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  1. k(k+1)(2k+1)/6 must be divisible by 100 means k(k+1)(2k+1) divisible by 600 which is 5x5x2x2x2x3 2k+1 is odd, and only one of k or k+1 can be even so you need k or k+1 divisible by 8 one of k, k+1, and 2k+1 will always be divisible...

    4 Answers · Science & Mathematics · 18/12/2013

  2. synthetic division 4 | 1 -k k 8 ---------------- .....1 4 | 1.. -k....... k............ 8 ...____4____4(4-k)_____4[k+4(4-k)] .....1.. 4-k....k+4(4-k)....0 According to last column, 8 + 4[k+4(4-k)] = 0 4[k+4(4-k)] = -8 4k + 16(4 - k) = -8 4k + 64 - 16k = -8 -12k = -72...

    1 Answers · Science & Mathematics · 09/11/2012

  3. ...you have written is evaluated like this: .030 = .22*(e^(-k))*(.005) I think you intended the exponent to be ((-k)*.005) so you should...

    1 Answers · Science & Mathematics · 28/09/2013

  4. y = x/(1 + kx) (1 + kx)y = x y + kxy = x kxy = x - y k = (x - y)/(xy) Say, this formula looks familiar!

    7 Answers · Science & Mathematics · 08/02/2010

  5. Showing K is a subgroup of G it's easy, try it. [Simply use the subgroup...normal in G: Let e_H be the identity element of H. K = { g in G: φ(g) = e_H}. To show K is normal...

    1 Answers · Science & Mathematics · 16/11/2009

  6. Decomposing 1/[k(k+1)(k+2)], you have: A/k + B/(k+1) + C/(k+2) = 1/[k(k+1.... Therefore: A(k+1)(k+2) + B(k)(k+2) + C(k)(k+1) = 1 which must hold true for all valid values of k. To determine...

    2 Answers · Science & Mathematics · 08/09/2007

  7. y = x/(1 + kx) y(1 + kx) = x y + kxy = x kxy = x - y k = (x - y)/(xy) I could have sworn I already answered this one!

    4 Answers · Science & Mathematics · 08/02/2010

  8. Note that Σ(k = 5 to ∞) C(k-1, k-5) k^3/2^k...to ∞) (k-1)(k-2)(k-3)(k-4) * k^3 x^k {at x = 1/2} = (1/24) x^5 * Σ(k...%29+C%28k-1%2C+k-5%29+k^3%2F2^k I hope...

    1 Answers · Science & Mathematics · 29/07/2014

  9. 0=< k =<4 points of intersection are (-1,1) and (-3,1...solve it with 2 quadratic equations, one for each point. You will get k. For (-1,1), k is between (0,2) For (-3,1), k is...

    2 Answers · Science & Mathematics · 02/09/2008

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