Container A and container B have leaks.Container A has 815 gallons of water and is leaking 10 gallons per minute.Container B has 1100 gallons and is leaking 15 gallons per minute.Let m represent the number of minutes the contains have been leaking.How many minutes, m, will it take for the two containers to have the same amount of water? How much water will each have? Write an equation that represents this situation. Use any method (guess and check, reasoning, or algebraic) to determine the answer. Show your work below that supports your answer.

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kobenhavn kobenhavn · Answer 1 · 2021-10-04T12:05:10+02:00

The equation that represents the situation is [tex]815-10m = 1100-15m[/tex].

After [tex]57[/tex] minutes, both containers will have [tex]245[/tex] liters of water.

Volume of water in container A [tex]= 815[/tex] gallons.

Rate of leaking [tex]= 10[/tex] gallons per minute.

Volume of water in container B [tex]= 1100[/tex] gallons.

Rate of leaking [tex]= 15[/tex] gallons per minute.

Let [tex]m[/tex] represent the number of minutes the contains have been leaking.

Volume of water in container [tex]A[/tex] after [tex]m[/tex] minutes [tex]= 815-10m[/tex].

Volume of water in container [tex]B[/tex] after [tex]m[/tex] minutes [tex]= 1100-15m[/tex].

Let there be same amount of water in both the containers after [tex]m[/tex] minutes.

So, [tex]815-10m = 1100-15m[/tex] is the equation that represents the situation.

[tex]-10m+15m=1100-815[/tex]

[tex]5m=285[/tex]

[tex]m=\frac{285}{5}[/tex]

[tex]m=57[/tex].

So, after [tex]57[/tex] minutes, there is equal volume of water in both the containers.

Volume in container [tex]A = 815-10m = 815-10\times 57 =815-570=245[/tex] liters.

Volume in container [tex]B = 1100-15m = 1100-15\times 57 =1100-855=245[/tex] liters.

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