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.

Answer 1

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

1. Initial height: The squirrel starts at a perch that is 100 feet above the ground.
2. Final height: The squirrel moves down to its acorns stored in a hole that is 45 feet above the ground.

Now, we can calculate the change in height:

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

Next, the time taken for this change is 5.25 seconds.

Now, we can find the rate of change of the squirrel's height by dividing the change in height by the time:

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

Calculating this gives:

\[ \text{Rate of change} \approx -10.48 \text{ feet/second} \]

Since the result is negative, it indicates that the squirrel is descending (losing height).

Therefore, the correct 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.