##### Ad

related to**K.C**

Sort by

- Relevance
- |Time

Using the binomial expansion of (3+1)^n = ∑( k=0 until n) 3^

**k****C**(n,k).1 Answers · Science & Mathematics · 18/06/2011

c = k ⊕ m c ⊕ m = k ⊕ m ⊕ m = k So

**k**=**c**⊕ m = 1001010001010111 ⊕ 0010010000101100 = 10110000011110111 Answers · Science & Mathematics · 12/03/2013

... in the formula: since k = n - (n-k), we have: C(n, k) = n! / (k! (n-

**k**)! )**C**(n, n-k) = n! / ((n-k)! (n - (n-k))! = n! / ( (n-k)! k! )1 Answers · Science & Mathematics · 14/03/2012

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

If x = -b, you have a 0 in the denominator. Division by 0 is undefined. Can you explain why?

1 Answers · Science & Mathematics · 20/08/2008

**k**- 10d =**c**/4 4k - 40d =**c**or**c**= 4k - 40d5 Answers · Science & Mathematics · 23/04/2011

Assuming that

**k**and**c**are just constants, the equation is separable, divide both sides by (**c**- y) and multiply the differential across: 1/(**c**- y) * dy =**k*** dx Integrate both sides: -ln(**c**- y) = kx +**C**Make y the subject, divide both...3 Answers · Science & Mathematics · 17/08/2013

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)2 Answers · Science & Mathematics · 24/04/2010

First off, note that ln(1) is an actual constant and simplifies to 0. The problem is now reduced to (VKC)/F=(1+E)*0-EX So now, you have a variable multiplied by X. To get rid of it, you want to divide both...

2 Answers · Science & Mathematics · 12/07/2009

Firstly, solving the original equation. x/3 = y + 2/3. Multiplying everything by 3 leaves x = 3y + 2. Now that we have an equation for x, we can solve the expression (x + 7)/3. Using the formula for x above, x + 7 = (3y + 2) + 7 = 3y + 9. Then, (x + 7)/3 = (3y + 9)/3 = y + 3...

5 Answers · Science & Mathematics · 24/06/2010

##### Ads

related to**K.C**