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:
18.75%
Step-by-step explanation:
(round answer if needed)
Answer: 0.364
The reason that it would be 0.364 is because the thousandths place is the third decimal place, which would leave you with a 3 but since there is a 6 after the 3, the 3 would be rounded up to a 4.
Answer:
Step-by-step explanation:
Vertex A of the triangle ABC when rotated by 90° counterclockwise about the origin,
Rule to be followed,
A(x, y) → P(-y, x)
Therefore, A(1, 1) → P(-1, 1)
Similarly, B(3, 2) → Q(-2, 3)
C(2, 5) → R(-5, 2)
Triangle given in second quadrant will be the triangle PQR.
If the point P of triangle PQR is reflected across a line y = x,
Rule to be followed,
P(x, y) → X(y, x)
P(-1, 1) → X(1, -1)
Similarly, Q(-2, 3) → Y(3, -2)
R(-5, 2) → Z(2, -5)
Therefore, triangle given in fourth quadrant is triangle XYZ.