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March 11[edit]

Please help me with this linear equation problem[edit]

The length of a rectangle is twice it's width. If it's perimeter is 54cm, find its length. Explain! — Preceding unsigned comment added by 49.137.31.101 (talk) 09:57, 11 March 2014 (UTC)[reply]

Consider how the length of the perimeter of a rectangle is related to its length and width. Rojomoke (talk) 11:14, 11 March 2014 (UTC)[reply]

This is classic homework. Check out rectangle for formulae. You'll end up setting up two equations with two unknowns, and solving them simultaneously. One equation will explain what you know about the perimeter, the other will explain what you know about the relationship between length and width. See System_of_linear_equations#Elementary_example for a worked example. If you show some of your attempts and explain where you are stuck, you might get more help. SemanticMantis (talk) 15:37, 11 March 2014 (UTC)[reply]

hyperbolic triangle inradius[edit]

What is the radius of a circle inscribed in a triangle in the hyperbolic plane, in terms of the angles? This will help me improve the code that generated these images. —Tamfang (talk) 11:22, 11 March 2014 (UTC)[reply]

Is the Problem of Apollonius relevant perhaps? --catslash (talk) 17:03, 16 March 2014 (UTC)[reply]

A slight problem of ignorance[edit]

Hi. I use FX-991ES Plus

When I type 10^-3 I get:

1x10^-3, or 1/1000

It's a bit weird because in other computations (Google calculator tool for example) I got:

1/10^3 or 0.001

Why is that? It (Just) seems as 2 different answers - will be gladful for your help ! — Preceding unsigned comment added by 79.179.193.243 (talk) 13:32, 11 March 2014 (UTC)[reply]

They're all the same. 1x10^-3 = 1/1000 = 1/10^3 = 0.001

Negating the power is the same as taking the reciprocal. Rojomoke (talk) 14:46, 11 March 2014 (UTC)[reply]

Yep, they are the same. See Fraction_(mathematics)#Converting_between_decimals_and_fractions and decimal for more info. SemanticMantis (talk) 15:40, 11 March 2014 (UTC)[reply]

Okay, Thank you. Could you also please help me understand why each one of the tools give another set of 2 answers and how could I bring my aforementioned Calculator to show the second set of answers which is allegedly simpler for me to work with (1/10^3 | 0.001)? That would also be of great help ! Thanks again guys. 79.179.193.243 (talk) 20:57, 11 March 2014 (UTC)[reply]

The reason is that your calculator defaults to fraction output, while google defaults to decimal output. I don't know if it will work, but I googled /FX-991ES decimal output/ and got this answer, which may help [1]. SemanticMantis (talk) 22:48, 11 March 2014 (UTC)[reply]

Math puzzle[edit]

I saw this math puzzle on the Internet today. I can't figure it out. Can someone let me in on the joke? Thanks. It said: "2 + 2 = fish; 3 + 3 = eight; 7 + 7 = triangle. Only smart people will figure this out." Any ideas? Thanks. — Preceding unsigned comment added by 75.44.113.200 (talk) 15:33, 11 March 2014 (UTC)[reply]

It appears that you can use the two terms of each equation to make a picture of the "sum." That is, draw two number 2s so they overlap, and it looks like a fish; two number 3s make an 8; and two number 7s make a triangle. OldTimeNESter (talk) 15:41, 11 March 2014 (UTC)[reply]

Oh, OK, I think I get it now. You don't really mean "overlap", though, do you? If you overlap, you are simply drawing a "2" right on top of another "2". Which simply looks like a "2". Right? If I understand correctly, you draw one figure, and right next to it, draw its mirror image (reflection). And it is that that creates the shape of the fish, figure eight, and triangle. Right? Thanks. 75.44.113.200 (talk) 17:05, 11 March 2014 (UTC)[reply]

Yes, your way of describing it is more accurate. OldTimeNESter (talk) 18:43, 11 March 2014 (UTC)[reply]

Agreed, but "overlap" doesn't necessarily mean the two are precisely aligned on top of each other. As long as any part of one copy is over any part of the other, they overlap (have an intersection). StuRat (talk) 20:50, 11 March 2014 (UTC)[reply]

@StuRat:: Yes, fair enough. An "overlap" does not necessarily have to overlap completely or 100%. As you say, there needn't be a precise alignment of the shapes. True. However, let's say that: (a) you draw a figure; and (b) right next to it, you draw the mirror image of that figure. Technically, there is no overlap whatsoever. Correct? Let's say that you "drew" the first figure on a piece of 8.5 by 11 paper; then you "drew" the mirror image of that figure on a separate piece of 8.5 by 11 paper; then, you put both pieces of paper very close to one another, but not quite "touching" one another. There is no overlap or intersection. Right? Thanks. Joseph A. Spadaro (talk) 22:55, 12 March 2014 (UTC)[reply]

Correct, although mathematicians might say "there is always an intersection, and in this case the intersection consists of the empty set". StuRat (talk) 12:43, 17 March 2014 (UTC)[reply]

This proof uses covariance under rotations. Count Iblis (talk) 16:05, 11 March 2014 (UTC)[reply]

Is that the correct link? That shows illustrations that do not correspond to the original question above. Are you sure that's the right link? Or am I missing something about the relationship between the illustrations in this link and the original question? Thanks. 75.44.113.200 (talk) 17:05, 11 March 2014 (UTC)[reply]

The link is a joke "proof" based on the same idea as the math puzzle in your question, viz. basing conclusions on the way math symbols look (rather than their symbolic content). Good times! OldTimeNESter (talk) 18:43, 11 March 2014 (UTC)[reply]

OK, got it. Yes, that makes sense. Thanks. Joseph A. Spadaro (talk) 22:50, 12 March 2014 (UTC)[reply]

That proof is awesome.Naraht (talk) 17:01, 11 March 2014 (UTC)[reply]