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...standard representation is " bold v " = (a_1) j_1 + (a_2) j_2 + ... + (

**a**_**n**)**j**_n where j_i are unit vectors in independent directions ...1 Answers · Science & Mathematics · 29/01/2013

**n**(**j**(x)) = 5/((-6 + 20/(-2 + x)) (-2 + x)) = 5/(32 - 6 x)1 Answers · Science & Mathematics · 30/07/2015

...of 1I_J ) P( X € A ) = Integral ( x €

**A****n****J**) dx finally, if you have understood all this (this is what i hope...1 Answers · Science & Mathematics · 02/03/2014

...from N. Then, if i > j, then n(i) > n(j), and so a(n(i)) >=

**a**(**n**(**j**)). Thus, a(n(m)) is monotone increasing. Decreasing...1 Answers · Science & Mathematics · 04/02/2012

...i) = 1 * (2^(n - j + 1) - 1) / (2 - 1) = 2^(n - j + 1) - 1 Therefore:

**a**_**n**= ∑ (**j**= 1 to n) ∑ (i = j to n) 2^(n - i) = ∑ (j = 1 to n) [2^(n - j + 1) - 1] = ∑ (j...1 Answers · Science & Mathematics · 30/06/2013

(x + 1/x)^6 = Σ [k=0 to 6] C(6,k) x^(6−k) (1/x)^k To find term with x^3, then find k so that exponent of x is 3 in x^(6−k) (1/x)^6 x^(6−k) (1/x)^k = x^3 x^(6−k) / x^k = x^3 x^(6−k−k) = x^3 6−2k = 3 No integer value of k gives us an exponent of 3. So (x + 1/x)^6...

1 Answers · Science & Mathematics · 09/05/2017

...identity matrix: its value 0. It is created by the sum of

**A**(**n**,**j**)*B(j,n-1) = 0 As all elements of the last row of A...1 Answers · Science & Mathematics · 07/04/2014

If A, B and N are given, why don't you give them to me so I can work with them!?!?!? I = (N-K+

**A**)/**N****J**= (N-L+B)/N1 Answers · Science & Mathematics · 05/09/2012

### If

**A**is**a**fixed nxn matrix, and T(B)=AB, prove that if**A**is diagonalizable then T is diagonalizable?...matrix. That is, for any j: a(1, j)b(1, j) + a(2, j)b(2, j) + ... +

**a**(**n**,**j**)b(n, j) = 0 But then, all the action happens in the jth column, where...1 Answers · Science & Mathematics · 11/03/2014

... the n-th root of the product of all n of them: GM(a[1], ...,

**a**[**n**]) = (∏[**j**=1,n] a[j])^(1/n) For n = 2, this is GM(a, b) = √(ab...3 Answers · Science & Mathematics · 15/04/2011