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

$$\text{Rate of change}=\frac{\text{change in height}}{\text{time taken}}$$

1. Initial height: The squirrel starts at a height of 100 feet.
2. Final height: The squirrel moves down to a height of 45 feet.

Now, calculate the change in height:

$$\text{Change in height}=\text{final height}-\text{initial height}=45\text{ feet}-100\text{ feet}=-55\text{ feet}$$

Next, we convert this change in height to a rate:

$$\text{Time taken}=5.25\text{ seconds}$$

So the rate of change is:

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

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

Therefore, the answer is:

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