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  1. (sin x + cos x)^2 = k sin x ^2 + cos x ^ 2 + 2*sin x * cos x = k (sin x ^2 + cos x ^ 2) + 2*sin x * cos x = k 1 + 2*sin x * cos x  = k 1 + sin 2x = k sin 2x = k - 1 let k - 1 = j find sin 2x = j where  0<...

    1 Answers · Science & Mathematics · 02/08/2020

  2. f(x) = (x^2-16)^6 Use derivative od composite function f'(x) = 6(x^2-16)^5 * 2x Equalize this to zero 6(x^2-16)^5 * 2x = 0 First solution is x=0 Other solutions are found from 6(x^2-16)^5=0 6((x+4)(x-4))^5 = 0 (x+4)^5 (x-4)^5 = 0 Two more...

    1 Answers · Science & Mathematics · 03/08/2020

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