Here's a question for all you math lovers! Best answer gets 10 points!!?
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Firstly, we have to multiply all of the eqautions together.\n\nxy*yz*zx \n\n= 6*9*24 \n\n= (xyz)^2 \n\n= 1296\n\nNext step is to find the square root. (Let V = square root)\n\nV1296\n\n= 36\n\n= xyz
116787
Euclid! Come and help us please!!
653235
It's not square *roots*... it is the series of PERFECT SQUARES!!! How many times are people going to ask this? And isn't it obvious? This is part of some Google DaVinci Code contest, isn't it?\n\nWhy don't you just search for the same question because it has been asked about 31 times now!!!
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This is the third time that this question has been asked!?!?!\n\nIt's a tough one, too.\n\nI'm taking it to work with me and asking my colleges to help me out on it.\n\nThis is as far as I got:\n(2B+1)^(D-E) - F^(3G+G2) = (10J)/A\n\nBut someone "did" crack this CHALLENGE Question.\n\nSo here's the answer. . .\n\nhttp://answers.yahoo.com/question/index;_ylt=AvnsMpyNhtwrHV1JVMvXrE4jzKIX?qid=20061130045858AAB57A7\n\nANSWER:  ABCDEFGHJ = 25,693,148\n\nA=2\nB=5 \nC=6\nD=9\nE=7\nF=3\nG=1\nH=4\nJ=8
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http://www.mathnstuff.com/math/spoken/here/1words/p/p7.htm
175327
Try this, maybe?\n\nhttp://mathforum.org/library/drmath/drmath.elem.html
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I am stuck on this one. I really would like to know the best answer.
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I did a search here's where I found the answer\nhttp://www.jcu.edu/math/isep/Quincunx/Quincunx.html
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Here are two sites that look into the problem:\n\nhttp://www.kw.igs.net/~jackord/bp/f6.html\nhttp://math.fullerton.edu/mathews/N310/projects/e6.htm
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i wont answer.... log onto www.mathguru.com and discover solutions with explanations.
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This is the third time that this question has been asked!?!?!\n\nIt's a tough one, too.\n\nI'm taking it to work with me and asking my colleges to help me out on it.\n\nBut someone "did" crack this CHALLENGE Question.\n\nSo here's the answer. . .\n\nhttp://answers.yahoo.com/question/index;_ylt=AvnsMpyNhtwrHV1JVMvXrE4jzKIX?qid=20061130045858AAB57A7\n\nANSWER:  ABCDEFGHJ = 25,693,148\n\nA=2\nB=5 \nC=6\nD=9\nE=7\nF=3\nG=1\nH=4\nJ=8
105824
Trigonometry.  The good old days when I still loved math... ahhh!
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I think a calculator would be great for this one! It sounds perfect for a math wiz to try...Good luck! MayMay
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(9^9)!\n\ni.e. factorial of 9 to the power 9\n\n\n\n\n\n\n--- ► λ ◄
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Depends on how fast you're counting...\n\nhttp://hub1.worlded.org/docs/maththing/ny1p5.htm\nhttp://mathforum.org/library/drmath/view/58739.html
812449
1/n! - 1/(n+1)! =  (n+1)!/[(n!)(n+1)!] - n!/[(n!)(n+1)!]\n                        = 1/[(n!)(n+1)!] X {(n!)(n+1-1)}\n                        = n*n!/[(n!)(n+1)!]                                 (divide by n)\n                        = n!/[(n-1)!(n+1)!]                  (divide by n!)\n                        =  1/[(n-1)!(n+1)]
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so far, so good\n\nn!/(n-4)! = 56 [n!/(n-2)!] \n\nmultiply both sides by (n-2)!/n!\n\n[n!(n-2)!] / [n!(n-4)!] = 56\n\n(n-2)! / (n-4)! = 56\n\n(n-2)(n-3) = 56\n\nYou could multiply this out and solve the quadratic, in which case you will get one positive and one negative root, or if you remember that 56 = 8*7, then that quickly leads to n = 10