbers will be counted as a single action. Swapping the location of two variables will meanwhile be considered THREE actions, one to store a variable to be swapped in a temporary storage location, another to swap the first of the variables to the second location, and a third to move the variable from the temporary storage location back to where it’s going to go.
So to summarize, comparing two objects in the array will count as ONE action, and swapping two objects in the array will count as THREE actions.
Your results should not only display the sorted version of the array, but also a count of the total number of actions that took place while sorting. Please sort the following three lists.
{1,3,7,5,2,4,6,8,9,10}
{10,9,8,7,6,5,4,3,2,1}
{1,10,2,9,3,8,4,7,5,6}
An example output might look like:
Sorted List: (1,2,3,4,5,6,7,8,9,10) 124 Actions Performed. | 370 | 1,452 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.578125 | 3 | CC-MAIN-2022-49 | latest | en | 0.934445 |
https://math.stackexchange.com/questions/2099679/why-cant-we-remove-the-sqrt-from-rms/2101819 | 1,586,363,124,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585371818008.97/warc/CC-MAIN-20200408135412-20200408165912-00055.warc.gz | 554,110,168 | 38,457 | # Why can't we remove the sqrt from rms?
In chemistry, we define the root-mean-square speed as
$\sqrt{\bar{u^2}}$ = $\sqrt{\frac{3\text{RT}}{\text{M}}}$
A student asked me why we can't just remove the square root symbol. And aside from "because this is how we define it", I didn't actually have a reason.
So, I'm hoping someone can shed some light on why the above equation is used and not:
$\bar{u^2} = \frac{3\text{RT}}{\text{M}}$
In case it is important, we use this equation to determine the rms speed of a gas. It depends on the temperature (T) and the molecular mass of the gas (M). R is a constant value. I understand we don't just use the average because in a set of gases, they move in a random direction so the average is 0. But, by squaring isn't that issue resolved, without the square root?
• $\bar{u^2}$ does not have units of velocity. – Michael McGovern Jan 16 '17 at 4:11
• The cheeky answer is "because then it isn't RMS, it's just MS". – Tim Pederick Jan 16 '17 at 8:04
• @Omry: How is $m^2/s^2$ the right units for velocity? It's the right units for velocity squared, which I'm sure is Michael's point... – Tim Pederick Jan 16 '17 at 8:12
• see also: Root mean square vs average absolute deviation?, stats.stackexchange.com/questions/7447/… – Florian Castellane Jan 16 '17 at 10:47
• The answer you accepted explains nothing. We want to know the RMS not because we want units of velocity! The RMS is not a kind of velocity whatsoever. The next-top-voted answer is wrong; see my comment. Juris' answer is the most accurate explanation (agreeing with mine). If you don't believe me, ask a professional mathematician in real life and not over the internet. – user21820 Jan 18 '17 at 12:50
$T$ has units of Kelvin (K). The gas constant $R$ has units of Joule/mole/K. The molecular mass $M$ has units of kg/mole. Also remember that a Joule is $\mathrm{kg.m^2/s^2}$. So the units of $3RT/M$ are $$\mathrm{\frac{kg\,m^2\,K}{s^2\,mole\,K\,kg\,mole^{-1}}=\frac{m^2}{s^2}}$$ which is a velocity squared. So taking the square root gives the correct units for a velocity.
• This is true, but I think it's worth noting that you can't take the square root, but you can take it squared. i.e., since $\bar{u}$ is an average, $\bar{u}^2 \neq \bar{(u^2)}$, but you can get $\bar{u^2}$ without the root. – Omry Jan 16 '17 at 4:31
• I always tell students to write out units, I suppose I should have done the same :) – J M Jan 16 '17 at 4:31
• This answer is not wrong, but it misses the broader context of the issue, which is that RMS is a concept defined and used way more broadly than this one formula. See also my answer for that part. – Brick Jan 17 '17 at 17:28
• Simple is good if it's correct. In this case it seems like you're feeding them a line that's convenient for you rather than helpful to them. Certainly dimensional analysis is helpful and important, which is why I said this answer isn't quite wrong. (I mentioned the consistency of units as well.) The answer is still misleading. I've taught freshman physics, so I both understand and abhor the point you're making here since you seem to be shirking your responsibility to educate rather than appease. @JM – Brick Jan 20 '17 at 20:56
• @David My objection is to everyone taking a narrow view to their own domain of something used ubiquitously across many, many fields. RMS is defined in ALL fields by the left-side of that equation. The fact there is a square root on the right side is an accident of this application. The students should at least be made aware of connections across fields. This is exacerbated by the phrasing of the question, which makes it appear that the equation stated "defines" RMS velocity. That's not true. RMS velocity is defined by the left side of the equation only - The right side is derived from physics. – Brick Jan 23 '17 at 0:49
While working out the units points it out nicely, one can also consider that in:
$\sqrt{\bar{u^2}}$ = $\sqrt{\frac{3\text{RT}}{\text{M}}}$
The value $\sqrt{\bar{u^2}}$ in the whole is the actual result, and the square root is just part of the way "RMS" is written out. The value $\bar{u^2}$ is uninteresting for practical purposes.
It may be more clear if the definition is spelled out explicitly, for example like this:
1. The speed $u$ of individual gas molecules is essentially random. However, there are useful statistical properties for the root mean square speed of a large set of molecules. We can call this root mean square speed $r$, and define it as $r = \sqrt{\bar{u^2}}$
2. If we know the temperature and properties of the gas, we can calculate $r$ as $r = \sqrt{\frac{3\text{RT}}{\text{M}}}$.
Which removes the temptation to "simplify" the definition of RMS.
I think this question is more like a physics question. If you wanted the mathematics, it's already answered - you can square both sides and the formula would still be correct. But why don't we do that?
If you have a bunch of gas particles you might want to somehow describe their average speed with a single number. Well, maybe not really the average but a speed that is characteristic to that particular gas in the state considered. How can we do that?
• Average velocity. As you correctly mention, averaging over all velocities you get 0 in many cases. This is useful in some cases, for example you can use $\bar{\boldsymbol{v}}=(0,0,0)$ as a boundary condition when deriving velocity distribution.
Let's look at the distribution of how many molecules move with certain speed. You can see that actually almost none of the particles have zero speed or velocity. You might conclude that the average velocity is not that great metric to characterize a state of a gas. If the gas fills a container, it's average velocity doesn't care about the state, only about the movement of the container. So let's just average over the graph linked above, shall we?
• Aveage speed. The speed is magnitude (absolute value) of velocity. You could take the average (mean) speed of gas molecules.
• Most probable speed. When you looked at the graph... you might actually want to use the speed where the peak (maximum) is. That would describe the curve well, wouldn't it?
• Root mean square speed. Let's take the square of speed. Find the mean value. Take the root. Sounds like a nightmare, but this is actually the most useful metric. It is the speed that characterizes the energy of the gas.
The molecules of gas moves with different speeds. Each of the molecules have some kinetic energy. You could calculate the total or the average energy if you could measure each speed and do quite a lot of maths (not doable in a lifetime for a reasonable container of gas). However, if all of the gas molecules were moving with a certain speed that we call root mean square speed their total and average kinetic energy would be the same as it really is. As energy is usually what we care about the most in physics/chemistry, this is the speed that describes the speeds of a gas in a way that is useful for us.
If you care about the others:
• Average speed describes the momentum in similar way (total/average magnitude of momentum would stay the same if the molecules would all move with the mean speed).
• Average velocity describes the motion as a whole (motion of the center of mass if all the particles are of equal masses, also the momentum of the system).
• Most probable speed says that more molecules have about that speed than any other speed. To be precise you should choose interval, let's say consider number of molecules having a speed +/- 10m/s. Than the number pf molecules having speed in that interval will be the highest if the speed is the most probable speed. That is the best usefulness for this number that I can come up with.
Some stuff for further (yet introductory) reading on wikipedia.
• "you can't always drop the SQRTs" What does this mean? Could you give an example? – quid Jan 17 '17 at 22:47
• That $A=B$ is not equivalent to $A^2 = B^2$ is clear to me. But drop the roots is not the same as square both sides. Except if you meant "omit/drop squares" rather than square roots. I am curious about $\sqrt{x}= \sqrt{-5}$ what $x$ other than $-5$ does make it true? – quid Jan 17 '17 at 23:08
• "Dropping the roots is a jargon for squaring both sides" So if you know that $2a= x+y$ and you conclude from this that $4a^2= x^2 +2xy +y^2$ then you'd refer to this as drop the square roots? Really. Seems rather backwards to me then. Why not just say "square both sides"? If the roots made sense to begin with you can always drop them. (The sole issue might be if the root is there to indirectly restrict the domain.) By the way can I now conclude $x=-5$ if I know $\sqrt{x}=\sqrt{-5}$. It seems to me I can. – quid Jan 18 '17 at 0:16
• What is your definition of $\sqrt{x}$? And what is the definition of $\sqrt{-5}$? As I commented elsewhere "I think if you have any sane notion of √ such that √x and √y are well-defined to begin with, then √x=√y will be equivalent to x=y. (I'd be curious to hear a counter-example.) " – quid Jan 18 '17 at 0:43
• Maybe to be more clear, if you define $\sqrt$ as some principal branch, say in polar coordinates with angle in $[0,2\pi)$, you just half the angle, which is compatible with you saying $\sqrt{-5}= i \sqrt{5}$ , then there is just no complex number whatsoever whose square root is $-i\sqrt{5}$, as they all have argument less than $\pi$. Yet, if you define $\sqrt{x}$ as any number whose square is $x$, then it is not clear why $\sqrt{-5}$ cannot be $-i \sqrt{5}$ too. At which point $\sqrt{-5}$ might not equal $\sqrt{-5}$! – quid Jan 18 '17 at 0:58
The simple answer is because RMS is "Root of Mean value of Squares", it's the definition.
The reason you can't skip the square root is because it will not be a good representative of the concept of mean value, and that's because one have squared the quantities in the first place.
More deeper explanation why one does this is because the velocity squared is proportional to the kinetic energy of the particle. If they have the same mass the RMS of the velocities correspond to the average energy of the particles. So the RMS of the velocities is the velocity a particle with average energy has.
Similar reason is behind RMS value of voltage. Because the power produced is $P=UI$ and $I$ is proportional to $U$ according to Ohm's law the power is proportional to $U^2$ and therefore the RMS value corresponds to the (constant) voltage that results in the same power as the average power.
It depends on what you are trying to do.
If you are, for example, using the RMSE as a cost function that you are trying to minimise, then it is indeed a waste of computational resources and human brain power to take the square root. You can minimise the mean square error instead. The position at which the mean square error is minimal is identical to the position at which the root mean square error is minimal.
However, if you are trying to communicate an error or uncertainty, chances are they are more familiar with 10 m/s than with 100 m²/s².
The RMS of $u$ is defined by the left side of the equation, for whatever physical quantity $u$ you're considering. It's just coincidence in this case that the right hand side of the equation, which expresses this in terms of physical parameters of the system, also has a square root. You could find many examples where that's not the case and this particular "simplification" wouldn't make sense. As other have noted, the "root" and the "square" part of the definition are chosen to ensure that you get an answer in the same units as $u$ itself.
Your student is quite right in seeing that the formula $\sqrt{x}=\sqrt{y}$ is equivalent to the formula $x=y$.
The question is, which of these two equivalent formulas is easiest to use?
In this case we are interested in calculating an average speed. And the Root-Mean-Square is the type of average we are going to use.
We are not interested in the Mean-Square on its own and there is no point in calculating it.
Therefore the formula gives the RMS speed directly, even if it looks a bit redundant.
A different question is which formula is easiest to remember. If the student thinks that it more intuitive and easier without the square root signs, then they can go ahead and learn it that way. They just have to be aware that the exam questions will ask for the ROOT-Mean-Square. If they answer with just the Mean-Square that will be a wrong answer.
• The student is NOT correct, and nor are you or the OP, because the formula is actually "sqrt(x^2) = sqrt(y^2)". For a single value, this gives x=abs(y); and for the mean you need to integrate positive and negative portions separately. The key to RMS is that it copes with the signal going negative, which a simple "average" fundamentally will not do. – Graham Jan 16 '17 at 11:30
• @Graham Neither $R$ nor $T$ nor $M$ are ever going negative. And as $u$ is considered positive and directionless, neither does it. The main point to keep in mind is that the formula is about (the square root of) $\overline{ u^2}$ and not $\overline u$ or $\overline{\vec u}$ (which would certainly be $0$) – Hagen von Eitzen Jan 17 '17 at 7:15
• @HagenvonEitzen Thanks for that context on the original equation. I'm coming from a background in electronics and engineering, where negative values are not just possible but expected. – Graham Jan 17 '17 at 10:42
• @Graham the negative numbers are tangential. I think if you have any sane notion of $\sqrt{}$ such that $\sqrt{x}$ and $\sqrt{y}$ are well-defined to begin with, then $\sqrt{x}=\sqrt{y}$ will be equivalent to $x= y$. (I'd be curious to hear a counter-example.) – quid Jan 17 '17 at 23:32
• @quid: Logically, if you have the assumption that there exists $r,s$ such that $r^2 = x$ and $s^2 = y$, and $\sqrt{x},\sqrt{y}$ is one such $r,s$ respectively, then indeed the equivalence is correct. But without specifying the assumptions one would be leading students into a pit. This incorrect attitude in teaching is far more pervasive and damaging than you might think. Just look at all the posts on MathEd SE and you will be forced to conclude that nearly all of students' illogical thinking arises from such inadequate instruction. – user21820 Jan 18 '17 at 3:26
Because if you remove 'Square Root', its unit will no longer be speed ( in this case ) i.e ${m^2}$/$sec^2$
This term $\bar{u^2}$ represents 'Mean of Squared Component', aka 'ms' from 'rms'. Hence to get the 'root mean square' value, you need to take square of 'ms'. | 3,668 | 14,556 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5625 | 4 | CC-MAIN-2020-16 | latest | en | 0.895591 |
https://votetinadavis.com/6th-grade-math-integers-worksheets.html | 1,620,512,598,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243988927.95/warc/CC-MAIN-20210508211857-20210509001857-00141.warc.gz | 644,544,374 | 11,672 | # 6th Grade Math Integers Worksheets
### Common Core Math Worksheets Sixth Grade Math Integer Preview 1 Math Integers Sixth Grade Math 7th Grade Math Worksheets
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Free Math Worksheets for Grade 6. These worksheets are. Some of the worksheets for this concept are Grade 6 integers work Basic integral representations and absolute value state Adding and subtracting integers Addingsubtracting integers date period Integers Grade 6 integers work Dividing integers date period Adding integers a.
### 3
Source : pinterest.com | 295 | 1,549 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2021-21 | latest | en | 0.809101 |
http://www.edupil.com/question/weight-of-a/ | 1,477,311,779,000,000,000 | text/html | crawl-data/CC-MAIN-2016-44/segments/1476988719566.74/warc/CC-MAIN-20161020183839-00565-ip-10-171-6-4.ec2.internal.warc.gz | 426,932,805 | 18,398 | # What will be the Weight of A? If
The average weight of 3 Men A, B and C is 84 kg. Another man D joins the group and the average now becomes 80 kg. If another man E whose weight is 3 kg more than that of D, replaces A, then the average weight of B, C, D and E becomes 79 kg. Then weight of A is:
1. 75 kg
2. 76 kg
3. 72 kg
4. 74 kg
Monis Rasool Professor Asked on 28th June 2015 in
Explanation:-
The average weight of A,B and C = 84Kg
The total weight of A, B and C = 84 x 3 = 252Kg.
The average weight of A, B , C and D = 80 kg
The total weight of A, B, C and D = 80 x 4 = 320 kg
The weight of D = 320 – 252 = 68 kg
The weight of Q = 68 + 3 = 71 kg
The average weight of B, C ,D and E = 79 kg
The total weight of B, C, D and E = 79 x 4 = 316 kg
The total weight of A,B, C and D – the total weight of B,C,D and E = 320 – 316 = 4 kg
A – E = 4
A = 4 + E
A = 4 + 71
A = 75 Kg
Hence, the weight of A is 75 Kg.
Anurag Mishra Professor Answered on 28th June 2015. | 360 | 963 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2016-44 | longest | en | 0.904097 |
https://brainmass.com/math/basic-calculus/finding-equations-represent-exponential-functions-563770 | 1,708,950,017,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474659.73/warc/CC-MAIN-20240226094435-20240226124435-00308.warc.gz | 142,784,434 | 7,008 | Purchase Solution
# Finding equations that represent exponential functions
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Write an equation using the following information:
Each represents an exponential function with base 2 or 3 translated and/or reflected
1) (-3,0), (-2,1), (0,7)
Equation of horizontal asymptote: y=-1
2) (-1,3), (0,4), (-3,-3)
Equation of horizontal asymptote: y=5
##### Solution Summary
The given points, together with the equations of the horizontal asymptotes, are used to determine properties of the equations that represent exponential functions whose graphs pass through those points. The actual equations are derived from those properties.
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Some questions on probability | 368 | 1,760 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.3125 | 3 | CC-MAIN-2024-10 | latest | en | 0.855315 |
http://sudoku.com.au/2E4-9-2012-1-sudoku.aspx | 1,571,416,123,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570986684226.55/warc/CC-MAIN-20191018154409-20191018181909-00127.warc.gz | 175,358,923 | 17,808 | Timer 00:00:00 Sudoku Sudoku
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Pumpkin Carving Pattern Free halloween pumpkin carving patterns. | 2,212 | 7,547 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.96875 | 3 | CC-MAIN-2019-43 | latest | en | 0.885569 |
https://uk.mathworks.com/matlabcentral/cody/players/11340134/solved | 1,611,026,269,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703517559.41/warc/CC-MAIN-20210119011203-20210119041203-00087.warc.gz | 595,689,242 | 22,095 | Cody
# Daniel Fry
Rank
Score
1 – 50 of 163
#### Problem 43269. Get all prime factors
Created by: Jamil Kasan
#### Problem 1826. Find vampire numbers
Created by: Andrew Newell
#### Problem 191. Proper Factors
Created by: @bmtran (Bryant Tran)
#### Problem 1066. Multiples of a Number in a Given Range
Created by: @bmtran (Bryant Tran)
#### Problem 234. Project Euler: Problem 3, Largest prime factor
Created by: Doug Hull
#### Problem 1925. Smith numbers
Created by: Sky Sartorius
Tags sum, factors
#### Problem 148. Factorize THIS, buddy
Created by: the cyclist
#### Problem 2869. There are 10 types of people in the world
Created by: James
#### Problem 2678. Find out sum and carry of Binary adder
Created by: Pritesh Shah
Tags binary
#### Problem 2891. Binary code (array)
Created by: Yannick
#### Problem 43977. Converting binary to decimals
Tags binary, basics
#### Problem 1547. Relative ratio of "1" in binary number
Created by: Marek Kuklis
Tags matlab, binary
#### Problem 2522. Convert given decimal number to binary number.
Created by: Pritesh Shah
Tags numbers
#### Problem 1295. Bit Reversal
Created by: @bmtran (Bryant Tran)
#### Problem 108. Given an unsigned integer x, find the largest y by rearranging the bits in x
Created by: AMITAVA BISWAS
Tags binary
#### Problem 661. Spot the outlier
Created by: Ned Gulley
Tags asee, lines, points
#### Problem 722. Make a run-length companion vector
Created by: Ned Gulley
Tags vector, asee, best
#### Problem 730. How many trades represent all the profit?
Created by: Doug Hull
#### Problem 729. Flag largest magnitude swings as they occur
Created by: Doug Hull
Tags asee, peaks, change
#### Problem 731. Given a window, how many subsets of a vector sum positive
Created by: Doug Hull
Tags asee, good
#### Problem 751. Implement simple rotation cypher
Created by: Doug Hull
#### Problem 44334. Sums of Multiple Pairs of Triangular Numbers
Created by: James
#### Problem 44350. Breaking Out of the Matrix
Created by: James
#### Problem 44372. Polarisation
Created by: Jan Orwat
#### Problem 44352. The Top 5 Primes
Created by: Peng Liu
Tags cody5
#### Problem 44309. Pi Digit Probability
Created by: Mehmet OZC
#### Problem 44307. The glass half full
Created by: HH
Tags contest, cody5
#### Problem 44339. Recaman Sequence - II
Created by: Mehmet OZC
Created by: goc3
Created by: goc3
#### Problem 44338. Recaman Sequence - I
Created by: Mehmet OZC
#### Problem 44319. Write c^3 as sum of two squares a^2+b^2
Created by: Mehmet OZC
#### Problem 44375. Missing five
Created by: Alfonso Nieto-Castanon
Tags cody5
Created by: goc3
#### Problem 44321. Van Eck's Sequence's nth member
Created by: Mehmet OZC
Created by: goc3
#### Problem 44314. A Simple Tide Gauge with MATLAB
Created by: AMITAVA BISWAS
#### Problem 44342. Spot the First Occurrence of 5
Created by: Peng Liu
Tags cody5
#### Problem 44370. Octoberfest festival
Created by: Andriy Kavetsky
Created by: goc3
#### Problem 44363. Is this is a Tic Tac Toe X Win?
Created by: Doug Hull
#### Problem 369. Basic electricity in a dry situation
Created by: AMITAVA BISWAS
#### Problem 44384. Find the nearest prime number
Created by: Matthew Eicholtz
Created by: goc3
#### Problem 44385. Extra safe primes
Created by: Matthew Eicholtz
#### Problem 44349. Tick. Tock. Tick. Tock. Tick. Tock. Tick. Tock. Tick. Tock.
Created by: James
Tags time, cody5
#### Problem 2736. Pernicious Anniversary Problem
Created by: Jan Orwat
#### Problem 361. Energy of a photon
Created by: AMITAVA BISWAS
#### Problem 564. How to subtract?
Created by: AMITAVA BISWAS | 1,047 | 3,665 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.171875 | 3 | CC-MAIN-2021-04 | latest | en | 0.719008 |
https://excelexamples.com/post/how-to-use-not-equal-to-in-excel/ | 1,656,521,841,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103640328.37/warc/CC-MAIN-20220629150145-20220629180145-00089.warc.gz | 308,246,409 | 7,349 | In excel we are aware of operators like arithmetic and logical operators. Logical operators are mostly used in any organization to compare the two values either it is true or false.
This logical operator is also called Boolean operator (true or false) because it allows us to find out whether the values are not equal to one another, if the values are equal it will return false or else it will return true.
In Excel, <> means not equal to. The <> operator in Excel checks if two values are not equal to each other.
1. The formula in cell C1 below returns TRUE because the text value in cell A1 is not equal to the text value in cell B1.
2. The formula in cell C1 below returns FALSE because the value in cell A1 is equal to the value in cell B1.
3. The IF function below calculates the progress between a start and end value if the end value is not equal to an empty string (“”), else it displays an empty string (see row 5).
4. The COUNTIF function below counts the number of cells in the range A1:A5 that are not equal to “red”. | 231 | 1,037 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2022-27 | longest | en | 0.811492 |
https://www.statistics-lab.com/%E8%AE%A1%E7%AE%97%E6%9C%BA%E4%BB%A3%E5%86%99%E6%9C%BA%E5%99%A8%E5%AD%A6%E4%B9%A0%E4%BB%A3%E5%86%99machine-learning%E4%BB%A3%E8%80%83probability-theory/ | 1,716,504,576,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058671.46/warc/CC-MAIN-20240523204210-20240523234210-00213.warc.gz | 859,939,413 | 39,854 | ### 计算机代写|机器学习代写machine learning代考|Probability Theory
statistics-lab™ 为您的留学生涯保驾护航 在代写机器学习machine learning方面已经树立了自己的口碑, 保证靠谱, 高质且原创的统计Statistics代写服务。我们的专家在代写机器学习machine learning代写方面经验极为丰富，各种代写机器学习machine learning相关的作业也就用不着说。
• Statistical Inference 统计推断
• Statistical Computing 统计计算
• Advanced Probability Theory 高等概率论
• Advanced Mathematical Statistics 高等数理统计学
• (Generalized) Linear Models 广义线性模型
• Statistical Machine Learning 统计机器学习
• Longitudinal Data Analysis 纵向数据分析
• Foundations of Data Science 数据科学基础
## 计算机代写|机器学习代写machine learning代考|Set Theory
A set describes an ensemble of elements, also referred to as events. An elementary event $x$ refers to a single event among a sampling space (or universe) denoted by the calligraphic letter $\mathcal{S}$. By definition, a sampling space contains all the possible events, $E \subseteq \mathcal{S}$. The special case where an event is equal to the sampling space, $E=\mathcal{S}$, is called a certain event. The opposite, $E=\emptyset$, where an event is an empty set, is called a null event. $E$ refers to the complement of a set, that is, all elements belonging to $\mathcal{S}$ and not to $E$. Figure $3.3$ illustrates these concepts using a Venn diagram.
Let us consider the example, ${ }^{5}$ of the state of a structure following an earthquake, which is described by a sampling space,
\begin{aligned} \mathcal{S} &={\text { no damage, light damage, important damage, collapse }} \ &={N, L, I, C} . \end{aligned}
In that context, an event $E_{1}={\mathbf{N}, \mathrm{L}}$ could contain the no damage and light damage events, and another event $E_{2}={C}$ could contain only the collapsed state. The complements of these events are, respectively, $\overline{E_{1}}={\mathrm{I}, \mathrm{C}}$ and $\overline{E_{2}}={\mathrm{N}, \mathrm{L}, \mathrm{I}}$.
The two main operations for events, union and intersection, are illustrated in figure 3.4. A union is analogous to the “or” operator, where $E_{1} \cup E_{2}$ holds if the event belongs to either $E_{1}, E_{2}$, or both. The intersection is analogous to the “and” operator, where $E_{1} \backslash E_{2} \equiv E_{1} E_{2}$ holds if the event belongs to both $E_{1}$ and $E_{2}$. As a convention, intersection has priority over union. Moreover, both operations are commutative, associative, and distributive.
Given a set of $n$ events $\left{E_{1}, E_{2}, \cdots, E_{n}\right} \in \mathcal{S}, E_{1}, E_{2}, \cdots, E_{n}$,the events are mutually exclusive if $E_{i} E_{j}=\emptyset, \forall i \neq j$, that is, if the intersection for any pair of events is an empty set. Events $E_{1}, E_{2}, \cdots, E_{n}$ are collectively exhaustive if $\cup_{i=1}^{n} E_{i}=\mathcal{S}$, that is, the union of all events is the sampling space. Events $E_{1}, E_{2}, \cdots, E_{n}$ are mutually exclusive and collectively exhaustive if they satisfy both properties simultaneously. Figure $3.5$ presents examples of mutually exclusive (3.5a), collectively exhaustive (3.5b), and mutually exclusive and collectively exhaustive $(3.5 \mathrm{c}-\mathrm{d})$ events. Note that the difference between (b) and (c) is the absence of overlap in the latter.
## 计算机代写|机器学习代写machine learning代考|Probability of Events
$\operatorname{Pr}\left(E_{i}\right)$ denotes the probability of the event $E_{i}$. There are two main interpretations for a probability: the Frequentist and the Bayesian. Frequentists interpret a probability as the number of occurrences of $E_{i}$ relative to the number of samples $s$, as $s$ goes to $\infty$,
$$\operatorname{Pr}\left(E_{i}\right)=\lim {s \rightarrow \infty} \frac{#\left{E{i}\right}}{s} .$$
For Bayesians, a probability measures how likely is $E_{i}$ in comparison with other events in $\mathcal{S}$. This interpretation assumes that the nature of uncertainty is epistemic, that is, it describes our knowledge of a phenomenon. For instance, the probability depends on the available knowledge and can change when new information is obtained. Throughout this book we are adopting this Bayesian interpretation.
By definition, the probability of an event is a number between zero and one, $0 \leq \operatorname{Pr}\left(E_{i}\right) \leq 1$. At the ends of this spectrum, the probability of any event in $\mathcal{S}$ is one, $\operatorname{Pr}(\mathcal{S})=1$, and the probability of an empty set is zero, $\operatorname{Pr}(\emptyset)=0$. If two events $E_{1}$ and $E_{2}$ are mutually exclusive, then the probability of the events’ union is the sum of each event’s probability. Because the union of an event and its complement are the sampling space, $E \cup E=\mathcal{S}$ (see figure $3.5 \mathrm{~d}$ ), and because $\operatorname{Pr}(\mathcal{S})=1$, then the probability of the complement is $\operatorname{Pr}(E)=1-\operatorname{Pr}(E)$.
When events are not mutually exclusive, the general addition rule for the probability of the union of two events is
$$\operatorname{Pr}\left(E_{1} \cup E_{2}\right)=\operatorname{Pr}\left(E_{1}\right)+\operatorname{Pr}\left(E_{2}\right)-\operatorname{Pr}\left(E_{1} E_{2}\right) .$$
This general addition rule is illustrated in figure 3.6, where if we simply add the probability of each event without accounting for the subtraction of $\operatorname{Pr}\left(E_{1} E_{2}\right)$, the probability of the intersection of both events will be counted twice.
## 计算机代写|机器学习代写machine learning代考|The probability of a single e
The probability of a single event is referred to as a maryinal probability. A joint probability designates the probability of the intersection of events. The terms in equation $3.1$ can be rearranged to explicitly show that the joint probability of two events $\left{E_{1}, E_{2}\right}$ is the product of a conditional probability and its associated marginal,
\begin{aligned} \operatorname{Pr}\left(E_{1} E_{2}\right) &=\operatorname{Pr}\left(E_{1} \mid E_{2}\right) \cdot \operatorname{Pr}\left(E_{2}\right) \ &=\operatorname{Pr}\left(E_{2} \mid E_{1}\right) \cdot \operatorname{Pr}\left(E_{1}\right) . \end{aligned}
In cases where $E_{1}$ and $E_{2}$ are statistically independent, $E_{1} \perp E_{2}$,
Note: Statis conditional probabilities are equal to the marginal, tween a pair
$$E_{1} \perp E_{2} \begin{cases}\operatorname{Pr}\left(E_{1} \mid E_{2}\right)=\operatorname{Pr}\left(E_{1}\right) & \text { that learning } \ \operatorname{Pr}\left(E_{2} \mid E_{1}\right)=\operatorname{Pr}\left(E_{2}\right) & \text { other. }\end{cases}$$
In the special case of statistically independent events, the joint probability reduces to the product of the marginals,
$$\operatorname{Pr}\left(E_{1} E_{2}\right)=\operatorname{Pr}\left(E_{1}\right) \cdot \operatorname{Pr}\left(E_{2}\right)$$
The joint probability for $n$ events can be broken down into $n-1$ conditionals and one marginal probability using the chain rule,
\begin{aligned} \operatorname{Pr}\left(E_{1} E_{2} \cdots E_{n}\right) &=\operatorname{Pr}\left(E_{1} \mid E_{2} \cdots E_{n}\right) \operatorname{Pr}\left(E_{2} \cdots E_{n}\right) \ &=\operatorname{Pr}\left(E_{1} \mid E_{2} \cdots E_{n}\right) \operatorname{Pr}\left(E_{2} \mid E_{3} \cdots E_{n}\right) \operatorname{Pr}\left(E_{3} \cdots E_{n}\right) \ &=\operatorname{Pr}\left(E_{1} \mid E_{2} \cdots E_{n}\right) \operatorname{Pr}\left(E_{2} \mid E_{3} \cdots E_{n}\right) \cdots \operatorname{Pr}\left(E_{n-1} \mid E_{n}\right) \operatorname{Pr}\left(E_{n}\right) \end{aligned}
Let us define $\left{E_{1}, E_{2}, E_{3}, \cdots, E_{n}\right} \in \mathcal{S}$, a set of mutually exclusive and collectively exhaustive events, that is, $E_{i} E_{j}=\emptyset, \forall i \neq$ $j, \cup_{i=1}^{n} E_{i}=\mathcal{S}$ – and an event $A$ belonging to the same sampling
space, that is, $A \in \mathcal{S}$. This context is illustrated using a Venn diagram in figure 3.7. The probability of the event $A$ can be obtained by summing the joint probability of $A$ and each event $E_{i}$,
$$\operatorname{Pr}(A)=\sum_{i=1}^{n} \underbrace{\operatorname{Pr}\left(A \mid E_{i}\right) \cdot \operatorname{Pr}\left(E_{i}\right)}{\operatorname{Pr}\left(A E{i}\right)} .$$
## 计算机代写|机器学习代写machine learning代考|Probability of Events
\operatorname{Pr}\left(E_{i}\right)=\lim {s \rightarrow \infty} \frac{#\left{E{i}\right}}{s} 。\operatorname{Pr}\left(E_{i}\right)=\lim {s \rightarrow \infty} \frac{#\left{E{i}\right}}{s} 。
## 有限元方法代写
tatistics-lab作为专业的留学生服务机构，多年来已为美国、英国、加拿大、澳洲等留学热门地的学生提供专业的学术服务，包括但不限于Essay代写，Assignment代写，Dissertation代写，Report代写，小组作业代写，Proposal代写，Paper代写，Presentation代写，计算机作业代写，论文修改和润色，网课代做，exam代考等等。写作范围涵盖高中，本科，研究生等海外留学全阶段，辐射金融，经济学，会计学，审计学，管理学等全球99%专业科目。写作团队既有专业英语母语作者，也有海外名校硕博留学生，每位写作老师都拥有过硬的语言能力，专业的学科背景和学术写作经验。我们承诺100%原创，100%专业，100%准时，100%满意。
## MATLAB代写
MATLAB 是一种用于技术计算的高性能语言。它将计算、可视化和编程集成在一个易于使用的环境中，其中问题和解决方案以熟悉的数学符号表示。典型用途包括：数学和计算算法开发建模、仿真和原型制作数据分析、探索和可视化科学和工程图形应用程序开发，包括图形用户界面构建MATLAB 是一个交互式系统，其基本数据元素是一个不需要维度的数组。这使您可以解决许多技术计算问题，尤其是那些具有矩阵和向量公式的问题，而只需用 C 或 Fortran 等标量非交互式语言编写程序所需的时间的一小部分。MATLAB 名称代表矩阵实验室。MATLAB 最初的编写目的是提供对由 LINPACK 和 EISPACK 项目开发的矩阵软件的轻松访问，这两个项目共同代表了矩阵计算软件的最新技术。MATLAB 经过多年的发展，得到了许多用户的投入。在大学环境中，它是数学、工程和科学入门和高级课程的标准教学工具。在工业领域，MATLAB 是高效研究、开发和分析的首选工具。MATLAB 具有一系列称为工具箱的特定于应用程序的解决方案。对于大多数 MATLAB 用户来说非常重要，工具箱允许您学习应用专业技术。工具箱是 MATLAB 函数（M 文件）的综合集合，可扩展 MATLAB 环境以解决特定类别的问题。可用工具箱的领域包括信号处理、控制系统、神经网络、模糊逻辑、小波、仿真等。 | 3,172 | 9,252 | {"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 | 4 | CC-MAIN-2024-22 | latest | en | 0.693953 |
https://www.spss-tutorials.com/comparing-metric-variables/ | 1,712,938,593,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296816024.45/warc/CC-MAIN-20240412132154-20240412162154-00554.warc.gz | 890,199,945 | 19,121 | SPSS TUTORIALS BASICS ANOVA REGRESSION FACTOR CORRELATION
# Comparing Metric Variables
## Summary
This tutorial shows how to create proper tables and means charts for multiple metric variables. If statistical assumptions are satisfied, these may be followed up by a repeated measures ANOVA. We'll use income 2010 through income_2014 from freelancers.sav throughout this tutorial.
## SPSS FREQUENCIES for Inspecting Histograms
We'll first inspect the data in income_2010 through income_2014 by running basic histograms. This is done very easily with FREQUENCIES as shown in the syntax below.
Note in the output viewer window that some of the histograms look very weird due to extreme values.
## SPSS FREQUENCIES Syntax for Histograms
*Run basic histograms for inspecting plausibility of values in income variables.
frequencies income_2010 to income_2014
/format notable
/histogram.
## Specifying User Missing Values
Note that some extreme values occur in the data. These presumably don't reflect actual yearly incomes but rather codes to indicate missing values. We'll specify them as user missing values with the syntax below. Reinspecting the histograms shows that the remaining values show plausible distributions.
*1. Specify extreme values as user missing.
missing values income_2010 to income_2014 (99999997 thru hi).
*2. Rerun histograms. Results now look ok.
frequencies income_2010 to income_2014
/format notable
/histogram.
## SPSS DESCRIPTIVES Table
The basic answer to our question regarding the comparison of mean incomes over years is answered by a basic DESCRIPTIVES table. Because it tends to have excessive decimal places, we'll suppress those by hiding the decimals or our values with FORMATS.
A nicer alternative here is a very simple tool, discussed in Set Decimals for Output Tables Tool. However, we'll stick with our quick and dirty solution for now.
## SPSS DESCRIPTIVES Syntax
*1. Hiding decimals for values suppresses excessive decimals in output table.
formats income_2010 to income_2014 (dollar8).
*2. Run descriptives table.
descriptives income_2010 to income_2014.
Conclusion: mean yearly incomes seem to gradually rise over the years. Starting from \$45,000,- they increase with some \$5,000,- per year on average.
## SPSS Bar Chart for Multiple Variables
We'd now like to visualize these mean incomes with a chart. An appropriate choice here is a bar chart, as basic version of which is utterly easy to construct. The screenshots below walk you through.
In the left window of the screenshot below, select income_2010 through income_2014 by using Shift. Move the variables to Bars Represent. By default, SPSS will presume you want to plot the means (rather than some other statistic) of selected variables so this section doesn't need further editing.
## SPSS Bar Chart for Multiple Variables Syntax
Completing the steps shown in the previous screenshots results in the syntax below. Running it creates the desired bar chart.
*Bar chart for means of multiple variables.
GRAPH
/BAR(SIMPLE)=MEAN(income_2010) MEAN(income_2011) MEAN(income_2012) MEAN(income_2013) MEAN(income_2014)
/MISSING=LISTWISE
/TITLE='Mean incomes 2010 - 2014 (N between 39 and 40)'.
## Styling the Bar Chart
The bar chart we've come up with so far provides insight into the mean incomes over 2010 through 2014. However, it could use some styling. The best way for doing so is building and applying an SPSS chart template (.sgt) file.
In order to create more space for the variable labels, our chart template transposes (“puts on its side”) the chart. Since this reverses the order of the variables in our chart, it's nice to circumvent this out by also reversing the variable order in the chart syntax as shown below.
Second, we'd like to display \$40,000,- as \$40 K. We can set “K” as a suffix for the x-axis with our chart template. DO REPEAT provides a fast way for dividing the data values of the income variables by 1000. We don't need to reverse this if we precede it with TEMPORARY.
Finally, we can highlight differences in mean income by having the x-axis run from 40 (now representing \$40,000,-) through 70. We'll do so with our chart template as well. After adding minor additional tweaks, the screenshot below shows our final result.
## SPSS Bar Chart Syntax Example
*1. Indicate that following data modifications must be reversible.
temporary.
*2. Divide income_2010 through income_2014 by 1000.
do repeat income = income_2010 to income_2014.
compute income = income / 1000.
end repeat.
*3. Reverse order of variables and rerun graph. This reverses previous data modification as well. Transpose chart with SPSS chart template.
GRAPH
/BAR(SIMPLE)=MEAN(income_2014) MEAN(income_2013) MEAN(income_2012) MEAN(income_2011) MEAN(income_2010)
/MISSING=LISTWISE
/TITLE='Mean incomes 2010 - 2014 (N between 39 and 40)'.
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# THIS TUTORIAL HAS 1 COMMENT:
• ### By adam magombe on November 1st, 2019
wish to learn about various data packages | 1,211 | 5,087 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-18 | longest | en | 0.777186 |
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### What sampling frequency should I use if Nyquist is not available?
HINT When you sample at below the Nyquist rate, aliasing happens. That means frequencies higher than half the sampling rate get folded back down to below half the sampling rate. Have a read about ...
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### What sampling frequency should I use if Nyquist is not available?
As correctly stated in Peter K.'s answer, this question is about aliasing. Since you can't sample at a rate that is sufficiently high to avoid aliasing - i.e., $f_s>50\textrm{ kHz}$ - you have to ...
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### Why is $A\cos(2\pi f_ct)$ a non-stationary process?
A random process is a collection of random variables, one random variable for each time instant. It is best to write the random process as $$\{X(t)\colon -\infty < t < \infty\} \tag{1}$$ where ...
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### What will be the filtered output?
First note that: $$\cos(2\pi 50 t) \longleftrightarrow 0.5 \delta(f+50) + 0.5\delta(f-50)$$ $$\sin(2\pi 150 t) \longleftrightarrow 0.5 j \delta(f+150) -j 0.5\delta(f-150)$$ Hence the baseband ...
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### Transmitting over Unknown Fading Channel
The strategy depends heavily your "realistic scenario", e.g. which estimator, which equalizer, which estimation error, etc. In general scenario, you cannot trust the estimate of $h$, it means that ...
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The standard meaning of white noise includes an insistence (whether implicit or explicit) that the mean is $0$. Thus, what you want to prove is trivially true: since $$Y[n] = \sum_{k=-\infty}^\infty ... • 20.6k 4 votes ### Fourier components of \cos(2\pi f_1t) HINT: Going from your last equation,$$\frac{\sqrt{T}}{2}\bigg(\frac{e^{j2\pi (f_1T-n)}-1}{j2\pi (Tf_1-n)} + \frac{e^{-j2\pi (f_1T+n)}-1}{-j2\pi (Tf_1+n)}\bigg)$$This can be simplified further down ... • 3,416 4 votes Accepted ### Expressing 2N point DFT in terms of N point DFT Consider N-point sequence x[n] whose N-point DFT is X[k]. To compute 2N-point DFT, X_2[k], of x[n], one should append N-zeroes to x[n], and make it a 2N-point sequence, denoted x_2[n]. ... • 28.3k 4 votes ### Difference between causality and memorylessness A memoryless system's output is determined by the current input value only, hence, every memoryless system must also be causal (a system is causal if its output does not depend on the future input ... • 28.3k 4 votes ### Finite impulse response FIR filters In order for H(z) to be a linear phase filter, it must zeros both on the inside of the unit circle and at the complementary locations (1/z) which are outside the unit circle. Therefore a linear phase ... • 53.3k 4 votes Accepted ### Discrete-time Fourier transform of \frac{\cos(\frac{n\pi} 6)}{(n+3)\pi} HINT:$$\frac{\cos(n\pi /6)}{(n+3)\pi}=\frac{\cos[(n+3)\pi/6 -\pi/2]}{(n+3)\pi}=\frac{\sin[(n+3)\pi/6]}{(n+3)\pi}
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Purely by inspection of the block diagram the system is causal, because the output is the sum of the current input sample and stuff that's delayed -- there's no $z$ blocks in there to predict the ... | 922 | 3,068 | {"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} | 3.109375 | 3 | CC-MAIN-2024-33 | latest | en | 0.820242 |
https://www.hackmath.net/en/math-problem/5987 | 1,611,573,108,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703565541.79/warc/CC-MAIN-20210125092143-20210125122143-00450.warc.gz | 812,660,452 | 13,291 | # Trapezoid - intersection of diagonals
In the ABCD trapezoid is AB = 8 cm long, trapezium height 6 cm, and distance of diagonals intersection from AB is 4 cm. Calculate the trapezoid area.
Correct result:
S = 36 cm2
#### Solution:
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The ABCD trapezoid with the parallel sides of the AB and the CD and the E point of the AB side. The segment DE divides the trapezoid into two parts with the same area. Find the length of the AE line segment. | 1,094 | 3,810 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.3125 | 4 | CC-MAIN-2021-04 | longest | en | 0.885887 |
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1 Dimensional Analysis 1. Convert one unit of measure to another. Math 98 Supplement 2 LEARNING OBJECTIVE Often measurements are taken using different units. In order for one measurement to be compared to another, it is necessary to convert one unit of measurement to another. For instance, suppose you are visiting Bellingham from Canada. You see the speed limit sign indicating 60mph. How fast should you driving in kilometers per hour? Or, if your height is 165centimeters how tall are in in inches? I am 165 cm. How tall am I in inches? This problem is an example of unit conversion. You are trying to change one unit (centimeters) to another (inches). To convert a unit, you need conversion equations. These equations tell you how different units are related to one another. For example, these are all examples of conversion equations: From these equations we get what are called conversion factors. Note that these conversion factors (fractions) all equal 1 since the numerator and the denominator of each fraction are equal. You will multiply the quantity with one unit by these conversion factors so that you can eliminate the unwanted units and go to the unit you wish to obtain without changing the original quantity described. (Multiplying a quantity by 1 will not change the original quantity, right? ) Let s try this problem of converting 165cm to inches using conversion factor!!! 1
2 Example cm = inches. (Use the conversion equation, 2.54 cm = 1 inch.) Step 1: Write the quantity you are trying to convert as a fraction. That is, write 165cm as Step 2: Decide by which conversion factor you need to multiply the above fraction in order to eliminate the cm from the numerator. In this problem, the conversion equation relating centimeters and inches is given above. Note that from each conversion equation, you get two possible conversion factors. You task is to decide which one is the appropriate one to use. In this case you have to choose between ( Which one should you use and why? Step 3: Multiply by the conversion factor you chose in Step 2. Cancel the unwanted unit and note that you are left with the unit that you needed to obtain. Now simply evaluate the operation as follows. So, the final answer is inches. Do not forget to give the correct units with your final answer! Try this! (Round your answer to two decimal places if necessary.) a. 55 kg = pounds b. 50 liters = gallons (Use 1 pound = kg) (Use 1 gallon = liters) Answer: a pounds b gallons 2
3 For the next example, let s consider something just a bit more involved. Example cm = feet? Note that this problem is asking for feet in the final answer. So, in addition to the conversion equation we used in Example 1, use 1 foot = 12 inches. This is an example of a multiple-step problem. In the first example, we converted centimeters to inches in one step. If we stop here; however, your final answer will be in inches. This isn t what you want. Now you need to take one more step to go from inches to feet. (Note! Every time you multiply by a conversion factor, cancel the unwanted unit at each step in order to keep track of where you are in the process.) The final answer is 5.41 feet. Try this! (Round your answer to the nearest tenth if necessary.) a. 55 kg = ounces b. 250 liters = quarts (1 pound = kg, 1 pound = 16 oz.) (1 gal = liters, 1 gal = 4 quarts) Answer: b ounces b quarts Some units such as square feet or cubic meters incorporate arithmetical operations. For instance, cubic meters means. How can we covert these units using conversion factors? To see how this is done, consider the following example. 3
4 Example 3 A room is measured to be 115 square meters. What is this are in square feet? Use the conversion equation, 1 meter = 3.28 feet. Be careful with square or cubic units!!! 115 square meters means So, as you can see below, using the conversion factor once is not going to eliminate both factors of meters. Note that in the above step, only one of the two factors of meters is canceled. In order to eliminate both of them, you will need to multiply by the conversion factor one more time. So, 115 square meters is about square feet. Try this! (Round your answer to two decimal places if necessary.) a. 150 sq ft = sq yards b cm 3 = ft 3 (1 yard = 3 feet) (1 inch = 2.54 cm, 1 foot = 12 inches) Careful! This is a multiple-step problem. Answer: a square yards b cubic feet Now let s consider some examples where the quantities you wish to convert are rates. The Earth completes its orbit around the Sun roughly in 365days. If the orbital distance is about million kilometers, what is Earth s orbital speed in miles per hour? We will try to do this conversion using conversion factors as follows. First let s consider all the conversion equations we will need to use. We are given that the Earth travels around the Sun million km per 365 days. That is, we need to convert into miles per hour. 4
5 Example 4 Let s first consider all the conversion equations necessary. Note that for this problem, we need to covert kilometers to miles and days to hours. We will use the following equations: In the original fraction, km is in the numerator and days is in the denominator. So, we will need the conversion factors, in order to eliminate these unwanted units. Now we can obtain our answer as follows: So, the Earth is going around the Sun at the speed of 10, mph. Did you know that we were traveling around the Sun at this high speed??? Try this! (Round your answer to two decimal places if necessary.) a. 600 inches/min = ft/sec b. 450 cm/min = mph (1 in = 2.54 cm, 1 ft = 12 in, 5,280 ft = 1 mile) Answer: a feet per second b miles per hour 5
6 TOPIC EXERCISES (Refer to the list of conversion equations attached at the end of this packet.) Convert each of the following using conversion factors. Round your answer to two decimal places if necessary Answer each of the following questions using conversion factors. Round your answer to two decimal places if necessary. 21. A new born baby weighs 2900 grams. How many pounds does this baby weigh? Also, how many pounds, how many ounces is it? 22. The gas tank of a car holds 65 liters of gas. How many gallons of gas does this tank hold? 23. If a recipe requires 2 ¼ cups of milk, how much milk is this in ml? 24. A house is built on a 0.27 acre of land. How many square miles is this land? 25. Is it reasonable for a new born human baby to be 1500mm in length? Explain your answer. 26. A river boat requires water depth of 2 fathoms. If the depth of the water is 2.5 meters, can this boat safely operate in this river? 27. One serving of dessert contains 5 teaspoons of sugar. If you need to prepare 250 servings of this dessert for a party, how many cups of sugar do you need? serving of ice cream is 1/2 cup and contains 285 calories. If you eat 25 teaspoons of this ice cream, how many calories do you consume? 6
7 29. You need to rent a moving truck which has a capacity of 750 cubic feet. If a truck has 23 cubic meters of capacity, is this a big enough truck for your move? Explain your answer. 30. A company from U.K. needs to purchase a parcel of land which is at least 2.5 square km. If an U.S. company offers to sell a parcel of land with an area of square feet, is this big enough for the U.K. company to consider purchasing? 31. A marathon is a race over km. If a runner runs this distance in 2 hours, how fast does he run in feet per second? 32. In one city in Florida, it rained 2.2 inches in 50 minutes one afternoon. How many millimeters per hour is this? ANSWERS
8 Conversion Equations: Length/Distance 1 m = 100 cm 1 mile = 8 furlong 1 m = 1000 mm 1 mile = 5280 feet 1 km = 1000 m 1 mile = km 1 inch = 2.54 cm 1 km = mile 12 inches = 1 foot 1 fathom = 6 feet 1 yard = 3 feet 1 nautical mile = 1852 m 1 furlong = 220 yards 1 light year miles Volume 1 L (liter) = 1000 cm 3 1 oil barrel = 42 gallons 1 L = 1000 ml 1 teaspoon = 5 ml 1 gallon = L 1 tablespoon = 3 teaspoon 1 pint = 16 fl oz 1 cup = 16 tablespoon 1 quart = 2 pint 1 cup = 240 ml 1 gallon = 4 quart 1 bushel = 32 quart 1 pint = 2 cups Area 640 acres = 1 square mile 1 hectare = m 2 1 hectare = acres Mass/Weight 1 gram = 1000 mg 1 pound = 16 ounces 1 kg = 1000 grams 1 ton = 2000 pounds 1 metric ton = 1000 kg 1 pound = kg 1 kg = pound 8
Dimensional Analysis is a simple method for changing from one unit of measure to another. How many yards are in 49 ft?
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https://www.blfilm.com/2020/10/04/how-do-i-calculate-retained-earnings/ | 1,670,375,417,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446711121.31/warc/CC-MAIN-20221206225143-20221207015143-00806.warc.gz | 719,598,838 | 16,324 | ## How do I calculate retained earnings?
How to find retained earnings. Retained earnings are shown in two places in your business’ financial statements: On the bottom line of your Income Statement (also called the Profit and Loss Statement) In the shareholder’s equity section of your Balance Sheet.
How do I calculate retained earnings on a balance sheet?
To calculate retained earnings subtract a company’s liabilities from its assets to get your stockholder equity, then find the common stock line item in your balance sheet and take the total stockholder equity and subtract the common stock line item figure (if the only two items in your stockholder equity are common …
How do you calculate retained earnings in first year?
### What is accumulated retained earnings on a balance sheet?
Accumulated retained earnings, which are represented on a company’s balance sheet as stockholders’ equity, are company profits that are held instead of distributing to shareholders.
Is retained earnings on the income statement?
Since the statement of retained earnings is such a short statement, it sometimes appears at the bottom of the income statement after net income.
What is in the statement of retained earnings?
The statement of retained earnings is a financial statement prepared by corporations that details changes in the volume of retained earnings over some period. Retained earnings are profits held by a company in reserve in order to invest in future projects rather than distribute as dividends to shareholders.
#### Which of the following equation is true for balance sheet?
Solution(By Examveda Team) The balance sheet is based on the fundamental equation: Assets = Liabilities + Equity(Capital).
How do you find retained earnings after closing entries?
In short, the change to retained earnings in each period is equal to that period’s net income minus the dividends declared for that period. Calculate the business’s net income for the period in question. Net income is equal to revenues minus expenses and can be found on the income statement.
Is retained earnings the same as accumulated earnings?
Retained earnings, also known as Accumulated Earnings or Accumulated Earnings and Profits, can be defined as a company’s accumulated surplus or profits after paying out the dividends to shareholders. Generally, Retained earnings represents the company’s extra earnings available at management’s disposal.
## How do you record retained earnings for a journal entry?
If the organization experiences a net loss, debit the retained earnings account and credit the income account. Conversely, if the organization experiences a profit, debit the income account and credit the retained earnings account.
Where is retained earnings in financial statements?
equity section
Retained earnings are listed on a company’s balance sheet under the equity section. A balance sheet provides a quick snapshot of a company’s assets, liabilities, and equity at a specific point in time. It helps business owners and outside investors understand the health and liquidity of the business.
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https://advancesincontinuousanddiscretemodels.springeropen.com/articles/10.1186/s13662-015-0542-3 | 1,725,804,884,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651002.87/warc/CC-MAIN-20240908115103-20240908145103-00269.warc.gz | 70,231,635 | 78,600 | Theory and Modern Applications
# On some classes of difference equations of infinite order
## Abstract
We consider a certain class of difference equations on an axis and a half-axis, and we establish a correspondence between such equations and simpler kinds of operator equations. The last operator equations can be solved by a special method like the Wiener-Hopf method.
## 1 Introduction
Difference equations of finite order arise very often in various problems in mathematics and applied sciences, for example in mathematical physics and biology. The theory for solving such equations is very full for equations with constant coefficients [1, 2], but fully incomplete for the case of variable coefficients. Some kinds of such equations were obtained by the second author by studying general boundary value problems for mode elliptic pseudo differential equations in canonical non-smooth domains, but there is no solution algorithm for all situations [3–5]. There is a certain intermediate case between the two mentioned above, namely it is a difference equation with constant coefficients of infinite order. Here we will briefly describe these situations.
The general form of the linear difference equation of order n is the following [1, 2]:
$$\sum_{k=0}^{n}a_{k}(x)u(x+k)=v(x), \quad x\in M\subset{\mathbb{R}},$$
(1)
where the functions $$a_{k}(x)$$, $$k=1,\ldots,n$$, $$v(x)$$ are defined on M and given, and $$u(x)$$ is an unknown function. Since $$n\in{\mathbf{N}}$$ is an arbitrary number and all points $$x, x+1,\ldots,x+n$$, $$\forall x\in M$$, should be in the set M, this set M may be a ray from a certain point or the whole R.
A more general type of difference equation of finite order is the equation
$$\sum_{k=0}^{n}a_{k}(x)u(x+ \beta_{k})=v(x),\quad x\in M\subset{\mathbb{R}},$$
(2)
where $$\{\beta_{k}\}_{k=0}^{n}\subset{\mathbb{R}}$$.
Further, such equations can be equations with a continuous variable or a discrete one, and this property separates such an equation on a class of properly difference equations and discrete equations. In this paper we will consider the case of a continuous variable x, and a solution on the right-hand side will be considered in the space $$L_{2}({\mathbb {R}})$$ for all equations.
### 1.1 Difference equation of a finite order with constant coefficients
This is an equation of the type
$$\sum_{k=0}^{n}a_{k}u(x+ \beta_{k})=v(x), \quad x\in{\mathbb{R}},$$
(3)
and it easily can be solved by the Fourier transform:
$$(Fu) (\xi)\equiv\tilde{u}(\xi)=\int_{-\infty}^{+\infty }e^{-ix\cdot\xi}u(x) \, dx.$$
Indeed, applying the Fourier transform to (3) we obtain
$$\tilde{u}(\xi)\sum_{k=0}^{n}a_{k}e^{ i\beta_{k}\xi}= \tilde {v}(\xi),$$
or renaming
$$\tilde{u}(\xi)p_{n}(\xi)=\tilde{v}(\xi).$$
The function $$p_{n}(\xi)$$ is called a symbol of a difference operator on the left-hand side (3) (cf. [6]). If $$p_{n}(\xi )\neq0$$, $$\forall\xi\in{\mathbb{R}}$$, then (3) can easily be solved,
$$u(x)=F^{-1}_{\xi\to x} \bigl(p_{n}^{-1}(\xi) \tilde{v}(\xi) \bigr).$$
### 1.2 Difference equation of infinite order with constant coefficients
The same arguments are applicable for the case of an unbounded sequence $$\{\beta_{k}\}_{-\infty}^{+\infty}$$. Then the difference operator with complex coefficients
$$\mathcal{D}: u(x)\longmapsto\sum_{-\infty}^{+\infty }a_{k}u(x+ \beta_{k}), \quad x\in{\mathbb{R}}, a_{k}\in{\mathbb{C}},$$
(4)
has the following symbol:
$$\sigma(\xi)=\sum_{-\infty}^{+\infty}a_{k}e^{i\beta_{k}\xi}.$$
### Lemma 1
The operator $$\mathcal{D}$$ is a linear bounded operator $$L_{2}({\mathbb {R}})\to L_{2}({\mathbb{R}})$$ if $$\{a_{k}\}_{-\infty}^{+\infty}\in {l}^{1}$$.
### Proof
The proof of this assertion can be obtained immediately. □
If we consider the operator (4) for $$x\in{\mathbb{Z}}$$ only
$$\mathcal{D}: u(x_{d})\longmapsto\sum _{-\infty}^{+\infty }a_{k}u(x_{d}+ \beta_{k}), \quad x_{d}\in{\mathbb{Z}},$$
(5)
then its symbol can be defined by the discrete Fourier transform [7, 8]
$$\sigma_{d}(\xi)=\sum_{-\infty}^{+\infty}a_{k}e^{i\beta_{k}\xi }, \quad \xi\in[-\pi,\pi].$$
### 1.3 Difference and discrete equations
Obviously there are some relations between difference and discrete equations. Particularly, if $$\{\beta_{k}\}_{-\infty}^{+\infty }={\mathbb{Z}}$$, then the operator (5) is a discrete convolution operator. For studying discrete operators in a half-space the authors have developed a certain analytic technique [9–11]. Below we will try to enlarge this technique for more general situations.
## 2 General difference equations
We consider the equation
$$(\mathcal{D}u) (x)=v(x), \quad x\in{\mathbb{R}}_{+},$$
(6)
where $${\mathbb{R}}_{+}=\{x\in{\mathbb{R}}, x>0\}$$.
For studying this equation we will use methods of the theory of multi-dimensional singular integral and pseudo differential equations [3, 6, 12] which are non-usual in the theory of difference equations. Our next goal is to study multi-dimensional difference equations, and this one-dimensional variant is a model for considering other complicated situations. This approach is based on the classical Riemann boundary value problem and the theory of one-dimensional singular integral equations [13–15].
### 2.1 Background
The first step is the following. We will use the theory of so-called paired equations [15] of the type
$$(aP_{+}+bP_{-})U=V$$
(7)
in the space $$L_{2}(\mathbb{R})$$, where a, b are convolution operators with corresponding functions $$a(x)$$, $$b(x)$$, $$x\in{\mathbb{R}}$$, $$P_{\pm }$$ are projectors on the half-axis $${\mathbb{R}}_{\pm}$$. More precisely,
$$(aP_{+}U) (x)=\int_{0}^{+\infty}a(x-y)U(y)\, dy, \qquad (bP_{-}U) (x)=\int_{-\infty}^{0}b(x-y)U(y)\, dy.$$
Applying the Fourier transform to (7) we obtain [12] the following one-dimensional singular integral equation [13–15]:
$$\tilde{a}(\xi) (P\widetilde{U}) (\xi)+\tilde{b}(\xi) (Q\widetilde {U}) (\xi)=\widetilde{V}(\xi),$$
(8)
where P, Q are two projectors related to the Hilbert transform
\begin{aligned}& (Hu) (x)=\mathrm{v.p.}\, \frac{1}{\pi i}\int_{-\infty}^{+\infty} \frac {u(y)}{x-y}\, dy, \\& P=\frac{1}{2}(I+H), \qquad Q=\frac{1}{2}(I-H). \end{aligned}
Equation (8) is closely related to the Riemann boundary value problem [13, 14] for upper and lower half-planes. We now recall the statement of the problem: finding a pair of functions $$\Phi^{\pm}(\xi )$$ which admit an analytic continuation on upper ($${\mathbb{C}}_{+}$$) and lower ($${\mathbb{C}}_{-}$$) half-planes in the complex plane $$\mathbb {C}$$ and of which their boundary values on $$\mathbb{R}$$ satisfy the following linear relation:
$$\Phi^{+}(\xi)=G(\xi)\Phi^{-}(\xi)+g(\xi),$$
(9)
where $$G(\xi)$$, $$g(\xi)$$ are given functions on $$\mathbb{R}$$.
There is a one-to-one correspondence between the Riemann boundary value problem (9) and the singular integral equation (8), and
$$G(\xi)=\tilde{a}^{-1}(\xi)\tilde{b}(\xi), \qquad \widetilde{V}(\xi )= \tilde{a}^{-1}(\xi)g(\xi).$$
### 2.2 Topological barrier
We suppose that the symbol $$G(\xi)$$ is a continuous non-vanishing function on the compactification $$\dot{\mathbb{R}}$$ ($$G(\xi)\neq0$$, $$\forall\xi\in\dot{\mathbb{R}}$$) and
$$\operatorname{Ind} G\equiv\frac{1}{2\pi}\int _{-\infty}^{+\infty }d\arg G(t)=0.$$
(10)
The last condition (10), is necessary and sufficient for the unique solvability of the problem (9) in the space $$L_{2}({\mathbb{R}})$$ [13, 14]. Moreover, the unique solution of the problem (9) can be constructed with a help of the Cauchy type integral
$$\Phi^{+}(t)=G_{+}(t)P \bigl(G_{+}^{-1}(t)g(t) \bigr),\qquad \Phi ^{-}(t)=-G_{-}^{-1}(t)Q \bigl(G_{+}^{-1}(t)g(t) \bigr),$$
where $$G_{\pm}$$ are factors of a factorization for the $$G(t)$$ (see below),
$$G_{+}(t)=\exp \bigl(P \bigl(\ln G(t) \bigr) \bigr),\qquad G_{-}(t)=\exp \bigl(Q \bigl(\ln G(t) \bigr) \bigr).$$
### 2.3 Difference equations on a half-axis
Equation (6) can easily be transformed into (7) in the following way. Since the right-hand side in (6) is defined on $$\mathbb{R}_{+}$$ only we will continue $$v(x)$$ on the whole $${\mathbb {R}}$$ so that this continuation $$\mathit{lf}\in L_{2}({\mathbb{R}})$$. Further we will rename the unknown function $$u_{+}(x)$$ and define the function
$$u_{-}(x)=(\mathit{lf}) (x)-(\mathcal{D}u_{+}) (x).$$
Thus, we have the following equation:
$$(\mathcal{D}u_{+}) (x)+u_{-}(x)=(\mathit{lf}) (x),\quad x\in \mathbb{R},$$
(11)
which holds for the whole space $${\mathbb{R}}$$.
After the Fourier transform we have
$$\sigma(\xi)\tilde{u}_{+}(\xi)+\tilde{u}_{-}(\xi)=\widetilde {\mathit{lv}}(\xi),$$
where $$\sigma(\xi)$$ is called a symbol of the operator $$\mathcal{D}$$.
To describe a solving technique for (11) we recall the following (cf. [13, 14]).
### Definition
A factorization for an elliptic symbol is called its representation if it is in the form
$$\sigma(\xi)=\sigma_{+}(\xi)\cdot\sigma_{-}(\xi),$$
where the factors $$\sigma_{+}$$, $$\sigma_{-}$$ admit an analytic continuation into the upper and lower complex half-planes $${\mathbb{C}}_{\pm}$$, and $$\sigma^{\pm1}_{\pm}\in L_{\infty}({\mathbb{R}})$$.
### Example 1
Let us consider the Cauchy type integral
$$\frac{1}{2\pi i}\int_{-\infty}^{+\infty}\frac {u(t)}{z-t}\, dt\equiv\Phi(z), \quad z=x\pm iy.$$
It is well known this construction plays a crucial role for a decomposition $$L_{2}({\mathbb{R}})$$ on two orthogonal subspaces, namely
$$L_{2}({\mathbb{R}})=A_{+}({\mathbb{R}})\oplus A_{-}({\mathbb{R}}),$$
where $$A_{\pm}({\mathbb{R}})$$ consists of functions admitting an analytic continuation onto $${\mathbb{C}}_{\pm}$$.
The boundary values of the integral $$\Phi(z)$$ satisfy the Plemelj-Sokhotskii formulas [13, 14], and thus the projectors P and Q are corresponding projectors on the spaces of analytic functions [15].
The simple example we need is
$$\exp(u)=\exp(Pu)\cdot\exp(Qu).$$
### Theorem 2
Let $$\sigma(\xi)\in C(\dot{\mathbb{R}})$$, $$\operatorname{Ind}\sigma =0$$. Then (6) has unique solution in the space $$L_{2}({\mathbb {R}}_{+})$$ for arbitrary right-hand side $$v\in L_{2}({\mathbb{R}}_{+})$$, and its Fourier transform is given by the formula
$$\tilde{u}(\xi)=\frac{1}{2}\sigma^{-1}(\xi)\widetilde{ \mathit{lv}}(\xi )+\frac{\sigma^{-1}_{+}(\xi)}{2\pi i}\, \mathrm{v.p.}\int_{-\infty }^{+\infty} \frac{\sigma^{-1}_{-}(\eta)\widetilde{\mathit{lv}}(\eta)}{\xi -\eta}\, d\eta.$$
### Proof
We have
$$\sigma_{+}(\xi)\tilde{u}_{+}(\xi)+\sigma^{-1}_{-}(\xi)\tilde{u}_{-}(\xi)= \sigma^{-1}_{-}(\xi)\widetilde{\mathit{lv}}(\xi).$$
Further, since $$\sigma^{-1}_{-}(\xi)\widetilde{\mathit{lv}}(\xi)\in L_{2}({\mathbb{R}})$$ we decompose it into two summands
$$\sigma^{-1}_{-}(\xi)\widetilde{\mathit{lv}}(\xi)=P \bigl( \sigma^{-1}_{-}(\xi )\widetilde{\mathit{lv}}(\xi) \bigr)+Q \bigl( \sigma^{-1}_{-}(\xi)\widetilde {\mathit{lv}}(\xi) \bigr)$$
and write
$$\sigma_{+}(\xi)\tilde{u}_{+}(\xi)-P \bigl(\sigma^{-1}_{-}(\xi )\widetilde{ \mathit{lv}}(\xi) \bigr)=Q \bigl(\sigma^{-1}_{-}(\xi)\widetilde { \mathit{lv}}(\xi) \bigr)-\sigma^{-1}_{-}(\xi)\tilde{u}_{-}(\xi).$$
The left-hand side of the last quality belongs to the space $$A_{+}({\mathbb{R}})$$, but the right-hand side belongs to $$A_{-}({\mathbb {R}})$$, consequently these are zeros. Thus,
$$\sigma_{+}(\xi)\tilde{u}_{+}(\xi)-P \bigl(\sigma^{-1}_{-}(\xi )\widetilde{ \mathit{lv}}(\xi) \bigr)=0$$
and
$$\tilde{u}_{+}(\xi)=\sigma^{-1}_{+}(\xi)P \bigl( \sigma^{-1}_{-}(\xi )\widetilde{\mathit{lv}}(\xi) \bigr),$$
(12)
or in the complete form
$$\tilde{u}_{+}(\xi)=\frac{1}{2}\sigma^{-1}(\xi)\widetilde{ \mathit{lv}}(\xi )+\frac{\sigma^{-1}_{+}(\xi)}{2\pi i}\, \mathrm{v.p.}\int_{-\infty }^{+\infty} \frac{\sigma^{-1}_{-}(\eta)\widetilde{\mathit{lv}}(\eta)}{\xi -\eta}\, d\eta.$$
 □
### Remark 1
This result does not depend on the continuation lv. Let us denote by $$M_{\pm}(x)$$ the inverse Fourier images of the functions $$\sigma ^{-1}_{\pm}(\xi)$$. Indeed, (12) leads to the following construction:
\begin{aligned} u_{+}(x)&=\int_{-\infty}^{+\infty}M_{+}(x-y) \biggl(\int _{0}^{+\infty}M_{-}(y-t) (\mathit{lv}) (t)\, dt \biggr)\, dy \\ &=\int_{-\infty}^{+\infty}M_{+}(x-y) \biggl(\int _{0}^{+\infty}M_{-}(y-t)v(t)\, dt \biggr)\, dy. \end{aligned}
### Remark 2
The condition $$\sigma(\xi)\in C(\dot{\mathbb{R}})$$ is not a strong restriction. Such symbols exist for example in the case that $$\sigma (\xi)$$ is represented by a finite sum, and $$\beta_{k}\in{\mathbb {Q}}$$. Then $$\sigma(\xi)$$ is a continuous periodic function.
## 3 General solution
Since $$\sigma(\xi)\in C(\dot{\mathbb{R}})$$, and Indσ is an integer, we consider the case $$\ae\equiv\operatorname{Ind} \sigma\in{\mathbb{N}}$$ in this section.
### Theorem 3
Let $$\operatorname{Ind} \sigma\in{\mathbb{N}}$$. Then a general solution of (6) in the Fourier image can be written in the form
$$\tilde{u}_{+}(\xi)=\sigma^{-1}_{+}(\xi)\omega^{-\ae}(\xi)P \bigl( \sigma^{-1}_{-}(\xi)\widetilde{\mathit{lv}}(\xi) \bigr)+( \xi-i)^{-\ae }\sigma^{-1}_{+}(\xi)P_{\ae-1}(\xi),$$
and it depends on æ arbitrary constants.
### Proof
The function
$$\omega(\xi)=\frac{\xi-i}{\xi+i}$$
has the index 1 [13–15], thus the function
$$\omega^{-\ae}(\xi)\sigma(\xi)= \biggl(\frac{\xi-i}{\xi+i} \biggr)^{-\ae}\cdot\sigma(\xi)$$
has the index 0, and we can factorize this function
$$\omega^{-\ae}(\xi)\sigma(\xi)=\sigma_{+}(\xi)\sigma_{-}(\xi).$$
Further, we write after (11)
$$\omega^{\ae}(\xi)\omega^{-\ae}(\xi)\sigma(\xi)\tilde{u}_{+}(\xi)+ \tilde{u}_{-}(\xi)=\widetilde{\mathit{lv}}(\xi),$$
factorize $$\omega^{-\ae}(\xi)\sigma(\xi)$$, and rewrite
$$\omega^{\ae}(\xi)\sigma_{+}(\xi)\tilde{u}_{+}(\xi)+\sigma ^{-1}_{-}( \xi)\tilde{u}_{-}(\xi)=\sigma^{-1}_{-}(\xi)\widetilde {\mathit{lv}}(\xi).$$
Taking into account our notations we have
\begin{aligned}& \sigma_{+}(\xi)\tilde{u}_{+}(\xi)+\omega^{-\ae}(\xi)\sigma ^{-1}_{-}(\xi)\tilde{u}_{-}(\xi) \\& \quad =\omega^{-\ae}(\xi)P \bigl(\sigma ^{-1}_{-}(\xi)\widetilde{ \mathit{lv}}(\xi) \bigr) +\omega^{-\ae}(\xi)Q \bigl(\sigma^{-1}_{-}( \xi)\widetilde{\mathit{lv}}(\xi ) \bigr), \end{aligned}
(13)
because
$$\sigma^{-1}_{-}(\xi)\widetilde{\mathit{lv}}(\xi)=P \bigl( \sigma^{-1}_{-}(\xi )\widetilde{\mathit{lv}}(\xi) \bigr)+Q \bigl( \sigma^{-1}_{-}(\xi)\widetilde {\mathit{lv}}(\xi) \bigr)$$
and
\begin{aligned}& \sigma_{+}(\xi)\tilde{u}_{+}(\xi)-\omega^{-\ae}(\xi)P \bigl( \sigma ^{-1}_{-}(\xi)\widetilde{\mathit{lv}}(\xi) \bigr) \\& \quad =\omega^{-\ae}(\xi)Q \bigl(\sigma^{-1}_{-}(\xi)\widetilde{ \mathit{lv}}(\xi ) \bigr)-\omega^{-\ae}(\xi)\sigma^{-1}_{-}(\xi) \tilde{u}_{-}(\xi) \end{aligned}
(14)
and we conclude from the last that the left-hand side and the right-hand side also are a polynomial $$P_{\ae-1}(\xi)$$ of order $$\ae-1$$. It follows from the generalized Liouville theorem [13, 14] because the left-hand side has one pole of order æ in $${\mathbb{C}}$$ in the point $$z=i$$. So, we have
$$\tilde{u}_{+}(\xi)=\sigma^{-1}_{+}(\xi)\omega^{-\ae}(\xi)P \bigl( \sigma^{-1}_{-}(\xi)\widetilde{\mathit{lv}}(\xi) \bigr)+( \xi-i)^{-\ae }\sigma^{-1}_{+}(\xi)P_{\ae-1}(\xi).$$
 □
### Remark 3
This result does not depend on the choice of the continuation l.
### Corollary 4
Let $$v(x)\equiv0$$, $$\ae\in{\mathbb{N}}$$. Then a general solution of the homogeneous equation (6) is given by the formula
$$\tilde{u}_{+}(\xi)=(\xi-i)^{-\ae}\sigma^{-1}_{+}( \xi)P_{\ae-1}(\xi).$$
## 4 Solvability conditions
### Theorem 5
Let $$-\operatorname{Ind} \sigma\in{\mathbb{N}}$$. Then (6) has a solution from $$L_{2}({\mathbb{R}}_{+})$$ iff the following conditions hold:
$${\int_{-\infty}^{+\infty}(\xi-i)^{k-1} \sigma^{-1}_{-}(\xi )\widetilde{\mathit{lv}}(\xi)\, d\xi}=0, \quad k=1,2, \ldots,|\ae|.$$
### Proof
We argue as above and use the equality (14); we write it as
\begin{aligned}& (\xi-i)^{\ae}\sigma_{+}(\xi)\tilde{u}_{+}(\xi)-( \xi+i)^{\ae }P \bigl(\sigma^{-1}_{-}(\xi)\widetilde{\mathit{lv}}( \xi) \bigr) \\& \quad =(\xi+i)^{\ae}Q \bigl(\sigma^{-1}_{-}(\xi)\widetilde{ \mathit{lv}}(\xi) \bigr)-(\xi+i)^{\ae}\sigma^{-1}_{-}(\xi) \tilde{u}_{-}(\xi). \end{aligned}
(15)
Since we work with $$L_{2}({\mathbb{R}})$$ both the left-hand side and the right-hand side are equal to zero at infinity, hence these are zeros, and
$$\tilde{u}_{+}(\xi)=\sigma^{-1}_{+}(\xi) \biggl(\frac{\xi-i}{\xi +i} \biggr)^{-\ae}P \bigl(\sigma^{-1}_{-}(\xi)\widetilde{\mathit{lv}}( \xi ) \bigr).$$
But there is some inaccuracy. Indeed, this solution belongs to the space $$A_{+}({\mathbb{R}})$$, but more exactly it belongs to its subspace $$A^{k}_{+}({\mathbb{R}})$$. This subspace consists of functions analytic in $${\mathbb{C}}_{+}$$ with zeros of the order −æ in the point $$z=i$$. To obtain a solution from $$L_{2}({\mathbb{R}}_{+})$$ we need some corrections in the last formula. Since the operator P is related to the Cauchy type integral we will use certain decomposition formulas for this integral (see also [12–14]).
Let us denote $$\sigma^{-1}_{-}(\xi)\widetilde{\mathit{lv}}(\xi)\equiv g(\xi)$$ and consider the following integral:
$$\int_{-\infty}^{+\infty}\frac{g(\eta)\, d\eta}{z-\eta }, \quad z=\xi+i\tau \in{\mathbb{C}}.$$
Using a simple formula for a kernel
$$\frac{1}{z-\eta}=\sum_{k=1}^{|\ae|} \frac{(\eta -i)^{k-1}}{(z-i)^{k}}+\frac{(\eta-i)^{|\ae|}}{(z-i)^{|\ae|}(z-\eta)}$$
we obtain the following decomposition:
$$\int_{-\infty}^{+\infty}\frac{g(\eta)\, d\eta}{z-\eta}=\sum _{k=1}^{|\ae|}{(z-i)^{-k}} {\int _{-\infty}^{+\infty }(\eta-i)^{k-1}g(\eta)\, d\eta}+ \int_{-\infty}^{+\infty }\frac{(\eta-i)^{|\ae|}g(\eta)\, d\eta}{(z-i)^{|\ae|}(z-\eta)}.$$
So, we have a following property. If the conditions
$${\int_{-\infty}^{+\infty}(\eta-i)^{k-1}g(\eta)\, d \eta }=0, \quad k=1,2,\ldots,|\ae|,$$
hold, then we obtain
$$\int_{-\infty}^{+\infty}\frac{g(\eta)\, d\eta}{z-\eta }=(z-i)^{\ae} \int_{-\infty}^{+\infty}\frac{(\eta-i)^{|\ae |}g(\eta)\, d\eta}{z-\eta}.$$
Hence the boundary values on $${\mathbb{R}}$$ for the left-hand side and the right-hand one are equal, and thus
$$P \bigl(g(\xi) \bigr)=(\xi-i)^{\ae}P \bigl((\xi-i)^{|\ae|}g(\xi ) \bigr).$$
Substituting the last formula into the solution formula we write
$$\tilde{u}_{+}(\xi)=\sigma^{-1}_{+}(\xi) (\xi+i)^{\ae}P \bigl((\xi -i)^{|\ae|}\sigma^{-1}_{-}(\xi)\widetilde{\mathit{lv}}(\xi) \bigr).$$
 □
## 5 Conclusion
It seems this approach to difference equations may be useful for studying the case that the variable x is a discrete one. We have some experience in the theory of discrete equations [9–11], and we hope that we can be successful in this situation also. Moreover, in our opinion the developed methods might be applicable for multi-dimensional difference equations.
## References
1. Milne-Thomson, LM: The Calculus of Finite Differences. Chelsea, New York (1981)
2. Jordan, C: Calculus of Finite Differences. Chelsea, New York (1950)
3. Vasil’ev, VB: Wave Factorization of Elliptic Symbols: Theory and Applications. Introduction to the Theory of Boundary Value Problems in Non-Smooth Domains. Kluwer Academic, Dordrecht (2000)
4. Vasilyev, VB: General boundary value problems for pseudo differential equations and related difference equations. Adv. Differ. Equ. 2013, 289 (2013)
5. Vasilyev, VB: On some difference equations of first order. Tatra Mt. Math. Publ. 54, 165-181 (2013)
6. Mikhlin, SG, Prößdorf, S: Singular Integral Operators. Akademie Verlag, Berlin (1986)
7. Sobolev, SL: Cubature Formulas and Modern Analysis: An Introduction. Gordon & Breach, Montreux (1992)
8. Dudgeon, DE, Mersereau, RM: Multidimensional Digital Signal Processing. Prentice Hall, Englewood Cliffs (1984)
9. Vasilyev, AV, Vasilyev, VB: Discrete singular operators and equations in a half-space. Azerb. J. Math. 3(1), 84-93 (2013)
10. Vasilyev, AV, Vasilyev, VB: Discrete singular integrals in a half-space. In: Current Trends in Analysis and Its Applications, Proc. 9th Congress, Krakow, Poland, August 2013, pp. 663-670. Birkhäuser, Basel (2015)
11. Vasilyev, AV, Vasilyev, VB: Periodic Riemann problem and discrete convolution equations. Differ. Equ. 51(5), 652-660 (2015)
12. Eskin, G: Boundary Value Problems for Elliptic Pseudodifferential Equations. Am. Math. Soc., Providence (1981)
13. Gakhov, FD: Boundary Value Problems. Dover, New York (1981)
14. Muskhelishvili, NI: Singular Integral Equations. North-Holland, Amsterdam (1976)
15. Gokhberg, I, Krupnik, N: Introduction to the Theory of One-Dimensional Singular Integral Equations. Birkhäuser, Basel (2010)
## Acknowledgements
The authors are very grateful to the anonymous referees for their valuable suggestions. This work is supported by Russian Fund of Basic Research and government of Lipetsk region of Russia, project No. 14-41-03595-a.
## Author information
Authors
### Competing interests
The authors declare that they have no competing interests.
### Authors’ contributions
All authors contributed equally to the writing of this paper. All authors read and approved the final manuscript.
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Vasilyev, A.V., Vasilyev, V.B. On some classes of difference equations of infinite order. Adv Differ Equ 2015, 211 (2015). https://doi.org/10.1186/s13662-015-0542-3 | 7,456 | 21,200 | {"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} | 3.03125 | 3 | CC-MAIN-2024-38 | latest | en | 0.887152 |
http://slideplayer.com/slide/1663967/ | 1,529,305,202,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267860089.11/warc/CC-MAIN-20180618051104-20180618071104-00528.warc.gz | 292,721,741 | 20,126 | Algebra - Complex Numbers Leaving Cert Helpdesk 27 th September 2012.
Presentation on theme: "Algebra - Complex Numbers Leaving Cert Helpdesk 27 th September 2012."— Presentation transcript:
Algebra - Complex Numbers Leaving Cert Helpdesk 27 th September 2012
Why do we need Complex Numbers?
Where are Complex Numbers used? Electronic Engineering Aircraft Design Medicine Movie/Computer game graphics
Adding and Subtracting Complex Numbers Group the real parts together and group the imaginary parts together. Examples:
Multiplication of Complex Numbers
Examples of Multiplication
The Complex Conjugate
Example of Division
Dividing one Complex Number by Another
Examples of the use of Complex Numbers
Argand Diagram
Modulus of a Complex Number
Polar Form of a Complex Number
Converting a Complex Number to Polar Form
De Moivres Theorem
De Moivres Theorem – Ex. 1
De Moivres Theorem - Example 2
Roots of a Complex Number
Proof by Induction
Proving De Moivres Theorem by Induction
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https://uk.mathworks.com/matlabcentral/cody/problems/2241-compare-two-input-matrices/solutions/2141444 | 1,586,094,310,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585371604800.52/warc/CC-MAIN-20200405115129-20200405145629-00204.warc.gz | 737,316,333 | 15,926 | Cody
# Problem 2241. Compare two input matrices
Solution 2141444
Submitted on 26 Feb 2020
This solution is locked. To view this solution, you need to provide a solution of the same size or smaller.
### Test Suite
Test Status Code Input and Output
1 Pass
A=[ 1 2 3 4 5; 5 9 3 0 9] B=[ 1 2 4; 8 9 2] y_correct = 1; assert(isequal(CompareMatrix(A,B),y_correct))
A = 1 2 3 4 5 5 9 3 0 9 B = 1 2 4 8 9 2 flag = logical 1
2 Fail
A=[ 1 2 3 ; 5 9 3 ] B=[ 1 2 4 2; 8 9 2 5] y_correct = 0; assert(isequal(CompareMatrix(A,B),y_correct))
A = 1 2 3 5 9 3 B = 1 2 4 2 8 9 2 5 flag = logical 1
Assertion failed.
3 Fail
A=magic(10) B=magic(7) y_correct = 1; assert(isequal(CompareMatrix(A,B),y_correct))
A = 92 99 1 8 15 67 74 51 58 40 98 80 7 14 16 73 55 57 64 41 4 81 88 20 22 54 56 63 70 47 85 87 19 21 3 60 62 69 71 28 86 93 25 2 9 61 68 75 52 34 17 24 76 83 90 42 49 26 33 65 23 5 82 89 91 48 30 32 39 66 79 6 13 95 97 29 31 38 45 72 10 12 94 96 78 35 37 44 46 53 11 18 100 77 84 36 43 50 27 59 B = 30 39 48 1 10 19 28 38 47 7 9 18 27 29 46 6 8 17 26 35 37 5 14 16 25 34 36 45 13 15 24 33 42 44 4 21 23 32 41 43 3 12 22 31 40 49 2 11 20 flag = logical 0
Assertion failed. | 615 | 1,178 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.90625 | 3 | CC-MAIN-2020-16 | latest | en | 0.351939 |
https://www.eeeguide.com/four-phase-stepper-motor/ | 1,720,801,391,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514450.42/warc/CC-MAIN-20240712161324-20240712191324-00825.warc.gz | 588,633,913 | 49,357 | Four Phase Stepper Motor:
Four Phase Stepper Motor – The stepper motor is a special type of synchronous motor which is designed to rotate through a specific angle (called a step) for each electrical pulse received by its control unit. Typical step sizes are 7.5°, 15° or larger. The stepper motor is used in digitally controlled position control systems in open loop mode. The input command is in the form of a train of pulses to turn a shaft through a specified angle.
There are two advantages in using stepper motors. One is their compatibility with digital systems and secondly no sensors are needed for position and speed sensing as these are directly obtained by counting input pulses and periodic counting if speed information is needed. Stepper motors have a wide range of applications; paper feed motors in typewriters and printers, positioning of print heads, pens in XY-plotters, recording heads in computer disk drives and in positioning of work tables and tools in numerically controlled machining equipment. The range of applications of these motors is increasing as these motors are becoming available in larger power ratings and with reducing cost.
Elementary operation of a Four Phase Stepper Motor with a two-pole rotor can be illustrated through the diagram of Fig. 10.32.
Let us assume that the rotor is permanent magnet excited. Such a rotor aligns with the axis of the stator field with torque being proportional to the sin θ, θ being the angle of displacement between the rotor axis and stator field axis. The torque-angle characteristics of Four Phase Stepper Motor is drawn in Fig. 10.33(a) with phase a excited and also with phase b excited. It is easily observed that the stable position of the rotor corresponds to that angle at which the torque is zero and is positive for smaller angles and negative for larger angles. Thus with phase a excited, the stable (or locked) position is θ = 0° but not θ = 180° (unstable) and the torque has a maximum positive value at θ = 90°. It is therefore easily concluded that each excitation pattern of phases corresponds to a unique position of the rotor. Therefore the excitation sequence a, b, c, d, a … causes the rotor to move in positive sequence in steps of 90°.
With rotor in position θ = 0 and a and b both excited the rotor will move to 45°, which is a stable position (net torque due to a and b zero and torque-angle slope negative). So excitation sequence a, a + b, b, b + c, c … make the rotor to move forward in steps of 45°. Patterns for phase winding excitations can be easily visualized for steps of 22.5°, 11.25°, and smaller per pulse input to the circuit. Another feature of a PM stepper motor is that, when excited, it seeks a preferred position which offers advantage in certain applications.
Consider now that the rotor (projecting pole) is made of just ferromagnetic material (no permanent magnet). The device now behaves as a variable-reluctance motor. The ferromagnetic rotor seeks the position which presents minimum reluctance to the stator field, i.e. the rotor axis aligns itself to the stator field axis. In Fig. 10.32 with phase a excited, this happens at θ= 0° as well as θ = 180° in which positions the torque on the rotor is zero. At θ = 90° the rotor presents infinite reluctance to the phase ‘a’ axis and so the torque has also a zero there. Thus the rotor torque is a function of sin 2θ as drawn in Fig. 10.33(b). For a reductance stepper motor there are two possible stable positions for a given excitation or set of excitations. | 786 | 3,537 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-30 | latest | en | 0.941776 |
http://math.stackexchange.com/questions/144693/showing-sum-k-1-1-k2-pi2-6 | 1,469,614,011,000,000,000 | text/html | crawl-data/CC-MAIN-2016-30/segments/1469257826759.85/warc/CC-MAIN-20160723071026-00059-ip-10-185-27-174.ec2.internal.warc.gz | 159,998,924 | 16,995 | # Showing $\sum _{k=1} 1/k^2 = \pi^2/6$ [duplicate]
I read my book of EDP, and there appears the next serie $$\sum _{k=1} \dfrac{1}{k^2} = \dfrac{\pi^2}{6}$$ And, also, we prove that this series is equal $\frac{\pi^2}{6}$ for methods od analysis of Fourier, but...
Do you know other proof, any more simple or beautiful?
-
## marked as duplicate by David Mitra, Henry, Asaf Karagila, Martin Sleziak, Nate EldredgeMay 13 '12 at 20:35
That should be $\pi^2/6$. – David Mitra May 13 '12 at 19:31
See here – David Mitra May 13 '12 at 19:33
@Martin: This question asks for methods to calculate the sum, not to prove its convergence. This is at least how I read this question. – Asaf Karagila May 13 '12 at 19:55
@Asaf I think you' right. Although, the body is different from the title. Which explains why I thought that the OP asks about convergence only. (It's not that important now, since we found duplicates for both possible meanings.) – Martin Sleziak May 13 '12 at 19:57
The title originally said Prove that this series converges – Henry May 14 '12 at 10:05
If you just want to show it converges, then the partial sums are increasing but the whole series is bounded above by $$1+\int_1^\infty \frac{1}{x^2} dx=2$$ and below by $$\int_1^\infty \frac{1}{x^2} dx=1,$$ since $\int_{k}^{k+1} \frac{1}{x^2} dx \lt \frac{1}{k^2} \lt \int_{k-1}^{k} \frac{1}{x^2} dx$. | 457 | 1,366 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.796875 | 4 | CC-MAIN-2016-30 | latest | en | 0.88315 |
http://www.justskins.com/forums/bit-shifts-in-cocoa-148244.html | 1,604,132,750,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107916776.80/warc/CC-MAIN-20201031062721-20201031092721-00233.warc.gz | 155,188,692 | 11,492 | # Bit shifts in Cocoa - Mac Programming
Does anyone know how to do bit shifts in Cocoa? I have declared an integer and I want to shift it one bit to the right (same as dividing by 2). How can this be done in Cocoa? I already tried "x >> 1", but it did not recognize the ">>" syntax. I'm also interested in information about other mathematical logical syntax in Cocoa. I didn't find anything about that on the apple developer web site. Any help is appreciated. -Korbin...
1. ## Bit shifts in Cocoa
Does anyone know how to do bit shifts in Cocoa? I have declared an
integer and I want to shift it one bit to the right (same as dividing
by 2). How can this be done in Cocoa? I already tried "x >> 1", but
it did not recognize the ">>" syntax. I'm also interested in
information about other mathematical logical syntax in Cocoa. I
didn't find anything about that on the apple developer web site. Any
help is appreciated.
-Korbin
Korbin Meiser Guest
2. ## Re: Bit shifts in Cocoa
On 10 Sep 2003, Korbin Meiser wrote:
> Does anyone know how to do bit shifts in Cocoa? I have declared an
> integer and I want to shift it one bit to the right (same as dividing
> by 2). How can this be done in Cocoa? I already tried "x >> 1", but
> it did not recognize the ">>" syntax. I'm also interested in
> information about other mathematical logical syntax in Cocoa. I
> didn't find anything about that on the apple developer web site. Any
> help is appreciated.
First of all, your question as asked doesn't make sense. Cocoa is not a
programming langauge, it is an API. You probably meant Objective-C, which
is the langauge normally used with Cocoa.
Objective-C is just C with some OO stuck on. So all the basic math
operators are the same. Your "x >> 1" statement should work. If it
doesn't, it's either because you have another error, or you're not using
Objective-C.
--
"From now on, we live in a world where man has walked on the moon.
And it's not a miracle, we just decided to go." -- Jim Lovell
Mike Ash - <http://www.mikeash.com/>, <mailto:mailmikeash.com>
Michael J Ash Guest
3. ## Re: Bit shifts in Cocoa
[email]cryptofishhotmail.com[/email] (Korbin Meiser) wrote:
> Does anyone know how to do bit shifts in Cocoa? I have declared an
> integer and I want to shift it one bit to the right (same as dividing
> by 2). How can this be done in Cocoa? I already tried "x >> 1", but
> it did not recognize the ">>" syntax. I'm also interested in
> information about other mathematical logical syntax in Cocoa. I
> didn't find anything about that on the apple developer web site. Any
> help is appreciated.
It would help if you told us what programming language you're using. :-T
The languages that have Cocoa bindings that I know of include
Objective C
Java
Python
Perl
AppleScript
and there are probably just as many others. Some of them support
bit-shifts, others don't. Objective C (which is the language you'll want
to be using if you're doing anything as advanced as a bit-shift)
supports the << and >> operators just fine.
Cheers,
-- M. Uli Kusterer
[url]http://www.zathras.de[/url]
Uli Kusterer Guest
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• | 831 | 3,258 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.734375 | 3 | CC-MAIN-2020-45 | latest | en | 0.94697 |
http://steamcommunity.com/sharedfiles/filedetails/comments/101604583?l=hungarian | 1,508,805,987,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187827662.87/warc/CC-MAIN-20171023235958-20171024015958-00016.warc.gz | 324,965,324 | 14,755 | Portal 2
### Portal 2
Conviction
30 megjegyzés
Barbos [készítő] 2016. febr. 3., 11:22
2 ASTERION
That's a very nice and tricky solution :) I think there's no need to change anything, especially because it's much harder to execute than intended way. Thanks anyway!
ASTERION 2016. febr. 2., 13:09
I found a solution with some different steps, maybe you want to fix it, maybe you don't care, anyway, here it is: SPOILERS https://youtu.be/6E_2aUEOJpI I think the fact that you can shoot the funnel from the other side breaks some part of the puzzle
Cool map :)
Spaceminnow 2015. okt. 18., 19:27
It was fun playing this one. I watch the video after I solved it. The solution video link didn't look like a link...
stormsend 2015. jún. 8., 17:31
I thoroughly enjoyed solving this one. Not extremely hard but not so easy you walk right through it. I watched both the videos below and I seem to have done it the right way. The chamber is well laid out and looks good. Excellent use of the elements and a nice variety. I'll be looking at your workshop. I'm going to favorite this one. Thanks.
Barbos [készítő] 2014. dec. 21., 0:35
2 RedSilencer
Wow! Thats very cool! I'll add you
RedSilencer 2014. dec. 20., 22:52
I'm remaking this puzzle in 2 parts. Both parts are modified to be more difficult than their original part. As of writinging this comment, I've finished the first one. If you want to try it out, Barbos, add me and you'll have access to those maps and many others on my workshop! :D
mebe 2013. nov. 25., 10:30
Nice, especially the first (cube) room.
eMbAh 2013. okt. 20., 0:18
Good puzzle!
[Tom K] 2013. szept. 30., 13:23
Here is my solution, hope you enjoy. http://www.youtube.com/watch?v=dgNnS8Aorzk
Fairly well designed, even though I don't have the slightest idea how to solve it.
Stagger Lee 2013. jan. 14., 13:50
Not overly hard but I did start to solve it incorrectly at first before I hit the "Aha" moment in getting the cube that was over the goo. Very enjoyable chamber though.
YUKI_67 2013. jan. 1., 8:58
I apprécis much your style, really, very nice work, I think I'll play all your testing room, continue like that!
Barbos [készítő] 2012. nov. 30., 0:56
2 toncica
well, i really like your solution :) Unfortunatly, can't change turrets AI, but the way you recieve the cube was a pleasant surprise, like it. Still, relying on the original portal puzzles, for many players this level may be hard.
toncica 2012. nov. 30., 0:24
For some reason I found this to be rather easy. Good puzzle nevertheless and enjoyable. You say you fixed several issues like bypassing the turrets and not using the light bridge. I was still able to do both things. It's not breaking the puzzle or making the way significantly shorter though. Just to let you know.
http://youtu.be/d8Nigf4nbJ8
Aardvark00 2012. nov. 18., 6:58
Excellent puzzle! Very satisfying to solve after a lot of thought along the way. Thanks for that one :)
SkyRoots 2012. okt. 16., 20:29
I agree that is is now fixed. =)
Barbos [készítő] 2012. okt. 16., 19:39
2 SkyRoots
Fixed again, now it must be impossible to avoid turrets by your way :) Also another fix for alternate solution.
SkyRoots 2012. okt. 16., 18:21
I can still solve it the same way. I posted a screenshot of my solution in my workshop.
SkyRoots 2012. okt. 16., 17:40
When the turrets are in "searching" mode, they swing back and forth. If you time it right, it is possible to run towards them and quickly shoot a portal underneath each one before they get a shot off. If you time it wrong, they shoot you dead.
Barbos [készítő] 2012. okt. 15., 21:14
2 SkyRoots
made a quick fix to this solution, check it please. Don't know how you passed the turrets without getting shot :)
SkyRoots 2012. okt. 15., 20:18
Great (hard) puzzle! I defeated the two turrets before getting the first cube and without getting shot once. I guess I must have found a "trick" because I solved the puzzle without ever using the top weighted button (by the orange beam). Can you figure out how I did it?
NyanNyanKoneko 2012. okt. 11., 15:03
That was a great puzzle. Very challenging yet of a simple design. Great job!
Linrox 2012. okt. 11., 10:53
Loved it. Keep me thinking. to the end. now try mine.
Afterbirth 2012. okt. 11., 9:12
best one of the day
ost-hanz 2012. okt. 11., 6:10
vielen dank war schön übersichtlich bringt einen zum grübren nach sowas sucht man lange
[GWJ] Kexx 2012. okt. 10., 19:41
Another great one!! Man, you really do master lasers. Congrats on that! Not just anybody can pull that off.
Bensenhaver 2012. okt. 9., 6:30
Yeah it is fixed, at least I couldn't make it my way and had to use the light bridge :)
Bensenhaver 2012. okt. 9., 6:11
Gonna try again then :)
Barbos [készítő] 2012. okt. 9., 6:02
Thanks for feedback, Bensengaver! I made quick fix to this solution, hope now it's fixed :)
Bensenhaver 2012. okt. 9., 5:17
Nice puzzle, kept me thinking for a while. Thumbs up! Never needed the light bridge though, got the cube by quickly shooting portals. | 1,537 | 4,991 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.03125 | 3 | CC-MAIN-2017-43 | latest | en | 0.924888 |
https://www.jiskha.com/questions/1006802/if-a-test-has-25-questions-with-4-possible-answers-per-questions-how-many-possible-answer | 1,576,038,120,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575540529745.80/warc/CC-MAIN-20191211021635-20191211045635-00068.warc.gz | 739,864,499 | 5,328 | # math
If a test has 25 questions with 4 possible answers per questions, how many possible answer combinations are there?
1. 👍 0
2. 👎 0
3. 👁 89
Hope this helps David.(:
1. 👍 0
2. 👎 0
posted by Jazzy
2. I believe there are 4^25 possible combinations.
consider that for each question there are 4 possible answers. It's like figuring the number of license plates where there are 25 letters, each of which may be A-D.
1. 👍 0
2. 👎 0
posted by Steve
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More Similar Questions | 917 | 3,526 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2019-51 | latest | en | 0.963298 |
https://www.java-forums.org/new-java/52033-dice-help-posting-number-times-number-rolled.html | 1,490,278,907,000,000,000 | text/html | crawl-data/CC-MAIN-2017-13/segments/1490218187113.46/warc/CC-MAIN-20170322212947-00007-ip-10-233-31-227.ec2.internal.warc.gz | 903,677,370 | 17,988 | Thread: Dice help. posting the number of times a number is rolled.
1. Member
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Dice help. posting the number of times a number is rolled.
What I need is a program to roll a die 10000 times and count the number of 1s rolled, 2s rolled, etc.. After that I need the program to use JOptionPane to post the total tally for each of the rolls in one window.
so far this is what I have:
Java Code:
```import javax.swing.*;
02
public class Dice
03
04
{
05
public static void main(String[] args)
06
{
07
08
09
int[] A = new int[10000];
10
11
for(int i=0; i<A.length; i++)
12
A[i] = 1 + (int) (6*Math.random());
13
14
for(int i=0; i<A.length; i++)
15
System.out.println(A[i]);
16
17
}
18
}```
What I need now is to somehow tally the numbers and post them in the JOptionPane window. I'm pretty sure I need to create a string but I don't know where to start.
2. Re: Dice help. posting the number of times a number is rolled.
Why do you have a 10000 long array? Do you plan on keeping every number rolled? If so why? I think a better idea would be to have an array to store the counts of each number rolled, if it is 1 increment the 1 count, if it was 2 increment the 2 count etc.
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Re: Dice help. posting the number of times a number is rolled.
Why do you have a 10000 long array? Do you plan on keeping every number rolled? If so why? I think a better idea would be to have an array to store the counts of each number rolled, if it is 1 increment the 1 count, if it was 2 increment the 2 count etc.
Well the assignment calls for a total of 10,000 rolls, but if there is an easier way to do this while still using Math.random and the 10,000 rolls i'm all ears.
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Re: Dice help. posting the number of times a number is rolled.
I've got similar program to make... i'll link u to a post about it:
http://www.java-forums.org/new-java/...m-2-cubes.html
hope it helps!
5. Re: Dice help. posting the number of times a number is rolled.
Make an array of size x, where x is the number of sides on the die. From there you increment the correct slot in the array each time you roll. Junky hinted at this as well. Give it a shot and see how it works out.
6. Re: Dice help. posting the number of times a number is rolled.
Your program only requires how many times something has been rolled. IE for a dice your options are 1,2,3,4,5,6. That means your array only needs to be 6 slots.
Run your Math.Random, then do a logical statement to detect what digit has been rolled. Assign each array slot a integer variable initialized to zero and assign it to a number corresponding to a side on the die. Then increment the variable in the array slot every time the side of the die is rolled.
Simple really, just got to put your noggin into it. You may also be able to use a primitive in the array too, depending on what you plan on doing after the dice are rolled. If all you are doing is rolling, storing and then print after 10,000 rolls then use a primitive.
7. Re: Dice help. posting the number of times a number is rolled.
Well the assignment calls for a total of 10,000 rolls, .
Where did I say "Don't roll 10,000 times"? What I did say was don't store those 10,000 rolls. Only store the current roll and 6 counts (one for each number). Just like you would on paper if I were to call out the 10,000 rolls to you.
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Re: Dice help. posting the number of times a number is rolled.
Ok so while I was in class the other day I asked some peers if they had any input. They told me that I need to include a do/while method. I took that input and tried to build off of it. So far I've managed to put together this jumble of nonfunctional code.
Java Code:
```import javax.swing.*;
public class DiceW
{
public static void main(String[] args)
{
int num1, num2, num3, num4, num5, num6;
int num1 = 0;
int num2 = 0;
int num3 = 0;
int num4 = 0;
int num5 = 0;
int num6 = 0;
int total = 10000;
String choice="";
do
{
int[] A = new int[10000];
for(int i=0; i<A.length; i++)
A[i] = 1 + (int) (6*Math.random());
for(int i=0; i<A.length; i++)
{
dice = (int)(Math.random90*6+1);
sum = (dice);
switch (sum)
{
case 1 : num1++;
case 2 : num2++;
case 3 : num3++;
case 4 : num4++;
case 5 : num5++;
case 6 : num6++;
break;
}
System.out.println(A[i]);
JOptionPane.showMessageDialog(null, "The dice has been rolled ten-thousand times, here is the statistics" +
"\nThe number of 1s" + "=" + num1 +
"\nThe number of 2s" + "=" + num2 +
"\nThe number of 3s" + "=" + num3 +
"\nThe number of 4s" + "=" + num4 +
"\nThe number of 5s" + "=" + num5 +
"\nThe number of 6s" + "=" + num6);
choice = JOptionPane.showInputDialog("Continue?");
} while(choice.equals("yes") | choice.equals("yes"));
}
}
}```
The Choice to continue at the end is one of the requirements of the assignment so i'm pretty sure that I have to use the do/while method and build off that. Hope that makes sense. I only get 4 errors when compiling so I think im on the right track.
9. Re: Dice help. posting the number of times a number is rolled.
I still do not understand why you want a 10,000 long array to store all the numbers. You need to revise switch statements. If you are getting error messages then you should post the exact messages here since we do not read minds.
10. Re: Dice help. posting the number of times a number is rolled.
You honestly don't even need switch statements. Here is some pesudocode:
Java Code:
```declare array of length n //n is the number of sides a die has
loop
increment a random element in the array
end loop```
The whole thing should be possible in ~ 10 lines of code.
11. Re: Dice help. posting the number of times a number is rolled.
Originally Posted by sunde887
You honestly don't even need switch statements. .
Yep that was my advice in the first reply but it seems all advice is being ignored.
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Re: Dice help. posting the number of times a number is rolled.
Its not that your advice is being ignored its that I do not understand java therefore, I do not know how to follow your advice. You need to be really specific with me and include examples otherwise, I have no clue what your talking about.
as far as the errors I receive they are as follows:
I:\Java\Project 2\DiceW.java:63: error: while expected
}
^
I:\Java\Project 2\DiceW.java:66: error: illegal start of expression
}
^
I:\Java\Project 2\DiceW.java:66: error: reached end of file while parsing
}
^
I:\Java\Project 2\DiceW.java:67: error: reached end of file while parsing
4 errors
Tool completed with exit code 1
13. Re: Dice help. posting the number of times a number is rolled.
When you declare an array you specify the length as well
Java Code:
`int[] x = new int[50];`
In this case, the array x would have a length of 50.
Since there are only 6 sides on the die you will only need an array which is 6 ints long. After creating this, start a loop which loops 10,000 times and increment the correct spot based on a random number. When indexing into an array you are able to index with the return value of Random.nextInt(...)
14. Re: Dice help. posting the number of times a number is rolled.
Its not that your advice is being ignored its that I do not understand java
In my first post I advised that you cahnge the 10,000 long array to one that was only 6 long. Yet in your code you have not done that. Since you have managed to declare an array then the only conclusion I can arrive at is that you understand enough Java to do this and therefore my advice is being ignored. If at any time you did not understand the point I was trying to make then ask don't ignore and continue stumbling along without a clue. | 2,084 | 7,816 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2017-13 | longest | en | 0.8748 |
http://mathhelpforum.com/differential-equations/164372-finding-greens-function-ode-print.html | 1,498,567,675,000,000,000 | text/html | crawl-data/CC-MAIN-2017-26/segments/1498128321410.90/warc/CC-MAIN-20170627115753-20170627135753-00654.warc.gz | 256,152,646 | 3,341 | # Finding Green's Function for ODE
• Nov 25th 2010, 10:17 AM
Creebe
Finding Green's Function for ODE
$$u''+u = f(x)$$
$0
$u(0) = 0 u(R) = 0$
• Nov 26th 2010, 11:55 PM
mukmar
1. Solve homogeneous equation
$u'' + u = 0 \Rightarrow u = a\cos x + b\sin x$
2. $LG(x,s) = 0$ for $x \neq s$
$x < s, G_1(x,s) = a_1(s)\cos x + b_1(s) \sin x, G(0, s) = 0 \Rightarrow a_1(s) = 0$
$x > s, G_2(x, s) = a_2(s)\cos x + b_2(s) \sin x, G(R, s) = 0 \Rightarrow a_2(s)\cos R + b_2(s) \sin R = 0$
We note $R \neq n\pi$ to ensure linear independence of solutions as the Wronskian is $\sin R$.
Hence $G_2(x, s) = a_2(s) (\cos x - \sin x \cot R)$
3. $G(x, s)$ continuous at $x = s$
$G_1(s, s) = G_2(s, s) \Rightarrow b_1(s) \sin s = a_2(s) (\cos s - \sin s \cot R)$
Therefore Eq. 1,
$
b_1(s) = a_2(s) (\cot s - \cot R)
$
4. Jump Condition
$G'(x)\Big{|}^{x = s^+}_{x = s^-} = 1 \Rightarrow -a_2(s)(\sin s + \cos s \cot R) - b_1(s)\cos s = 1$
Which gives:
Eq. 2
$
-a_2(s)(\sin s + \cos s \cot R) - b_1(s)\cos s = 1
$
Now Eq. 1 and Eq. 2 together give:
$-a_2(s)(\sin s + \cos s \cot R + \cos s\cot s - \cos s\cot R ) = 1$
$\therefore a_2(s) = -\sin s$
Now using Eq. 1 again, we know $b_1(s) = - \sin s(\cot s - \cot R) = \frac{\sin(s - R)}{\sin R}$
Also we can simplify: $G_2(x, s) = -\sin s(\cos x - \sin x \cot R) = \sin s \frac{\sin(x - R)}{\sin R}$
So putting this all together, we get our Green's function:
$G(x, s) =
\begin{cases}
\frac{\sin(s - R)}{\sin R} \sin x& x < s \\\\
\sin s \frac{\sin(x - R)}{\sin R} & x > s
\end{cases}$
From here you should be able to represent the solution as a sum of two integrals that utilize the Green's function. Good luck :)
P.S. Let me know if I made any mistakes!
• Nov 27th 2010, 08:28 AM
Creebe | 761 | 1,735 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 21, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.125 | 4 | CC-MAIN-2017-26 | longest | en | 0.632866 |
https://apple.stackexchange.com/questions/321030/macos-battery-estimation-question | 1,563,215,323,000,000,000 | text/html | crawl-data/CC-MAIN-2019-30/segments/1563195523840.34/warc/CC-MAIN-20190715175205-20190715200238-00033.warc.gz | 310,146,651 | 33,674 | MacOS battery estimation question
Why do you think the OS estimates a battery percentage a little above the actual value? I noticed it is always between 0% and 5% more. In particular the two percentage gets closer when the remaining battery gets lower. Maybe this way the device seems to get faster to 100%, since the 95%-100% takes a lot of time...?
Some details. I can find the percentage two ways:
• From the battery menu bar icon or with `pmset -g batt` (eg. now they say 96%)
• By calculating it manually from System Information.app or with information from `ioreg -l -n AppleSmartBattery -r` (eg. now I get: MaxCapacity: 4181, CurrentCapacity: 3813; meaning 91.2%)
• It's not so simple. See this video: youtube.com/watch?v=R8hTQXqURB4 – Allan Mar 31 '18 at 12:51
• Is this pirely a how does the battery chemistry map to measured values or is there something practical/specific you wish to do here once this is explained? – bmike Mar 31 '18 at 13:45
• Thank you @Allan, I'll surely watch it as soon as I have some spare time. – Diego Apr 1 '18 at 13:22
• @bmike, the question is also a curiosity, but I'm mainly making an Alfred script to get the remaining battery and I wasn't sure which one was the "right" one – Diego Apr 1 '18 at 13:23
2 Answers
If you were to script this - keep in mind. One lithium cell charges very non-linearly. And a battery of individual cells will charge like a mess of non-linear units - none of them in phase.
Your data from `ioreg` is far superior to the processed guess of percentage from `pmset` and I would only use the power manager tool to log and process changes - not to do the calculations you seek.
From the registry - you can focus on these three items in my opinion:
• "DesignCapacity" = 5297
• "CurrentCapacity" = 2331
• "MaxCapacity" = 5059
So in my case, the theoretical design maximum milli amp hours is 5297 and I can expect from the readings to only store 5059 mAh of which I have a measured 2331 left to use before I'm flat. You may or may not want to try and incorporate the amperage draw or temperature depending on how sophisticated your app will be or if it just will report the calculated time to empty listed as "InstantTimeToEmpty"
• Thank you. I've also seen the video in one of the comments and I understand the matter is complicated... I wouldn't know how to incorporate other informations like amperage or temperarure, so I think I'll just compute the percentage with those three values and a remaining time with "InstantTimeToEmpty". – Diego Apr 2 '18 at 8:07
My guess is that with AppleSmartBattery, you are calculating the percentage based on the batteries brand new, fully charged state. Whereas the OS is basing the percentage on what the battery is currently capable of at full charge, which will slowly degrade over time.
• mmmh but actually with ioreg there's also `DesignCapacity = 4315`, so I think `MaxCapacity: 4181` takes into account battery ageing... (sorry I haven't mentioned that) – Diego Mar 31 '18 at 11:26
• Well perhaps the rate at which the battery drains isn't quite constant and changes based on its charge, so the raw mah value might need to be adjusted slightly to get a percentage that is more linear based on time. – l008com Mar 31 '18 at 11:36 | 824 | 3,250 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2019-30 | latest | en | 0.925099 |
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At the end of Ode to a Nightingale, the speaker is left A. settled B. comforted C. resolved D. shaken
At the end of Ode to a Nightingale, the speaker is left SHAKEN.
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Asked 2/11/2013 12:45:10 PM
Updated 2/14/2015 12:00:02 PM
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Edited by yumdrea [2/14/2015 11:59:59 AM], Confirmed by yumdrea [2/14/2015 12:00:02 PM]
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Home | Contact | Blog | About | Terms | Privacy | © Purple Inc. | 744 | 2,030 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.71875 | 3 | CC-MAIN-2020-40 | latest | en | 0.837752 |
http://math.stackexchange.com/questions/200742/proving-z-n-converge-to-z/200744 | 1,467,153,671,000,000,000 | text/html | crawl-data/CC-MAIN-2016-26/segments/1466783397213.30/warc/CC-MAIN-20160624154957-00098-ip-10-164-35-72.ec2.internal.warc.gz | 206,294,625 | 19,281 | # proving $z_n$ converge to $z$
Suppose $b_n$ is a sequence $>0$ and $b>0$ where $b_n$ converge to $b$. Suppose $z_n=\log b_n$ and $z=\log b$, prove that $z_n$ converge to $z$. I know the definition of limit but not sure how to satisfy the condition
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Do you know that $\log x$ is a continuous function defined on $(0,+\infty)$? What do you know about the erlation of convergent sequences and continuous functions? – Pedro Tamaroff Sep 22 '12 at 16:06
prove by definition definition, we need to find a $N$ from sequence $z_n$ st the condition satisfies by not sure how to find $N$, the only hints is taht we can find such $N$ from sequence $b_n$ – Mathematics Sep 22 '12 at 16:08
In order to be able to help we have to know the rules of the game. What is your definition of $\log$, and what properties of $\log$ (or of sequences, for that matter) are you allowed to use? – Christian Blatter Sep 22 '12 at 18:07
Hint use $\log \space a - \log\space b = \log \space a/b$ and $\log 1 = 0$
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well, of course i did use but cannot really show the inequality – Mathematics Sep 22 '12 at 15:36
Z - Zn = log b/bn as bn converges to b, zn - z tends to 0 – Ram Sep 22 '12 at 15:39
ya, but from the definition, you gotta show that there should exist an integer $N$ s.t. for all integers $n$, where $n>N, |z_n-z|<\epsilon$ for all $\epsilon >0$ – Mathematics Sep 22 '12 at 15:42
if |zn−z|<ϵ implies |log b/bn| <ϵ => b/bn = e^ϵ (or 10^ϵ depending on log or ln) bn = be^-ϵ – Ram Sep 22 '12 at 15:47
yup, but we have no idea even we got $b_n$ to find the $N$ from sequence {$z_n$} – Mathematics Sep 22 '12 at 15:50
I don't know if the hypothesis that the logarithm is continuous is the best choice, so I'll add something. From any definition of the logarithm, you'll extract that
$$\log x - \log y = \log \frac{x}{y}$$
and that
$$1 - \frac{1}{x} < \log x < x - 1$$
for $x\neq 1$. If $x=1$, we have equalities. From $(2)$, we have that, for $x\neq 1$,
$$\frac{1}{x} < \frac{{\log x}}{{x - 1}} < 1$$
From the squeeze theorem it follows that
$$\mathop {\lim }\limits_{x \to 1} \frac{{\log x - \log 1}}{{x - 1}} = 1$$
from where the logarithm is differentiable at $x=1$, and thus continuous at $x=1$. But the fact that it is continuous at $x=1$ means it is continuous for every $x>0$.
Indeed, pick any sequence $a_n>0$ that converges to $a(>0)$. Then
$$\displaylines{ \mathop {\lim }\limits_{n \to \infty } \log {a_n} = \log a \cr \Leftrightarrow \mathop {\lim }\limits_{n \to \infty } \left( {\log {a_n} - \log a} \right) = 0 \cr \Leftrightarrow \mathop {\lim }\limits_{n \to \infty } \log \frac{{{a_n}}}{a} = 0 \cr}$$
But $\frac {a_n}{a}\to 1$ and $\log 1=0$.
You have that $(b_n)$ is a sequence of positive numbers, that is, $b_n>0\;\forall n$, that converges to $b$. This means that for every $\epsilon>0$ there is an $N_0$ such that, whenever $n\geq N_0$, $|b-b_n|<\epsilon$.
Now, we're setting $z_n=\log \; b_n$. This makes sense for each $n$ for $b_n>0$. Now, we want to prove that, $z_n\to z=\log b$. This means that, for every $\epsilon >0$, there is an $N_1$ such that, whenever $n\geq N_1$, $|z_n-z|<\epsilon$, that is
$$|\log b_n-\log b|<\epsilon$$
But $\log x$ is continuous for $x>0$, this means that for any $\epsilon>0$ there is a $\delta >0$ such that, for all $x$,
$$|x-a|<\delta\implies |\log x-\log b|<\epsilon$$
But then, since $b_n$ converges to $b$, for any $\delta >0$, there will be an $N_\delta$ for which
$$|b-b_n|<\delta$$
and consequently
$$|\log b_n-\log b|<\epsilon$$
Thus, we can take $N=N_\delta$. This means that for any $\epsilon >0$, whenever $n\geq N_\delta$ we'll have $$|z-z_n|=|\log b_n-\log b|<\epsilon$$ as desired.
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Setting $$x_n=\min\{b,b_n\},\ y_n=\max\{b,b_n\},$$ we have $$y_n+x_n=b+b_n,\ y_n-x_n=|b-b_n|.$$ Therefore $$\lim_{n\to \infty}(y_n+x_n)=2b,\ \lim_{n\to \infty}(y_n-x_n)=0,$$ and we deduce that $$\lim_{n\to \infty}y_n=\lim_{n\to \infty}x_n=b.$$ Since $b>0$, we have $$\lim_{n \to \infty}\frac{y_n}{x_n}=1.$$ Thus, given $\varepsilon>0$ there is an $N=N(\varepsilon) \in \mathbb{N}$ such that $$\frac{y_n}{x_n}-1\le \varepsilon \quad \forall\ n\ge N.$$ Hence, for every $n \ge N$ we have $$|z-z_n|=\left|\int_b^{b_n}\frac{dt}{t}\right|=\int_{x_n}^{y_n}\frac{dt}{t}\le \frac{y_n-x_n}{x_n}=\frac{y_n}{x_n}-1 \le \varepsilon,$$ i.e. $z_n \to z$ as $n \to \infty$.
- | 1,599 | 4,329 | {"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} | 3.84375 | 4 | CC-MAIN-2016-26 | latest | en | 0.856894 |
http://db.naturalphilosophy.org/abstract/?abstractid=5906 | 1,508,323,872,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187822930.13/warc/CC-MAIN-20171018104813-20171018124813-00770.warc.gz | 84,857,273 | 6,328 | Enter the content which will be displayed in sticky bar
# Abstract
Remarks on the Foundation of Special Relativity
Mario Ludovico
Year: 2006 Pages: 25
Keywords: special relativity
At variance with a largely shared opinion, both the foundation and the logical structure of Special Relativity (SR) have substantially been laid by Hendrik Lorentz and by Henri Poincar?, not by Albert Einstein. Yet, the mathematical generalization of SR comes from Hermann Minkowski, who in 1907 proposed the spacetime reference frame in its current notation, though the first mathematical formulation and use of a spacetime reference frame was clearly made by Poincar? in June 1905. (?Spacetime? is also referred to as ?chronotope?).
In this paper, which forms a \"Special Appendix\" to the book \"Vacuum, Vortices and Gravitation\" (fully and freely readable online at www.mario-ludovico.com), questionable points of Einstein\'s special relativity are given evidence. In particular, the well-known mass-energy equivalence equation is discussed in the light of Lorentz\'s theoretical analysis concerning the motion of a material body with respect to the ether. It is in fact remarked that the mass-energy equivalence equation is not an achievement of Einstein\'s special relativity. That equation is intrinsically inherent in the Lorentz\'s definition of \"transverse mass\", when the body\'s relative speed with respect to the ether is nil.
Perhaps, in a view to attaining - by his own - the \"equivalence\" relationship between mass and energy previously and differently formulated by Poincar?, Einstein published in September 1905 a very short paper, in which - starting from his precedent paper on special relativity - he \"proves\" the equation E = mc2 through the introduction of an unexpected simplification-approximation of the Lorentz\'s factor 1/(1-v2/ c2)1/2 , which Einsteins equals to 1+v2/2c2 cutting the relevant series at the second order term. Should one consider such a formal expedient as logically acceptable and appliable to all special relativity equations, the theory would take (particularly from the experimental standpoint) a \"physical\" significance remarkably different from that conventionally celebrated.
Accounting for the possible existence of the \"ether\" (or \"plenum\" in the author\'s terminology) the same mass-energy equivalence can be obtained analytically, with no use of relativistic paradigms. It is also observed that the introduction of Minkowski\'s \"chronotope\" has actually involved the mass-energy equivalence as an axiom proper to the spacetime paradigm. | 567 | 2,597 | {"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} | 2.609375 | 3 | CC-MAIN-2017-43 | longest | en | 0.927167 |
https://www.i-programmer.info/news/105-artificial-intelligence/16392-ai-invents-first-new-sorting-algorithm.html | 1,721,667,646,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763517890.5/warc/CC-MAIN-20240722160043-20240722190043-00388.warc.gz | 700,712,230 | 10,898 | AI Invents First New Sorting Algorithm
Wednesday, 21 June 2023
... well not quite. There have been similar headlines elsewhere, but the truth is more subtle and we certainly don't have a rival to "Quicksort".
One of the problems with the impressive success of AI is that news of it has to be increasingly impressive-sounding. The latest report from Google DeepMind is impressive but not as impressive as you might think if you misunderstand it, even slightly.
What they have done is to take AlphaZero, their neural network that learns by reinforcement learning RL and applied it to the problem of improving the speed of sort algorithms. Of course, you know how important sorting speed is and you know that there are fundamental algorithms like Quicksort and merge sort and practical algorithms like Timsort and Powersort. If not all of them are explained elsewhere on this site, see the list of Related Articles.
Finding faster sorting algorithms is very fundamental and what could be better than to turn AlphaZero loose on the problem and see if it can do better than generations of creative programmers.
To get reinforcement learning into the procedure, the task of finding a faster sorting algorithm was turned into a game. Then AlphaDev, the customized RL agent, was trained while playing the game and it worked:
"AlphaDev discovered small sorting algorithms from scratch that outperformed previously known human benchmarks. These algorithms have been integrated into the LLVM standard C++ sort library."
Now, if you are anything like me, or like a lot of programmers out there, you will want to know what this amazing algorithm is. After all, it has been a long time since Quicksort was invented and that was a truly mind-boggling algorithm - this could be fun.
However when you start to read the small print beyond the Abstract of the paper, things don't look quite so exciting. It isn't so much that AlphaDev discovered a new algorithm, more that it found an optimization of the machine code implementing the original algorithm.
Basically AlphaDev was presented with the assembly language that implemented a small sort - sorting 3 items into order say. Then it varies the code and it is tested and AlphaDev gets a reward that is based on the correctness and the speed of the new code. The speed of the code was taken to be an average speed across 100 different machines.
AlphaDev found an optimization of the code that involved deleting a mov instruction that wasn't doing anything. The fact that it isn't doing anything isn't easy to see and as such it is a worthwhile improvement to the code. It's an improvment' but not to the sorting algorithm in use. After finding the C++ code that makes LLVM produces the same assembly language the team had the code incorporated into the standard library.
A second improvement was to a sort 4 algorithm - usually this is just done by testing how many items are to be sorted and using the optimum sort for that number. AlphaDev came up with a more interesting decomposition of the problem. If there are 2 things then perform an optimal sort 2. If there are 3 or more perform a sort 3 and if there are 4 perform a sort 4 using the fact that the first three elements are sorted. However, on closer inspection the advantage of the new algorithm is that it results in a better assembly language implementation.
As I have said, optimizing assembler isn't as creative as inventing a new algorithm to challenge Quicksort, say, but it is interesting. It is in the same spirit as using simpler machine learning and statistics to optimize code and this has been going on for a long time. In fact, whenever I look at optimized machine code that has been derived from say a few lines of simple C I am often impressed by how "clever" the compiler is. So much so that I almost feel challenged by its efforts. Of course, I don't feel so impressed when the same compiler removes some vital bit of code because it has decided that it is undefined behavior. I guess all programmers have a love hate relationship with their compilers.
Perhaps in the not too distant future the role of AI in code will mean that the relationship can blossom into something deeper, more meaningful and more productive... I'm joking...
Faster sorting algorithms discovered using deep reinforcement learning
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https://gbee.edu.vn/isosceles-and-equilateral-triangles-z8oa49rl/ | 1,695,891,460,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510368.33/warc/CC-MAIN-20230928063033-20230928093033-00499.warc.gz | 298,357,246 | 12,418 | Types of Triangle | Isosceles Triangle | Base Angle /Base | Vertex Angle | Maths Tricks #shorts
Types of Triangle | Isosceles Triangle | Base Angle /Base | Vertex Angle | Maths Tricks #shorts
1
Isosceles and Equilateral Triangles
2
4.5 Isosceles and Equilateral Triangles
Isosceles Triangle: Vertex Angle Leg Leg Base Angles Base *The Base Angles are Congruent*
3
Isosceles Triangles Theorem 4-3 Isosceles Triangle Theorem
If two sides of a triangle are congruent, then the angles opposite those sides are congruent B <A = <C A C
4
Isosceles Triangles Theorem 4-4 Converse of the Isosceles Triangle Theorem If two angles of a triangle are congruent, then the sides opposite those angles are congruent B Given: <A = <C Conclude: AB = CB A C
5
Isosceles Triangles Theorem 4-5
The bisector of the vertex angle of an isosceles triangle is the perpendicular bisector of the base B Given: <ABD = <CBD Conclude: AD = DC and BD is ┴ to AC A C D
6
Equilateral Triangles
If a triangle is equilateral, then the triangle is equiangular If a triangle is equiangular, then the triangle is equilateral
7
Using Algebra Find the values of x and y: M ) ) y° N x° 63° O L
8
Proof of the Isosceles Triangle Theorem
Statements Reasons 1. 2. 3. 4. 5.
9
Find the values of the variables.
10
Find the values of the variables.
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https://www.numbersaplenty.com/138231313 | 1,679,962,228,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296948708.2/warc/CC-MAIN-20230327220742-20230328010742-00647.warc.gz | 986,097,203 | 3,180 | Search a number
138231313 = 1129577751
BaseRepresentation
bin10000011110100…
…11111000010001
3100122002212201021
420033103320101
5240341400223
621414435441
73265642513
oct1017237021
9318085637
10138231313
1171034260
123a362b81
132283b29c
1414503bb3
15c20765d
hex83d3e11
138231313 has 16 divisors (see below), whose sum is σ = 156476160. Its totient is φ = 120960000.
The previous prime is 138231239. The next prime is 138231319. The reversal of 138231313 is 313132831.
It is a cyclic number.
It is not a de Polignac number, because 138231313 - 29 = 138230801 is a prime.
It is a Duffinian number.
It is not an unprimeable number, because it can be changed into a prime (138231319) by changing a digit.
It is a pernicious number, because its binary representation contains a prime number (13) of ones.
It is a polite number, since it can be written in 15 ways as a sum of consecutive naturals, for example, 183688 + ... + 184438.
It is an arithmetic number, because the mean of its divisors is an integer number (9779760).
Almost surely, 2138231313 is an apocalyptic number.
It is an amenable number.
138231313 is a deficient number, since it is larger than the sum of its proper divisors (18244847).
138231313 is a wasteful number, since it uses less digits than its factorization.
138231313 is an odious number, because the sum of its binary digits is odd.
The sum of its prime factors is 1368.
The product of its digits is 1296, while the sum is 25.
The square root of 138231313 is about 11757.1813373784. The cubic root of 138231313 is about 517.0534946087.
The spelling of 138231313 in words is "one hundred thirty-eight million, two hundred thirty-one thousand, three hundred thirteen". | 503 | 1,709 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2023-14 | latest | en | 0.847336 |
https://www.thestudentroom.co.uk/showthread.php?t=4051735 | 1,606,447,220,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141189038.24/warc/CC-MAIN-20201127015426-20201127045426-00243.warc.gz | 879,584,063 | 37,256 | # English Language AQA GCSE
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#1
I didnt do so well in my controlled assessments, would i still be able to achieve an A in my final grade?
0
4 years ago
#2
(Original post by siddique247)
I didnt do so well in my controlled assessments, would i still be able to achieve an A in my final grade?
Depends what you got in your controlled assessments?!!
0
4 years ago
#3
(Original post by _Xenon_)
Depends what you got in your controlled assessments?!!
I did not expect to see YOU here...friend...
0
4 years ago
#4
(Original post by Wolfram Alpha)
I did not expect to see YOU here...friend...
I'm asking for help on differentiation. :P
So whilst waiting for responses I'm here lol
0
4 years ago
#5
(Original post by _Xenon_)
I'm asking for help on differentiation. :P
So whilst waiting for responses I'm here lol
Ah, I see.
With differentiation, you...well, here is an example.
2x^2 - 3X + 4
You get the power and multiply 2x by it which gives 4x. The new power will be 2-1 which is one so the first part is now 4x. -3X has no power so you simply remove the X. Now you have 4X - 3. Any number without an unknown is ignored. Thus dy/dx = 4X - 3. Do you understand?
0
#6
I got Cs
Posted from TSR Mobile
0
4 years ago
#7
(Original post by Wolfram Alpha)
Ah, I see.
With differentiation, you...well, here is an example.
2x^2 - 3X + 4
You get the power and multiply 2x by it which gives 4x. The new power will be 2-1 which is one so the first part is now 4x. -3X has no power so you simply remove the X. Now you have 4X - 3. Any number without an unknown is ignored. Thus dy/dx = 4X - 3. Do you understand?
Yep thanks man and PRSOM.
I will do some Chemistry today too!
I've printed loads of past papers ready to complete any time for science. I got 2 past papers for each unit so 18 exams to practise + all the ones for B2 so over 20+ in total. (As we did not finish B2 and I need to work on it)...
I have study leave atm so plenty of time - 8 hrs a day?
0
4 years ago
#8
(Original post by _Xenon_)
Yep thanks man and PRSOM.
I will do some Chemistry today too!
I've printed loads of past papers ready to complete any time for science. I got 2 past papers for each unit so 18 exams to practise + all the ones for B2 so over 20+ in total. (As we did not finish B2 and I need to work on it)...
I have study leave atm so plenty of time - 8 hrs a day?
That sounds pretty good. If you revise like that until examinations I have no doubt you will achieve your desired grades.
0
#9
Posted from TSR Mobile
0
4 years ago
#10
(Original post by Wolfram Alpha)
That sounds pretty good. If you revise like that until examinations I have no doubt you will achieve your desired grades.
Yep thanks!
0
#11
Keep on revising, make it a normal thing, don't give up after doing it for one day. The more you revise like that, the more better grades you'll get and the more you will get used to revision.
Posted from TSR Mobile
1
4 years ago
#12
(Original post by siddique247)
Posted from TSR Mobile
you could get a B but A might be difficult
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https://math.stackexchange.com/questions/996983/limit-lim-n-rightarrow-infty-a-1n-a-kn1-n-max-a-1-a | 1,566,533,983,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027317847.79/warc/CC-MAIN-20190823041746-20190823063746-00515.warc.gz | 558,319,491 | 28,479 | # Limit $\lim_{n \rightarrow \infty} (A_1^n + … A_k^n)^{1/n}= \max\{ A_1, …, A_k\}$ [duplicate]
I have the following question. I was asked to compute the following limit: Let $A_1 ... A_k$ be positive numbers, does exist:
$$\lim_{n \rightarrow \infty} (A_1^n + ... A_k^n)^{1/n}$$ My work: W.L.O.G let $A_1= \max{ A_1, ..., A_k}$, so I have $$A_1^n \leq A_1^n + ... A_k^n \leq kA_1^n$$ so that
$$A_1 = \lim_{n \rightarrow \infty} (A_1^n)^{1/n} \leq \lim_{n \rightarrow \infty}(A_1^n + ... A_k^n)^{1/n} \leq \lim_{n \rightarrow \infty} (kA_1^n)^{1/n} = kA_1$$
Can I do something else to sandwich the limit?
## marked as duplicate by Guy Fsone, Fabio Lucchini, muaddib, Parcly Taxel, Hans EnglerJan 28 '18 at 13:59
You have made a little mistake. Correction: $$A_1 = \lim_{n \rightarrow \infty} (A_1^n)^{1/n} \leq \lim_{n \rightarrow \infty}(A_1^n + ... A_k^n)^{1/n} \leq \lim_{n \rightarrow \infty} (kA_1^n)^{1/n} = \lim_{n \rightarrow \infty} A_1{\color{blue}{k^{1/n}}}=A_1$$
Hint: $$(ab_n)^{1/n} = \exp \left(\frac 1n \log b_n + \frac 1n \log a \right)$$ has the same limit when $n\to\infty$ (if any) as $b_n^{1/n}$. | 470 | 1,122 | {"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} | 3.890625 | 4 | CC-MAIN-2019-35 | latest | en | 0.540828 |
http://www.chegg.com/homework-help/write-english-phrase-algebraic-expression-simplify-expressio-chapter-1.6-problem-94e-solution-9780132356794-exc | 1,475,242,346,000,000,000 | text/html | crawl-data/CC-MAIN-2016-40/segments/1474738662197.73/warc/CC-MAIN-20160924173742-00233-ip-10-143-35-109.ec2.internal.warc.gz | 380,578,908 | 19,403 | # Introductory Algebra for College Students (5th Edition) View more editions Solutions for Chapter 1.6 Problem 94EProblem 94E: Write the English phrase as an algebraic expression. Then si...
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Chapter: Problem:
Write the English phrase as an algebraic expression. Then simplify the expression. Let x represent the number.
The quotient of −7 and a number, subtracted from the quotient of −12 and the number
STEP-BY-STEP SOLUTION:
Chapter: Problem:
Corresponding Textbook
Introductory Algebra for College Students | 5th Edition
9780132356794ISBN-13: 0132356791ISBN: Robert F. BlitzerAuthors:
Alternate ISBN: 9780023108532, 9780130102270, 9780136029243, 9780136029281, 9780136031161, 9780136031468, 9780136031642, 9780136032083, 9780321563910, 9780321565648, 9780321565778, 9780321578624, 9780321589200, 9780321831385, 9780536191540 | 303 | 1,029 | {"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} | 2.578125 | 3 | CC-MAIN-2016-40 | latest | en | 0.704401 |
https://www.aldeid.com/wiki/Category:Drones/Battery | 1,726,403,659,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651630.14/warc/CC-MAIN-20240915120545-20240915150545-00344.warc.gz | 600,420,721 | 9,744 | # Category:Drones/Battery
You are here Drones Battery
# LiPo?
• LiPo batteries are Lithium polymer batteries utilising a lithium-ion technology. They are commonly used for drones because they are commonly light weight, support large capacity and discharge rate.
• For long range, you can use Lithium-Ion (li-ion) batteries which support lower discharge voltages (down to 2.5V/cell)
# Cells and voltage
The number of cells is indicated as number of "S". It refers to cells in Serial (S). Old LiPo batteries were found in Parallel (P).
# cells Nominal voltage Min voltage* Max voltage
1S 3.7 3.3 4.2
2S 7.4 6.6 8.4
3S 11.1 9.9 12.6
4S 14.8 13.2 16.8
5S 18.5 16.5 21.0
6S 22.2 19.8 25.2
(*) the minimal voltage is 3.0V but I consider 3.3V since it is commonly accepted to never go below this value
# Capacity, discharge rate and burst
## Capacity (mAh)
The capacity of a LiPo is expressed in milliamps per hour (mAh). In racing/freestyle, 1300-1600mAh LiPo batteries are commonly used. Higher capacity LiPo batteries will increase the flight time but will also mean a bigger size and heavier batterie, what will affect the manoeuvrability. Higher capacity LiPo batteries are rather used for stable aerial photography drones.
## Discharge rate (C)
The discharge rate of a LiPo (C rating) is the number of amps the LiPo can give instantaneously without damaging it. To calculate it, you have to multiply the capacity in amps by the C rating.
For a 1300mAh Lipo with a discharge rate of 45C, the current (Amps) that you can safely draw continuously is:
```1.3A * 45C = 58.5A
```
## Burst
The burst rating is the instantaneous current that you can get from the LiPo during a maximum of 10 seconds.
In our previous example (1300mAh 45C that delivers 58.5A), if the burst rating is 75C, you can get a maximum of:
```1.3A * 75C = 97.5A (during 10 seconds)
```
# Connectors and balance plug
## Connector
Connector Current Rating (A) Burst Current (A)
EC2 20 30
EC3 60 70
EC5 120 170+
XT-60 60 70
XT-90 90 120
Mini-Deans 15 20
Deans 60 75
JST 5 8
Tamiya 15 25
Traxxas 70 100
4mm Bullet 75 100
6.5mm Bullet 200 250+
8mm Bullet 300 350+
MPX 35 ?
The most commonly used connectors in racing/freestyle drones is XT60. JST connectors are commonly used as 5V/12V connectors for various equipments (LED, camera, buzzer, ...).
## Balance plug
LiPo with several cells have a balance plug that is used:
• during flight: to monitor the voltage of each cell and trigger an alarm when 1 cell reaches a given threshold
• during charge: to balance voltage between cells (it is critical that you don't have a too important difference of current between cells)
There are 3 different balancing connectors depending on battery brands:
• Thunder Power (TP)
• JST-XH
• Hyperion (Polyquest)
# Control lipo voltage
You can use a lipo tester like this one:
This is very convenient because it will warn you once a cell reaches a given threshold (I set it to 3.60V), independantly from the total voltage of the LiPo.
# How to choose your LiPo?
• Max supported voltage supported by your ESC
• Max supported voltage supported by your motors
• Flight style (racing, freestyle, photography)
• Capacity
• Number of C / burst
# Charging
To charge LiPo batteries, you need a good charger. I use a Pro-Tronik AP681BLC.
I highly recommend that you charge your batteries using the BALANCE mode instead of the CHARGE mode. This mode will not only charge your batteries, it will also ensure that cells are balanced.
Here are the menus to select:
• Choose LiPo battery:
```┌──────────────────┐
│ PROGRAM SELECT │
│ Li-Po BATT │
└──────────────────┘
```
• Then select the BALANCE mode. Also choose the appropriate amps (e.g. 1.3A) and number of cells (e.g. 4S):
```┌──────────────────┐
│ LiPo BALANCE │
│ 1.3A 14.8V(4S) │
└──────────────────┘
```
• Long press ENTER:
```┌──────────────────┐
│ BATTERY CHECK │
│ WAIT.... │
└──────────────────┘
```
• It will check that the number of cells is correct. Press ENTER to confirm:
```┌──────────────────┐
│ R: 4SER S: 4SER │
│ CONFIRM (ENTER) │
└──────────────────┘
```
• During the charge, it will display a similar screen:
```┌──────────────────┐
│ Li4S 1.3A 16.42V │
│ BAL 025:03 00563 │
└──────────────────┘
```
where:
• Li4S: LiPo 4S battery
• 1.3A: current amps charged to battery (will decrease until it reaches 0)
• 16.42V: current voltage in battery (will increase until battery is fully charged)
• BAL: Balance mode
• 025:03: time since battry is charging
• 00563: number of cycles (increases until battery is fully charged)
# First uses
For the 5 to 10 first uses, it is recommended to:
• 1st charge your battery (it comes only half charged)
• charge at 0.5C (e.g. charge at 650mAh for a 1300mAh LiPo) until it reaches 70% (16.2V for a 4S), then charge at 0.2C (e.g. 260mAh for a 1300mAh LiPo)
• Have easy flights and don't go below 50% of LiPo capacity (3.85V/cell) | 1,454 | 4,938 | {"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} | 2.796875 | 3 | CC-MAIN-2024-38 | latest | en | 0.883417 |
https://se.mathworks.com/matlabcentral/answers/1780090-delete-duplicate-cell-in-a-cell-of-complex-double?s_tid=prof_contriblnk | 1,675,864,489,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764500813.58/warc/CC-MAIN-20230208123621-20230208153621-00528.warc.gz | 540,617,583 | 30,469 | # Delete duplicate cell in a cell of complex double.
7 views (last 30 days)
NHJ on 16 Aug 2022
Commented: NHJ on 16 Aug 2022
Hi, how to delete duplicate cell in a cell using Matlab? I can run using single cell, but I can't when I change the input to cell in a cell.
Below is my code:
% Create cell in cell
F = cell(1,3)
F(1,1)={cell(1,1)}
F(1,2)={cell(1,8)}
F(1,3)={cell(1,8)}
F{1,1}(1,1)= {[0.04 0.2 0.56;
0.31 0.67 0.22]}
F{1,2}(1,1)= {[6+6j 7+3j 8-6j;
6+8j 7-6j 3-3j]}
F{1,2}(1,2)= {[5+6j 8+5j 3-6j;
6+8j 7-6j 2+3j]}
F{1,2}(1,3)= {[3-6j 2+1j 8-6j;
5+6j 7+6j 9-3j]}
F{1,2}(1,4)= {[1-1j 7+5j 8+6j;
9+8j 8-6j 9-2j]}
F{1,2}(1,5)= {[6+6j 7+3j 8-6j;
6+8j 7-6j 3-3j]}
F{1,2}(1,6)= {[5+6j 8+5j 3-6j;
6+8j 7-6j 2+3j]}
F{1,2}(1,7)= {[3-6j 2+1j 8-6j;
5+6j 7+6j 9-3j]}
F{1,2}(1,8)= {[1-1j 7+5j 8+6j;
9+8j 8-6j 9-2j]}
F{1,3}(1,1)= {[6+6j 7+3j 8-6j;
6+8j 7-6j 3-3j]}
F{1,3}(1,2)= {[5+6j 8+5j 3-6j;
6+8j 7-6j 2+3j]}
F{1,3}(1,3)= {[3-6j 2+1j 8-6j;
5+6j 7+6j 9-3j]}
F{1,3}(1,4)= {[1-1j 7+5j 8+6j;
9+8j 8-6j 9-2j]}
F{1,3}(1,5)= {[6+6j 7+3j 8-6j;
6+8j 7-6j 3-3j]}
F{1,3}(1,6)= {[5+6j 8+5j 3-6j;
6+8j 7-6j 2+3j]}
F{1,3}(1,7)= {[3-6j 2+1j 8-6j;
5+6j 7+6j 9-3j]}
F{1,3}(1,8)= {[1-1j 7+5j 8+6j;
9+8j 8-6j 9-2j]}
% Delete duplicate cells
for s=1:length(F)
for w=2:length(F{s})
for ii = numel(F{w}):-1:2
for jj = 1:ii-1
if isequal((F{s}{ii}),(F{s}{jj}))
F{s}{ii} = {};
break
end
end
end
end
end
F{:}
The output should be: remain value in F{1}{1}(1,1), F{1,2} (1,1)-(1,4) and F{1,3}(1,1)-(1,4). The rest cell (duplicate cells) will be deleted.
##### 2 CommentsShowHide 1 older comment
Jan on 16 Aug 2022
Edited: Jan on 16 Aug 2022
By the way:
F{1,1} = cell(1,1)
% is more efficient than:
F(1,1) = {cell(1,1)}
In the second case, Matlab creates a cell and inserts it in a cell array. In the first case the array is inserted as cell element directly. The same for:
F{1,1}{1,1} = [0.04 0.2 0.56;
0.31 0.67 0.22]
Does the show code work? I get the error message: "Index exceeds array bounds."
What does "deleted" mean? Do you want empty cells or remove the cells?
Stephen23 on 16 Aug 2022
Edited: Stephen23 on 16 Aug 2022
As far as I can tell, you only want to remove local duplicate (i.e. only checking within the same cell of F):
F = {{[0.04,0.2,0.56;0.31,0.67,0.22]},{...
[6+6j,7+3j,8-6j;6+8j,7-6j,3-3j],...
[5+6j,8+5j,3-6j;6+8j,7-6j,2+3j],...
[3-6j 2+1j 8-6j;5+6j,7+6j,9-3j],...
[1-1j,7+5j,8+6j;9+8j,8-6j,9-2j],...
[6+6j,7+3j,8-6j;6+8j,7-6j,3-3j],...
[5+6j 8+5j 3-6j;6+8j,7-6j,2+3j],...
[3-6j,2+1j,8-6j;5+6j,7+6j,9-3j],...
[1-1j,7+5j,8+6j;9+8j,8-6j,9-2j]},{...
[6+6j,7+3j,8-6j;6+8j,7-6j,3-3j],...
[5+6j,8+5j,3-6j;6+8j,7-6j,2+3j],...
[3-6j,2+1j,8-6j;5+6j,7+6j,9-3j],...
[1-1j,7+5j,8+6j;9+8j,8-6j,9-2j],...
[6+6j,7+3j,8-6j;6+8j,7-6j,3-3j],...
[5+6j,8+5j,3-6j;6+8j,7-6j,2+3j],...
[3-6j,2+1j,8-6j;5+6j,7+6j,9-3j],...
[1-1j,7+5j,8+6j;9+8j,8-6j,9-2j]}} % more efficient way to create that array
F = 1×3 cell array
{1×1 cell} {1×8 cell} {1×8 cell}
F{:}
ans = 1×1 cell array
{2×3 double}
ans = 1×8 cell array
{2×3 double} {2×3 double} {2×3 double} {2×3 double} {2×3 double} {2×3 double} {2×3 double} {2×3 double}
ans = 1×8 cell array
{2×3 double} {2×3 double} {2×3 double} {2×3 double} {2×3 double} {2×3 double} {2×3 double} {2×3 double}
for kk = 1:numel(F)
for ii = numel(F{kk}):-1:2
for jj = 1:ii-1
if isequal((F{kk}{ii}),(F{kk}{jj}))
F{kk}(ii) = [];
break
end
end
end
end
F{:}
ans = 1×1 cell array
{2×3 double}
ans = 1×4 cell array
{2×3 double} {2×3 double} {2×3 double} {2×3 double}
ans = 1×4 cell array
{2×3 double} {2×3 double} {2×3 double} {2×3 double}
Note that this is the same as my other answer, just replacing F with F{kk} and the corresponding outer loop:
NHJ on 16 Aug 2022
Yes, this is what I want. Remove the local duplicate. I was quite confused when it became a cell in a cell. By the way, thank you very much
Jan on 16 Aug 2022
Edited: Jan on 16 Aug 2022
% Create cell in cell
F = cell(1,3);
F{1,1} = {[0.04 0.2 0.56; 0.31 0.67 0.22]};
F{1,2} = {[6+6j 7+3j 8-6j; 6+8j 7-6j 3-3j], ...
[5+6j 8+5j 3-6j; 6+8j 7-6j 2+3j], ...
[3-6j 2+1j 8-6j; 5+6j 7+6j 9-3j], ...
[1-1j 7+5j 8+6j; 9+8j 8-6j 9-2j], ...
[6+6j 7+3j 8-6j; 6+8j 7-6j 3-3j], ...
[5+6j 8+5j 3-6j; 6+8j 7-6j 2+3j], ...
[3-6j 2+1j 8-6j; 5+6j 7+6j 9-3j], ...
[1-1j 7+5j 8+6j; 9+8j 8-6j 9-2j]};
F{1,3} = {[6+6j 7+3j 8-6j; 6+8j 7-6j 3-3j], ...
[5+6j 8+5j 3-6j; 6+8j 7-6j 2+3j], ...
[3-6j 2+1j 8-6j; 5+6j 7+6j 9-3j], ...
[1-1j 7+5j 8+6j; 9+8j 8-6j 9-2j], ...
[6+6j 7+3j 8-6j; 6+8j 7-6j 3-3j], ...
[5+6j 8+5j 3-6j; 6+8j 7-6j 2+3j], ...
[3-6j 2+1j 8-6j; 5+6j 7+6j 9-3j], ...
[1-1j 7+5j 8+6j; 9+8j 8-6j 9-2j]};
% Delete duplicate cells
for iF = 1:numel(F)
G = F{iF}; % Process current cell of F
nG = numel(G);
keep = true(1, nG); % Elements to keep
for iG = 1:nG
for iG2 = iG+1:nG % Check following elements of G
if keep(iG2) % Do not test, if excluded before
keep(iG2) = ~isequal(G{iG}, G{iG2});
end
end
end
F{iF} = G(keep);
end
F{:}
ans = 1×1 cell array
{2×3 double}
ans = 1×4 cell array
{2×3 double} {2×3 double} {2×3 double} {2×3 double}
ans = 1×4 cell array
{2×3 double} {2×3 double} {2×3 double} {2×3 double}
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• learning algebra poem | 1,036 | 4,250 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2017-51 | longest | en | 0.908302 |
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## Explore printable Probability & Combinatorics worksheets for 12th Class
Probability & Combinatorics worksheets for Class 12 are essential resources for teachers who want to help their students master these complex mathematical concepts. These worksheets provide a wide range of problems and exercises that challenge students to apply their knowledge of probability and combinatorics in various real-life situations. By incorporating these worksheets into their lesson plans, teachers can ensure that their Class 12 students have a solid foundation in these critical areas of math. Moreover, these worksheets are designed to cater to different learning styles and abilities, making them an invaluable tool for teachers who want to differentiate their instruction and meet the diverse needs of their students. In conclusion, Probability & Combinatorics worksheets for Class 12 are indispensable for teachers who want to equip their students with the skills necessary to excel in advanced mathematics courses.
Quizizz is an excellent platform that offers a variety of Probability & Combinatorics worksheets for Class 12, along with other valuable resources for teachers. This platform allows teachers to create engaging quizzes and interactive lessons that can be easily integrated into their existing curriculum. In addition to Probability & Combinatorics worksheets, Quizizz also provides access to a vast library of resources covering other areas of math, making it a one-stop-shop for teachers looking to enhance their instruction. Furthermore, Quizizz offers real-time feedback and analytics, enabling teachers to monitor their students' progress and identify areas where they may need additional support. By utilizing Quizizz and its extensive collection of Probability & Combinatorics worksheets for Class 12, teachers can create a dynamic and engaging learning environment that fosters mathematical proficiency and success. | 689 | 2,862 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-18 | latest | en | 0.911449 |
https://kidsworksheetfun.com/how-to-explain-3-digit-subtraction/ | 1,708,526,591,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947473518.6/warc/CC-MAIN-20240221134259-20240221164259-00410.warc.gz | 373,363,872 | 25,084 | # How To Explain 3 Digit Subtraction
How To Explain 3 Digit Subtraction. Other names used in subtraction are minus, less, difference, decrease, take away, deduct. In this instance, we have three columns to subtract:
The number that is to be subtracted. In this instance, we have three columns to subtract: A 3 digit number minus a smaller 3 digit number and a 4 digit number minus a 3 digit number.
### Let’s Take This Step By Step.
When you subtract, always start on the right (ones place) and move left. Regrouping may be required in the tens place, the hundreds place, or both. The ones place for our example tells us to.
### The Result Of Subtracting One Number From Another.
And 900 minus 600 is 300. 11 minus 9 is 2. Make enough copies so that each student will have his own.
### Now, Let's Look At Our Previous Examples.
Make sure as you give students problems to solve that you listen very carefully to how they solve them. Now subtract the hundreds place, which in. And now we can actually subtract.
### So This Subtraction Should Result In 300 Plus 10 Plus 2, Which Is 312.
Try the free mathway calculator and problem solver below to practice various math topics. The names of the numbers in a subtraction fact are: 3 digits three digit subtraction:
### The Next Step In Our Subtraction Problem Is To Borrow.
Just like with addition strategies, there are multiple ways for students to show what they’re thinking. Separate the numbers into ones, tens, hundreds and thousands. I like to cut apart and laminate these break it down subtraction cards. | 352 | 1,568 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.28125 | 4 | CC-MAIN-2024-10 | latest | en | 0.886083 |
https://www.teacherspayteachers.com/Product/MULTIPLICATION-AND-DIVISION-ZAP-MATH-CENTER-GAME-PAIR-PACK-BUNDLE-5068672 | 1,575,597,597,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575540482954.0/warc/CC-MAIN-20191206000309-20191206024309-00544.warc.gz | 890,062,337 | 28,583 | # MULTIPLICATION AND DIVISION ZAP MATH CENTER GAME PAIR PACK BUNDLE!!
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1. Multiplication fluency center game. Practice multiplication while playing this fun ZAP game! Perfect game for partners, small groups and homework fun! Learn and have fun, WIN WIN!Each Set Includes:**Bright colored multiplication and ZAP cards for color copying.**Black and white multiplication and ZA
2. DIVISION fluency center game. Practice division while playing this fun ZAP game! Perfect game for partners, small groups and homework fun! Learn and have fun, WIN WIN!Each Set Includes:**Bright colored division and ZAP cards for color copying.**Black and white division and ZAP cards to print on colo
• Bundle Description
• StandardsNEW
MULTIPLICATION AND DIVISION fluency ZAP center games. Practice multiplication and division while playing this fun ZAP game! Perfect game for partners, small groups and homework fun! Learn and have fun, WIN WIN!
Each Set Includes:
**Bright colored multiplication and division and ZAP cards for color copying.
**Black and white multiplication and division and ZAP cards to print on colored cardstock.
**Directions how to play ZAP.
**A Division Table chart
**Recording sheet
How to play ZAP!
1. Shuffle cards and lay them face down in a pile.
2. Each child takes a turn taking a card. If the child can solve the math fact, they keep the card. If they can’t, it gets put on the bottom of the pile.
3. If a child gets a ZAP card, they have to return ALL of their cards to the bottom of the pile.
4. Play continues until time is up (teacher determined).
More Math Goodies!
NUMBER CARDS 1-1000
MATH FACTS TO 20 ZAP ADDITION MATH GAME
MATH FACTS TO 10 ZAP ADDITION MATH GAME
ROLL AND COVER 1-6
ROLL AND COLOR 1-6
ROLL AND COVER 2-12 NUMBER SENSE ADDITION CENTER
ROLL AND COLOR 2-12 NUMBER SENSE ADDITION CENTER
SPRING 120 CHARTS WITH MISSING NUMBERS DIFFERENTIATED
SUMMER 120 CHARTS WITH MISSING NUMBERS DIFFERENTIATED
FALL 120 CHARTS WITH MISSING NUMBERS DIFFERENTIATED
120 CHARTS WITH MISSING NUMBERS DIFFERENTIATED-THE BUNDLE!
HALLOWEEN 120 CHARTS WITH MISSING NUMBERS DIFFERENTIATED
THANKSGIVING 120 CHARTS WITH MISSING NUMBERS DIFFERENTIATED
DIFFERENTIATED 120 CHARTS WITH MISSING NUMBERS
DOUBLE TEN FRAMES TO PRACTICE TEEN NUMBERS
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Fluently multiply and divide within 100, using strategies such as the relationship between multiplication and division (e.g., knowing that 8 × 5 = 40, one knows 40 ÷ 5 = 8) or properties of operations. By the end of Grade 3, know from memory all products of two one-digit numbers.
Understand division as an unknown-factor problem. For example, find 32 ÷ 8 by finding the number that makes 32 when multiplied by 8.
Apply properties of operations as strategies to multiply and divide. Examples: If 6 × 4 = 24 is known, then 4 × 6 = 24 is also known. (Commutative property of multiplication.) 3 × 5 × 2 can be found by 3 × 5 = 15, then 15 × 2 = 30, or by 5 × 2 = 10, then 3 × 10 = 30. (Associative property of multiplication.) Knowing that 8 × 5 = 40 and 8 × 2 = 16, one can find 8 × 7 as 8 × (5 + 2) = (8 × 5) + (8 × 2) = 40 + 16 = 56. (Distributive property.)
Interpret products of whole numbers, e.g., interpret 5 × 7 as the total number of objects in 5 groups of 7 objects each. For example, describe a context in which a total number of objects can be expressed as 5 × 7.
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https://www.coursehero.com/file/p5q25cgf/To-include-people-with-zero-incomes-This-assumes-including-people-with-zero/ | 1,585,732,325,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585370505550.17/warc/CC-MAIN-20200401065031-20200401095031-00018.warc.gz | 858,980,586 | 739,108 | To include people with zero incomes This assumes including people with zero
# To include people with zero incomes this assumes
• Notes
• 9
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To include people with zero incomes. This assumes including people with zero income is actually useful and reasonable in this context… 5. Why did I not include a dummy for people who have not completed high-school? When regressing across a set of complementary and mutually exclusive category dummies, I need to keep one group as my reference group (the mean of which is absorbed in the constant term for the model). 6. Why do you think I included these variables? Can you think of other variables that would be important for explaining variation in income across individuals? These seem to be the ‘usual suspects’ in terms of explaining variation in income across individuals: we expect people with higher education to have higher incomes, we know that income tends to increase with age (due to experience/tenure), and definitely with hours worked. It might also vary depending on whether you are partnered or not, and have children, though potentially differently for men and women. Week 6 Pre-Lab Exercise Questions: 7. What is the relationship between education and income? How does it differ between males and females? Income does improve with education, though more strongly for men than for women. This appears to show that the financial returns of education is higher for males! To find out of this is true we’d have to also check for occupation. Nonetheless, it seems that the difference between the bottom and top education groups is almost twice as high for males as for females. Why is this? Jobs dominated by women earn less than jobs dominated by men. Glass ceilings. The jury is still out… Specifically, the difference for males can be worked out as follows: ln ??? = 9.44 + 0.597𝑃? So when PG=0: ln ??? = 9.44 ; which means that Income = e 9.44 = \$12,581 When PG=1 ln ??? = 9.44 + 0.597 = 10.037 ; which means that Income = e 10.037 = \$22,857 Thus , the predicted “returns” to a postgrad degree, compared to no qualification, is \$10,276, when we compare two males who are the same age, civil status (including kids), and work the same hours per week. | 506 | 2,258 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.65625 | 3 | CC-MAIN-2020-16 | latest | en | 0.976762 |
https://oeis.org/A241720 | 1,586,189,732,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585371637684.76/warc/CC-MAIN-20200406133533-20200406164033-00158.warc.gz | 625,422,732 | 3,659 | The OEIS Foundation is supported by donations from users of the OEIS and by a grant from the Simons Foundation.
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A241720 Number of compositions of n into distinct parts with exactly one descent. 2
1, 1, 2, 6, 7, 11, 16, 31, 36, 55, 71, 101, 147, 188, 245, 327, 421, 540, 732, 899, 1154, 1445, 1831, 2253, 2846, 3545, 4352, 5352, 6576, 7993, 9778, 11826, 14314, 17430, 20866, 25076, 30043, 35974, 42776, 51053, 60463, 71772, 85094, 100370, 118155, 139331 (list; graph; refs; listen; history; text; internal format)
OFFSET 3,3 LINKS Alois P. Heinz, Table of n, a(n) for n = 3..1000 EXAMPLE a(3) = 1: [2,1]. a(4) = 1: [3,1]. a(5) = 2: [3,2], [4,1]. a(6) = 6: [4,2], [5,1], [2,3,1], [3,1,2], [1,3,2], [2,1,3]. a(7) = 7: [4,3], [5,2], [6,1], [2,4,1], [4,1,2], [1,4,2], [2,1,4]. a(8) = 11: [6,2], [5,3], [7,1], [1,4,3], [3,1,4], [2,1,5], [3,4,1], [2,5,1], [5,1,2], [1,5,2], [4,1,3]. a(9) = 16: [5,4], [8,1], [7,2], [6,3], [2,6,1], [3,5,1], [6,1,2], [1,6,2], [2,1,6], [3,4,2], [4,2,3], [5,1,3], [2,4,3], [1,5,3], [3,2,4], [3,1,5]. CROSSREFS Column k=1 of A241719. Sequence in context: A088227 A231500 A174000 * A224082 A051678 A079906 Adjacent sequences: A241717 A241718 A241719 * A241721 A241722 A241723 KEYWORD nonn AUTHOR Alois P. Heinz, Apr 27 2014 STATUS approved
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Last modified April 6 12:08 EDT 2020. Contains 333273 sequences. (Running on oeis4.) | 768 | 1,666 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.484375 | 3 | CC-MAIN-2020-16 | latest | en | 0.662587 |
https://www.dreamwings.cn/hdu6121/4921.html | 1,721,849,581,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763518427.68/warc/CC-MAIN-20240724171328-20240724201328-00867.warc.gz | 640,782,181 | 12,572 | # HDU 6121 Build a tree （技巧）
## Description
HazelFan wants to build a rooted tree. The tree has $n$ nodes labeled $0$ to $n−1$ , and the father of the node labeled $i$ is the node labeled $\lfloor\frac{i-1}{k}\rfloor$. HazelFan wonders the size of every subtree, and you just need to tell him the XOR value of these answers.
## Input
The first line contains a positive integer $T(1≤T≤5)$ , denoting the number of test cases.
For each test case:
A single line contains two positive integers $n,k(1≤n,k≤10^{18})$ .
## Output
For each test case:
A single line contains a nonnegative integer, denoting the answer.
2
5 2
5 3
7
6
## 思路
• $cnt[i]$ ，第 $i$ 层的节点个数
• $sum[i]$ ，高为 $i$ 的满 $k$ 叉树节点个数
• $xor[i]$ ，高为 $i$ 的满 $k$ 叉树所有子树异或和
## AC 代码
#include<bits/stdc++.h>
typedef __int64 LL;
using namespace std;
const int maxn =105;
LL cnt[maxn]; // 第 i 层的节点个数
LL sum[maxn]; // 高为 i 的满 k 叉树节点个数
LL xxor[maxn]; // 高为 i 的满 k 叉树所有子树异或和
LL n,k;
inline LL F(LL a,LL b)
{
return b&1?a:0;
}
{
if(dep==0)return 0;
}
LL solve()
{
if(k==1) // 对 1 特判
{
int mo = n%4;
if(mo==0)
return n;
else if(mo==1)
return 1;
else if(mo==2)
return n+1;
else
return 0;
}
int dep=1;
sum[dep]=cnt[dep]=xxor[dep]=1;
while((n-sum[dep])/k>=cnt[dep]) // 枚举深度（所有满节点的层）
{
dep++;
cnt[dep] = cnt[dep-1]*k;
sum[dep] = sum[dep-1] + cnt[dep];
xxor[dep] = sum[dep]^F(xxor[dep-1],k);
}
return dfs(dep,n-sum[dep]);
}
int main()
{
int T;
cin>>T;
while(T--)
{
LL ans;
cin>>n>>k;
ans=solve();
cout<<ans<<endl;
}
return 0;
} | 589 | 1,511 | {"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} | 3.25 | 3 | CC-MAIN-2024-30 | latest | en | 0.169437 |
https://www.physicsforums.com/threads/analytically-continue-rzf-using-gamma-function-step-by-step-guide.466823/ | 1,708,743,648,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474482.98/warc/CC-MAIN-20240224012912-20240224042912-00693.warc.gz | 968,458,932 | 17,208 | # Analytically Continue RZF Using Gamma Function: Step By Step Guide
• camilus
In summary, the conversation discusses the theory of the Riemann zeta function and its analytical continuation using the gamma function. It also mentions the Jacobi theta function and Riemann's use of the functional equation to equate two expressions. The main focus is on resolving an error in the book that caused confusion in understanding the last equation in Riemann's manuscript.
#### camilus
theory of Riemann zeta function question
analytically continuing the Riemann zeta function (RZF) using the gamma function leads to this identify:
$$n^{-s} \pi^{-s \over 2} \Gamma ({s \over 2}) = \int_0^{\infty} e^{-n^2 \pi x} x^{{s \over 2}-1} dx$$ ________(1)
and from that we can build a similar expression incorporating the RZF:
$$\pi^{-s \over 2} \Gamma ({s \over 2}) \zeta (s) = \int_0^{\infty} \psi (x) x^{{s \over 2}-1} dx$$ ________(2)
where
$$\psi (x) = \sum_{n=1}^{\infty} e^{-n^2 \pi x}$$
is the Jacobi theta function.
Then Riemann proceeds to use the functional equation for the theta function:
$$2\psi (x) +1 = x^{-1 \over 2} (2 \psi ({1 \over x})+1)$$
to equate (2) with:
$$\pi^{-s \over 2} \Gamma ({s \over 2}) \zeta (s) = \int_1^{\infty} \psi (x) x^{{s \over 2}-1} dx + \int_1^{\infty} \psi ({1 \over x}) x^{-1 \over 2} x^{{s \over 2}-1} dx+{1 \over 2} \int_0^{1} (x^{-1 \over 2} x^{{s \over 2}-1} - x^{{s \over 2}-1}) dx$$
This is the step I am stuck on, I am trying to figure out what he did to get that last equation. Any help would be greatly appreciated.
Last edited:
I found an error, and the reason why I've been stuck. There is a typo in my book, maybe I should contact the publisher. I had to check Riemann's original handwritten manuscript in german to see that the second integral is from 0 to 1.
the second integral is not
$$\int_1^{\infty} \psi ({1 \over x}) x^{-1 \over 2} x^{{s \over 2}-1} dx$$
but
$$\int_0^1 \psi ({1 \over x}) x^{-1 \over 2} x^{{s \over 2}-1} dx$$
problem resolved. I verified and indeed
$$\pi^{-s \over 2} \Gamma ({s \over 2}) \zeta (s) = \int_1^{\infty} \psi (x) x^{{s \over 2}-1} dx + \int_0^1 \psi ({1 \over x}) x^{-1 \over 2} x^{{s \over 2}-1} dx+{1 \over 2} \int_0^{1} (x^{-1 \over 2} x^{{s \over 2}-1} - x^{{s \over 2}-1} ) dx$$
=>
$$\zeta (s) = {\pi^{s \over 2} \over \Gamma (\frac{s}{2})} \left( \int_1^{\infty} \psi (x) x^{{s \over 2}-1} dx + \int_0^1 \psi ({1 \over x}) x^{-1 \over 2} x^{{s \over 2}-1} dx+{1 \over 2} \int_0^{1} (x^{-1 \over 2} x^{{s \over 2}-1} - x^{{s \over 2}-1} ) dx \right)$$
=>
$$\zeta (s) = {\pi^{s \over 2} \over \Gamma (\frac{s}{2})} \left( {1 \over s(s-1)} + \int_1^\infty \psi(x) \left( x^{{s \over 2}-1} + x^{-{s+1 \over 2}} \right) dx \right)$$
The error I am talking about can be found in the middle of page 3 not counting the cover page.
http://www.claymath.org/millennium/Riemann_Hypothesis/1859_manuscript/EZeta.pdf [Broken]
Riemann's manuscript, bottom of page 2, where you can see the real version.
http://www.claymath.org/millennium/Riemann_Hypothesis/1859_manuscript/riemann1859.pdf [Broken]
Last edited by a moderator:
I can say that this is a very interesting and complex topic in mathematics. The Riemann zeta function is a fundamental object in number theory and its analytical continuation has important implications in various areas of mathematics and physics. The use of the gamma function in this context is a powerful tool that allows us to extend the domain of the RZF and study its behavior in a wider range of values.
In this step-by-step guide, you have shown the process of analytically continuing the RZF using the gamma function and deriving the identity (1). This identity is then used to build a similar expression (2) incorporating the RZF. However, to equate (2) with the functional equation for the theta function, Riemann makes use of a clever substitution and some algebraic manipulation to arrive at the last equation.
This type of mathematical reasoning and manipulation is often necessary in order to make connections and derive new identities. It is a crucial part of the scientific process and shows the creativity and ingenuity of mathematicians in solving complex problems.
Overall, this guide provides a clear and detailed understanding of the steps involved in analytically continuing the Riemann zeta function using the gamma function. It is a valuable resource for anyone interested in this topic and showcases the beauty and complexity of mathematical concepts.
## 1. What is Analytically Continue RZF using Gamma Function?
Analytically Continue RZF (Riemann Zeta Function) using Gamma Function is a mathematical method used to extend the Riemann Zeta function to the entire complex plane. This method uses the properties of the Gamma function to analytically continue the Riemann Zeta function, which is defined only for real numbers greater than 1, to the entire complex plane.
## 2. Why is Analytically Continue RZF using Gamma Function important?
The Riemann Zeta function is an important tool in number theory and has applications in various fields of mathematics. By extending the Riemann Zeta function to the entire complex plane, we can better understand its properties and use it in a wider range of mathematical problems.
## 3. What are the steps involved in Analytically Continue RZF using Gamma Function?
The steps involved in Analytically Continue RZF using Gamma Function include defining the Riemann Zeta function, using the Gamma function to analytically continue the Riemann Zeta function, and then using this extended function to solve mathematical problems. This process can be complex and often requires advanced mathematical knowledge.
## 4. What are the applications of Analytically Continue RZF using Gamma Function?
Analytically Continue RZF using Gamma Function has various applications in mathematics, including in number theory, complex analysis, and physics. It is also used in solving problems related to the distribution of prime numbers and the Riemann Hypothesis, which is one of the most famous unsolved problems in mathematics.
## 5. Are there any limitations to Analytically Continue RZF using Gamma Function?
While Analytically Continue RZF using Gamma Function has many applications, it also has some limitations. This method can only be applied to certain types of functions, and the process can be complex and time-consuming. Additionally, the accuracy of the results may depend on the initial assumptions made during the process. | 1,839 | 6,524 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.734375 | 4 | CC-MAIN-2024-10 | latest | en | 0.810528 |
http://mizar.uwb.edu.pl/version/current/html/proofs/transgeo/75 | 1,568,628,409,000,000,000 | text/plain | crawl-data/CC-MAIN-2019-39/segments/1568514572517.50/warc/CC-MAIN-20190916100041-20190916122041-00174.warc.gz | 127,693,550 | 1,839 | let OAS be OAffinSpace; :: thesis: for p, q, x, y being Element of OAS
for f being Permutation of the carrier of OAS st f is dilatation & f . p = p & q <> p & Mid q,p,f . q & not p,x,y are_collinear holds
x,y // f . y,f . x
let p, q, x, y be Element of OAS; :: thesis: for f being Permutation of the carrier of OAS st f is dilatation & f . p = p & q <> p & Mid q,p,f . q & not p,x,y are_collinear holds
x,y // f . y,f . x
let f be Permutation of the carrier of OAS; :: thesis: ( f is dilatation & f . p = p & q <> p & Mid q,p,f . q & not p,x,y are_collinear implies x,y // f . y,f . x )
assume A1: f is dilatation ; :: thesis: ( not f . p = p or not q <> p or not Mid q,p,f . q or p,x,y are_collinear or x,y // f . y,f . x )
assume A2: ( f . p = p & q <> p & Mid q,p,f . q ) ; :: thesis: ( p,x,y are_collinear or x,y // f . y,f . x )
then Mid x,p,f . x by ;
then A3: x,p // p,f . x by DIRAF:def 3;
Mid y,p,f . y by A1, A2, Th60;
then A4: y,p // p,f . y by DIRAF:def 3;
x,y '||' f . x,f . y by ;
hence ( p,x,y are_collinear or x,y // f . y,f . x ) by ; :: thesis: verum | 405 | 1,070 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.953125 | 3 | CC-MAIN-2019-39 | latest | en | 0.556255 |
http://www.computational-cad.com/help/NetHelp/Documents/theory.htm | 1,653,386,051,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662570051.62/warc/CC-MAIN-20220524075341-20220524105341-00565.warc.gz | 70,669,003 | 3,123 | Theory
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## Theory
A 2d Delaunay triangulation (DT) for a set P of points in the plane is a triangulation such that no point in P is inside the circumcircle of any triangle in the triangulation. It can be shown that for all possible triangulations of P, a Delaunay triangulation maximizes the minimum angle of all angles of the triangles in the triangulation.
Thus, a Delaunay triangulation tends to avoid “skinny” triangles. This property makes it the triangulation of choice for many purposes, including Digital Terrain Modelling (DTM).
Figure 31: A Delaunay triangulation with all circumcircles and their centres. Image available under GNU license at http://en.wikipedia.org/wiki/File:Delaunay_circumcircles_centers.png
A Delaunay triangulation is unique in a general case. It is not unique if four triangulation points lie on the same circle.
A 2d conforming Delaunay triangulation (CDT) is a Delaunay triangulation that respects constraints (edges). This is done by iteratively inserting additional points (called Steiner points) until no triangle crosses a constraint.
Computational power
In general, triangulating pure point data is much faster than triangulation pure constraint data. This is because constraints are processed iteratively until no constraint crosses a triangle. Benchmarks for pure constraint data cannot be given because the number of iterations depends on the distribution of the constraints.
On an average 32bit machine, ComputationalCAD for AutoCAD triangulates about 0.5M to 1M points in about one minute. This may extend to several millions of points on a 64bit machine. The computational complexity is , resulting in near-linear computation time over the number of triangulation points. However, the maximum number of triangulation points is limited by the available amount of memory.
Figure 32: Triangulation timings | 407 | 1,873 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.625 | 3 | CC-MAIN-2022-21 | longest | en | 0.855232 |
https://www.road-map.in/Cutting-steel/ccce2fab736935.html | 1,627,654,091,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046153966.60/warc/CC-MAIN-20210730122926-20210730152926-00510.warc.gz | 1,030,161,952 | 7,352 | ### 01.05.04:Building Model Bridges Following the Engineering
Jan 05, 2004 · "Building Model Bridges Following the Engineering Process" is a curriculum unit designed to be used with students in grades 5 through 8. I am a Magnet School Facilitator at East Rock Global Magnet School where I run an after school program called CPEP. AH36, SS400 Steel Plate-AH36 Steel Supplier10CrMo9-10, 1.7380, A335 P22, A182 F22 steelLow-alloy steel 10CrMo9-10, 1.7380 - description and s
### Arch architecture Britannica
Arch, in architecture and civil engineering, a curved member that is used to span an opening and to support loads from above. The arch formed the basis for the evolution of the vault. Arch construction depends essentially on the wedge. If a series of wedge-shaped blocksi.e., ones in which the Bridges:Types, Span and Loads Civil EngineeringWeight per metre of trusses and bracings = 0.785 A Newton/metre. Where, A = Maximum net area of the tension chord (b) Fullers Formula:Weight per metre of truss bridges = (150 L + 5500) Newton/metre (spans 30 m to 90 m) Weight per metre of plate girders = (200 L + 1000) Newton/metre (spans 10 m to 30 m) where L = span of the bridge in metres Building Bridges 4-H STEM Lab Activity 4-HIn this activity, kids will learn about the engineering design process and the basic mechanics behind building bridges while designing their own bridge. This activity is a lot of fun, and was created by Arianna Smith, a 4-H teen leader in STEM, in collaboration with the University of Tennessee.
### China Ss400 JIS Standard Steel Channel with Different Size
China Ss400 JIS Standard Steel Channel with Different Size, Find details about China Channel Bar, Ss400 Channel from Ss400 JIS Standard Steel Channel with Different Size - Tangshan Honory International Trading Co., Ltd. China Wide Flange H Beam (SS400 Q235 A36) - China H H beam is widely used in various building structure and engineering structure:A). Used for the plant, high-rise building construction B). Used for the bridge, shipment building C). Used for lifting and transportation machinery, equipment manufacturing base building D). Used for the support, foundation pile manufacturing How to Build a Weighing Scale Using a Load Cell and a
• LCD DisplayLoad CellInstrumentation AmplifierMaking The Connections / SchematicConfiguring The Microcontroller Development KitAttaching The Load Cell Base PlatesThe FirmwareBuilding and Loading The Code, and Verifying The Accuracy of The ScaleNext Steps For Making A Real ProductBridges Science Projects & Science Lessons:Early ElementaryTo make the cables used in bridges, thousands of strong steel wires are twisted and bound together to make them even stronger. The Brooklyn Bridge in New York is one of the most famous bridges, now not so much for its size as its beauty. Built in 1883, it was the first great suspension bridge and a feat of engineering.
### Materials Used for Bridge Construction
Materials Used in Bridge Construction Stones, Timber, Concrete and Steel are the traditional materials that are used to carry out bridge construction. During the initial period, timber and stones were used in the construction, as they are directly obtained from nature and easily available. Q235 SS400 A36 Q345 steel plate used in steel structures You can also choose from q235, q235b, and ss400. As well as from hot rolled, welded. And whether a36 mild steel h beams is steel workshop, structure pipe, or steel fabricated house.[plate]Q345b sheet used in bridge engineering supplier[steel]Q345b sheet used in bridge engineering supplier Straw Bridges - Activity - TeachEngineeringApr 30, 2021 · Working as engineering teams, students design and create model beam bridges using plastic drinking straws and tape as their construction materials. Their goal is to build the strongest bridge with a truss pattern of their own design, while meeting the design criteria and constraints. They experiment with different geometric shapes and determine how shapes affect the strength of
### Steel Bridges - an overview ScienceDirect Topics
Weiwei Lin, Teruhiko Yoda, in Bridge Engineering, 2017. 7.1 Introduction. Steel bridges are widely used around the world in different structural forms with different span length, such as highway bridges, railway bridges, and footbridges. The main advantages of structural steel over other construction materials are its strength, ductility, easy fabrication, and rapid construction. | 973 | 4,464 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2021-31 | latest | en | 0.880929 |
https://answersdrive.com/what-is-the-meaning-of-confidence-level-in-statistics-5924350 | 1,586,267,381,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585371799447.70/warc/CC-MAIN-20200407121105-20200407151605-00531.warc.gz | 347,543,189 | 8,918 | # What is the meaning of confidence level in statistics?
A confidence level refers to the percentage of all possible samples that can be expected to include the true population parameter. For example, suppose all possible samples were selected from the same population, and a confidence interval were computed for each sample.
A.
### What does the 95% confidence interval tell us?
A confidence interval does not quantify variability. A 95% confidence interval is a range of values that you can be 95% certain contains the true mean of the population. This is not the same as a range that contains 95% of the values.
• #### What effect does the sample size have on the confidence level?
Increasing the sample size decreases the width of confidence intervals, because it decreases the standard error. c) The statement, "the 95% confidence interval for the population mean is (350, 400)", is equivalent to the statement, "there is a 95% probability that the population mean is between 350 and 400".
• #### Is a higher confidence interval better?
Apparently a narrow confidence interval implies that there is a smaller chance of obtaining an observation within that interval, therefore, our accuracy is higher. Also a 95% confidence interval is narrower than a 99% confidence interval which is wider. The 99% confidence interval is more accurate than the 95%.
• #### Which confidence interval is wider 90 or 95?
A 99 percent confidence interval would be wider than a 95 percent confidence interval (for example, plus or minus 4.5 percent instead of 3.5 percent). A 90 percent confidence interval would be narrower (plus or minus 2.5 percent, for example).
B.
### What is the confidence level in stats?
In statistics, a confidence interval (CI) is a type of interval estimate, computed from the statistics of the observed data, that might contain the true value of an unknown population parameter. Most commonly, the 95% confidence level is used.
• #### What is the level of confidence?
It is expressed as a percentage and represents how often the true percentage of the population who would pick an answer lies within the confidence interval. The 95% confidence level means you can be 95% certain; the 99% confidence level means you can be 99% certain. Most researchers use the 95% confidence level.
• #### What is the effect size?
Effect size is a simple way of quantifying the difference between two groups that has many advantages over the use of tests of statistical significance alone. Effect size emphasises the size of the difference rather than confounding this with sample size.
• #### What is ap value and what does it tell us?
When you perform a hypothesis test in statistics, a p-value helps you determine the significance of your results. The p-value is a number between 0 and 1 and interpreted in the following way: A small p-value (typically ≤ 0.05) indicates strong evidence against the null hypothesis, so you reject the null hypothesis.
C.
### What is a 95% confidence limit?
Confidence limits are the numbers at the upper and lower end of a confidence interval; for example, if your mean is 7.4 with confidence limits of 5.4 and 9.4, your confidence interval is 5.4 to 9.4. Most people use 95% confidence limits, although you could use other values.
• #### How do confidence levels compared to significance levels?
So, if your significance level is 0.05, the corresponding confidence level is 95%. If the P value is less than your significance (alpha) level, the hypothesis test is statistically significant. If the confidence interval does not contain the null hypothesis value, the results are statistically significant.
• #### How do you find the z score for confidence intervals?
Step 1: Divide your confidence level by 2: .95/2 = 0.475. Step 2: Look up the value you calculated in Step 1 in the z-table and find the corresponding z-value. The z-value that has an area of .475 is 1.96. Step 3: Divide the number of events by the number of trials to get the “P-hat” value: 24/160 = 0.15.
• #### What is the p value?
The p-value is the level of marginal significance within a statistical hypothesis test representing the probability of the occurrence of a given event. The p-value is used as an alternative to rejection points to provide the smallest level of significance at which the null hypothesis would be rejected.
Updated: 6th December 2019 | 949 | 4,381 | {"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} | 4.0625 | 4 | CC-MAIN-2020-16 | longest | en | 0.909551 |
https://www.coursehero.com/file/6557250/test1-in-class/ | 1,524,340,446,000,000,000 | text/html | crawl-data/CC-MAIN-2018-17/segments/1524125945317.36/warc/CC-MAIN-20180421184116-20180421204116-00571.warc.gz | 765,079,839 | 45,998 | {[ promptMessage ]}
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test1_(in_class)
# test1_(in_class) - a n 1 = 1 1 1 a n n ∈ N where a 1 = 1...
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Test 1 (In class part), MAA4226 Fall 2010 Student Name: 60 minutes. Closed book. This part weighs 40% of the total weight of Test 1. 1: State and prove the extreme value theorem. 2: Let { a n } be a sequence such that lim n →∞ a 2 n = lim n →∞ a 2 n +1 = l . Show that lim n →∞ a n = l . 3: Let { a n } be the sequence defined by the recurrence relation
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Unformatted text preview: a n +1 = 1 + 1 1 + a n , n ∈ N , where a 1 = 1. Use problem 2 above to show that { a n } is convergent and find its limit. 3: Let f : [0 , 1] → R be continuous and one-to-one. Show that f is strictly monotone. 1...
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http://www.devsplanet.com/question/35263633 | 1,498,200,042,000,000,000 | text/html | crawl-data/CC-MAIN-2017-26/segments/1498128320023.23/warc/CC-MAIN-20170623063716-20170623083716-00100.warc.gz | 515,749,328 | 4,308 | # Developers Planet
teo93 February 2016
### Babylonian Algorithm with C++ unable to get loop to run
Is the refusal for the program to produce the result a consequence of it unable to get into the loop?
``````#include <iostream>
#include <cmath>
#include <iomanip>
using namespace std;
int main()
{
double x, squareroot, oldsquareroot, counter=0;
cout << "Enter a positive number: " ;
cin >> x ;
{
for (oldsquareroot = x/2; abs(squareroot-oldsquareroot) >= 1e-9 && abs(squareroot - oldsquareroot)>=(1E-8*x);counter++ )
{
squareroot = (squareroot+x/squareroot)/2;
}
}
cout << setprecision(15);
cout << "The value of the squareroot is:" << squareroot << endl;
cout << "The number of interations required for the computation is:" << counter << endl;
return 0;
}
``````
ErmIg February 2016
There were some mistakes. Now it works fine.
``````#include <iostream>
#include <cmath>
#include <iomanip>
using namespace std;
int main()
{
double x, squareroot, oldsquareroot, counter = 0;
cout << "Enter a positive number: ";
cin >> x;
{
squareroot = x; // fixed error 1: using of uninitialized variable.
for (oldsquareroot = x / 2; abs(squareroot - oldsquareroot) >= 1e-9 && abs(squareroot - oldsquareroot) >= (1E-8*x); counter++)
{
oldsquareroot = squareroot;// fixed eror 2: oldsquareroot is not changed in the loop and can't exit.
squareroot = (squareroot + x / squareroot) / 2;
}
}
cout << setprecision(15);
cout << "The value of the squareroot is:" << squareroot << endl;
cout << "The number of interations required for the computation is:" << counter << endl;
return 0;
}
`````` | 475 | 1,589 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2017-26 | longest | en | 0.717969 |
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# What is the greatest 6-digit number without repeating any digit?
Updated: 9/25/2023
Wiki User
9y ago
I believe it is:
987,654
Leslie Stehr
Lvl 9
2y ago
Earn +20 pts
Q: What is the greatest 6-digit number without repeating any digit?
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987
9876543
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9,876,543
987654321
### How do you find the greatest 6digit number divisible by 12?
Multiples of 12 are even. Start with 999998. It's not divisible by 12. Try 999996. It is divisible by 12. Stop there.
### What is the smallest 6digit number?
The smallest positive integer six-digit number is 100000.
### What are the greatest possible number whose digits are all even numbers from 1 to 9?
The greatest possible number is 888... (repeating).
0
### When you divide a 2 digit number by 6 what is the greatest remainder you can have?
5, or .8 with a three repeating.
### What would be the greatest 9 digit whole number without repeating a digit?
Since we want our number to be as large as possible, we have to put the largest digits in the front, in the largest places so they have more value. Therefore, our answer would be 987,654,321.
1,023,456,789 | 315 | 1,177 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.890625 | 4 | CC-MAIN-2024-30 | latest | en | 0.866288 |
https://hossfeld.github.io/performance-modeling/chapter4_markovianSystems/ch4-1-mmn-loss.html | 1,679,371,635,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296943625.81/warc/CC-MAIN-20230321033306-20230321063306-00237.warc.gz | 354,668,477 | 119,499 | # M/M/n-0 Loss System¶
(c) Tobias Hossfeld (Aug 2021)
This script and the figures are part of the following book. The book is to be cited whenever the script is used (copyright CC BY-SA 4.0):
Tran-Gia, P. & Hossfeld, T. (2021). Performance Modeling and Analysis of Communication Networks - A Lecture Note. Würzburg University Press. https://doi.org/10.25972/WUP-978-3-95826-153-2
The M/M/n loss system represents a fundamental queueing system. The steady state probabilities $x(i)$ for $i=0,...,n$ can be derived based on the analysis of Markov systems. The system consists of $n$ parallel servers, but no waiting room. Customers arrive according to a Poisson process with rate $\lambda$. The service time $B$ is exponentially distributed with mean $E[B]=1/\mu$. The M/M/n-0 system is called the Erlang loss model. The blocking probability $p_B = x(n)$ due to the PASTA property of the system, i.e., the probability of the state as seen by an outside random observer is the same as the probability of the state seen by an arriving customer: $x_A(i)=x(i)$. Hence, $p_B=x_A(n)=x(n)$.
## Direct computation of the blocking probability¶
The Erlang-B formula (or Erlang's loss formula) is
$p_B = \frac{\frac{a^n}{n!}}{\sum_{i=0}^n\frac{a^i}{i!}}$
with the offered load $a=\lambda E[B]=\frac{\lambda}{\mu}$. For the direct computation, the factorial needs to be computed which may lead to numerical issues.
## Iterative computation of Erlang's loss formula¶
A convenient formula for the efficient computation and the avoidance of numerical issues (up to a certain point), is based on the recurrence formula.
$\frac{1}{B(n,a)}=1+\frac{n}{a}\frac{1}{B(n-1,a)}$ with the initial condition $B(0,a)=1$.
This can also be expressed as
$B(n,a) = \frac{aB(n-1,a)}{n+aB(n-1,a)}$
The blocking probability is then $p_B = B(n,a)$ for the M/M/n system with offered load $a=\lambda E[B]$.
## Example¶
Changing the offered load will show numerical issues with the direct computation approach, while the recurrence formula is more robust.
## Economy of Scale for the M/M/n Loss System¶
The economy of scale describes the phenomenon that if many server groups are merged together into a larger server group the resulting QoS - in terms of blocking probability - is increasing. A larger server group thus operates more economically. With the Erlang formula we can show this effect quantitatively.
Let us consider
• system A: M/M/n-0 with rate $\lambda$
• system B: M/M/2n-0 with rate $2\lambda$
Both systems have the same normalized offered traffic $a/n$. | 675 | 2,550 | {"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} | 3.296875 | 3 | CC-MAIN-2023-14 | latest | en | 0.844525 |
https://study-assistantph.com/math/question2382713 | 1,656,162,812,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103035636.10/warc/CC-MAIN-20220625125944-20220625155944-00747.warc.gz | 616,823,741 | 61,477 | • Accueil
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• Which set/sets of ordered pairs is/are functions?...
# Which set/sets of ordered pairs is/are functions? explain.
• Réponse publiée par: homersoncanceranguiu
ordered pairsThe first set of ordered pairs is a function, because no two ordered pairs have the same first coordinates with different second coordinates for example ( -1, 2), ( 1, 0), (2, 1) . The second example is not a function, because it contains the ordered pairs (1,2) and (1,4). the first set is repeated . These have the same first coordinate and different second coordinates. Read the details about the set of ordered pairs in The range of a function is the complete set of all possible resulting values of the dependent variable (y, usually), after we have substituted the domain. The range is the resulting y-values we get after substituting all the possible x-values. Read the details about the definition of range in In the ordered pairs we find the domain and range. Ordered pair of numbers used to locate a point on a coordinate plane is called an ordered pair. An ordered pair is written in the form (x, y) where x is the x-coordinate and y is the y-coordinate.
1. An ordered pair contains the coordinates of one point in the coordinate system. A point is named by its ordered pair of the form of (x, y). The first number corresponds to the x-coordinate and the second to the y-coordinate. Read the details about an example of ordered pairs in
• Réponse publiée par: elishakim80 | 338 | 1,476 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.84375 | 4 | CC-MAIN-2022-27 | latest | en | 0.806881 |
http://nasawavelength.org/resource-search?educationalLevel=&n=100&qq=Magnetism&facetSort=1&instructionalStrategies%5B%5D=Nonlinguistic+representations&instructionalStrategies%5B%5D=Cues%2C+questions%2C+and+advanced+organizers&instructionalStrategies%5B%5D=Discussions | 1,430,749,494,000,000,000 | text/html | crawl-data/CC-MAIN-2015-18/segments/1430454443858.51/warc/CC-MAIN-20150501042723-00056-ip-10-235-10-82.ec2.internal.warc.gz | 133,055,800 | 17,981 | ## Narrow Search
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# Jump Rope Generator
In this activity, students will demonstrate the generator effect, which is due to electromagnetic induction when a conductor (a long metal wire) moves through a magnetic field. Materials required to this activity include: a 100-foot extension cord... (View More)
Keywords: Magnetism
Audience: High school
Materials Cost: Over \$20 per group of students
# Mapping Magnetic Field Lines
This is a lesson about the magnetic field of a bar magnet. The lesson begins with an introductory discussion with learners about magnetism to draw out any misconceptions that may be in their minds. Then, learners freely experiment with bar magnets... (View More)
Audience: Middle school
# Magnetic Energy and the Cause of Solar Flares
This is a lesson about how magnetism causes solar flares. Learners will set up an electrical circuit with magnets to examine magnetic fields and their similarities to magnetic fields seen on the Sun. Learners should have a conceptual understanding... (View More)
# Electric Current Generated with a Moving Magnet
This is an activity about magnetic induction. Learners will induce a flow of electricity in a wire using a moving bar magnet and measure this flow using a galvanometer, or Am meter. Through discussion, this activity can then be related to magnetic... (View More)
# Compass Needles around a Simple Circuit
This is an activity about electromagnetism. Learners will set up a simple circuit using a battery, wire, and knife switch, and then use a compass to map the magnetic field lines surrounding the wire. Next, they will add a coil of wire to the simple... (View More)
# Exploring Magnetic Fields in Your Environment
This is a lesson about measuring magnetic field directions of Earth and in the environment. First, learners go outside, far away from buildings, power lines, or anything electrical or metal, and use compasses to identify magnetic North. Next, they... (View More)
Keywords: electromagnetism
Audience: Middle school
# Mapping Magnetic Field Lines
This is an activity about electromagnetism. Learners will use a compass to map the magnetic field lines surrounding a coil of wire that is connected to a battery. This activity requires a large coil or spool of wire, a source of electricity such as... (View More)
# Induction in an Aluminum Can
This activity demonstrates Lenz's Law, which states that an induced electromotive force generates a current that induces a counter magnetic field that opposes the magnetic field generating the current. In the demonstration, an empty aluminum can... (View More)
Keywords: Magnetism; Lenz's Law
Audience: High school
Materials Cost: \$1 - \$5 per group of students
# What is Magnetism?
This is an activity about magnetism and magnetic forces. Learners will explore objects to which a magnet is attracted or repelled and record information in a learning log. Additionally, learners will identify magnetic devices or phenomena they... (View More)
Audience: High school
# Playing with Magnetism
This is an activity about magnetism and magnetic forces. Learners will explore objects to which a magnet is attracted or repelled, and investigate the attractive and repulsive forces of two like and two unlike magnetic poles. This is the first... (View More)
Audience: Elementary school
# Science Conference
This is an activity about sunspots and solar flares. Learners will work collaboratively to create abstracts detailing their knowledge of sunspots and solar flares and then present their work to the entire audience. A background understanding of our... (View More)
# Science and Engineering Conference
This is an activity about how to measure the interplanetary magnetic field, or IMF. Learners will act as scientists and engineers at a conference to explain their discoveries from earlier activities of the larger resource where they designed their... (View More)
Audience: Middle school
Materials Cost: Free
# Learning About Space Weather
This is an activity about the magnetic fields of the Sun and Earth, and the interplanetary magnetic field, or IMF. Learners will engage in a question and answer dialogue, make connections using bar magnet examples and overhead transparencies, and... (View More)
Audience: Middle school
Materials Cost: \$1 - \$5 per group of students
# Measuring the IMF
This is an activity about measuring the interplanetary magnetic field, or IMF. Learners will utilize cardboard boxes with a magnet inside to design a spacecraft, and experiment with ways to attach a magnetometer that will measure the IMF rather than... (View More)
Audience: Middle school
Materials Cost: \$5 - \$10 per group of students
# Universal Time
This is a lesson plan for an activity that explores time zone math. Learners will translate their local time to times in other zones around the world and work with the concept of Universal Time, specifically in reference to the reporting,... (View More)
# The Wandering Magnetic Pole
This is an activity about Earth's magnetism. Using polar coordinates and several sets of provided information, learners will plot the position of the magnetic north pole to investigate its movement over time. This is the sixth activity in the... (View More)
Audience: Middle school
# Mapping Magnetic Fields
This is an activity about magnetism. Using bar magnets, classroom materials, and a compass, learners will explore how bar magnets interact with one another and with other materials, use a compass to find the direction north, and use various... (View More)
# Model a Magnet
This is a summative activity about magnets. Learners will observe a demonstration of the action of a magnet on a test tube of iron filings, answer questions, and, using the concepts learned in previous activities, write an essay about their... (View More)
# Journey to the Stars Educator's Guide: Activities for Grades 9-12
This is an educators guide to accompany the Journey to the Stars planetarium show. The materials for grades 9-12 include suggestions for activities to do before and after viewing the planetarium show, including class discussions about units of... (View More)
# What are Magnetic Fields?
This is an activity about magnetic fields. Using iron filings, learners will observe magnets in various arrangements to investigate the magnetic field lines of force. This information is then related to magnetic loops on the Sun's surface and the... (View More)
Audience: High school
# Exploring Magnetic Fields
This is an activity about magnetic fields. Using iron filings, learners will observe magnets in various arrangements to investigate the magnetic field lines of force. This information is then related to magnetic loops on the Sun's surface and the... (View More)
Audience: Elementary school
# Electrolysis of Salt Water
This is a hands-on lab activity about the chemical composition and conductivity of water. Working in groups, learners will: conduct an experiment involving the process of electrolysis, prepare an experiment to better understand the process of ion... (View More)
Keywords: Electrolytes; Salinity
Audience: High school
Materials Cost: \$1 - \$5 per group of students
# Journey to the Stars Educator's Guide: Activities for Grades 6-8
This is an educators guide to accompany the Journey to the Stars planetarium show. The materials for grades 6-8 include suggestions for activities to do before and after viewing the planetarium show, including class discussions about our Solar... (View More)
Audience: Middle school, Informal education
Materials Cost: Free
# Our Magnetic Sun Outreach Toolkit
This is a collection of outreach resources about the Sun that are meant to be used in informal education settings. This toolkit was originally designed for NASA Night Sky Network member clubs and the Astronomical Society of the Pacific's Astronomy... (View More)
Audience: Informal education
1 | 1,690 | 8,282 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2015-18 | longest | en | 0.846242 |
https://www.quantumstudy.com/the-moment-of-inertia-of-a-metre-scale-of-mass-0-6-kg-about-an-axis-perpendicular-to-the-scale-and-located-at-the-20-cm-position-on-the-scale/ | 1,657,180,183,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656104683708.93/warc/CC-MAIN-20220707063442-20220707093442-00248.warc.gz | 979,210,738 | 12,408 | # The moment of inertia of a metre scale of mass 0.6 kg about an axis perpendicular to the scale and located at the 20 cm position on the scale in kg m^2 is
Q: The moment of inertia of a metre scale of mass 0.6 kg about an axis perpendicular to the scale and located at the 20 cm position on the scale in kg m^2 is (Breadth of the scale is negligible)
(a) 0.074
(b) 0.104
(c) 0.148
(d) 0.208
Ans: (b) | 126 | 406 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.578125 | 3 | CC-MAIN-2022-27 | latest | en | 0.884165 |
http://www.newton.dep.anl.gov/askasci/phy99/phy99x25.htm | 1,386,496,346,000,000,000 | text/html | crawl-data/CC-MAIN-2013-48/segments/1386163059081/warc/CC-MAIN-20131204131739-00035-ip-10-33-133-15.ec2.internal.warc.gz | 423,888,555 | 4,550 | Pendulum Swings ```Name: N/A Status: N/A Age: N/A Location: N/A Country: N/A Date: N/A ``` Question: A grandfather clock with a metal pendulum is located on a person's deck. Can you explain why the clock loses time in the summer and gains time in the winter. Replies: If the grandfather clock mechanism is of the relatively inexpensive variety, there is no temperature correction for the pendulum. In the summer when it is warmer, the pendulum expands and is longer. This causes a slightly longer period, hence the clock runs slow (literally). The converse is true in the colder winter weather. The pedulum contracts just a little bit, and has a shorter period. Precision pendulum clocks have glass enclosed collumns of mercury on the base of the swinging platform of the pendulum. When the pendulum cable would expand, lowering the center of mass, the mercury collumn would also expand, but upward. This upward expansion raises the center of mass just enough to cancel the lowering of the center of mass due to the cable's expansion. l l l ---- cable l l l l l l l l l l l l l ---- mercury collumn l l l l---l---l e tolman If you're thinking the different temperature changes the length of the pendulum, and this changes the rate of the clock -- forget it. A pendulum takes the same amount of time to tick-tock whatever its length. I don't think you have done enough fact-collecting yet to theorize. You clearly suspect the change in temperature is the cause. Test this. Move the clock indoors and see if the change persists. You don't have to do it for a full year: just move it inside for a few days during the summer, when the inside is bound to be cooler, and see if the clock starts speeding up. Or else look at how fast the clock goes during the day and during the night on any day, since the night will typically cooler than the day. There are certainly a lot of parts in a clock which change shape and size slightly in response to temperature, and these could affect the clock's accuracy. But there are other possibilities, too. It could be humidity, for example, which will change from summer to winter. Perhaps dew condenses on the clock, and is burned off faster by the earlier rising sun in the summer. And so on. Dr. c. Grayce This sounds like a homework problem. No sane person would keep such a valuable piece of furniture outside in the weather. Most likely, it would rust in the spring, and never keep time again. So that you can answer this question yourself, what determines the period of a pendulum? Its length? Its mass? Its kinetic energy? What would temperature changes effect? Richard Barrans Jr., Ph.D. The period (or time) of a pendulum swing is related to the mass of the pendulum and the length of the pendulum. The mass will not change during the seasons; however, the length does. In winter, the length shortens and the pendulum swings faster. The opposite occurs in the summer when the pendulum increases in length and the pendulum takes longer to swing. I suggest you go to the library and look in a physics book on the laws governing the period of oscillation of a pendulum (there is an equation that will tell you in mathematical terms what I just mentioned). Galilao is probably the most famous scientist who investigated the laws. Have fun. Dr. C. Murphy Click here to return to the Physics Archives
NEWTON is an electronic community for Science, Math, and Computer Science K-12 Educators, sponsored and operated by Argonne National Laboratory's Educational Programs, Andrew Skipor, Ph.D., Head of Educational Programs.
For assistance with NEWTON contact a System Operator (help@newton.dep.anl.gov), or at Argonne's Educational Programs | 820 | 3,734 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.3125 | 3 | CC-MAIN-2013-48 | longest | en | 0.921044 |
https://forexwaihui.com/blog/2019/08/29/triangle-and-wedge-patterns-in-technical-analysis/ | 1,660,921,784,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882573699.52/warc/CC-MAIN-20220819131019-20220819161019-00513.warc.gz | 256,795,995 | 23,694 | The Triangle pattern appears on different charts rather frequently. The classic technical analysis considers it a pattern signifying the continuation of the trend; however, in my opinion, this pattern may equally work in line with or against the existing trend.
## Description of the Triangle pattern
There are several types of the Triangle, each of them having its own specific features. On the chart, a Triangle is composed of the converging (less often diverging) support and resistance lines. To draw a Triangle, four points are to be marked on the chart, which are two subsequent maximums and two subsequent minimums; through these points, the sides of the Triangle (the support and resistance lines) are drawn. As a rule, five waves form inside the Triangle before it is broken through. After the price breaks one of the sides of the Triangle away, there is likely to appear a strong impulse (movement) towards the breakaway. It is similar to a spring that is squeezed inside the Triangle tighter and tighter until it shoots up or down.
## Types of the Triangle and their execution
There are four types of Triangle in technical analysis.
### Equilateral, or Symmetrical Triangle
A universal pattern that can either continue the trend or go against it. It is formed by the converging support and resistance lines. The following picture emerges: the bears are gradually pushing the price downwards while the bulls are pushing it upwards from the support line. In the end, some of them turn out stronger, and the price breaks through the border of the Symmetrical Triangle, gathering Stop Losses and pending orders on its way.
A position should be opened in the direction of the breakaway after the price closes outside the borders of the Symmetrical Triangle. If the upper border is broken away, we buy, placing a limiting Stop Loss on the closest minimum of the Triangle; the execution is most likely to be the size of the base of the Triangle (the biggest wave) — let us call it H (the value in points). If the lower border of the Triangle is broken away, we sell, placing a Stop Loss on the closest maximum of the Triangle; the size of the execution of the pattern — the H value in points — will be the base of the Triangle (the biggest wave).
### Ascending Triangle
This is the chart pattern of the uptrend continuation, though a reversal execution is sometimes possible. The Ascending Triangle forms between the horizontal resistance level and the ascending support line. In the uptrend, the bulls run into a strong resistance level that they fail to overcome at once. From this level, the price makes pullbacks downwards, which form the waves of the Ascending Triangle. Gradually, they become weaker, and at some moment the bulls, having bought all the bearish Sell orders, break this level away upwards, gathering Stop Losses and pending Buy orders.
Upon a breakaway of the upper border of the Ascending Triangle buying is recommended; the Stop Loss is placed below the closest minimum of the Triangle, the execution is the size of the Triangle base H (in points), which is the biggest wave.
### Descending Triangle
This is the chart pattern continuing a downtrend, though it may sometimes execute against the trend. It is formed by the descending resistance line and the horizontal support level. In a downtrend, the bears bump into a strong support level, which they fail to break through at once. Then several pullbacks from this level upwards follow, forming the Descending Triangle. In the end, the bears sweep all buying orders of the bulls away and break the support level through top-down, gathering Stop Losses and pending Sell orders. After the lower border of the Descending Triangle is broken through it is recommended to sell; Stop Loss is placed above the closest maximum of the Triangle; the size of the execution (H) is the size of the Triangle base, which is its biggest wave.
### Diverging Triangle
This chart pattern is opposite to the Symmetrical Triangle; it may signal either the end of the current trend or vice versa, demonstrate a correction, after which the trend will resume. It is formed by the diverging support and resistance lines; the angle is looking to the left. The pattern is traded same as the Converging Triangles: to my mind, it is not really wise to trade it at the border breakaway as the latter happens at the base of the Triangle after the biggest wave, and a technically correct Stop Loss for the closest maximum/minimum will be too large, the stop/profit ratio almost 1 : 1.
So, a position should be entered in advance: if we are expecting growth, we should enter upon the bounce off the support line; if we are waiting for a decline, then — after a bounce off the resistance line. In this case the stop/profit ratio will be comfortable because the Stop Loss at the closest minimum/maximum will be small while the execution (profit) — much bigger, at least, spreading to the opposite border of the Diverging Triangle, and if it is broken through — as big as the latest biggest wave of the Triangle (base) as well. For a better understanding of trading in diverging price formations (such as the Diverging Triangle), I can recommend the article: "The Wolfe Waves: Description And Trading Strategy".
## Description and execution of the Wedge pattern
Normally, the Wedge is considered a reversal pattern, forming on maximums and minimums of a price chart in an up- or downtrend. A Wedge is quite similar to a Triangle, forming between the two converging support and resistance lines. The main difference between the two patterns is the inclination of the two lines and the pattern itself: all the lines are inclined either upwards or downwards. In accordance with the inclination of the sides, the Wedge may be of two types: an ascending and descending ones.
### Ascending Wedge
It is formed by the two converging and ascending lines of support and resistance. If the Ascending Wedge forms on the maximums of a price chart in an uptrend, it signals a probable reversal or correction. Upon a breakaway of the lower border of the Wedge, selling is recommended with a Stop Loss above the closest maximum of the Wedge and the execution sized as the H base (the biggest wave at the base of the pattern).
If an Ascending Wedge forms in a downtrend upon the renewal of the minimums, it signifies a correction, and in case the lower border of the pattern is broken through the downtrend is likely to continue. After a breakaway of the lower border of the Wedge selling is recommended, a Stop Loss is placed above the closest maximum, execution is sized as the H base of the Wedge (the biggest wave at the base of the pattern).
### Descending Wedge
This Wedge pattern is formed by the two converging and descending lines of support and resistance. If a Descending Wedge forms on the minimums of a price chart in a downtrend, it signifies a possible correction or even a reversal. In case the upper border of the pattern (the resistance line) is broken away, buying is recommended, with a Stop Loss below the closest minimum. A profit is locked after growth to the size of the H base (the biggest wave at the base of the pattern).
If Descending Wedge forms after renewing the maximums in an uptrend, it may signal the continuation of the trend. In this case a Descending Wedge is a correction after the growth; in case the upper border is broken away, buying is recommended, with a Stop Loss below the closest minimum of the pattern; the execution is expected at least the size of the H base (the biggest wave at the base of the pattern).
## Summary
The Triangle and Wedge chart patterns of technical analysis are rather frequent to appear on charts and may be rather helpful in assessing the perspectives of future price movements. The probability of their execution seems to me rather high, and they are worth including into the portfolio. Only, practice is needed in finding patterns on the price chart and reacting on all other factors, such as the current trend, the stop/profit ratio, and fundamental factors.
Material is prepared by
#### Victor Gryazin
Has traded in financial markets since 2004. The knowledge and experience he has acquired constitute his own approach to analyzing assets, which he is happy to share with the listeners of RoboForex webinars. | 1,718 | 8,337 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2022-33 | latest | en | 0.947513 |
https://bainite.wordpress.com/2008/12/18/emergency-number/ | 1,534,370,635,000,000,000 | text/html | crawl-data/CC-MAIN-2018-34/segments/1534221210362.19/warc/CC-MAIN-20180815220136-20180816000136-00690.warc.gz | 637,037,814 | 18,680 | • ## Bainite
It's not pearlite or martensite. A blog written by Mathew Peet.
## Emergency Number
My department has two emergency numbers, one for office hours (8.45-5.15) and one for the rest of the time. The numbers are 67760 for office hours and 31818 for the rest of the time. In lunch time I’m not sure what we should do.
If we add the number we get 67760 + 31818 = 99578
which is 421 short of 99999. 99999 and 421 are easy to remember. So if you can remember this helpful other numbers and one of the emergency numbers you should easily be able to work out the other one. So now you only have to remember either 31818 or 67760. | 173 | 637 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2018-34 | longest | en | 0.917593 |
https://electronics.stackexchange.com/questions/671633/why-are-coils-of-wire-used-in-heating-vs-a-straight-wire/671634 | 1,713,224,753,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817033.56/warc/CC-MAIN-20240415205332-20240415235332-00755.warc.gz | 206,749,275 | 44,595 | # Why are coils of wire used in heating vs a straight wire?
Attached to this you will find a picture of a bathroom heating element. I noticed that the wires that emit heat are coils. From what I understand this effectively makes them inductors. Why is an inductor preferred over a long straight wire?
• Filament lightbulbs have to fit even more wire into an even smaller space, so they use coiled coils. Jun 24, 2023 at 22:46
• Why is an inductor preferred? ... please do not make an assumption that inductance is the desired property Jun 24, 2023 at 23:24
Three benefits to coiled resistance wire used in heating:
1. Thermal expansion. There is no strain on the terminals.
2. Easier to tune the resistance to the required voltage and power requirements for almost any application.
3. Easier to distribute heat over a surface because of the coil diameter.
There are applications where straight wires are used. Bread toasters is an example.
You are correct in observing that the inductance of the coil increases. However a coil is not required for inductance. Only current is required. Coiling increases the inductance in a compact form.
In the case for resistance heaters, the resistance is the dominant parameter. When applying pulse width modulation, the inductive kick must be handled.
• This is a good answer, but I think your last sentence could do with some clarification as there are both resistive and inductive heaters. I believe that it's resistance that causes the heat in both cases, but in the case of inductive heating it's the resistance of the object in which the current is induced rather than in the coil itself. Jun 26, 2023 at 13:36
• Sure, but the OP is clearly confused about inductive vs resistive heating so I think it's worth the clarification. It's your answer, so it's up to you though. Jun 26, 2023 at 15:35
• Thermal expansion is a big point. When I had to make a linear wire heater, I had to add a spring to maintain tension as it heated. That wasn't even all that hot, but it did need decent tension (I needed a hot wire cutter to cut 150mm-thick foam in such a hurry I could only make it from my stash of junk) Jun 26, 2023 at 16:16
• I had forgotten about those cutters. I have seen stationary thermal wire cutters for foam board that use a counter weight on a polley system to maintain a precise tension You can use the position of the counter weight to measure temperature. @ChrisH. Jun 26, 2023 at 16:22
• @RussellH that's a neat idea. I used a tent pole spring on the other end of a lever, which along with the guide fence was made from scrapwood. But it wasn't table-mounted (a mattress would need a big table). Jun 26, 2023 at 18:43
• From what I understand this effectively makes them inductors.
Not at line frequencies. At 50 or 60Hz the contribution of their inductance to their behavior is so minuscule as to only be theoretical. Moreover, because it's resistance wire, at whatever radio frequencies that their inductance does become meaningful, they wouldn't be very high quality inductors.
Mostly, they're coiled for mechanical convenience, as mentioned in @RusselH's answer.
They are not inductors, they are just coiled so they can fit a longer wire. If it was straight from A to B it would draw too much current. They coil it so they can make them the correct resistance (and therefore power) for whatever power rating the element is designed for.
• Thinner wire could also reduce the current, but very thin wire is fragile. Jun 24, 2023 at 22:46
• A short length of straight wire would have a very small surface area. It would need to be heated to incandescence to radiate any significant heat, which would shorten its life and illuminate the room rather than heat it. Jun 25, 2023 at 22:31
• @Paul_Pedant but a light bulb filament is a coiled coil Jun 26, 2023 at 19:34
• @DDuck Sure, but the question is about a heating element, and that needs to output radiation in the infrared range. A lamp needs to emit mainly white light. For any resistive load, there is a whole balance between supply voltage, unit resistivity, cross-sectional area, length, and required temperature (and note the resistivity is extremely temperature-sensitive). Once you get that stuff optimally worked out, you fit the element in the available space by coiling it as much as you need. Inductance and capacitance have no significant effect here. Jun 27, 2023 at 8:23
A coiled wire makes things more thermally and mechanically stable. Under heat, wire tends to expand. A straight wire would become slack and in danger of touching the next wire. A coil under tension will be able to contain the slack. If one area of the coil gets hotter than another, it will become looser, and with the windings gaining distance, the temperature goes down due to less inter-coil heating.
As a rule, at mains frequency (and voltage), any significantly effective inductivity will require an iron-plated core; the "inductivity" of the depicted coils will become noticeable only at RF frequencies.
Langmuir in 1912 worked out that flow of gas over the filament cooled it, and that there was a stationary layer of gas surrounding it. (Langmuir sheath)
A coil has more filament inside the sheath and less convective cooling and glows hotter. The hotter filament is a better radiator of heat (Stefan-Boltzmann's law) and perhaps a better match to the bathroom type of heater shown in the OP's photo. A radiant heat source (over a convective) is also better for items such as toasters.
The heater coil is typically mounted horizontally so the amount of cooling is uniform over its length.
The amount of surface area per unit length of heater coil is lower than a straight heater element and requires less support points for the same radiant output.
• But coiled heater wires are also used in fan heaters and freestanding convection heaters, both of which rely on heating the air to heat the room. Heat guns and hair dryers too. Jun 27, 2023 at 14:07 | 1,387 | 5,968 | {"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} | 2.609375 | 3 | CC-MAIN-2024-18 | latest | en | 0.96199 |
https://proficientessayhelp.com/tco-b-suppose-the-governor-of-california-has-proposed-decreasing-toll-rates-on-californias-toll-roads-and-has-presented-two-possible-scenarios-to-implement-these-increases-the-following-are/ | 1,611,842,688,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610704847953.98/warc/CC-MAIN-20210128134124-20210128164124-00117.warc.gz | 472,634,047 | 22,976 | ###### Topic 1: Business Ethics: Contracts with Foreign Businesses Currently, limited international laws exist in areas such as copyright/patent protection…
September 8, 2020
###### One futures contract is traded where both the long and short parties are closing out existing positions. what is the resultant change in the open interest?
September 9, 2020
TCO B) Suppose the governor of California has proposed decreasing toll rates on California’s toll roads and has presented two possible scenarios to implement these increases. The following are projected data for the two scenarios for the California toll roads. (20 points)
Scenario 1: Toll rate in 2012: \$10.00. Toll rate in 2016: \$8.00.
#### How many pages is this assigment?
For every 100 cars using the toll roads in 2012, 140 cars will use the toll roads in 2016.
Scenario 2:
Toll rate in 2012: \$10.00. Toll rate in 2016: \$9.00.
For every 100 cars using the toll roads in 2012, 120 cars will use the toll roads in 2016.
(Part A) Using the midpoint formula, calculate the price elasticity of demand for Scenario 1 and Scenario 2. (10 points)
(Part B) Assume 50,000 cars use California toll roads every day in 2012. What would be the daily total revenue received for each scenario in 2012 and in 2016? (6 points)
(Part C) Is demand under Scenario 1 and under Scenario 2 price elastic, inelastic, or unit elastic? Briefly explain.
## “Looking for a Similar Assignment? Get Expert Help at an Amazing Discount!”
The post TCO B) Suppose the governor of California has proposed decreasing toll rates on California’s toll roads and has presented two possible scenarios to implement these increases. The following are project appeared first on Graduate Paper Help. | 407 | 1,731 | {"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} | 2.8125 | 3 | CC-MAIN-2021-04 | latest | en | 0.898076 |
https://www.physicsforums.com/threads/mechanical-advantage-help.885565/ | 1,508,663,378,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187825154.68/warc/CC-MAIN-20171022075310-20171022095310-00287.warc.gz | 1,005,164,177 | 17,628 | 1. Sep 16, 2016
### Armando Valle
What is the mechanical advantage of a nail puller where you exert a force 45 cm from the pivot and the nail is 1.8 cm on the other side? What minimum force must you exert to apply a force of 1250 N to the nail?
2. Relevant equations
2. Relevant equations
MA= FO/FI * LI/LO
3. The attempt at a solution
Ive tried to make the problem but i dont understand it, can someone help me? thanks!
2. Sep 16, 2016
### Staff: Mentor
What have you tried? What's the geometry of a nail puller? Have you drawn a diagram and labelled the forces? A "pivot" implies that torques may be involved. What have you done to investigate this?
3. Sep 16, 2016
### Armando Valle
Ive tried to use the MA equation but i don´t find a relationship with the forces included on the problem. this are the images of the problems, thanks!
#### Attached Files:
• ###### Screen Shot 2016-09-16 at 6.36.46 PM.png
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4. Sep 16, 2016
### Staff: Mentor
I don't understand your relevant equation ("MA equation") as presented. Can you define the variables?
5. Sep 16, 2016
### Armando Valle
this are the equations
#### Attached Files:
• ###### Screen Shot 2016-09-16 at 6.57.44 PM.png
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6. Sep 16, 2016
### Staff: Mentor
does not reflect what is in your latest image. It should be:
MA= FO/FI = LI/LO
Note the "=" between the terms. So you're dealing with ratios.
Can you pair up the given values in your problem with the variables in your equation?
7. Sep 16, 2016
### SammyS
Staff Emeritus
The figure for OP's problem together with said equations. | 456 | 1,620 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.046875 | 3 | CC-MAIN-2017-43 | longest | en | 0.886705 |
http://bootmath.com/find-the-limit-lim_x-to-1-leftfracp1-xp-fracq1-xqright-p-q-0.html | 1,529,824,161,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267866888.20/warc/CC-MAIN-20180624063553-20180624083553-00404.warc.gz | 48,333,213 | 5,908 | # Find the limit $\lim_{x \to 1} \left(\frac{p}{1-x^p} – \frac{q}{1-x^q}\right)$ $p ,q >0$
I Know series expansion and L’Hospital’s rule . But here both of them are not of any help.
#### Solutions Collecting From Web of "Find the limit $\lim_{x \to 1} \left(\frac{p}{1-x^p} – \frac{q}{1-x^q}\right)$ $p ,q >0$"
$$x^m-1=(x-1)(x^{m-1}+x^{m-2}+\ldots+x+1)\implies$$
$$\lim_{x\to1}\frac p{1-x^p}-\frac q{1-x^q}=\lim_{x\to1}\frac{\frac p{x^{p-1}+\ldots+x+1}-\frac q{x^{q-1}+\ldots+x+1}}{1-x}=$$
$$\stackrel{\text{l’Hospital}}=-\lim_{x\to1}\left(-p\frac{(p-1)x^{p-2}+\ldots+1}{(x^{p-1}+\ldots+x+1)^2}+q\frac{(q-1)x^{q-2}+\ldots+1}{(x^{q-1}+\ldots+x+1)^2}\right)=$$
$$=\frac{p\frac{(p-1)p}2}{p^2}-\frac{q\frac{(q-1)q}2}{q^2}=\frac{p-1}2-\frac{q-1}2=\frac{p-q}2$$
If $p=q$ then the result will be $0$.
If $p \neq q$, then applying L’Hospital rule twice,
\begin{align} &\lim _{x\rightarrow 1}{\frac {p}{1-{x}^{p}}}-{\frac {q}{1-{x}^{q}}}\\ =&\lim_{x \to 1}{\frac {p-p{x}^{q}-q+q{x}^{p}}{ \left( -1+{x}^{p} \right) \left( -1+{x}^{q} \right) }}\\ =&\lim_{x \to 1}{\frac {xpq \left( {x}^{q-1}-{x}^{p-1} \right) }{{x}^{p}p-{x}^{p+q}p+{x}^{q}q-{x}^{p+q}q}}\\ =&\lim_{x \to 1}{\frac {xpq \left( q{x}^{q-1}-p{x}^{p-1} \right) }{{x}^{p}{p}^{2}-{x}^{p+q}{p}^{2}-2\,{x}^{p+q}pq+{x}^{q}{q}^{2}-{x}^{p+q}{q}^{2}}}\\ =& \frac{p-q}{2}. \end{align}
Hint $\lim_{x \to 0} \frac{a^x-1}{x}=\ln(a)$ can you some how edit it to use it to get the resulting answer.
Given limit can be set equivalent to by using $x=e^{t}$ and bit of simplification.
$$\lim_\limits{\large{t\to0}}\frac{1}{t}\left(\dfrac{(e^{pt}-1)}{tp}-\dfrac{e^{qt}-1}{tq}\right)$$
Let $$l=\lim_{x\rightarrow 1}\left(\frac{p}{1-x^p}-\frac{q}{1-x^q}\right)\;, p,q>0$$
Now put $\displaystyle x= \frac{1}{t}\;,$ Then $$l=\lim_{t\rightarrow 1}\left(-\frac{pt^p}{1-t^p}+\frac{qt^q}{1-t^q}\right) = \lim_{t\rightarrow 1}\left[\frac{p(1-t^p-1)}{1-t^p}-\frac{q(1-t^q-1)}{1-t^q}\right]$$
So $$l=(p-q)-l\Rightarrow l = \frac{p-q}{2}$$
Here’s a rather standard series expansion method: rewriting $x$ as $1+(x-1)$, and using Newton’s generalised binomial expansion,
\begin{align}
\frac{p}{1-x^p}-\frac{q}{1-x^q} & = \frac{p}{1-(1+(x-1))^p}-\frac{q}{1-(1+(x-1))^q} \\
& = \frac{p}{1-(1+p(x-1)+\frac{p(p-1)}{2}(x-1)^2+O(|x-1|^3))} \\
&-\frac{q}{1-(1+q(x-1)+\frac{q(q-1)}{2}(x-1)^2+O(|x-1|^3))} \\
& = \frac{1}{(x-1)+\frac{q-1}{2}(x-1)^2+O(|x-1|^3)} \\
&-\frac{1}{(x-1)+\frac{p-1}{2}(x-1)^2+O(|x-1|^3)} \\
& = \frac{1}{x-1}(\frac{1}{1+\frac{q-1}2(x-1)+O(|x-1|^2)}-\frac{1}{1+\frac{p-1}2(x-1)+O(|x-1|^2)})\\
& = \frac1{x-1}(\frac{p-1}2(x-1)-\frac{q-1}{2}(x-1)+O(|x-1|^2)) \\
& = \frac{p-q}2.
\end{align}
Note that this requires no condition on $p,q$. They can even be negative. | 1,371 | 2,709 | {"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} | 4.375 | 4 | CC-MAIN-2018-26 | latest | en | 0.33628 |
https://people.sc.fsu.edu/~jburkardt/f_src/pyramid_jaskowiec_rule/pyramid_jaskowiec_rule.html | 1,713,620,191,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817650.14/warc/CC-MAIN-20240420122043-20240420152043-00354.warc.gz | 412,833,522 | 2,027 | pyramid_jaskowiec_rule, a Fortran90 code which returns symmetric quadrature rules, with exactness up to total degree 20, over the interior of a pyramid in 3D, by Jan Jaskowiec, Natarajan Sukumar.
The integration region is:
``` - ( 1 - Z ) <= X <= 1 - Z
- ( 1 - Z ) <= Y <= 1 - Z
0 <= Z <= 1.
```
When Z is zero, the integration region is a square lying in the (X,Y) plane, centered at (0,0,0) with "radius" 1. As Z increases to 1, the radius of the square diminishes, and when Z reaches 1, the square has contracted to the single point (0,0,1).
### Languages:
pyramid_jaskowiec_rule is available in a C version and a C++ version and a Fortran90 version and a MATLAB version and an Octave version and a Python version.
### Related Data and Programs:
pyramid_exactness, a Fortran90 code which computes the monomial exactness of a quadrature rule over the interior of a pyramid in 3D.
pyramid_felippa_rule, a Fortran90 code which returns a Felippa quadrature rule for approximating integrals over the interior of a pyramid in 3d.
pyramid_grid, a Fortran90 code which computes a grid of points over the interior of the unit pyramid in 3D;
pyramid_integrals, a Fortran90 code which returns the exact value of the integral of any monomial over the interior of the unit pyramid in 3d.
pyramid_monte_carlo, a Fortran90 code which uses the Monte Carlo method to estimate the integral of a function over the interior of the unit pyramid in 3d.
pyramid_rule, a Fortran90 code which computes a conical product quadrature rule over the interior of the unit pyramid in 3D;
pyramid_witherden_rule, a Fortran90 code which returns a Witherden quadrature rule, with exactness up to total degree 10, over the interior of a pyramid.
### Reference:
High order cubature rules for tetrahedra and pyramids,
International Journal of Numerical Methods in Engineering,
Volume 121, Number 11, pages 2418-2436, 15 June 2020. | 509 | 1,915 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-18 | latest | en | 0.831416 |
https://20somethingfinance.com/cost-of-air-conditioning-vs-a-fan/ | 1,720,780,173,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514387.30/warc/CC-MAIN-20240712094214-20240712124214-00749.warc.gz | 51,235,138 | 32,702 | # The Cost of Air Conditioning Vs. a Fan
## An American Love Affair with Air Conditioning (AC)
Summer is here. Time to rev up the air conditioning units. Or is it? Americans are addicted to AC. We DEMAND it in our homes, our offices, our shopping malls, our cars. According the U.S. Dept. of Energy, about 88% of U.S. homes have air conditioning. By contrast, only 47% of U.S. homes had AC in 1973.
Admittedly, I’ve become a bit spoiled when it comes to air conditioning. I have a very hard time sleeping comfortably unless it’s about 70 degrees or less. It can get pretty muggy in Michigan. Ever since I was about 10 years old, I have lived with air conditioning. The one exception was in college when I was in the dorms for two miserably hot, muggy, and often sleepless years.
But an article in the New York Times that got me thinking a bit about our wasteful ways, my wasteful ways – “Bringing in the Big Fans“. And it made me want to run some numbers on the cost of air conditioning versus a simple old fan.
## The Cost of Air Conditioning Vs. Fans
According to the article:
• a good window AC unit runs on 1.2 kilowatts and costs 14 cents an hour to run
• a three-ton central air unit (a common cooling system), runs on about 3 kilowatts and costs about 36 cents an hour to run
• a good ceiling fan? It draws only 30 watts to run, costing about 1 cent per three hours of use
One penny for every three hours?!
I knew that fans used less energy than AC units, but to use less than 1% of the energy? Hmmm….. maybe I need to re-think my love for air conditioning.
## Running the AC versus Fan Cost Figures Over a Month
Putting these hourly costs into more practical terms made sense, so I decided to run the math. If you assume that your unit is running half of the day during summer months,
• a window AC unit would cost \$50.40 per month
• a central AC unit would cost \$129.60 per month
• each ceiling fan would cost \$1.20 per month
I could potentially be saving over \$128 per month by putting up a ceiling fan in my bedroom and weening myself off of AC? Amazing!
Even just installing a window unit in my room so that I can sleep well (the biggest reason I’m addicted to AC in the summer) would save me \$78 per month, probably more if I only use it at night.
## Additional Benefits to Getting Rid of AC
I love AC, don’t get me wrong. But lately I have been opening the windows more often and going without. I like the outdoor sounds and really like the fresh air. And, according to the EPA, “studies found levels of about a dozen common organic pollutants to be 2 to 5 times higher inside homes than outside, regardless of whether the homes were located in rural or highly industrial areas.”
And when you consider that 17% of all electricity consumed at home in the U.S. is to power AC units and that most of that electricity is coming from carbon-rich coal, maybe it’s time we re-thought our love for AC. It’s hard to justify AC if you call yourself an environmental steward.
Whether you switch to a ceiling fan, a window unit, a whole house fan, or some combination of, you may want to start thinking about your love of air conditioning. I will. Stay tuned.
## Air Condition Versus Fans Discussion:
• How do you keep yourself cool, especially at night?
• Have you switched from AC to another method of staying cool? What did you do?
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33. tyrone loukas | 1,001 | 3,908 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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 | 0.946331 |
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8_Primes II (Posted on 2013-10-15)
No Solution Yet Submitted by brianjn No Rating
Comments: ( Back to comment list | You must be logged in to post comments.)
computer solution Comment 1 of 1
The program:
DEFDBL A-Z
OPEN "8primesb.txt" FOR INPUT AS #1
'OPEN "8primes2.txt" FOR OUTPUT AS #2
mindiffdg = 999
CLS
DO
LINE INPUT #1, l\$
a\$ = MID\$(l\$, 2, 3)
b\$ = MID\$(l\$, 7, 3)
c\$ = MID\$(l\$, 12, 3)
a1 = VAL(LEFT\$(a\$, 1)): a2 = VAL(MID\$(a\$, 2, 1)): a3 = VAL(RIGHT\$(a\$, 1))
b1 = VAL(LEFT\$(b\$, 1)): b2 = VAL(MID\$(b\$, 2, 1)): b3 = VAL(RIGHT\$(b\$, 1))
c1 = VAL(LEFT\$(c\$, 1)): c2 = VAL(MID\$(c\$, 2, 1)): c3 = VAL(RIGHT\$(c\$, 1))
IF a3 + b2 + c1 = 4 THEN
IF a1 + a2 + a3 = 11 THEN
IF b1 + b2 + b3 = 17 THEN
IF c1 + c2 + c3 = 5 THEN
PRINT a\$: PRINT b\$: PRINT c\$
ct = ct + 1
END IF
END IF
END IF
END IF
LOOP UNTIL EOF(1)
CLOSE 1
'CLOSE 2
PRINT ct
goes through the output from the original program and finds:
641
809
311
Note that it was not necessary to check every total. The requirements were added as necessary to narrow the answer to one solution.
Posted by Charlie on 2013-10-15 20:42:09
Search: Search body:
Forums (0) | 494 | 1,254 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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 | latest | en | 0.699194 |
https://socratic.org/questions/5a40cfb97c014951ff09988f | 1,627,574,340,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046153860.57/warc/CC-MAIN-20210729140649-20210729170649-00585.warc.gz | 526,150,832 | 8,447 | # Question 9988f
Dec 26, 2017
Here's what I got.
#### Explanation:
The idea here is that the nitrite anion, ${\text{NO}}_{2}^{-}$, acts as a weak base in aqueous solution, so right from the start, you should expect to find
$\text{pH" > 7" " and " " "pOH} < 7$
The nitrite anions are delivered to the solution by the soluble sodium nitrite in a $1 : 1$ mole ratio, so you know that you start with
["NO"_ 2^(-)]_0 = "0.125 M"
Now, when the nitrite anion is present in aqueous solution, the following equilibrium is established
${\text{NO"_ (2(aq))^(-) + "H"_ 2"O"_ ((l)) rightleftharpoons "HNO"_ (2(aq)) + "OH}}_{\left(a q\right)}^{-}$
The base dissociation constant, ${K}_{b}$, of the nitrite anion can be calculated by using the fact that an aqueous solution at ${25}^{\circ} \text{C}$ has
$\textcolor{b l u e}{\underline{\textcolor{b l a c k}{{K}_{a} \cdot {K}_{b} = 1 \cdot {10}^{- 14}}}}$
In your case, the acid dissociation of the nitrous acid, ${\text{HNO}}_{2}$, is equal to $6.5 \cdot {10}^{- 4}$, so you can say that the nitrite anion will have
${K}_{b} = \frac{1 \cdot {10}^{- 14}}{6.5 \cdot {10}^{- 4}} = 1.54 \cdot {10}^{- 11}$
Now, by definition, the base dissociation constant is equal to
${K}_{b} = \left(\left[{\text{HNO"_2] * ["OH"^(-)])/(["NO}}_{2}^{-}\right]\right)$
Notice that the nitrite anions react in $1 : 1$ mole ratios to produce nitrous acid and hydroxide anions, so if you say that $x$ $\text{M}$ represents the equilibrium concentrations of the two products, you can say that the equilibrium concentration of the nitrite anions will be equal to
${\left[{\text{NO"_ 2^(-)] = ["NO}}_{2}^{-}\right]}_{0} - x$
This basically means that in order for the reaction to produce $x$ $\text{M}$ of nitrous acid and hydroxide anions, it must consume $x$ $\text{M}$ of nitrite anions.
This means that you have
${K}_{b} = \frac{x \cdot x}{0.125 - x}$
Since the value of the base dissociation constant is so small compared with the initial concentration of the nitrite anions, which means that the equilibrium lies far to the left, you can use the approximation
$0.125 - x \approx 0.125$
You will thus have
${K}_{b} = {x}^{2} / 0.125$
which will get you
x = sqrt( (0.125 * 1.54 * 10^(-11))
$x = 1.39 \cdot {10}^{- 6}$
Since $x$ represents the equilibrium concentration of the hydroxide anions, you can say that you have
["OH"^(-)] = color(darkgreen)(ul(color(black)(1.39 * 10^(-6)color(white)(.)"M")))
Using the fact that an aqueous solution at ${25}^{\circ} \text{C}$ has
color(blue)(ul(color(black)(["H"_3"O"^(+)] * ["OH"^(-)] = 1 * 10^(-14)color(white)(.)"M"^2)))
you can say that this solution will have
["H"_3"O"^(+)] = (1 * 10^(-14) "M"^color(red)(cancel(color(black)(2))))/(1.39 * 10^(-6)color(red)(cancel(color(black)("M")))) = color(darkgreen)(ul(color(black)(7.19 * 10^(-9)color(white)(.)"M")))
Consequently, you will have
"pOH" = - log(["OH"^(-)])
$\text{pOH} = - \log \left(1.39 \cdot {10}^{- 6}\right) = \textcolor{\mathrm{da} r k g r e e n}{\underline{\textcolor{b l a c k}{5.86}}}$
and
"pH" = - log (["H"_3"O"^(+)])
$\text{pH} = - \log \left(7.19 \cdot {10}^{- 9}\right)$
$\text{pH} = \textcolor{\mathrm{da} r k g r e e n}{\underline{\textcolor{b l a c k}{8.14}}}$
Notice that you get the same result by using
$\textcolor{b l u e}{\underline{\textcolor{b l a c k}{\text{pH + pOH} = 14}}}$
The equilibrium concentrations are rounded to three sig figs, but I'll leave the $\text{pH}$ and the $\text{pOH}$ of the solution rounded to two decimal places instead of three.
As predicted, the $\text{pH}$ of the solution is indeed $> 7$, which is consistent with the fact that the nitrite anion acts as a weak base in aqueous solution. You can also say that sodium nitrite is a basic salt.
$\textcolor{w h i t e}{\frac{a}{a}}$
A MORE ACCURATE APPROACH
An interesting thing to keep in mind here is that pure water already contains hydronium and hydroxide anions $\to$ think the auto-ionization of water here.
At ${25}^{\circ} \text{C}$, pure water has
color(blue)(ul(color(black)(["H"_3"O"^(+)] = ["OH"^(-)] = 1 * 10^(-7)color(white)(.)"M")))
Now, notice that the equilibrium concentration of hydroxide anions that we calculated
["OH"^(-)] = 1.39 * 10^(-6)color(white)(.)"M"
is actually quite small and comparable to the initial concentration of hydroxide anions, i.e. $1 \cdot {10}^{- 7}$ $\text{M}$.
To account for this, you can use the fact that the solution already contains hydroxide anions when you dissolve the salt. So if you use $x$ $\text{M}$ to represent the equilibrium concentration of nitrous acid, you can say that the equilibrium concentration of hydroxide anions will be
$1 \cdot {10}^{- 7} + x$
This means that the concentration of hydroxide anions starts at $1 \cdot {10}^{- 7}$ in pure water and increases by $x$ $\text{M}$ when $x$ $\text{M}$ of nitrite anions react.
The base dissociation constant will now be
${K}_{b} = \frac{x \cdot \left(1 \cdot {10}^{- 7} + x\right)}{0.125 - x}$
Using the same approximation as before, you will end up with
${x}^{2} + 1 \cdot {10}^{- 7} \cdot x - 0.125 \cdot 1.54 \cdot {10}^{- 11} = 0$
This quadratic equation will produce two solutions, one positive and one negative. Since we're looking for concentration here, you can discard the negative solution to say that
$x = 1.34 \cdot {10}^{- 6} \textcolor{w h i t e}{.} \text{M}$
This time, you will have
["OH"^(-)] = 1.34 * 10^(-6)color(white)(.)"M"
["H"_3"O"^(+)] = 7.46 * 10^(-9)color(white)(.)"M"
$\text{pOH} = 5.87$
$\text{pH} = 8.13$
Now, because the percent error that you get by disregarding the initial concentration of hydroxide anions is < color(red)(5%)
"% error" = (1.39 * color(red)(cancel(color(black)(10^(-6)color(white)(.)"M"))) - 1.34 * color(red)(cancel(color(black)(10^(-6)color(white)(.)"M"))))/(1.34 * color(red)(cancel(color(black)(10^(-6)color(white)(.)"M")))) * 100% = 3.73% < color(red)(5%)#
you can use the values that you got without accounting for the initial concentration of the hydroxide anions. | 2,031 | 6,022 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 64, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.578125 | 4 | CC-MAIN-2021-31 | latest | en | 0.841204 |
https://origin.geeksforgeeks.org/breaking-cryptography/ | 1,680,161,941,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296949107.48/warc/CC-MAIN-20230330070451-20230330100451-00085.warc.gz | 489,123,794 | 36,813 | Open in App
Not now
# Breaking Cryptography
• Last Updated : 17 Jun, 2022
In this article, we will discuss the overview of symmetric encryption protocols and how the need for Asymmetric Encryption came into the picture. Here we will discuss the overview, cryptography examples, and rules for breaking cryptography. Let’s discuss them one by one.
Overview :
A key scrambles the messages and that very key can decrypt the messages. Then, Alice and Bob can exchange messages once they agree on a key. But how can they both agree on a secret key over a network that is monitored by eavesdroppers? Let’s assume that the eavesdropper is Eaves! Now, to resolve the problem we use the asymmetric key instead.
Example –
Alice encrypts her message with Bob’s public key because only Bob can decrypt the message with his very own private key. Similarly, Bob encrypts his message with the public key of Alice can only be decrypted by Alice’s private key. Now, Alice can also use this method to authenticate herself to Bob. She can encrypt her message with her private key and then encrypt that result with Bob’s public key. When Bob decrypts the message with his own private key, he sees that there already exists an inner layer of encryption. Now, if this can be encrypted with Alice’s public key then he’ll be sure that it was Alice who encrypted it.
Generating a private key :
Let’s now get into the essence of how these keys were generated in the first place as follows.
Approach –
1. Generating a private key is computationally very difficult to produce. It simply cannot be produced by reverse-engineering the public key produced by an algorithm called the RSA algorithm.
2. What is the protocol of the RSA algorithm?
3. We produce a number(N) which is a product of two prime numbers(let’s say p and q)
4. Now, the number produced is quite large (nearly a hundred to a thousand digits).
`p ∗q = N`
Example –
Let’s understand it by one simple example as follows.
Step-1 :
By using the above protocol let us define n using the multiplication of two prime numbers. (Taking a prime number directly can make the calculations complex.)
```p=61, q =53.
compute N.
N = p*q
N = 3233
Applying Euler's Totient function,
∅(n) = ∅(p*q)
= ∅(p) * ∅(q)
= (p-1)*(q-1)
= 60*52
= 3120```
Step-2 :
We didn’t take N directly because it would have become complex for us to compute a number of numbers that are co-prime to it.
```Choose 'e' ; 1<=e<= ∅(n), co prime to ∅(n)
Going by that, if we take e = 17
Then, Gcd(17,3120) should be = 1
(that means the numbers should be co-prime to each other.)
our public key is :(e,N)= (17,3233)```
Step-3 :
Now let’s determine the private key : (d, N)
```e d = 1 mod ∅(n)
here e and ∅(n) are co primes to each other
Now going by the concept of Multiplicative Inverse we have,
d = e-1 mod ∅(n)
Finding d:
d is calculated using the formula:
= [(∅(n) * i)+1]/e```
Step-4 :
We keep on calculating it by changing the values of i. Here we took I as 1,2,3…, etc. till we reach an answer where we get a non-decimal number.
```at i = 15, we get d as 2753
hence our private key becomes:
(d,N)=(2753,3233)```
Step-5 :
After generating the public and private keys we perform encryption and decryption.
```(m e)d ≡ m mod N,
where
m - It is the message. Such computation is a one way function.```
The strength of the RSA algorithm lies in N. The larger the number N, the slower it gets to take out its prime factors.
Breaking Cryptography :
Having discussed this, let’s get into how can we break cryptography!
1. Cracking open secure systems would be easy if we knew how to factor in huge numbers.
2. RSA uses these prime factors as keys to decrypt these messages. For that, we’ll need to find out the prime factors of a really huge number!
3. Now, this can get quite boring and time-consuming.
4. We’ll need to have a strategic plan to factor in them. Euler, a famous mathematician came to the rescue. He contemplated over prime numbers and
5. It gave various postulates like modular arithmetic. Basically, he gave all the math under RSA cryptography.
6. Modular Arithmetic:
7. Here we consider the remainder when dividing two numbers.
```a ≡ x mod N
when we divide a/N, the remainder is --> x
2 × 3 = 6
6/5 = 1
so,
2 × 3 mod 5 is 1
its written as:
2*3 ≡ 1 mod 5```
Relation between exponentiation and modular arithmetic :
Let’s look at the relation between exponentiation and modular arithmetic as follows.
Let’s understand with the help of examples.
1. If we take powers of 3 (3k) we’ll get the numbers as 3,9.27,81,243
2. Now, if we take the mod of the numbers with 10, we get a sequence as 9,7,1,3,9,7,1,3…so on.
3. We observe that the sequence of power increases, but the modular cycle repeats itself.
4. We also observe that the last digit of the pattern is always 1.
5. Now, as long as x and N are co primes, x mod N, x2 mod N, x3 mod N … all of these will hold this property.
6. We call the length of the repeating pattern period.
7. Here, the period of 3 mod 10 is 4 (as after every fourth digit the pattern repeats).
8. The period is the smallest number ‘r’, such that,
`x r ≡ 1 mod N`
Calculating the factor of a number :
Now let’s see how modular arithmetic, exponentiation, and period are related to calculating factors of the number.
Let’s say you are given a number N and you are also given its factors, p, and q. Nothing is known about p and q.
`N = p .q`
Step-1 :
Pick up any number smaller than N. Let’s call it ‘a’. Check whether a and N co primes. We check this by taking the GCD of both. GCD(a, N)=1, then they are co-prime to each other.
Step-2 :
The second step is to compute the period of a mod N. In this example, we take N = 32, and we assume a = 8 (since they both are co-prime to each other.) Let’s assume the period to be ‘r’.
```powers of 8 = 8,64,512,4096.....
mod 35 = 8,29,22,..1
We compute r to be as 4.```
To make the arithmetic workout, we’ll need to divide r by 2. So, we need to know that r is even. Later, we’ll also need to know that ar/2 + 1 ≇ 0 mod N. If either of these things fails, we’ll need to pick a different ‘a’ in the starting step.
Step-3 :
For step 3, we have to do some algebra as follows. Let’s start with the things that we know.
```ar ≡ 1 mod N
which on subtracting 1, gives
ar - 1 ≡ 0 mod N```
Saying something as above is equivalent to saying that it is a multiple of N(since the remainder obtained here is 0.) So, there must exist some integer ‘k’ such that:
```ar - 1 = k .N
Since we assumed 'r' as an even number here, we can write it as :
(a r/2 - 1 )(a r/2 +1 )= k. N
Since N=p .q, so we replace N with p . q
(a r/2 - 1 ) (a r/2 +1 ) = k. p .q
Since, we have the period of 8 mod 35 as 4, we can write it as:
84 ≡ 1 mod 35
so, 84 - 1 ≡ 0 mod 35 (its equivalent to saying that its a multiple of N)
So we can re write it as: 84 - 1 = k . 35
rewriting this as (since r is even here):
(8r/2 - 1)(8r/2 + 1) = k. p .q (where p .q are the prime factors of N)```
Step-4 :
Now, step 4 is to calculate for p & q. We assume here that GCD((ar/2 – 1), N) is one of the prime factors. Let’s call it p. And, GCD((ar/2 + 1), N) is another prime factor. Let’s call it q. In equation (ar/2 – 1)(ar/2+ 1) = k.p. q we make the assumption that p must divide one of the factors on the left side and q must also divide one of the factors on the left side. Both of them can’t divide the same factors on the left side since that factor would be divisible by N.
Shor’s Algorithm :
We make assumptions as follows.
```p divides (ar/2 - 1) and q divides (ar/2 + 1)
We get, p = Gcd(63,35) = 7
and q is the Gcd(65,35) = 5```
Let’s have a quick recap of what we did as follows.
1. Take ‘a’ less than N.
2. Find the period of a mod N
3. Rearrange with algebra
4. Let p = Gcd ((ar/2– 1), N) and q = Gcd ((ar/2 + 1), N)
And you are done! You have destroyed the security system. But we know that RSA still works and taking out the period takes a long time. But Here, comes the role of Quantum Computers. They are excellent at calculating the period. The above was the outline of Shor’s Algorithm.
My Personal Notes arrow_drop_up
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http://www.convexoptimization.com/wikimization/index.php?title=Gurobi_Mex:_A_MATLAB_interface_for_Gurobi&diff=1867&oldid=1866 | 1,545,166,451,000,000,000 | text/html | crawl-data/CC-MAIN-2018-51/segments/1544376829812.88/warc/CC-MAIN-20181218204638-20181218230638-00468.warc.gz | 346,849,465 | 8,360 | # Gurobi Mex: A MATLAB interface for Gurobi
(Difference between revisions)
Revision as of 21:28, 19 January 2010 (edit)m (→Tested platforms)← Previous diff Revision as of 21:31, 19 January 2010 (edit) (undo)m (→Input Description)Next diff → Line 70: Line 70: * '''A''': constraint coefficient matrix, double, sparse. * '''A''': constraint coefficient matrix, double, sparse. * '''b''': constraint right-hand side vector, double. * '''b''': constraint right-hand side vector, double. - * '''contypes''': constraint types. Char array of '>', '<', '='. Example: '>><=' means the first two constraints have ">= "signs, the third has "<=" sign, and the last is an equality constraint. Warning: '>=' means two constraints instead of one an inequality constraint. + * '''contypes''': constraint types. Char array of '>', '<', '='. + :Example: '>><=' means the first two constraints have ''greater or equal to'' signs, the third has ''less than or equal to'' sign, and the last is an equality constraint. + :Warning: '>=' means two constraints instead of one an inequality constraint. * '''lb''': variable lower bound vector, double. Empty means 0 lower bound. * '''lb''': variable lower bound vector, double. Empty means 0 lower bound. * '''ub''': variable upper bound vector, double. Empty means no (or inf) upper bound. * '''ub''': variable upper bound vector, double. Empty means no (or inf) upper bound. - * '''vartypes''': variable types. Char array of chars 'C', 'B', 'I', 'S', 'N'. C for continuous; B for binary; I for integer; S for semi-continuous; N for semi-integer. Empty means all variables are continuous. Example: 'CCCCC' stands for five continuous variables. Note that semi-continuous variables are variables that must take a value between their minimum and maximum or zero. Semi-integer variables are similarly defined. + * '''vartypes''': variable types. Char array of chars 'C', 'B', 'I', 'S', 'N'. C for continuous; B for binary; I for integer; S for semi-continuous; N for semi-integer. Empty means all variables are continuous. + :Example: 'CCCCC' stands for five continuous variables. + :Note that semi-continuous variables are variables that must take a value between their minimum and maximum or zero. Semi-integer variables are similarly defined. * '''options''': ''optional'' structure that may contain one or more of the following fields: (see [http://www.gurobi.com/html/doc/refman/node378.html Gurobi's parameter help] for their allowed values. Also, see [[Gurobi_Mex:_A_MATLAB_interface_for_Gurobi#Three_Examples|examples]] below.) * '''options''': ''optional'' structure that may contain one or more of the following fields: (see [http://www.gurobi.com/html/doc/refman/node378.html Gurobi's parameter help] for their allowed values. Also, see [[Gurobi_Mex:_A_MATLAB_interface_for_Gurobi#Three_Examples|examples]] below.) ** '''options.IterationLimit''': see Gurobi's parameter help ** '''options.IterationLimit''': see Gurobi's parameter help
## Revision as of 21:31, 19 January 2010
Gurobi Mex is a MATLAB interface for Gurobi 2 written by Wotao Yin. It calls Gurobi to solve linear/mixed-integer optimization problems. Gurobi is one of the leading linear and mixed integer programming solvers.
The interface is open source and subject to Creative Commons Attribution-Share Alike 3.0 United States License. It is a tool for MATLAB users to quickly call Gurobi, and its source code serves as a start point for those who want to develop a customized MATLAB interface for Gurobi.
Its current version is 1.05 published on Jan 2, 2010.
## Model
```min/max c'x
subject to
Ax [>= / <= / =] b,
lb <= x <= ub,
x(i) is [continuous / binary / integer / semi-continuous / semi-integer].
```
### History
v1.05 Major bug fix: char array of constraint sense has been fixed
v1.04 support writing model to files in various formats such as MPS, REW, LP, ...
v1.03 support log file
v1.02 fixed a memory leak issue
v1.01 update: support output dual solution lambda; allow vartypes to be empty (for all continuous variables).
v1.00 initial version.
### Building Gurobi Mex in MATLAB
#### Under Windows
```mex -O -I"<gurobi include path>" gurobi_mex.c "<absolute path/gurobi20.lib>"
```
#### Under Unix
```mex -O -I"<gurobi include path>" gurobi_mex.c -L"<gurobi lib path>" -lgurobi20
```
#### Tested platforms
• Windows 32-bit and gcc (included in free Mingw/GnuMex)
• Ubuntu Linux 9.10 64-bit and gcc.
MATLAB's built-in lcc cannot link with Gurobi.
Please make sure the dynamic library of Gurobi 2.x is in system path.
## Syntax
x = gurobi_mex(c, objtype, A, b, contypes, lb, ub, vartypes);
x = gurobi_mex(c, objtype, A, b, contypes, lb, ub, vartypes, options);
[x,val] = gurobi_mex(...);
[x,val,flag] = gurobi_mex(...);
[x,val,flag,output] = gurobi_mex(...);
[x,val,flag,output,lambda] = gurobi_mex(...);
## Input Description
• c: objective coefficient vector, double.
• objtype: 1 (minimization) or -1 (maximization).
• A: constraint coefficient matrix, double, sparse.
• b: constraint right-hand side vector, double.
• contypes: constraint types. Char array of '>', '<', '='.
Example: '>><=' means the first two constraints have greater or equal to signs, the third has less than or equal to sign, and the last is an equality constraint.
Warning: '>=' means two constraints instead of one an inequality constraint.
• lb: variable lower bound vector, double. Empty means 0 lower bound.
• ub: variable upper bound vector, double. Empty means no (or inf) upper bound.
• vartypes: variable types. Char array of chars 'C', 'B', 'I', 'S', 'N'. C for continuous; B for binary; I for integer; S for semi-continuous; N for semi-integer. Empty means all variables are continuous.
Example: 'CCCCC' stands for five continuous variables.
Note that semi-continuous variables are variables that must take a value between their minimum and maximum or zero. Semi-integer variables are similarly defined.
• options: optional structure that may contain one or more of the following fields: (see Gurobi's parameter help for their allowed values. Also, see examples below.)
• options.IterationLimit: see Gurobi's parameter help
• options.FeasibilityTol: see Gurobi's parameter help
• options.IntFeasTol: see Gurobi's parameter help
• options.OptimalityTol: see Gurobi's parameter help
• options.LPMethod: see Gurobi's parameter help
• options.Presolve: see Gurobi's parameter help
• options.Display: controls the level of screen output. 0 for no output; 1 for no output unless error; 2 (default) for normal output.
• options.UseLogfile: this field (no matter what value is assigned to it) lets Gurobi generate a log file. The file name is "gurobi_mex.log" by default. To change the name, modify field LOGFILENAME in the source file.
• options.WriteToFile: char array; contains the name of the file to which optimization data is written. See Gurobi C-Reference entry GRBwrite for supported formats. This option helps one verify whether the model is correctly passed to Gurobi.
## Output Description
x: primal solution vector; empty if Gurobi encounters errors or stops early (in this case, check output flag).
val: optimal objective value; empty if Gurobi encounters errors or stops early.
flag: value meanings:
```1 for not started
2 for optimal
3 for infeasible
4 for infeasible or unbounded
5 for unbounded
6 for objective worse than user-specified cutoff
7 for reaching iteration limit
8 for reaching node limit
9 for reaching time limit
10 for reaching solution limit
11 for user interruption
12 for numerical difficulties
```
output: structure contains the following fields
```output.IterCount: number of Simplex iterations
output.Runtime: running time in seconds
output.ErrorMsg: contains Gurobi error message, if any
output.lambda: Lagrange multipliers; DO NOT ask for this output for MIPs.
```
## Three Examples
### Example 1
#### A linear program (borrowed from MATLAB's linprog help)
```min –5 x1 – 4 x2 –6 x3,
subject to
x1 – x2 + x3 ≤ 20
3 x1 + 2 x2 + 4 x3 ≤ 42
3 x1 + 2 x2 ≤ 30
0 ≤ x1, 0 ≤ x2, 0 ≤ x3.
```
MATLAB code:
```c = [-5; -4; -6];
objtype = 1;
A = sparse([1 -1 1; 3 2 4; 3 2 0]);
b = [20; 42; 30];
lb = zeros(3,1); % same as lb = [];
ub = [];
contypes = '<<<';
vtypes = []; % same as vtypes = 'CCC'; empty means 'C...C'
```
```clear opts
opts.IterationLimit = 20;
opts.FeasibilityTol = 1e-6;
opts.IntFeasTol = 1e-5;
opts.OptimalityTol = 1e-6;
opts.LPMethod = 1; % 0 - primal, 1 - dual
opts.Presolve = -1; % -1 - auto, 0 - no, 1 - conserv, 2 - aggressive
opts.Display = 1;
opts.UseLogfile = []; % as long as opts.UseLogfile exists, a log file is created
opts.WriteToFile = 'test_gurobi_mex_LP.mps';
```
```[x,val,exitflag,output,lambda] = gurobi_mex(c,objtype,A,b,contypes,lb,ub,vtypes,opts);
```
Results:
```x' =
0 15 3
val =
-78
exitflag =
2
output =
IterCount: 2
Runtime: 0
ErrorMsg: []
lambda' =
0 -1.5000 -0.5000
Log file: gurobi_mex.log.
MPS file: test_gurobi_mex_LP.mps.
```
### Example 2
#### Integer program (borrowed from mip1_c.c of Gurobi 2.0)
```max x + y + 2z,
subject to
x + 2 y + 3 z <= 4
x + y >= 1
x, y, z binary.
```
MATLAB code:
```c = [1; 1; 2];
objtype = -1; % 1 for minimize, -1 for maximize
A = sparse([1 2 3; 1 1 0]);
b = [4; 1];
lb = [];
ub = [];
contypes = '<>';
vtypes = 'BBB';
```
```clear opts
opts.IterationLimit = 20;
opts.FeasibilityTol = 1e-6;
opts.IntFeasTol = 1e-5;
opts.OptimalityTol = 1e-6;
opts.LPMethod = 1; % 0 - primal, 1 - dual
opts.Presolve = -1; % -1 - auto, 0 - no, 1 - conserv, 2 - aggressive
opts.Display = 1;
opts.UseLogfile = []; % as long as opts.UseLogfile exists, a log file is created
opts.WriteToFile = 'test_gurobi_mex_MIP.mps';
```
```[x,val,exitflag,output] = gurobi_mex(c,objtype,A,b,contypes,lb,ub,vtypes,opts);
```
Gurobi does not give lambda (Pi, or Lagrange multipliers) for MIPs, without calling modelfix().
Results:
```disp('Solution:');disp(x')
disp('Optimal obj value:');disp(val)
disp('Exit flag:');disp(exitflag)
disp('Optimization info:');disp(output)
```
```Solution:
1 0 1
Optimal obj value:
3
Exit flag:
2
Optimization info:
IterCount: 0
Runtime: 0
ErrorMsg: []
Log file: gurobi_mex.log.
MPS file: test_gurobi_mex_MIP.mps.
```
### Example 3
A compressive sensing example. See example m-file test_gurobi_mex_CS.m in the package.
## Feedback
I would be delighted to hear from you if you find Gurobi Mex useful, or if you have any suggestions, contributions, or bug reports. Please send these to Wotao Yin (wotao.yin AT rice.edu)
## How to cite
Wotao Yin. Software: Gurobi Mex, URL: www.caam.rice.edu/~wy1/gurobi_mex, 2009-2010. | 3,064 | 10,596 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.5625 | 3 | CC-MAIN-2018-51 | latest | en | 0.852845 |
https://www.sidefx.com/docs/houdini/nodes/chop/vector.html | 1,723,347,552,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640843545.63/warc/CC-MAIN-20240811013030-20240811043030-00052.warc.gz | 755,154,078 | 15,038 | # Vector channel node
Performs vector operations on a set or sets of channels.
Vector math has several useful operations for use with channels, such as projections, dot products and distance/length calculations. This can be used for determining whether objects are facing each other, how far they are apart, or the angle between them.
This CHOP separates input channels into vectors based on the Vector Mask parameter, and applies a vector operation to each set individually. Two inputs can be used for binary vector operations. If the second input does not have as many vectors as the first, Vector cycles through the second input’s vectors.
Some vector operations produce a scalar result. In this case, the vector channels are removed and replaced with a single channel, whose name depends on the operation performed (such as 'dist', 'dot' or 'angle').
Additionally, an optional test can be done, comparing the vector operation results to a constant value or other vector channels. The result of this test can be used to remove vector channels or to produce zero/one channels representing where these vectors fail or pass the test. The Test tab contains the parameters for the vector test.
## Vector ¶
Operation
Specifies which vector operation to perform on the vectors.
Magnitude of A
Computes the magnitude of A and replaces the vector channels with a single scalar channel, 'len'. If more than one vector is processed, the wildcard replacement is appended to 'len' (ie, (tx3,ty3,tz3) would produce len3). Only one input is required for this operation; the second input is ignored.
Normalize A
Reduces the vector to a unit-length vector. Only one input is required for this operation; the second input is ignored.
Distance between A and B
Computes the straight-line distance between A and B, as if they were points. This replaces the vector channels with a single channel named 'dist'.
A dot B
Computes the dot product of a vector from input 1 (A) with a vector from input 2 (B). If more vectors exist in A than B, the B vectors cycle back to the first vector. This replaces the vector channels with a single channel named 'dot'.
Normalized A dot B
Computes the dot product of A and B, but normalizes A and B first. This replaces the vector channels with a single channel named 'dot'.
Angle between A and B
Computes the angle between the two vectors, A and B, in degrees. This replaces the vector channels with a single channel named 'angle'.
A cross B
Computes the cross product of A and B. Both vectors must have exactly
three components. The cross product is the vector that is orthogonal to both A and B.
Scalar Project B onto A
The scalar projection of B onto A determines the length of the vector produced when B is projected onto A. This replaces the vector channels with a single channel named 'proj'.
Vector Project B onto A
Projects the vector B onto the vector A, so that the result is parallel to A, and the magnitude is proportional to the angle between A and B.
Adds vector A to B.
Subtract
Subtracts vector A from B.
Reverse Subtract
Subtracts vector B from A.
No operation
Passes vector A through as-is. Generally only useful when used with Perform Vector Test to compare vectors.
A set of patterns which describe the members of the vectors. A wildcard `*` can be used to match more than one vector. A single vector will contain one channel from each pattern with the same wildcard replacement (ie, `tx* ty* tz*` would process vectors like `(tx1,ty1,tz1)` and `(txa,txb,txc)`.
An optional parameter which sets the vector mask for vectors in the second input (B vectors), in case they happen to represent a different vector quantity (such as N, P or V). If disabled, the Vector Mask is used.
The vector mask for the third comparison input, similar to the above two. If disabled, the Vector Mask is used.
Align
For two-input vector operations, this parameter specifies which frame range to use if the inputs' frame ranges differ, and how to sample the inputs.
Extend to Min/Max
Find the earliest start and latest end, and extend all inputs to that range using the extend conditions.
Stretch to Min/Max
Find the earliest start and latest end, and stretch every input channel’s start and end to that range.
Shift to Minimum
Find the earliest start and shift all input channels so they all start at that index. All channels are extended to the length of the longest one.
Shift to Maximum
Find the latest end and shift all input channels so they all end at that index. Extend all channels to the length of the longest one.
Shift to First Interval
Shift all input channels to the start of the first channel and sample all inputs using the first input’s range.
Trim to First Interval
Trim all input channels to first channel’s range.
Stretch to First Interval
Stretch all input channels to the first channel’s range.
Trim to Smallest Interval
Trim all input channels to the smallest start/end interval. The start and end values may not come from the same channel.
Stretch to Smallest Interval
Stretch all channels to the smallest start/end interval. The start and end values may not come from the same channel.
## Test ¶
A comparison test can be run after the initial vector operation, which can remove vectors that fail the test or produce a channel containing the samples which passed or failed for each vector.
Perform Vector Test
Enables the vector test on the result of the vector operation, which compares the result to a constant value or another input.
Result Vector Test
Describes how the vector is compared, with one of the inequality operations listed.
Test Against
The vector result is compared to a constant (Vector Constant) or vector/scalar channels from one of the three inputs.
Vector Constant
This parameter is the constant when Test Against is set to Constant. It will either have all fields enabled (if the vector result is a vector quantity) or only the first field enabled (if the result is a scalar).
Test Result
The output of the vector test can be the original vectors, modified vectors, or an on/off channel containing the test results.
Keep Original Vector A
If the test passes, the original A Vector is kept in the output. If the test failed, the vector channels are removed. This is useful for eliminating points or directions that fail the test in order to thin out the number of channels for faster downstream cooking.
Keep Result Vector
If the test passes, the result of the vector operation is kept in the output. If the test failed, it is removed.
Pass/Fail Channel
A single channel is created called 'test' for each vector, with the suffix of the vector appended to it. This channel is 1 for samples where the test passed, and 0 for where it failed. This can be used for controlling a Blend or Switch CHOP.
Pass If
When determining whether to keep or discard a channel (when Test Result is Keep Original Vector or Keep Result Vector), the test may pass for some samples and fail for others. Since a channel can only be entirely removed or entirely kept, vector channels that pass on some samples and fail on others must be resolved.
All Samples Pass
All samples in the vector test must pass the test in order for the vector channels to be kept.
One Sample Passes
Only one sample needs to pass the vector test in order for the vector channels to be kept.
A Fraction of the Samples Pass
A specified fraction (from 0.0 to 1.0) of the samples must pass the test in order for the vector channels to be kept.
Pass Fraction
If Pass If is set to 'Fraction of the Samples Pass', this sets the fraction of samples which must pass in order for the test to be considered successful. Zero means none need to pass, while 1.0 means that all samples must pass.
## Common ¶
Some of these parameters may not be available on all CHOP nodes.
Scope
To determine the channels that are affected, some CHOPs have a scope string. Patterns can be used in Scope, for example `*` (match all), and `?` (match single character).
The following are examples of possible channel name matching options:
`chan2`
Matches a single channel name.
`chan3 tx ty tz`
Matches four channel names, separated by spaces.
`chan*`
Matches each channel that starts with `chan`.
`t?`
The `?` matches a single character. `t?` matches two-character channels starting with t.
`blend[3-7:2]`
Matches number ranges, giving `blend3`, `blend5`, and `blend7`.
`blend[2-3,5,13]`
Matches channels `blend2`, `blend3`, `blend5`, `blend13`.
`t[xyz]`
`[xyz]`matches three characters, giving channels `tx`, `ty` and `tz`.
Sample Rate Match
The Sample Rate Match options handle cases where multiple input CHOPs’ sample rates are different.
Resample At First Input’s Rate
Use the rate of the first input to resample the others.
Resample At Maximum Rate
Resample to the highest sample rate.
Resample At Minimum Rate
Resample to the lowest sample rate.
Error if Rates Differ
Does not accept conflicting sample rates.
Units
The units of the time parameters.
For example, you can specify the amount of time a lag should last for in seconds (default), frames (at the Houdini FPS), or samples (in the CHOP’s sample rate).
Time Slice
Time slicing is a feature that boosts cooking performance and reduces memory usage. Traditionally, CHOPs calculate the channel over its entire frame range. If the channel needs to be evaluated every frame, then cooking the entire range of the channel is unnecessary. It is more efficient to calculate only the fraction of the channel that is needed. This fraction is the Time Slice.
Causes the memory consumed by a CHOP to be released after it is cooked, and the data passed to the next CHOP.
Export Prefix
The Export Prefix is prepended to CHOP channel names to determine where to export to.
For example, if the CHOP channel was named `geo1:tx`, and the prefix was `/obj`, the channel would be exported to `/obj/geo1/tx`.
Graph Color
Every CHOP has this option. Each CHOP gets a default color assigned to it for display in the graph, but you can override the color with the Graph Color. There are 36 RGB color combinations in the palette.
Graph Color Step
When the graph displays the animation curves, and a CHOP has two or more channels, this defines the difference in color from one channel to the next, giving a rainbow spectrum of colors.
## Locals ¶
C
The current channel (0 to NC-1).
NC
The total number of channels.
# Channel nodes
• Design audio filters and sound materials for the spatial audio system.
• Imports an animation clip from an agent primitive.
• Calculates the area under a channel’s graph, which is the same as calculating the integral of a channel, or integrating the channel.
• Adds, removes or updates attributes of the input chop.
• Receives audio input from the analog audio ports or the digital port.
• A 14-band equalizer which filters audio input channels in the same way that a conventional band equalizer uses a bank of sliders to filter fixed-frequency bands of sound.
• Manually tap the beat of a piece of music, and automatically generate a repeating ramp or pulse that continues to keep time with the music after the taps stop.
• Combines two or more chops in input 2, 3 and so on, by using a set of blending channels in input 1.
• Performs multi-dimensional, example-based interpolation of channels.
• Creates channels from the value of its parameters.
• Contains a VOP network that can manipulate channel data.
• Runs a VEX snippet to modify channel data.
• Layers (blends) the channels of one CHOP on the channels of another CHOP.
• Create up to forty new channels.
• Combines two or more CHOP inputs using a list of weights specified as parameters.
• Returns an Object Local Transform.
• Returns an Object Parent Transform.
• Returns an Object World Transform.
• Constrains rotation so it always points toward a target position.
• Compares two objects and returns information on their relative positions and orientations.
• Compares two objects and returns information on their relative positions and orientations.
• Returns an Object Pretransform.
• Applies an transformation offset after evaluating a constraint.
• Reparent an object.
• Position an object on a path and orient it to the path’s direction.
• Position and Orient an object using point positions from a geometry.
• Combines multiple chops by blending the inputs in sequence.
• Combines two chops by using a single weight specified as a parameter.
• Position and Orient an object using the surface of a geometry.
• Takes translate, rotate, and/or scale channels and transforms them.
• Produces multiple copies of the second input along the timeline of the first input.
• Counts the number of times a channel crosses a trigger or release threshold.
• Creates cycles.
• Delays the input, and can be run in normal or time-sliced mode.
• Removes channels coming from its input.
• Turns data from device inputs into transform data
• Time-warps the node’s first input (source clip) using its second input (reference clip) as a reference.
• Extracts any information from a DOP simulation that is accessible through the dopfield expression function.
• Outputs the maximum amplitude in the vicinity of each sample of the input.
• Fixes discontinuity of rotation data after cracking matrices
• A convenient tool for exporting channels.
• Export Constraints Network on any object
• Export Transforms to Constraints Network of many objects
• Modify input channels by using expressions.
• Only sets the extend conditions of a chop, which determines what values you get when sampling the CHOP before or after its interval.
• (Deprecated) Extracts the current world or local space bone transforms from a geometry object.
• Extracts locomotion from an animation clip.
• (Deprecated) Creates channels from the specified derived transforms, node parameters and CHOP channels for pose-space deformation.
• Reads in channel data from an FBX file.
• Used for controlling other CHOPs.
• Get the state of a chop as it was one frame or time slice ago.
• Imports channels from other CHOPs.
• Imports channels from other OPs.
• Reads in channel and audio files for use by chops.
• Smooths or sharpens the input channels.
• Computes when position channels are stationary.
• Divides the input channels into groups, cooking the contained network for each group.
• Provides more complicated math functions than found in the Math CHOP such as trigonometic functions, logarithmic functions, and exponential functions.
• Turns input values for the gamepad or joystick device into channel outputs.
• Uses a geometry object to choose a sop from which the channels will be created.
• The engine which drives Inverse Kinematic solutions using the Handle object.
• Sample and hold the value of the first input.
• Solves inverse kinematics rotations for bone chains.
• Returns an identity transform.
• Converts rows and/or columns of pixels in an image to CHOP channels.
• Treats its multiple-inputs as keyframes and interpolates between them.
• Generates channels for bone objects based on a bone chain and an end affector.
• Returns an invert transform of the input.
• Creates a jiggling effect in the translate channels passed in.
• Turns key presses into channel output.
• Adds lag and overshoot to channels.
• Mix weighted layers of keyframed animation from multiple Channel CHOPs to a base Channel CHOP.
• Provides a variety of functions to limit and quantize the input channels.
• Converts channels of all its input chops into binary channels and combines them using a variety of logic operations.
• Uses a channel in the first input to index into a lookup table in the second input, and output values from the lookup table.
• The MIDI In CHOP reads Note events, Controller events, Program Change events, and Timing events from both midi devices and files.
• The MIDI Out CHOP sends MIDI events to any available MIDI devices.
• Perform a variety of arithmetic operations on and between channels.
• Takes multiple inputs and merges them into the output.
• Outputs X and Y screen values for the mouse device.
• Turns input values for the Connexion space mouse into channel outputs.
• Post multiplies all the input transformations.
• Similar to the Pipe In/Out CHOPs in Network mode.
• Makes an irregular wave that never repeats, with values approximately in the range -1 to +1.
• Used as a place-holder and does not have a function of its own.
• Compares two objects and returns information on their relative positions and orientations.
• Creates channels representing the transforms for a chain of objects.
• Generates sounds in two ways.
• Marks the output of a sub-network.
• Filters an audio clip, and then applies other audio effects.
• Produces translate and rotate channels to move Objects according to the positions of particles in a POP Network.
• Filters audio input using one of four different filter types.
• Translates english text into a series of phonetic values.
• Pipes data from custom devices into a CHOP, without needing the Houdini Developers' Kit or knowledge of Houdini internals.
• Transmit data out of Houdini to other processes.
• Attempts to extract the fundamental pitch of a musical tone from the input audio.
• Store a transform pose for later use by evaluating the input.
• Computes the difference between two poses.
• Takes translate, rotate, and/or scale channels and transforms them using the pretransform of the given object.
• Generates pulses at regular intervals of one channel.
• Renames channels.
• Reorders the first input CHOP’s channels by numeric or alphabetic patterns.
• Resamples an input’s channels to a new rate and/or start/end interval.
• Takes all its inputs and appends one chop after another.
• This time-shifts a CHOP, changing the start and end of the CHOP’s interval.
• Reorganizes a list of channels.
• Calculates the slope (or derivative) of the input channels.
• The rendering engine for producing 3D audio.
• Calculates the frequency spectrum of the input channels, or a portion of the channels.
• Edit the channel data by using direct manipulation of cubic or Bezier handles in the graph of the CHOP.
• Creates vibrations influenced by the input channels, as if a mass was attached to a spring.
• Caches the input motion in the node on command, and then uses it as the node’s output.
• (Deprecated) Stashes the bone transforms and pose-drivers for use by the Pose-Space Deform SOP and Pose-Space Edit SOP nodes.
• Preserves the shape of channels and the sampling rate, but resamples the channels into a new interval.
• Allows for the simplification of complex networks by collapsing several CHOPs into one.
• Control the flow of channels through a CHOPnet.
• This converts an input node in Current Frame mode to a Time Range mode by re-cooking it multiple times.
• This time-shifts a CHOP, re-cooking the node using different time.
• Takes translate, rotate, and/or scale channels and transforms them.
• Contains a VOP network that can manipulate transform data.
• Combines a chain of translate, rotate, and/or scale channels.
• Adds an audio-style attack/decay/sustain/release (ADSR) envelope to all trigger points in the input channels.
• Shortens or lengthens the input’s channels.
• This function is a sub-set of the waveform CHOP.
• Performs vector operations on a set or sets of channels.
• The Voice Split CHOP takes an audio track and separates words out into different channels.
• The Voice Sync CHOP detects phonemes in an audio channel given some audio phoneme samples and pro…
• Time-warps the channels of the first input (the Pre-Warp Channels) using one warping channel in the second input.
• Creates a waveform that is repeated. | 4,197 | 19,845 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-33 | latest | en | 0.924682 |
https://h1ros.github.io/posts/precision/ | 1,632,529,514,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780057584.91/warc/CC-MAIN-20210924231621-20210925021621-00716.warc.gz | 335,628,097 | 5,504 | # Precision
## Goal¶
This post aims to introduce one of the model evaluation metrics, called Precision score. Precision score is used to measure the prediction ratio of how many of predictions were correct out of the total number of the predictions. As the precision score is higher, the prediction would be high likely true whenever such prediction is made.
Precision score is defined as the following equations:
$${\displaystyle {\text{Precision}}={\frac {True\;Positive}{True\;Positive + False\;Positive}}\,} = \frac {True \;Positive}{total\;\#\,of\;samples\;predicated\;as\;True}$$
Reference
## Libraries¶
In [2]:
from sklearn.metrics import precision_score
import pandas as pd
## Create a prediction and ground truth¶
In [21]:
df_prediction = pd.DataFrame([0, 1, 0, 1, 1 ,1, 1, 1],
columns=['prediction'])
df_prediction
Out[21]:
prediction
0 0
1 1
2 0
3 1
4 1
5 1
6 1
7 1
In [22]:
df_groundtruth = pd.DataFrame([0, 0, 0 , 0, 1, 1, 1, 1],
columns=['gt'])
df_groundtruth
Out[22]:
gt
0 0
1 0
2 0
3 0
4 1
5 1
6 1
7 1
## Compute F1 Score¶
In [24]:
precision_score(y_true=df_groundtruth,
y_pred=df_prediction, average='binary')
Out[24]:
0.6666666666666666
### double check by precision and recall¶
In [19]:
TP = (df_prediction.loc[df_prediction['prediction']==1,'prediction'] == df_groundtruth.loc[df_prediction['prediction']==1,'gt']).sum()
TP
Out[19]:
4
In [25]:
FP = (df_prediction.loc[df_prediction['prediction']==1,'prediction'] != df_groundtruth.loc[df_prediction['prediction']==1,'gt']).sum()
FP
Out[25]:
2
In [26]:
TP / (TP + FP)
Out[26]:
0.6666666666666666 | 495 | 1,584 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.28125 | 3 | CC-MAIN-2021-39 | latest | en | 0.547133 |
https://www.gradesaver.com/textbooks/math/prealgebra/prealgebra-7th-edition/chapter-1-review-page-91/38 | 1,537,843,938,000,000,000 | text/html | crawl-data/CC-MAIN-2018-39/segments/1537267160923.61/warc/CC-MAIN-20180925024239-20180925044639-00200.warc.gz | 757,984,620 | 14,195 | ## Prealgebra (7th Edition)
Published by Pearson
# Chapter 1 - Review - Page 91: 38
50
#### Work Step by Step
$4\underset\uparrow5$ Since the ones place is not less than five, the tens place is inceased by 1 when rounding. The ones place is zeroed out.
After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback. | 109 | 418 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.125 | 3 | CC-MAIN-2018-39 | longest | en | 0.917814 |
https://codeforces.com/problemset/problem/217/E | 1,660,528,049,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882572089.53/warc/CC-MAIN-20220814234405-20220815024405-00137.warc.gz | 191,276,070 | 14,562 | E. Alien DNA
time limit per test
2 seconds
memory limit per test
256 megabytes
input
standard input
output
standard output
Professor Bajtocy is conducting experiments on alien DNA. He has discovered that it is subject to repetitive mutations — each mutation happens in the same way: some continuous subsequence of the alien DNA becomes active, copies itself, the copy gets mangled and inserts itself right after the original subsequence. The mangled copy of the activated continuous subsequence is formed by first joining all the elements at the even positions in that subsequence, and then joining all the elements at the odd ones at the end. That is, if the activated subsequence consists of 11 elements and represented as s1s2... s11, its mangled copy is s2s4s6s8s10s1s3s5s7s9s11.
For example, if the original sequence was "ACTGG" and the mutation happened on the segment [2, 4] (that is the activated subsequence is "CTG"), the mutated DNA is: "ACTGTCGG". The mangled copy of the activated subsequence is marked with bold font.
Professor Bajtocy has written down the original DNA sequence and the mutations that sequentially happened to it, and he now asks you to recover the first k elements of the DNA sequence after all the mutations.
Input
The first line of input contains the original DNA sequence, consisting only of letters "A", "C", "T" and "G" and not exceeding 3·106 in length.
The second line contains a single integer k (1 ≤ k ≤ 3·106).
The third line contains a single integer n (0 ≤ n ≤ 5000) — the number of mutations. The next n lines describe the mutations in chronological order — each mutation is described by two numbers li and ri (1 ≤ li ≤ ri ≤ 109), meaning that the continuous subsequence [li, ri] has become active and cloned itself, joining itself with the mangled copy.
It is guaranteed that the input data is correct, that is, no mutation acts on non-existing elements of the DNA sequence, and the resulting DNA sequence has at least k elements.
Assume that the DNA elements are indexed starting from 1 and that the notation [l, r] meaning the continuous subsequence of DNA sequence that consists of r - l + 1 elements starting at the l-th DNA sequence element and ending at the r-th DNA sequence element.
Output
Output a single line, containing the first k letters of the mutated DNA sequence.
Examples
Input
GAGA40
Output
GAGA
Input
ACGTACGT1621 22 8
Output
ACCAGTACCGACATCG
Note
In the second example, after the first mutation the sequence is "ACCAGTACGT". After the second mutation it's "ACCAGTACCGACATCGT". | 632 | 2,555 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 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} | 2.546875 | 3 | CC-MAIN-2022-33 | latest | en | 0.93949 |
http://www.ck12.org/algebra/Horizontal-and-Vertical-Line-Graphs/ | 1,490,834,284,000,000,000 | text/html | crawl-data/CC-MAIN-2017-13/segments/1490218191405.12/warc/CC-MAIN-20170322212951-00067-ip-10-233-31-227.ec2.internal.warc.gz | 452,649,019 | 16,360 | <img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />
# Horizontal and Vertical Line Graphs
## Line y = # is parallel to x-axis and line x = # is parallel to y-axis
Levels are CK-12's student achievement levels.
Basic Students matched to this level have a partial mastery of prerequisite knowledge and skills fundamental for proficient work.
At Grade (Proficient) Students matched to this level have demonstrated competency over challenging subject matter, including subject matter knowledge, application of such knowledge to real-world situations, and analytical skills appropriate to subject matter.
Advanced Students matched to this level are ready for material that requires superior performance and mastery.
## Horizontal and Vertical Line Graphs
by CK-12 //at grade
Write and graph lines of the form y = a constant value and x = a constant value.
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1
• PLIX
## Horizontal and Vertical Line Graphs
by CK-12 //at grade
Horizontal and Vertical Line Graphs Interactive
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## Horizontal and Vertical Line Graphs
by CK-12 //at grade
Horizontal and Vertical Line Graphs Interactive
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• Video
## Graphs of Horizontal and Vertical Lines: A Sample Application
by CK-12 //at grade
This video demonstrates a sample use of graphs of horizontal and vertical lines.
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## Graphs of Horizontal and Vertical Lines: An Explanation of the Concept
by CK-12 //at grade
This video provides an explanation of the concept of graphs of horizontal and vertical lines.
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• Critical Thinking
## Horizontal and Vertical Line Graphs Discussion Questions
by CK-12 //at grade
A list of student-submitted discussion questions for Horizontal and Vertical Line Graphs.
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• Real World Application
## Doritos Locos Tacos
by CK-12 //at grade
Students will apply their understanding of horizontal and vertical linear graphs by looking at the relationship between the unit cost and amount sold of the most popular Taco Bell product.
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• Real World Application
## Flatlined
by CK-12 //at grade
Find out why "flatlined" is synonymous with "dead."
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0 | 599 | 2,684 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2017-13 | latest | en | 0.851883 |
http://ariya.blogspot.com/2009/08/how-far-is-that-plane.html | 1,475,323,454,000,000,000 | text/html | crawl-data/CC-MAIN-2016-40/segments/1474738662856.94/warc/CC-MAIN-20160924173742-00075-ip-10-143-35-109.ec2.internal.warc.gz | 14,648,721 | 16,738 | My blog has been moved to ariya.ofilabs.com.
## Sunday, August 09, 2009
### how far is that plane
This did happen in my last flight. I consider making it a quick math quiz for any candidate I have to interview in the future (to see whether it takes him 5 minutes or 5 hours to analyze).
A few minutes after the captain announced that we were flying at forty thousand feet above the sea level, cruising at the speed of 900 km/h, through the window we saw another airplane flying at roughly the opposite direction, but at a slightly lower altitude. Of course, the obvious question was asked: How far is the other plane from us? This is better than the typical Fermi problem. Assuming that that airplane was visible in our 90 degrees field of view for 4 seconds and it flew just as fast as our plane, I did a quick calculation and came up with an answer, the flight trajectory of that plane was 1 km apart from ours.
The other guy was impressed, but he was not convinced. Now, since my math skills are rusty, I am sure I missed a thing or two. Now, what is your take?
(Picture rendered from Jarno's public-domain vectorized aircraft).
Dion Moult said...
I got 2km. But don't take it from me I'm not too confident in my math either :P
In fact, I think I'm completely missing the question here. All I did was relative speed = 1800kmph for 4 seconds.
I'm probably wrong.
Beat Wolf said...
i got 1km.
@Dion: what you calculated was the length of your 90degree viewfield. Which is one side of a triangle, the other thing you know is one angle (90) of that triangle, which means the two other sides are the same length.. and yeah, then you go on until you got the height of that triangle(which is the distance to the plane after 2 seconds of seeing it, the question did not specify at what point you want the distance)
Enrico said...
1km for me too ((900+900)/3600*(4/2)/tan(45))
Dion Moult said...
@BeatWolf: ah yes, sorry misunderstood the question ;) Thanks!
Anonymous said...
Since you estimated relative radial velocity (pi/8 rad/s): r = 2*900/3600 *8/pi = 4/pi = 1km.
Zarin said...
Why even use trig? The fact that we know that our triangle edge has an angle of 90' and the opposite side is parallel means we can easily subdivide by hand.
If both planes are travelling at 900km/h in opposite directions then the relative velocity is 1800km/h. Divide by the time taken to pass to get the length of that side of the triangle: 1800/60/60*4 = 2km
As the opposite is parallel we know we have an isosceles triangle, thus we can split the triangle equally length-wise easily to form another triangle. Splitting will get us an angle of 90'/2=45' on our edge, 90' on the far edge and 180'-90'-45'=45' on the side edge... We have another isosceles triangle! As the length of both edges on the isosceles triangle that are adjacent to the far 90' angle are the same length the distance between the two planes is 2km/2 = 1km
Diagram 'cause I suck at descriptions: http://img15.imageshack.us/img15/4717/omgplanes.png
Zarin said...
Correction: I meant multiply by time instead of divide. A speed is distance/time so to cancel the time you need to multiply.
Of course this calculation ignores that the altitudes are slightly different. As the error would probably only be ~10% at most it can be ignored for your use case, however if you want to say shoot out one of the windows of the other plane with a sniper rifle you would need to take altitudes into account (As well as air pressure, wind, bullet velocity/spin, plane vibration and other aerodynamic stuff).
Anselmo Melo said...
I got 1 km too. 1800km/h for 4 seconds give us 2km as the hipotenuse of a triangle with the rectangle angle placed in our aircraft. The height of that triangle is the minimum distance, 1km. Of course it's only an approximate result =)
Anonymous said...
for sure it's 1km..
Anonymous said... | 959 | 3,887 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.921875 | 4 | CC-MAIN-2016-40 | longest | en | 0.974281 |
https://www.tlsdowntown.org/30-of-350000/ | 1,568,642,584,000,000,000 | text/html | crawl-data/CC-MAIN-2019-39/segments/1568514572744.7/warc/CC-MAIN-20190916135948-20190916161948-00486.warc.gz | 1,049,514,310 | 50,269 | # 30% Of 350000
### Contents
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Mortgage Costs for a \$350,000 Home – Amortization Table – For a \$350,000 home, a 20% down payment would be \$70,000. Home Purchasing Fees The buyer of a home will usually be required to pay for an inspection, closing costs and other fees during the closing process.
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What is 30 percent of 300000 – step by step solution – 2) x=30% where left sides of both of them have the same units, and both right sides have the same units, so we can do something like that: 300000/x=100%/30% 6. Now we just have to solve the simple equation, and we will get the solution we are looking for. 7. solution for what is 30% of 300000 300000/x=100/30
soluciones capitulo 6-1 – 1 Gary's Pipe and Steel company expects. – . Economy Sales Probability Expected Outcome Strong \$800,000 .20 \$160,000 Steady 500,000 .50 250,000 Weak 350,000 .30 105,000 Expected level of sales .
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Mortgage Interest Deduction Income Tax Savings Benefit Calculator.estimated mortgage interest Tax Deduction Savings on a 4.5% APR 30-Year. \$50,000 adding a garage then your cost basis on the home becomes \$350,000. | 1,061 | 4,421 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.796875 | 3 | CC-MAIN-2019-39 | longest | en | 0.91464 |
https://philosophy.stackexchange.com/questions/67143/kants-notion-of-synthetic-a-prioiri-as-logical-entailment | 1,627,342,479,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046152156.49/warc/CC-MAIN-20210726215020-20210727005020-00393.warc.gz | 460,758,268 | 38,863 | # Kant's Notion of Synthetic A Prioiri as Logical Entailment
Is there something wrong about interpreting Kant's notion of synthetic a priori statements to be logical entailments?
I understand, I think, that Kant didn't want to say such statements (e.g math theorems) were a priori because the truth of the predicate was not in the subject. The truth of the propositions follows because of the way that humans think (synthetic), but don't require additional experience (a priori).
Now while Kant didn't mean something like logical entailment, that's how I want to see what Kant was doing. He probably didn't see such a thing because he was still using Aristotelian logic.
The reason I want to say such a thing is because, at least for the cases of math theorems. It doesn't follow from the definition of a triangle that its angles add up to 180 degrees (on a simple definition of triangle). One also requires some Euclidean axioms. So it's those axioms together that entail the conclusion.
In summary, I want to understand a priori propositions as true in virtue of a single definition where synthetic a priori deductions require several propositions.
Is there something wrong with viewing Kant's ideas like this?
• The problem is that such interpretation is neither Kantian nor synthetic. This was more or less Frege's view of mathematics, later elaborated by Russell and others. But they explicitly disagreed with Kant, and called mathematics analytic, not synthetic (although Frege left room for some axioms themselves to be synthetic). It is true that Kant needed synthetic judgments to make up for the poverty of Aristotelian logic, but while what they accomplished can be reconstructed the Frege-Russell way, it is very doubtful that mathematics in practice functions that way. – Conifold Sep 18 '19 at 4:57
• "In summary, I want to understand a priori propositions as true in virtue of a single definition where synthetic a priori deductions require several propositions." I'm probably being slow but I can't make sense of this. Deductions in general for Kant require several propositions, so what distinction are you making here? It's actually modern logical entailment that doesn't require multiple propositions, where 'A, therefore A' is valid whereas it is not for Kant. – transitionsynthesis Sep 18 '19 at 5:39
• So a case of an a priori proposition, I am thinking something like "All bachelors are unmarried men." That bachelors are unmarried men follows from the definition of "bachelor". But Kant won't say that a mathematical theorem like that all triangles have a total of 180 degrees is a prior because that isn't part of the definition of a triangle. For Kant, you need to do some mental work to get to that. What I want to say is that Kant's "mental work" is really just logical entailment (of more than 1 premise). You can take the bachelors are unmarried men to be a 1-premise entailment. – user32564 Sep 18 '19 at 5:46
• @transitionsynthesis To say a bit more, Kant will say that the mental work always works as such because it's the way all humans are configured to perceive things, but I want to say he's really just hitting the notion of logical entailment. That triangles have 180 degrees doesn't follow because of the way humans must perceive the world. It's an entailment from several Euclidean propositions. – user32564 Sep 18 '19 at 5:55
• 'All bachelors are unmarried men' counts as analytic, I'd say, rather than a priori. – Geoffrey Thomas Sep 18 '19 at 8:32 | 787 | 3,498 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.953125 | 3 | CC-MAIN-2021-31 | longest | en | 0.969586 |
https://www.khanacademy.org/math/ap-statistics/analyzing-categorical-ap/stats-two-way-tables/v/interpreting-two-way-tables | 1,709,591,105,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947476532.70/warc/CC-MAIN-20240304200958-20240304230958-00110.warc.gz | 831,702,727 | 102,067 | If you're seeing this message, it means we're having trouble loading external resources on our website.
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### Course: AP®︎/College Statistics>Unit 1
Lesson 3: Representing two categorical variables
# Interpreting two-way tables
Two-way tables let us sort a group in two ways. For example, we see how men and women voted in the 2012 US presidential election. We can compare the percentages of men and women who voted for each candidate. Two-way tables help us understand how categories relate.
## Want to join the conversation?
• Correct me if I'm wrong, but, if 52% of Men voted for Romney, and 43% of women did, that would make 95% as a total. Whereas, Obama only got 94%. How does that make Obama win?
• I know this comment is five years old, but this table is talking about percentages not the number of people. In this case, it would mean more females voted than males.
• I am having a hard time with interpreting two way tables. In some of the questions there are like 2 or 3 right answers to choose. i don't understand how to single out one of the choices to get the right answer. plz help.
• me too am having the same problem. Sal only discussed a little part of it and he doesn't explain it detailed since he just said percent women and percent men and then says yes to the question
• Does relative frequency have to be a decimal or can it be a fraction? I've seen people do it different ways, but I'm not sure if that is the proper way?
• Yeah this video doesn't prepare you at all for the practice questions. The video features a simple format then the practice segment goes off the rails.
• The practice problems are a lot more harder & difficult to calculate than this.
• Okay but i still don't understand the difference between the column and row relative frequency
• So men and women are columns of the table.
Obama, Romney, Other and Column Total form the rows of the table.
So basically the difference is the orientation
• I've been getting the following feedback a lot during the exercises: "We only know the column relative frequencies, not the row relative frequencies, so we cannot make this claim." Wouldn't that apply to this video as well? | 509 | 2,284 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.0625 | 4 | CC-MAIN-2024-10 | latest | en | 0.947682 |
https://certifiedcalculator.com/t-test-sample-size-calculation/ | 1,723,676,968,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641137585.92/warc/CC-MAIN-20240814221537-20240815011537-00642.warc.gz | 135,344,547 | 13,772 | # T Test Sample Size Calculation
Introduction: The T Test Sample Size Calculator aids researchers in determining the appropriate sample size for a t-test, ensuring the study’s statistical power to detect meaningful effects.
Formula: The calculator employs the formula based on Cohen’s d, alpha (significance level), power, and sides (two-sided or one-sided), yielding the optimal sample size.
How to Use:
1. Input the effect size (Cohen’s d).
2. Specify the significance level (α).
3. Specify the desired statistical power.
4. Choose between two-sided or one-sided test.
5. Click the “Calculate” button to obtain the required sample size.
Example: For instance, with an effect size of 0.5, significance level of 0.05, power of 0.8, and a two-sided test, the calculator may yield a recommended sample size of 64.
FAQs:
1. What is Cohen’s d effect size?
• Cohen’s d quantifies the standardized difference between two means, providing a measure of practical significance.
2. Why is statistical power important?
• Statistical power represents the probability of detecting a true effect, minimizing the risk of Type II errors in hypothesis testing.
3. What is a two-sided test vs. one-sided test?
• In a two-sided test, the researcher examines if there is a significant difference in either direction, while a one-sided test focuses on a difference in a specific direction.
Conclusion: The T Test Sample Size Calculator enhances experimental design by ensuring that studies have an adequate sample size to detect meaningful effects with appropriate statistical power. Researchers can optimize their resources and maximize the reliability of their findings. | 337 | 1,659 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.3125 | 3 | CC-MAIN-2024-33 | latest | en | 0.885507 |
https://www.internet4classrooms.com/printables/common_core/math_mathematics_3rd_third_grade/description_download_29595-CCSS.Math.Content.3.NF.A.3.d.htm | 1,675,243,766,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764499919.70/warc/CC-MAIN-20230201081311-20230201111311-00829.warc.gz | 836,019,306 | 8,201 | @internet4classr I4C
# Description/Download - Worksheet #29595, CCSS.Math.Content.3.NF.A.3.d Grade 3 Mathematics Common Core
(Page last edited 10/16/2017)
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Problem Type: Order three fractions from smallest to largest.
This worksheet supports the following Common Core State Standard:
CCSS.Math.Content.3.NF.A.3.d
Compare two fractions with the same numerator or the same denominator by reasoning about their size. Recognize that comparisons are valid only when the two fractions refer to the same whole. Record the results of comparisons with the symbols >, =, or <, and justify the conclusions, e.g., by using a visual fraction model.
Publisher: National Governors Association Center for Best Practices, Council of Chief State School Officers, Washington D.C.
Internet4classrooms is a collaborative effort by Susan Brooks and Bill Byles. | 426 | 1,833 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.953125 | 3 | CC-MAIN-2023-06 | longest | en | 0.801733 |
https://www.snapxam.com/problems/13460674/-csca-2-1-0-csca-1-1-sina-1-0 | 1,600,486,527,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400189928.2/warc/CC-MAIN-20200919013135-20200919043135-00331.warc.gz | 1,067,178,347 | 7,844 | # Step-by-step Solution
Go!
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=
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sin
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cot
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csc
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## Step-by-step explanation
Problem to solve:
$\frac{\csc^2a-1}{\csc a+1}=\frac{1}{\sin a}-1$
We couldn't solve this problem right now. Please check back later.
$\frac{\csc^2a-1}{\csc a+1}=\frac{1}{\sin a}-1$
### Main topic:
Polynomial factorization
0
### Time to solve it:
~ 3.05 s (SnapXam) | 281 | 567 | {"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} | 3.5 | 4 | CC-MAIN-2020-40 | latest | en | 0.490947 |
https://multi-converter.com/liters-to-quarts-uk | 1,721,078,576,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514713.74/warc/CC-MAIN-20240715194155-20240715224155-00807.warc.gz | 366,885,953 | 6,788 | # Liters to Quarts (UK)
Change to Quarts (UK) to Liters
Share:
## How to convert Liters to Quarts (UK)
1 [Liters] = 0.87987699319635 [Quarts (UK)]
[Quarts (UK)] = [Liters] / 1.1365225
To convert Liters to Quarts (UK) divide Liters / 1.1365225.
## Example
90 Liters to Quarts (UK)
90 [l] / 1.1365225 = 79.188929387672 [qt (UK)]
## Conversion table
Liters Quarts (UK)
0.01 l0.0087987699319635 qt (UK)
0.1 l0.087987699319635 qt (UK)
1 l0.87987699319635 qt (UK)
2 l1.7597539863927 qt (UK)
3 l2.6396309795891 qt (UK)
4 l3.5195079727854 qt (UK)
5 l4.3993849659818 qt (UK)
10 l8.7987699319635 qt (UK)
15 l13.198154897945 qt (UK)
50 l43.993849659818 qt (UK)
100 l87.987699319635 qt (UK)
500 l439.93849659818 qt (UK)
1000 l879.87699319635 qt (UK) | 312 | 744 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-30 | latest | en | 0.460706 |
https://imtd.in/education-news/cbse-class-12-maths-2021-answer-key/ | 1,660,679,780,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882572515.15/warc/CC-MAIN-20220816181215-20220816211215-00739.warc.gz | 295,983,691 | 16,615 | # CBSE Class 12 Maths 2021 Answer Key
CBSE Class 12 Maths: Today the Board category twelve arithmetic test for Term-1 was conducted on 06th Dec 2021 and lakhs of twelfth category students have appeared within the testbecause the test is over currently, students should be dashing to understand the right responses to the Maths MCQ queries asked in today’s testwe have a tendency to area unit providing the solution Key of the category twelve Maths test ready by our knowledgeable college here so the scholars will match their responses tried within the test. With the right answers that area unit provided here, you’ll be able to calculate what number marks will be expected in your term-1 board test.
## CBSE Maths Class 12 Answer Key 2021 Question Paper with Answer Key
The students WHO appeared within the Maths category twelve MCQ examination will currently check their Answer Key for sophistication twelve Maths Paper that is provided here by our knowledgeable school
Q2. The number of all possible matrices of order 2 x 3 with each entry 1 or 2 is
(a) 16
(b) 6
(c) 64
(d) 24
Q3. A function f: R → R is defined as f(x) = x³ + 1. Then the function has
(a) no minimum value
(b) no maximum value
(c) both maximum and minimum values
(d) neither maximum value nor minimum value
Q4. If sin y = xcos (a +y), then dx/dy is
(a) cos a/cos² (a+y)
(b) – cos a/cos² (a+y)
(c) cos a/sin² y
(d) – cos a/sin² y
Q5. The points on the curve x²/9 + y²/25 + 1, where tangent is parallel to x-axis are
(a) (±5, 0)
(b) (0, ±5)
(c) (0, ±3)
(d) (±3, 0)
Q6. Three points P(2x, x + 3), Q(0, x) and R(x + 3, x + 6) are collinear, then x is equal to
(a) 0
(b) 2
(c) 3
(d) 1
Q7. The principal value of cos^-1 (1/2) + sin^-3 (-1/√2) is
(a) 𝞹/12
(b) 𝞹
(c) 𝞹/3
(d) 𝞹/6
Q8. If (x²+ y²)² = xy, then dy/dx is
(a) y + 4x (x²+ y²)/4y (x²+ y²) – x
(b) y – 4x (x²+ y²)/x + 4(x²+ y²) – x
(c) y – 4x (x²+ y²)/4y (x²+ y²) – x
(d) 4y(x²+ y²)- x/y-4x (x²+ y²)
Q9. If a matrix A is both symmetric and skew-symmetric, then A is necessarily
(a) Diagonal matrix
(b) Zero square matrix
(c) Square matrix
(d) Identity matrix
Q10. Let set X = {1, 2, 3} and a relation R is defined in X as : R = {(1, 3), (2, 2), (3, 2)}, then minimum ordered pairs which should be added in relation R to make it reflexive and symmetric are
(a) {(1, 1), (2, 3), (1, 2)}
(b) {(3, 3), (3, 1), (1, 2)}
(c) {(1, 1), (3, 3), (3, 1), (2, 3)}
(d) {(1, 1), (3, 3), (3, 1), (1, 2)}
Q11. A Linear Programming Problem is as follows:
Minimise z = 2x + y
subject to the constraints
x ≥3 , x ≤9, y ≥ 0
x -y ≥ 0, x + y≤14
The feasible region has
(a) 5 corner points including (0, 0) and (9,5)
(b) 5 corner points including (7.7) and (3, 3)
(c) 5 corner points including (14, 0) and (9, 0)
(d) 5 corner points including (3, 6) and (9,5)
Q12. The function f(x) = e^3x – e^5x/x, if x 0 & k, if x = 0 is continuous at x = 0 for the value of k as
(a) 3
(b) 5
(c) 2
(d) 8
Q14. The function of y = x² e^x is decreasing in the interval
(a) (0,2)
(b) (2, ∞)
(c) (-∞, 0)
(d) (-∞, 0) (2, ∞)
Q15. If R= {(x, y); x, y € z, x² + y² < 4} is a relation in set Z, then domain of R is
(a) {0, 1, 2}
(b) {-2, -1, 0, 1, 2}
(c) {0,-1, -2}
(d) {-1, 0, 1}
Q16. The system of linear equations
5x + ky = 5,
3x + 3y = 5;
will be consistent if
(a) k ≠ -3
(b) k = -5
(c) k = 5
(d) k ≠ 5
Q17. The equation of the tangent to the curve y (1 + x²) = 2 – x, where it crosses the x-axis is
(a) x – 5y = 2
(b) 5x – y = 2
(c) x+ 5y = 2
(d) 5x + y = 2
Q19. The principal value of tan^-1 (tan 9𝞹/8) is –
(a) 𝞹/8
(b) 3𝞹/8
(c) -𝞹/8
(d) -3𝞹/8
Section B
Q21. The function f(x) = 2x³ – 15x² + 36 x + 6 is increasing in the interval
(a) (∞, 2) u (3, ∞)
(b) (-∞,2)
(c) (-∞, 2) u (3,∞)
(d) (3, ∞)
Q22. If x = 2 cosθ – cos 2θ and y = 2 sinθ – sin 2θ, then dy/dx is
(a) cos θ+cos 2θ/sin θ- sin 2θ
(b) cos θ-cos 2θ/sin θ- sin 2θ
(c) cos θ-cos 2θ/sin θ- sin 2θ
(d) cos θ-cos 2θ/sin 2θ+sin θ
Q23. What is the domain of the function cos^-1 (2x – 3) ?
(a) [-1, 1]
(b) (1, 2)
(c) (-1, 1)
(d) [1,2]
Q24. The number of elements in A which are more than 5, is
(a) 3
(b) 4
(c) 5
(d) 6
Q27. Let X = {x² : x € N} and the function f:N → X is defined by f(x) = x², x € N. Then this function is
(a) injective only
(b) not bijective
(c) surjective only
(d) bijective
Q28. The corner points of the feasible region for a Linear Programming problem are P(0, 5). Q(1, 5), R(4. 2) and S(12, 0). The minimum value of the objective function Z = 2x + 5y is at the point
(a) P
(b) Q
(c) R
(d) S
Q29. The equation of the normal to the curve ay^2 = x^3 at the point (am^2, am^3) is
(a) 2y – 3mx + am^3 = 0
(b) 2x + 3my – 3am^4 – am^2 = 0
(c) 2x + 3my + 3am^4 – 2am^2 = 0 (d) 2x + 3my – 3am^4 – 2am^2 = 0 | 2,050 | 4,780 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.375 | 4 | CC-MAIN-2022-33 | latest | en | 0.460396 |
https://forums.geocaching.com/GC/index.php?/topic/66247-how-do-i-figure-this-one-out/#comment-832701 | 1,686,140,805,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224653764.55/warc/CC-MAIN-20230607111017-20230607141017-00750.warc.gz | 288,076,864 | 20,952 | # How Do I Figure This One Out?
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I really want to find this cache but I am having some trouble.
I don't want anyone to give me the answer but I would like to know exactly what program/map type/geek tools I need to figure this one out. Can anyone help me here?
THANKS!
Math was always my worst subject. I am a computer/caching geek, not a math geek. This cache has been sitting in my 'not found' list for a while and I am sick of seeing it there!
Edited by Imajika
Being horrible at math also, my only thought would be to, on a mapping program on the computer, plot the center points, and convert the acres to miles, or something, so you can measure the distance on the mapping program, and just mess with it, until you find 3 points that intersect, or are very close to eachother.
Other than that, I would email the people who have already found it, and ask them for help!
Maybe the name of the cache, Pies Are Square has something to do with it? I know there is that formula, I think it shows the area of a circle, which is PI r squared. You know, PI=3.14, r= radius, and then squared=to the 2nd power. Thats all I got!
Edited by SBPhishy
The first step is to find the radius of each circle--the distance from the center to the rim. The formula for that is in the cache name
Area= pi * (radius)2 <--the constant pi (3.14) times the radius squared.
You have the Area and pi, so you solve for the radius using simple arithmetic.
Then you need to plot them on a map, or eye-ball it to see where they should touch. I think the maps at topo zone could help you there.
EDIT:spelling
Edited by GeoWorms
Hey one of my ideas was actually right!
If pi = 3.14 still leaves too much fuzziness in your answer, use this value for pi:
3.14159263589793238...
You don't have to use all the decimal places, but 3.14 may be too few. If your calculator or spreadsheet has a built in constant go ahead and use that one.
Another constant you may need is one acre = 43,560 square feet or 4046.856421 sq. meters.
Instead of mapping, you could solve this with straight math. Switch the coordinates to UTM (because everything is in linear meters and a lot easier to calculate distances. After you do the calculations mentioned above, then you have the three radii and the three centers. You also have two unknowns: the easting and northing of the final point. That means you can use Pythagorean's Theorem: A**2 + B**2 = C**2.
A is the difference between the unknown point's easting coordinate and the easting of the center of one of the circles. B is the difference between the northing coordinates. C is the radius of that circle. You can solve for one of the unknown coordinates in terms of the other (A**2 = C**2 - B**2. When you substitute that into the equation using another circle, you get two potential northings and eastings, because you are dealing with squares, and numbers may be plus or minus (essentially on one side of a circle's center or on the other). That's why you have the third circle, because only one of the answers will be the proper radius from the that third center.
i'm so confused. the only post i understood was the first one.
My brain hurts...I've actually started drawing circles on my computer screen with a compass..The three waypoint are in the center of lakes..Pi r squared...mary had a little lamb...I've got it narrowed down to an area within 36 sq miles...I hate math caches!!!
Talk to Jim at the C.A.C.H.E. mtg. he should be there.
Here is a map, disregard the Header its for another map that I'm working on.
Edited by Tahosa and Sons
I think I am going to have to talk to Jim at the meeting. This one is making my brain hurt. I am going to ask a friend in Texas who is a math genius to help me with it this week. Maybe I can actually find it before the meeting. I think that's my only hope!
I think I am going to have to talk to Jim at the meeting. This one is making my brain hurt. I am going to ask a friend in Texas who is a math genius to help me with it this week. Maybe I can actually find it before the meeting. I think that's my only hope!
Now that a bunch of us are "hooked" (and I'm a very long way from Colorado) I'd appreciate it if you would ask "at the meeting" what the author means by "There is one point where all three circles touch". Geometrically speaking, the "point" at which three circles "touch" isn't physically possible. If he means "intersect", that's another story. Or perhaps I slept throught that lecture.
Oh yeah, one more thing. If you're into EQOA and photography you certainly have a head for math... just turn it on cuz your smarter than you might believe.
edit for the closing comment
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Only 75 emoji are allowed. | 1,197 | 4,988 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.65625 | 4 | CC-MAIN-2023-23 | latest | en | 0.963623 |
https://www.jiskha.com/math/statistics/?page=8 | 1,516,582,866,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084890928.82/warc/CC-MAIN-20180121234728-20180122014728-00351.warc.gz | 941,165,453 | 11,701 | # Statistics
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Roulette wheels in Nevada have 38 pockets. They are all numbered 0, 00, and 1 through 36. Of all 38 pockets, there are 18 red, 18 are black, and 2 are green. Each time the wheel is spun, a ball lands in one of the pockets, and each pocket is equally likely. If you spin the ...
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θ=0.5 θ=1 θ=2 θ=3 λ = 2 θ = 0.5 μ = 0.5 σ = 0.5 θ=0.5 0 2 1 0.270670566 2 0.036631278 3 0.004957504 4 0.000670925 5 9.07999E-05 draw the graph for θ=1 θ=2 θ=3 The first θ=0.5 graph was drawn for you so follow the...
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Among 15 metal parts produced in a machine shop, 4 are defective. If a random sample of five of these metal parts is selected, find a) The probability that this sample will contain at least three defectives. b) The probability that this sample will contain at most two defective.
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A bolt manufacturer is using a hypothesis test with α = 0.01 to see if the diameter of their 0.75 cm diameter bolts are being manufactured properly. The goal is to have the average bolt diameter be 0.75 ± 0.0075 cms. | 5,128 | 22,289 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2018-05 | latest | en | 0.917768 |
https://math.stackexchange.com/questions/3313603/differential-equation-for-bacterial-growth | 1,571,614,478,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570987750110.78/warc/CC-MAIN-20191020233245-20191021020745-00241.warc.gz | 578,935,289 | 32,130 | Differential equation for bacterial growth
I am assigned with a question which states the rate of a microbial growth is exponential at a rate of (15/100) per hour. where y(0)=500, how many will there be in 15 hours?
I know this question is generally modelled as:
$$y=y_0*e^{kt}$$
However, my solution ended up being modelled as :
$$y=e^{kt}*e^{c}$$ via $$y'=ky$$
The resulting equation was:
$$y=e^{kt}*e^{ln500}$$
I ended up getting $$y(15)=4743.86$$ which is the same answer for both methods.
I'm wondering how the general equation was modelled, and if someone could explain how I could tidy up my equations.
Thanks.
• You mean y'=ky instead of y'=kx – dallonsi Aug 4 at 20:33
• Is there a typo? Should be $y'=ky,$ because in the exponential growth is the speed $y'$ of the growth proportional to the amount $y.$ – user376343 Aug 4 at 20:33
• Yes, oops, my bad – persimonns Aug 4 at 20:34
• My end value would be $500\cdot 1.15^{15}=4068.53$. – Christian Blatter Aug 5 at 8:13
$$e^{\ln 500}$$ is equal to 500. Notice that if $$y = e^x$$, taking the natural log of both sides gives you $$\ln y = \ln (e^x) = x$$. Thus, to undo this operation, take each side as the power of $$e$$ to get $$e^{\ln y} = e^x$$, which must be $$y = e^x$$. | 396 | 1,246 | {"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} | 3.59375 | 4 | CC-MAIN-2019-43 | latest | en | 0.914535 |
http://eprisner.de/MAT103/Pro27.html | 1,695,913,170,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510412.43/warc/CC-MAIN-20230928130936-20230928160936-00475.warc.gz | 11,726,656 | 1,282 | MAT103
Franklin College
Erich Prisner
# Triangular Mixture Problem
You have three large 2 liter containers, named A, B, and C, and a small cup of yet unknown volume. At the beginning you have 0.9 liter of 30% acid solution in container B, and 1.2 liter of 50% acid solution in container C. In container A you have 1.2 liter of an acid solution of unknown concentration.
Now we take one cup of the solution in container A and put it into container B, one cup of the resulting solution in container B and put it into container C, and one cup of the resulting solution in container C and put it into container A.
We know that the acid concentration in container A after this procedure is three fourth of its concentration before. Determine the necessary size of the cup used as a function of the concentration of the solution in container A. What is the domain of this function?
Erich Prisner, August 2003 | 212 | 908 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.4375 | 3 | CC-MAIN-2023-40 | latest | en | 0.912619 |
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# Problem 29. Nearest Numbers
Solution 1952170
Submitted on 28 Sep 2019 by Jossy Russell
This solution is locked. To view this solution, you need to provide a solution of the same size or smaller.
### Test Suite
Test Status Code Input and Output
1 Pass
A = [30 46 16 -46 35 44 18 26 25 -10]; correct = [8 9]; [i1 i2] = nearestNumbers(A); assert(isequal([i1 i2],correct))
mindist = 16 mindist = 16 index1 = 1 index2 = 2 mindist = 14 index1 = 1 index2 = 3 mindist = 5 index1 = 1 index2 = 5 mindist = 4 index1 = 1 index2 = 8 mindist = 2 index1 = 2 index2 = 6 mindist = 2 index1 = 3 index2 = 7 mindist = 1 index1 = 8 index2 = 9
2 Pass
A = [1555 -3288 2061 -4681 -2230 -4538 -4028 3235 1949 -1829]; correct = [3 9]; [i1 i2] = nearestNumbers(A); assert(isequal([i1 i2],correct))
mindist = 4843 mindist = 4843 index1 = 1 index2 = 2 mindist = 506 index1 = 1 index2 = 3 mindist = 394 index1 = 1 index2 = 9 mindist = 112 index1 = 3 index2 = 9
3 Pass
A = [-1 1 10 -10]; correct = [1 2]; [i1 i2] = nearestNumbers(A); assert(isequal([i1 i2],correct))
mindist = 2 mindist = 2 index1 = 1 index2 = 2
4 Pass
A = [0 1000 -2000 1001 0]; correct = [1 5]; [i1 i2] = nearestNumbers(A); assert(isequal([i1 i2],correct))
mindist = 1000 mindist = 1000 index1 = 1 index2 = 2 mindist = 0 index1 = 1 index2 = 5
5 Pass
A = [1:1000 0.5]; correct = [1 1001]; [i1 i2] = nearestNumbers(A); assert(isequal([i1 i2],correct))
mindist = 1 mindist = 1 index1 = 1 index2 = 2 mindist = 0.5000 index1 = 1 index2 = 1001
6 Pass
% Area codes A = [847 217 508 312 212]; correct = [2 5]; [i1 i2] = nearestNumbers(A); assert(isequal([i1 i2],correct))
mindist = 630 mindist = 630 index1 = 1 index2 = 2 mindist = 339 index1 = 1 index2 = 3 mindist = 291 index1 = 2 index2 = 3 mindist = 95 index1 = 2 index2 = 4 mindist = 5 index1 = 2 index2 = 5
7 Pass
% Zip codes A = [60048 61802 01702 60601 10001]; correct = [1 4]; [i1 i2] = nearestNumbers(A); assert(isequal([i1 i2],correct))
mindist = 1754 mindist = 1754 index1 = 1 index2 = 2 mindist = 553 index1 = 1 index2 = 4
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# m14 q18
new topic post reply Question banks Downloads My Bookmarks Reviews Important topics
Author Message
Intern
Joined: 25 Jan 2011
Posts: 15
Concentration: Marketing, Entrepreneurship
Schools: Foster (M)
GPA: 3.62
WE: Design (Other)
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Kudos [?]: 7 [0], given: 2
### Show Tags
31 Oct 2011, 12:44
I'm pretty confused on the question below...
If set $$S$$ consists of distinct numbers such that the difference between any two different elements of set $$S$$ is an integer, how many elements does set $$S$$ contain?
1 The difference between any two different elements of set $$S$$ is 2.
2 The range of set $$S$$ is 2.
Statement (1) by itself is sufficient. S1 says that there are only two different elements in the set. As all elements in the set are distinct, we can conclude that set $$S$$ contains 2 elements.
Statement (2) by itself is insufficient. Consider $$(-1, 0, 1)$$ and $$(0, 2)$$ .
The correct answer is A.
Am I missing something here? Where on earth does it say in S1 that there are only two elements in the set? To me, it says that the difference between ANY two elements is 2. Which means that the set could be (0,2,4,6,8,12) or just (2,4).
Can anyone please offer an explanation?
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Posts: 2013
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Kudos [?]: 1822 [0], given: 376
Re: m14 q18 [#permalink]
### Show Tags
31 Oct 2011, 13:30
ModRos wrote:
I'm pretty confused on the question below...
If set $$S$$ consists of distinct numbers such that the difference between any two different elements of set $$S$$ is an integer, how many elements does set $$S$$ contain?
1 The difference between any two different elements of set $$S$$ is 2.
2 The range of set $$S$$ is 2.
Statement (1) by itself is sufficient. S1 says that there are only two different elements in the set. As all elements in the set are distinct, we can conclude that set $$S$$ contains 2 elements.
Statement (2) by itself is insufficient. Consider $$(-1, 0, 1)$$ and $$(0, 2)$$ .
The correct answer is A.
Am I missing something here? Where on earth does it say in S1 that there are only two elements in the set? To me, it says that the difference between ANY two elements is 2. Which means that the set could be (0,2,4,6,8,12) or just (2,4).
Can anyone please offer an explanation?
ANY two means ANY two, not just the consecutive two:
(0,2,4,6,8,12)
0-2=2
2-4=2
4-6=2
OKAY
how about 0-8=8{:Falsifies the statement.}
_________________
Intern
Joined: 25 Jan 2011
Posts: 15
Concentration: Marketing, Entrepreneurship
Schools: Foster (M)
GPA: 3.62
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Kudos [?]: 7 [0], given: 2
Re: m14 q18 [#permalink]
### Show Tags
31 Oct 2011, 13:48
Oh geez. That was so obvious, my brain just didn't "see" it for some reason. Thank you!
Re: m14 q18 [#permalink] 31 Oct 2011, 13:48
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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®. | 1,179 | 4,103 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.3125 | 4 | CC-MAIN-2017-22 | longest | en | 0.91385 |
https://money.stackexchange.com/questions/91035/how-is-interest-calculated-on-pcp-agreements/91045 | 1,660,426,761,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882571987.60/warc/CC-MAIN-20220813202507-20220813232507-00581.warc.gz | 376,252,625 | 68,400 | # How is interest calculated on PCP agreements?
in December 2018, I'll be at the end of my PCP agreement for my current car and I'm weighing up my options. I've never really paid attention to the details about my previous finance agreements - I just knew I needed a car that wouldn't break down and signed the dotted line.
I was wondering if, given the information I know about my current agreement, if it's possible to calculate the interest I'm due over the duration of my PCP deal?
I've been trying to get do it on Google Sheets for a few days and although I get close, I can't quite match the figure I was given by the finance company. Even when I phoned the finance company for a breakdown, they told me I shout get a financial advisor.
If I use my real life figures as an example, I feel as if I should be able to calculate amount of interest I've been quoted to pay back but I can't - surely it's a science and they don't just pluck numbers out the air? :D
Car Price: £14,299
Deposit: £700
Credit Required: £13,599
Interest Rate: 3.5%
APR: 7.8%
Total Interest: £2,579.44
Repayment:
First Payment: £276.53 (includes £40 "Credit Facility Fee")
47 payments of: £236.53
Final Payment (GFMV): £4,825
Completion Fee: £299
I just can't figure out how, given the information above, the total interest is £2,579.44 and I'm worried it's been miscalculated?
Absolutely any information would be much appreciated, my finance company have been useless and I'm struggling to wrap my head around this.
• Does your contract say anything about including the £40 Credit Facility Fee and the £299 Completion Fee as part of the Credit Required, or have you paid it upfront and separately? Feb 19, 2018 at 15:25
• @theonlydanever my first payment was £276.53 which is the regular payment of £236.53 + the £40 facility fee. The completion fee is paid at the end along with the final payment, which takes that payment to £5,124 Feb 19, 2018 at 17:04
You might be finding it difficult to arrive at the exact number for a few reasons:
• they may be using a different interest frequency. For example, they could be working it out daily, monthly or yearly (even though you're paying monthly), but it will change the amount due.
• you only pay interest on the capital outstanding, so the amount of interest you pay will reduce over time. However, all your payments are equal - this is called amortisation. There are several amortisation methods, so you want to make sure you're using the same method as the credit supplier.
I can get within £5 of your £2,579.44 figure, but only when the £40 and £299 fees are included - I think the Credit Facility Fee and Completion Fee is where the problem lies. This article might give explain the APR a little better than I can (look at the sections on Multiple definitions of effective APR and Additional considerations) - it gives a little insight in to how even a small amount can massively affect the payment.
Also , just to give you some peace of mind - you're protected by the Financial Conduct Authority (FCA) in the UK and so, if the interest was found to be incorrect, even just a little bit, then they'd come down hard on the company you're leasing from... so there's a very, very good chance it's correct.
Hope this helps!
• thank you very much for your reply. I guess I take comfort in the fact there appears to be some "variable" aspect then so that explains why I can't quite calculate it! Feb 19, 2018 at 17:07
• Compounding frequency should not affect the interest or the amount due. As the link you included states: For Europe and the UK "A single method of calculating the APR was introduced in directive 98/7/EC", that being the effective interest rate. This is borne out by the equation in the link you provided, e.g. `(1 + APR/100)` etc. Nominal interest rates cannot be used in this manner. Feb 20, 2018 at 9:24
• @ChrisDegnen This article states that "The effect of compounding depends on the frequency interest is compounded". There's also this table on the link you posted (regarding effective interest rates) that shows the rate changing based on the compounding frequency? With all that said, OP is likely paying simple interest anyway, which is charged on a daily basis. Feb 20, 2018 at 10:18
• @theonlydanever The article you have quoted is appropriate for the US where they use nominal rates with specified compounding periods. By EU law the interest rate in the EU and UK is the effective interest rate. This does not depend upon the compounding period. The daily rate can be calculated from the effective rate. Clarifying the rate issue is quite important for this question and worthwhile settling. The best proof I know of is the equation in the EU section of the APR link which only works for an effective APR. Feb 20, 2018 at 11:58
• @theonlydanever You might find it interesting to look up loan calculators on US and UK sites. You will find the US sites calculate using the nominal rate method while the UK sites use the effective rate method. Feb 20, 2018 at 12:04
As theonlydanever mentions, you can get somewhere if you add the credit facility fee and the completion fee. However, I have had to add them to the balloon payment to reach the loan figure. This doesn’t account for the credit facility fee also being paid with the first payment. Also the completion fee is included in the loan and generates an interest charge.
Using the formula derived here: https://money.stackexchange.com/a/76041/11768
``````L = (B + (M ((1 + R)^N - 1)/R))/(1 + R)^N
where
L = present value of loan
M = periodic repayment
R = periodic rate
B = balloon payment
N = number of periods
``````
Taking the APR as the effective interest rate. (See note 1)
When
``````M = 236.53
R = (1 + 7.8/100)^(1/12) - 1 = 0.00628 (approx)
B = 4825 + 299 + 40 = 5164
N = 48
L = (B + (M ((1 + R)^N - 1)/R))/(1 + R)^N = 13599.38
``````
I don't know if this is the calculation your finance company is using, but it's odd how the figures come out.
The interest works out fairly simply using the following calculation
``````N M + 4825 - 13599 = 48*236.53 + 4825 - 13599 = 2579.44
``````
However, if the previous figures are used, with `B = 5164`
``````N M + B - L = 48*236.53 + 5164 - 13599.38 = 2918.06
``````
although this includes the closing fee.
The calculation can be illustrated like so, as the interest on the principal remaining month by month, up to the penultimate month.
The principal remaining at month `n` is given by `P[n]`
``````P[n_] := (M + (1 + R)^n (R L - M))/R
``````
and the interest payable at the end of each month is `P[n] R`, e.g.
the first month's interest is `P[0] R = 85.40` and the last is `P[47] R = 33.70`.
The formula is derived here: https://money.stackexchange.com/a/73683/11768
Note 1 EU regulations specify APR is the effective annual rate, as defined in the above link. This is evident from the included equation:
EU regulations also require that financial services providers provide a representative example, the basic figures for which have been provided. A more illustrative example would help since the figures don't seem to agree. | 1,800 | 7,124 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.75 | 3 | CC-MAIN-2022-33 | latest | en | 0.980185 |
https://destinypapers.net/i-need-help-with-my-project/ | 1,670,600,946,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446711417.46/warc/CC-MAIN-20221209144722-20221209174722-00672.warc.gz | 229,962,031 | 13,912 | # Project part 1
For
these project assignments throughout the course you will need to
it here.
In
this data set – the ROI data set – for 2 different majors (Business
and Engineering), you are given a sample of the 20 best colleges
according to ROI (ROI = Return on Investment) and their ‘School
Type’, ‘Cost’, ’30-Year ROI’, and ‘Annual % ROI’.
1. The
Week 8 project is completed in part every week. If you do each
week’s project, your Week 8 project will go a lot better than if you
do not. Open the spreadsheet. Explain what ROI is.
2. From
the spreadsheet, define what each column is telling you. Go beyond
giving the name of the column. For example, what is a 30-year
ROI?
3. For
each of the 2 majors create a pie chart using the column ‘School
Type’. Make sure that the title clearly explains what you
are looking at.
4. For
each of the 2 majors create a frequency distribution and histogram
using the column ‘Annual % ROI’. Group with starting at 6%
(0.06), ending at 11% (0.11), and go by 0.5% (0.005). Make
sure that there are NO overlapping bars. For example, if one
5. In
a highlighted box, interpret your results. Do the samples
represent public and private schools equally? What does this
mean in terms of the population of schools? Is there a certain
histogram bar that is higher than all the others? What does
this mean in terms of ROI? These are not the only questions
you can use to interpret the results, but they are a push in the
right direction. Keep going. What else do you see?
# Project part 2
For
these project assignments throughout the course you will need to
it here.
Using
the ROI data set:
1. For each of
the 2 majors calculate the mean, median, minimum, maximum, range,
and standard deviation for the columns ‘Cost’ and ’30-Year
ROI’.
2. Define each
of the terms in #1, and explain how each can be used to provide
3. By hand or
with Excel, for each of the 2 majors calculate the probability that
a college picked from the column for ‘School Type’ is ‘Private’.
4. By hand or
with Excel, for each of the 2 majors find the probability that a
college with the ‘School Type’ ‘Private’ has a ’30-Year
ROI’ between \$1,500,000 and \$1,800,000.
5. This is
explanations in #2 to analyze the data. What have you learned
about the data set of each major? What does it mean if there
is a greater range in terms of ROI? What does it mean if there
is a greater standard deviation in terms of ROI? How do minimum and
maximum factor into the question? Which major is looking like it has
the better ROI right now? Why? | 656 | 2,539 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.359375 | 3 | CC-MAIN-2022-49 | latest | en | 0.906864 |
https://boards.straightdope.com/t/volts-vs-amps-confused/78642 | 1,669,900,882,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710813.48/warc/CC-MAIN-20221201121601-20221201151601-00682.warc.gz | 172,248,142 | 13,931 | # Volts vs. Amps 😕
Can someone explain that old rule of thumb “it’s the amps, not the volts”…or do I have it backwards?
Let’s use a human body as a resistance. Let’s assume it’s some constant value of some resistance “R”. Then, isn’t it obvious that the greater the volts “V”, the greater the resulting current “I”? (Knowing V=IR)
So, again I ask…is it the volts or the amps that’s the worry?
I am assuming the current flows through the human evenly, if that matters, although studies show teh current sometimes will only flow across the surface of the skin if the voltage is within some range which I forget.
Working with only one hand, as a good electrician should,
• Jinx
Er, actually it’s both of them together.
Most often*, the damage done by electricity to a body is proportional to the power expended, which is equal to volts times amps (V x A) and is measured in watts (W)**.
This explains how you can touch one of those electrostatic charge generators*** at the Science Centre at 250 000 V, and not be harmed: there’s negligible current, so almost no power. 250 000 V x 0.000 000 01 A = 0.025 W.
[sub]numbers pulled out of the air[/sub]
[sub]*I’m ignoring any voltage breakdown effects you might get in areas of high field strengh (where different parts of a body might be at dramatically different voltages), and also things like dissimilar metal migration in areas of high current flow.
**I’m assuming a purely resistive sircuit and ignoring reactance here.
***I’m also ingnoring frequency effects, such as the current-flowing-across-the-skin effect you mentioned.
Van de Graaf generators*.
*****Also the name of a band.
What, you want a complete recap of 2 years of electronics school?[/sub]
As I understand it (get someone who actually knows the details in here, since my knowledge comes from a lab I took 6 years ago and paid minimal attention to in any event), it is indeed true that the current is what matters.
You see, while a higher voltage will of course result in a higher current, where you err is in thinking that the higher the current, the worse the effects. For small enough currents, you won’t even notice a thing, then as current becomes larger, it can do some bad things to you, then you die, and at even stronger currents, the effects become so strong that things sort of lock down before they become fatal. Of course, you need to be disconnected from the current and revived before you die from heart failure (I THINK that’s what I was told), but the actual damage isn’t irreversible.
Or so I dimly recall from something I read a long time ago. The more detailed answer is anxiously awaited, but I wanted to be high up on that list of people who answered for once…
Or, on preview, Sunspace said something before me. But I wanted to give my 1/2 cent’s worth anyway.
Hmm… on further thought, in the case of a human body, I seem to remember that relatively small amounts of current (0.25 A?) will do damage (possibly because they disrupt existing currents in, say, the heart), but the human body can withstand quite a lot of voltage (i.e. the VdG generator from my previous post).
But for things like resistors, which get damaged by heat, the power is what counts.
Maybe an EMT will happen by to clarify things.
“One day, I heard a hospital intern claim that 110 volts ac from the wall socket is not dangerous because they told him in medical school that it’s not the voltage that kills, it’s the current.”
“…110 volts ac from residential wall sockets is the most common cause of electrocution in the United States. In addition, medical studies reveal that the 50 to 60 Hz frequency used in ac power distribution almost worldwide is the most dangerous of frequencies.”
“Higher and lower ac frequencies are less dangerous than 60 Hz ac (but not safe!) According to medical experts who have studied electrical shock, the kiling factor is * current density * in a certain area in the right atrium of the heart called the * sinoatrial node. *”
"Current flowing through that section of the heart can induce fatal V. Fib. In general, for limb-contact electrical shocks through intact skin (macroshcok), the following approximations are accepted:
1-5 mA Level of perception
10 mA Level of pain
100 mA Severe muscular contraction
100-300mA Electrocution
Keep in mind these are approximate, and not to be taken as guidelines to approximate ‘assumed risk.’ Death can occur under certain circumstances with considerably lower levels of current. For example, when you are sweating, or standing in salt water, then risks escalate tremendously."
Is High Current, Low Voltage Safe?
“…an electronics technician I know recently injured himself severely when he cut himself on a +5 Vdc, 30-ampere computer power-supply terminal. A large amount of current flowed in his arm, and caused severe pain and some physical damage. Second, high current is extremely dangerous if you happen to be wearing jewelry… … the large current turned the watchband red hot, and gave him serious second and third degree burns.”
Carr, Joseph J.
• Old time radios! : restoration and repair *
1990
>>Can someone explain that old rule of thumb “it’s the amps, not the volts”…or do I have it backwards?<<
One easy way to remember this is “West Virginia”, or “WVA”; where Wattage=Voltage X Amperage. Think of electricity like water flowing through a hose. The voltage is the amount of water, while the amperage is the rate of flow. The wattage is the measure of both combined. In order to deliver a large payload wattage, the hose must be able to accomodate the volume and rate simultaneously. To address your question more directly, a large amount of voltage can be dealt with by the human body if it is delivered at a slow rate (a trickle or light spray from the nozzle of a small hose.) If you increase the rate of flow, and in turn the size of the hose, we move toward the force of a firehose. This, at the very least, will get your attention, and may become injurious to sensitive parts of the body. When we crank it up to say the rate of flow seen in a real lightning bolt, (as opposed to one of those plasma globes) the force is so tremendous that its introduction to the body can be similar to being hit by a cannon ball.
>>Let’s use a human body as a resistance. Let’s assume it’s some constant value of some resistance “R”. Then, isn’t it obvious that the greater the volts “V”, the greater the resulting current “I”? (Knowing V=IR)
So, again I ask…is it the volts or the amps that’s the worry?<<
The resistance of the body has a great deal to do with the damage that is potentially done, but the aforementioned must be considered before voltage can be thought of as a determining factor.
In an electrical circuit, a resistor is designed to allow a specific amount of current to pass, while dissipating the remainder as heat. The non-conductive materials used in resistors are designed according to the specifations required by the application. Since people aren’t “designed” for such instances, the human body can be fall anywhere within the range of “decent to poor” resistors. If you get a “zap” from touching the hot lead of a regulated power source, you will most likely live to tell the tale. On the other hand, if you are wading in a pool of water in the basement while working on the main circuit breaker of your house (unregulated 220v), and accidentally touch the hot lead, you will, most likely, find yourself a smoldering corpse in a pool of recently-warmed water; as you, as a resistor, failed to be made up of the sufficient materials required to deal with such an electrical discharge.
>>I am assuming the current flows through the human evenly, if that matters, although studies show the current sometimes will only flow across the surface of the skin if the voltage is within some range which I forget.<<
Again, it is not the voltage that dictates the characteristics of the flow. Electricity will tend to find the path of least resistance from its origin to the nearest ground. Since the surface of the skin often provides this path, many have found themselves lucky to be spared the organ disruption that heavy current can cause to vital organs.
>>Working with only one hand, as a good electrician should<<
And finally, the idea behind this notion is that if you are connected to a ground by the right hand, and come into contact with a live wire by the left, the current can make its “path of least resistance” loop though the chest–you know, the area where the heart is…
I think you get the idea!
Hope this helps,
Al
One important distinction between voltage and current is that voltage is merely potential. Amperage represents flowing electrical current. Exposing onesself to a voltage is essentially meaningless; you must have amperage to experience current flow (by definition).
Nice post, Al. I would offer, however, that voltage is more akin to pressure in the water flow analogy.
Since finding theremin plans and deciding in an LED-intensive costume for Halloween, I’ve been studying basic electronics. But all the sites I’ve found are either very basic (Ohm’s law gets 3 pages plus 2 more of diagrams) or very advanced (Now as we all know a selenium PNP transponder works best when combined with a flyback-quantum-mosfet). Can any Dopers recommend a good site or book?
``I am NOT doing any thing involving the wiring of my house and am NOT building any weapons or other illegal projects (My place already looks like a bomb went off. Building an actual bomb , which of coures would detonate on the workbench, would thus be pointless). HMM...no smiley with charred face and wild hair ala Wile E Coyote.``
Ok…
I have seen several amounts of current that represent the minimum amount to kill a person. Typically, it is listed around 2ma (2/1000 of an amp). To start with, the minimum amount to kill will not be a “likely” amount to kill. Anyway, we will use 2ma for this.
To kill anyone with a minimum amount of current requires that you pass that current through the chest cavity and cause the heart to misbehave. (i.e. fibrillate ) For this argument, we will say that 2ma is the least amount of current that has been demonstrated to sometimes cause fibrillation. Obviously, higher currents would increase the odds. With a high enough current, it would be a near certainty. Interestingly enough, the easiest way to save someone dying after this type of shock, is to shock them again until the heart picks its normal rhythm back up.
So, having established that passing some current through a person’s heart is the way to use the minimum current, how is that accomplished? The first rule is that the current is going to flow from the entry point to the exit point (source to drain). A person can safely handle ANY amount of voltage if there is no discharge path (witnessed by the fact that birds safely sit on high tension wires with no ill effects). The next rule is that the path the current follows between the source and drain should cross the chest cavity and through the heart. The more directly the current crosses the heart the more dangerous it is. So, ways to achieve this would be:
1. Applying the conductors one each to each side of the chest. (best)
2. From on hand/arm to the other
3. From arm to leg. (Left arm to right leg is good)
4. From head to leg… etc.
Now the interesting part…… How much voltage is required to drive 2ma of current through a human body? As mentioned before, the amount of current flowing through a circuit depends upon the voltage and the total resistance, following the law of I (current in amps) = E (electromotive force in Volts) / R (resistance in ohms). In this case, essentially all of the resistance is from the contact between the person’s skin and the conductor. For dry clean skin, the resistance value of the human body is around 100000 ohms. This would indicate that it takes about 200 volts to be dangerous with only dry skin contact. And this again assumes that current is flowing across the heart. But, there are a number of ways to reduce the resistance of the junction between the skin and a conductor.
1. Make the conductor contact a very large section of skin.
2. Use a conductive jelly or fluid at the contact point (i.e. saltwater)
3. Break the skin, or insert the conductor into the skin.
Any of these would reduce the amount of voltage necessary. Of course, there are ways of accidentally doing the same thing. Having sweaty skin, or standing in the rain leaps to mind. Being very sweaty could lower your resistance to 25000 ohms making it conceivable to cause electrocution with 50 volts. Of course standing in the shower or tub can make house voltage (110) quite dangerous. (Don’t use a hair dryer while still in the shower kids…… The warning labels on hair dryers REALLY say that. Of course, if you were dumb enough to want to use a hair dryer while still in the shower, you probably aren’t bright enough to read the warning label, eh?)
So, house current is typically quite safe, even 208/220 is unlikely to kill a person. However, under wet conditions, people have been killed by as little as 48 volts. People have also been exposed to voltages well into the 10’s of thousands of volts and survived. (I’ve witnessed this personally working on a RADAR) For example, if the current enters and exits on the same limb, it is very unlikely to be fatal, but can do extensive local damage (burns to the skin or worse) and be truly painful. I’ve personally been in contact with 220 a number of times, and power supplies of 1500 volts a few times. (1500 volts hurts enough to quickly condition a person not to repeat it.)
In summary:
So, while it is true that the current is what kills you…. It takes enough voltage to drive that current through your body’s resistance. There are ways of reducing a body’s typical resistance that can vastly lower the amount of voltage required to drive that much current also. A typical car batter can deliver 400+ amps through a low resistance load. But, touching the terminals of the battery with your dry hands won’t even push enough current through you (0.12 ma) for you to even feel.
I’ve always disliked the “it’s the amps that kill you” line because it’s fairly meaningless. It’s kind of like saying - “when you burn yourself, it’s not the temperature that hurts you, it’s the rate heat flows into you.” True, but big deal.
We’re used to encountering fixed voltage sources such as mains voltage and batteries. The current they will pass through you is indeed governed by Ohm’s law, V=IR. scotth’s post covered this very nicely.
Confusion arises when we consider static electricity with its potentials of thousands of volts. The explantion frequently given for the harmlessness of static is that “the currents are so small”. Why are they so small? Why doesn’t ohm’s law apply? It DOES, and the currents from static discharge are large, but they only last for a tiny instant of time. It’s the amount of available energy that’s small - the capacitance involved in most static discharges is miniscule.
It’s analogous to the sparks coming off a grinder which don’t burn you - the sparks are orange-hot but they cool down instantly on touching your skin because they are so small. Same with 10000V static electricity - the voltage drops to zero instantly on contact, because the charge “reservoir” is tiny. Try charging a big capacitor to 10000V and you still have “static” electricity, but discharge it through you and it could kill you.
Worry about the voltage. If you have a low voltage supply which can supply very high currents such as scotth’s car battery, it can’t push enough current through you to hurt you. If you have a high voltage supply which can only supply 0.1 amp, that’s enough to be trouble.
I see a lot of talk about “what the source voltage is” and “what the source current is” etc. This type of talk can be very misleading to a lot of people.
So I’ll start at the beginning…
Let’s say there is an electrical power source. This can be a wall outlet, output of a power supply, battery – anything.
You walk up and touch one of the terminals. Is it going to hurt you? Kill you?
First of all, we must determine what components make up the “circuit.” They are as follows:
1. Electrical power source
3. Resistor
Electrical power source
As mentioned above, the electrical power source can be pretty much anything. For the sake of discussion, let’s assume it’s a DC supply w/ two terminals ( “+” and “-” ). Let’s also assume it’s a non-isolated supply, i.e. the “-” output is tied to earth ground. Furthermore, the electrical power source can be modeled as an ideal voltage source in series with a known resistance. (This is a 1st-order model, but good enough for this analysis.) The voltage source is really called the Thevenin equivalent voltage source, and the series resistance is called the equivalent source resistance or, more accurately, *Thevenin resistance. *
During an electrocution incident, your body can be 1st-order modeled as a 2-terminal resistor. Of course, the actual value of resistance depends on a lot of things, and can vary wildly from person-to-person. Very dry non-broken skin can be as high as 500,000 ohms from hand-to-hand; wet skin can result in a hand-to-hand resistance less than 1,000 ohms. Sweaty hands typically register between 20,000 and 40,000 ohms. For broken skin, an arm or leg is typically 500 ohms, while the trunk is 100 ohms. Again, this depends on a lot of factors, and actual resistance may be very different from the values I stated. Also, I believe your body resistance is somewhat non-linear (i.e. the resistance is not constant with current), but we’ll ignore this factor.
Resistor
This is the resistance between your body and the “-” terminal of the power supply. I’ll call this the “ground resistance.” It can be very low, e.g. if you’re standing barefoot in a puddle. In can even be zero, e.g. if your left hand touches the “+” side of the power source, and your right hand touches the “-” side. Of course, it can also be very high, e.g. if you’re wearing rubber-soled shoes.
So here’s what the circuit model looks like:
Positive terminal of Thevenin voltage source connected to Thevenin resistor.
Other end of Thevenin resistor connected to your body.
Other end of your body connected to ground resistor.
Other end of ground resistor connected to negative terminal of power supply.
With me so far?
Now the risk factor to your body depends on the current through your body. This is because body resistance can be just about anything (see above), thus voltage value alone is doesn’t tell you a whole lot. (I = V/R). Currents above 10 mA are considered “bad” for an adult, though it is possible to receive a lethal shock from currents below 10 mA. The GFCI outlet in your bathroom is designed to break the circuit when it detects a leakage of current of 4 to 6 mA.
So given the model I described above, what determines the current through your body?
1. The value of the Thevenin voltage source = Vs (volts)
2. The value of the Thevenin resistance = Rth (ohms)
3. Your body resistance = Rb (ohms)
4. The value of the ground resistance = Rg (ohms)
The current through your body (Ib, in mA) is thus:
Ib = 1000 * Vs / (Rth + Rb + Rg)
Therefore, for a given voltage source, the way to decrease the current through your body is to increase Rb and/or Rg. (You usually don’t have much control over Rth.)
Consequently, the voltage across your body (Vb, in volts) is:
Vb = (Vs * Rb) / (Rth + Rb + Rg)
So what does all of this mean? Let’s look at some examples.
Let’s say you put one hand on the “+” terminal of a car battery, and the other hand on the “-” terminal. Vs = 12.6 V, Rth = 0.01 ohms, Rb = 100,000 ohms, Rg = 0 ohms. The current through your body is 0.126 mA. No problem here. Notice that the maximum current capability of the car battery (about 500 A) makes no difference, except for the fact that more current capability = lower Thevenin resistance.
Let’s say you have a skin break in each hand, and do the same as the previous example. Vs = 12.6 V, Rth = 0.01 ohms, Rb = 1000 ohms, Rg = 0 ohms. The current through your body is 12.6 mA. Not good.
Let’s say you’re very well grounded and you touch a Van de Graaff generator. (This type of device generates a “static” voltage using a motor-driven belt and combs. As a result, it can generate fairly high voltages, but the Thevenin resistance is extremely high. It’s also very non-linear, but we’ll ignore that fact for this analysis.) Let’s say it can generate up to 10,000 volts… Vs = 10,000 V, Rth = 1,000,000,000 ohms, Rb = 100,000 ohms, Rg = 0 ohms. (I’m guessing at these values.) The current through your body is 0.01 mA. Definitely no problem here. Also notice that the voltage across your body is only 1 V. Because the Thevenin source resistance is so high, 10,000 V is never actually across your body. In other words, before you touch the Van de Graff generator, the voltage is indeed 10,000 V. But it drops to 1 V as soon as you touch it. Of course, this is with Rg = 0. To get the notorious “hair raising effect,” Rg must be very high. (Wear rubber-soled shoes!) This means your entire body will “float” at a fairly high voltage above ground, but the current through your body is (still) extremely small.
Now let’s talk about AC. Assume you have rubber-soled shoes on and you touch the hot side of a 120 VAC outlet. Vs = 173 V (max), Rth = 0.001 ohms, Rb = 100,000 ohms, Rg = 10,000,000 ohms (due to the shoes). The maximum instantaneous current through your body is 0.017 mA; you probably won’t feel a thing. (But because you’re a capacitor, you might feel a charging current. But that’s another topic.) Now let’s say you’re standing barefoot in a puddle and you touch the hot side of a 120 VAC outlet. Vs = 173 V (max), Rth = 0.001 ohms, Rb = 1000 ohms, Rg = 100 ohms. The maximum instantaneous current through your body is 157 mA. You’re dead. (I have also read that 60 Hz is more lethal than DC.)
So there you have it. To re-cap: | 5,138 | 22,058 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.234375 | 3 | CC-MAIN-2022-49 | latest | en | 0.918623 |
https://es.mathworks.com/matlabcentral/cody/problems/854-find-the-list-of-all-open-files/solutions/1227994 | 1,600,507,551,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400191160.14/warc/CC-MAIN-20200919075646-20200919105646-00357.warc.gz | 406,907,212 | 16,592 | Cody
# Problem 854. Find the list of all open files
Solution 1227994
Submitted on 10 Jul 2017
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### Test Suite
Test Status Code Input and Output
1 Fail
fclose('all'); fid = fopen('titi.txt','w+') assert(isequal(your_fcn_name, {'titi.txt'}))
fid = 3
Undefined function or variable 'fid'. Error in your_fcn_name (line 3) [C_filenames,~,~,~] = fopen(fid); Error in TestPoint1 (line 3) assert(isequal(your_fcn_name, {'titi.txt'}))
2 Fail
fclose('all'); fid = fopen('titi.txt','w+') fid = fopen('cloclo.mat3','w+') assert(isequal(your_fcn_name, {'titi.txt','cloclo.mat3'}))
fid = 3 fid = 4
Undefined function or variable 'fid'. Error in your_fcn_name (line 3) [C_filenames,~,~,~] = fopen(fid); Error in TestPoint2 (line 4) assert(isequal(your_fcn_name, {'titi.txt','cloclo.mat3'}))
3 Fail
fclose('all'); assert(isequal(your_fcn_name, {}))
Undefined function or variable 'fid'. Error in your_fcn_name (line 3) [C_filenames,~,~,~] = fopen(fid); Error in TestPoint3 (line 2) assert(isequal(your_fcn_name, {})) | 356 | 1,125 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2020-40 | latest | en | 0.515237 |
https://fourmilab.ch/rpkp/experiments/pipeorgan/ | 1,621,102,903,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243990551.51/warc/CC-MAIN-20210515161657-20210515191657-00602.warc.gz | 289,046,034 | 4,399 | # The Probability Pipe Organ
Your browser does not support HTML5 canvas.
The Java version of this page may work for you.
Runs: Flips/run:
This page demonstrates how experiments with random data converge toward the predictions of probability theory as more and more experiments are run.
The panel above initially shows the bell-curve normal distribution approximating the binomial distribution for an experiment consisting of 16 coin flips. To see the bell curve for experiments with different numbers of flips, enter a value between 4 and 1024 inclusive in the “Flips/run” box before pressing “Start”. For large numbers of flips, the bell curve will be very narrow, since the probability of a large excess of heads or tails is very low. In every case, the peak of the bell curve, the most probable result, represents an equal numbers of heads and tails.
Pressing the “Start” button begins a series of simulated experiments, each consisting of the number of flips specified by “Flips/run”. The number of heads are tallied and displayed as a histogram superimposed on the normal curve. As more and more experiments are run, the scale of the histogram bars is adjusted so the tallest bar remains as high as the peak in the normal curve. Press “Stop” to suspend the running of experiments; press “Run” to resume after a pause. “Reset” stops the experiment if running and clears the result for a new run. If you like, you can enter a new value for “Flips/run” whenever the experiment is stopped; previous results will be cleared. The “Step” button runs one experiment and updates the histogram each time it is pressed.
Observe how at the outset, especially for experiments with relatively few flips, results may appear to depart substantially from the chance expectation, but as more and more experiments are run, the results converge ever more closely on the prediction from probability theory. The initial outlying results “scroll down” as more and more experiments produce the most probable outcomes.
by John Walker | 408 | 2,020 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2021-21 | latest | en | 0.915434 |
https://zcos.in/reasoning/words-logical-order/2 | 1,701,907,630,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100626.1/warc/CC-MAIN-20231206230347-20231207020347-00527.warc.gz | 1,204,718,294 | 7,164 | ## Reasoning Logical Order Of WordsPage 2
11.
Arrange the words in a meaningful, logical order and then select the appropriate sequence from the alternatives given below. 1. South America 2. Africa 3. Europe 4. North America (a)2, 3, 1, 4 (b)3, 1, 4, 2 (c)2, 4, 3, 1 (d)2, 4, 1, 3
Answer is: D2, 4, 1, 3
Africa – North America – South America - Europe
12.
Arrange the words in a meaningful, logical order and then select the appropriate sequence from the alternatives given below. 1. Farmer 2. Seed 3. Food 4. Cultivation (a)1, 3, 4, 2 (b)1, 4, 3, 2 (c)1, 2, 4, 3 (d)1, 2, 3, 4
Answer is: C1, 2, 4, 3
Farmer – Seed – Food - Cultivation
13.
Arrange the words in a meaningful, logical order and then select the appropriate sequence from the alternatives given below. 1. Tree 2. Wood 3. Seed 4. Plant (a)3, 1, 2, 4 (b)3, 4, 1, 2 (c)2, 3, 1, 4 (d)4, 1, 2, 3
Answer is: B3, 4, 1, 2
Seed – Plant – Tree - Wood
14.
Arrange the words in a meaningful, logical order and then select the appropriate sequence from the alternatives given below. 1. Vegetable 2. Cutting 3. Cooking 4. Food (a)2, 3, 1, 4 (b)1, 4, 3, 2 (c)1, 2, 3, 4 (d)4, 1, 2, 3
Answer is: C1, 2, 3, 4
Vegetable – Cutting – Cooking - Food
15.
Arrange the words in a meaningful, logical order and then select the appropriate sequence from the alternatives given below. 1. Honey 2. Flower 3. Honey Bee 4. Wax (a)2, 3, 1, 4 (b)3, 1, 4, 2 (c)4, 1, 2, 3 (d)1, 2, 3, 4
Answer is: A2, 3, 1, 4
Flower – Honey Bee – Honey - Wax
16.
Arrange the words in a meaningful, logical order and then select the appropriate sequence from the alternatives given below. 1. Weaving 2. Cotton 3. Cloth 4. Thread (a)2, 3, 1, 4 (b)2, 3, 4, 1 (c)2, 4, 3, 1 (d)2, 4, 1, 3
Answer is: D2, 4, 1, 3
Cotton – Thread – Weaving - Cloth
17.
Arrange the given words as per order in the dictionary. 1. Zephyr 2. Zodiac 3. Zenith 4. Zig-zag (a)3, 1, 2, 4 (b)3, 1, 4, 2 (c)2, 3, 1, 4 (d)4, 1, 2, 3
Answer is: B3, 1, 4, 2
Zenith – Zephyr – Zig zag - Zodiac
18.
Arrange the given words as per order in the dictionary. 1. Chemistry 2. Chamber 3. Cheap 4. Cheerful (a)2, 3, 1, 4 (b)2, 3, 4, 1 (c)1, 2, 4, 3 (d)1, 2, 3, 4
Answer is: B2, 3, 4, 1
Chamber – Cheap – Cheerful - Chemistry
19.
Arrange the given words as per order in the dictionary. 1. Genesis 2. Gender 3. Gently 4. General (a)2, 3, 1, 4 (b)2, 4, 1, 3 (c)4, 1, 2, 3 (d)3, 1, 2, 4
Answer is: B2, 4, 1, 3
Gender – General – Genesis - Gently
20.
Arrange the given words as per order in the dictionary. 1. Billion 2. Bifurcate 3. Bilateral 4. Bilirubin (a)4, 1, 2, 3 (b)2, 3, 4, 1 (c)3, 1, 2, 4 (d)1, 2, 3, 4
Answer is: B2, 3, 4, 1
Bifurcate – Bilateral – Bilirubin - Billion
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# A Survey of Mathematics with Applications Expanded (8th Edition)Solutions for Chapter 2.3 Problem 61EProblem 61E: In Exercise, letU = {a, b, c, d, e, f, g, h, i, j, k}A = {a,...
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Chapter: Problem:
In Exercise, let
U = {a, b, c, d, e, f, g, h, i, j, k}
A = {a, c, d, f, g, i}
B = {b, c, d, f, g}
C = {a, b, f, i, j}
Determine the following.
A C
STEP-BY-STEP SOLUTION:
Chapter: Problem:
Corresponding Textbook
A Survey of Mathematics with Applications Expanded | 8th Edition
9780321501080ISBN-13: 032150108XISBN: Authors:
Alternate ISBN: 9780321510891, 9780321510914, 9780321528605, 9780321563163, 9780321571106, 9780321583062, 9780321589132, 9780321589262, 9780321690708, 9781256348887 | 330 | 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} | 2.578125 | 3 | CC-MAIN-2016-36 | latest | en | 0.645755 |
https://www.physicsforums.com/threads/special-relativity-problem.239848/ | 1,696,248,253,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510994.61/warc/CC-MAIN-20231002100910-20231002130910-00624.warc.gz | 1,010,648,237 | 18,379 | # Special relativity problem
• Goccam
#### Goccam
A little problem with special relativity.
The equation for general relativity E^2= M^2c^4 + P^2c^2 is usually reduced in special relativity using the argument:
If the mass is at rest, momentum is zero, therefore
E^2 = 0 + M^2c^4
Therefore
E = Mc2.
Similarly if M is zero,
E^2 = p^2c^2 + 0
And therefore E = pc
The problem with this is that pc is only = to mc^2 if m is at rest.
If M is in motion there is an additional kinetic energy component E= 1/2mv^2, therefore the total energy is more than pc.
As a consequence I think that the intrinsic momentum, as represented by the massless photon, has to be considered separately from inertial momentum of a particle.
??!
1) Why are you posting this 3 times??
2) Why are you posting this in "High energy, Nuclear, Particle physics"-forum?
pc can NEVER be equal to mc^2 if the particle is at rest, if the particle is at rest, then pc=0.
Also kinetic energy is not equal to 1/2Mv^2 in special relativity..
The equation for general relativity E^2= M^2c^4 + P^2c^2 is usually reduced in special relativity using the argument:
That equation is a result of special relativity.
If the mass is at rest, momentum is zero, therefore
E^2 = 0 + M^2c^4
Therefore
E = Mc2.
OK. That's the energy of a particle with mass M at rest (its rest energy). Meaningless for photons.
Similarly if M is zero,
E^2 = p^2c^2 + 0
And therefore E = pc
OK. This applies for photons (and other massless particles).
The problem with this is that pc is only = to mc^2 if m is at rest.
Huh? If you're talking about a massive particle at rest, pc = 0, not mc^2. (Again, meaningless for photons.)
If M is in motion there is an additional kinetic energy component E= 1/2mv^2, therefore the total energy is more than pc.
Huh? A true statement would be that for a massive particle in motion, the total energy is greater than mc^2 (the rest energy) by a kinetic energy term. That kinetic energy term is not simply 1/2mv^2 (that's only true nonrelativistically).
As a consequence I think that the intrinsic momentum, as represented by the massless photon, has to be considered separately from inertial momentum of a particle.
Sorry. No idea what you're talking about here.
Edit: I merged the multiple threads you created. Once is enough!
Last edited:
What is:
"intrinsic momentum" and "inertial momentum" ?
A little problem with special relativity.
The equation for general relativity E^2= M^2c^4 + P^2c^2 is usually reduced in special relativity using the argument:
If the mass is at rest, momentum is zero, therefore
E^2 = 0 + M^2c^4
Therefore
E = Mc2.
Similarly if M is zero,
E^2 = p^2c^2 + 0
And therefore E = pc
The problem with this is that pc is only = to mc^2 if m is at rest.
If M is in motion there is an additional kinetic energy component E= 1/2mv^2, therefore the total energy is more than pc.
As a consequence I think that the intrinsic momentum, as represented by the massless photon, has to be considered separately from inertial momentum of a particle.
The total energy of a particle is $$E^2= p^2c^2 + m_0^2c^2$$ where m0 is the rest mass. A photon has a rest mass of zero so for a photon E=pc. You might argue that since p=mv and the mass of a photon is zero then p=0 and E=0 but that would be confusing rest mass with inertial mass. The momentum in terms of rest mass is $$p=\frac{m_0c}{\sqrt{(1-v^2/c^2)}}$$
so for a photon p=0/0=inderminate. That means there is insufficient information to determine momentum for a photon using that equation. The required information is that the energy of a photon is E= hf where h = Planck's constant and f=frequency. So for a photon hf=pc and momentum p=hf/c=mc and m=hf/c^2 and E=mc^2=pc=hfc^2/c^2=hf. In fact it is a mute point whether it is valid to ask what the rest mass of a photon is, when there is no valid reference frame where the photon is at rest.
For a particle the rest mass is given by:
$$m0 = \frac{E}{\sqrt{c^4+\frac{v^2}{(1-v^2/c^2)}}}$$
and when v=c it can be seen that m0=0. In short a particle like a photon without rest mass still has inertial mass, momentum and kinetic energy. In relativity, Total Energy = Kinetic Energy + Rest Energy and in the case of a photon the Total Energy is all due to Kinetic Energy. In short a photon has inertial mass, momentum and Kinetic Energy despite having zero rest mass.
First, apologies with the multiple posting, I had finger trouble with the module.
I obviously did not ask the question correctly , by pc = to Mc^2 I was talking about the equivalence,
In relativity, Total Energy = Kinetic Energy + Rest Energy and in the case of a photon the Total Energy is all due to Kinetic Energy. In short a photon has inertial mass, momentum and Kinetic Energy despite having zero rest mass.
And of course if the mass is at rest, the total energy is the rest mass energy, and if p = mc,
Then (mc)c is the same as mc^2. And of course the other equivalent forms of writing energy, hf, frequency, wavelength, electron volts, are all related to rest mass
However if the mass is not at rest, the total energy = kinetic energy + rest energy, and in this case the momentum is mv, and the kinetic energy is 1/2mv^2. which is where I came in.
However if the mass is not at rest, the total energy = kinetic energy + rest energy,
Right.
and in this case the momentum is mv, and the kinetic energy is 1/2mv^2. which is where I came in.
Not right. Those expressions for momentum and kinetic energy are only valid non-relativistically: for speeds much lower that the speed of light. (In his post, kev gives the relativistic expression for the momentum of a massive particle.)
Welcome to PF!
If M is in motion there is an additional kinetic energy component E= 1/2mv^2, therefore the total energy is more than pc.
As a consequence I think that the intrinsic momentum, as represented by the massless photon, has to be considered separately from inertial momentum of a particle.
Hi Goccam! Welcome to PF!
No, the "additional kinetic energy component" is:
E = 1/2mv^2 - 1/24mv^4 + 1/720mv^6 - …
"intrinsic momentum" is not a vector or a tensor or anything else geometric.
Nor is "inertial momentum".
Only by adding the two do we get a vector!
We add the two only because the result fits into a mathematical category (of vectors).
It is virtually impossible to talk about anything which fits into no mathematical category … and why would one want to?
Does that help?
...
However if the mass is not at rest, the total energy = kinetic energy + rest energy, and in this case the momentum is mv, and the kinetic energy is 1/2mv^2. which is where I came in.
In relativity Kinetic Energy = Total energy -Rest energy = $$\frac{m_0c^2}{\sqrt{(1-v^2/c^2)}}-m_oc^2$$
which is not the same as 1/2mv^2 except at very low velocities.
Last edited:
However if the mass is not at rest, the total energy = kinetic energy + rest energy, and in this case the momentum is mv, and the kinetic energy is 1/2mv^2. which is where I came in.
no, the momentum is:
$$p = \frac{m_0v}{\sqrt{1-(v/c)^2}}$$
And kinetic energy is:
$$E_k = (\gamma - 1)m_0c^2$$
Where: $$\gamma = \frac{1}{\sqrt{1-(v/c)^2}}$$
Now to your "question", can you defice intrinsic vs. inertial momentum for us?
You seems to be stuck in the old Newtonian system and its definition of linear momentum...
OK
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# Maximums Intensities of an Image
Posts: 18Member
Hi to all!
If I want to find the maximum intensity of an image I write
max_int = max(Image(:))
If I want to find the second maximum intensity of an image (or the third, fourth etc. ) how can I write it?
Thank you in advance
· ·
## Comments
• Posts: 217Member
If you have a uint8 image with values in the range [1 255] you can try to use the imhist function with the parameter n=256 (see the matlab help)
In this way every luminance value of your image is "counted".
[counts,x] = imhist(I,256)
then the last values of x [u]x(end)[/u] is the maximum value of the image (you have to check that the corresponding count is not null) and [u]counts(end)[/u] is the number of pixels with that value. Then, if you want to find the positions of this values in your image:
pos=find(Image==x(end));
Then you can loop this procedure checking [u]x(end-1)[/u], [u]x(end-2)[/u], etc. Always checking if the corresponding count is not null.
· ·
• Posts: 18Member
thank you!
· ·
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