A squirrel has stored its acorns in a hole that is 45 feet from the ground in a tall tree. The squirrel starts on a perch 100 feet above the ground. The squirrel moves from the perch down to its stored acorns in 5.25 seconds. Interpret the quotient to describe the rate of change in the squirrel’s height above the ground. Give your answer to the nearest hundredth.(1 point)

Question

A squirrel has stored its acorns in a hole that is 45 feet from the ground in a tall tree. The squirrel starts on a perch 100 feet above the ground. The squirrel moves from the perch down to its stored acorns in 5.25 seconds. Interpret the quotient to describe the rate of change in the squirrel’s height above the ground. Give your answer to the nearest hundredth.(1 point)
Responses

The quotient that describes the rate of change in the squirrel’s height above the ground is −10.48 feet/second.
The quotient that describes the rate of change in the squirrel’s height above the ground is negative 10.48 feet/second.

The quotient that describes the rate of change in the squirrel’s height above the ground is 10.48 feet/second.
The quotient that describes the rate of change in the squirrel’s height above the ground is 10.48 feet/second.

The quotient that describes the rate of change in the squirrel’s height above the ground is −19.05 feet/second.
The quotient that describes the rate of change in the squirrel’s height above the ground is negative 19.05  feet/second.

The quotient that describes the rate of change in the squirrel’s height above the ground is 19.05 feet/second.
The quotient that describes the rate of change in the squirrel’s height above the ground is 19.05 feet/second.
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GPT-4o mini · Answer

To find the rate of change in the squirrel’s height above the ground, we need to calculate the change in height and then divide that by the time taken.

Initial height of the squirrel: 100 feet (on the perch)
Final height (height of stored acorns): 45 feet
Time: 5.25 seconds

Now, calculate the change in height:

\[ \text{Change in height} = \text{Final height} - \text{Initial height} = 45 \text{ feet} - 100 \text{ feet} = -55 \text{ feet} \]

Now, using the change in height and the time, we can find the rate of change:

\[ \text{Rate of change} = \frac{\text{Change in height}}{\text{Time}} = \frac{-55 \text{ feet}}{5.25 \text{ seconds}} \approx -10.48 \text{ feet/second} \]

Thus, the interpretation of the quotient is:

The quotient that describes the rate of change in the squirrel’s height above the ground is −10.48 feet/second.