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.

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

26 is 33.33% of 78

step-by-step explanation:

26 is what percent of 78?

26 is p% of 78

equation: y = p% * x

solving our equation for p

p% = y/x

p% = 26/78

p = 0.3333

convert decimal to percent:

p% = 0.3333 * 100 = 33.33%

no

step-by-step explanation: