Consider a tank used in certain hydrodynamic experiments. After one experiment the tank contains 200 L of a dyesolution with a concentration of 1 g/L. To prepare for the next experiment, the tank is to be rinsed with fresh water owing in at the rate of 2 L/min, the well-stirred solution owing out at the same rate.
a. Find the time that will elapse before the concentration of dye in the tank reaches 1% of its original value?

Time,t that it will elapse before the concerntration of the dye in the tank reaches1% of its original value is 460.52 seconds

Step-by-step explanation:

Let Q(t) = amount of dye for all time,t.

Let Q'= rate in - rate out.

But Q/200 = concerntration of dye

Therefore rate out= Q/200 ×2.

Q'/Q = - 1/100

Dividing by Q gives

Ln/Q/ + c = -1/100t + c1

Integrating both sides with respect to t

Ln/Q/ = -1/100t + c2.

c1 and c1 are just another constants

Q= c3e^-t/100

Exponentiating both sides will cause the absolute values to get absorbed into the constants giving rise to a new constant c3

200= c3-0/100

c3= 200

2= 200e^-t/100

Ln1/100= - t/100

Ln0.01 = - t/ 100

Cross multiply

t= Ln0.01 × 100

t = 4.6052 ×100

t = 460.52seconds

d, x=8

step-by-step explanation:

simplifying

4x + -5 = 2x + 11

reorder the terms:

-5 + 4x = 2x + 11

reorder the terms:

-5 + 4x = 11 + 2x

solving

-5 + 4x = 11 + 2x

solving for variable 'x'.

move all terms containing x to the left, all other terms to the right.

add '-2x' to each side of the equation.

-5 + 4x + -2x = 11 + 2x + -2x

combine like terms: 4x + -2x = 2x

-5 + 2x = 11 + 2x + -2x

combine like terms: 2x + -2x = 0

-5 + 2x = 11 + 0

-5 + 2x = 11

add '5' to each side of the equation.

-5 + 5 + 2x = 11 + 5

combine like terms: -5 + 5 = 0

0 + 2x = 11 + 5

2x = 11 + 5

combine like terms: 11 + 5 = 16

2x = 16

divide each side by '2'.

it is not that hard all u have to do is plot the graph

step-by-step explanation: