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 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 find the rate of change in the squirrel's height above the ground, we can use the formula for the average rate of change, which is given by:

\[ \text{Rate of change} = \frac{\text{Change in height}}{\text{Change in time}} \]

First, we need to determine the change in height as the squirrel moves from its perch to the hole where the acorns are stored.

1. Initial height (perch): 100 feet

2. Final height (hole): 45 feet

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

4. Change in time: 5.25 seconds

Now we can plug these values into the formula:

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

Thus, the interpretation of this quotient is that the squirrel's height above the ground is decreasing at a rate of approximately -10.48 feet per second.

Therefore, the correct response is:

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