A rock is thrown from one side of a river to another. The function h(t) = −16t2 + 80t + 30 h(t) = −16t2 + 80t + 30 gives the height in inches of the rock t seconds after it has been thrown.(3 points) Put responses in the correct input to answer the question. Select a response, navigate to the desired input and insert the response. Responses can be selected and inserted using the space bar, enter key, left mouse button or touchpad. Responses can also be moved by dragging with a mouse. a) Find h(2.5)h(2.5). Response area b) What does h(2.5)h(2.5) represent in the context of this problem? Response area c) Does it make sense to find h(−10)h(−10) in the context of this problem?

Let's evaluate the function \( h(t) = -16t^2 + 80t + 30 \).

a) **Finding \( h(2.5) \)**

To find \( h(2.5) \), we can substitute \( t \) with \( 2.5 \):

\[
h(2.5) = -16(2.5)^2 + 80(2.5) + 30
\]
\[
= -16(6.25) + 80(2.5) + 30
\]
\[
= -100 + 200 + 30
\]
\[
= 130
\]

So, \( h(2.5) = 130 \).

**Response area for a)**: 130

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b) **What does \( h(2.5) \) represent in the context of this problem?**

\( h(2.5) \) represents the height of the rock above the ground, measured in inches, at \( t = 2.5 \) seconds after it has been thrown.

**Response area for b)**: The height of the rock above the ground at 2.5 seconds after it has been thrown.

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c) **Does it make sense to find \( h(-10) \) in the context of this problem?**

Finding \( h(-10) \) does not make sense in this context because negative time (before the rock was thrown) is not applicable to the scenario. The function is only relevant for \( t \geq 0 \).

**Response area for c)**: No, it does not make sense to find \( h(-10) \) because negative time is not applicable in this scenario.