Answer:
Rate of change = 3 gallons/minute
Starting volume = 30 gallons
Step-by-step explanation:
A graph of the pool's water level would be useful. The slope of the line would represent the rate of change for filling the pool. It will also help us determine the amount of water that was in the pool to start. Although it is tempting to say the rate of fill is 45gallons/5 minutes, or 9 gallons/minute, it is not correct. The pool already started with an unknown volume of water. The rate of change after 5 minutes, however, can be used. The problem states it is being added at a constant rate. Let's find an equation for a line, in the format of y=mx+b, where m is the slope (rate of change) and b is the y-intercept.
We'll let y represent the volume of water in the pool, and x will be the time in minutes. The y-intercept will give us the amount of water in the pool at the start (x = 0).
The slope can be determined between the two given points:
(5,45) and (30,120):
Slope = Rise/Run [The change in y divided by the change in x].
Rise = (120 - 45) = 75
Run = (30-5) = 25
Slope = 75/25 or 3
The line, so far, becomes y=3x + b
The slope of 3 has units of gallons/minute. It is the rate at which the pool is being filled.
We can find b, the starting volume of water, by using one of the two points in the equation and solving for b:
y=3x + b [I'll use (5,45)]
45 = 3*(5) + b
b = 30 gallons
The pool had 30 gallons to start (at x = 0(minutes))
See the graph of the line, attached.
