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]

17   x 256=4352

sorry if this is wrong, i'm new.

step-by-step explanation:

solution:

the given function is ,

      =

the three roots of this quadratic function is , 0 ,8 and 10.all the roots are real and rational.

if a quadratic equation is of the form ,

sum of the roots = a + b +c= 0+8+10=18=  

product of roots taken two at a time = a b + b c+ c a = 0+80+0=80 =  

product of roots = a b c = 0=  

there are 11 doors in a hall starting from 0 to 11.the entry in the hall is permitted through gate number 0 and exit is allowed through only gate number 8 and 10.so, three roots(0,8,10) of quadratic function is given by entry and exit route of this hall.