Question 924628
time differential between the first and second measurement:
dt = 10 - 5 = 5 minutes
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volume differential between the first and second measurement:
dv = 260 - 300 = -40 gallons
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drain rate:
r = dv/dt
r = -(-40)/5 = 8 gal/min
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at the first measurement the drain had been open 5 minutes ...
how much water was drained in that time?
r = v/t
v = rt
v = 8*5
v = 40
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the tank's initial water volume was: 300 + 40 = 340
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time to empty the tank:
r = v/t
t = v/r
t = 340/8
t = 42.5 minutes
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