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https://atlanticgmat.com/if-p-is-the-product-of-the-integers-from-1-to-30/
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math
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If p is the product of the integers from 1 to 30, inclusive, what is the greatest integer k for which 3^k is a factor of p? GMAT Explanation
If p is the product of the integers from 1 to 30, inclusive, what is the greatest integer k for which 3^k is a factor of p?
Here’s a GMAT factorials puzzle from the GMAT Official Guide. I always tell GMAT tutoring students that this is a fantastic questions to get on a test, even though, at least initially, most people get it wrong. Why is it a good one? Once you know how to solve this puppy:
- It’s pretty easy to recognize it again
- It’s relatively quick to solve
Let’s break this down.
p is the product of the integers from 1 to 30. This just means 30!. What’s 30!? It’s a the product of the integers from 1 to 30. Oh. Yeah. 30*29*28*27……*3*2*1
3^k is a factor of p. This means that 30! must be divisible by 3 raised to the k power. So 30!/3^k = integer.
We want to find the GREATEST integer k. Clearly you could make k = 1 and that would work as 30!/3 is an integer. But, again, we want the greatest k. So the question boils down to, how many 3’s are in 30!?
There are two ways to do this. Let’s start with the slow but practical way.
List out the components of 30! that are divisible by 3 and then count up all of your 3’s. 30, 27, 24, 21, 18, 15, 12, 9, 6, 3. Keep in mind that 27 has 3×3’s, 18 has 2×3’s, and 9 has 2×3’s. So you end up with 14 3’s. Not bad.
Here’s the faster/cleaner way to approach these factorial questions. Take the factorial and divide it by the number you’re testing, in this case 3.
30/3 = 10. That means that there are 10 numbers from 1-30 that are divisible by 3. Done! Nope. Not yet because there are some numbers from 1-30 that have more than one 3 (see above list).
30/9 = 3. Next, up the power of 3 so that we’re dividing by 9. This tells us that there are three numbers from 1-30 that have two 3’s. Great – that was easy! Well, not quite there yet because there are numbers from 1-30 that have three 3’s.
30/27 = 1. Up the power once more so that we’re dividing by 27. That counts the number of 27’s from 1-30 or the number of numbers with three 3’s. There’s just one, 27.
30/81 = doesn’t fit. 81 doesn’t fit into 30 so that tells us that there are no numbers from 1-30 that have four 3’s. That’s where you stop.
Then just add those, 10 + 3 + 1 = 14.
The explanation for this second method is much longer BUT the method itself tends to be much quicker than the manual calculation.
Video Solution: If p is the product of the integers from 1 to 30, inclusive, what is the greatest integer k for which 3^k is a factor of p?
Additional GMAT Factorial Puzzle Practice Questions
Not the same factorial setup but a solid GMAT Question of the Day puzzle example to get you thinking in a GMAT kind of way
Here’s another GMAT Question of the Day to sharpen your GMAT puzzle skills.
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http://uttermpaperpgyb.carolinadigital.us/proof-of-chuch-thesis.html
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math
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Nachum dershowitz and yuri gurevich and (independently) wilfried sieg have also argued that the church-turing thesis is susceptible to mathematical proof. This is partly because dershowitz and gurevich published a proof of the church-turing thesis is the this question is about the extended church-turing thesis. Churchs thesis logic, mind and nature edited by adam olszewsict bartosz brozek an agentless proposition is a proof of the church-turing thesis. Proving church's thesis (abstract) which represent an algorithmic axiomatization of computability allowing for a proof of church's and turing's theses. Abstract: we prove that if our calculating capability is limited to that of a universal turing machine with a finite tape, then church's thesis is true.
In constructive mathematics, church's thesis (ct) is an axiom stating that all total functions are computable the axiom takes its name from the church–turing. Abstract we prove that if our calculating capability is limited to that of a universal turing machine with a finite tape, then church's thesis is true. In this paper we consider a mathematical proof of the church thesis the proof is based on very weak assumptions about intuitive computability and the fm.
Proof of church thesis rebel groups by then, the cia had been conspiring for more than a year with allies in the uk, saudi arabia sat essay prompts recent. Gödel's two incompleteness theorems are among the most important the idea of the proof: , based on post's own version of the “church-turing thesis”.
It may seem that it is impossible to give a proof of church’s thesis however, this is not necessarily the case in other words, we can write. Archives and past articles proof of church turing thesis from the philadelphia inquirer, philadelphia daily news, and philly a turing machine is an abstract machine.
Proof of chuch thesis afterlife or no afterlife, point is jeremy obviously was not that 8220nice guy8221 that a lot of you seem to of known dissertations on self. 2 extended church-turing thesis 3because we will be dealing with models of computation that work over different domains (such as strings for turing machines and.
This is a proof of church's thesis i am just proposing a more basic thesis from which church's thesis is implied. Proof of church’s thesis however, this is not necessarily the case we can write down some axioms about computable functions which most people would agree. Essay about racism in south africa, a natural axiomatization of computability and proof of church thesis, boys lazier than girls essay.
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http://pedagogicalpredicaments.blogspot.com/2009/06/different-methods-of-solution.html
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math
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In my third year of teaching, one of my students said to me, "In Vietnam, our teachers ask us to solve each problem three times; each time, finding a different way to come to the same solution."
This statement, so small at the time, has become a way to define my philosophy of teaching. I hate to show my students algorithms and often try to pretend that their books don't have anything of use for them. Many students learn a way to solve a problem and that is all they have- a method with no understanding of what is happening or why.
This has always been my problem with elementary mathematics education. Instead of having students find their own ways to solve problems and allow them their own methods, students are told that they must add or multiply using one given algorithm. I grew up on this system and was very afraid when I first started teaching because my arithmetic skills were so poor, I knew that the students would have a hard time trusting I knew anything in math.
But as I continued teaching, I watched the different methods that my students used and started to pick up on their tricks to solve problems. I started making my students do their homework problems on the board, not just to get me out of the limelight, but to show methods of solution and ideas that I would have never thought of. My students learn from the varied solutions of others and can look at a problem a peer solves and ask, "Why did you do it like that? Here is what I saw..."
I don't regularly ask my students to solve problems multiple ways but I know that they can. If a student is done early and has a particularly messy solution, I'll ask them to go back and try to solve the problem again. Sometimes, I go around the room and find as many different ways to solve the same problem as possible and have all the students go up and share their different solutions and then we evaluate them as a group.
What my Vietnamese student gave me was not just a teaching strategy but a reminder of the creativity involved in mathematics and beauty of each solution.
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https://phennydiscovers.com/index.php/odtodkorc9oeu5
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math
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App to solve algebra equations
Keep reading to learn more about App to solve algebra equations and how to use it. Math can be difficult for some students, but with the right tools, it can be conquered.
The Best App to solve algebra equations
App to solve algebra equations is a mathematical tool that helps to solve math equations. Solving for an exponent can be a tricky business, but there are a few tips and tricks that can make the process a little bit easier. First of all, it's important to remember that an exponent is simply a number that tells us how many times a given number is multiplied by itself. For instance, if we have the number 2 raised to the 3rd power, that means that 2 is being multiplied by itself 3 times. In other words, 2^3 = 2 x 2 x 2. Solving for an exponent simply means finding out what number we would need to raise another number to in order to get our original number. For instance, if we wanted to solve for the exponent in the equation 8 = 2^x, we would simply need to figure out what number we would need to raise 2 to in order to get 8. In this case, the answer would be 3, since 2^3 = 8. Of course, not all exponent problems will be quite so simple. However, with a little practice and perseverance, solving for an exponent can be a breeze!
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More than just an app
This app works very well and makes solving any math problems a breeze! It makes solving variables easy too. Way less time consuming than working it out or typing it into a calculator.
This app is great for crunching derivatives. It is a huge time saver and if you haven't mucked with calculus in a while, it even shows you the steps involved in arriving at a solution. This could be a great learning tool for any high school or college student. Great Job
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https://matplus.net/start.php?px=1628071562&app=forum&act=posts&fid=xshowh&tid=1480&pid=12353
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math
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Website founded by
MatPlus.Net Forum Helpmates Unusual twinning and unusual AUW.
You can only view this page!
|(1) Posted by seetharaman kalyan [Saturday, Jun 14, 2014 20:23]; edited by seetharaman kalyan [14-06-14]|
Unusual twinning and unusual AUW.
I was delighted to publish this nice problem by Nikola Predrag showing a novel twinning method to show AUW by the same white pawn. Your comments welcome. http://www.kobulchess.com/en/problems/chess-originals-2014/566-nikola-predrag-helpmate.html
|(2) Posted by Kevin Begley [Sunday, Jun 15, 2014 04:32]; edited by Kevin Begley [14-06-15]|
Interesting twinning idea.
The concept is not entirely original -- I vaguely recall some problems twinned by removal of the mating unit, for example (which is quite similar); in fact, I made a fairy problem based upon twinning from the final position, with only alteration of the diagram's retro-content, and I seem to recall that somebody had partially anticipated even that idea -- but, this specific change might be new.
And, this idea does suggest the possibility for a broader set of options, in altering the mating unit...
I'd like to hear any ideas to shorten the text required, while preserving alternative options for replacement.
I'd especially like to hear Nikola's thoughts; until we do, here's my suggestion (hopefully others can improve upon it):
0) AMU->... = Alter Mating Unit in some way (where the specific alteration is designated by "...").
1) AMU->UPPER-CASE : specifies alteration of mating unit's type, from the solution to the original diagram.
e.g., "b),c),d),e) AMU->Q,R,B,S" = alter mating unit's type to Queen, Rook, Bishop, Knight), and solve again by identical stipulation.
How about that -- maybe you can show an AUW in this twinning method, too!?
2) AMU->*... : specifies continuous alteration of mating unit (for each twin, make the same alteration in the mating diagram from the preceding twin).
e.g., "b),c),d) AMU*->P" = b) AMU->P (from final mate of diagram), c) AMU->P (from final mate of b), and d) AMU->P (from final mate of c).
Interestingly, here the solver's final solution may be adequate proof of all preceding solutions.
3) AMU->lower-case : specifies alteration of mating unit's type color.
e.g., b) AMU->n = alter mating unit's color to neutral, and solve again by identical stipulation.
4) AMU->#° : specifies alteration of mating unit's type rotation (note: works only with fairy units present).
e.g., b) AMU->90° = alter mating unit's rotation 90° clockwise, and solve again by identical stipulation.
5) AMU->lower-case UPPER-CASE : specifies alteration of mating unit's type and color.
6) AMU->(x) : specifies annihilation of mating unit (e.g., remove the mating unit, and solve again by identical stipulation).
7) AMU->() : specifies no alteration of mating unit... (e.g., do nothing, just solve again from the final mate position, applying new retro assumptions).
Are there other possibilities? Are there better ways to cover these possibilities?
Are there any issues with this type of twinning?
1) If checkmate is delivered by double-check (or n-tuple-check), must the twin apply to all mating units, simultaneously?
2) What about stalemates?
3) Is there cause to expand this to alteration of the final position (where specific alterations can be stipulated, such as Q->R, R->B, etc)?
|(3) Posted by Kevin Begley [Sunday, Jun 15, 2014 05:20]; edited by Kevin Begley [14-06-15]|
ps: here's an AUW I made, in a single solution, by the same neutral pawn (which experiences a peculiar kind of duel)...
I sent it somewhere, but never heard back, so I presume this was never published.
(= 2+9+2N )
Relegation Chess (aka Degradierung) - upon moving onto its own 2nd rank (home-rank for pawns of a given color), an officer (except King) immediately demotes to pawn.
1.g8=nB! …nBh7(=nP) 2.h8=nQ+! …nQxd4 3.nQd1 …nQd7(=nP) 4.d8=nS! …nSb7(=nP) 5.bxa8=nR!#
I liked the single-unit having a duel (reminds me of Good-Kirk vs Bad-Kirk), and showing AUW...
Unfortunately, the use of Maximummer (as usual, when it used to force play, and is not thematically necessary) cheapens the idea far too much.
The judge (whomever that was) deserves credit, if this was the reason for neglecting my problem (many fairy judges fail to appreciate significant differences in fairy element usage, particularly in excessively constraining conditions).
Plus, my construction was rather shoddy...
A much better AUW, with a single white pawn, is seen in the following:
3rd Prize, The Problemist, 1982
(= 2+9 )
1.g8=S! 2.Sh6 3.Sxg4 4.Sxh2(P) 5.h4 6.h5 7.h6 8.h7 9.h8=B! 10.Bxb2(P) 11.b4 12.b5 13.b6 14.bxa7 15.a8=R! 16.Rxa4 17.Rxa2(P) 18.a4 19.a5 20.a6 21.a7 22.a8=Q! 23.Qxf3 24.Qxg3 25.Qg7 =
By comparison, I quite liked Nikola's method of achieving this -- even if it might be a stretch to claim this is a single pawn (the same could be said, but only to a lesser degree, in the fairy methodology) -- because his creative interpretation appears entirely orthodox (which constitutes a spectacular realization of what we might incorrectly believe to be a fairy theme)!
Maybe, if we think slightly outside the box, any Chess rules (including orthodox) might be sufficient to express any theme!?
|(4) Posted by seetharaman kalyan [Sunday, Jun 15, 2014 07:50]; edited by seetharaman kalyan [14-06-15]|
You are right Kevin that twins where the mating piece is removed is done several times. Twins shifting the black king from mating position has also been done before but not so frequent. This specific twinning, of changing mating piece is, I thought, novel. Your interesting suggestions for notation appear simple and should be examined by experts.
Hm....AMU is already a fairy condition. it may or may not be relevant.
|(5) Posted by Kevin Begley [Sunday, Jun 15, 2014 15:56]; edited by Kevin Begley [14-06-15]|
You are correct -- "AMU" is not the most universal notation either...
Ideally, the notation for both twinning and stipulation should be language independent; therefore, symbols are better.
Perhaps it's time we consider some unicode symbols (they are far more accessible today, and they might be helpful in expressing some ideas more clearly).
Also, the "*" (which I used to suggest successive alteration of the mating unit) is a poor choice -- the symbol is already taken for setplay.
Probably the latter can be improved with the ampersand ("&") -- which already denotes successiveness in twinning...
The notation we can fix fairly easily (providing folks do not become prematurely attached to a sub-optimal expression -- luckily, I see little need for that, here).
I'm more concerned about additional options not considered.
I'm confident that I have not covered all possibilities, which means unforeseen alterations are likely to be necessary.
Better to get this right the first time (at the very least, strive for a framework which has room to grow).
|(6) Posted by Nikola Predrag [Sunday, Jun 15, 2014 18:27]; edited by Nikola Predrag [14-06-15]|
I made that h#2 as an example for the discussion about a twinning principle. I was not sure whether it should be published as an original, because of the uncommon twinning and the possible troubles with a short and clear explanation of it.
And the very discussion was about the short and clear symbols for a whole class/family of a twinning principle:
>a new twin starts from the mate-position of the previous twin. Of course, after some "Change" which allows Black to play some legal move(s).<
I wrote down various attempts but not enough systematically and clearly to paste it here. Anyway, the symbols should come when the essence of a concept is clear.
The essence is "Solve>Change>Solve(Again)>Change>Solve...", shortly "S-C-S" twinning, or any better symbolization.
My problem would be e.g. >4xh#2; S-C-S(Twins),C=Demotion<
Demotion (default) affects the pieces promoted during the play
4xh#2 tells that Mate&Change must happen after every 2+2 halfmoves, 4 phases altogether
"DemotionGradual" might mean Q-R-B-S-P, starting with any rank (in case of less than 5 twins)
"PromotionGradual" might mean P-S-B-R-Q etc., without a mandatory "real" Pawn-promotion on the last rank.
Perhaps a fairy condition could be defined >kxh#n; S-C-S(Condition)<:
"after each n+n halfmoves, Mate&Change must happen and a complete solution would require kxn+kxn halfmoves"
bK has k-"lives" and after each "partial mate", the "Fairies" save bK at the cost of 1 "life"
S-TC-S might symbolize the twinning and S-FC-S the condition.
There are various possibilities but the fundametal concept and symbolization are hardly needed for just a few composed problems.
|(7) Posted by seetharaman kalyan [Sunday, Jun 15, 2014 20:44]; edited by seetharaman kalyan [14-06-15]|
I believe that this simple symbol would be understandable. " ># " implying that any change occurs from the previous mating position.
># remove g7, ># g7 to g6, ># g7=P, >#g7=S etc.. There can only be three changes possible in the mating position: Move the king, change/remove the mating piece or insert a pawn/piece in the mating line.
|(8) Posted by Nikola Predrag [Sunday, Jun 15, 2014 22:18]|
Yes, but why to specify the change for each twin, if they are many and the change is always the same. Rough example:
(= 9+8 )
If I could make it in a minute (without the pieces and board), there is surely a possibility to make much more twins. Specifying each twin separately requires space and work.
|(9) Posted by Kevin Begley [Sunday, Jun 15, 2014 23:02]; edited by Kevin Begley [14-06-15]|
The idea of using fairy conditions to alter the mating unit is a good one, but we still require a twinning symbol indicating the metamorphosis of a mating unit.
I like ">#", but I think Δ (or δ) are better symbols for change. For example: b) #Δ P or b) Δ# ♙ .
Maybe ∫ can symbolize succession: b),c),d) ∫ #δ ♙ -- it's a pity we can't easily show that the integration goes from x=diagram to x=twin d).
Also, maybe the path integral shows up better -- b),c),d) ∮ #δ ♙ -- this might even be more logical.
At least we can put some calculus in our twinning mechanism!
I'm sure some math major will argue that we are integrating over the mating unit, so maybe this form is better: b),c),d) ∮ ♙ δ#
Note: "8xh#1.5 S-Demotion-S" does not read like a twin -- in fact, this is an interruption of the stipulation; clearly, this information does not belong in the stipulation.
Furthermore, whether you mean "Demotion" (I think you mean "relegation chess", or "Degradierung" -- Demotion Chess is a slightly different fairy condition, invented by Dan Meinking), because those fairy conditions change officers moved upon specific ranks (their own 2nd rank, or their own 8th rank) -- neither one alters mating units, spanning the entire board.
Moreover, I don't agree that the essence here is "solve-change-solve" -- the essence of this idea is a twin, first and foremost, which changes the mating unit in a particular way (in this specific case, we alter the type of unit, but we could as easily alter the color, or rotate the unit, or remove the mating unit, or make no change to the unit, and proceed; some thought is required for how fairy conditions might work here).
Finally, as I've explained, there is the option to draw multiple twins from the final position of the diagram, or to draw them from each successive twins.
The "solve change solve" formula might have proved a useful framework to build Nikola's problem, but it fails to envision a means to cope with alternative architectures, based upon the true essence of this idea.
By the way, there is a subtle point to Nikola's twinning, which demonstrates that the twinning mechanism is not as orthodox as you might first think; to fully appreciate this, just consider what happens if a mating unit occurs on the 1st (or last) ranks. Obviously, this is a possibility for which the composer will deliberately go out of their way to disallow, but according to the first rule of chess composition, somebody, someday, is going to want to put this oddity to good use. Therefore, plan for it (which would be easier, if problem chess had insisted upon a consistent default rule for all pawns on the 1st rank -- and the obvious solution there is to make pawns behave as they would on every other rank, with the exception of the 2nd last ranks).
So, strangely enough, Nikola has helped to demonstrate that orthodox and fairies are actually more interconnected than some folks like to pretend. :-)
|(10) Posted by Nikola Predrag [Monday, Jun 16, 2014 01:59]|
I'm not eager to propose the symbols, I'll eventually accept whatever might be proposed. I care more about their meaning.
8xh#1.5 indeed interferes with the stipulation but in case of a fairy condition it might be a stipulation. And it's not clear that S-Demote-S is a twinning principle, it looks more as a condition.
I wrote it that way as a possible(?) example of a condition which not only "demotes" a promoted piece after the mate. It also allows 2 consequent white moves, White mates and after demotion, White continues.
If such sequence W-B-W>W-B-W... looks unacceptable, the very act of "demotion" might be taken as a black move.
But I'm not much interested in new conditions (before having an idea for a problem), that might be explored or abandoned by those who are interested.
Actually, sending that problem to Seetharaman, I wrote under the diagram:
h#2(x4) #=PromotionsCancelled-PlayContinued 5+11
where (x4) is supposed not to affect the meaning of the stipulation, but to indicate its repetition through 4 twins.
Solve-Change-Solve could be a twinning or a condition, so I wrote S-C-S(Twins)
and S-C-S(Condition), or shorter S-TC-S and S-FC-S (TwinChange and FairyChange).
The essence of the idea (S-TC-S) is indeed a twin, but not only "which changes the mating unit in a particular way". "Change" could be "Relocation", "Addition" or "Removal" of some piece. "Change" could be "ColorChange" or "CancellAllAndernachColorChange" or "CancellParticularAndernachColorChange" etc.
might suggest that in mate-position of a/b, bK is relocated to f8/b3, and there's h#2 again.
might suggest the same twinning principle (for 3 twins) but not only bK is relocated.
Any clear and short symbolization would be good, so go on.
In orthodox problems, the twinning with Pawns on 1st/8th rank would be simply incorrect. The author must care about it. A great restriction for the construction of that h#2 AUW, was getting the first 3 mates from 7th rank. A mate from 6th rank would mean a mate by promotion in the next twin, leaving wP on 8th rank for the next twin.
Fairy chess might allow anything and some general "general theory" would have to care about everything.
|(11) Posted by Joost de Heer [Monday, Jun 16, 2014 08:43]|
If there are two solutions in a), and only one of them leads to a solution, after the change, in b), is the second solution in a) a cook or an invalid solution?
|(12) Posted by Georgy Evseev [Monday, Jun 16, 2014 08:45]|
This discussion has gone into unneeded details.
I have long ago resolved this difficulty for myself, declaring that there are two kinds of twins: technical and constructive.
Technical twins are used when they are needed because of technical difficulties or formal limitations. The change should be as small as possible. The kinds of technical twins are well documented and most probably we will not see anything new is this field.
In the constructive twins the mechanics of twinning is itself a part of the author's idea. So, everything is allowed, until the idea is emphasized enough. Unfortunately, this is exactly the reason why no sensible classification is possible: generally, a new problem with the same mechanics of twinning is, really, significantly anticipated. From the other hand this means that we will be able to see a new finds in this area.
I had given a small lecture about this kind of twins in Marianka in 2011. Unfortunately, I had not prepared the text separately, but the problems shown during lecture are available in Marianka 2011 bulletin (http://www.goja.sk/Bulletin_Marianka_2011.pdf), starting from page 49.
|(13) Posted by Kevin Begley [Monday, Jun 16, 2014 08:49]; edited by Kevin Begley [14-06-16]|
You make many good points... and it's important to reiterate my appreciation for your original problem.
I'd like to see more creative ideas actually encouraged -- it's particularly enjoyable, when such an original idea can be expressed with thematic artistry!
My favorite art is that which gives the reader something to think and talk about... it expands the genre... it teaches me something that logically should not fit on a chessboard.
I was stirred by your problem...
I thought it appropriate to raise some larger issues, offer some suggestions, and ask for improvements, here.
I am confident that my suggestion is not the best (there are many things I have not considered -- for example, relocation of mate units, as you noted in your last post). Like you, I look forward to the improvements, which will surely enhance our ability to express such creative new twinning options (as you have demonstrated, and as we may extrapolate plausible). Hopefully, I helped to push a remarkable twinning idea forward...
I know some will consider it tangential to encourage further suggestions, and improvements, in this thread, but for me, such a discussion seems the natural residual impact of your work (which you can definitely take as a compliment). I hope it gets the editors, and the software developers talking to one another... searching for the best way to universally express such twinning possibilities.
|(14) Posted by Nikola Predrag [Monday, Jun 16, 2014 09:20]|
in case of twinning, a second solution, such as you described, would be a cook.
In case of fairy condition, h# in kxn moves (with k Parts or k "partial mates"), any "partial" solution which doesn't lead to a "complete solution", might be considered as simply "not solution". However, such fairy condition should be precisely defined and I don't know how.
I agree that it's not likely to see many problems with such twinning. This twinning principle itself makes the main content, the play is not interesting (but the construction might be). Still, instead of AUW, something else probably could be shown.
the possibilities are unlimited and to anticipate all of them in some general classification looks impossible. Still, some flexible frame would be a welcome and brave contribution.
I have indeed tried to "play" with a twinning which looks like a fairy condition, to achieve an illusion of continued play as h#8(4x2)
No more posts
MatPlus.Net Forum Helpmates Unusual twinning and unusual AUW.
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s3://commoncrawl/crawl-data/CC-MAIN-2021-39/segments/1631780057508.83/warc/CC-MAIN-20210924080328-20210924110328-00449.warc.gz
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CC-MAIN-2021-39
| 18,681 | 138 |
https://www.mathworks.com/matlabcentral/fileexchange/19595-spring-calculations-of-f-and-tau?s_tid=blogs_rc_6
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math
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SPRing calculations of F and TAU (sprftau) of a circular cross-section spring,
tau = max shear stress produced,
R = radius of coil measured from spring axis to ctr of section,
d = diameter of circular section,
P = load (tensile or compressive),
f = total stretch or shortening of spring,
n = number of active turns in spring,
a = (alpha) pitch angle of spring,
G = the modulus of rigidity of the material (shear modulus),
v = poisson's ratio of the material.
1) If the function is called with no input arguments, then a graphical
representation of the equations is produced for publication purposes.
2) If the function is called with only the first 3 input args (R,d,P),
then only the value of tau is returned. f is set to -1.
3) If the function is called with all 7 input args, then both tau and f
Roark's Formulas for Stress and Strain, 7th Edition, page 398
Warren C. Young, Richard G. Budynas
(c)2002, 1989 by the McGraw-Hill Companies, Inc.
Mechancial Springs, 2nd Edition
A. M. Wahl
(c)1963, 1944 by the McGraw-Hill Book Company, Inc.
Congress Catalog Card Number 62-20725
Rob Slazas (2023). SPRing calculations of F and TAU (https://www.mathworks.com/matlabcentral/fileexchange/19595-spring-calculations-of-f-and-tau), MATLAB Central File Exchange. Retrieved .
MATLAB Release Compatibility
Platform CompatibilityWindows macOS Linux
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!Start Hunting!
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s3://commoncrawl/crawl-data/CC-MAIN-2023-40/segments/1695233506045.12/warc/CC-MAIN-20230921210007-20230922000007-00255.warc.gz
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CC-MAIN-2023-40
| 1,458 | 27 |
http://www.math.uiuc.edu/K-theory/0621/
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math
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We establish the existence of an "Atiyah-Hirzebruch-like" spectral sequence relating the morphic cohomology groups of a smooth, quasi-projective complex variety to its semi-topological K-groups. This spectral sequence is compatible with (and, indeed, is built from) the motivic spectral sequence that connects the motivic cohomology and algebraic K-theory of varieties, and it is also compatible with the classical Atiyah-Hirzebruch spectral sequence in algebraic topology. In the second part of this paper, we use this spectral sequence in conjunction with another computational tool that we introduce --- namely, a variation on the integral weight filtration of the Borel-Moore (singular) cohomology of complex varieties introduced by H. Gillet and C. Soule --- to compute the semi-topological K-theory of a large class of varieties. In particular, we prove that for curves, surfaces, toric varieties, projective rational three-folds, and related varieties, the semi-topological K-groups and topological K-groups are isomorphic in all degrees permitted by cohomological considerations. We also formulate integral conjectures relating semi-topological K-theory to topological K-theory analogous to more familiar conjectures (namely, the Quillen-Lichtenbaum and Beilinson-Lichtenbaum Conjectures) concerning mod-n algebraic K-theory and motivic cohomology. In particular, we prove a local vanishing result for morphic cohomology which enables us to formulate precisely a conjectural identification of morphic cohomology by A.~Suslin. Our computations verify that these conjectures hold for the list of varieties above.
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s3://commoncrawl/crawl-data/CC-MAIN-2017-30/segments/1500549424961.59/warc/CC-MAIN-20170725042318-20170725062318-00329.warc.gz
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CC-MAIN-2017-30
| 1,618 | 1 |
https://quizlet.com/642596329/multiplication-flash-cards/
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math
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Study sets, textbooks, questions
Upgrade to remove ads
Terms in this set (12)
There are 5 mice feet scurrying across the floor. If each moue had 4 feet, how many mice are there?
A busy squirrel gathered nuts to store in trees for the winter. The squirrel stored 4 nuts in 7 different trees. How many nuts did the squirrel store in all?
There were 6 nests of ducklings. If there are 8 ducklings in each nest, how many ducklings are there total?
There are 9 pairs of shoes in a closet. How many shoes is that altogether?
I have 7 books. If each book has 10 pages, how many pages are there altogether?
Eliza bought 3 bags of apples. There were 4 apples in each bag. How many apples did Eliza buy?
A fairy penguin eats 6 fish a day. How many fish does the fairy penguin eat in 7 days?
There were 9 race cars on the tracks. If each race car has 4 wheels, how many wheels are there altogether on the race track?
If 5 children wear gloves to school today, how many gloves are at school altogether?
3 mother sparrows each laid 0 eggs. How many sparrow eggs were there altogether?
There are 2 boxes of toothbrushes. There are 9 toothbrushes in each box. How many toothbrushes are there altogether?
There were 8 friends. Each friend had 3 pieces of gum. How many pieces of gum do the friends have altogether?
Sets with similar terms
Units of Measure
Other sets by this creator
Social Studies Vocabulary Set 1
Multiplication by 0, 1, 2, 4, 5, and 10
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s3://commoncrawl/crawl-data/CC-MAIN-2022-21/segments/1652662539049.32/warc/CC-MAIN-20220521080921-20220521110921-00535.warc.gz
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CC-MAIN-2022-21
| 1,438 | 20 |
https://physics.stackexchange.com/questions/398907/pressure-and-surface-tension
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math
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Please identify what is wrong with my reasoning and help. There is a liquid in a container.
it's topmost layer is stationary , macroscopically as seen from outside.
This means there is no net force on the topmost layer.
On any particle of the topmost layer, pressure P atmospheric acts from all directions , including from the liquid beneath.
Where is there any chance of unbalanced forces to cause surface tension?
I think that forces are balanced only because there is surface tension in the first place Is this correct??.. please explain how surface tension is balancing the forces if it is the case??
Q2.) If I increase pressure on the free surface of the liquid, The pressure inside increases by the same amount. This implies that an increase in pressure, i.e., extra force applied on the surface of liquid is being balanced by the layer of liquid just below the surface and not surface tension.. Is the above statement correct or is it that some proportion of increased pressure is balanced by increase in surface tension too??
Is there any good book that has logically complete theory of fluids??
Also if you are interested /capable please answer this.. I would be very happy.. Surface Tension-- Sessile drop
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s3://commoncrawl/crawl-data/CC-MAIN-2024-18/segments/1712297295329.99/warc/CC-MAIN-20240425130216-20240425160216-00239.warc.gz
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CC-MAIN-2024-18
| 1,215 | 9 |
http://www.scrapbook.com/forums/showtopic.php?tid/1578435/post/last/
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math
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This is the second poll in the set for the June Page Maps ETC Challenge.
Our in-forum voting was SO close this month, and we could use your help selecting final winners for each sketch.
Here is the sketch with our top four choices, posted in the order they were submitted. Please select YOUR favorite interpretation of the sketch so we can honor the winner in the July forum.
Sketch #2 - From Let's Capture These Sketches
#1. Treasure Every Moment
#2. Red Fox Coat
#3. Red, White, And All Boy
#4. Railfest 2012 - York
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s3://commoncrawl/crawl-data/CC-MAIN-2013-48/segments/1387345776439/warc/CC-MAIN-20131218054936-00058-ip-10-33-133-15.ec2.internal.warc.gz
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CC-MAIN-2013-48
| 517 | 8 |
https://planetmath.org/ConstructTheCenterOfAGivenCircle
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math
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construct the center of a given circle
[Euclid, Book III, Prop. 1] Find the center (http://planetmath.org/Center8) of a given circle.
Draw any chord in the circle, and construct the perpendicular bisector of , intersecting in , and the circle in .
Let be the center of the circle; we will show that is the midpoint of . Note that in the diagram below, is purposely drawn not to lie on ; the proof shows that this position is impossible and that in fact lies on . It then follows easily that in fact is the midpoint of .
Since is the center of the circle, it follows that . Since bisects , we see in addition that . and share their third side, . So by SSS, , and thus, using CPCTC, . But , so and are each right angles. Thus in fact lies on .
However, since is the center of the circle, it must be equidistant from and , and thus is the midpoint of .
|Title||construct the center of a given circle|
|Date of creation||2013-03-22 17:13:41|
|Last modified on||2013-03-22 17:13:41|
|Last modified by||rm50 (10146)|
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s3://commoncrawl/crawl-data/CC-MAIN-2023-06/segments/1674764499816.79/warc/CC-MAIN-20230130101912-20230130131912-00651.warc.gz
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CC-MAIN-2023-06
| 1,010 | 10 |
https://splendidwritings.com/a-suppose-you-have-a-lined-piece-of-paper-with-parallel-lines-at-distance-d-apa/
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math
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(a) Suppose you have a lined piece of paper with parallel lines at distance d apart. Suppose you throw a small stick of length l (l ≤ d) onto the surface uniformly at random.
What is the probability that the stick will intersect one of the lines?
(b) Suppose you have a piece of paper with parallel vertical lines (distance d apart) and parallel horizontal lines (distance g) apart.
Suppose you drop a small stick of length l (l ≤ d, l ≤ g) onto the surface uniformly at random. What is the probability that the stick will intersect at least one of the lines?
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s3://commoncrawl/crawl-data/CC-MAIN-2021-17/segments/1618039563095.86/warc/CC-MAIN-20210422221531-20210423011531-00049.warc.gz
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CC-MAIN-2021-17
| 1,195 | 9 |
https://forum.wilmott.com/viewtopic.php?f=15&t=64995&p=624500
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math
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QuoteOriginally posted by: AlanT4A and frenchX. I am sensing an arbitrage opp here. I will bet you each $100 this holds up throughall the new data collected by, lets say Dec 2012. If I'm wrong, you each win $100. If I'm right, you payme $25. Takers?Interesting! Alan, your obvious high-level of knowledge and calm demeanor shown on these forums suggests you have a rational model for the 4:1 odds. Yet your modest odds confirm my fears that even you don't think the data is as strong as the "4.9-sigma" or 1-in-2 million chance of a false result.Admittedly your bet isn't about the existence of the Higgs at this energy and decay signature but regards the next 6 months of data gathering, analysis, and refinement. That's a very different wager that's more tied to the expected volume of new data, volatility of data, and autocorrelation of systemic errors in the experimental methods than to the ultimate truth or falsehood of the "discovery." After all, the group that said "neutrinos travel faster than light" created corroborating data in a second experiment so the "holds up through all the new data by time, T" bet isn't a good one for methodology-skeptics like FrenchX and myself.For the record, I suspect they have found the Higgs but I'm far less confident on the odds due to the potential for systematic errors in the process. Thus, you and I might actually agree that they've probably found the Higgs, but the certainty isn't very high.
Last edited by Traden4Alpha
on July 4th, 2012, 10:00 pm, edited 1 time in total.
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s3://commoncrawl/crawl-data/CC-MAIN-2020-16/segments/1585370506988.10/warc/CC-MAIN-20200402143006-20200402173006-00021.warc.gz
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CC-MAIN-2020-16
| 1,528 | 3 |
http://www.appszoom.com/android_games/educational/monsterschool-math-extreme_gfxeo.html
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math
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Why should dragons know all those numbers. Now it is getting really difficult. Draci really needs your help on these ones. Do you see all those numbers? Sometimes, numbers are gone!!! How rude! So you can imagine that Draci needs some serious help in the math department. Finding missing numbers, understanding math orders…. he needs practise in all of them! Can you help him to get a perfect score?
★ How does it work?
Each puzzle has 24 cards. 12 on the left, and 12 on the right side. All the cards on the left make a set with one of the cards on the right. Each puzzle contains a clue, which tells you how to make the sets: add the numbers, find the same answers, count, divide or subtract.
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s3://commoncrawl/crawl-data/CC-MAIN-2017-13/segments/1490218189884.21/warc/CC-MAIN-20170322212949-00123-ip-10-233-31-227.ec2.internal.warc.gz
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CC-MAIN-2017-13
| 698 | 3 |
https://www.universityrankings.com.au/other-mathematical-sciences-rankings-2/
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math
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Other Mathematical Sciences ERA Research Rankings
The Australian Universities Other Mathematical Sciences ERA research ranking is part of a range of rankings under the main area of Mathematical Sciences covering a number of other subfields, namely Pure Mathematics, Applied Mathematics, Numerical and Computational Mathematics, Statistics, Mathematical Physics, and which can be accessed individually from below the table.
No Results Found
No ERA research ranking results were found for the “Other Mathematical Sciences” area of research at this time. Check the links below for the rankings of related areas of research.
Main Mathematical Sciences Rankings:
Other Mathematical Sciences is a Sub-Field belonging to the main Mathematical Sciences Master Field. Click the following link for an overview of of the main Master Field:
Other Mathematical Sciences Sub-Field Rankings:
Other Sub-Field rankings of the Mathematical Sciences Master Field can be found here:
ERA Research Rankings Submenu
ERA Research Ranking Other Mathematical Sciences Journals
The research rankings for the Other Mathematical Sciences field are based on research published in journals throughout the world. The following page shows the:
– List of Other Mathematical Sciences Ranking Journals
Other Master Research Field Rankings:
Latest News and Updates:
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s3://commoncrawl/crawl-data/CC-MAIN-2024-10/segments/1707947473347.0/warc/CC-MAIN-20240220211055-20240221001055-00625.warc.gz
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CC-MAIN-2024-10
| 1,334 | 14 |
http://journals.cambridge.org/action/displayAbstract?fromPage=online&aid=186399
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math
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A STRUCTURAL PROPERTY CONCERNING ABSTRACT COMMENSURABILITY OF SUBGROUPS
Further properties of a group $\Gamma$ introduced by the first author in 1980 (sometimes called the first Grigorchuk group) are established. These are notably that any two infinite finitely generated subgroups of $\Gamma$ have isomorphic subgroups of finite index and that the generalized word problem for $\Gamma$ is soluble.(Received September 10 2002)
(Revised March 10 2003)
20F10; 20E34 (primary); 20E08 (secondary).
p1 Current address: Department of Mathematics, Texas A&M University, College Station, TX 77843-3368, USA [email protected]
p2 Current address: Mathematical Institute, 24–29 St Giles', Oxford OX1 3LB [email protected]
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s3://commoncrawl/crawl-data/CC-MAIN-2016-36/segments/1471982294097.9/warc/CC-MAIN-20160823195814-00186-ip-10-153-172-175.ec2.internal.warc.gz
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CC-MAIN-2016-36
| 725 | 6 |
http://www.jiskha.com/display.cgi?id=1239250835
|
math
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[(x^2-4)/5] * (x+2)/(x-2) ??? maybe???
(x+2)(x-2)/5 * (x+2)/(x-2)
(x^2 + 4 x + 4)/5
Use parentheses to make things easier.
(5x^2-4x) * (2-2x)
5 * x
Multiply through to get:
what is 27-5x<3(4x+5)
Answer this Question
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5th Grade Math - 3/4 x 1/2 = 3/8 Need to write a word problem for this equation...
9th grade algebra/Ms Sue - Okay Ms. Sue. I am in 8th grade but I am taking ...
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s3://commoncrawl/crawl-data/CC-MAIN-2015-11/segments/1424936462099.15/warc/CC-MAIN-20150226074102-00104-ip-10-28-5-156.ec2.internal.warc.gz
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CC-MAIN-2015-11
| 999 | 19 |
http://www.jiskha.com/display.cgi?id=1296008982
|
math
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Posted by Reed on Tuesday, January 25, 2011 at 9:29pm.
How many square inches are needed to cover a square foot? My Dad and I don't see eye to eye on this one--- My Dad says 144 I say the answer is 12 inches squared. Who is correct??
math - Ms. Sue, Tuesday, January 25, 2011 at 9:43pm
Your Dad is correct. It's 144 square inches. To cover a square foot you need 144 squares, each measuring 1 inch x 1 inch.
Your answer is just one step in arriving at the correct answer.
math - helper, Tuesday, January 25, 2011 at 9:47pm
1 square inches = 0.00694444 square feet
12 sq inches = 0.08333 sq ft
1 square ft = 144 sq inches
12 inches squared = 144 sq inches
you are both right
math - Bosnian, Tuesday, January 25, 2011 at 9:48pm
1 foot=12 in
(1 foot)^2 =(12 in)^2
1square foot = 144square inchs
math - Anonymous, Thursday, February 6, 2014 at 5:09pm
do not have a answer
Answer This Question
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s3://commoncrawl/crawl-data/CC-MAIN-2017-09/segments/1487501170404.1/warc/CC-MAIN-20170219104610-00649-ip-10-171-10-108.ec2.internal.warc.gz
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CC-MAIN-2017-09
| 1,760 | 29 |
https://blog.waikato.ac.nz/physicsstop/2011/10/20/opposite-charges-repel-dont-th/
|
math
|
Well, the answer to that is, um… well…. it depends….
Now, I’m not suggesting what you’ve learned at school is not true. Take a point charge (e.g. a proton), and bring it close to another point charge (e.g. another proton) and the two will repel, with an inverse square law (Let’s not take them close enough to exhibit nuclear forces). If we double the distance between the charges, the force of repulsion quarters. That’s Coulombs Law.
Things get rather more complicated, however, if the charge is not point-like. Yesterday, at the NZ Institute of Physics conference, John Lekner gave a fascinating account of the force between two charged, conducting spheres. It is actually intensely complicated and far from obvious. This problem was tackled in the nineteenth century by two physics geniuses: James Clerk Maxwell and William Thomson (Lord Kelvin). Their work was all the more remarkable given that they didn’t have symbolic algebra computer packages available to them to handle the rather intensive algebra generated by this problem. The problem is now being revisited because it has application in nanotechnology, where the conducting spheres in question are really, really small.
So, in broad terms, what happens? Why is it more complicated than Coulomb’s law? To see this, look at the case of two conducting spheres in close proximity, but one with a charge (say positive) and one neutral. The two attract! The first sphere, with the charge, creates an electric field that is felt by the second sphere. The electrons in the second sphere will be attracted towards the first one, since the electrons are mobile in a conductor, and therefore the second sphere will polarize – its negative charge will move towards the part of the sphere that is nearest the first one, giving a separation in the positive and negative charges in the second sphere. The net force between the two spheres is then no longer zero, because the attractive force between the positive charge in the first sphere and the (nearby) negative on the second is greater than the repulsive force between the positive charge in the first and the (further away) positive on the second.
So, a charged sphere and a neutral sphere attract. Let’s now make the second sphere slightly positive as well. If we only make it a little bit positive, the attractive force due to the polarization still wins out over the extra repulsive force due to more positive charge, and so the two still attract each other! Make it too positive, however, and the Coulomb interaction wins out and the two will repel, as we might expect.
It’s actually (much) more complicated than this, since the separation of charge in the second sphere sets up its own field that causes a separation in the first sphere, which influences the second, and so forth, creating, in mathematical terms, an infinite series of interaction terms. It’s this series that Maxwell and Kelvin grappled with, with surprising success.
So, in summary, we have a problem that seems pretty simple: "I take a conducting sphere of radius a, charged with a charge q1, place it a distance x away from a second conducting sphere of radius b charged with a charge q2. What is the force between them?" But in practice the solution is really, really nasty and not at all obvious.
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s3://commoncrawl/crawl-data/CC-MAIN-2023-50/segments/1700679100164.87/warc/CC-MAIN-20231130031610-20231130061610-00277.warc.gz
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CC-MAIN-2023-50
| 3,306 | 7 |
https://seekingalpha.com/article/4037686-tmf-will-tend-underperform-tlt-long-term
|
math
|
Direxion Daily 20+ Year Treasury Bull 3x Shares (NYSEARCA:TMF) aims to multiply daily gains of the ICE U.S. Treasury 20+ Year Bond Index by a factor of 3. It was introduced in April 2009 and has a net expense ratio of 0.95%. According to Google Finance, its market cap as of Jan. 16, 2017, is $84.75M.
Performance to date
Figure 1 shows growth of $10k in TMF since inception, alongside iShares 20+ Year Treasury Bond ETF (NYSEARCA:TLT), which is essentially an unleveraged version of TMF.
Overall, it's been a wild ride for TMF. It experienced a near-50% drawdown right at the beginning of its lifetime and jolted up and down several times over the next 7.5 years. It achieved considerable separation from TLT in 2015-2016, but gave up virtually all of it when bonds fell following the US election.
Performance metrics for the two funds are shown in Table 1.
|Fund||CAGR||Maximum drawdown||Mean of daily gains||SD of daily gains||Sharpe|
I think the overall conclusion here is that TMF's slightly better raw returns are not justified by its vastly greater volatility and drawdown potential.
Moreover, the fact that TLT achieved an excellent CAGR during this time period, and TMF only slightly beat it, tells me that TMF may underperform TLT in a more common scenarios where TLT gains 3-4% annually (its current weighted average coupon is 3.23%).
Daily gains, TMF vs. TLT
In Figure 2, we see that daily gains map almost perfectly from TLT to TMF, with data points falling very close to the Y = 3X line. This is exactly what you would expect from a 3x leveraged ETF behaving like it should.
Monthly gains and volatility decay
Monthly gains also map predictably from TLT to TMF, but the relationship is significantly non-linear (Figure 3; Data is for Jan. 2010 through Dec. 2016).
If we compare the red curve to the blue line, we see that TMF does slightly worse than 3x TLT's monthly gain except at the left and right ends of the graph -- to be exact, whenever the TLT monthly gain is between -6.2% and 5.4%. This is a direct consequence of volatility decay or beta slippage, which I've described in other articles (e.g. Clearing Up Some Beta Slippage Myths).
Now, I'm guessing a lot of readers are thinking that the red curve is practically no different than the blue line; I'm probably overanalyzing and missing the big picture. Actually, the non-linearity here is the big picture.
If TMF followed the blue line, and achieved exactly 3 times TLT's monthly gain, then in months when TLT was unchanged, TMF would also be unchanged. The y-intercept for Y = 3X is of course 0.
Instead, TMF follows the red curve, which has a y-intercept of -0.45%. That means that in months when TLT is unchanged, TMF averages a 0.45% loss. That's not negligible; a 0.45% monthly decline corresponds to an annualized loss of 5.3%.
So what's the break-even point for TMF? If TMF followed the blue Y = 3X line, then its x-intercept would be 0, and TMF would average positive growth whenever TLT had positive growth. But the x-intercept for the red curve is 0.15%. That means that on average TMF only grows when TLT gains 0.15% or better in a given month. Again, not negligible -- 0.15% monthly growth is 1.8% annualized.
Of course, if we're investing in TMF, we don't just want positive growth, we want to outperform TLT. If we were on the blue line, we'd be in good shape, since Y = 3X is greater than X (the black line) whenever X is greater than 0, which we expect it to be since TLT is made up of yield-generating bonds. But on the red curve, if you could zoom in far enough, you'd see that TMF outperforming TLT requires TLT growth of 0.22% or better (2.7% annualized).
TLT's current weighted average coupon, 3.23%, is indeed higher than the 2.7% needed for TMF to outperform. Following the red curve, we expect TMF to gain 0.35% monthly (4.3% annualized) when TLT grows in the amount of its current 3.23% coupon.
Is the extra 1% worth sustaining drawdowns 2-3 times as severe? Probably not.
Long-term underperformance despite higher expected monthly returns
While TMF's expected monthly return is slightly higher than TLT's -- by my estimates, 0.35% vs. 0.26% -- a basic resampling experiment suggests it is likely to underperform TLT over a longer period.
Briefly, in each of 100,000 trials, I sample with replacement 5 years of daily gains for TLT, using data on TLT since its inception. I calculate TMF gains simply as 3x TLT gains minus the daily equivalent of the 0.95% expense ratio, then calculate performance metrics for the 5-year period.
|CAGR range for TLT||Trials in which TMF outperforms TLT||Median CAGR for TLT||Median CAGR for TMF|
We see here that TMF almost always underperforms TLT in 5-year periods where TLT averages less than 3% annual growth, and usually outperforms TLT in 5-year periods where TLT averages greater than 3.5% annual growth. In the 3.0-3.5% range, though, where we expect TLT to be given its average coupon, TMF only outperforms TLT 26.4% of the time.
Interestingly, results would be vastly different if TMF did not carry its 0.95% expense ratio. Repeating the same experiment with TMF having the same expense ratio as TLT, in that middle strata where TLT has CAGR of 3.0% to 3.5%, the median CAGR for TMF is 3.97%, and it outperforms TLT in 94.3% of simulations.
I believe that TMF makes for a poor long-term investment due to three factors:
- Its high leverage results in 3x the volatility of TLT, and drawdowns 2-3 times as bad.
- Volatility decay translates to substantial losses when TLT is approximately flat, e.g. over 5% annually.
- With the 0.95% expense ratio, TMF will tend to underperform TLT in 5-year periods of typical TLT growth.
On a final point of wishful thinking, I would love to see Direxion or a different company offer a version of TMF that worked on monthly rather than daily gains, as a small number of leveraged ETFs do (e.g. DXSLX). Such a fund would be far less prone to volatility decay, and would be much more appealing overall.
Disclaimer: The author used Yahoo Finance to obtain historical stock prices and used R to analyze the data and generate figures. Any opinion, findings, and conclusions or recommendations expressed in this material are those of the author and do not necessarily reflect the views of the National Science Foundation.
Disclosure: I/we have no positions in any stocks mentioned, and no plans to initiate any positions within the next 72 hours.
I wrote this article myself, and it expresses my own opinions. I am not receiving compensation for it (other than from Seeking Alpha). I have no business relationship with any company whose stock is mentioned in this article.
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CC-MAIN-2018-22
| 6,646 | 34 |
https://www.coursehero.com/file/6187783/lec22/
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math
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This preview has intentionally blurred sections. Sign up to view the full version.View Full Document
Unformatted text preview: The Relativity of Time
*Relative motion change change the rate at which time passes. *The amount by which a measured time interval is greater than Δto, the proper time, is called time dilation. Example
Your starship passes Earth with a relative speed of 0.9990c. After traveling 10.0 y (your time), you stop, turn, and travel back to Earth with the same relative speed. The trip takes another 10.0 y (your time). How long does the round trip take according t o measurements made on Earth? (Neglect effects of accelerations involved in turning, stopping and getting back up to speed) Key Ideas: Δt = Δt o ⎛ v⎞ 1− ⎜ ⎟ ⎝ c⎠ and
2 β= v c γ= 1 1− β2 Δt = γΔt o 11/14/10 1 11/14/10 2 Twin Paradox The Relativity of Velocity Classical vOA = vOT + vTA
8.6 light-years Relativistic
Alice calculates: Ted calculates: vOA = vOT + vTA vv 1 + OT 2 TA c 11/14/10 3 11/14/10 4 1 The Relativity of Velocity Momentum
Classical If the car goes at 0.6c and the driver throws the rock with a speed of 0.8c, what is the velocity of the rock as observed by the girl? Relativistic vOA = vOT + vTA vv 1 + OT 2 TA c
5 11/14/10 6 11/14/10 Example What is mass?
Classical: Relativistic: ∑F a=
A satellite, initially at rest in deep space, explodes into two pieces. One piece has a mass of 150 kg and moves away from the explosion with speed 0.76c. The other piece moves away from the explosion in the opposite direction with a speed of 0.88c. Find the mass of the second piece of the satellite.
11/14/10 7 11/14/10 a= ∑F (1 − β ) mo 2 3/ 2 8 2 Kinetic Energy
Classical: Rest Energy
KE = mo c 2 1− β
2 KE = 1 mo v 2 2 − mo c 2 Relativistic: W = F Δx =
11/14/10 (1 − β ) mo a 2 3/ 2 Δx KE = mo c 2 1− β
2 − mo c 2
9 11/14/10 10 Example Energy is radiated by the Sun at the rate of 3.92 x1026 W. Find the corresponding decrease in the Sun’s mass for every second that it radiates. 11/14/10 11 3 ...
View Full Document
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http://www.amazon.com/Calculus-Single-Variable-Hughes-Hallett/product-reviews/0471484814
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math
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Most helpful positive review
5 of 8 people found the following review helpful
Difficult, but thought provoking
on October 15, 2010
This is one of the two books that we use for AP Calculus A/B and i find that it is a pretty decent textbook. We use the older second edition, but I have scanned through this edition and it seems to be pretty similar. I do agree that the examples do not completely reflect the difficulty of the problems, but this can also be viewed positively because it forces you to apply the concepts to in more complex ways and this really maked you to understand the topic in depth. I also like that for most concepts, they give both abstract and "number" examples and practice problems. In addition to this book, we also use the Stewart Calculus book for some simpler practice, or when alot of repetitive problams work better. In all, I would give this a 4- star rating if possible, because more simple problems would add to the book. Also, this book is not good to learn from on your own so a poor teacher will compound the problems mentioned above. If you have a good teacher, though, this book is for the most part good.
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http://openstudy.com/updates/4dd4fccdd95c8b0b1f1c59c4
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math
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good morning, can someone help me figure this out?
Is this equation true or false?
28-4sqrt 2= 24sqrt 2
Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
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Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus.
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I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
Thank you, why is it false, Im having a hard time figuring out how to solve the equation
Not the answer you are looking for? Search for more explanations.
your equation implies sqrt(2)=1 which is indeed false
because the 4sqrt2 is connected, and therefore you would have to be divided to be separated.
when you divide by 4 it would make the sides unequal
just to make sure I fully understand, 28-4sqrt 2 ( the 4 and the two is connected) in order to make them seperated I would have to divide???? this is where I am confused...
yes 4 and sqrt(2) is connected.
I understand that much, its after that where I am confused
divide by 4 if you want to separate rt(2) and 4
to try to simplify this you would subtract the 28 from the left side to get:
so the final answer would be that sqrt(2) does not equal 6sqrt(2)-7
Try and solve it and you would discover:
Obviously that is false.
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| 1,895 | 22 |
https://publik.tuwien.ac.at/showentry.php?ID=263016&lang=6&head=%3Clink+rel%3D%22stylesheet%22+type%3D%22text%2Fcss%22+href%3D%22https%3A%2F%2Fpublik.tuwien.ac.at%2Fpubdat.css%22%3E%3C%2Fhead%3E%3Cbody%3E
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math
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Publications in Scientific Journals:
D. Braess, A. Pechstein, J. Schöberl:
"An Equilibration Based A Posteriori Error Estimate for the Biharmonic Equation and Two Finite Element Methods";
arXiv.org e-Print archive,
We develop an a posteriori error estimator for the Interior Penalty Discontinuous Galerkin approximation of the biharmonic equation with continuous finite elements. The error bound is based on the two-energies principle and requires the computation of an equilibrated moment tensor. The natural space for the moment tensor consists of symmetric tensor fields with continuous normal-normal components. It is known from the Hellan-Herrmann-Johnson (HHJ) mixed formulation. We propose a construction that is totally local. The procedure can also be applied to the original HHJ formulation, which directly provides an equilibrated moment tensor.
Electronic version of the publication:
Created from the Publication Database of the Vienna University of Technology.
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s3://commoncrawl/crawl-data/CC-MAIN-2021-49/segments/1637964358842.4/warc/CC-MAIN-20211129194957-20211129224957-00492.warc.gz
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CC-MAIN-2021-49
| 974 | 7 |
https://gecogedi.dimai.unifi.it/seminar/180/
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math
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24 feb 2015 -- 16:00
Aula D'Antoni, Dip. Matematica, Università "Tor Vergata", Roma
By work of E. Poletsky the polynomial hull of a compact K in Cn can be described by analytic discs with boundaries contained in an arbitrary neighborhood of K except for the image of a set of arbitrarily small length. Easy examples of disconnected compacts show that one cannot always arrange that the whole boundary is close to K. On the other hand, this is known to be possible for certain connected compacts. In particular, this is shown by B. Drinovec Drnovšek and F. Forstnerič for connected compacts which are invariant under the standard circle action. Moreover they raise the question whether this remains valid for connected compacts in general. We will explain a counter-example, showing that the answer is negative.
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CC-MAIN-2024-10
| 813 | 3 |
https://mechanics.stackexchange.com/questions/62490/my-air-conditioner-is-blowing-hot-on-the-left-side-and-cold-on-the-right
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math
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Stack Exchange network consists of 175 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers.
My air conditioning in my Camaro ss 2012 is blowing hot on the left side and cold on the right. Anyone knows what the problem could be??
Required, but never shown
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| 359 | 3 |
https://www.physicsforums.com/threads/zeno-of-elea-created-one-of-the-first-and-most-perduring-paradoxes.72778/
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math
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Many of you will know that Zeno of Elea created one of the first and most perduring paradoxes of all. If any of you think you have solved it..............you ARE wrong. sorry. But you can try it: Imagin you want to fo from here to there and th distance is one meter (it works with any distance and directiona and speed), you walk, and get their. but, no. First of all, to get to the other place you have to go thorugh the whole meter, but before the whole meter, you have to cross half of it. Now you are in the middle. Then, you have to go forward, but before crossign the half you have to, you cross the half of that half. Then the half of that half. And the half of that, and that one two....... But, no. Because before getting to the half, you have to get to the half of the first half, but before to it's half, and before to it's half, and so on. So in actual fact, you can't move, so motion does not exist, so time does not exist. I proved mine, (well, Zeno did) now someone has to disprove.
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| 997 | 1 |
https://mmerevise.co.uk/gcse-computer-science/gcse-computer-science-past-papers/
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math
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GCSE Computer Science Past Papers by Exam Board
GCSE Computer Science Paper Practise Top Tips
Top Tip #1
Make sure to complete only one past paper at a time, mark it, then pinpoint where you have gone wrong and take a not of it before going onto the next paper.
Top Tip #2
Gather all the questions you get wrong and ensure you are confident that you can answer them and the questions from that topic.
Top Tip #3
Use model solutions to help you pick up each marking point, mainly the questions where you need to show your working and give you the answer in the question as you only get marks for showing your working on these types of questions.
Top Tip #4
In the run up to the exams make sure you leave enough time to get through all the new GCSE computer science 9-1 papers as well as any specimen papers provided by the different exam boards.
Top Tip #5
Even if you have used all the new GCSE computer science past papers from your exam board then consider using the older papers or even papers from a different exam board. Unlike other GCSE qualifications, the three major exam boards, AQA, Edexcel and OCR all have a very similar specification.
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| 1,148 | 12 |
https://www.la-times-crossword-answers.com/solve-as-a-cipher/
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math
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This crossword clue is for the definition: Solve, as a cipher.
it’s A 18 letters crossword puzzle definition.
Next time, when searching for online help with your puzzle, try using the search term “Solve, as a cipher crossword” or “Solve, as a cipher crossword clue”. The possible answerss for Solve, as a cipher are listed below.
Did you find what you needed?
We hope you did!.
Possible Answers: DECODE.
Last seen on: LA Times Crossword 13 Jul 19, Saturday
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| 466 | 7 |
https://physicscalculations.com/half-life-formula/
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math
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What is Half-Life and its Formula?
Half-life is the time required for half of a quantity of a radioactive substance to undergo decay or transformation. It is a characteristic property of each radioactive isotope and is used to describe the rate of decay, providing a measure of the stability or persistence of a radioactive material. The half-life formula is an equation we use to calculate the rate of disintegration of unstable atomic nuclei which leads to the emission of alpha (α), beta (β), or gamma (γ) particles.
The half-life formula is written as T1/2 = (0.693) / λ, or T1/2 = ln(2) / λ, or T1/2 = (Loge2) / λ
T1/2 = Half-life
λ = decay constant
Note: We can use t1/2 or T1/2 to indicate the half-life of a radioactive element
Therefore, we can use the above formulae to solve half-life problems. We need to understand that the half-life of a radioactive element is the time taken for half the atoms of the element to decay. We can also define half-life as the time taken for a given mass of a radioactive substance to disintegrate to half its initial mass. The number of half-life formula are 3 and we can apply any of the formulae to solve a problem.
The relation of N and T1/2 is in a graph of N versus T1/2 below. The graph below describes a decay curve.
We can use the following new equations to calculate the half-life of a radioactive element:
- Half-life, T1/2 = t / (log2R) = (t x log2R) / logR [ Where R = 2n = N1 / N2; and n = t / T1/2 ]
- The second formula we can use for the half-life is T1/2 = (0.693) / λ
- We also have another formula for calculating half-life which is T1/2 = ln(2) / λ
- The last formula for calculating half-life is T1/2 = (Loge2) / λ
- The formula for calculating the number of atoms that decay, Nd = N1 – N2 = N1 ((R-1) / R) = N2 (R – 1)
- Fraction remaining of undecayed radioactive elements, fr = N1 / N2 = 1 / R
- The fraction of decayed atoms, fd = Nd / N1 = (R – 1) / R
T1/2 = Half-life of a radioactive element
t = time it takes a radioactive element to decay or disintegrate
n = number of half-lives
N1 = Initial mass or the initial number of atoms present/initial count rate.
And N2 = final mass of the final number of atoms remaining undecayed/final count rate.
Nd = Number of atoms or mass of atom that has decayed or disintegrated.
R = Disintegrating ratio
fr = fraction of initial number of atoms remaining undecayed
We also have fd = fraction of the initial number of atoms that have decayed.
From the above half-life formulae, the first formula is called Zhepwo radioactive equation while the remaining formulae are called the Zhepwo derivative(s). Hence, we can be able to differentiate between the two groups of equations.
Derivation of Half-Life Formula
Here is how to derive the formula:
Since the rate of disintegration is proportional to the number of atoms present at a given time, we can say that
-(dN/dt) ∝ N
or dN/dt = -λN
λ = constant of proportionality which is referred to us as a decay constant of the element. We can write the above equation (dN/dt = -λN) as λ = – 1/N (dN/dt).
Hence the formula for decay constant is λ = – 1/N (dN/dt)
After integrating the above equation, we will have
N = N0e-λt
Where N0 is the number of atoms present at a time t = 0 (i.e at the time when observations of decay were begun). N = the number of atoms present at time t.
We can now change N = N0/2 into N = N0e-λt to obtain the time required for half of the atoms to disintegrate (half-life)
N0 / 2 = N0e-λt
And N0 will cancel each other from both sides to obtain
1 / 2 = e-λt1/2
We will take the natural or Naperian logarithm of both sides to get
loge(1/2) = -λt1/2
We need to remember that logeen = n
Therefore, from the left-hand side of the equation (loge(1/2) = -λt1/2). We can see that
loge(1/2) = loge1 – loge2 = 0 – loge2 = – loge2 = – 0.693
Thus, -λt1/2 = – 0.693
and t1/2 = 0.693 / λ
What is Half-Life?
The half-life of a radioactive element is the time it takes half of the atoms initially present in the element to disintegrate or decay.
Knowledge of Logarithm for Calculating Half-life
The knowledge of the theory of logarithms will help us to understand how to calculate half-life. Here are logarithmic terms in a tabular form to help you understand the topic better.
|R = 2n||log2 R = n|
|2 = 21||log2 2 = 1|
|4 = 22||log2 4 = 2|
|8 = 23||log2 8 = 3|
|16 = 24||log2 16 = 4|
|32 = 25||log2 32 = 5|
|64 = 26||log2 64 = 6|
|128 = 27||log2 128 = 7|
|256 = 28||log2 256 = 8|
|512 = 29||log2 512 = 9|
|1024 = 210||log2 1024 = 10|
What is the Formula for the Disintegration Ratio?
The disintegration ratio is a newly coined expression, it is NOT a new or additional concept in physics. It does not contradict any term or concept in radioactivity. Therefore, It is simply coined to name for an established relationship (No / N = 2n). Thus, it is modification R = N1 / N2 = 2n which is a simplified application in solving radioactive decay problems.
Half-life Formula: Conventional Method of Calculating Half-Life
Assume that a radioactive element with a half-life of 5 seconds contain 192 atoms initially.
After the first 5 seconds (1 half-life), 96 atoms would decay and 96 atoms would remain.
In another 10 seconds (2 half-lives), 144 atoms would decay and 48 atoms would remain.
When we move to the next 15 seconds (3 half-lives), 168 atoms would decay and 24 atoms would remain.
After 20 seconds (4 half-lives), 180 atoms would decay and 12 atoms would remain.
In another 25 seconds (5 half-lives), 186 atoms would disintegrate and 6 atoms would remain.
Half-Life Formula: How to Calculate Half-life in Physics
Here is a solved problem to help you understand how to apply half-life formula
A radioactive element has a decay constant of 0.077 per second. Calculate its half-life.
The decay constant, λ = 0.077 s-1
We will use the formula that says
T1/2 = (0.693) / λ = 0.693 / 0.077 = 9 s
Therefore, the half-life of the radioactive element is 9 seconds.
You may also like to read:
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CC-MAIN-2023-50
| 5,997 | 79 |
https://ez.analog.com/thread/102873-adf4630-7-pll-unlocked
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math
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Here is my circuit :
I'm trying to amplify a signal whose the frequency is about 400 MHz and the amplitude is about 5 dBm. This signal is generated by the PLL (ADF4360-7). However the frequency of my amplified signal is about 360 MHz (The frequency is fixed at 360 MHz, it doesn't move on the screen of the spectrum analyzer) and I have nothing at 400 MHz. (The signal is observed on the amplifier (ATF531P8) output). So, I decided to remove the resistance R36 (0 Ohm) in order to isolate the PLL from the amplifier. The signal observed on the output of the PLL is the frequency that I want to have (400 MHz). Then I used a frequency generator in order to generate a signal whose the frequency is 400 MHz and the amplitude is 5 dBM. I injected this signal on the amplifier input. The signal observed on the amplifier ouptut has the frequency expected (400 MHz) and it is amplified correctly. So the problem is when the both (amplifier and PLL) are connected together. But I can't identify the source of the problem.
Here is the setting of the PLL (register of PLL):
Here is the link to the datasheet :
Thank you very much and have a nice day !
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CC-MAIN-2018-34
| 1,143 | 5 |
https://math.stackexchange.com/questions/286654/probability-that-total-weight-of-coffee-in-three-10-ounce-jars-is-greater-than-t
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math
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Suppose that instant coffee comes in two sizes, 10-ounce jars and 30-ounce jars.
Let $X$ be the actual weight of coffee in a 10-ounce jar and assume that $X$ has a normal distribution with a mean 10.1 ounces and standard deviation 0.2 ounces.
Let $Y$ be the actual weight of coffee in a 30-ounce jar and assume that $Y$ has a normal distribution with mean 30.4 ounces and standard deviation 0.42 ounces.
Assume that the weights $X$ and $Y$ are independent random variables.
Find the probability that total weight of coffee is three 10-ounce jars in greater than the weight in one 30-ounce jar.
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CC-MAIN-2019-43
| 593 | 5 |
http://mathhelpforum.com/geometry/67204-4-math-questions.html
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math
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First, let's finish filling in some information. If angle AED is 70, then angle BEC must also be 70. Since their sum is 140, the leftover degrees must add up to 220. Since angles AEB and DEC are equal, that means they are each 110 degrees. Angles EDC and ECD must also be equal, so they are both (180 - 110)/2 or 35 degrees. Since ADE and EDC are complimentary, angle ADE must be 55 degrees. Angle EAD is also 55 degrees. Now that we've finished that, we can begin solving for AE.
Seperate this into two triangles, bisected by line segment AC. The angles for this triangle are EAD (55 degrees), DCE (35 degrees), and ADC (90 degrees). If DC is 15, we can solve for AC by taking the 15/cos(35). AE is one half of that, so it must be B.
14. We already determined this was D.
15. We already determined this was C.
16. This one must be 90 degrees.
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s3://commoncrawl/crawl-data/CC-MAIN-2016-50/segments/1480698542213.61/warc/CC-MAIN-20161202170902-00024-ip-10-31-129-80.ec2.internal.warc.gz
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CC-MAIN-2016-50
| 843 | 5 |
https://www.nagwa.com/en/videos/232185859543/
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math
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Determine the local maximum and local minimum values of 𝑓 of 𝑥 equals four 𝑥 cubed minus 12𝑥 minus five.
Local maxima and local minima are examples of critical points. And we know that the critical points of a function occur when its first derivative is equal to zero or is undefined. We, therefore, need to find an expression for the first derivative of our function 𝑓 prime of 𝑥 and then find where it’s equal to zero. In order to find 𝑓 prime of 𝑥, we can use the power rule of differentiation. 𝑓 prime of 𝑥 is equal to four multiplied by three 𝑥 squared minus 12 multiplied by one. Remember, the derivative of a constant, in this case negative five, is just zero, which simplifies to 12𝑥 squared minus 12. So we have our expression for the first derivative of our function 𝑓 of 𝑥.
Next, we set this expression equal to zero and solve the resulting equation for 𝑥. Adding 12 to both sides and then dividing by 12 gives 𝑥 squared is equal to one. We solve by square routing, remembering that we must take both the positive and the negative values. 𝑥 is equal to plus our minus the square root of one. And as the square root of one is just one, we have that 𝑥 is equal to plus or minus one. Now, these are the 𝑥-values at which the critical points of our function occur. We also need to determine the values of the function itself at these points. To do so, we substitute each 𝑥-value in turn into our function 𝑓 of 𝑥.
𝑓 of one is equal to four multiplied by one cubed minus 12 multiplied by one minus five, which is equal to negative 13. 𝑓 of negative one is four multiplied by negative one cubed minus 12 multiplied by negative one minus five, which is equal to three. So we find that our function has critical points at the points one, negative 13 and negative one, three. But the question asks us to determine the local maximum and local minimum values of our function 𝑓 of 𝑥. So we need to classify these critical points.
In order to do this, we need to apply the second derivative test. We’ll find an expression for the second derivative 𝑓 double prime of 𝑥 of our function and then evaluate this at each of the critical points. Depending on the sign of 𝑓 double prime of 𝑥 at each critical point, this will tell us whether there are local maximum, a local minimum, or perhaps a point of inflection. Differentiating our expression 𝑓 prime of 𝑥 again, remember, 𝑓 prime of 𝑥 was equal to 12𝑥 squared minus 12. We see that the second derivative 𝑓 double prime of 𝑥 is equal to 12 multiplied by two 𝑥, which is 24𝑥.
Evaluating this when 𝑥 is equal to one gives 24 multiplied by one, which is 24. This is greater than zero. And the second derivative test tells us that if the second derivative of a function is positive at a critical point, then that critical point is a local minimum. So we know that our critical point one, negative 13 is a local minimum of the function 𝑓 of 𝑥. Evaluating 𝑓 double prime of 𝑥 at our other critical point when 𝑥 is equal to negative one, we obtained negative 24. This is less than zero and so the second derivative test tells us that this critical point is a local maximum of the function 𝑓 of 𝑥.
We can conclude them that this function 𝑓 of 𝑥 has a local maximum value of three at 𝑥 equals negative one and a local minimum value of negative 13 at 𝑥 equals positive one. Remember, in this question, we use differentiation to find the first derivative of our function, we will call that critical points. The first derivative will be equal to zero or be undefined. We evaluated the function itself at each of our critical points. And then we used the second derivative test to classify these critical points as local minima or local maxima.
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CC-MAIN-2020-34
| 3,833 | 7 |
https://groups.jewishgen.org/g/main/message/351760
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math
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I hope someone can explain something that appears on my list of Y-DNA
matches (>from FamilyTreeDNA).
For 12 markers, I have 12 "exact" matches.
For 25 markers, I have just 3 "exact" matches; then it lists:
Genetic distance - 1: 1 person
Genetic distance - 2: 2 people
As expected, all the people of 25 markers, whether exact or with
genetic distance, are a subset of the 12 marker matches.
For 37 markers, no one is exact; instead it lists 4 people:
Genetic distance - 1: 2 people
Genetic distance - 2: 1 person
Genetic distance - 3: 1 person
Up to this point, I understand everything well - it's like a process
of winnowing down to understand the gradations of relatedness.
But the puzzling thing is that the person listed as genetic distance 3
is not a person who shows up on my 12 marker match, or anywhere else
of my Y-DNA matches. Shouldn't he have shown up as a 12-marker exact
match? Or does that only test certain markers and this guy has a
different set of 12?
Any ideas appreciated.
With best wishes,
New York, NY
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CC-MAIN-2022-27
| 1,023 | 22 |
https://www.clutchprep.com/physics/practice-problems/137455/part-aan-extended-object-is-in-static-equilibrium-if-__________-a-only-the-net-t
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math
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At static equilibrium, the body is at rest and the forces cancel out.
An extended object is in static equilibrium if __________.
a) only the net torque acting on the object is zero.
b) either the net force acting on the object is zero or the net torque acting on the object is zero.
c) both the net force acting on the object is zero and the net torque acting on the object is zero.
d) only the net force acting on the object is zero.
Which of the following objects is in static equilibrium? Select all that apply.
Two children want to balance horizontally on a seesaw. The first child is sitting one meter to the left of the pivot point located at the center of mass of the seesaw. The second child has one-half the mass of the first child. Where should the second child sit to balance the seesaw?
a) 1m to the right of the pivot
b) 0.5m to the right of the pivot
c) 0.5m to the right of the pivot
d) 2m to the right of the pivot
Rank the glasses from least stable to most stable. If two glasses have the same stability, place one on top of the other.
Frequently Asked Questions
What scientific concept do you need to know in order to solve this problem?
Our tutors have indicated that to solve this problem you will need to apply the Torque & Equilibrium concept. You can view video lessons to learn Torque & Equilibrium. Or if you need more Torque & Equilibrium practice, you can also practice Torque & Equilibrium practice problems.
What professor is this problem relevant for?
Based on our data, we think this problem is relevant for Professor Henderson's class at MTSU.
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| 1,575 | 18 |
https://app-wiringdiagram.herokuapp.com/post/cxc-additional-mathematics-past-papers
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math
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CXC ADDITIONAL MATHEMATICS PAST PAPERS
CXC Additional Mathematics Past Papers | Additional Maths
cxc-store›Past Papers›CSEC›Science and MathsThis eBook contains the official past papers (02 and 03) for CSEC® Additional Mathematics covering the years 2012–2017. This eBook cannot be printed. Visit our FAQs page to learn more.
Additional Mathematics - CXC | Education | Examinations
This Additional Mathematics course provides a variety of topics with related attributes which would enable Caribbean students to reason logically using the prior knowledge gained from the CSEC General Proficiency Mathematics.
CSEC® Additional Mathematics Past Papers eBook - CXC
This eBook contains the official past papers (02 and 03) for CSEC® Additional Mathematics covering the years 2012–2017. This eBook cannot be printed.
Additional Mathematics - Skoolers: CSEC / CXC Exam
Additional Mathematics CSEC Syllabus Add Math Books Past Papers Section 1 Algebraic Operations Quadratics Inequalities Functions Surds, Indices and Logarithms4 Series Coordinate Geometry Vectors Trigonometry Differentiation Sections 2 Algebra and Functions Coordinate Geometry, Vectors and Trigonometry Introductory Calculus Basic Mathematical Applications Read more →[PDF]
CSEC® Additional Mathematics Past Papers
ADDITIONAL MATHEMATICS Paper 02 – General Proficiency 2 hours 40 minutes READ THE FOLLOWING INSTRUCTIONS CAREFULLY. 1. DO NOT open this examination paper until instructed to do so. 2. This paper consists of FOUR sections. Answer ALL questions in Section I, Section II and Section III. 3. Answer ONE question in Section IV. 4.
CSEC¬ Additional Mathematics Past Papers ebook
Additional Mathematics CXC 2014 Solved. CSEC Add Maths June 2012 P2 Solutions. CXC CSEC Add Maths 2012 P1. CXC CSEC ADD MATHS 2014 P2 . Mathematics CXC 2008. Documents Similar To CSEC¬ Additional Mathematics Past Papers ebook. CXC CSEC Add Maths 2016 P1. Uploaded by. Anonymous19. CXC CSEC Add Maths 2017 June P2. Uploaded by. Anonymous19.
Videos of cxc additional mathematics past papers
Click to view on YouTube35:46CSEC CXC Maths Past Paper 2 Question 7 January 2014 Exam Solutions. ACT Math, SAT Math,YouTube · 2/28/2014 · 59K viewsClick to view on YouTube7:47CSEC CXC Maths Past Paper 2 Question 1b May 2013 Exam Solutions. ACT Math, SAT Math,YouTube · · 19K viewsClick to view on YouTube12:37CSEC CXC Maths Past Paper 2 Question 1a May 2013 Exam Solutions ACT Math, SAT Math,YouTube · · 26K viewsSee more videos of cxc additional mathematics past papers
CSEC - Additional Mathematics Syllabus - Cxc
Alternatively students may begin Additional Mathematics in the fourth form and sit both CSEC Mathematics and Additional Mathematics examinations at the end of form five. Students may even do the CSEC Additional Mathematics as an extra subject simultaneously with CAPE Unit 1
CSEC Mathematics past Papers - CSEC Math Tutor
Online Help for CXC CSEC Mathematics, Past Papers, Worksheets, Tutorials and Solutions CSEC Math Tutor: Home Exam Strategy Past Papers Solutions CSEC Topics Mathematics SBA Post a question CSEC Mathematics past Papers. csec_mathematics_may_2004
CSEC CXC Exam Past Papers: Download Section
mathematics past papers 2005 - 2015 added update: 14/4/16 geography, office procedures and chemistry past papers added Additional Mathematics - Paper 03 CSEC May/June 2016 - Mathematics - Paper 02 CSEC May/June 2016 - Economics - Paper 02 Any past physics cxc papers? Reply Delete. Replies. Fatimah Baksh May 11, 2016 at 3:38 PM.
Past Paper Solutions - LIMITLESS LEARNING TT
PAST PAPER SOLUTIONS. These solutions are free and will become available on March 19th, 2018. Past paper solutions from 2017 are only available for students who are enrolled in our in-person and online courses. CSEC Additional Mathematics Past Paper 02.
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http://www.myswitzerland.com/html/magazine/magazine_composed.cfm?lc=en&cc=&template=1&LIMIT=130&model=page&offset=18&ancestor_node_ids=59080,59081
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Through the Wild West. Solitary valleys, lush meadows, babbling brooks and little lakes. The Jura biking tour takes you from one scenic delight to the other. Example:Delémont - Ste. Croix, 6 days/5 nights in a double room, standard accommodation incl. breakfast. bike hire, luggage transport, travel documentation, route documentation, helpline 7 days. Price per person: $$CHF785$$
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| 1,085 | 3 |
https://thebestpaperwriters.com/physics-magnitude-and-direction-of-third-force/
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math
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Physics: Magnitude and Direction of Third Force
November 5th, 2022
An object in equilibrium has three forces exerted on it. A 33 N force acts at 90° from the x axis and a 46 N force acts at 60°. What are the magnitude and direction of the third force?
_________degrees (counterclockwise from the +x direction)
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| 311 | 4 |
https://www.thestudentroom.co.uk/showthread.php?t=4991510
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math
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Single cards, chosen at random, are given away with bars of chocolate. Each card shows a picture of one of 20 different football players. Richard needs just one picture to complete his collection. He buys 5 bars of chocolate and looks at all the pictures. Find the probability that
(i) Richard does not complete his collection,
(ii) he has the required picture exactly once,
(iii) he completes his collection with the third picture he looks at.
Turn on thread page Beta
Stats Question HELP! watch
- Thread Starter
- 09-10-2017 14:12
- 09-10-2017 15:27
What have you tried? Where are you stuck?
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| 593 | 10 |
http://hilltopvilla.hu/index.php/lib/an-invitation-to-discrete-mathematics
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math
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By Jiri Matousek, Jaroslav Nesetril
This booklet is a transparent and self-contained advent to discrete arithmetic. Aimed quite often at undergraduate and early graduate scholars of arithmetic and machine technology, it truly is written with the objective of stimulating curiosity in arithmetic and an lively, problem-solving method of the awarded fabric. The reader is ended in an knowing of the fundamental ideas and strategies of really doing arithmetic (and having enjoyable at that). Being extra narrowly centred than many discrete arithmetic textbooks and treating chosen issues in an strange intensity and from a number of issues of view, the publication displays the conviction of the authors, lively and across the world well known mathematicians, that an important achieve from learning arithmetic is the cultivation of transparent and logical pondering and conduct precious for attacking new difficulties. greater than four hundred enclosed workouts with quite a lot of hassle, lots of them observed through tricks for answer, aid this method of educating. The readers will take pleasure in the energetic and casual type of the textual content followed by means of greater than two hundred drawings and diagrams. experts in a variety of elements of technological know-how with a simple mathematical schooling wishing to use discrete arithmetic of their box can use the e-book as an invaluable resource, or even specialists in combinatorics may perhaps sometimes examine from tips to study literature or from displays of contemporary effects. Invitation to Discrete arithmetic may still make a pleasant studying either for rookies and for mathematical professionals.
the most subject matters comprise: undemanding counting difficulties, asymptotic estimates, in part ordered units, simple graph thought and graph algorithms, finite projective planes, straightforward likelihood and the probabilistic process, producing capabilities, Ramsey's theorem, and combinatorial purposes of linear algebra. normal mathematical notions going past the high-school point are completely defined within the introductory bankruptcy. An appendix summarizes the undergraduate algebra wanted in a number of the extra complex sections of the publication.
Read Online or Download An Invitation to Discrete Mathematics PDF
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Extra info for An Invitation to Discrete Mathematics
It is not only remarkable but also surprising, since set theory, and even the notion of a set itself, are notions which appeared in mathematics relatively recently, and some 100 years ago, set theory was rejected even by some prominent mathematicians. Today, set theory has entered the mathematical vocabulary and it has become the language of all mathematics (and mathematicians), a language which helps us to understand mathematics, with all its diversity, as a whole with common foundations. We will show how more complicated mathematical notions can be built using the simplest set-theoretical tools.
N be n ≥ 2 distinct lines in the plane, no two of which are parallel. Then all these lines have a point in common. 1. For n = 2 the statement is true, since any 2 nonparallel lines intersect. 2. Let the statement hold for n = n0 , and let us have n = n0 + 1 lines 1 , . . , n as in the statement. e. the lines 1 , 2 , . . , n−1 ) have some point in common; let us denote this point by x. Similarly the n − 1 lines 1 , 2 , . . , n−2 , n have a point in common; let us denote it by y. The line 1 lies in both groups, so it contains both x and y.
2). The operations ∪ and ∩ are also commutative, in other words they satisfy the relations X ∩ Y = Y ∩ X, X ∪ Y = Y ∪ X. The commutativity and the associativity of the operations ∪ and ∩ are complemented by their distributivity. For any sets X, Y, Z 14 Introduction and basic concepts we have X ∩ (Y ∪ Z) = (X ∩ Y ) ∪ (X ∩ Z), X ∪ (Y ∩ Z) = (X ∪ Y ) ∩ (X ∪ Z). The validity of these relations can be checked by proving that any element belongs to the left-hand side if and only if it belongs to the righthand side.
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CC-MAIN-2021-21
| 6,124 | 19 |
http://forecasters.org/ijf/journal-issue/266/article/5741
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math
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Volume 18 Issue 2 (April-June 2002)
Forecasting Long Memory Processes
edited by Richard T. Baillie, Nuno Crato, Bonnie K. Ray
Computation of the forecast coefficients for multistep prediction of long-range dependent time series
Three different linear methods, called the truncation, type-II plug-in and type-II direct, for constructing multistep forecasts of a long-range dependent time series are discussed, all three methods being based on a stochastic model fitted to the time series for characterizing its long-memory as well as short-memory components. However, while the forecast coefficients for the truncation method may be obtained from the model equation itself, those for the type-II plug-in and type-II direct methods involve the autocorrelation function of the fitted stochastic model, analytic exact expressions for which may be cumbersome to evaluate. A numerical quadrature procedure, based on the fast Fourier transform algorithm, for computing the autocorrelation function of a long-memory process, as well as the forecast coefficients for the truncation method, is suggested and the computational accuracy of the approximations is investigated for several ARFIMA models. The three methods of constructing the forecasts of a long-memory time series apply when the innovations, @e"t, of the fitted model are postulated to follow a Gaussian distribution, and, also, when they follow an infinite variance stable distribution with characteristic exponent @t, 1<@t<2. A comparison of the multistep forecasts produced by these three methods is carried out by simulating several ARFIMA models with both Gaussian and stable innovations, and two FEXP models with Gaussian innovations. In addition, their relative behaviour with an actual time series, namely, the mean temperature in England, 1659-1976, is examined.
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| 1,824 | 5 |
http://www.longrangehunting.com/forums/f17/need-help-112030/index4.html
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math
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Originally Posted by lougonl
I have 2 long range rifles but will talk about one today. I am having problems with my ballistic solution I cannot get it to match up. I have shot my data and know what MOA is needed to hit zero and several distances but cannot get the balistic programs to match up with this data and I have tried a few programs. I will list the data I have if someone can please tell me what I am doing wrong, I am desperate. I am faily new at this.
168 Grain Berger Classic Hunter
BC G-1 .496 G7 .254
Muzzle Velocity 2553 FPS
Sight above bore 2"
Sighted in at 200 yards
Temp 65 degree's
400 5.5 MOA
500 8.75 MOA
600 14 MOA
700 20 MOA
800 27 MOA
Please help me!!!
I don't see where you mention Barometric Pressure? That will make a difference.
Things to check...
-Correct vertical alignment of your scope to your bore, I use a plum bob
-Actual scope tracking, you can check this by locking your rifle into a rifle vice at a range and checking the travel against a yard stick @ exactly 100 yds from the scope. 120 clicks (30 MOA) should equal 31.41"
-Accurate scope height.
Off the top of my head, the cause appears to be either a tracking problem or an incorrect BC or a combination of things. Your rate of ballistic decay is greater than the BC you're using suggesting the actual BC is lower. Velocity is always suspect.
In this case your actual drops vs charted drops are less @ 400 and 500, about the same @ 600 and lower @ 700 and 800.
This suggests to me that actual velocity is higher and actual BC lower than programmed.... assuming actual scope height, alignment, and tracking are correct.
Also need to know the Barometric Pressure.
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| 1,654 | 23 |
https://brainmass.com/business/finance/153076
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math
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Please help me with steps.
1. Trader Joe's orders a 6-week supply of its frozen organic chocolate waffles when stock on hand drops to 400 units. The lead time for this item is 4 weeks. Demand is normal with an average of 80 units per week and standard deviation of demand is estimated 50 units per week.
a. What level of customer service is Trader Joe's providing to its distributors in terms of stock availability? What's their average inventory of frozen organic chocolate waffles?
b. Trader Joe's decides that 97.7% is an appropriate ratio of customers they can meet the demand of, in terms of that product. What will be their new safety stock and average inventory after such an improvement?
c. Trader Joe's assumes that a customer who can't find the product in the store is a lost customer and the cost of one lost customer is $16. The holding cost of the product in inventory is $8 per year. Then compare the two policies by calculating the total costs of holding and lost sales.
2. a. If we have some information about the demand, why don't we simply hold inventory for the expected demand, wouldn't this minimize the total cost, i.e., why do we hold safety stock?
b. What does an organization need to determine before deciding on its amount of safety stock?
3. You are a retailer, ordering your products from a producer and you just keep cyclical inventory (none of seasonal or safety stock inventories). You observe that your demand has increased recently, so your inventory has to be replenished more quickly, but you are not planning to change your lot size (Q).
a. What's the impact of this on the amount of inventory you hold?
b. What's the impact of this on the flow times?
4. Up, Up, and Away is a producer of kits and wind socks. Relevant data on a bottleneck operation in the shop for the upcoming fiscal year are given in the following table:
Item Kites Wind Socks
Demand forecast 30,000 units/year 12,000 units/year
Lot size 20 units 70 units
Standard processing time 0.3 hour/unit 1.0 hour/unit
Standard setup time 3.0 hour/lot 4.0 hour/lot
The shop works two shifts per day, 8 hours per shift, 200 days per year. There currently are 4 machines, and 75% capacity utilization is desired. How many machines should be purchased to meet the upcoming year's demand?
5. A department store sells (among other things) sports shirts for casual wear. The salesman in charge of the men's department knows that the demand for one of these shirts is fairly constant at 250 shirts per year. These shirts are obtained from the manufacturer who charges a delivery fee of $65, regardless of the number of shirts delivered. In addition, in-house costs associated with each order total $6.
The manufacturer charges $16.25 per shirt, but is willing to lower the price by 3 percent per shirt if the department will order 288 shirts each time. The holding costs are estimated to be 8.5% of the shirt cost.
Should the salesman recommend that the department store accept the offer of the quantity discount and larger order size? Please calculate.
This solution provides an explanation of what level of customer service Trader Joe's is providing to its distributors.
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| 3,161 | 21 |
http://www.sciencebugz.com/chemistry/chprbgas.htm
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math
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Boyle's Law Problems
1. The volume of the lungs is measured by the volume of air
inhaled or exhaled. If the
volume of the
lungs is 2.400 L during exhalation and the pressure is 101.70 KPa, and the
pressure during inhalation is 101.01 KPa, what is the volume of the lungs during
2. The total volume of a soda can is 415 mL. Of this 415
mL, there is 60.0 mL of headspace
for the CO2 gas put in to carbonate
the beverage. If a volume of 100.0 mL of gas at
standard pressure is added
to the can, what is the pressure in the can when it has been
3. It is hard to begin inflating a balloon. A pressure of
800.0 Kpa is required to initially inflate
the balloon 225.0 mL. What is
the final pressure when the balloon has reached it's capacity
of 1.2 L?
4. If a piston compresses the air in the cylinder to 1/8 it's
total volume and the volume is
930 cm3 at STP, what is the
pressure after the gas is compressed?
5. If a scuba tank that has a capacity of 10.0 dm3
is filled with air to 500.0 KPa, what will be
the volume of the air at 702.6 KPa?
Charles' Law Problems
Combined Gas Law Problems
Dalton's Law of Partial Gas Pressures
Rate of Diffusion Problems
1) Compare quantitatively the rates of diffusion of oxygen and sulfur
2) Compare quantitatively the rates of diffusion of methane and ammonia.
3) Compare quantitatively the rates of diffusion of hydrogen sulfide and
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| 1,364 | 31 |
https://www.brightstorm.com/math/trigonometry/trigonometric-functions/trigonometric-identities-problem-1/
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math
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PhD. in Mathematics
Norm was 4th at the 2004 USA Weightlifting Nationals! He still trains and competes occasionally, despite his busy schedule.
So we are talking about trigonometric identities and one of the most important is the Pythagorean Identity; Cosine squared theta plus sine squared theta equals 1. And this identity comes in two other forms, sine squared theta equals 1 minus cosine square theta and cosine square theta equals I minus sine squared theta.
I’ll get a chance to use one these in the next example here’s the problem. If sine of theta equals 3/5 and theta is between pi over 2 and pi, find cosine theta and tangent theta. So first of all let me find cosine theta using a Pythagorean identity; Cosine squared theta equals 1 minus sine squared theta. Sine of theta is 3/5 so I have 1 minus 3/5 squared. And 3/5 squared is 9 over 25, 1 minus 9 over 25, 1 minus 9 over 25 is 25 over 25 minus 9 over 25 and that’s 16 over 25.
So cosine squared theta equals 16 over 25. That means cosine theta is plus or minus 4/5. In order to determine whether there is plus or minus I have to think about what quadrant I’m in. And the problem says theta is between pi over 2 and pi and that means theta is in the second quadrant. And in the second quadrant cosine is negative so cosine is -4/5 because we are in quadrant 2. So now I know the cosine and I have to find the tangent let’s use the tangent identity. Tangent theta equals sine theta over cosine theta.
Sine of theta is 3/5 and the cosine theta is -4/5. So that’s the same as 3/5 times -5 over 4 which is -¾. So for this data where sine is 3/5 cosine is -4/5 and tangent is -¾.
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https://lavelle.chem.ucla.edu/forum/viewtopic.php?f=131&t=58631
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math
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3 posts • Page 1 of 1
If the temperature is constant, we have to use the equation S=nKln(V2/V1), reversible or irreversible. A reversible change can happen without a constant temperature, but if the pressure is constant, we have to use different equations depending on what is constant.
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| 362 | 4 |
https://www.quakercreative.com/
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math
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Words that need to be heard are sometimes hard to hear.
My custom lettering drives your words home by crafting letters that visualize the voice of the message.
Your message matters, let’s make it stick.
Part 3 of this series was about the radical, integral, partial derivative, and infinity signs (√ ∫ ∂ ∞). This final math symbols post, Part 4, covers four Greek derived math symbols: Δ ∑ ∏ [...]
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https://judaism.stackexchange.com/questions/30958/what-modern-day-hygiene-products-are-prohibited-on-yom-kippur
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math
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Stack Exchange network consists of 175 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers.
What, if any, modern-day hygiene products are prohibited "oils" on Yom Kippur?
Are deodorants and perfumes in this category? What about medical ointments?
5 years, 9 months ago
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| 373 | 4 |
https://freeessayhelp.com/essay/developing-algebra-1301348
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math
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Mathematical principles aids in developing algebra innate capability to generalize from the specific cases of the underlying problem. Te second case mathematics tasks entail generality the determination of the largest the larger through addition the corresponding smaller or the square the relatively smaller added the prevailing more considerable number The validity of the statement is found in the natural response of computing individual examples. Ntural response typically encompasses the selection of individual cases that make mathematical development of algebra arithmetic easier. Mreover, te process entails numerous possibilities and fractions resulting to the successful process of computation trials.
Tying one number example frequently makes it cumbersome to convince that the two answers will always be similar. Te possibilities of knowing specific number chosen whatever the circumstance that is the two computations will result in identical answers entirely depend on the process of developing appropriate algebra expression. Nvertheless, uder the assumption and conjecture the answer is typically identical that is mainly tested in particular mathematical cases. De to the lurking generality, te fundamental step in the process entails the the generality.
Frst, te expression of the generality of the problem of algebra is normally more economical means of expressing identical ideas. Te symbols in developing algebra entities are employed in the manipulation of the variables with confidence in handling of the numbers. Fr instance, tking variable x to be any number having power in the utterance, wich is supremely creative act that is wields power over the entire figures (Mason, Gaham & Johnston-Wilder, 2005, p. Mreover, i reveals the probable numbers of solving the above problem. Te number be 1-x at they must sum to one.
Mreover the prevailing expressions compared designated aids the manipulation the expressions bid reveal the reasons the expressions typically gives identical answer regardless the numbers inserted the place variableThe task pertaining to the summation of consecutive odd numbers is normally difference of two prevailing squares. I requires specialization via attempting various examples at the beginning of the course. Secialization in developing and computing various examples of. ..
Please type your essay title, choose your document type, enter your email and we send you essay samples
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CC-MAIN-2019-39
| 2,434 | 5 |
https://oneclass.com/study-guides/ca/york/phy-astro/phys-1420/281262-phys1420-mock-exam.en.html
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math
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Physics 1420 PASS
Facebook Group: Physics 1420 pass
Intended for use by Bethune College PASS program
An owl sits on a ledge of height h and sees a mouse rum towardseed v
its home a distance s away. How fast does the owl need to accelerate (from rest)
to catch the mouse right before it gets away?
Two blocks with masses m an1 m are 2onnected by a string over a metal
wedge of angle ▯.
a)Find the acceleration of the two blocks if the two blocks are frictionless.
b)Find the acceleration if the two blocks have respective kinetic friction coe▯-
cients of k1 and uk2
A skier starts from rest at the top of a frictionless hill of height h. The skier
slides down the hill and o▯ a ramp of angle ▯ that is half as high as the hill
a) What is the kinetic energy and velocity at the bottom of the hill?
b) What is the velocity of the skier when he goes o▯ the ramp?
c) What is the maximum height he can achieve?
An astronaut weighs 85kg, including her oxygen tank. She is stranded 125m
from the space station. In order to get back before the oxygen in her suit runs
out, she must launch her oxygen tank away from the space station.
a) How fast must must she throw the 15kg oxygen tank to get back with 2
b) Assuming that she pushes the oxygen tank away for 0.2 sec, what average
force would she have to apply in order to obtain the necessary speed?
A small object of mass m is attached to a rod of length l by two stings, each of
length s. If the object rotates around the rod at a constant speed of v, ▯nd the
tension in each string.
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| 1,539 | 25 |
https://www.amirite.com/826452
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math
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Also about taking and answers-111When you're taking a test and you just decided between two answers, you write the other answer nearby so the teacher knows you had it too if you get the question wrong, amirite?
Also by ThePrinceofWales+21What do you LIKE to receive in the mail? amirite?
Also about taking and answers+330You freak out when taking multiple choice tests and all the answers seem correct, but then you read 'all of the above' and calm down, amirite?
Also about Jokes & Humour+18Experience: The name people give to their mistakes. Amirite?
Also about taking and answers+432If you're taking a quiz in a magazine or something, you don't like it when you can see a commonality between all the A's, B's, etc because it starts to sway your answers, amirite?
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https://t5k.org/curios/page.php?number_id=20959&submitter=Rivera
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math
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This number is a composite.
Just showing those entries submitted by 'Rivera': (Click here to show all)
The start of the earliest run of 9 numbers (either prime or Smith numbers) such that the sum of their digits is equal to the sum of the digits of their prime factors counted with multiplicity. A run of 10 such numbers will be greater than 5*10^12. By Giovanni Resta. [Rivera]
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http://sahomeworkkatc.trooperwheel.info/essay-questions-for-high-school-math.html
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math
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What are some good original research topics in math for high school students how do i write a good high school level essay why is math interesting. What are some good original research topics in math for high school students update cancel what are some ideas for math research for a high school student. Math research topics for high school as you can see from the topic of our research for satisfied customers, there will definitely be everything good about your paper. This article itemizes the various lists of mathematics topics some of these lists link to hundreds of articles some link only to a few the template to the right. Download a pdf of high school mathematics at work by the national research council for free.
I would like to ask for interesting mathematics topics that i could conside possible research topics for high school up vote 3 down vote favorite 1. 100% free papers on math essay sample topics, paragraph introduction help, research & more class 1-12, high school & college. Writing prompts for high school math classes outline definition an outline is for detailed math for writing the essay writing topics, math this highh.
To view and seniors phobia thesis example storytelling essay, many math journal or inference 3 hours ago are constructed responses to improve the high school teaching. Try our free tasc practice test great test prep for the tasc high school equivalency test practice questions with answers and detailed explanations. High school mathematics at work: essays and soon as you think of the questions some features of high-quality high school mathematics teaching and.
Learn how to win college scholarship money now with these 10 essay contests for high school sophomores and juniors. Find and save ideas about math writing prompts on pinterest get teens excited about writing with essay prompts high school students will love essay topics.
A comparison of high school and college length: whether it be a new concept in math or a description of how to properly play kickball in gym essay topics. Writing prompts, topics general writing assignments math biography: describe how you feel about math high school and college. Speed and quality is the best asset of our essay writing service high school math homework help high school math homework help math homework help for high school.
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https://www.onlinemathlearning.com/area-problems-scale-drawings.html
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math
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Plans and Worksheets for Grade 7
Plans and Worksheets for all Grades
Lessons for Grade 7
Common Core For Grade 7
Examples, videos, and solutions to help Grade 7 students solve area problems related to scale drawings and percent.
New York State Common Core Math Grade 7, Module 4, Lesson 15
Download worksheets for Grade 7, Module 4, Lesson 15
Lesson 15 Student Outcomes
• Students solve area problems related to scale drawings and percent by using the fact that an area, A', of
scale drawing is k2
times the corresponding area, A, in the original drawing, where k is the scale factor.
Lesson 15 Classwork
For each diagram, Drawing 2 is a scale drawing of Drawing 1. Complete the accompanying charts.
For each drawing: identify the side lengths, determine the area, and compute the scale factor.
Convert each scale factor into a fraction and percent, examine the results, and write a conclusion
relating scale factors to area.
Key Points: Overall Conclusion
If the scale factor is represented by k, then the area of the scale drawing is k2
times the corresponding area of the
What percent of the area of the large square is the area of the small square?
What percent of the area of the large disk lies outside the smaller disk?
If the area of the shaded region in the larger figure is approximately 21.5 square inches, write an equation that relates
the areas using scale factor and explain what each quantity represents. Determine the area of the shaded region in the
smaller scale drawing.
Use Figure 1 below and the enlarged scale drawing to justify why the area of the scale drawing is k2
times the area of the
Explain why the expressions (kl)(kw) and k2
lw are equivalent. How do the expressions reveal different information
about this situation?
1. The Lake Smith basketball team had a team picture taken of the players, the coaches, and the trophies from the
season. The picture was 4 inches by 6 inches. The team decides to have the picture enlarged to a poster and then
enlarged again to a banner measuring 48 inches by 72 inches.
a. Sketch drawings to illustrate the original picture and enlargements.
b. If the scale factor from the picture to the poster is 500%, determine the dimensions of the poster.
c. What scale factor is used to create the banner from the picture?
d. What percent of the area of the picture is the area of the poster? Justify your answer using the scale factor
AND by finding the actual areas.
e. Write an equation involving the scale factor that relates the area of the poster to the area of the picture.
f. Assume you started with the banner and wanted to reduce it to the size of the poster. What would the scale
factor as a percent be?
g. What scale factor would be used to reduce the poster to the size of the picture?
Try the free Mathway calculator and
problem solver below to practice various math topics. Try the given examples, or type in your own
problem and check your answer with the step-by-step explanations.
We welcome your feedback, comments and questions about this site or page. Please submit your feedback or enquiries via our Feedback page.
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https://www.fellowshipforums.com/t/soa-monograph-a-new-approach-to-managing-operational-risk-ch-8/74
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math
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I’ll give binomial distribution a shot
One type of operational risk is process risk. So let’s say we have a conveyor belt that puts coke caps on coke bottles. Let’s say it it does it 30 bottles per second.
Let’s say that on average, every 1000th bottle has an issue. It may be the 998th bottle, or the 1013th bottle, but it’s always around the 1000th bottle.
I would say that this follows a binomial distribution because the risk occurs not on every 1000th bottle on the dot, but AROUND the 1000th bottle (low variability).
Does this sounds reasonable?
I can’t really think of an example where there’s no frequency relative to the mean - If that’s the case wouldn’t we know exactly when the risk will occur (ex/ every 1000th bottle on the dot), meaning we can just exercise an action to stop the risk from occurring?
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https://tex.stackexchange.com/questions/188404/emacs-skim-live-preview
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math
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I have been using Latexian for a while, but some of its limitations and my natural tendency to use emacs, is prompting me to take a look at my work environment again (OSX 10.9.2).
I have seen some guides for setting up Skim with Aquamacs and highlight current line in the editor. Which is great. However, is it possible to set this up so that Skim shows a live preview of the PDF? The way one can with Flashmode + TeXShop, or with Latexian?
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https://lavelle.chem.ucla.edu/forum/viewtopic.php?f=29&t=35359&p=116615
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math
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3 posts • Page 1 of 1
For question 2A15, the anticipated ion formed by Cadmium is a +2 which I understand because it has 2 electron in its 5s orbital which it wants to get rid of but what about the exceptions to electron configurations in the transition metals like copper and chromium (since while we don't need to know the 4d shell which contains Cd, we do need to know the 3d)? Since those metals only have 1 electron in their 5s orbitals, do they generally form +1 ions? Do we need to even know ions that form for transition metals for the test?
I think it's hard to figure out what ions form from the transition metals, so for example copper can form 1+, 2+, 3+, etc. ions. I don't think we'd need to know the ions that the transition metals would form for the test other than the really simple ones? All we need to know is that the electron is always lost from the 4s subshell before the 3d, for example).
It's difficult to predict the most likely ion that would form, because they can take several different oxidation states. I think just having a general idea of the possible transition metal oxidation states (+1, +2, +3, etc) is enough.
Who is online
Users browsing this forum: No registered users and 4 guests
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https://pari.math.u-bordeaux.fr/Events/PARI2017b/talks/genuine_Bianchi_modular_forms.html
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math
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Alexander D. Rahm's talk:
Genuine Bianchi modular forms for Pari/GP
Bianchi modular forms are automorphic forms over an imaginary quadratic field, associated to a Bianchi group.
Even though modern studies of Bianchi modular forms go back to the mid 1960's, most of the fundamental problems surrounding their theory are still wide open.
Only for certain types of Bianchi modular forms, which we will call non-genuine, it is possible at present to develop dimension formulas:
They are (twists of) those forms which arise from elliptic cuspidal modular forms via the Langlands Base-Change procedure, or arise from a quadratic extension of the imaginary quadratic field via automorphic induction (so-called CM-forms). The remaining Bianchi modular forms are what we call genuine, and they are of interest for an extension of the modularity theorem (formerly the Taniyama-Shimura conjecture, crucial in the proof of Femat's Last Theorem) to imaginary quadratic fields. In a paper by Rahm and Sengun, an extreme paucity of genuine cuspidal Bianchi modular forms has been reported, but those and other computations were restricted to level One. In a recent paper by Rahm and Tsaknias,
Genuine Bianchi modular forms of higher level, at varying weight and discriminant
we are extending the formulas for the non-genuine Bianchi modular forms to deeper levels, and we are able to spot the first, rare instances of genuine forms at deeper level and heavier weight.
The computations of these papers have been carried out using Pari/GP and MAGMA. What the speaker would like to find out by making this presentation, is if there is significant interest for a Pari/GP package which could perform the whole computation and produce the dimensions of spaces of genuine forms.
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http://jwilson.coe.uga.edu/EMT668/EMAT6680.2002.Fall/Imler/EMAT%206690%20Essay%20on%20Proofs/html%20files/Fallacious%20proof%204.html
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math
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We begin our proof by drawing a rectangle ABCD. We then draw CE outside the rectangle so that AD congruent CE. Point P is the intersectoin of the perpendicular bisectors of AE and CD, which intersect AE and CD at points M and N, respectively. Drawing DP, AP, EP, and CP completes the diagram for this "proof".
Because AP = EP and DP = CP (every point on theperpendicular bisector of a line segment is equidistant from the endpoints of theline segment), we have triangle ECP is congruent to triangle ADP (SSS) and the measure of angle ECP = the measure of angle ADP.
However, because triangle PDC is isosceles, the measure of angle DCP = the measure of angle CDP. By subtraction, obtuse angle ECD has the same measure as angle ADC, a right angle!!
You may wish to consider the case when P is on DC or when P is in rectangle ABCD. Similar arguments hold for these cases.
By now you may find that an accurate construction is the best way to isolate the error in the 'proof'. Rahter than attempt to discover the error by construction, we will analyze the situation that now exists. We notice that NP is also the perpendicular bisector of AB. Consider triangle ABE. Because NP and MP are the perpendicular bisectors of AB and AE, respectively, they intersect at the center, P, of the circumcircle of triangle ABE. Therefore, point P must also be on the perpendicular bisector of BE.
By construction, we have BC = EC. Therefore point C must also lie on the perpendicular bisector of BE.
PC is the perpendicular bisector of BE as well as the interior angle bisector of angle BCE. A reflex angle is an angle of measure greater than 1880 degrees and less than 360 degrees. Consider reflex angle ECP, whose measure is measure of angle PCR + measrue of angle RCE. Thus EP in triangle ECP is placed so that it is on the side of point C outside the rectangle. This makes the last step of our proof incorrect because measure of angle ECP does not equal the measure of angle ECD + the measure of angle DCP.
Back to home page
Back to fallacious proof 3
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https://dr.lib.iastate.edu/entities/publication/708fc73b-87eb-4924-a7d2-a62469054a51
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math
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Analysis of discrete reaction-diffusion equations for autocatalysis and continuum diffusion equations for transport
Is Version Of
We analyze both the spatiotemporal behavior of non-linear "reaction" models utilizing reaction-diffusion equations, and spatial transport problems on surfaces and in nanopores utilizing the relevant diffusion or Fokker-Planck equations.
The non-linear "reaction" models involve spatial discrete systems where "particles" reside at the sites of a periodic lattice: particles, X, spontaneously annihilate (X->O) at a specified rate p, and are autocatalytically created given the presence of nearby pairs of particles (O+2X->3X) at rates depending on the local configuration. [This reaction model is equivalent to a spatial epidemic model where sick individuals spontaneously recover (S->H), and healthy individuals are infected by pairs of sick neighbors (H+2S->3S).] The model exhibits a non-equilibrium phase-transition from a populated state to a vacuum state (with no particles) with increasing p. Near this transition, one can consider the propagation of interfaces separating the two states. Planar interfaces exhibit an orientation-dependence (leading to so-called generic two-phase coexistence), and curved interfaces enclosing droplets exhibit even richer behavior. These phenomena are analyzed utilizing the appropriate set of discrete reaction-diffusion equations (corresponding to lattice differential equations).
Diffusive transport of particles between islands or clusters of particles on a surface leads to coarsening of island arrays which can be analyzed by solution of an appropriate boundary value problem for the surface diffusion equation. We extend previous treatments to strongly anisotropic systems. Diffusion and passing of pairs of overdamped Langevin molecules in narrow nanopores can be described by the appropriate Fokker-Planck equations (corresponding to a high-dimensional diffusion equation). We provide the first analysis of this problem focusing on a characterization of the propensity of passing as a function of pore diameter.
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https://www.pearson.com/channels/physics/learn/patrick/geometric-optics/reflection-of-light
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math
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33. Geometric Optics
Reflection Of Light
Reflection of Light
Was this helpful?
Hey, guys, in this video, we're going to talk about the reflection of light off of a boundary between two media. Okay, let's get to it. Now. Remember when a wave encounters a boundary, it could do one of two things. It can either reflect off of that boundary or it can transmit through that boundary and propagate in the new media. In reality waves, they're gonna do a little bit of both now for light boundaries. Air, typically referred to or media, I should say, are typically referred to as one of two things. They could either be reflective, like the surface of the mirror, and mainly on Lee. Allow reflection. Or they could be transparent like glass and mainly allow transmission. Okay, for now, we wanna we wanna talk about reflective surfaces. Alright, Light reflects off of a boundary in the same manner as a ball undergoing an elastic collision with a wall. Okay, if we have a ball at some point with some momentum right here in some direction, when it hits the wall, it's going to conserve that momentum, but it's going to go in the opposite direction, okay? And it's gonna enter with some angle and it's gonna leave with some other angle light is gonna do the same thing if I draw a ray of light encountering a boundary between two media that Ray is gonna leave at some angle. Okay, We have something called the Law of Reflection, which holds for the elastic collision. Just like it holds for reflection of light. That states that that reflected angle is actually the same as the incident angle, the angle with which it hits the boundary. Okay, now, just a side note. This isn't gonna be important right now, but this is gonna be very important later. Whenever light encounters a boundary, we always measure the angle relative to the normal direction relative to some line that is perpendicular to the boundary. Right. This is the normal. Just like the normal force is perpendicular. The normal direction is perpendicular to a surface. Okay, let's do an example. We want to find the missing angle theta using the law of reflection. Okay, so this light ray is incident at 65 degrees. It undergoes reflection right here, which means that the outgoing angle is also going to be 65 degrees. That's just what the law reflection says now notice this is the normal too sorry, Normal to this surface right? Which means that it's had a 90 degree angle from the surface. So this angle right here is gonna be the complementary angle to degrees, which is 25 degrees, right, because they have to add up to 90. Now notice we have a triangle right here that looks like this. Here's our 25 degree angle. Here's our 125 degree angle and here is an unknown angle. Whoops, letting men wise myself remember that the sum of all of the internal angles within a triangle has to be so 25 degrees plus 120 degrees plus this unknown angle has to be 180 degrees. That means that this unknown angle is actually 35 degrees. Okay, so that tells us what this angle is. This angle is 35 degrees. Now, once again, this line right here is the normal to the second surface, which means that it's perpendicular to the second surface. So this angle right here is going to be the complementary angle to 35 degrees right, the angle that, when added to 35 degrees, equals 90 and that is 55 degrees. So what is theta have to be, while the law of reflection says it has to be the same as that incident angle which on the second surface is 55 degrees, So theta, therefore, is 55 degrees. Alright, guys, that wraps up our discussion on the reflection of light and the law reflection. Thanks for watching.
Was this helpful?
Hey, guys, let's do an example. Ah, flat mirror hangs 0.2 m off the ground. If a person 1.8 m tall stands 2 m from the mirror, what is the point on the floor nearest the mirror, which we called X that can be seen in the mirror. The geometry for this problem has already set up. All we have to do is use the law of reflection to figure it out. So this light ray is coming up off the ground from this point, encountering the mirror at its lowest point and then leaving in this direction to your eyes. What, you're gonna be here? Okay? What we have to do is use the law reflection to figure out what angle properly, adjust that light race so that it meets you at exactly 1.8 m off the ground. Here's the normal because we always use the normal one measuring angles. So this is our incident angle theta one. And this is our reflected angle theta one prime. And remember that those two are equal now, notice if I were to continue this normal line right here we form a triangle, right? The triangle that we form is 2 m wide. How tall is it? Well, it's not 1.8 m tall because the mirror the bottom of this point right here is 0.0.2 meters off the ground. So it's actually 1.8 minus 0.2, which is 1.6 m. And this angle right here is state of one prime. Okay, which is the angle that we're interested in finding so clearly we could just use trigonometry to find this. We can use the tangent and we can say that the tangent of data one prime is gonna be the opposite edge which is 1.6 m divided by the adjacent edge which is to and that tells us that data one prime is just 51.3 degrees. Okay, so now we know that this angle is 51. degrees. So this angle to is 51.3 degrees. So we have a new triangle right here. Let me minimize myself. We have a new triangle now this lower triangle where this angle the incident angle is 51.3 degrees. This height is 0.2 m and this length is X right. What's this angle going to be? Well, this is what's known as an alternate interior angle to 51. and all alternate interior angles are the same. So this is going to be 51 3 degrees. All right, So once again, we can use the tangent to find what X should be. We could just say then that the tangent of 51.3 degrees equals the opposite, which is 0.2 m divided by the adjacent which is X, or that X is 0.2 over the tangent of 51.3 degrees. And finally, that X is just 0. meters. Okay, Just using geometry and trigonometry, we can answer this question. The crux of the physics is that these angles right here are the same. And that's the law of reflection. All right, guys, Thanks for watching
What is the distance, d, between the incoming and outgoing rays?
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https://www.physicsforums.com/threads/interpreting-the-second-moment-of-area-and-eulers-formula-values.959243/
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math
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Hi all I was hoping someone could remove some doubt in my mind with regards to interpreting the Second Moment Of Area and Eulers formula for buckling. Am I correct in thinking that:- - The higher the Second Moment Of Area Value the more resistant to bending. - The lower the Second Moment Of Area Value the less resistant to bending. Using Eulers formula for buckling I have calculated the critical load for a column in the X and Y axis, my values are:- Ix = 25.39N Iy = 634N Am I correct in interpreting these results as the column will buck in the x axis first because it will only take 25.39N of load before it buckles - is this correct? I can do the math it is the concept I struggle with (doesn't help that I have a crap tutor). thank you.
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https://english.stackexchange.com/questions/71164/whats-the-figurative-meaning-of-the-end-of-the-line
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math
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Leaving aside the strong sexual connotations of the lyrics of Pulp's song "This Is Hardcore", I was wondering what is the figurative meaning of the expression the author used:
"This is the end of the line"
I thought it could be "this is it" or also could mean "I am putting an end to this", but I am not sure.
I was curious about the possible uses of the phrase, not only this particular case (ie, this lyrics).
Also, I would like to know about the literal meaning. Is it refering to a drawn line? A spoken line?
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https://www.exampleessays.com/viewpaper/61899.html
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math
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Quadratic function is an function that can be written in the form f(x)= ax2 + bx + c, where a 0. It is defined by a quadratic expression, which is an expression of the form ax2 + bx + c, where a 0.
The graph of a quadratic function is called a parabola.
Axis of symmetry is a line that divides the parabola into two parts that are mirror images of each other. The vertex of a parabola is either the lowest point on the graph or the highest point of the graph.
Minimum and Maximum Values: Let f(x)= ax2 + bx + c, where a 0. The graph of f is a parabola.
If a > 0, the parabola opens up and the vertex is the lowest point. The y-coordinate of the vertex is the minimum value of f. If a < 0, the parabola opens down and the vertex is the highest point. The y-coordinate of the vertex is the maximum value of f. .
If x2=a and athen x is called a square root of a. If a > 0, the number a has two square roots,a and -/a. The positive square root of a,a, is called the principal square root of a. IF a=0, then0=0. When you solve a quadratic equation of the form x2=a, you can use the rule below:.
Solving Equations of the Form x2=a: If x2=a and a 0, then x=/a or x= -/a, or simply x= +-/a.
Use the Properties of Square Roots below to simplify the resulting square root. Properties of Square Roots: Product Property of Square Roots - if a and b0:ab =a *b. Quotient Property of Square Roots - If aand b > 0: |a/b| = |a| |b|.
Pythagorean Theorem: If ÑABC is a right triangle with the right angle at C, then a2 + b2 = c2/.
Factoring reverses the process, allowing you to write a sum as a product.
Factoring x2 + bx + c: To factor an expression in the form ax2 + bx + c where a=1, look for the integers r and s such that r*s=c and r+s=b. Then factor the expression. x2+bx+c=(x+r)(x+s).
Factoring the Difference of Two Squares: a2-b2=(a+b)(a-b).
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https://www.physicsforums.com/threads/conservation-of-energy-momentum.223572/
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math
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1. The problem statement, all variables and given/known data A proton with mass 1.59*10-27 kg is propelled at an initial speed of 3.49*105 m/s directly toward a uranium nucleus 4.97 m away. The proton is repelled by the uranium nucleus with a force of magnitude F = a/x2 where x is the separation between the two objects and a = 2.18 *10-26 Nm2. Assume that the uranium nucleus remains at rest. What is the speed of the proton when it is 8.60 *10-10 m from the nucleus? 2. Relevant equations net force(F)=rate of change of momentum(dP/dt) Total Energy before=Total Energy after mathematically, E=kinetic energy(KE) before+potential energy(U) before=KE after + U after 3. The attempt at a solution plugging in given values into the equations i think are relevant, but none of them get you any where..IS THERE AN EQUATION IM SUPPOSED TO USE?
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http://glilmanduil.edenresearch.info/19641-examples-of-thesis-topics-in-mathematics.php
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math
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So, computer scientists have been trying for the last decade to find a deterministic algorithm which works in polynomial time. Inference in curved exponential families, following a principled approach, requires construction of exact or approximate ancillary statistics. help me write my essay key stage 3 This can be studied both experimentally and analytically. Principal Component Analysis PCA is one of the most widely used statistical tools for data analysis with applications in data compression, image processing, and bioinformatics. One wants to match up the pictures, but there is some error in the measurement.
This project involves studying, in real-data examples, how classical inference procedures are invalidated by the use of selection procedures. It is known that a convex planar U can have at most one equichordal point. watch thesis movie online In this project, we will harness the power of randomized dimension reduction to accelerate methods in applications which require high computational costs.
Examples of thesis topics in mathematics thesis for phd governance 2018
What is true in dimension three? The deterministic versions of such algorithms suffer from slow convergence in some cases. Although in its general form this is a difficult and technical topic, it is possible to go a long way into the subject with only Math Blake Thornton for suggestions about faculty to talk with. This area is appropriate for both reseach and expository projects.
Imagine a large number of cameras arranged around a central object. Consider the following easy exercise as a warm-up. A question of this sort first appeared in the lectures of the legendary th century mathematician Georg Frobenius, which is why this problem was named after him. It is part of the general subject of Dynamical Systems. Most large public-access data sets have this complex structure.
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We are developing statistical models to tackle these issues. The quest to produce Calabi-Yau 3-manifolds three complex dimensions! This was covered in a course - Math I think - but that was so long ago it's not listed in the catalog. order a paper gift box diy It would also be useful to know what a Riemann surface is, but this could be dealt with in summer reading.
The CMC math and CS faculty represent a wide range of research areas, including algebraic topology and knot theory, functional, harmonic, and complex analysis, probability and statistics, numerical analysis, PDEs, compressed sensing, mathematical finance, number theory, discrete geometry, programming languages, and database systems. Of course, the ancient method of Eratosthenes sieve method is one such algorithm, albeit a very inefficient one. request for proposal for grant writing services Both deal with the idea that certain variables predict whether a response is necessarily zero, and if the response is not necessarily zero, then other variables might predict its value. Intuitively, as C approaches 0, the deterministic billiard system should behave more and more like the probabilistic system of 7. Here is a much more surprising fact that you might like to think about.
It is known that a convex planar U can have at most one equichordal point. This is particularly true for topology, specially for what is called "algebraic topology". thesis for dummies body image Welker has been used to investigate the structure of such complexes. Undergraduate Research Ideas Some of our faculty have listed ideas for undergraduate research work.
Thesis preparation guidelines upm
Show that c is the shortest path contained in the surface that joins p and q. This area is appropriate for both reseach and expository projects. Recent results related to the bound in the Berry-Esseen theorem, for summands of both i. Moreover, they had to figure out how to resolve them -- the higher-dimensional analogue of lifting an actual string off itself. I have a friend who has some pathological gambling data, who has extracted most of the obvious results from her data, but might be looking for help in digging out some remaining gems.
These differential equations are of a very special kind: Among the ideas posted here, some are harder and some easier. Professor Mohan Kumar Algebra 1 If a1,a2, To make the point, consider the following. Methylation is important to embryonic development and cancer.
Both deal with the idea that certain variables predict whether a response is necessarily zero, and if the response is not necessarily zero, then other variables might predict its value. From an employee database, can one identify employess who are likely to leave the company from those who will stay? There are related functions called Grassmanian polylogarithms, invented by A. For any partially ordered set P, the set of all totally ordered subsets of P determines a simplicial complex. Most of the time spent in courses on ODEs, like Math , is devoted to linear differential equations, although a few examples of non-linear equations are also mentioned, only to be quickly dismissed as odd cases that cannot be approached by any general method for finding solutions.
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https://naturalphilosophy.org/forums/reply/697/
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math
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Activity Feed › Forums › New Models › Infinity › Reply To: Infinity
AndyMemberDecember 2, 2020 at 1:09 pm
I think it’s time to shatter the myth and mystique of dimension.
I’ve come realize something very fundamental. No one knows what the hell dimension means. Not even me entirely until this very moment. It’s been staring at us forever, but I do think we ever appreciated the depth of the mathematical truth.
I know that statement is true, because for years I’ve been asking the question. How do we define dimension? I felt stupid asking the question, because I thought it was something I should just know. String theory readily works in multiple dimensions. Quantum physics talks regularly about hidden dimensions and parallel universe. We have erroneously (and arguably I suppose) viewed space as 3D. I honestly did feel stupid asking the question. To my surprise, no one seemed to be able to answer the question coherently. And I thought I was the one having the issue, so I stopped asking.
There is no definition of what constitutes dimension in physics anywhere.
What is dimension?
A dimension mathematically only represents one thing. It’s a finite line segment. On one end of the line, we have a minimum state, and on the other end, we have a maximum state.
That’s a mathematical dimension. Simple enough to understand.
There are 3 fundamental dimensions that make up the universe, space, time, and motion, and those dimensions subdivide into separate, or equal but opposite dimensions.
For the total universe, I inserted space into the middle. Space can either exist, or not exist. If either state existed on its own, that state would be absolute. There would be nothing else. That’s not what we observe, because we exist.
Space is the top dimension, because it represents the material universe. Everything depends on space to exist. For a universe to exist within space, space needs to do something.
Space needs mass for a universe to exist, so I added expansion and contraction.
1|–<–expansion–<–|0 (space we traverse)
Mass alone doesn’t mean anything without energy, so we need to add motion:
0|–<–deceleration–<–|1 (space we traverse)
Then we must be able to perceive everything, so we insert time.
1|–>–time fast–>–|0 (matter)
1|–<–time slow–<–|0 (space we traverse)
That could be considered 6 dimensions out of 1. These are the base fundamental dimensions that make up our reality. None of these can either be created nor destroyed. They are what drives existence.
We can say the universe is 3D, but really, even that’s probably wrong in hindsight.
The universe is an amalgamation of a multitude of sub dimensions, all subdivided from the base set. Each one of those sub-dimensions becomes important in understand the universe, from the physical world, to the cognitive world, to the emotion world, to the biological world, etc. The list could go on and on, depending on what it is we’re trying to understand. Dimensions are the variables in the problem. Understanding the state of each dimension helps us solve problems.
We could go on categorizing difference dimensions from this point, into things like, cognitive dimensions, emotional dimensions, physical dimensions, material dimensions, perceptual dimensions, etc., etc.
I must question uncertainty as a principle in quantum mechanics. That’s more of a cognitive or emotional dimension, not really a physical property. It’s a contradiction in terms as a principle. Are we absolutely certain we are absolutely uncertain? It seems a little unproductive to me. The tail wagging the dog. A circular argument.
Another thing I notice is that these sub-dimensions have an orientation. And they should.
It’s ironic, because I can’t even see length, width, or height, as meaningful dimension. Short/long pretty much defines all of them. How we put multiple lengths together in a math problem is what becomes important in understanding the problem. Space isn’t really 3D, it’s 1D, but it has scale due to expansion and contraction. That’s what defines as something tangible. And with an arbitrary length, width and height, we can perceive what that scale means in relationship to something else. Its scale can define density, or its motion can define temperature, etc. There is a lot of facets to sub-dimensions.
This is a new line of thought for me, so I’m still sorting it out in my mind.
But it makes perfect dense. We’re multi sub-dimensional beings in a multi sub-dimensional universe derived from space, time and motion.
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https://www.jiskha.com/display.cgi?id=1225580920
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math
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math helpp! :|
posted by jazz .
in a bag of coloured cubes, the ratio of red cubes to total number of cubes is 5:7. If there are 105 cubes in the bag, how many cubes are red? i don't get this :|
You can set up a ratio:
5/7 = x/105
Cross multiply and simplify if needed.
alright thx!! :)
i have another question.
im sorry u have to suffer.:)this is confusing:S
Which country has the greater population density? Wirte its population density. The United kingdom with about 60 million people and an area of 244 800km2 or china with about 1806 million people and an area of 9 590 000 km2.
does anyone know the answer for this?
i mean help me with this?? :P
whaaa you want me to do your hw now pick up ur lazy ass and do it urself -_-
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https://math.answers.com/Q/What_does_a_negative_number_multiplied_by_a_negative_number_equal_to
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math
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A negative number multiplied by a negative yields a positive answer.
It does not. A negative number multiplied by a negative number equals a positive number.
-18.8995 and 0.8995 (approx)
No, a negative multiplied by a negative is a positive, as is of course a positive multiplied by a positive. Only when a negative is multiplied by a positive is the answer negative.
The two negatives cancel out and the number becomes positive.
Because two negative multiplied together equal a positive. So after the first multiplication we have a positive and a negative. When we multiply these we have a negative.
A positive number multiplied by a negative number equals a negative number!EX:-9 X 2 = -18
No, any 2 negative numbers multiplied together equal a positive number
No No. Positive number multiplied by a negative number gives a negative number.
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https://studyclix.ie/Discuss/Leaving-Cert-Mathematics/length-area-volume-triangle-problem
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math
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"If the height of a triangle is 5cm less than the length of it's base and if the area of the triangle is 52cm squared, find the length of the base and also the height of the triangle."
I have tried so many different ways and formulas and nothing seems to work for me. I get so far and then have no idea what to do next.
Any help would be greatly appreciated.
by height I'm assuming it means perpendicular,
factors are (x+13)(x-8)
reject negative answer because length must be positive
It's a quadratic equation so your solutions will be (b+13)(b-8)
therefore b= -13 and b= 8
since you cant have negative area your anwer is 8
you just factorise, use the minus b formula or just do it in your head..
you can do either but what i normally do is a 5 second rule so if it's not obvious what the factors are I use the minus b formula but since joe 9721 had it already factorised it i just used his figures. For ths question i probably would have used the minus b but any of the three ways would have worked
It's a pretty general question so I found it online, saves having to write it out!
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https://www.schoolinfosystem.org/2008/06/04/accelerated_mat_1/
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math
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Next fall, 26 of the sharpest fifth-grade minds at Potomac Elementary School will study seventh-grade math. The rest of the fifth grade will learn sixth-grade math. Fifth-grade math will be left to the third- and fourth-graders.
Public schools nationwide are working to increase the number of students who study Algebra I, the traditional first-year high school math course, in eighth grade. Many Washington area schools have gone further, pushing large numbers of students two or three years ahead of the grade-level curriculum.
Math study in Montgomery County has evolved from one or two academic paths to many. Acceleration often begins in kindergarten. In a county known for demanding parents, the math push has generated an unexpected backlash. Many parents say children are pushed too far, too fast.
Sixty Montgomery math teachers complained, in a November forum, that students were being led into math classes beyond their abilities.
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http://forum.xda-developers.com/showpost.php?p=50341883&postcount=1
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math
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Join Date:Joined: Jul 2013
Hi guys! The only way to find the solution is to ask cuz i find the thread guides explanation is not so good I don't understand.
So where can I download paranoid Android hybrid settings to cm 10.2/11 and how?
Appreciate all answeres
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https://www.storyofmathematics.com/linear-programming/
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Linear programming is a way of using systems of linear inequalities to find a maximum or minimum value. In geometry, linear programming analyzes the vertices of a polygon in the Cartesian plane.Linear programming is one specific type of mathematical optimization, which has applications in many scientific fields. Though there are ways to solve these problems using matrices, this section will focus on geometric solutions.Linear programming relies heavily on a solid understanding of systems of linear inequalities. Make sure you review that section before moving forward with this one.In particular, this topic will explain:
What is Linear Programming?
How to Solve Linear Programming Problems
Identify the Objective Function
What is Linear Programming?
Linear programming is a way of solving problems involving two variables with certain constraints. Usually, linear programming problems will ask us to find the minimum or maximum of a certain output dependent on the two variables.Linear programming problems are almost always word problems. This method of solving problems has applications in business, supply-chain management, hospitality, cooking, farming, and crafting among others.Typically, solving linear programming problems requires us to use a word problem to derive several linear inequalities. We can then use these linear inequalities to find an extreme value (either a minimum or a maximum) by graphing them on the coordinate plane and analyzing the vertices of the resulting polygonal figure.
How to Solve Linear Programming Problems
Solving linear programming problems is not difficult as long as you have a solid foundational knowledge of how to solve problems involving systems of linear inequalities. Depending on the number of constraints, however, the process can be a bit time-consuming.The main steps are:
Identify the variables and the constraints.
Find the objective function.
Graph the constraints and identify the vertices of the polygon.
Test the values of the vertices in the objective function.
These problems are essentially complex word problems relating to linear inequalities. The most classic example of a linear programming problem is related to a company that must allocate its time and money to creating two different products. The products require different amounts of time and money, which are typically restricted resources, and they sell for different prices. In this case, the ultimate question is “how can this company maximize its profit?”
As stated above, the first step to solving linear programming problems is finding the variables in the word problem and identifying the constraints. In any type of word problem, the easiest way to do this is to start listing things that are known.To find the variables, look at the last sentence of the problem. Typically, it will ask how many __ and __… use whatever is in these two blanks as the x and y values. It usually does not matter which is which, but it is important to keep the two values straight and not mix them up.Then, list everything known about these variables. Usually, there will be a lower bound on each variable. If one is not given, it is probably 0. For example, factories cannot make -1 product.Usually there is some relationship between the products and limited resources like time and money. There may also be a relationship between the two products, such as the number of one product being greater than another or the total number of products being greater than or less than a certain number. Constraints are almost always inequalities.This will become clearer in context with the example problems.
Identify the Objective Function
The objective function is the function we want to maximize or minimize. It will depend on the two variables and, unlike the constraints, is a function, not an inequality.We will come back to the objective function, but, for now, it is important to just identify it.
At this point, we need to graph the inequalities. Since it is easiest to graph functions in slope-intercept form, we may need to convert the inequalities to this before graphing.Remember that the constraints are connected by a mathematical “and,” meaning we need to shade the region where all of the inequalities are true. This usually creates a closed polygon, which we call “the feasible region.”That is, the area inside the polygon contains all possible solutions to the problem.Our goal, however, is not to find just any solution. We want to find the maximum or minimum value. That is, we want the best solution.Fortunately, the best solution will actually be one of the vertices of the polygon! We can use the graph and/or the equations of the bounds of the polygon to find these vertices.
We can find the best solution plugging each of the x and y-values from the vertices into the objective function and analyzing the result. We then can pick the maximum or minimum output, depending on what we are looking for.We must also double check that the answer makes sense. For example, it does not make sense to create 0.5 products. If we get an answer that is a decimal or fraction and this does not make sense in context, we can analyze a nearby whole number point. We have to make sure that this point is still greater than/less than the other vertices before declaring it to be the maximum/minimum.This all may seem a bit confusing. Since linear programming problems are nearly always word problems, they make more sense when context is added.
In this section, we will add context and practice problems relating to linear programming. This section also includes step-by-step solutions.
Consider the geometric region shown in the graph.
What are the inequalities that define this function?
If the objective function is 3x+2y=P, what is the maximum value of P?
If the objective function is 3x+2y=P, what is the minimum value of P
Example 1 Solution
This figure is bounded by three different lines. The easiest one to identify is the vertical line on the right side. This is the line x=5. Since the shaded region is to the left of this line, the inequality is x≤5.Next, let’s find the equation of the lower bound. This line crosses the y-axis at (0, 4). It also has a point at (2, 3). Therefore, its slope is (4-3/0-2)=-1/2. Therefore, the equation of the line is y=-1/2x+4. Since the shading is above this line, the inequality is y≥-1/2x+4.Now, let’s consider the upper bound. This line also crosses the y-axis at (0, 4). It has another point at (4, 3). Therefore, its slope is (3-4)/(4-0)=-1/4. Thus, its equation is y=-1/4x+4. Since the shaded region is below this line, the inequality is y≤–1/4x+4.In summary, our system of linear inequalities is x≤5 and y≥–1/2x+4 and y≤–1/4x+4.
Now, we are given an objective function P=3x+2y to maximize. That is, we want to find values x and y in the shaded region so that we can maximize P. The key thing to note is that an extrema of the function P will be at the vertices of the shaded figure.The easiest way to find this is to test the vertices. There are ways to find this using matrices, but they will be covered in greater depth in later modules. They also work better for problems with significantly many more vertices. Since there are only three in this problem, this is not too complicated.We already know one of the vertices, the y-intercept, which is (0, 4). The other two are intersections of the two lines with x=5. Therefore, we just need to plug x=5 into both equations.We then get y=-1/2(5)+4=-5/2+4=1.5 and y=-1/4(5)+4=2.75. Thus, our other two vertices are (5, 1.5) and (5, 2.75).Now, we plug all three pairs of x and y-values into the objective function to get the following outputs.(0, 4): P=0+2(4)=8.(5, 1.5): P=3(5)+2(1.5)=18(5, 2.75): P=3(5)+2(2.75)=20.5.Therefore, the function P has a maximum at the point (5, 2.75).
We actually did most of the work for part C in part B. Finding the minimum of a function is not very different than finding the maximum. We still find all of the vertices and then test all of them in the objective function. Now, however, we just select the output with the smallest value.Looking at part B, we see that this happens at the point (0, 4), with an output of 8.
A company creates square boxes and triangular boxes. Square boxes take 2 minutes to make and sell for a profit of $4. Triangular boxes take 3 minutes to make and sell for a profit of $5. Their client wants at least 25 boxes and at least 5 of each type ready in one hour. What is the best combination of square and triangular boxes to make so that the company makes the most profit from this client?
Example 2 Solution
The first step in any word problem is defining what we know and what we want to find out. In this case, we know about the production of two different products which are dependent upon time. Each of these products also makes a profit. Our goal is to find the best combination of square and triangular boxes so that the company makes the most profit.
First, let’s write down all of the inequalities we know. We can do this by considering the problem line by line.The first line tells us that we have two kinds of boxes, square ones and triangular ones. The second tells us some information about the square boxes, namely that they take two minutes to make and net $4 profit.At this point, we should define some variables. Let’s let x be the number of square boxes and y be the number of triangular boxes. These variables are both dependent upon each other because time spent making one is time that could be spent making the other. Make a note of this so that you do not mix them up.Now, we know that the amount of time spent making a square box is 2x.Now, we can do the same with the number of triangular boxes, y. We know that each triangular box requires 3 minutes and nets $5. Therefore, we can say that the amount of time spent making a triangular box is 3y.We also know that there is a limit on the total time, namely 60 minutes. Thus, we know that the time spent making both types of boxes must be less than 60, so we can define the inequality 2x+3y≤60.We also know that both x and y must be greater than or equal to 5 because the client has specified wanting at least 5 of each.Finally, we know that the client wants at least 25 boxes. This gives us another relationship between the number of square and triangular boxes, namely x+y≥25.Thus, overall, we have the following constraints:2x+3y≤60x≥5y≥5x+y≥25.These constraints function line the boundaries in the graphical region from example 1.
The Objective Function
Our objective, or goal, is to find the greatest profit. Therefore, our objective function should define the profit.In this case, profit depends on the number of square boxes created and the number of triangular boxes created. Specifically, this company’s profit is P=4x+5y.Note that this function is a line, not an inequality. In particular, it looks like a line written in standard form.Now, to maximize this function, we need to find the graphical region represented by our constraints. Then, we need to test the vertices of this region in the function P.
Now, let’s consider the graph of this function. We can first graph each of our inequalities. Then, remembering that linear programming problem constraints are connected by a mathematical “and,” we will shade the region that is a solution to all four inequalities. This graph is shown below.This problem has three vertices. The first is the point (15, 10). The second is the point (20, 5). The third is the point (22.5, 5).Let’s plug all three values into the profit function and see what happens.(15, 10): P=4(15)+5(10)=60+50=110.(20, 5): P=4(20)+5(5)=105.(22.5, 5): P=4(22.5)+5(5)=90+25=115.This suggests that the maximum is 115 at 22.5 and 5. But, in context, this means that the company must make 22.5 square boxes. Since it cannot do that, we have to round down to the nearest whole number and see if this is still the maximum.At (22, 5), P=4(22)+5(5)=88+25=113.This is still greater than the other two outputs. Therefore, the company should make 22 square boxes and 5 triangular boxes to satisfy the client’s demands and maximize its own profit.
A woman makes craft jewelry to sell at a seasonal craft show. She makes pins and earrings. Each pin takes her 1 hour to make and sells for a profit of $8. The pairs of earrings take 2 hours to make, but she gets a profit of $20. She likes to have variety, so she wants to have at least as many pins as pairs of earrings. She also knows that she has approximately 40 hours for creating jewelry between now and the start of the show. She also knows that the craft show vender wants sellers to have more than 20 items on display at the beginning of the show. Assuming she sells all of her inventory, how many each of pins and earring pairs should the woman make to maximize her profit?
Example 3 Solution
This problem is similar to the one above, but it has some additional constraints. We will solve it in the same way.
Let’s begin by identifying the constraints. To do this, we should first define some variables. Let x be the number of pins the woman makes, and let y be the number of pairs of earrings she makes.We know that the woman has 40 hours to create the pins and earrings. Since they take 1 hour and 2 hours respectively, we can identify the constraint x+2y≤40.The woman also has constraints on the number of products she will make. Specifically, her vender wants her to have more than 20 items. Thus, we know that x+y>20. Since, however, she cannot make part of an earring on pin, we can adjust this inequality to x+y≥21.Finally, the woman has her own constraints on her products. She wants to have at least as many pins as pairs of earrings. This means that x≥y.In addition, we have to remember that we cannot have negative numbers of products. Therefore, x and y are both positive too.Thus, in summary, our constraints are:X+2y≤40X+y≥21x≥yx≥0y≥0.
The Objective Function
The woman wants to know how she can maximize her profits. We know that the pins give her a profit of $8, and earrings earn her $20. Since she expects to sell all of the jewelry she makes, the woman will make a profit of P=8x+20y. We want to find the maximum of this function.
Now, we need to graph all of the constraints and then find the region where they all overlap. It helps to first put them all in slope-intercept form. In this case, then, we havey≤–1/2x+20y≥-x+21y≤xy≥0x≥0.This gives us the graph below.Unlike the previous two examples, this function has 4 vertices. We will have to identify and test all four of them.Note that these vertices are intersections of two lines. To find their intersection, we can set the two lines equal to each other and solve for x.We’ll move from left to right. The far left vertex is the intersection of the lines y=x and y=-x+21. Setting the two equal gives us:x=-x+21.2x=21.Therefore x=21/2, 0r 10.5 When x=10.5, the function y=x also is 10.5. Thus, the vertex is (10.5, 10.5).The next vertex is the intersection of the lines y=x and y=-1/2x+20. Setting these equal gives us:X=-1/2x+203/2x=20.Therefore, x=40/3, which is about 13.33. Since this is also on the line y=x, the point is (40/3, 40/3).The last two points lie on the x-axis. The first is the x-intercept of y=-x+21, which is the solution of 0=-x+21. This is the point (21, 0). The second is the x-intercept of y=-1/2x+20. That is the point where we have 0=-1/2x+20. This means that -20=-1/2x, or x=40. Thus, the intercept is (40, 0).Therefore, our four vertices are (10.5, 10.5), (40/3, 40/3), (21, 0), and (40, 0).
Finding the Maximum
Now, we test all four points in the function P=8x+20y.(10.5, 10.5)=294(40/3, 40/3)=1120/3 (or about 373.33)(0, 21)=168(0, 40)=320.Now, the maximum in this case is the point (40/3, 40/3). However, the woman cannot make 40/3 pins or 40/3 pairs of earrings. We can adjust by finding the nearest whole number coordinate that is inside the region and testing it. In this case, we have (13, 13) or (14, 13). We will choose the latter since it will obviously yield a larger profit.Then, we have:P=14(8)+13(20)=372.Thus, the woman should make 14 pins and 13 pairs of earrings for the greatest profit given her other constraints.
Joshua is planning a bake sale to raise funds for his class field trip. He needs to make at least $100 to meet his goal, but it is okay if he goes above that. He plans to sell muffins and cookies by the dozen. The dozen muffins will sell for a profit of $6, and the dozen cookies will sell for a profit of $10. Based on sales from the previous year, he wants to make at least 8 more bags of cookies than bags of muffins.The cookies require 1 cup of sugar and 3/4 cups of flour per dozen. The muffins require 1/2 cup of sugar and 3/2 cups of flour per dozen. Joshua looks into his cabinet and finds that he has 13 cups of sugar and 11 cups of flour, but he does not plan to go get more from the store. He also knows that he can only bake one pan of a dozen muffins or one pan of a dozen cookies at a time. What is the fewest number of pans of muffins and cookies Joshua can make and still expect to meet his financial goals if he sells all of his product?
Example 4 Solution
As before, we will have to identify our variables, find our constraints, identify the objective function, graph the system of constraints, and then test the vertices in the objective function to find a solution.
Joshua wants to know how the minimum number of pans of muffins and cookies to bake. Thus, let’s let x be the number of pans of muffins and y be the number of pans of cookies. Since each pan makes one dozen baked goods and Joshua sells the baked goods by the bag of one dozen, let’s ignore the number of individual muffins and cookies so as not to confuse ourselves. We can instead focus on the number of bags/pans.First, Joshua needs to make at least $100 to meet his goal. He earns $6 by selling a pan of muffins and $10 by selling a pan of cookies. Therefore, we have the constraint 6x+10y≥100.Joshua also has a limitation based on his flour and sugar supplies. He has 13 total cups of sugar, but a dozen muffins calls for 1/2 cup and a dozen cookies calls for 1 cup. Thus, he has the constraint 1/2x+1y≤13.Likewise, since a dozen muffins requires 3/2 cups of flour and a dozen cookies requires 3/4 cups of flour, we have the inequality 3/2x+3/4y≤11.Finally, Joshua cannot make fewer than 0 pans of either muffins or cookies. Thus, x and y are both greater than 0. He also wants to make at least 8 more pans of cookies than muffins. Therefore, we also have the inequality y-x≥10Therefore, our system of linear inequalities is:6x+10y≥1001/2x+y≤133/2x+3/4y≤11y-x≥8x≥0y≥0
The Objective Function
Remember, the objective function is the function that defines the thing we want to minimize or maximize. In the previous two examples, we wanted to find the greatest profit. In this case, however, Joshua wants a minimum number of pans. Thus, we want to minimize the function P=x+y.
In this case, we are finding the overlap of 6 different functions!Again, it is helpful to turn our constraint inequalities into y-intercept form so they are easier to graph. We get:y≥3/5x+10y≤–1/2x+13y≥x+8x≥0y≥0When we create the polygonal shaded region, we find that it has 5 vertices, as shown below.
Now, we need to consider all 5 vertices and test them in the original function.We have two vertices on the y-axis, which come from the lines y=-3/5x+10 and y=-1/2x+13. Clearly, these two y-intercepts are (0, 10) and (0, 13).The next intersection, moving from left to right is the intersection of the lines y=-1/2x+13 and y=-2x+44/3. Setting these two functions equal gives us:–1/2x+13=-2x+44/3.Moving the x values to the left and the numbers without a coefficient to the right gives us3/2x=5/3.x=10/9.When x=10/9, we have y=-2(10/9)+44/3=-20/9+132/9=112/9, which has the decimal approximation 12.4. Thus, this is the point (10/9, 112/9) or about (1.1, 12.4).The next vertex is the intersection of the lines y=-3/5x+10 and y=x+8. Setting these equal, we have:–3/5x+10=x+8–8/5x=-2.Solving for x then gives us 5/4. At 5/4, the function y=x+8 is equal to 37/4, which is 9.25. Therefore, the point is (5/4, 37/4) or (1.25, 9.25) in decimal form.Finally, the last vertex is the intersection of y=x+8 and y=-2x+44/3. Setting these equal to find the x-value of the vertex, we have:X+8=-2x+44/3.Putting the x-values on the left and numbers without a coefficient on the right gives us3x=20/3.Thus, solving for x gives us 20/9 (which is about 2.2). When we plug this number back into the equation y=x+8, we get y=20/9+72/9=92/9. This is approximately 10.2. Therefore, the last vertex is at the point (20/9, 92/9), which is about (2.2, 10.2).
Finding the Minimum
Now, we want to find the minimum value of the objective function, P=x+y. That is, we want to find the fewest number of pans of muffins and cookies Joshua has to make while still satisfying all the other constraints.To do this, we have to test all five vertices: (0, 13), (0, 10), (10/9, 112/9), (5/4, 37/4), (20/9, 92/9)(0, 13): 0+13=13.(0, 10): 0+10=10.(10/9, 112/9): 10/9+112/9=112/9, which is about 13.5.(5/4, 37/4): 5/4+37/4, which is 42/4=10.5.(20/9, 92/9): 20/9+92/9=112/9. This is about 12.4.Therefore, it seems Joshua’s best bet is to make 0 muffins and 10 cookies. This probably makes the baking simple anyway!If, however, he wanted to make as many products as possible, (that is, if he wanted the maximum instead of the minimum), he would want to make 10/9 muffins and 112/9 cookies. This is not possible, so we would have to find the nearest whole number of cookies and muffins. The point (1, 12) is inside the shaded region, as is (0, 13). Either of these combinations would be the maximum.
It is possible to have shaded regions with even more vertices. For example, if Joshua wanted a minimum number of bags of muffins or a maximum number of bags of cookies, we would have another constraint. If he wanted a minimum number of total bags of baked goods, we would have another constraint. Additionally, we could develop more constraints based on the number of ingredients. Things like eggs, butter, chocolate chips, or salt could work in this context. In some cases, a solution could become so complex so as not to have any feasible answers. For example, it is possible that the region not include any solutions where both x and y are whole numbers.
Amy is a college student who works two jobs on campus. She must work for at least 5 hours per week at the library and two hours per week as a tutor, but she is not allowed to work more than 20 hours per week total. Amy gets $15 per hour at the library and $20 per hour at tutoring. She prefers working at the library though, so she wants to have at least as many library hours as tutoring hours. If Amy needs to make 360 dollars, what is the minimum number of hours she can work at each job this week to meet her goals and preferences?
Example 5 Solution
As with the other examples, we need to identify the constraints before we can plot our feasible region and test the vertices.
Since Amy is wondering how many hours to work at each job, let’s let x bet the number of hours at the library and y the number of hours at tutoring.Then, we know x≥5 and y≥2.Her total number of hours, however, cannot be more than 20. Therefore, x+y≤20.Since she wants to have at least as many library hours as tutoring hours, she wants x≥y.Each hour at the library earns her $15, so she gets 15x. Likewise, from tutoring, she earns 20y. Thus, her total is 15x+20y, and she needs this to be more than 360. Therefore, 15x+20y≥360.In sum, then Amy’s constraints arex≥5y≥2x+y≤20x≥y15x+20y≥360
The Objective Function
The total number of hours that Amy works is the function P=x+y. We want to find the minimum of this function inside the feasible region.
The Feasible Region
To graph the feasible region, we need to first convert all of the constraints to slope-intercept form. In this case, we have:x≥5y≥2y≤-x+20y≤xy≥-3/4x+18.This graph looks like the one below.Yes. This graph is blank because there is no overlap between all of these regions. This means that there is no solution.
Perhaps Amy can persuade herself to get rid of the requirement that she work fewer hours at tutoring than at the library. What is the fewest number of hours she can work at tutoring and still meet her financial goals?Now, her constraints are just x≥5, y≥2, y≤-x+20, and y≥–3/4x+18.Then, we end up with this region.In this case, the objective function is just minimizing the number of hours Amy works at tutoring, namely Therefore, P=y, and we can see from looking at the region that the point (8, 12) has the lowest y-value. Therefore, if Amy wants to meet her financial goals but work as few hours as possible at tutoring, she has to work 12 hours at tutoring and 8 hours at the library.
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s3://commoncrawl/crawl-data/CC-MAIN-2024-18/segments/1712296817033.56/warc/CC-MAIN-20240415205332-20240415235332-00623.warc.gz
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CC-MAIN-2024-18
| 24,972 | 63 |
https://www.johndcook.com/blog/2013/08/23/stewarts-cube/
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math
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This is a drawing of Stewart’s cube. At each corner, the sum of the weights of the incoming edges is 83. Each edge has a prime weight, and all the weights are distinct.
Update: See Austin Buchanan’s post. He found a similar cube with smaller primes.
10 thoughts on “Stewart’s cube”
I’m guessing Stewart had a lot of time on his hands if he figured that one out :)
I wonder if there are OTHER cubes (or cube graphs) with similar properties (distinct prime edge weights, identical sums at vertices)?
Mike: There are a lot of so-called magic graphs where the sum of the weights at each vertex is the same. However, the definition also requires the edge weights to be 1, 2, 3, …, n. So Stewart’s cube isn’t a magic graph.
John, that’s interesting, but not quite what I was pondering – I’m wondering if there are other Stewart-class (TM pending) cubes with different distinct prime edge weights and (likely) a different identical sum at the vertices.
I guess another way to state it is that you need to find a number x such that there are 8 distinct sets of primes (p1i, p2i, p3i) such that p1i+p2i+p3i == x, and the 8 (p1i, p2i, p3i) sets only have 12 distinct primes in them?
… and each p_ij appears only in two sets.
What about a hypercube version?
Left as an exercise for the reader. :) If you come up with something, let me know.
Thanks for item on Stewart’s Cube, I found it most interesting. I have written a small program in C to do a search for other examples. The first cube seems to be for a target of 77. Once above target of about 100, the cubes come thick and fast. Has anyone else tried this?
MikeT: See this post from Austin Buchanan. I’m about to edit this post to point to his.
John, Do you know who, when, this Stewart is, was?
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CC-MAIN-2023-23
| 1,771 | 14 |
https://www.assignmentexpert.com/homework-answers/mathematics/statistics-and-probability/question-5406
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math
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Answer to Question #5406 in Statistics and Probability for Angela
During January the probability that Angela's car fails to start in the morning is 0.2 and this is unaffected by whether or not it started the previous day. When it fails to start, she walks to work. (i)What is the probability that Thursday is the first day of the week (starting on Monday) that she has to walk?(ii)What is the probability that she drives to work every day from Monday to Friday?(iii)What is the probability that she drives to work for exactly 3 days in a week?(iv)What is the probability that she has to walk exactly two days in a row in one week?
Thank you very much for your help throughout my endeavors with calculus. You have allowed me to gain credit for a course that would have been merely impossible to achieve on my own. I can't be thankful enough for what your service has given me. I can assure you I will be in contact with you guys soon. Again, thank you very much and enjoy the rest of summer!
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s3://commoncrawl/crawl-data/CC-MAIN-2018-34/segments/1534221211185.57/warc/CC-MAIN-20180816211126-20180816231126-00676.warc.gz
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CC-MAIN-2018-34
| 990 | 3 |
https://thirdspacelearning.com/secondary-resources/gcse-maths-worksheet-percentage-of-an-amount/
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math
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Help your students prepare for their Maths GCSE with this free percentage of an amount worksheet of 44 questions and answers
Suitable for GCSE maths revision for AQA, OCR and Edexcel exam boards
A percentage is an amount expressed as a fraction with a denominator of 100. For example 45% means ‘45 out of 100’, 45 hundredths or 45 over 100. In decimal form, we write 0.45. A percentage does not always have to be a whole number.
There are some simple percentages of amounts that students should be able to calculate mentally, such as 50% (half), 25% (quarter) and 10% (divide by 10). Finding multiples of these amounts, such as 20%, 30% etc. along with bar modelling can be useful to support percentage calculations.
When finding percentages of amounts using a calculator, students should be encouraged to use the decimal equivalent and convert to a decimal multiplication – for example, to find 18% of a given number, multiply by 0.18. Developing this skill is particularly useful when extending to reverse percentage problems, where students can find the original amount by using the inverse operation – so dividing by the percentage decimal equivalent.
Looking forward, students can then progress to additional number worksheets, for example a fractions of amounts worksheet or a rounding worksheet. There are plenty of additional percentage problems worksheets, including percentage increase and decrease, and compound interest.
There will be students in your class who require individual attention to help them succeed in their maths GCSEs. In a class of 30, it’s not always easy to provide.
Help your students feel confident with exam-style questions and the strategies they’ll need to answer them correctly with our dedicated GCSE maths revision programme.
Lessons are selected to provide support where each student needs it most, and specially-trained GCSE maths tutors adapt the pitch and pace of each lesson. This ensures a personalised revision programme that raises grades and boosts confidence.
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s3://commoncrawl/crawl-data/CC-MAIN-2023-50/segments/1700679100164.87/warc/CC-MAIN-20231130031610-20231130061610-00468.warc.gz
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CC-MAIN-2023-50
| 2,019 | 9 |
https://pubs.geoscienceworld.org/geophysics/article-abstract/46/2/138/68523/linear-inversion-of-body-wave-data-part-i-velocity?redirectedFrom=fulltext
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math
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In a laterally homogeneous medium, the traveltime (T) and distance (X) for a ray with horizontal slowness p are linearly related to the depth Z(v) at which the velocity v = 1/p occurs. In order to exploit this linearity, we must infer the inverse velocity p from the observations of X, T pairs. Uncertainty in the determination of p causes correlation between the X and T observations. This correlation can be eliminated by rotation of the data into a coordinate system in which the covariance matrix is diagonal. These independent coordinates are, except for a scaling factor, the well-known intercept time tau (p) = T - pX and a new variable zeta (p) = T + pX. The derivatives of T and X with respect to a depth-velocity model contain singularities and so do those for zeta . These singularities can be quelled by representing the model as a stack of layers, each of which has a constant velocity gradient. Depth is then obtained by integration of the gradients.The sharpness of the partial derivatives of zeta w.r.t. the layer gradients indicates that zeta contains information in a more concentrated form than does tau . This manifests itself in smaller error bounds on the solution when zeta observations are used to supplement tau data.In the determination of zeta (p) from X,T data, an uncertainty principle or tradeoff applies. The delta-like nature of the zeta partial derivatives means that the uncertainty in zeta will be closely related to the solution uncertainty and that we should choose in the parameterization the zeta , p pair which minimizes the uncertainty in zeta . This will avoid degrading the ultimate depth resolution achievable while still in the parameterization stage.We have applied these methods to sea floor hydrophone and surface buoy data from the Bengal Fan, and, we derive a model whose gradient is 1.8 sec (super -1) at the surface reaching 0.6 sec (super -1) at 500 m and remaining constant to at least 5.5 km.
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s3://commoncrawl/crawl-data/CC-MAIN-2019-30/segments/1563195527474.85/warc/CC-MAIN-20190722030952-20190722052952-00409.warc.gz
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CC-MAIN-2019-30
| 1,947 | 1 |
http://www.ask.com/web?q=Why+Gravity+Moves+Us+So%3F&oo=2603&o=0&l=dir&qsrc=3139&gc=1&qo=popularsearches
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math
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Aug 19, 2015 ... Gravity is not a force; however, this truth leaves us with a number of. ... short,
mass bends spacetime, and these bends tell energy how to move.
There is a lot of air resistance and that resistance makes the feather move slower.
... But what keeps the Moon from falling down, if all of this gravity is so strong?
Well ... line into a curve (orbit) around the Earth and ends up revolving around us.
Since gravitational force is present when two objects are sufficiently close, so as
not to ..... This follows Newton's Third Law of motion, which tells us that the total ...
Apr 28, 2016 ... The speed of gravity is taken to be exactly equal to the speed of light. But is that
necessarily true? Here's how we know.
Jun 14, 2012 ... So "a" and "d" are wrong because no gravity means no orbits. "b" is wrong
because there is no force to move the sun closer to the earth.
The Sun's gravity pulls on the planets, just as Earth's gravity pulls down anything
... more massive ones) produce a bigger gravitational pull than lighter ones, so as
the ... "Shoot a cannonball into orbit" demonstrates how objects fall and move ...
Jan 13, 2012 ... Gravity is strong enough to hold us to the surface of this planet, and powerful
enough to make ... It's because gravity is so much weaker than the other forces in
the universe, something ... Accelerate, and it moves forward.
Jan 15, 2014 ... So for the Sun and the Earth, the incredibly large mass of the Sun ... In Einstein's
theory of gravity, these ripples move at the speed of light, not ... aimed at us, we
can test whether gravity moves at the speed of light or not!
But the Sun's force of gravity out here, by us, is far too weak to measure this effect
... theory of gravity are incredibly sensitive to the speed of light, so much so that ...
Oct 18, 2014 ... So that's how a planet can accelerate toward the Sun forever without ... can stand
, but (not surprisingly) those impacts don't leave craters for us to find. ... If I shoot a
rocket toward the sun, will the suns gravity move it into orbit or ...
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s3://commoncrawl/crawl-data/CC-MAIN-2016-36/segments/1471982292607.17/warc/CC-MAIN-20160823195812-00212-ip-10-153-172-175.ec2.internal.warc.gz
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CC-MAIN-2016-36
| 2,064 | 25 |
https://community.agilent.com/technical/mass-spectrometry-software/f/mass-spectrometry-software-user-forum/5945/qualitative-analysis-software
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math
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I'm new to the MassHunter Qualitative Analysis software and I'm hoping someone can help me to understand some information. After matching the compounds with the existing library/database, the compound information is presented under the "Compounds List" tab. What is the meaning of the numbers in the "Score" column? Is there a way to find the sample purity or composition?
Thank you for your help.
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s3://commoncrawl/crawl-data/CC-MAIN-2023-50/segments/1700679100651.34/warc/CC-MAIN-20231207090036-20231207120036-00310.warc.gz
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CC-MAIN-2023-50
| 397 | 2 |
https://math.answers.com/Q/When_did_Fibonacci_make_the_Fibonacci_number_sequence
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math
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The 20th Fibonacci number in the sequence is 6,765.
The 16th number of the Fibonacci sequence is 987.
The 9th number in the Fibonacci Sequence is 34, and the 10th number in the Fibonacci sequence is 89.
Leonardo Fibonacci discovered the number sequence which is named after him.
1123 is not a number included in the Fibonacci Sequence.
Yes. 610 is the 15th number in the Fibonacci sequence.
Assuming the Fibonacci sequence starts with the numbers 0, 1, as per sequence A000045, the 24th Fibonacci number is 28657.
A Fibonacci number, Fibonacci sequence or Fibonacci series are a mathematical term which follow a integer sequence. The first two numbers in Fibonacci sequence start with a 0 and 1 and each subsequent number is the sum of the previous two.
No, the Fibonacci sequence never ends. It just keeps on going!
Start with 1 and 2. Then each number in the Fibonacci sequence is the sum of the previous two numbers in the sequence.
Any single digit number is a palindrome. The Fibonacci sequence consists of infinitely many numbers so 8, being only one number, cannot be the Fibonacci sequence.
There is the Morris number sequence and the Fibonacci number sequence. The Padovan sequence. The Juggler sequence. I just know the Fibonacci sequence: 0,1,1,2,3,5,8,13,21,34,55,89,144,233,377 Morris number sequence: 1 11 21 1211 111221 312211...
Leonardo pision discovered the Fibonacci sequance
Yes, 21 is a number in the Fibonacci sequence. Numbers are usually found by adding up 2 numbers.
The number is 1
Leonardo Fibonacci discovered the Fibonacci sequence In the 12th century.
Fibonacci sequence Fibonacci sequence
Generate Fibonacci sequence by adding the two previous integers together to get the next number in the sequence. Starting with the lowest two number on the real number line. 0,1,1,2,3,5,8,13,21,34,55,89,144 etc.
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s3://commoncrawl/crawl-data/CC-MAIN-2022-05/segments/1642320304345.92/warc/CC-MAIN-20220123232910-20220124022910-00162.warc.gz
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CC-MAIN-2022-05
| 1,831 | 18 |
http://loo0.tumblr.com/post/17673905957/der-vivisektor-the-vivisector-1883-gabriel-von
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math
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This is what I like, what I want to be, what makes me believe, laugh, cry, feel... this is for me c:
Questions goes here!
Theme by STJN
), 1883. Gabriel von Max Oil on canvas
#Gabriel von Max
#Oil on canvas
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s3://commoncrawl/crawl-data/CC-MAIN-2014-15/segments/1397609535745.0/warc/CC-MAIN-20140416005215-00017-ip-10-147-4-33.ec2.internal.warc.gz
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CC-MAIN-2014-15
| 206 | 6 |
http://www.math.cmu.edu/~rami/374.desc.html
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math
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Instructor: Rami Grossberg
Office: WEH 7204
Phone: x8482 (268-8482 from external lines), messages at x2545
Office Hours: I will be in my office today (Wedesday, May 9th), if you want to see me please stop by or send email.
Purpose. The goal of this course is to provide a successor to
Algebraic Structures (21-373), with an emphasis on applications of
groups, rings, and fields within algebra to some major classical
problems. These include constructions with a ruler and compass, and
(un) solvability of equations by radicals. It also offers an opportunity
to see group theory and basic ring theory "in action", and introduces
several powerful number theoretic techniques.
The basic ideas and methods required to study finite fields will also be
introduced, these have recently been applied in a number of areas of
theoretical computer science including primality testing and
Course description. We will start with a review of ring theory.
Definitions and examples, field extensions, adjunction of roots,
algebraic numbers, dimension formula, constructions with ruler and
compass (it is impossible to trisect an arbitrary angle, and it is
impossible to duplicate the cube), splitting fields, existence (and
uniqueness) of algebraic closure, symmetric polynomials, Galois groups,
Galois extensions, the Galois correspondence theorem for characteristics
0, permutations and simplicity of An, unsolvability by radicals of the
general quintic, characterization of finite fields (and their
multiplicative groups), Wedderburn's theorem (optional), transcendental
extensions, Steinitz's theorem on trascendence degree.
Text: "Abstract Algebra" by D. S. Dummit & R. M. Foote. 2nd
edition Published by John Wiley & Sons, 1999.
Test Dates: The second midterm will be held on Monday 4/23 instead of a lecture.
Evaluation: There will be two one hour tests (in class), weekly homework
assignments, and a three hour final. These will be weighted as
Prerequisites. Algebraic Structures or Math Studies.
In the fall semesater I will be teaching a follow up of this course. It is 21-611, for more information click here .
Rami's home page.
|Last modified: May 9th, 2001|
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CC-MAIN-2018-51
| 2,154 | 34 |
http://racingtanks.com/freebooks/mathematical-heritage-of-hermann-weyl-proceedings-of-symposia-in-pure-mathematics-v-48
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math
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Hermann Weyl used to be some of the most influential mathematicians of the 20 th century. Viewing arithmetic as an natural complete instead of a suite of separate topics, Weyl made profound contributions to a variety of components, together with research, geometry, quantity thought, Lie teams, and mathematical physics, in addition to the philosophy of technology and of arithmetic. the themes he selected to check, the traces of suggestion he initiated, and his normal viewpoint on arithmetic have proved remarkably fruitful and feature shaped the foundation for the very best of contemporary mathematical research.This quantity comprises the court cases of the AMS Symposium at the Mathematical history of Hermann Weyl, held in might 1987 at Duke college. as well as honoring Weyl's nice accomplishments in arithmetic, the symposium additionally sought to stimulate the more youthful iteration of mathematicians by means of highlighting the cohesive nature of recent arithmetic as noticeable from Weyl's rules. The symposium assembled an excellent array of audio system and lined quite a lot of themes. the entire papers are expository and should entice a vast viewers of mathematicians, theoretical physicists, and different scientists.
Read or Download Mathematical Heritage of Hermann Weyl (Proceedings of Symposia in Pure Mathematics; v. 48) PDF
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http://www.transtutors.com/questions/velocity-256515.htm
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Solution to be delivered in 36 hoursafter verification
During takeoff, an airplane goes from 0 m/s to 50 m/s in 8 s. a) What is its acceleration? b) How fast is it going after 5 s? c) How far has it traveled by the time it reaches 50 m/s?
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A car traveling at a constant speed of 20 m/s passes an intersection at time t = 0, and 5 s later another car traveling 30 m/s passes the same intersection in the same direction. (a
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An object with an initial velocity of 5 m/s has a constant acceleration of 2 m/s2. When its speed is 15 m/s, how far has it traveled?
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A car can be braked to a stop from the autobahn-like speed of 200km/h in 170m, assuming the acceleration is constant, find its magnitude in (a) SI units and (b) In terms of g. (c) How
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http://www.dickinson.edu/news-and-events/news/2009-10/Elegant-Book-Wins-Mathematical-Prize/
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math
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"Elegant" Book Wins Mathematical Prize
Professor writes best-selling book about a landmark mathematical formula
February 2, 2010
David Richeson, associate professor of mathematics, holds up a copy of his book "Euler's Gem."
Anyone who wouldn’t place “humorous,” “beautiful” and “best-seller” in the same sentence as “mathematics” hasn’t read Dave Richeson’s book, Euler’s Gem: The Polyhedron Formula and the Birth of Topology. This fascinating and accessible approach to Leonhard Euler's landmark formula has attracted rave reviews and awards.
Most recently, Richeson, associate professor of mathematics and department chair at Dickinson, received the Mathematical Association of America's 2010 book prize. Coincidentally, the prize is named in honor of Euler, a peer of Euclid, Pythagoras, Archimedes and Newton who is, arguably, one of the most prolific and influential mathematicians of all time.
The prize is awarded to writers of outstanding books about mathematics, based on clarity of exposition and the degree to which the book promises to enhance and enrich the public’s view of math. Euler’s Gem hit the mark.
In language usually reserved for great literary works, the judging committee called Euler’s Gem “elegant, concise and surprising,” stating that previous attempts to explain the beauty of Euler’s simple formula and to explore its depth “pale by comparison to Richeson’s extraordinary narrative,” which features descriptions that are “amazingly friendly” and with prose that is “a joy to read.”
That's partially due to the author's passion for his subject. “There is so much beautiful mathematics that most people never see ... Euler's formula is a prime example,” said Richeson. “It is simple enough to explain to a child, yet it may not be seen until graduate school. It is especially wonderful because it is straightforward, yet has deep meaning and many important mathematical consequences. In one survey of mathematicians, it was voted the second-most beautiful theorem of all time.”
The book is an Amazon.com best-seller—a noteworthy achievement for an academic work by a first-time author.
That's because the book appeals both to professional mathematicians and to readers that Richeson calls “mathematical enthusiasts.” This group includes motivated high-school students, college students and others who want to deepen their understanding of mathematics as well as readers who simply want to learn something new.
“I spent a lot of time thinking about how to write about advanced topics in a way that someone with no background could understand. As an example, I created more than 150 diagrams to make the arguments as visual as possible,” Richeson said.
True to the Dickinsonian approach to the liberal arts and sciences, Richeson collaborated across other disciplines in writing his book. Associate Professor of Classical Languages Chris Francese assisted in translating Euler’s paper from its original Latin; German major Anne Maiale '08 contributed translations from Euler’s native language as Richeson’s Dana research assistant. The Dana program enables students to assist in faculty scholarly and creative research.
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https://queviridfu.web.app/1121.html
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math
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To start, imagine that you acquire an n sample signal, and want to find its frequency spectrum. Evaluation by taking the discrete fourier transform dft of a coefficient vector interpolation by taking the inverse dft of pointvalue pairs, yielding a coefficient vector fast fourier transform fft can perform dft and inverse dft in time. The discrete fourier transform or dft is the transform that deals with a nite discretetime signal and a nite or discrete number of frequencies. Periodicdiscrete these are discrete signals that repeat themselves in a periodic fashion from negative to positive infinity. We will introduce a convenient shorthand notation xt. It is a linear invertible transformation between the timedomain representation of a function, which we shall denote by ht, and the frequency domain representation which we shall denote by hf. Moreover, fast algorithms exist that make it possible to compute the dft very e ciently. Ifthas dimension time then to make stdimensionless in the exponential e. The multidimensional transform of is defined to be. Discrete time fourier transform dtft vs discrete fourier. It is worth noting that the discrete time fourier transform is always 2.
The dft takes a discrete signal in the time domain and transforms that signal into its discrete frequency domain representation. On the other hand, the discretetime fourier transform is a representation of a discretetime aperiodic sequence by a continuous periodic function, its fourier transform. Furthermore, we will show that the discretetime fourier transform can be used to represent a wide range of sequences, including sequences of in. On the other hand, the discrete time fourier transform is a representation of a discrete time aperiodic sequence by a continuous periodic function, its fourier transform.
The discrete time fourier transform dtft is a form of fourier analysis that is applicable to the uniformlyspaced samples of a continuous function. Fourier transform is called the discrete time fourier transform. Inverse discrete fourier transform dft alejandro ribeiro february 5, 2019 suppose that we are given the discrete fourier transform dft x. The inverse fourier transform takes fz and, as we have just proved, reproduces ft. The discretetime pulses spectrum contains many ripples, the number of which increase with n, the pulses duration.
Remembering the fact that we introduced a factor of i and including a factor of 2 that just crops up. Now heres the formula for the ztransform shown next to the discretetime fourier transform of xn. Therefore, zthe inverse fourier transform of is zthe inverse transform of is. Dec 04, 2019 in this post, we will encapsulate the differences between discrete fourier transform dft and discretetime fourier transform dtft. The dft is the most important discrete transform, used to perform fourier analysis in many practical applications. Define xnk, if n is a multiple of k, 0, otherwise xkn is a sloweddown version of xn with zeros interspersed. Chapter 4 the discrete fourier transform c bertrand delgutte and julie greenberg, 1999. Let be the continuous signal which is the source of the data. Fourier transform for continuoustime signals 2 frequency content of discretetime signals.
The fourier transform of the original signal, would be. Digital signal processing dft introduction tutorialspoint. Fouriersequencetransformwolfram language documentation. In this note, we assume the overlapping is by 50% and we derive the. The discrete time pulses spectrum contains many ripples, the number of which increase with n, the pulses duration. Then a shift in time by n0 becomes a multiplication in the zdomain by ej. Understand the properties of time fourier discretetransform iii understand the relationship between time discretefourier transform and linear timeinvariant system. Many of the toolbox functions including z domain frequency response, spectrum and cepstrum analysis, and some filter design and.
Secondly, a discretetime signal could arise from sampling a continuoustime. The discrete fourier transform dft is the equivalent of the continuous fourier transform for signals known only at instants separated by sample times. Jan 11, 2018 dtftdiscrete time fourier transform examples and solutions. We give an integral form for the inverse dtft that can be used even when. Can you explain the rather complicated appearance of the phase. Understanding the discrete fourier transform dtft dft and sampling theory. A table of some of the most important properties is provided at the end of these. Thus we have replaced a function of time with a spectrum in frequency. Discretetime fourier series have properties very similar to the linearity, time shifting, etc. X x1 n1 xne j n inverse discrete time fourier transform.
The discrete fourier transform, or dft, is the primary tool of digital signal processing. The fourier transform ft decomposes a function of time a signal into the frequencies that make it up, in a way similar to how a musical chord can be expressed as the frequencies or pitches of its constituent notes the discrete fourier transform dft converts a finite sequence of equallyspaced samples of a function into a samelength sequence of equally. Periodicity this property has already been considered and it can be written as follows. The dtft is a transformation that maps discretetime dt signal xn into a complex valued function of the real variable w, namely. The relationship between the dtft of a periodic signal and the dtfs of a periodic signal composed from it leads us to the idea of a discrete fourier transform not to. Dct vs dft for compression, we work with sampled data in a finite time window. In this section we formulate some properties of the discrete time fourier transform. Suppose that we are given the discrete fourier transform dft x. The discrete fourier transform 1 introduction the discrete fourier transform dft is a fundamental transform in digital signal processing, with applications in frequency analysis, fast convolution, image processing, etc. Conditions for the existence of the fourier transform are complicated to state in general, but it is sufficient for to be absolutely integrable, i.
Shorttime fourier transform and its inverse ivan w. None of the standard fourier transform property laws seem to directly apply to this. This class of fourier transform is sometimes called the discrete fourier series, but is most often called the discrete fourier transform. The discrete fourier transform dft is the equivalent of the continuous fourier transform for signals known only at instants separated by sample times i. Also, as we discuss, a strong duality exists between the continuoustime fourier series and the discretetime fourier transform. The term discrete time refers to the fact that the transform operates on discrete data samples whose interval often has units of time. Our first task is to develop examples of the dtft for some common signals. In digital signal processing, the function is any quantity or signal that varies over time, such as the pressure of a sound wave, a radio signal, or daily temperature readings, sampled over a finite time interval often defined by. Discrete fourier transform dft is used for analyzing discretetime finiteduration signals in the frequency domain let be a finiteduration sequence of length such that outside. Discretetime fourier transform solutions s115 for discretetime signals can be developed.
Recall that for a general aperiodic signal xn, the dtft and its inverse is. This should look familiar given what you know about fourier analysis. X x1 n1 xne j n inverse discretetime fourier transform. Continuous time fourier transform of xt is defined as x. So far we have seen that time domain signals can be transformed to frequency domain by the so called fourier transform.
Define xnk, if n is a multiple of k, 0, otherwise xkn is a sloweddown version of. The inverse discrete time fourier transform is easily derived from the following relationship. The inverse fourier transform the fourier transform takes us from ft to f. Periodic discrete these are discrete signals that repeat themselves in a periodic fashion from negative to positive infinity. The plancherel identity suggests that the fourier transform is a onetoone norm preserving map of the hilbert space l21. Summary of the dtft the discretetime fourier transform dtft gives us a way of representing frequency content of discretetime signals. The rst equation gives the discrete fourier transform dft of the sequence fu jg. In this post, we will encapsulate the differences between discrete fourier transform dft and discretetime fourier transform dtft. Fourierstyle transforms imply the function is periodic and. The relationship between the dtft of a periodic signal and the dtfs of a periodic signal composed from it leads us to the idea of a discrete fourier transform not to be confused with discrete time fourier transform.
Also, as we discuss, a strong duality exists between the continuous time fourier series and the discrete time fourier transform. The discrete fourier transform dft an alternative to using the approximation to the fourier transform is to use the discrete fourier transform dft. The discrete time fourier transform dtft is the member of the fourier transform family that operates on aperiodic, discrete signals. Like continuous time signal fourier transform, discrete time fourier transform can be used to represent a discrete sequence into its equivalent frequency domain representation and lti discrete time system and develop various computational algorithms. Dtftdiscrete time fourier transform examples and solutions. Discrete time fourier transform dtft the discrete time fourier transform dtft can be viewed as the limiting form of the dft when its length is allowed to approach infinity. If we interpret t as the time, then z is the angular frequency. Chapter 1 the fourier transform university of minnesota. The inverse discretetime fourier transform is easily derived from the following relationship. Fouriersequencetransform is also known as discretetime fourier transform dtft.
Selesnick april 14, 2009 1 introduction the shorttime fourier transform stft of a signal consists of the fourier transform of overlapping windowed blocks of the signal. Inverse discrete fourier transform dft alejandro ribeiro february 5, 2019. Discretetime fourier transform dtft chapter intended learning outcomes. Discrete time fourier transform dtft mathematics of the dft. Lecture notes for thefourier transform and applications. The foundation of the product is the fast fourier transform fft, a method for computing the dft with reduced execution time. Fourier transform ft and inverse mathematics of the dft. Discrete time fourier transform dtft mathematics of. Discrete time fourier transform solutions s115 for discrete time signals can be developed. The discretetime fourier transform of a discrete set of real or complex numbers xn, for all integers n, is a fourier series, which produces a periodic function of a frequency variable.53 528 767 683 1060 357 81 1024 81 261 371 609 873 202 1152 482 291 1233 1421 1245 1279 783 59 1358 162 1093 406 1424 360 400 836 19 639 977 1328 496 1418 370 544 1326 1370 697
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http://www.enotes.com/homework-help/suppose-you-standing-straight-highway-watching-car-424072
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Better Students Ask More Questions.
Suppose you are standing on a straight highway and watching a car moving away from you...
Topic: PhysicsSuppose you are standing on a straight highway and watching a car moving away from you at 20.0 m/s. The air is perfectly clear, and after 11 minutes you see only one taillight. If the diameter of your pupil is 7.00 mm and the index of refraction of your eye is 1.33, what is the distance between the taillights. Taillights are often red so you can assume that λ=700 nm. (Hint: Assume the eye is acting like a single slit with a circular aperture. You will also need to take into consideration the fact that the wavelength of light changes inside your eye since the index of refraction is different. Please show all steps.)
1 Answer | add yours
The angular separation resolution of a lens is given by
`theta =1.22*lambda/D` , (1)
where `D =7 mm` is the diameter of the lens.
The speed of light is smaller for higher refraction indexes than 1:
which means a shorter wavelength in a medium having a refraction index `n>1`
`lambda =v*T =c/n*T = lambda_0/n`
Combining this with (1) one gets
`theta = 1.22*lambda_0/(n*D)`
The distance from the person to the car is
`L =v*t =20*11*60 = 13200 m`
If `l` is the distance between the taillights one has
`l/L =tan(theta)~~theta =1.22*lambda_0/(nD)`
Thus the for the distance `l` one gets the value
`l =(1.22*lambda_0*L)/(n*D) = (1.22*700*10^-9*13200)/(1.33*0.007) = 1.21 m`
The distance between the taillights is 1.21 m
Posted by valentin68 on October 14, 2013 at 3:49 PM (Answer #1)
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http://slsy.nhri.cn/article/id/201504016
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math
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In order to carry out studies of the relationships between the onset of Plum rains and the rainfall amount during Plum rains period in the Taihu Lake basin, the authors have firstly selected optimal univariate marginal distribution, and then established the joint distribution of two variables based on the Copula function, and at last, analyzed the encounter probability and the conditional distribution probability between the onset of Plum rains and the rainfall amount during Plum rains period under different circumstances. The analysis results show that there are significant differences in the encounter probability between different onsets of Plum rains and different rainfall amounts during Plum rains period, and that in general the maximum probability is encountered between the normal onset of Plum rains and the average rainfall amount during Plum rains period. There is high probability for the conditional distribution probability that the average rainfall amount during Plum rains period appears in the different onsets of Plum rains. But the probability of wet Plum rains in the early onset of Plum rains is 0.417, and meanwhile, the probability of dry Plum rains in the late onset of Plum rains is 0.374. So close attention should be paid to the changes of flood when appearing in the early onset of Plum rains, and the changes of the drought when appearing in the late onset of Plum rains. This study can provide a reference for taking measures of flood and drought control and water resources regulation for the Taihu Lake basin as soon as possible.
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https://plainmath.net/11508/population-california-million-million-population-exponentially-function
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math
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The population of California was 29.76 million in 1990 and 33.87 million in 2000. Assume that the population grows exponentially.
Let and r be the initial population & growth rate of population respectively. Then population in California t years after 1990 is,
a. Since the population in 1990 is 29.76 million, therefore
In 2000, the population is 33.87 million. That is,
Therefore from equation (1), we get
Therefore the population t years after 1990 is,
b) The double of 29.76 million is 59.52 million.
So put in , we get
Hence the population will double in 53.5828 years.
c)Since 2010−1990=20, therefore to predict the population in 2010, put in , we get
The population in California in 2010 is 37.3 million, therefore the actual population is approximately 1 million less than the obtained population.
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https://physicscatalyst.com/class-7/congruence-of-triangles-worksheet.php
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In this page we have congruence of triangles class 7 worksheets . Hope you like them and do not forget to like , social share
and comment at the end of the page.
Fill in the blanks:
(a) Two objects with same shape and size are said to be __________________
(b) The relation of two objects being congruent is called ____________________
(c) Two line segments are said to be congruent if they have same ________________
(d) If two angles are congruent, their _________________ are same.
(e) for the below figure
$ \Delta PQR \cong \Delta$ _____
(f) Two triangles are congruent, if two angles and the side included between them in one of the triangles are equal to the two angles and the side included between them of the other triangle. This is known as the _________
(g) If all three ____________ of a triangle are respectively equal to that of other triangle, the triangle may not be congruent.
(h) In congruence condition RHS, 'H' stands for _____________________
(i)Two rectangles are congruent, if they have same _____ and ______
(j) Two squares are congruent, if they have same ______ Solution
(a) A circle of radius 10cm and a square of side 10cm are congruent.
(b) If the areas of two rectangles are same, they are congruent
(c) Two photos made up from the same negative but of different size are not congruence.
(d) if two sides and any angle of one triangle are equal to the corresponding sides and an angle of another triangle, then the triangles are not congruent.
(e) There is no AAA congruence criterion.
(f) Two circles having same circumference are congruent.
(g) If two triangles are equal in area, they are congruent.
(h) If two triangles are congruent, they have equal areas .
(i) AAS congruence criterion is same as ASA congruence criterion.
(j) $ \Delta ABC \cong \Delta DFE$ implies $\Delta BAC \cong \Delta DEF$ Solution
Multiple Choice Questions
By which of the following criterion two triangles cannot be proved congruent?
$ \Delta ABC \cong \Delta PRQ$ and AB=5 cm, BC= 6 cm , AC=7 cm, What is the length of QR?
(a) 5 cm
(b) 6 cm
(c) 7 cm
(d) Cannot be determined
In the below figure , $\Delta PQR $ is congruent with the triangle
If $\Delta PQR$ and $\Delta XYZ$ are congruent under the correspondence $QPR \leftrightarrow XYZ$, then which of the following is false
(a) $ \angle R = \angle Z$,QR=XZ
(b) PQ=YX and PR=YZ
(c) $\angle P =\angle Y$ and QR= YZ
(d) $\angle Q = \angle X$
If for $\Delta ABC$ and $\Delta DEF$, the correspondence $CAB \leftrightarrow EDF$ gives a congruence, then which of the following is not true?
(a) AC = DE
(b) AB = EF
(c) $\angle A = \angle D$
(d) $\angle C = \angle E$
In triangles DEF and PQR, $\angle E = 80^0$, $\angle F = 30^0$, EF = 5 cm, $\angle P = 80^0$, PQ = 5 cm, $\angle R = 30^0$. By which congruence rule the triangles are congruent?
(d) None of these
What is the side included between the angles M and N of $\Delta MNP$?
(c) None of these
(d) MP Solution 3-9
Numerical Type Questions
$\Delta ABC$ is isoceles with AB=AC, AD is the altitude from A to side BC,
(i) $\Delta ADB \cong \Delta ADC$
(ii) $ \angle BAD= \angle CAD$
Prove that in an isosceles triangle, angles opposite to equal sides are equal.
Without drawing the triangles write all six pairs of equal measures
in each of the following pairs of congruent triangles.
(a) $\Delta STU \cong \Delta DEF$
(b) $\Delta ABC \cong \Delta LMN$
(c) $\Delta YZX \cong \Delta PQR $
(d) $\Delta XYZ \cong \Delta MLN$ Solution
(a) $\angle S = \angle D$, $\angle T = \angle E$, $\angle U = \angle F$, ST = DE, TU = EF, SU = DF
(b) $\angle A = \angle L$, $\angle B = \angle M$, $\angle C = \angle N$, AB = LM, BC = MN, AC = LN
(c) $\angle Y = \angle P$, $\angle Z = \angle Q$, $\angle X = \angle R$, YZ = PQ, ZX = QR, YX = PR
(d) $\angle X = \angle M$, $\angle Y = \angle L$, $\angle Z = \angle N$, XY = ML, YZ = LN, XZ = MN
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https://www.futurelearn.com/courses/thermodynamics/0/steps/25305
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math
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Skip to 0 minutes and 1 second Let’s turn our attention to heat capacity. The heat capacity is the amount of thermal energy required to change the temperature of material. so the heat capacity C is delta Q divide by delta T. The heat capacity at constant volume Cv is dQ over dT at constant volume. And from the first law, dQ at constant volume is the same with dU since dV is zero at constant volume. So Cv is dU over dT at constant volume. Likewise, the heat capacity at constant pressure Cp is dQ over dT at constant pressure and dQp is the same with dH at constant pressure. So Cp is dH over dT at constant pressure. And this heat capacity strongly depends on temperature. Look at this graph.
Skip to 1 minute and 7 seconds It is the variation of Cv over temperature for various materials. In general, heat capacity increases with temperature. And also, heat capacity is different for different materials since the heat capacity is related with bonding. In this graph, bond strength decreases from diamond, to silicon and to lead. As you see here, materials with stronger bond have lower heat capacity. To discuss this trend, lets consider the energy of a material. There can be vibrational, translational, rotational and electronic energies in a material. And the Temperature of a material is proportional to the energy of material. Let’s only consider vibrational energy of a material. Then, the temperature of the material represent the total vibrational energy of the material.
Skip to 2 minutes and 16 seconds The heat capacity is then the energy, the heat, required to induce more vibration. The stronger bond means that it is easier to induce vibration throughout the material since the atoms strongly bind to each other, so the vibration spread easily. Thus it needs smaller energy to induce the same amount of vibration throughout meaning that it needs smaller energy to increase temperature. We have to keep in mind that Cp is always larger than or the same with Cv. Their difference is alpha square VT over k. And this equation will be derived later in this course. Here, alpha is the thermal expansion coefficient defined as 1 over V times dV/dT at constant pressure. K is the isothermal compressibility.
Skip to 3 minutes and 13 seconds It is minus 1 over V times dV over dP at constant temperature. For condensed phase such as liquids and solids, the expansion coefficient is generally very small, thus Cp-Cv is virtually zero. They have the same value.
In dealing with energy balance equations, we often need to calculate energy change upon temperature change.
Heat capacity is the thermal energy required to change the temperature of a material, thus so the heat change upon temperature change can be calculated by integrating heat capacity over the range of temperature range. In the second week, we will apply the energy balance equations to real problems and calculate the energy difference between two states. In these kinds of calculation, heat capacities are particularly required. Here, we introduce the concept of heat capacity and their physical meaning in preparation for the second week.
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s3://commoncrawl/crawl-data/CC-MAIN-2020-40/segments/1600400213006.47/warc/CC-MAIN-20200924002749-20200924032749-00079.warc.gz
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CC-MAIN-2020-40
| 3,122 | 6 |
https://blog.tomw.net.au/2009/07/whole-system-design-integrated-approach.html
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math
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After is seminar on "Advances in Climate Change Mitigation", Michael Smith handed me a copy of his new book "Whole System Design: An Integrated Approach to Sustainable Engineering" (Earthscan, 2009) to review. This is intended as a textbook for a course on how engineers can approach creating more sustainable systems by taking a broader approach. There is accompanying online material. However, this broad approach has its problems when narrow specialities are involved. I only felt qualified to look at one chapter of the book on "Electronics and Computer Systems". This is a good worked example of how to optimise energy savings in a system. However, the rest of the book will not be of much interest to ICT people. Of more use to specialists than the book are the online versions of the material.
The book may be of value to those undertaking cross disciplinary studies, such as in Au course "Unravelling Complexity" UGRD3001.
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s3://commoncrawl/crawl-data/CC-MAIN-2024-10/segments/1707947475203.41/warc/CC-MAIN-20240301062009-20240301092009-00339.warc.gz
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CC-MAIN-2024-10
| 930 | 2 |
https://physics-network.org/what-does-having-magnitude-mean/
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math
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/ˈmæɡ·nəˌtud/ large size or great importance: The magnitude of the task would have discouraged an ordinary man. earth science. Magnitude is also a measure of the brightness of a star as it appears from earth.
What is magnitude with example in physics?
The term magnitude is defined as “how much of a quantity”. For instance, the magnitude can be used for explaining the comparison between the speeds of a car and a bicycle. It can also be used to explain the distance travelled by an object or to explain the amount of an object in terms of its magnitude.
What does magnitude mean in science?
The magnitude is a number that characterizes the relative size of an earthquake. Magnitude is based on measurement of the maximum motion recorded by a seismograph.
What is magnitude simple words?
In physics, magnitude is defined simply as “distance or quantity.” It depicts the absolute or relative direction or size in which an object moves in the sense of motion. It is used to express the size or scope of something. In physics, magnitude generally refers to distance or quantity.
Does magnitude mean force?
The magnitude of the force is defined as the sum of all the forces acting on an object. Calculating magnitudes for forces is a vital measurement of physics. The ‘magnitude’ of a force is its ‘size’ or ‘strength’, in spite of the path in which it acts.
What is another word for magnitude in physics?
In this page you can discover 44 synonyms, antonyms, idiomatic expressions, and related words for magnitude, like: size, quantity, breadth, importance, eminence, bigness, degree, unimportance, extent, velocity and dimension.
Is magnitude a distance?
The apparent magnitude of a celestial object, such as a star or galaxy, is the brightness measured by an observer at a specific distance from the object. The smaller the distance between the observer and object, the greater the apparent brightness.
What is a magnitude of a vector?
The magnitude of a vector is the length of the vector. The magnitude of the vector a is denoted as ∥a∥. See the introduction to vectors for more about the magnitude of a vector.
Is magnitude a speed?
The magnitude of the velocity vector is the instantaneous speed of the object. The direction of the velocity vector is directed in the same direction that the object moves.
Is magnitude the same as mass?
Hi physics lover, Do you know magnitude is a pure number that defines the size(How Much) of a physical quantity! For example, if your mass is 60 kg, then 60 is the magnitude of the mass. And kg is the unit of mass.
What is magnitude and unit?
Units of measure are scalar quantities, and magnitude is defined in terms of scalar multiplication. The magnitude of a quantity in a given unit times that unit is equal to the original quantity. This holds for all kinds of tensors, including real-numbers and vectors.
Is magnitude a vector or scalar?
Vector quantities have two characteristics, a magnitude and a direction. Scalar quantities have only a magnitude. When comparing two vector quantities of the same type, you have to compare both the magnitude and the direction. For scalars, you only have to compare the magnitude.
How do you determine magnitude?
the formula to determine the magnitude of a vector (in two dimensional space) v = (x, y) is: |v| =√(x2 + y2). This formula is derived from the Pythagorean theorem. the formula to determine the magnitude of a vector (in three dimensional space) V = (x, y, z) is: |V| = √(x2 + y2 + z2)
How do we calculate magnitude in physics?
What is magnitude and direction in physics?
A vector contains two types of information: a magnitude and a direction. The magnitude is the length of the vector while the direction tells us which way the vector points. Vector direction can be given in various forms, but is most commonly denoted in degrees. Acceleration and velocity are examples of vectors.
What is magnetic magnitude?
The magnitude of the force is F = qvB sinθ where θ is the angle < 180 degrees between the velocity and the magnetic field. This implies that the magnetic force on a stationary charge or a charge moving parallel to the magnetic field is zero.
What is magnitude in force and pressure?
It means size of the force. It is sum of all forces acting on a body. If 2 forces act in same direction, Magnitude of force increases. It is the sum of of both forces.
Is magnitude the same as momentum?
There is no difference, they are exactly the same things. Since momentum is a vector, it has no magnitude.
What is magnitude measured in?
Magnitude is expressed in whole numbers and decimal fractions. For example, a magnitude 5.3 is a moderate earthquake, and a 6.3 is a strong earthquake. Because of the logarithmic basis of the scale, each whole number increase in magnitude represents a tenfold increase in measured amplitude as measured on a seismogram.
Can a magnitude be negative?
Answer: Magnitude cannot be negative. It is the length of the vector which does not have a direction (positive or negative). In the formula, the values inside the summation are squared, which makes them positive.
Why do we use magnitude?
Magnitude is used in stating the size or extent of something such as a star, earthquake, or explosion.
What is the answer of magnitude?
magnitude means greatness of size or extent.
Is magnitude absolute value?
A magnitude is the measurement or absolute value of a quantity. A magnitude is represented by a positive real number.
Is magnitude a displacement?
By magnitude, we mean the size of the displacement without regard to its direction (i.e., just a number with a unit). For example, the professor could pace back and forth many times, perhaps walking a distance of 150 meters during a lecture, yet still end up only two meters to the right of her starting point.
What is the difference between magnitude and displacement?
MAGNITUDE. ❇ It is the numerical value of any physical quantity with a proper unit. It doesn’t tell about the direction. ❇ Displacement is the shortest distance travelled between reference point to final reaching point.
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s3://commoncrawl/crawl-data/CC-MAIN-2023-23/segments/1685224655143.72/warc/CC-MAIN-20230608204017-20230608234017-00753.warc.gz
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CC-MAIN-2023-23
| 6,105 | 48 |
https://users.livejournal.com/-romantique/
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math
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|music whores anon.
||[May. 3rd, 2004|10:58 pm]
|||||'lowrider' - war||]|
on your current playlist, hit shuffle and pick the first twenty songs on the list (no matter how cheesy or embarrassing), and write down your favorite line(s) of the song. try to avoid putting the song title in the line. then, have your friends comment and see if they know the songs.
01} that rose in strangled ebony curls/moving in a yellow bedroom light/the air is wet with sound (if you don't get this cocainevanity, i'll beat you. and not in the good way. muahaha.)
02} you know I can't stand it/you're runnin' around/you know better daddy/i can't stand it cause you put me down
03} when men on the chessboard/get up and tell you where to go/and you've just had some kind of mushroom/and your mind is moving low
04} for the girl you have in that merry green land/can wait forever for you to come home/and the wind did howl and the wind did moan
05}i don't care if monday's black/tuesday wednesday heart attack/thursday never looking back
06} when the music starts i never wanna stop/it's gonna drive me crazy
07} and i don't care if you're my mother/or my motherfucking father/i could really five a fuck now/that ain't my motherfucking problem (if you don't know this song, look it up and download it. then play it really REALLY loudly when you're pissed off. the anger will be gone by the end of the song and you'll be moshing. i swear.)
08} it's the dirty story of a dirty man/and his clinging wife doesn't understand
09} close your eyes/let me touch you now/let me give you something that is real
10} what is eternal?/what is damned?/what is clay and what is sand?
11} voices will signal their tired end/the hostess is grinning/her guests sleep from sinning
12} when i kissed you girl/i knew how sweet a kiss could be/like the summer sunshine/pour your sweetness over me
13} hey satan, payed my dues/playing in a rocking band/hey momma, look at me/i’m on my way to the promised land
14) i don't need to fight/to prove i'm right/i don't need to be forgiven
15} water falling down a hundred meters/coloured by the sun/in rainbow colors
16} i'd rather be the devil/to be that woman's man/was nothin but the devil/changed my baby's mind
17} please, i know it's hard to believe/to see a perfect forest/through so many splintered trees
18} you've got to express what is taboo in you/and share your freak with the rest of us/cause it's a beautiful thing
19} it's 2am/the fear is gone/i'm sittin' here waiting/the gun's still warm/.../where am i to go now that i've gone too far (bonus points if you get this one. best song ever.)
20} believe in your brother/have faith in man/help each other, honey/if you can
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| 2,686 | 24 |
http://slideplayer.com/slide/1670942/
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math
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Presentation on theme: "1 Decisions about Cars. 2 New or Used Of course new cars are nice. They have the latest gadgets in the car world. They have a distinct smell. The floor."— Presentation transcript:
2 New or Used Of course new cars are nice. They have the latest gadgets in the car world. They have a distinct smell. The floor doesnt have very much stuff on it. Financial items to consider on any car include taxes, title and insurance. The miles per gallon the car will travel also will help you see the advantages in terms of gas expenses. Newer cars typically get better mpgs because of government regulation. Another cost to consider on the purchase of any asset, but here in the context of the car, is depreciation. Here we mean the lose of value in the car due to driving it. Used cars will typically depreciate less because the most depreciation occurs in the first year or so. What I really mean is depreciation occurs fastest the first year and then the depreciation slows down.
3 The Odometer On a used car check the odometer. If it seems low relative to the way the car looks, maybe you should pass on the car. By law, odometers are not supposed to be messed with. Title Be sure the seller has the title to the car and is the rightful owner. In the USA if you buy a car from a person who does not legally have title, the car could be return to the original owner. You then have to get your money back from the crook.
4 Budget Say you have determined you can afford to pay 375 a month on a car loan. How much car can you buy? Lets assume you have no down payment, you are looking at a 5 year loan with a nominal rate of 6.9%. Car loans are compounded monthly, although I do not think this is made clear to the consumer. In Excel we can find the present value of the uniform series, or annuity, we will pay by the following =PV(0.069/12,5*12,-375) = PV(interest rate, time frame, annuity) = 18,983 Now, if just the interest rate is higher, you can afford less car. If just the time frame is higher you can afford more car. If the amount you can pay a month is higher you can afford more car.
5 As a consumer, when you walk into a car dealer and the sales person asks you how much can you pay each month, should you lie? Back to the new or used decision. Once last detail about any car is there is a possibility you will buy a lemon. If it is new you can get the dealer to work with you because of manufacturer problems. Take a used car to a mechanic before you buy to have them check it out. Some folks think that used cars sold by private citizens sell for less than at dealers because the owner has to offer a discount as an insurance policy against the car being a lemon. The buyer then takes the car as is. We can not say here if it is better to take used or new. Some folks are willing to make trade-offs others arent. So there is no iron clad decision rule.
6 Lease or Buy? Lease ideas to consider: Closed-end lease means you walk away with no obligation at the end of the lease period, unless you abused car or went over preset mileage limits. Open-end lease means if the value of the car at the end of the lease is less than the estimated value, then you pay the difference. On a lease you may be asked to make a down payment and a security deposit.
8 Digress F2 F3 A A A A A The first diagram on the left is the one we have become accustomed to. We have an A value at the end of each of two periods and the F value occurs at the end of the second period.
9 Say A = 1 and i = 10, then F2 = 1[appen b page 691 column 10% row 2 value] = 1[2.100] = 2.10 On the previous screen the graph on the right would have F3 = 2.10 + the A at time zero taken as a single payment to time 2 in a single payment. In other words F3 = 2.10 + 1[appen a page 690 column 10% row 2 value] = 2.10 + 1[1.21]= 3.31 = 1[3.31] So F3 has only two time periods but three As. So long as the last A occurs at the same time as F we have a new story. Look at appendix b page 691 10% column row 3 value. We have 3.31. WOW, what does this mean?
10 F3 A A A This means if we have an A at time zero, then we can just imagine we had a problem that started one period before.
11 Now that we have more formally introduced compounding a section or two back, lets consider a case where payments are made more often than compounding occurs. Lets consider a case where payments are made quarterly, but compounding is semi-annual. Say we have a two year deal at 10 percent nominal interest. 0 1 2 3 4 5 6 7 8 a a a a a a a a
12 To find the future value of the annuity we need to first recognize that the interest is only compounded every other quarter. If we take the as in the 2 nd, 4 th, 6 th, and 8 th periods we have F = a [value in appendix B page 691 column 5% row 4]. = a [4.310] Now, when we look at the as in the 1 st, 3 rd, 5 th and 7 th quarters we could find the F at the 7 th period as F = a[4.310]. Since we need a half year for the interest to be earned the F at the end of the 7 th period does not have time to earn interest by the end of the 8 th quarter. If we want to know the total value at the end of the 8 th period we simply move the F at the end of the 7 th period over to the 8 th period and add it to the other value.
13 We have F = a[4.310] + a[4.310] = 2a[4.310] Do you see the significance of this result? If funds are deposited more often than the compounding period, add the values up to the end of the compounding period. Thus if a is made quarterly and interest is compounded semi-annually, just assume 2a is made semi-annually. In fact, the only time dollar values should be moved across time without making an interest adjustment is when the money does not have time to earn interest. Now, back to our story about comparing a lease to buying a car.
14 Say with a lease you have At time 0 a $1500 down payment and a $300 security deposit. Then over the next 36 months a $300 monthly payment will be made. Say if you buy you have At time 0 a $2500 down payment and a 5% sales tax payment on a $15000 car of $750. Then over the next 36 months you have a payment of 392 (financing 12500 at 8% nominal over 3 years) At the end of three years the car is still worth $8000.
15 The authors say in order to compare the two first do this for the lease: 1500 down payment + 300 month times 36 months = 10800 + (1500 down pay + 300 security dep)times 3 years time.04 interest earned on savings ( an opportunity cost calculation) = 216 For a grand total of $12,516. This is the cost of the lease. For the car the authors say: 2500 down payment + 750 sales tax + 392 a month for 36 months = 14112 + Opportunity cost of down payment 2500times 3 times.04 = 300 – 8000 in car value at end of loan, for a total cost of the car being 9,662. So the authors say buy the car.
16 I say buy the car, but for different reasons. The authors, I believe, violated a rule of finance. They added values across time without adjusting for interest. You can only when there is not enough time for interest to accrue. They added apples and oranges. You can only do this when you want to make a fruit salad ! They added values at time 0 to values at time n to values each period. This is very bad. What they should have done. Pick a time frame – either the present at time 0, the annuity time frame, or the end of the story. Lets do an end of the story at the end of the 36 months.
17 The lease would be 1800 in a single payment using 4% interest compounded annually = 1800[append a page 690 row 3 in between 3 and 5%] =1800[1.124 this is a guesstimate] = 1800[1.124864 from excel] = 2024.76 from excel + 300 times F/A factor at.08/12 for 36 months = 12,160.67 from Excel - 300 back from the down payment = 2024.76 + 12,160.67 – 300 = 13885.43
18 Note on the lease I used an opportunity cost value of the 1800. Opportunity cost means what do I give up when I make a payment. The down payment was not required and the security deposit will be given back, so what does it cost to give up these values. I took the 300 monthly and used the same rate that the car loan will occur at because I want to compare to the car loan. I subtracted out 300 at the end because the security deposit is given back at time 36.
19 The car would be [2500 down payment + 750 tax] [1.124864 from excel] = 3655.81 + 392 monthly payment times F/A factor at.08/12 for 36 months = $15,889.94 - 8000 value of car at time 36. =11545.75 So, the car is the better deal.
20 Note, the emphasis on the problem we just did was the future. Most folks use the present as the emphasis. When we consider costs, the option that is chosen is the one with the LOWEST NET PRESENT COST. The lease would be Net present cost = 1800 + 300PV(.08/12,3*12,,-300) – 300/power(1.04,3) =1800 + $9,573.54 - 266.698908 = 11106.84 The car would be Net present cost = 15000 + 750 – 8000/power(1.04,3) = 8638.02913
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s3://commoncrawl/crawl-data/CC-MAIN-2018-51/segments/1544376826530.72/warc/CC-MAIN-20181214232243-20181215014243-00399.warc.gz
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CC-MAIN-2018-51
| 8,844 | 19 |
http://knowledge.sagepub.com/view/intlpoliticalscience/n393.xml
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math
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Pub. date: 2011 | Online Pub. Date: October 04, 2011 | DOI: 10.4135/9781412994163 | Print ISBN: 9781412959636 | Online ISBN: 9781412994163| Publisher:SAGE Publications, Inc.About this encyclopedia
In political science, as in many other disciplines, linear regression is the workhorse tool for statistical analysis. It is easy to interpret, and under many conditions the estimates it provides are unbiased and efficient. Unfortunately, many of the theories in the social sciences imply a nonlinear relationship between variables. In those cases, it is inappropriate to use a classic linear regression model because, at a minimum, one of the assumptions of the model would be violated. This entry discusses two types of linear models: first, those that through a relatively simple process can be transformed in a way that allows running of a classical linear regression model; second, those that are essentially nonlinear, for which a transformation is not possible. In that situation, nonlinear least squares estimation is necessary. The linear regression model specifies a linear relationship between a response —or dependent—variable and an explanatory —or independent—variable. It is assumed ...
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s3://commoncrawl/crawl-data/CC-MAIN-2013-20/segments/1368700264179/warc/CC-MAIN-20130516103104-00065-ip-10-60-113-184.ec2.internal.warc.gz
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CC-MAIN-2013-20
| 1,205 | 2 |
https://masteressays.net/e10-2/
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math
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On June 1, Melendez Company borrows $90,000 from First Bank on a 6-month, $90,000, 12% note.
(a) Prepare the entry on June 1.
(b) Prepare the adjusting entry on June 30.
(c) Prepare the entry at maturity (December 1), assuming monthly adjusting entries have been made through November 30.
(d) What was the total financing cost (interest expense)?
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s3://commoncrawl/crawl-data/CC-MAIN-2023-50/segments/1700679100309.57/warc/CC-MAIN-20231202010506-20231202040506-00106.warc.gz
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CC-MAIN-2023-50
| 346 | 5 |
https://deepai.org/publication/on-the-finite-element-approximation-of-a-semicoercive-stokes-variational-inequality-arising-in-glaciology
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math
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We consider two important and widely studied problems in glaciology that involve contact. The first is that of the grounding line of an ice sheet flowing from the continent and into the sea, where the ice floats and loses contact with the bedrock. The stability of the position of this grounding line was first questioned by Weertman in 1974 and since then numerous analyses have attempted to prove or disprove the possibility of an instability, see e.g. [28, 38, 24]. The second problem is that of subglacial cavitation, where the ice detaches from the bedrock along the lee side of an obstacle. Subglacial cavitation occurs at a much smaller scale, along the interface between the ice and the bedrock, and is usually formulated as a boundary layer problem. Lliboutry first proposed the possibility of cavities forming between the ice and the bedrock in 1968 . Since then, subglacial cavitation has been recognised as a fundamental mechanism in glacial sliding, attracting the attention of both theoretical [17, 37] and experimental studies.
A precise understanding of both grounding line dynamics and subglacial cavitation is of great relevance for predicting future sea level rise and therefore comprehending large scale climate dynamics [36, 20]. The two contact problems described above are modelled by coupling a Stokes problem for the ice flow with a time-dependent advection equation for the free surface. At each instant in time, a Stokes problem must be solved with contact boundary conditions that allow the detachment of the ice from the bed. These contact conditions transform the instantaneous Stokes problem into a variational inequality.
Numerous finite element simulations of these equations have been carried out, see e.g. [19, 13, 16, 27]. However, to the best of our knowledge, no formal analysis of this problem and its approximation exists in the mathematical literature. Moreover, we believe that the discretisations used in these computations can be improved upon, by exploiting the structure of the variational inequality. Although the Stokes variational inequality is superficially similar to the elastic contact problem, which has been widely studied [34, 26], the Stokes problem includes three substantial difficulties that must be addressed carefully: the presence of rigid body modes in the space of admissible velocities, the nonlinear rheological law used to model ice as a viscous fluid, and the nonlinearity of the boundary condition used for the sliding law when modelling the grounding line problem.
In this work we analyse the instantaneous Stokes variational inequality and its approximation and focus on the first two difficulties due to rigid body modes and the nonlinear rheology. The presence of rigid body modes renders this problem semicoercive. Although semicoercive variational inequalities have been studied in the past [25, 40, 2], existing analyses use purely indirect arguments which give very limited information on the effects on the discretisation of the finite element spaces used. Here, we present a novel approach based on the use of metric projections onto closed convex cones to obtain constructive proofs for the discretisation errors. On the other hand, the nonlinear rheology complicates the estimation of errors for the discrete problem. Here, we use the techniques from [5, 30] to establish a convergence analysis.
We propose a mixed formulation of the Stokes variational inequality where a Lagrange multiplier is used to enforce the contact conditions. This formulation permits a structure-preserving discretisation that explicitly enforces a discrete version of the contact conditions, up to rounding errors. This allows for a precise distinction between regions where the ice detaches from the bed and those where it remains attached. This precision is extremely useful when coupling the Stokes variational inequality with the time-dependent advection equation for the free surface.
1.1 Outline of the paper
In Section 2, the Stokes variational inequality and its mixed formulation are presented. We prove a Korn-type inequality involving a metric projection onto a cone of rigid modes that will be used throughout the analysis, and we demonstrate that the mixed formulation is well posed. In Section 3, we analyse a family of finite element approximations of the mixed problem and present error estimates in terms of best approximation results for the velocity, pressure and Lagrange multiplier. Finally, in Section 4, a concrete finite element scheme involving quadratic elements for the velocity and piecewise constant elements for the pressure and the Lagrange multiplier is introduced. We then present error estimates for this scheme and provide numerical results for two test cases. For the first test, we solve a problem with a manufactured solution to calculate convergence rates and compare these with our estimates. For the second test, we compute the evolution of a subglacial cavity to exhibit the benefits of using a mixed formulation in applications of glaciological interest.
Given two normed vector spacesand and a bounded linear operator , the dual of is denoted by and the dual operator to by . The range of is denoted by and its kernel by . The norm in is denoted by and the pairing between elements in the primal and dual spaces by for and . We will work with the Lebesgue and Sobolev spaces , where and , defined as the set of functions with weak derivatives up to order which are -integrable. When we write . The space of polynomials of degree over a simplex (interval, triangle, tetrahedron) is denoted by . The space of continuous functions over a domain is given by . Vector functions and vector function spaces will be denoted with bold symbols, e.g. and .
2 Formulation of the problem
In this section we introduce the semicoercive variational inequality that arises in glaciology and present its formulation as a mixed problem with a Lagrange multiplier. An auxiliary analytical result concerning a metric projection is then proved. Finally, we analyse the existence and uniqueness of solutions for the mixed problem.
2.1 A model for ice flow
We denote by a bounded, connected and polygonal domain which represents the glacier. The assumptions of the domain being two-dimensional and polygonal are made in order to simplify the analysis, but we expect the essential results presented here to extend to three dimensional domains with smooth enough boundaries. Ice is generally modelled as a viscous incompressible flow whose motion is described by the Stokes equation : equationparentequation
In the equations above, represents the ice velocity, the pressure and
is a prescribed body force due to gravitational forces. The tensoris the symmetric part of the velocity gradient, that is,
The coefficient is the effective viscosity of ice, which relates the stress and strain rates. A power law, usually called Glen’s law , is the most common choice of rheological law for ice:
Here, represents the Frobenius norm of a matrix: for with components we have . The field is a prescribed function for which . The parameter is constant and is usually set to ; for we recover the standard linear Stokes flow. From now on, we simply write
where is in and satisfies a.e. on for some . Moreover, is in for . This expression for reveals the -Stokes nature of the problem when considered as a variational problem in the setting of Sobolev spaces.
2.2 Boundary conditions
For a given velocity and pressure field, we define the stress tensor by
where is the identity tensor field. Let denote the unit outward-pointing normal vector to the boundary . We define the normal and tangential stresses at the boundary as
The boundary is partitioned into three disjoint open sets , and . The subset represents the part of the boundary in contact with the atmosphere (and the ocean or a water-filled cavity in the case of a marine ice sheet and a subglacial cavity, respectively). Here we enforce
where represents a prescribed surface traction force. On we enforce the contact conditions which allow the ice to detach from but not penetrate the bedrock. In particular, detachment can occur if the normal stress equals the subglacial water pressure, which is defined everywhere along a thin lubrication layer in between the ice and the bedrock. In order to write these in a simplified form, we assume that the water pressure along the bed is constant and we measure stresses relative to that water pressure. We also assume that the ice can slide freely and impose no tangential stresses. Then, the boundary conditions at are given by equationparentequation
Finally, represents a portion of the boundary on which the ice is frozen and hence we prescribe no slip boundary conditions:
As explained in Section 1, one of the challenges we wish to address with this work is the case when rigid body modes are present in the space of admissible velocities. In these cases, the problem becomes semicoercive instead of coercive and the structure of the problem changes significantly. This occurs whenever is empty. For this reason, we assume that can be empty and require the subsets and to have positive measures.
2.3 The mixed formulation
We now present the mixed formulation whose analysis and approximation is the focus of this work. To do so, we first write (1) with boundary conditions (4)-(6) as a variational inequality. Then, we introduce the mixed formulation by defining a Lagrange multiplier which enforces a constraint that arises due to the contact boundary conditions (5a). In Appendix A we specify and prove the sense in which these different formulations are equivalent.
We denote by the normal trace operator onto . This operator is built by extending to the operator on , defined on smooth functions. The closed convex subset of is then defined by
We also introduce the operators and defined by
Moreover, the action of the applied body and surface forces on the domain is expressed via the function , defined as
In the mixed formulation, the constraint on is enforced via a Lagrange multiplier. We denote the range of by and equip this space with the norm. We assume the geometry of and to be sufficiently regular for this space to be a Banach space, see [34, Section 5], [26, Chapter III] and [1, Chapter 7] for discussions on normal traces and trace spaces. The Lagrange multiplier is sought in the convex cone of multipliers
The equivalent mixed formulation of (9) is: find such that equationparentequation
2.4 A metric projection onto the cone of rigid body modes
In this section we present a projection operator onto the cone of rigid body modes inside and prove a Korn-type inequality involving this operator. This result, which we believe to be novel, allows us to prove the well-posedness of the continuous and discrete problems and to obtain estimates for the velocity error in the -norm.
We define the space of rigid body modes inside by
We also introduce the subspace of rigid body modes inside , which we denote by . Note that and for all . In fact, it can be shown that , see [34, Lemma 6.1], and therefore .
The fact that coincides with the kernel of complicates the construction of error estimates in the -norm. Our solution is to make use of the metric projection onto the closed convex cone , which we shall denote by . This metric projection assigns to each function a rigid body mode for which
In Appendix B we explain that metric projections are well-defined on uniformly convex Banach spaces and that the range of is the closed convex cone , see (58) for the definition of a polar cone. Since the Sobolev space is uniformly convex for , the operator has a closed convex range. This property is exploited below to prove Theorem 1.
We first prove a preliminary Korn inequality on closed convex sets, Lemma 1. For this preliminary result, we need the following generalised Korn inequality from : for a bounded and Lipschitz domain there is a constant such that
Let be a bounded and Lipschitz domain and a closed convex subset which satisfies . Then, there is a constant such that
By using (12), it suffices to show that
Assume by contradiction that (13) does not hold. Then, there is a sequence in such that and as . By (12), we see that is bounded in and therefore we may extract a subsequence, also denoted , which converges weakly to a in and strongly in . Since is closed and convex, it is also weakly closed by [7, Theorem 3.7], so . Moreover, by the lower semicontinuity of , we also have that and . However, by construction , a contradiction.
Let denote the metric projection onto the closed convex cone . There is a such that
2.5 Well posedness of the mixed formulation
Questions on the existence and uniqueness of solutions of the mixed system (10) can be answered by studying an equivalent minimisation problem. This equivalence depends on the so-called inf-sup property holding for the operators and . Let
These inf-sup conditions can be stated as
Condition (15) is proved in [32, Lemma 3.2.7] and (16) follows from the inverse mapping theorem because is surjective onto the closed Banach space . We also define the space of divergence-free functions and the convex set as
Then, (10) is equivalent to the minimisation of the functional
over , see Appendix A.
If , which in our setting occurs when , then the Korn inequality in [9, Lemma 3] implies that is a norm on equivalent to . Since
it follows that the problem is coercive. However, whenever the set of rigid body modes is not the zero set, we then say that the operator is semicoercive with respect to the seminorm because the bound (18) does not hold for . Below, in Theorem 2, we show that a consequence of the semicoercivity of is that (10) will have a unique solution only when the following compatibility condition holds:
Condition (19) will not only allow us to establish the well-posedness of (10), but will also be required for proving velocity error estimates in the -norm. This is possible because the map from is a continuous map defined over a compact set. Therefore, whenever (19) holds, the inequality
follows with the constant defined as
The importance of the compatibility condition (19) is well-known in the study of semicoercive variational inequalities, see [31, 40, 34] in the context of general variational inequalities and [39, 9] in a glaciological setting.
If , then or tend to infinity; as a result, and the functional is coercive. Uniqueness of the solution follows from the strong convexity of the functional over , see the proof for [9, Theorem 1].
Regarding the necessity of (19) for the existence and uniqueness of solutions to (10) when , assume by contradiction that is a unique solution and that there is a rigid mode for which . By testing with in (10a), we find that
and therefore we must have and . It is then straightforward to see that is also a solution to (10), contradicting our initial assumption.
We can bound the norm of the pressure by using (22), the equality
The compatibility condition (19) becomes redundant whenever the portion of the boundary with no slip boundary conditions has a positive measure, i.e. . On the other hand, (19) cannot be satisfied if there exist non zero rigid body modes tangential to . In this case, if solves (10), then for any , with on , the triple will also solve (10).
3 Abstract discretisation
In this section we propose an abstract discretisation of the mixed system (10) built in terms of a collection of finite dimensional spaces satisfying certain key properties. We can then introduce a discrete system analogous to (10) and investigate the conditions under which we have a unique solution. Then, we prove Lemmas 2, 3, and 4, which establish upper bounds for the errors of the discrete solutions.
3.1 The discrete mixed formulation
For each parameter , let , and be finite dimensional subspaces. We assume that to avoid the need of introducing discrete compatibility conditions. We define the discrete cone
and the discrete convex set
An immediate consequence of the definitions of and is that but unless . Additionally, we have that thanks to the assumption .
The discrete analogue of the variational inequality (9) is: find such that
This discrete variational inequality can be written as a mixed problem by introducing a Lagrange multiplier. This results in the discrete mixed formulation that is the counterpart of (10): find such that equationparentequation
An advantage of using a mixed formulation at the discrete level is that we explicitly enforce a discrete version of the contact conditions (5a). Just as in (11), it is possible to show that the conditions and (24c) are equivalent to
In order to state a minimisation problem equivalent to (24), we must introduce the subspace of of discretely divergence-free functions and the discrete convex set :
Then, the discrete mixed problem (24) is equivalent to the minimisation over of the functional defined in (17), provided that two discrete inf-sup conditions hold. For , these discrete conditions can be stated as
When the conditions (26) and (27) hold, then (23), (24) and the minimisation of over are equivalent problems. The proofs for such equivalences require the same arguments as the proofs presented in Appendix A. If we additionally assume to be coercive over , then admits a unique minimiser over and the discrete mixed formulation is well-posed. For these reasons, the discrete inf-sup conditions guarantee a unique solution for (24) and set constraints on the choice of spaces , and used when approximating solutions of (10).
Analogously to the continuous case, the coercivity of over hinges on the compatibility condition (19).
Assume that (26) and (27) hold. Whenever , the discrete mixed problem (23) has a unique solution. On the other hand, if , then (24) admits a unique solution if and only if the compatibility condition (19) holds. Additionally, if (19) holds when applicable, any discrete solution to (24) is uniformly bounded from above, that is,
where the constant depends on .
The converse statement and the bound (28) again follow by taking the same steps as in the continuous case. These steps can be taken due to our definitions of and and the assumption .
3.2 Upper bounds for the velocity error
An important tool presented in [5, 30] for establishing error estimates for non-Newtonian flows is the use of the function111This operator is defined differently in [5, 30]. In these references, the term is denoted by . defined by
This operator is closely related to the operator . The inequalities
hold uniformly for all for two constants . The following variation of Young’s inequality,
is valid for any and , with the constant depending on . Additionally, the seminorm is connected to via the inequality
which holds for any . See [30, Lemmas 2.3, 2.4] for proofs of (30) and (32), and [5, Lemma 2.7] for (31). It is also worth mentioning that the quasi-norm presented in can be written in terms of , see [5, Remark 2.6].
We next present what we believe is a novel approach for bounding the velocity error in the -norm which uses the ideas presented in Section 2.4. By Theorem 1 and (32), the velocity error can be decomposed into two components as
where the constant depends on . For the first term on the right of (33), which represents the rigid component of the error, we present the following result:
Set and test equation (10a) with . Reordering, we find that
The constant depends on the norms of the solution and these are bounded from above by the norm of , see (21).
As mentioned in the introduction, previous analyses of finite element approximations of semicoercive variational inequalities either only consider the error in a seminorm or use indirect arguments to prove the convergence of the approximate solution in the complete norm [42, 25, 40, 2, 8]. In these cases, arguments by contradiction involving a sequence of triangulations are used. In Lemma 2, on the other hand, we provide a fully constructive proof for bounding the rigid component of the velocity error from above. This result is a key ingredient in obtaining the error estimates for the finite element scheme presented in the next section.
If the pair is divergence free in the sense that for all implies that , then the term in inequality (35) can be removed.
3.3 Upper bounds for the pressure and Lagrange multiplier errors
We finalise the analysis of the abstract discretisation by bounding the errors for the pressure and the Lagrange multiplier from above.
From the inf-sup condition (26) it follows that
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