Question

Water is being drained out of a swimming pool at a constant rate of 780 gallons per hour. The swimming pool initially contained 45000 gallons of water. A chemical additive must be added to the pool when it has more than 15000 gallons of water remaining in the pool.A. Write an equation 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.

Answer 1

a) Amount left = 45000 - 780h

b) 45000 - 780h > 15000

c) No, it won't

Step-by-step explanation:

a) The water is drained at rate = 780 gallons per hour

Initial gallons of water = 45000

let the number of hours = h

Amount remaining after h hours:

`\text{Amount left = }45000\text{ - 780h}`
`\begin{gathered} b)\text{ The chemical must be added when gallons of water in the pool > 15000} \\ \text{Equation showing when it is to be added:} \\ 45000\text{ - 780h > 15000} \end{gathered}`
`\begin{gathered} c)To\text{ }\det er\min e\text{ the number of days or hours it will take before the chemical is added,} \\ we^(\prime)ll\text{ solve the equation above:} \\ 45000\text{ - 780h > 15000} \\ 45000\text{ - 15000 > 780h} \\ 30000\text{ > 780h} \\ (30000)/(780)\text{ > }(780h)/(780) \\ 38.46\text{ > h} \\ h\text{ < 38.46} \\ \end{gathered}`

Hence for the chemical to be added, it will take less than 38.46 hours

2 days in hours = 2(24) = 48 hours

48 hours > 38.46 hours

As a result, it would take less than two days before the chemical is added.

No, It won't