Answer
(a) $806,400x$
(b) $24,192,000\ gal$
(c) $40.32x$, $40$ pools.
(d) $25$ days.
Work Step by Step
(a) The minimum amount required each day is $1120\ gal/min$ for $12$ hours, for $x$ days, we have $A=1120(12\times60)x=806,400x$
(b) For 30 days, $x=30$, we have $A=806,400(30)=24,192,000\ gal$
(c) Since each pool holds $20,000\ gal$, the minimum number of pools needed after $x$ days will be $P=\frac{806,400x}{20,000}=40.32x$ and for each day $P(1)\approx40$ pools.
(d) To fill $1000$ pools, we have $40.32x=1000$, thus $x=\frac{1000}{40.32}\approx25$ days.
