A tank contains 22 gallons of water when all of a sudden the water begins draining at a constant rate of 2 gallons per hour. Let t represent the number of hours since the water begun draining and let v represent the volume of water in the tank. Write a formula that expresses y in terms of t .


Answer 1


V= 22-2t

Step-by-step explanation:

I guess the alphabet should be v instead of y. So I am working using v

The rate at which water is draining from the tank is 2gallons/hour. This is the rate of water removal from the tank. So after an hour, 2 × 1= 2 gallons would have drained. After 5 hours, 2×5 =10 gallons would have drained

Therefore to obtain the amount of water in gallons that have been removed from the tank, you will multiply the rate by the time in hours after which the draining started.

Amount (gallons) =2×t

The amount of water remaining in the tank will be obtained by subtracting the amount of water drained after some hour (2×t) from the initial amount of water in the tank (22)

Therefore, the amount of water present in the tank (v)= 22-2t or 2(11-t) gallons

Answer 2

Final answer:

The formula to express volume of the water v in terms of time t is v = 22 - 2t, where 22 is the initial volume and 2t represents the rate at which the water is draining.


This problem is a mathematical representation of a real-world scenario using a linear equation. The volume of water v in the tank can be represented in terms of time t through the equation v = 22 - 2t. This equation illustrates the initial volume of water in the tank (22 gallons) and accounts for the constant rate at which the water is draining (2 gallons per hour).

When time t = 0 (meaning no time has passed since the water started draining), the volume of water v = 22 (the initial volume). As time increases, the volume gradually decreases at a rate of 2 gallons per hour, represented by the term -2t.

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