Answer:
0
Step-by-step explanation:
h(x) = -16x^2 + 32x
Let x= 2
h(2) = -16(2)^2 + 32(2)
= -16(4) +64
=0
Answer: I(s,t) = {(6s, 5s+6t, 5t)| sER, tER }
Given the data in the question;
A vector is in the span of V1 and V2 only if it is a linear combination.
of V1 and V2.
so 'V' is in the span if;
V = SV1 + tV2 [ S and t are real numbers ]
V = [ 6si + 5sj ] + [ 6tj + 5tz]
V = 6s, 5s+6t, 5t
so
I(s,t) = {(6s, 5s+6t, 5t)| sER, tER }
x=-8
All you need to do is solve for x. The image shows all the work.
60 + 12 = 72
5x + x = 72
=> 6x = 72
=> x = 72 ÷ 6
=> x =12
∴ 5×12+12=72
=> 60 + 12 = 72
Answer: c
