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  1. I love K-Pop more than American Pop and most of the highest...K-Pop is crazy (in a good way), since there are so many new K-Pop groups and artists coming out so often, that ...

    14 Answers · Entertainment & Music · 26/11/2011

  2. -k^3 + k^2pk - pk +p^2 = 0 p^2 + pk (k^2 - 1) - k^3 = 0 Solve for p by using the quadratic... up my own problem: The second term "k^2pk" looks very strange, so I will suppose it ...

    1 Answers · Science & Mathematics · 06/10/2009

  3. 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 · 19/12/2013

  4. (k-1)x^2+2x-(k-3)=0 suppose k-1=a and k-3=b so,a-b=k-1-k+3=2 so the equation becomes ax^2+(a-b)x-b=0 => ax^2+ax-bx-b=0 =>ax(x+1)-b(x+1)=0 =>(ax-b)(x-1)=0 x=-1 or x=b/a=(k-3)/(k-1)

    5 Answers · Science & Mathematics · 31/01/2011

  5. k = rate/(mol/L)^3 If the rate is in mole/sec, then k = L^3/(mole.sec)

    4 Answers · Science & Mathematics · 17/05/2011

  6. k-8/7+k = -1/5 okay :P afraid? maths <3 awesome :P love XD 5(k-8) = -1 (7+k) 5k - 40 = -7 - k 5k+k= -7 + 40 6k = 33 k= 33/6

    2 Answers · Education & Reference · 07/10/2012

  7. Let k ≥ 2. Then p_(k - 1) + 3k - 2 = (k - 1)[3(k - 1) - 1]/2 + (6k - 4)/2 = [(k - 1)(3k - 3 - 1) + 6k - 4]/2 = [(k - 1)(3k - 4) + 6k - 4]/2 = (3k² - 7k...

    1 Answers · Science & Mathematics · 13/09/2009

  8. ...1 ... equation 1 (given)     - x  +         y  +  kz  =  k ... equation 3 (given)    ——————————<ADD...

    2 Answers · Education & Reference · 25/04/2013

  9. k-4/9=3 you want to isolate k to find the answer, so you have to do the operations in reverse. so k-4=3x9 k-4=27 k=27+4 k=31

    3 Answers · Education & Reference · 28/02/2010

  10. Let (x - k) / (x - 1) = y x - k = yx - y x = yx - y + k x - yx = y + k x(1 - y) = y + k x = (-y + k) / (1 - y) x = (y - k) / (y - 1) It would seem that the function is generally its own inverse...

    2 Answers · Science & Mathematics · 12/08/2010

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