Answer:
5 and -6
Step-by-step explanation:
Explanation:
Since {v1,...,vp} is linearly dependent, there exist scalars a1,...,ap, with not all of them being 0 such that a1v1+a2v2+...+apvp = 0. Using the linearity of T we have that
a1*T(v1)+a2*T(v1) + ... + ap*T(vp) = T(a1v19+T(a2v2)+...+T(avp) = T(a1v1+a2v2+...+apvp) = T(0) = 0.
Since at least one ai is different from 0, we obtain a non trivial linear combination that eliminates T(v1) , ..., T(vp). That proves that {T(v1) , ..., T(vp)} is a linearly dependent set of W.
Answer:
Option A, x ≥ -2
Step-by-step explanation:
<u>Since the line is not dotted, that means that -2 is included. Also, since the darkened part is toward positive infinity, that means that it is ≥.</u>
<u />
So, it is <em>x ≥ -2</em>
Answer: Option A, x ≥ -2
Answer:
36x^6/y^10
Step-by-step explanation:
First evaluate the parentheses, 6x^3 x 6x^3 = 36x^6
y^5 x y^5 = y^10 ; remember m^k x m^r = m^k+r
final answer should be 36x^6/y^10
hope this helps :')
