A. (7, -2), (4, -1), (3, 0), (4, 1), (7,2)

B. (-2, 1), (-1, 2), (0, 3), (1, 4) (2, 7)

C. (-2, 7), (-1, 4) (0, 3) (1, 4), (2, 7)

D. (1, -2), (2, -1), (3, 0), (4, 1), (7, 2)

Since someone answered wrongly I'll try to give the correct answer without the given xs.  It won't matter.  We're looking for a set of points which satisfy

y = f(x) = x² + 3

A. f(7)=7² + 3 = 52, not -2, so not A.

B. f(-2)=(-2)² + 3 = 7, not 1, so not B

C. f(-2)=7, that's good so far.  

f(-2)=(-1)²+3=1+3=4, good

f(0) = 0²+3 = 3, good

f(1)=1²+3 = 4, good

f(2) = 2²+3 = 7, good

C

We'll rule out D.  f(1)=1²+3=4, not -2, so not D.

a) 0

step-by-step explanation:

i^413 x i^0 = 0 as any number multiplied by a zero is

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