Question

1. water is being drained out of a swimming pool at a constant rate of 780 gallons per hour. the swimming pool initially contained 45,000 gallons of water. a chemical additive must be added to the pool when it has no more than 15,000 gallons of water remaining in the pool.

(a) write an expression for the amount of water remaining in the pool after h - hours.
(b) write an equation that could be solved to find the least number of hours before the chemical could be added.
(c) will it take longer than two days before the chemical can be added? justify your response.

Explanation:

Step1: Find the expression for remaining water

The initial amount of water is \(45000\) gallons and the rate of draining is \(780\) gallons per hour. After \(h\) hours, the amount of water drained is \(780h\) gallons. So the amount of water remaining \(W\) is \(W = 45000-780h\)

Step2: Write the equation for the least number of hours

We want to find when \(W = 15000\). Substitute \(W = 15000\) into the expression from step 1: \(15000=45000 - 780h\)

Step3: Solve for \(h\) in the equation \(15000=45000 - 780h\)

First, subtract \(45000\) from both sides: \(15000-45000=-780h\), so \(- 30000=-780h\). Then divide both sides by \(-780\): \(h=\frac{30000}{780}=\frac{500}{13}\approx38.46\) hours. Two days is \(2\times24 = 48\) hours. Since \(38.46<48\)

Answer:

a. The expression is \(45000 - 780h\)
b. The equation is \(15000=45000 - 780h\)
c. No, it will not take longer than two days. Because solving \(15000 = 45000-780h\) gives \(h=\frac{500}{13}\approx38.46\) hours and \(38.46<48\) (48 hours in two days)