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  1. k - 1 - 12/(k - 1) = 1 Multiply both sides by k - 1 to get.. k(k - 1) - k + 1 - 12 = k - 1 k² - k - k - 11 = k - 1 k² - 2k - 11 = k - 1 Bring...

    4 Answers · Science & Mathematics · 28/08/2011

  2. Definitely the K Blade Tour, even though my instructor ...an ease for you to "wrist snap" on your serves. K Pro is also very head light but its ...

    3 Answers · Sports · 07/12/2008

  3. 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

  4. ∫x^k * (lnx)^n dx is worked out by parts u = (lnx)^n, and dv/dx = x^k then du/dx = (n/x)* (lnx)^n-1 , v = x^(k+1 / k+1 so S(k,n) = (lnx...

    2 Answers · Science & Mathematics · 07/06/2007

  5. (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

  6. ...quot;Thanks". Anyway... = = = = = = = = = = = = = = = = = = y = x² + (k + 2)x + 2k + 1 If the curve touch the x-axis, then y must be nil: y = 0 ...

    3 Answers · Science & Mathematics · 17/09/2013

  7. -5 < k < -4 (a) Multiply all three parts by -1. ...the inequalities: 5 > -k > 4 or equivalently 4 < -k < 5 -k is between 4 and 5. (b) Add 5 to all three...

    1 Answers · Science & Mathematics · 27/12/2013

  8. a(n) = Σ(k = 0 to ∞) 10^(-k^2) is irrational (choice C, assuming you mean...bit more, showing that it is transcendental. Let p(n) = Σ(k = 0 to n) 10^(n - k^2) and q(n) = 10^n. Then, |Σ(k = 0 to ∞) 10...

    2 Answers · Science & Mathematics · 13/04/2013

  9. (1/k)= (1/a) +(1/b) +(1/c) Multiply both abc →abc(1/k)=abc{(1/a) +(1/b) +(1/c)} →abc(1/k)=bc +ac +ab Divide both sides...

    3 Answers · Science & Mathematics · 15/08/2012

  10. k for the expression kx² + 3kx + 9 which is a positive definite b^2 - 4ac =>0 (3k)^2 - 4 * k *9 = >0 9k^2 - 36 k = > 0 9k (k -4) => 0 k =0 ; k = >4 or k...

    4 Answers · Science & Mathematics · 19/04/2013

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