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  1. k=(-x)/(x^2+64)

    1 Answers · Science & Mathematics · 09/10/2007

  2. -7k + 5 + k = 47 -6k=47-5 k=42/(-6) k=-7

    2 Answers · Science & Mathematics · 06/11/2011

  3. (2k+1) * (2k) * (2k-1) * ... * (k+2) * (k+1) / 2 = (2k+1) * k * [(2k-1) * ... * (k+2) * (k+1)] In...ellipsis notation from becoming ambiguous. (Otherwise the k factor is out of sequence and makes that notation unclear.)

    1 Answers · Science & Mathematics · 18/07/2011

  4. You could solve k^2 - 3k + 5 = 0 and plug the two solutions into the second expression... for k. You can, however, use the original equation to simplify the k^4 expression: k^4 - 6k^3 + 9k^2 - 7 k^2(k^2 - 6k + 9) - 7 k^2(k^2 - 3k - 3k...

    4 Answers · Science & Mathematics · 26/11/2008

  5. k(x) = x² - 6 k(x - 1) = (x - 1)² - 6 k(x - 1) = x² - 2x + 1 - 6 k(x...2(2x + 4) + 4 h(h(x)) = 4x + 8 + 4 h(h(x)) = 4x + 12 h(x) = 2x + 4, k(x) = x² - 6 h(k(x)) = 2(x² - 6) + 4 h(k(x)) = x²...

    2 Answers · Science & Mathematics · 21/09/2012

  6. (k - 1)x² - (k + 1)x + 2 = 0 first off, notice that k <> 1 because... linear and can have only 1 solution now, proceed as you did until: (k - 3)² > 0 k > 0 ← this step is not justified...

    1 Answers · Science & Mathematics · 02/04/2011

  7. derivative of k(x)=20(2x+5)^8*(4x+1) TO THE SMART GUY THAT... A "BAD ANSWER" Using the chain rule: k'(x)=4(5+2x)^9 + 4x*9(5+2x)^8 *2 =4(5+2x)^9 + 72x(5+2x)^8 Putting...

    5 Answers · Science & Mathematics · 12/03/2007

  8. K-9/27 =7/9 Step 1, Multiply 27 to the whole entire problem. 27(K-9/27 =7/9) 27K - 9 = 21 Step 2, Add 9 and Divide...

    2 Answers · Science & Mathematics · 05/05/2012

  9. -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

  10. k ∫ (2kx - x^2)dx = 18 0 ....................k (kx^2 - x^3/3) | = 18 ....................0 k(k^2) - k^3/3 - (k(0^2) - 0^3/3) = 18...

    1 Answers · Science & Mathematics · 02/05/2011

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