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Qianfan is draining his pool at a rate of 11,6 gallons of water per minute. After 60 minutes, the pool has 18,504 gallons in it. If y is the gallons of water in the pool after x minutes, which equation models the situation?

User Divyum
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Firstly, we need to find the initial amount of water in the pool. At the rate of 11.6 gallons per minute, after 60 minutes, 696 gallons of water (11.6 * 60) would have been drained from the pool.

Given that there are still 18,504 gallons of water in the pool at this time, this implies that the initial volume of the pool before the draining started was 19,200 gallons (18,504 + 696).

Let's denote:
- y as the volume of water left in the pool after x minutes
- x as the minutes passed since the draining started

Both y and x aren't fixed and are variables that change with time.

Since the pool's volume decreases at a rate of 11.6 gallons per minute, we can write that as: 11.6 * x (because each minute, x, results in 11.6 fewer gallons of water).

If we start with the initial volume of 19,200 gallons and subtract the amount drained over x minutes, this gives us an equation that models the situation: y = 19200 - 11.6 * x.

Therefore, the equation that models the situation is y = 19,200 - 11.6 * x, where y is the volume of water in gallons left in the pool after x minutes.

User Andrei Gavrila
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