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  1. Meanings for K * In accounting and investment banking, K = 1000 (similar... between May 16 and 31 of a year. o in infrared astronomy, the K-band is the region around 2.2 µm wavelength. * In baseball, K...

    11 Answers · Business & Finance · 13/12/2006

  2. -k^3 + k^2 + 2*k + 4 = -k(k^2 -k - 2) + 4 = -k(k^2 -2k + k -2) + 4 = -k[k(k-2) + 1(k-2)] + 4 = -k(k+1)(k-2) + 4 How do you want us to calculate? What...

    1 Answers · Education & Reference · 25/01/2011

  3. (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

  4. (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

  5. Assume |k−√2| < ϵ. Need to prove that | (k+2)/(k+1)−√2... from |k−√2| < ϵ assemble an expression for (k+2)/(k+1)−√2 |k−√2| < ϵ → √2−ϵ &...

    1 Answers · Science & Mathematics · 28/02/2013

  6. 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

  7. a[k] = 1/ln(k)^ln(k) a[k+1] = 1/ln(k(1+1/k))^ln(k+1) = 1/( ln(k) + ln(1+1/k))^ln(k+1) = 1/ ( ln(k)^ln(k+1) ( 1+ (ln(1+1/k)/ln(k) )^ln(k+1) a[k]/a[k+1] =ln(k)(1+(ln...

    3 Answers · Science & Mathematics · 18/04/2007

  8. "Solve for k in terms of k" doesn't make sense: k=k is obvious. I...thing and see if we can get a cleaner statement of the relationship between x and k: log(base2)x+log(base2)(x+5)=k log(base2) [x(x+5)] = k {by properties of...

    2 Answers · Science & Mathematics · 19/10/2009

  9. ========= x^2 + k + 6 = kx +3x x^2 -x(k+3) + (k+6) = 0 The equation will have equal ... happens when the discriminant is zero. So, solve (k+3)^2 - 4(k +6) = 0 k^2 + 6k + 9 - 4k - 24 = 0 k^2 + 2k - 15 = 0 (k+5)(k-3...

    2 Answers · Science & Mathematics · 14/10/2012

  10. ... 6/(k+1) - 1/(k) = 1 ← note: k ≠ 0 or -1 or 6(k) - (k+1) = (k+1)(k) or 6k - k - 1 = k² + k or 0 = k² + k- 6k + k...4*3 ) ] / 2 k = [ 4 ± 2√3 ) ] / 2 k = 2 ± √3 k = { 2 - √3, 2 + √3 } k ~ { 0.2679, 3.7321 }

    3 Answers · Science & Mathematics · 11/09/2010

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