this is an identity question, anyone who can help, please give me a hand, i've been thinking about this questions for an hr and a half already, i really need to prove this~~ need to use sum to product method to prove it tho~
137467
Look at this link.\n\nUse the first two "sum to product" identities.\n\nOn the right you will have:\n\n= (2 / 2) (sin a / cos a) (cos b / sin b)\n\n= 1 (tan a) (cot b)\n\na = (x + y) / 2,  b = (x - y) / 2\n\nso\n\n= (tan (x+y)/2) (cot (x-y)/2)\n\nOh, darn!  You're right, the signs are reversed from this.  I wonder if the problem was given to you incorrectly.  I don't see how you could go from our result to the problem the way it was given.  Ugh!\n\nI'm sorry, but I can't go any further with this.  Good luck, and let us know if the problem was stated wrong, please.
198209
Assume there is a pair a_i, b_j with a non zero product.\nWrite fg= (a_i x^i + other terms) (b_j x^j + other terms) =\na_i b_j x^(i+j) is non zero.\n\nContradiction.
211922
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.
18132
Only one side is correct, the other is FALSE (for both sum and product)!\n\nYou can have n irrational numbers (n>1) with a rational sum:\nx1=x2=...=x(n-1)=sqrt(2), xn=-(n-1)*sqrt(2)\nx1+x2+...+xn=0\n\nYou can also have n irrational numbers (n>1) with a rational product:\nx1=x2=...=xn=2^(1/n)\nx1*x2*...*xn=2\n\nThe correct side (when all numbers are rational, their sum and product are rational) is trivial, you can prove it through induction.
227314
We have 3 products of single digit numbers (0-9) that all equate to the same value (call it x):\n\n1) abc = x\n2) bge = x\n3) def = x\n\nWe can't have one of the digits be 0 because its product would be 0.  To make the other products zero, one of their digits would also have to be zero, but no letter appears in all 3 equations.\n\nSimilarly, 5 cannot be one of the digits because that would require a five in all the other equations.  No other digit is divisible by 5 so there is no way to get a multiplier of 5 except from another 5.\n\nLikewise 7 cannot be one of the digits for the same reasons.  There are no other digits that are divisible by 7.\n\nThat leaves us with the set {1,2,3,4,6,8,9} for the seven letters a,b,c,d,e,f,g.  So we know the values of abcdefg, just need to figure out what equates to what.\n\nLet's think about the combinations for abc and def that will multiply to the same product:\n\nThe only combinations that work are:\n2x3x6 = 36\n1x4x9 = 36\n\nOr:\n3x4x6 = 72\n1x8x9 = 72\n\nIf we try to work with the first product (36), there is no way to form another product bge that is also 36.  Try it for yourself.\n\nHow else can we make 72 though?  Only one way.\n2x4x9 = 72\n\nThe one new digit used is 2.  So we are done.  We don't know the exact assignments of abc and def, but we do know that g must equal 2.\n\nThe products are:\n3x4x6 = 72\n1x8x9 = 72\n2x4x9 = 72\n\na is one of {1,3,6,8}\nb is one of {4,9}\nc is one of {1,3,6,8}\nd is one of {1,3,6,8}\ne is one of {4,9}\nf is one of {1,3,6,8}\ng is 2\n\nThe answer is g is 2.
10918
Yuo can't solve this.Because it is a formula and not an equation..If u assign some value equal to this formula such as 0 u have the solution.
497740
This question has been asked so many times!?!?!\n\nGood challenge question!\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\nSomeone "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
474840
we need to prove this by induction...\n\nfor n belonging to Z\n\nf(n)=n*(n+1)*(2n+1)/6\n\nfor n=1\n\nLHS:\n f(1)= 1^2 =1\nRHS\n1*(1+1)8(2*1+1)/6=6/6=1\n\nThus its true when n=1;\n\nAssume f(k) is true i.e. \n1^2+2^2+......+k^2=k(k+1)(2k+1).. Let this be equation 1\n\nNow we need to prove f(k+1) is true...\nie\nf(k+1)=(k+1)(k+2)(2(k+1)+1)/6\n          =(k+1)(k+2)(2k+3)/6\nL.h.s\n\nf(k+1)=1^2+2^2........+k^2+(k+1)^2\n          =k(k+1)(2K+1)/6 + (k+1)^2 from eq 1\n          =(k+1)(k(2k+1)/6 +k+1)\n          =(1/6)(k+1)(2k^2+k+6k+6)\n          =(1/6)(k+1)(2k^2+7k+6)\n          =(1/6)(k+1)(k+2)(2k+3)\n\nHence f(k+1) is true if f(k) is true....\n\nThus its true for all positive integers.......
184011
Add the two tops and add two bottoms.Then the sum is\n(Top)[1+0 , 3+0] and (Bottom)[2+0 , 6+0] = E\nE is any 2x2 matrix and O is the 2x2 zeromatrix.\nAlways is E+O = O+E = E.\nThis is also true for nxn matrices.
1418
1. It's one multiplied by ten to the twenty-third power.\n  -Otherwise ten to the twenty-third power.\n\n2. It is also an example of multiplicative identity.\n\nSorry if that is not what you are looking for.
476135
Ok, I think that I got it. on the third one you posted, it's about simplifying the binomials (2 numbers). So you add the exponents of the like variables (or coefficiants- there's no number, then it's assumed to be 1. Ex.\n(z^4)^-3 * (z^-2)^-5\n(z^2)^-8\n\nAre there no answers in the back of the book? I would email you, but it blocks your address, if I could see some of your notes it would be easier to explain I think.
421526
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.
422410
Do you know your log identities?  That would help you here a lot!\n\nUse these log identities to solve your problems:\n\n. log(a) + log(b) = log(ab)\n. log(a) - log(b) = log(a/b)\n. c*log(a) = log(a^c)\n\nSo, for problem 1 (I'm assuming the "a" means that "a" is your base) we get the following:\n\nlog a(y+5)+2log a (x+1) =\nloga(y+5) + loga((x+1)²) =\nloga((y+5)(x+1)²)  \n\nProblem 2:\n\nIn(x+5)-In(3)=In(x-3)\nln((x+5)/3) = ln(x-3)\n(x+5) / 3 = x-3\nx + 5 = 3x - 9\n2x = 14\nx = 7\n\nProblem 3:\n\n3^x+5=9^x\n\nIs 3 raised to the power of x only, or to the power of x+5?  It's hard to tell, and it does matter to the solution.\n\n~ ♥ ~
452228
I'll haev a go at number 2.\n\na) So we prove that sum(k=1 to n)k + sum(k=1 to n-1) = n^2.\nSince the sum of the first n natural numbers is n(n+1)/2, this is equal to\nn(n+1)/2 + (n-1)n/2\n= (n^2 + n + n^2 -n)/2\n= 2n^2/2\n= n^2.  QED.\n\nb) We prove that (x^0 + x^1 + x^2 + ... + x^n)^2 = x^0 + 2x +... + (n+1)x^n + nx^(n+1) + ... + 2x^(2n-1) + x^(2n).\nThe LHS will generate n^2 terms x^k where k is an integer.  How many identical terms x^k will there be for a given k?\n\nCase 1) If k<=n, there are no restrictions on k so there are k+1 multiplications resulting in x^k which are x^0*x^k, x*x^(k-1), x^2*x^(k-2), ... ,x^k*1.  The coefficient is k+1.\nCase 2) If n<k<=2n, the products resulting in x^k are x^n*x(k-n), x^(n-1)*x^(k-n+1), ... x^(k-n+1)*x^(n-1), x^(k-n)*x^n, a total of (2n-k+1) terms.  The coefficient is 2n-k+1.\n\nSo the coefficient is 1 for x^0, increases by 1 up to (n+1)x^n, and then decreases by one up to x^2n.\n\nSetting x=10 demonstrates 1*1=1, 11*11=121, etc.\n\n\nc) n^2-n is divisible by 2 for all n.  Proof: if n is even then n^2 is even and n^2-n is also even.  If n is odd then n^2 is odd and n^2-n is odd.\n\nNow n^3-n = n(n^2-1) = n(n+1)(n-1).  This is the product of 3 consecutive numbers, one of which must be a multiple of 3.\n\nThe property doesn't hold for n^4-n, for example 2^4-2 = 14 which isn't divisible by 4.
615567
lnx - 2lny + 5lnz\nlnx - lny^2 + lnz^5 [using the identitiy (b)log (a) = loga^b]\nln[(x)*(z^5)) - ln[(y^2)] [ using the identitiy log (a) + log (b) = log (ab)]\nln [((x)*(z^5))/(y^2)] [using the identitiy log(a) - log (b) = log (a/b)]\n\nFor a list of logarithmic identities, visit the link below or refer to your textbook.
400275
The problem is to show that f is identically 0 using the density of the rational numbers. I would strongly suggest to the asker that they do their own homework at this level. If you don't understand the questions or techniques, you will have a *lot* of trouble later on.
505962
Hello Rashej\n\nYes we can write that we can prove it by contradiction\nSuppose that our statement is not correct \nthus although 1^(2n+3) = 1^(3n+2)\n2n+3 - 3n+2 is not 0, for some n\nthus for some n , -n +5 , is not 0.\n\nFor n = 1 it is not 0 but clearly 1^(5) = 1^(5)\n\nThis contradicts our assumtion and thus the negation of our assumtion is valid.