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 .

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

55/i dont know i do it for the points

step-by-step explanation:

the answer is y < -3 or   y > 3

-3y > 7 + 2

-3 y > 9

y < -3