Read section 3.e and 3.f v) 1. let v be a vector space and t an operator on v (i. e., a linear map t: v -» suppose that t2 - 5t + 61 = 0, where i is the identity operator and 0 stands for the zero operator. let w3 ve v | tv =3v w2 {ve v | tu = 2v} (a) show that w3 is a subspace of v. (b) show that w3n w2 {0}. (c) for any vector v e v, let v3 = (tv - 2v) and v2 = (3v - tv). show that v3 e w3 and v2 e w2 (d) show that w3 w2 v. (hint: using the notation from the previous part, start by showing that v v32.) w3 w2 (e) use the results of the previous parts to prove that v =

see explanation

step-by-step explanation:

since the sequence is geometric there is a common ratio r between consecutive terms, that is

r = 4 ÷ - 2 = - 2

r = - 8 ÷ 4 = - 2

r = 16 ÷ 8 = - 2

the next term - 13 must be incorrect as it should be - 32 for the sequence to be geometric.

the first 4 terms are therefore geometric with common ratio r = - 2

answer: (c) m∠qpo + (2x + 16)° = 180°

step-by-step explanation:

the opposite angles of a quadrilateral are supplementary.

so, ∠o + ∠q = 180°   and   ∠p + ∠r = 180°

since ∠r = 2x + 16°, we can use substitution as follows:

∠p + ∠r = 180°

∠p + (2x + 16)° = 180°

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step-by-step explanation:

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