Given that v1, v2, · · · , vr are basis of a vector space v . suppose for some vector u in v , we have u = c1v1 + c2v2 + · · · + crvr, and the same vector u can also be expressed as u = k1v1 + k2v2 + · · · + krvr. show that ci must be equal to ki for all i = 1, 2, · · · r. do not cite any theorem, do it from definition, nothing fancy.

We have that v and

matching these two expression,

The last expression show the zero vector as linear combination of the vectors of the basis. But this vectors are linear independent, then the coefficients of the linear combination must be zero. That is,

This implies that

the image is given below

since the options are not given so we draw it from the equation

the given equation is

y = (x-1) (x+4)

y = y² + 3x - 4

the above equation is a quadratic equation which will have two roots. if we draw the graph for it, it will look like this

a ≈ 172.05

step-by-step explanation:

the pictures.