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A tank holding 5000 liters of water is leaking at a rate of 3 liters per hour. How many liters of water would be in the tank after one day?

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4,928 liters of water would be in the tank after one day.

Since it is a constant rate of change, the questions fits a linear function (y=mx+b).

Y can be used to represent how much is still in the tank. M can be used to represent the change happening, which is 3. Since the water is leaking/decreasing, it would be -3. X can be used to represent the amount of hours spent. Finally, since B is our starting point, this would be the initial liters of water the tank had. With those in mind, the equation should be y=-3x+5000.

Since we want to find how many liters of water would be in the tank after one day, we can’t put that in as x, as the x-value should be in hours. Instead, we would convert a day to hours (1 day = 24 hours). 24 hours would be our x, which makes our equation y=-3(24)+5000. We solve it (-3 x 24 = -72. -72 + 5000 = 4,928), which should make y be equal to 4,928 liters of water.
User Jamie Humphries
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