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**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

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

... 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

If m =

**k**/**c**If**k**is a constant, then Δm = (-**k**/**c**²)Δ**c**If**k**is not a constant, then Δm = (-**k**/**c**²)Δ**c**+ (1/**c**)Δ**k**1 Answers · Science & Mathematics · 04/12/2008

∑(

**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

[07] 24

**K**/**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

ƒ(

**k**) = √[**k**+ (**k**−**c**)^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(**k**−**c**)] ⋅ 1/2 ⋅ [**k**+ (**k**−**c**)^2]^(-1/2) ƒ'(**k**) = (1 + 2k − 2c) / (2√[**k**+ (**k**−**c**...3 Answers · Science & Mathematics · 23/02/2013

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

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

**K**= ab + (**c**/2) a = 3 b = 4**c**= 6**K**= (3)(4) + (6/2)**K**= 12 + 3**K**= 154 Answers · Science & Mathematics · 10/11/2010

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