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Question
Math
Posted 6 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.

How much water remains in the tank after the drain is closed?
\square gallons
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Answer from Sia
Posted 6 months ago
Solution by Steps
step 1
The rate of water drainage is given by the function on the graph, which starts at 10 gal/min at t=0t = 0 and decreases linearly to 2 gal/min at t=60t = 60 minutes
step 2
To find the total amount of water drained, we need to integrate the rate function over the interval from t=0t = 0 to t=60t = 60
step 3
The rate function can be approximated as a linear function: r(t)=10860t=10215tr(t) = 10 - \frac{8}{60}t = 10 - \frac{2}{15}t
step 4
Integrate r(t)r(t) from 00 to 6060: 060(10215t)dt \int_{0}^{60} \left(10 - \frac{2}{15}t\right) dt
step 5
Compute the integral: 06010dt060215tdt=[10t]060[215t22]060 \int_{0}^{60} 10 \, dt - \int_{0}^{60} \frac{2}{15}t \, dt = \left[10t\right]_{0}^{60} - \left[\frac{2}{15} \cdot \frac{t^2}{2}\right]_{0}^{60}
step 6
Evaluate the definite integrals: 10602156022=6002151800=600240=360 gallons 10 \cdot 60 - \frac{2}{15} \cdot \frac{60^2}{2} = 600 - \frac{2}{15} \cdot 1800 = 600 - 240 = 360 \text{ gallons}
step 7
Subtract the total drained water from the initial amount: 375360=15 gallons 375 - 360 = 15 \text{ gallons}
Answer
15 gallons
Key Concept
Integration of a linear rate function
Explanation
To find the remaining water in the tank, we integrated the rate of drainage over the given time interval and subtracted the total drained water from the initial amount.

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