A 200-liter tank initially full of water develops a leak at the bottom. Given that 30% of the water leaks out in the first 5 minutes, find the amount of water left in the tank 10 minutes after the leak develops if the water drains off at a rate that is proportional to the amount of water present.

the amount of water that is left in the tank after 10 min is 98 L

Step-by-step explanation:

since the water drains off at rate that is proportional to the water present

(-dV/dt) = k*V , where k= constant

(-dV/V) = k*dt

-∫dV/V) = k*∫dt

-ln V/V₀=k*t

or

V= V₀*e^(-k*t) , where V₀= initial volume

then since V₁=0.7*V₀ at t₁= 3 min

-ln V₁/V₀=k*t₁

then for t₂= 10 min we have

-ln V₂/V₀=k*t₂

dividing both equations

ln (V₂/V₀) / ln (V₁/V₀) =(t₂/t₁)

V₂/V₀ = (V₁/V₀)^(t₂/t₁)

V₂=V₀ *  (V₁/V₀)^(t₂/t₁)

replacing values

V₂=V₀ *  (V₁/V₀)^(t₂/t₁)  = 200 L * (0.7)^(10min/5min) = 98 L

then the amount of water that is left in the tank after 10 min is 98 L

sum= when you add is how much the whole amount is

distance= when you use angles and it means how far it goes

step-by-step explanation:

you answer is a

step-by-step explanation: