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A squirrel has stored its Ekans in a hole that is 45 feet from the ground in a tall tree. The squirrel starts on a perch 100 fe...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 acorn 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.
            
            
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                    Answered by
            GPT-4o mini
            
    To find the rate of change in the squirrel's height above the ground, we first need to determine the total change in height and then divide that by the time it took to make that change.
1. **Initial height of the squirrel:** 100 feet (the perch)
2. **Final height (height of acorns):** 45 feet
The change in height can be calculated as follows:
\[
\text{Change in height} = \text{Initial height} - \text{Final height} = 100 \text{ feet} - 45 \text{ feet} = 55 \text{ feet}
\]
Next, we find the rate of change using the given time of 5.25 seconds.
\[
\text{Rate of change} = \frac{\text{Change in height}}{\text{Time}} = \frac{55 \text{ feet}}{5.25 \text{ seconds}}
\]
Now we can perform the division:
\[
\text{Rate of change} \approx \frac{55}{5.25} \approx 10.48 \text{ feet per second}
\]
So, the rate of change in the squirrel's height above the ground is approximately **10.48 feet per second** (to the nearest hundredth).
    
1. **Initial height of the squirrel:** 100 feet (the perch)
2. **Final height (height of acorns):** 45 feet
The change in height can be calculated as follows:
\[
\text{Change in height} = \text{Initial height} - \text{Final height} = 100 \text{ feet} - 45 \text{ feet} = 55 \text{ feet}
\]
Next, we find the rate of change using the given time of 5.25 seconds.
\[
\text{Rate of change} = \frac{\text{Change in height}}{\text{Time}} = \frac{55 \text{ feet}}{5.25 \text{ seconds}}
\]
Now we can perform the division:
\[
\text{Rate of change} \approx \frac{55}{5.25} \approx 10.48 \text{ feet per second}
\]
So, the rate of change in the squirrel's height above the ground is approximately **10.48 feet per second** (to the nearest hundredth).
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