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.
Use slope formula:
m = (y2-y1) / (x2-x1)
-1/4 = 6-2 / 1/2 - x
-1/4 = 4 / 1/2 - x
-1(1/2 - x) = 4(4)
-1/2 + x = 16
x = 16 1/2 or 33/2
Answer:
C.E.: x <em>=/=</em> -4
Step-by-step explanation:
Answer:
a = -2
b = 1
c = 5
Step-by-step explanation:
Given:
2a + 4b + c = 5 ............(1)
a - 4b = - 6
or
a = 4b - 6 .............(2)
2b + c = 7
or
c = 7 - 2b ...........(3)
substituting 2 and 3 in 1, we get
2(4b - 6 ) + 4b + (7 - 2b) = 5
or
8b - 12 + 4b + 7 - 2b = 5
or
10b - 5 = 5
or
b = 1
substituting b in 2, we get
a = 4(1) - 6
or
a = -2
substituting b in 3, we get
c = 7 - 2(1)
or
c = 5
thus,
a = -2
b = 1
c = 5
