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.
$\mathrm{gal} / \mathrm{min}$
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 $10 - \frac{8}{60}t$

step 2

We integrate the rate function from $t = 0$ to $t = 60$:
$\int_0^{60} \left(10 - \frac{8}{60}t\right) dt$

step 3

Compute the integral:
$\int_0^{60} \left(10 - \frac{8}{60}t\right) dt = \left[10t - \frac{8}{60} \cdot \frac{t^2}{2}\right]_0^{60}$
$= \left[10t - \frac{4}{60}t^2\right]_0^{60}$
$= \left[10(60) - \frac{4}{60}(60)^2\right] - \left[10(0) - \frac{4}{60}(0)^2\right]$
$= 600 - 240 = 360$

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:
$375 - 360 = 15 \text{ gallons}$

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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