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(x+2)^2=4=>x+2=2+>x=0 or x+2=-2=>x=-4

3cos(2A)-5cos(A)+2=0. The first thing we do is use a trignometric identity, namely cos(2A) = 2(cos^2(A))-1. This gives us a new form of the original equation.

3(2(cos^2(A))-1)-5cos(A)+2=0: we expan...

Vector Equation So we know it contains three points so we can find two lines in the plane. 1) (1,2,3) + A((0,1,2) - (1,2,3)) = (1,2,3) + A(-1,-1,-1) 2) (1,2,3) + B...

Consider the Matrix M: M= a b       c d  First, find the determinant. To do this, we use the formula:  det(M) = ad - bc Then plug the information into the formula used to find the inverse: 1/det(m) x   c ...

Proof by induction

Base Case when n=1 

LHS = 1^3=1 RHS= 1/4(1^2)(1+1)^2=1/4(1)(2^2)=1/4(4)

Assume true for n=k 

so  ∑r^3= 1/4k^2(k+1)^2

For n=k+1 

∑r^3 = ∑k terms...