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Question
Math
Posted 4 months ago

A water tank holds 375 gallons. A drain is opened to let water out of the tank. The drain is closed 60 minutes later. The graph below shows the rate at which water drains from the tank as a function of time.
gal/min\mathrm{gal} / \mathrm{min}

How much water remains in the tank after the drain is closed?
\square gallons
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Answer from Sia
Posted 4 months ago
Solution by Steps
step 1
To determine how much water remains in the tank after the drain is closed, we first need to find the total amount of water drained over the 60 minutes. This can be done by integrating the rate function given by the graph. The rate function is 10860t10 - \frac{8}{60}t
step 2
We integrate the rate function from t=0t = 0 to t=60t = 60: 060(10860t)dt \int_0^{60} \left(10 - \frac{8}{60}t\right) dt
step 3
Compute the integral: 060(10860t)dt=[10t860t22]060 \int_0^{60} \left(10 - \frac{8}{60}t\right) dt = \left[10t - \frac{8}{60} \cdot \frac{t^2}{2}\right]_0^{60} =[10t460t2]060 = \left[10t - \frac{4}{60}t^2\right]_0^{60} =[10(60)460(60)2][10(0)460(0)2] = \left[10(60) - \frac{4}{60}(60)^2\right] - \left[10(0) - \frac{4}{60}(0)^2\right] =600240=360 = 600 - 240 = 360
step 4
The total amount of water drained is 360 gallons. Since the tank initially holds 375 gallons, the amount of water remaining in the tank after the drain is closed is: 375360=15 gallons 375 - 360 = 15 \text{ gallons}
Answer
15 gallons
Key Concept
Integration of a rate function
Explanation
To find the total amount of water drained, we integrate the rate function over the given time interval. The result of the integration gives us the total volume of water drained, which we subtract from the initial volume to find the remaining water.

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