find the set of values of k such that?
912600
-1 < k < 3\n\nthis monic quadratic is a parabola.\na parabola (in fact any equation) will be entirely positive when it is above the x-axis for all x (since it is monic it will open upwards and so it will be entirely positive when there are no real roots ) \nthere are no real roots when b^2-4ac < 0.\n(-2k)^2 - 4(3+2k) < 0\n4k^2 - 4(3+2k) < 0\nk^2 - 3k - 3 < 0\n(k-3)(k+1) < 0\n-1 < k < 3\n\nI hope this has helped you.  Please remember to choose a best answer for all of your questions.
513219
Perfect Squares\n\nEdit: First Four to be precise
924141
Keq = [C]/[A][B]\n[C] = Keq*[A]*[B]\n[C] = 0.76*[0.8]*[0.6]\n[C] = 0.36
913197
You have two equations:\nK+M = 4\nK = M+2\n\nSubtitute "M+2" for K in the first and you get M + 2 + M = 4, or 2M = 2, or M = 1.  Since K = M + 2, K = 3.\n\nKetchup = 3 pkg\nMustard = 1 pkg
738298
[k+2]=13\nk+2-2=13-2\n       k=11
147821
- -----------O\n------------ . .-----------  _\n  K+ ( O: Mn  :O )\n-------------. .\n------------ O\n\nomit the long dotted lines
826583
Yes, this is a difficult one.\nI'll keep trying, but don't hold your breath.\nAll I can say so far is that I could find\nonly one solution : (x, y) = (7,1).\nThere do not seem to be any more\nsolutions with x >= 8, and this is what\nneeds to be proved. So good luck with\nfinding someone who can do this.
1052440
The trivial zeros are at (-2K+0i), where K is a positive integer. The nontrivial zeros are all the others. It has been shown that the nontrivial zeroes must have a real part between 0 and 1.
1137927
I think the best you will be able to do is get a recurrence relation for the coefficients. Even that isn't going to be so easy. For example, it is clear that if we write\nt_k (x)=sum p_nk x^n,\nthen\na_n=sum p_nk  c_k.\n\nNotice that even a_0 will depend on infinitely many c_k (since t_k(0)=p_0k  is non-zero for all even k).\n\nAdded:\nIt looks like\na_0 = sum (-1)^k c_{2k}.\nThere is a recurrence relation among the p_nk which you might be able to exploit to get formulas for the other a_k. I'll think about it a bit more.\n\nAdded:\nOK. In an e-mail you suggested doing the integrals for the expnasions of x^alpha. Here's a suggestion:\nYou want to evaluate \nint_-1^1 x^n t_k(x)/sqrt(1-x^2) dx.\nif you let x=cos(A), then t_k(x)=cos(kA), so the integral transforms into\nint_-pi/2^pi/2 cos^n(A) cos(kA) dA.\nIf you now write this in terms of complex exponentials and expand by the binomial theorem, several terms will zero out. I'm not sure if this does what you want, but it makes some progress, I think.
842178
Start with the whole array as set Q.\nUse algorithm X to get the median M of Q.\nSet a = [size of Q]/2.\nIf a = k, stop; M is your answer.\nIf a > k, set Q = the values less than M, otherwise set Q = the values greater than M.\nRepeat.\n\nThis algorithm is a bit off because when [a > k], the comparison [a <=> k] is subsequently invalid.  But you get the gist of it.  Use X in a binary search mode.
943885
if you take the numbers one-by-one: 25! / ( (25-1)!*1! )\n\ncombination from n numbers taken by k:  n! / ( (n-k)!*k! )\n\nfor example:\n    -25 numbers, taken by 7: 25! / (18!*7!)\n    -25 numbers, taken by 4: 25!/ (21!*4!)
375635
Kp=Kc(RT)n\na.Kp=Kc(RT)-1/2\nb.Kp=KcRT\nc.Kp=KcRT\nd.Kp=Kc(RT)2
551213
x/(1-n)=kn\nx/k=n(1-n)\nx/k=n-n^2\nn^2-n+x/k=0\n\nUse the quadratic formula on n.
886843
(1)\nh = 3\nk = 4\n5 = a(0-3)² + 4\n5 = 9a + 4\n1 = 9a\n(1/9) = a\n\nY = (1/9)(x - 3)² + 4\n\n(3)\nY = (¼)x² + 4x -  7\nY = (¼)(x²+16x) - 7\n\n x² +16x = x² + 16x + (16/2)² - (16/2)²\n     = [x + (16/2)]² - (16/2)\n     = (x + 8)² - 8²\n     = (x + 8)² - 64\n\nY = (¼)[(x + 8)² - 64] - 7\nY = (¼)(x + 8)² - 16 - 7\nY = (¼)(x + 8)² - 23
522619
1 1 1 2\n1 1 2 1\n1 2 1 1\n2 1 1 1\n\nNo other case is possible, So n(A)=4\n\n\n1 1 1 3\n1 1 3 1\n1 3 1 1\n3 1 1 1\n\n1 1 2 2\n1 2 1 2\n1 2 2 1\n\n2 1 1 2\n2 1 2 1\n2 2 1 1\n\nNo other possibility\nn(B)=10
363034
m = kn\nk = m/n\n\nk = 144/8\nk = 18\n\nANS : 18
242099
Choose(n-k+1,k), n = 20, k = 5;\nlet f be the function f(n,k) -- same definition\n\nfor a set S of size n, u can either choose the largest or not.  If you choose the largest (ie n), then you can't have n-1.\n\nIf you dont you still have k left.\n\nHence the recursion is f(n,k) = f(n-2, k-1) + f(n-1, k).  Figure out the base cases, prove using induction.