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 show that ci must be equal to ki for all i = 1, 2, · · · r. do not cite any theorem, do it from definition, nothing fancy.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

(- 1, 0 )

step-by-step explanation:

an x- intercept is where the function crosses the x- axis.

to find the x- intercept/s look for the value of x when f(x) = 0

from the given table there are 3 values of x where f(x) = 0

when x = - 1, x = 2, x = 3

the one that is listed is (- 1, 0 )

1 3 5

step-by-step explanation:

i took the test: )