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  1. (i+1)^k = -(3^k) => ((1+i)/3)^k = -1 => ln[((1+i)/3)^k] = ln(-1) => k ln((1+i)/3) = ln(-1) => k [ (1/2...k/2) and the abs. v. of the right side is 3^k, and a power of 3 cannot be equal to a power of...

    2 Answers · Science & Mathematics · 06/05/2013

  2. k^2 - 4k - 12 > 0 k^2 - 6k + 2k - 12 > 0 k (k - 6) + 2 (k - 6) > 0 (k - 6) (k + 2) >...you can get a product greater than 0. So EITHER k - 6 > 0 AND k + 2 > 0 (both factors are positive...

    1 Answers · Science & Mathematics · 22/01/2009

  3. k^2 - 3k = 28 k^2 - 3k - 28 = 0 (k - 7)(k + 4) = 0 k = -4, 7 and if you're asked to find x: k = -4: -4 = (x + 3)/(x - 4) -4(x - 4) = x + 3 -4x + 16 = x + 3 16 - 3 = x...

    2 Answers · Science & Mathematics · 18/08/2019

  4. I assume you need to find k so that this function is continuous at x = -2. We need to show that lim...2+) f(x) = lim (x->-2-) f(x) lim (x->-2+) (1/2)kx = lim (x->-2-) (2x^2 + k) (1/2)(-2) k = 2(-2)^2 + k -k = 8 + k -2k = 8 k = -4 Then lim (x->...

    1 Answers · Science & Mathematics · 14/07/2011

  5. Question K-Form (Precalc/Calculus)? The directions...6k ((((THIS 6k SHOULD BE 2k(pi) )))) 2x=(5pi)/6 + 2k(pi) k= 0,1 x= pi/12 + k(pi) x=(5pi)/12 + k(pi) x= pi=12, 13(pi)/12, 5(pi)/12, 17(pi...

    1 Answers · Science & Mathematics · 07/12/2008

  6. (h, k) Explanation: y = a*(x – h)^2 + k If x = h, then y = a*(h – h)^2 + k y = a*(0)^2 + k...a > 0 i.e. the parabola has upward orientation, then a + k > k. Therefore, y = k is the lowest...

    3 Answers · Science & Mathematics · 08/06/2011

  7. ...e^(kx) - e^(-kx) ] / [ e^(3x) + e^(-3x) ] When x is large and assume k >= 0 sinh(kx) / cosh(3x) ≈ e^(kx) / e^(3x) = e^(x(k - 3)) For sinh(kx...

    1 Answers · Science & Mathematics · 21/02/2019

  8. [-k +/- SQRT (k^2 - 4)] / 2 Since this equation opens up, the vertex...to never cross the x-axis, then it must have imaginary roots. So.. when k^2 - 4 < 0 So, k must be between -2 and 2

    1 Answers · Science & Mathematics · 21/04/2007

  9. Suppose K is not compact. Then, by Heine Borel theorem, K...since K is unbounded, for every M > 0 there is x ∈ K such that f(x) = ||x|| > M, which shows f is a continuous unbounded...

    2 Answers · Science & Mathematics · 04/07/2010

  10. If k = 1, then this series diverges by the nth term test, because lim(n→∞) n! = ∞. ---------- Assuming k > 1, now we use the Ratio Test: r = lim(n→∞) [((n+1)!)^2 / (k...

    1 Answers · Science & Mathematics · 07/04/2013

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