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I'll denote dim(

**V**) by n, and write B = {b_1... is a basis for**V**there are scalars c... is a basis for**V**, B is linearly independent...1 Answers · Science & Mathematics · 11/09/2013

a)

**v**(t) =-0.5(t+5)^2 +2000**v**'(t) = -0.5 (2)(t+5) = -1(t+5) = -t-5**v**'(10) = -10-5 =-15 (instantaneous...t)=-10 -10 = -t-5 add 5 to both sides -5 = -t t = 5 c) [**v**(2)-**v**(0)]/(2-0) = [**v**(2)-**v**(0)] /2**v**(2) = -0.5(2+5)^2 +2000 = -0.5...1 Answers · Science & Mathematics · 02/09/2011

**v**=√(9900/11)**v**=√900**v**=+/-304 Answers · Science & Mathematics · 27/09/2009

**v**(t) = (t^7)^(1/2) - t^(1/2)**v**'(t) = 1/2(t^7)^(-1/2) * (7t^6) - (1/2)t^(-1/2)**v**'(t) = (7t^6)/2(t^7...2 Answers · Science & Mathematics · 14/10/2009

If

**v**has the same direction as u, it is a multiple of u. First we... u by √3, we can shorten it so that it's length is 2:**v**= u/√3 = (1, √3, 0) Hope this helps.2 Answers · Science & Mathematics · 20/05/2007

**v**=-5i +6j W=-2i -3j now A.B = (mag.A)(mag.B)cos...in dot product, i.i = 1,j.j = 1 but i.j = 0 (as cos 0 = 1 and cos 90 = 0) hence,**v**.w = (-5i +6j)(-2i -3j) = +10i.i + 15i.j - 12j.i - 18j.j = +10 + 0 - 0 - 18 = 10 - 18...5 Answers · Science & Mathematics · 01/08/2007

... show that L^-1(T) is a subspace of

**V**, we need to show three things: (1) The zero vector...we needed to prove. To prove (2), let u and**v**be in L^-1(T). Then we need to show...1 Answers · Science & Mathematics · 23/07/2012

195 = -

**v**+ 27 195-27 =-**v**168 = -**v****v**= -168.7 Answers · Science & Mathematics · 15/05/2010

Let y =

**v**z (where we use z instead of y1 for simplicity...39; =**v**' z +**v**z' and y '' =**v**'' z + 2**v**' z' +**v**z''... to u = (dv/dt) = 2k t^(-1/2) and so**v**= k t^(1/2) + C and therefore y =**v**z...1 Answers · Science & Mathematics · 19/07/2012

Suppose that u,

**v**is linearly independent but not spanning. Since they are... not spanning, there is a vector w so that a_1*u + a_2***v**≠ w for any choice of a_1 and a_2. Now, suppose that u,**v**,w are...2 Answers · Science & Mathematics · 29/04/2013