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


Over which interval does h have a negative average rate of change?


Choose 1 answer:


(Choice A)


−2≤t≤0


(Choice B)


1≤t≤4


(Choice C)


−4≤t≤−3


(Choice D)


−3≤t≤4

Respuesta :

sam4040

Answer:

Choice C

[tex]-4\leq t\leq-3[/tex]

Step-by-step explanation:

[tex]h(t)=(t+3)^2+5[/tex]

Rate of change [tex]=h'(t)[/tex]

[tex]h'(t)=\frac{d}{dt}h(t)\\ h'(t)=2(t+3)[/tex]

For negative rate of change [tex]h'(t)\leq0[/tex]

[tex]2(t+3)\leq0\\(t+3)\leq0\\t\leq-3[/tex]

Choice C satisfies this condition.

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