A tank contains 1,000 L of brine with 12 kg of dissolved salt. Pure water enters the tank at a rate of 10 L/min. The solution is kept thoroughly mixed and drains from the tank at the same rate. (a) How much salt is in the tank after t minutes

y = 12*e^[-t/100]

Step-by-step explanation:

We have to mix problems we use:

dy / dt = (rate in) - (rate out)

The water entering the tank at a speed of 10 L / min but it has no salt, therefore

rate in = 0

The tank has 1000 liters of brine with 12 kg of salt initially, therefore the concentration of salt at time "t" is:

y (t) / 1000

then rate out would be:

rate out = y (t) / 1000 * 10 = y (t) / 100

The difference equation would then be:

dy / dt = 0 - y (t) / 100

1 / y (t) * dy = 1/100 * dt

We integrate from both sides and we have:

ln y = - (1/100) * t + C1

y = e ^ [- (1/100) * t + C1]

y = e ^ [- (1/100) * t] * e ^ [C1]

We assume that C = e ^ [C1]

Thus:

y = C * e ^ [- (1/100) * t]

now y = 12 to t = 0, replacing:

12 = C * e ^ [- (1/100) * 0]

12 = C, therefore we would have:

y = 12 * e ^ [- t / 100]

∠acb = 8°

step-by-step explanation:

∠xba lies on the circle and is half the measure of the intercepted arc ax

hence

measure of arc ax = 2 × 42° = 84°

∠acb is outside the circle and is

∠acb = (arcab - arcax ), hence

∠acb = 0.5 (100 - 84) = 0.5 × 16° = 8°

show us the graphs dude

step-by-step explanation: