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July 23

Mathematics

Solv This Equation : 2x+5=35 —Preceding unsigned comment added by Kelukumar (talk • contribs) 03:05, 23 July 2010 (UTC)[reply]

This is very basic algebra. The normal method of solving this type of equation is to repeatedly "do the same thing" to both sides of the equation until it is in the form "x=...". -- SGBailey (talk) 04:00, 23 July 2010 (UTC)[reply]

it wouldn't be so basic if it were multivariable calculus. 92.230.65.204 (talk) 20:40, 23 July 2010 (UTC)[reply]

Really? 129.234.53.144 (talk) 23:50, 23 July 2010 (UTC)[reply]

First, subtract 5 from each side to get 2x = 30. Then, divide each side by 2 to get x = 15. --138.110.206.99 (talk) 12:59, 24 July 2010 (UTC)[reply]

Inverse Gabriel's Horn?

Gabriel's Horn is a figure with an infinite surface area and a finite volume. Is there any figure with the opposite: an infinite volume, but a finite surface area? --Carnildo (talk) 07:04, 23 July 2010 (UTC)[reply]

Interesting question. The whole of Eucledean 3-space has infinite volume and zero surface. Bo Jacoby (talk) 07:21, 23 July 2010 (UTC).[reply]

Straightforwardly no you can't because a sphere has the minimum area for a given volume so as the volume goes to infinity the area would too, however you could consider the surface of the earth as bounding all of space except the earth. If you consider filo pastry you can see how easy it is to have the area tend to infinity while the volume remains the same. Dmcq (talk) 09:00, 23 July 2010 (UTC)[reply]

Sure there is, technically speaking. Paul (Stansifer) 05:39, 14 August 2010 (UTC)[reply]

If we can be non-Euclidean about it, then it's easy: consider the rubber sheet diagram of a black hole, as in [1] (the blue grid). That is our non-Euclidean geometry. Now draw a circle around the hole. This is now a 1D perimeter over a 2D infinite surface area of the "horn". In other words, we just packaged an infinite amount of space into a finite part of our geometry. SamuelRiv (talk) 07:03, 17 August 2010 (UTC)[reply]

kind of a weird question...

About how many US $100 bills can you fit into a one foot cube? For what it is worth, the dimensions of a $100 bill are as follows: 6.14x2.61x.0043in. Also, while you could fold the notes to make them fit, they must remain intact. What would this cube weigh? Googlemeister (talk) 15:28, 23 July 2010 (UTC)[reply]

You would have to know the thickness of a bill to figure out its volume and answer the question.96.49.98.57 (talk) 15:32, 23 July 2010 (UTC)[reply]

I have found and added thickness dimension. Googlemeister (talk) 15:33, 23 July 2010 (UTC)[reply]

By my calculations it is 25,076. (12*12*12)/(6.14x2.61x.0043)= 25,076. You just take the total volume and divide it by the volume of one bill. Eiad77 (talk) 15:39, 23 July 2010 (UTC)[reply]

This doesn't take into account the shape of the bills. The actual number that could fit would have to be less than Eiad's number. You'd have a hard time finding the exact answer though. I'd say "around 25,000" is a good estimate without lots more work. Staecker (talk) 16:03, 23 July 2010 (UTC)[reply]

This here: http://www.stupidquestionsanswered.com/answered/dollar.htm says a dollar weighs .03 ounces. So 25,000 of them would be 46.8 pounds. 16:06, 23 July 2010 (UTC)

He didn't state that the bills need to be flat. They can be any shape. So, just keep shoving in bills and compressing it. You will get around 25,000 in the box, but they will likely be all crumpled up and impossible to completely separate. -- kainaw ™ 16:08, 23 July 2010 (UTC)[reply]

Wouldn't you fit in a larger number of them if they were flat rather than crumpled? Michael Hardy (talk) 23:57, 23 July 2010 (UTC)[reply]

Yes, fit as many as possible flat (new bills will fit better without having to compress them), then cut up more to fit the remaining space with flat pieces. If you object to cutting up $100 bills then fold to fit the space, but not quite as many will fit. 25,000 sounds a good estimate. Dbfirs 08:00, 25 July 2010 (UTC)[reply]

Here is a graphic representation that addresses your question. Wikiant (talk) 12:45, 25 July 2010 (UTC)[reply]

Not really related, but [2] (see question 13) is fun.—msh210℠ 19:50, 26 July 2010 (UTC)[reply]

advancing mathematics with little mathematics training

What is the field of mathematics someone is most likely (even if that is still just one in a million) to make an advance in with little mathematics training (say, high school algebra and the like a bit of calculus, and so on) merely by thinking hundreds of kilometers outside the box? 92.230.65.204 (talk) 20:17, 23 July 2010 (UTC)[reply]

I would say number theory. Eiad77 (talk) 21:38, 23 July 2010 (UTC)[reply]

Combinatorics, definitively. Number theory is known for generating easily stated, but tantalizingly difficult questions. Reaching the research frontier in algebraic number theory possibly requires the most knowledge of any field of math. Phils 23:01, 23 July 2010 (UTC)[reply]

I think that you'd have to think so far "out of the box" that you might actually fashion your own field. People have been doing mathematics for millenia. As such, all of the known fields have been studied intensively to varying degrees. There's probably nothing noteworthy that you could ever do in any known field with high school mathematics -- it'll have already been done; a million times over. If you did some how stumble upon a body of results that were noteworthy and new, then I reckon that you would have started a whole new field of research... topological number theory, anyone? ;-) — Fly by Night (talk) 23:13, 23 July 2010 (UTC)[reply]

If you are interested in mathematics your chances are better than if you are merely interested in being famous. Niels Henrik Abel is quoted for this recipe for being a mathematician: "by studying the masters, not their pupils". Bo Jacoby (talk) 09:20, 24 July 2010 (UTC).[reply]

You might be able to apply mathematics to a problem that no one has thought about mathematically before. The example that comes to mind is Britney Gallivan, who as a high school student examined the limits of folding a piece of paper in half multiple times. I've read the pamphlet she wrote about it—sure, the math is simple, but it was something that no one had looked at before. —Bkell (talk) 21:31, 24 July 2010 (UTC)[reply]

I agree. If you want to do original research in maths and you aren't in at least the second year of a maths PhD, then applying existing maths to a new problem is your best bet. --Tango (talk) 02:04, 25 July 2010 (UTC)[reply]

Alternatively, you could join a distributed computing project aimed at solving a mathematical problem (for example, the Great Internet Mersenne Prime Search, PrimeGrid, ABC@Home, or SZTAKI Desktop Grid). Participation would require no mathematics knowledge at all, and who knows? your computer might be the one that finds the crucial counterexample to an open conjecture. —Bkell (talk) 21:39, 24 July 2010 (UTC)[reply]