5. a tank initially contains 20 lb of salt dissolved in 200-gallon of water. starting at time t = 0, a solution containing 0.5 lb of salt per gallon enters the tank at a rate of 4 gallons per minute and the well-stirred solution is withdrawn at the same rate. (a) find the amount(in lb) of salt q in the solution as a function of t in minutes. (b) find the time at which the amount of salt in the tank reaches 50 lb. (c) find the quantity of salt in the solution as t + .

(a) Amount of salt as a function of time

(b) The time at which the amount of salt in the tank reaches 50 lb is 23.5 minutes.

(c) The amount of salt when t approaches to +inf is 100 lb.

Step-by-step explanation:

The rate of change of the amount of salt can be written as

Then we can rearrange and integrate

Then we have the model of A(t) like

(b) The time at which the amount of salt reaches 50 lb is

(c) When t approaches to +infinit, the term e^(-0.02t) approaches to zero, so the amount of salt in the solution approaches to 100 lb.

the smallest possible measurement of the width is 14 feet

step-by-step explanation:

let's assume length of parallelogram is 'l'

width of parallelogram is 'w'

the perimeter of a parallelogram must be no less than 40 feet

so, we get

we are given

length of parallelogram is 6 feet

so,

l=6

we can plug it

now, we can solve for w

subtract both sides by 12

divide both sides by 2

and we get

so,

the smallest possible measurement of the width is 14 feet

m∠a is 90°

m∠b is 40°