The only swimming pool at the el cheapo motel is outdoors. it is 5.0 m wide and 12.0 m long. if the weekly evaporation is 2.35 in., how many gallons of water must be added to the pool if it does not rain?

Answers

rounding to the nearest thousandth would be 452,000

Your cleary not see the answer is 25

946.10 gallons per week

Step-by-step explanation:

1 cm = 0.393701 inch

Width = 5.0m = 196.85 inch

Length = 12.0 m = 472.44 inch

The volume evaporated weekly is given by:

V = L*W*2.35 = 196.85*472.33*2.35\\V=218,550.12\ in^3

Converting to gallons:

1\ gal = 231\ in^3\\V=\frac{218,550.12}{231}\\V= 946.10\ gal

946.10 gallons of water must be added to the pool each week.

946.10 gallons per week

Step-by-step explanation:

1 cm = 0.393701 inch

Width = 5.0m = 196.85 inch

Length = 12.0 m = 472.44 inch

The volume evaporated weekly is given by:

V = L*W*2.35 = 196.85*472.33*2.35\\V=218,550.12\ in^3

Converting to gallons:

1\ gal = 231\ in^3\\V=\frac{218,550.12}{231}\\V= 946.10\ gal

946.10 gallons of water must be added to the pool each week.



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