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Pool A starts with 380 gallons of water. It has a leak and is losing water at a rate of 9 gallons of water per minute. At the same time, Pool B starts with 420 gallons of water and also has a leak. It is losing water at a rate of 13 gallons per minute. The variable t represents the time in minutes. After how many minutes will the two pools have the same amount of water? How much water will be in the pools at that time? ➡>​

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Answer:10 minutes

Step-by-step explanation:The amount of water in Pool A after t minutes can be represented by the function A(t) = 380 - 9t, where 9t is the amount of water lost due to the leak. The amount of water in Pool B after t minutes can be represented by the function B(t) = 420 - 13t, where 13t is the amount of water lost due to the leak.

To find when the two pools have the same amount of water, we need to solve the equation A(t) = B(t):

380 - 9t = 420 - 13t

4t = 40

t = 10

Therefore, the two pools will have the same amount of water after 10 minutes. To find how much water will be in the pools at that time, we can substitute t = 10 into either A(t) or B(t):

A(10) = 380 - 9(10) = 290

B(10) = 420 - 13(10) = 290

Therefore, both pools will have 290 gallons of water after 10 minutes.

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