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

+5h, left parenthesis, t, right parenthesis, equals, left parenthesis, t, plus, 3, right parenthesis, squared, plus, 5

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

Choose 1

Choose 1

(Choice A)

A

-2 \leq t \leq 0−2≤t≤0minus, 2, is less than or equal to, t, is less than or equal to, 0

(Choice B)

B

-4 \leq t \leq -3−4≤t≤−3minus, 4, is less than or equal to, t, is less than or equal to, minus, 3

(Choice C)

C

-3 \leq t \leq 4−3≤t≤4minus, 3, is less than or equal to, t, is less than or equal to, 4

(Choice D)

D

1 \leq t \leq 41≤t≤4

B

Step-by-step explanation:

h(t) = (t + 3)² + 5

A) h(t) = (t + 3)² + 5      -2 ≤ t ≤ 0

average rate of change in h over the change in t is given as:

It does not have a negative average rate of change

B) A) h(t) = (t + 3)² + 5      -4 ≤ t ≤ -3

average rate of change in h over the change in t is given as:

It have a negative average rate of change

c) A) h(t) = (t + 3)² + 5      -3 ≤ t ≤ 4

average rate of change in h over the change in t is given as:

It does not have a negative average rate of change

D) h(t) = (t + 3)² + 5      1 ≤ t ≤ 4

average rate of change in h over the change in t is given as:

It does not have a negative average rate of change

arrange the equations with like terms in columns.

analyze the coefficients of x or y.

add the equations and solve for the remaining variable.

substitute the value into either equation and solve.

check the solution.

step-by-step explanation:

the first one

step-by-step explanation:

just id the test.