Yahoo Web Search

  1. Ads
    related to K.C
  1. Sort by

  1. log k= log(v^c) +log p=log(p*v^c) so k=(v^c)*p. just remember : log(a^b)=b log(a) and log a+ log b= log(a*b)

    1 Answers · Science & Mathematics · 27/04/2012

  2. Take the expression n-k and put in in place of k. C(a, b) means take the formula for C(n, k), use a in place of n and b...

    2 Answers · Science & Mathematics · 05/05/2014

  3. ... through A and B is just: m = Δy/Δx = (k - 4)/(1 - 3) = (k - 4)/(-2) = (4 - k)/2. (2) Since ABCD is a...

    1 Answers · Science & Mathematics · 28/01/2013

  4. If m = k/c If k is a constant, then Δm = (-k/c²)Δc If k is not a constant, then Δm = (-k/c²)Δc + (1/ck

    1 Answers · Science & Mathematics · 04/12/2008

  5. ∑(k=0,n) C(n,k)exp(kix) = (1+exp(ix))ⁿ = (exp(ix/2))ⁿ(exp(ix/2)+exp(−ix/2))ⁿ = exp(nix/2)(2cos(x/2))ⁿ Equate imaginary parts for ∑(k=1,n)C(n,k)sin(kx) = 2ⁿsin(nx/2)cosⁿ(x/2)

    1 Answers · Science & Mathematics · 04/01/2018

  6. [07] 24K/C(n-1)=T 24K=T{C(n-1)}[multiplying both sides by C(n-1) K=T{C(n-1)}/24 [dividing both sides by 24]

    2 Answers · Science & Mathematics · 10/03/2008

  7. ƒ(k) = √[k + (kc)^2], assuming c is a constant, and k is the only variable. ƒ...v, by chain rule, we get: ƒ'(k) = u' ⋅ v ⋅ u^(v - 1) ƒ'(k) = [1 + 2(kc)] ⋅ 1/2 ⋅ [k + (kc)^2]^(-1/2) ƒ'(k) = (1 + 2k − 2c) / (2√[k + (kc...

    3 Answers · Science & Mathematics · 23/02/2013

  8. LHS 2 cos(x + 80) = 2(cos x cos 60 - sin x sin 60) = 2( cosx (1/2) - sinx (sqrt 3)(1/2)) = cosx - sqrt(3) sinx RHS k cosx - c sin (x + 60) = k cosx - c(2sinxcos60) = kcosx - c(2sinx (1/2)) = k cos x - c sinx k = 1, c = sqrt(3)

    1 Answers · Science & Mathematics · 25/06/2008

  9. t = t_start:h:t_end; y = {(F/M)/(k-c)^2+ω^2}*{exp(-k*.t) – exp(-c*t)*[cosω*.t-{(k-c)/(ω) ; plot(t,y); (t_start, t...

    1 Answers · Science & Mathematics · 08/11/2007

  10. K = ab + (c/2) a = 3 b = 4 c = 6 K = (3)(4) + (6/2) K = 12 + 3 K = 15

    4 Answers · Science & Mathematics · 10/11/2010

  1. Ads
    related to K.C
  1. Try asking your question on Yahoo Answers