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Water flows into a tank according to the rate F(t) = (6+t)/(1+t), and at the same time empties out at the rate E(t) = ln (t+2)/(t+1), with both F(t) and E(t) measured in gallons per minute. How much water, to the nearest gallon, is in the tank at time t = 10 minutes.

Respuesta :

Answer:

F(t) = 1.5 gallon /min

E(t) = 5.5 gallon/min

Step-by-step explanation:

The rate at which water flows into the tank = F(t)

The rate at which water is emptied from the tank = E(t)

F(t) = (6+t)/(1+t)

E(t) = ln (t+2)/(t+1)

When t = 10 mins

F(t) = (6 + 10)/(1+10)

= 16/11

F(t) = 1.5 gallon /min

E(t) = ln (10 +2)(10+1)

= ln(12)(20)

= ln(240)

E(t) = 5.5 gallon/min

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