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guapka [62]
4 years ago
10

H(t)=(t+3)^2+5

Mathematics
1 answer:
Neko [114]4 years ago
3 0

Answer:

C

Step-by-step explanation:

Find the average rate of change in each option:

A. For -2\le t\le 0:

\dfrac{h(0)-h(-2)}{0-(-2)}\\ \\=\dfrac{((0+3)^2+5)-((-2+3)^2+5)}{2}\\ \\=\dfrac{(9+5)-(1+5)}{2}\\ \\=\dfrac{14-6}{2}\\ \\=4

Positive

B. For 1\le t\le 4:

\dfrac{h(4)-h(1)}{4-1}\\ \\=\dfrac{((4+3)^2+5)-((1+3)^2+5)}{3}\\ \\=\dfrac{(49+5)-(16+5)}{3}\\ \\=\dfrac{54-21}{3}\\ \\=11

Positive

C. For -4\le t\le -3:

\dfrac{h(-3)-h(-4)}{-3-(-4)}\\ \\=\dfrac{((-3+3)^2+5)-((-4+3)^2+5)}{1}\\ \\=\dfrac{(0+5)-(1+5)}{1}\\ \\=\dfrac{5-6}{1}\\ \\=-1

Negative

D. For -3\le t\le 4:

\dfrac{h(4)-h(-3)}{4-(-3)}\\ \\=\dfrac{((4+3)^2+5)-((-3+3)^2+5)}{7}\\ \\=\dfrac{(49+5)-(0+5)}{7}\\ \\=\dfrac{54-5}{7}\\ \\=7

Positive

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Divide the minutes by 60 to find the decimal portion of the time:


21 minutes / 60 = 0.35

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