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...have ln(e^5) = k*2 + C. So we get ln(e^5) is 5, and 5 = 2*

**k**+**C**. With y(5) = e^2, we get 2 = 5***k**+**C**. Solving the 2 ...2 Answers · Science & Mathematics · 24/07/2016

... coefficient can be calculated as follows: a₀ = (

**k****C**0) = k! / (0! k!) a₁ = (**k****C**1) = k! / (1! (k-1)!) a₂ = (**k****C**...6 Answers · Science & Mathematics · 10/09/2011

...when there are no zeros b^2 - 4ac < 0 b = -4 a =

**k****c**= k (-4)^2 - 4(k)(k) < 0 k^2 > -4 For what number...1 Answers · Science & Mathematics · 26/02/2014

... its equation. For a plane equation in the form a(x-h) + b(y-

**k**) +**c**(z-m) = 0 the normal vector is n = <a, b, c>. So...2 Answers · Science & Mathematics · 27/05/2010

...fit the defining Pascal's Triangle property: C(m,

**k**) +**C**(m,k-1) = C(m+1,k) [0 < k <= m]. ...1 Answers · Science & Mathematics · 09/02/2020

...x)) / g(x) > 0 k (x + 3) (x − 4) / c > 0 This is true when

**k**/**c**> 0 So general solution is: g(x) = c f(x) = k (x + 3) (x − 4...1 Answers · Science & Mathematics · 23/01/2018

...prime, we may conclude that p|m. Writing m = cp for some integer c, we get (cp)^

**k**=**c**^k * p^k = p * n^k ==> n^**k**=**c**^k * p^(k-1). Since k > 1, we have...1 Answers · Science & Mathematics · 17/10/2009

...a+d, b+e, c+f) = T((a,b,c) + (d,e,f)), and k T(a, b, c) =

**k**(**c**, b, a) = (kc, kb, ka) = T(ka, kb, kc) = T(k(a,b,c)), as required...1 Answers · Science & Mathematics · 19/06/2013

... when the probability of success is p is: (p^k)*[(1-p)^(n-

**k**)]*[**C**(n,k)] (Note, tossing a coin is always a Bernoulli...2 Answers · Science & Mathematics · 08/08/2009

... and bars" along with inclusion-exclusion, # of ways = sum (-1)^

**k*****C**(24,**k**)***C**(38-3k,23), k = 0 to 4 = 2252099280<------- # of ...2 Answers · Science & Mathematics · 03/12/2013