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- 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 ...

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...

( 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)

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

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

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

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