Can someone PROVE hero's formula?
471779
We use the formula  area = (1/2)bc.sin(A)\n\nThe formula above can be "proven" easily by definition that a triangle's area is just 1/2 x "base" x "height".\nNote in above, base = b. and height = c sinA. \nIts one way of looking at it. Draw a diagram.\n\nWe'll use that and some basic trigonometry to prove "Hero's formula".  \n\nAnd sin(A) = 2 sin(A/2).cos(A/2)    \n\nSo area = bc.sin(A/2).cos(A/2)\n\nNext we find expressions for cos(A/2) and sin(A/2) in terms of \ns, a, b, c, where a, b, c are the sides of the triangle and s is \nthe semi-perimeter.\n\nSo  s = (a+b+c)/2\n\n                                    \ncos(A) \n\n=   2cos^2(A/2) - 1 \n\n= [ b^2 + c^2 - a^2 ] / [2bc]\n\nSo,\n                                  \n2cos^2(A/2) \n\n= 1 + [ b^2 + c^2 - a^2 ] / [2bc]\n\n= [(b+c)^2 - a^2 ]/[2bc] = [(b+c-a)(b+c+a)]/[2bc]\n\n= [(2s-2a)2s ]/[2bc]                          \n\nSo,\n              \ncos^2(A/2) =  [s(s-a)] / [bc]\n\n                                   b^2 + c^2 - a^2\nNext, \n\ncos(A) \n\n= 1 - 2 sin^2(A/2) \n\n= [b^2 + c^2 - a^2] / [2bc]\n\nSo,\n\n2sin^2(A/2) = 1 - [b^2 + c^2 - a^2] / [2bc]\n\n2sin^2(A/2) = [a^2 - (b-c)^2] / [2bc]\n\n2sin^2(A/2) = [(a-b+c)(a+b-c)] / [2bc]\n\n2sin^2(A/2) = [(2s-2b)(2s-2c)] / [2bc]\n \nHence,                                 \n                                    \nsin^2(A/2)  = [(s-b)(s-c)] / [bc] \n\n                                    \nWe can now return to the formula for the area of the triangle using the above:\n\n(1/2)bc.sin(A) = bc.sin(A/2).cos(A/2) \n\n(1/2)bc.sin(A) = [bc.sqrt[s(s-a)(s-b(s-c)]/[bc]\n\n(1/2)bc.sin(A) = sqrt[s(s-a)(s-b)(s-c)]\n\nDone.\n\nNote : The dots "." mean nothing just "spacing".\nRegards
910564
umm none that im aware of, what kind of mathamatics leval are you looking at?
314766
For A), use the Poisson distribution. M. Poisson was a French mathematician whose name also happens to mean "fish" in French. To those who disagreed with him, M. Poisson said, "One man's meat is another man's Poisson!" \n\nIf you tested 20 tubes many times, on average 4 of them would last more than 500 hours (.20 of 20). The formula you need for the probability of exactly 5 lasting more than 500 hours is\n\nP(5) = e^-4 * (4^5/5!)\n\nThis is not so hard to compute. 4^5 is 1024, and 5! is 120. So we have\n\nP(5) = e^-4 * 1024/120\n\nP(5) = 0.018 * 1024/120 = .156\n\nFor B), I don't whether (1 - x) is in the exponent or is a factor of multiplication. Please clarify with parentheses or asterisks.\n\nf(x) = Ax^[2 * (1 - x)] ?\n\nor\n\nf(x) = A * (1 - x) * x^2 ?\n\nI see that math girl is using the latter, wherein (1 - x) is a factor of the multiplication. In that case, I agree with her that A = 12, for the reason she gives that the integral from 0 to 1 is equal to 1. She gives you the expression for the integral so you should be able to find the value of it over the interval 1/3 to 1/2. If not, choose this answer as best and ask me to integrate your expression from 1/3 to 1/2 for you.
1148906
it was not shown in the movie.but it was definetly more than 100.he played right from the beginning to the end.anyway,tht was a movie.hero's should be a hero.right?so,they just show like tht.it was 122.tht was the exact score.
603804
apply formula for it\nits easy
471450
weezy is on the right track, that glow1989 is on the right track.  For everyone who thinks that writing 1+1=2 is so obvious that it constitutes a self-contained proof of itself(!)... if you undertake a serious study of mathematical proof, you might find that 357 pages is a good length for a quick introduction to the ideas that are involved in considering questions in this area.  Of course, I am talking about actually UNDERSTANDING something - very different to simply passing an exam, regardless of the 'level' of maths you are enrolled in.\n\nYes, as you mentioned, this is a hard question.  'Proving' that 1+1=2 is not as simple as it sounds, because of the inherent question of what we mean by 'prove'.  You might have noticed that some proofs you have used depend on other proofs, which in turn depend on other proofs, ... which ultimately rest on a handful of axioms.  It is the nature of operations that are allowed by any axiomatic system, that determine the characteristics of mathematical problems that can be said to be SOLVABLE.  It is no small task to understand the axioms that underlie our systems of mathematics.  For example, maths students learn: that division by zero is undefined; that any object raised to a null power is equal to unity; some differences between rational and irrational numbers; what we mean when we consider objects to exist in 'sets'; and so on.  Those who consider themselves to be excellent students are usually able to list these rules in rapid-fire style.  But if we are asked to 'prove' why those rules are the way they are, the questions start to seem fuzzy, and you typically run out of convenient formulas to rely on.  You could easily write 357 pages just to explain what you actually MEAN by the numbers 1 or 2, without worrying about the definition of the addition operation!  As usual, people might do better to stop and consider what the question is asking before booting in a smug answer.\n\nLook into Hilbert's Problem (David Hilbert - German mathematician), which was a more workable and compact form of a proposal by Russell and Whitehead to formalise mathematics on a general basis, according to well-established axioms and procedures.\n\nOther notable mathematicians to work in this area were Alan Turing, Alonzo Church (look for Church's Lambda Calculus), Emil Post, and the brilliant Kurt Gödel.  In more recent decades, look for Gandy (1988) and of course many others.\n\nLol, molawby is spot-on.  I wish he'd posted b4 me... would've saved me a bit of typing.  If all you need to do is learn to sing the right song, to pass your exams and get the job that stipulates that you need to have completed such-and-such a course, then you don't have to worry about any of this.  Of course, that doesn't mean you have to dismiss it either.  But if you are chasing a deeper understanding, it probably does take the 20 yrs, and you won't necessarily get paid for it, heheheh.\n\nFor bedside reading, I suggest Roger Penrose etc...
1082808
mathematical dilemma. there is no answer to the question. Well at least none that are known. Just like the elements we know there out there but we have to figure them out first.  just like Einstein theory of fusion. its passable just not proved.
794328
By "Vactorial" I assume you mean "factorial"?  Then you can just as well say (2n)! / (n!)^2 for the right hand side.\n\nI wasted an hour of my life trying induction to prove this but I'm going around in circles.  The ravages of aging...
467967
this is probably not a perfect answer but this site may give you some clues.\n\nhttp://www.gomath.com/Questions/question.php?question=4751\n\naltrnatively type 'prove v*v = 2gr*r/s+v0*v0-2gr' into a search engine and see what comes up.
483801
Obviously, proof is a little hard here. It is quite impossible for anyone to PROVE they know anything about this topic.\n\nTry reading Dante's "The Divine Comedy". It is quite intriguing to say the least.
316406
Depends on how fast you're counting...\n\nhttp://hub1.worlded.org/docs/maththing/ny1p5.htm\nhttp://mathforum.org/library/drmath/view/58739.html
950045
Correct formula should be as follows:\n\nE = 1/2*I*w^2
806805
You can prove it by induction.\n\nIt is obviously true when n = 1.\n\nInduction pypothesis:  Suppose it is true for n\n\nWe need to show  is true for (n+1).\n\nDoes 24 divide (5^(2*(n+1))-1))?\n\nWe can rewrrite this as:\n\n25*(5^2n)-1  Subtract 24 from this to get 25*(5^2n)-25 \n\nThis is just equal to 25*(5^2n)-1).\n\nThis has to be divisible by 24 -- since one of the factors is (5^2n)-1) and our induction hypothesis says that it is.  \n\nAdd 24 to it and we get something that is also divisible by 24 -- but it is (5^(2*(n+1))-1)) -- so our induction is done.
405714
I'm not so sure about that. I think, in order for someone to be counted as a "hero", that they would had to have actually 'saved' someone else. If you can come up with a premise under which she was able to save someone else from something, you might be able to make it work...
1102516
by any chance is this a question from your maths homework? lol :D\n\nI cant work it out myself because to put it simple, im crap at maths.
803702
Check out this web page: http://mathworld.wolfram.com/Net.html
502657
Try this link and read it all the way through.  It's short.\n\nhttp://mathcentral.uregina.ca/qq/database/QQ.09.00/tom3.html\n\nThis link is better:\n\nhttp://mathforum.org/library/drmath/view/56760.html\n\nJust scroll down.