A very large tank initially contains 100L of pure water. Starting at time t = 0 a solution with a salt concentration of 0.8kg/L is added at a rate of 5L/min. The solution is kept thoroughly mixed and is drained from the tank at a rate of 3L/min. 1. Let y(t) be the amount of salt (in kilograms) in the tank after t minutes. What differential equation does y satisfy?
2. How much salt is in the tank after 40 minutes?

1.

2.

Step-by-step explanation:

Given that:

A very large tank initially contains 100 L of pure water.

Starting at time t = 0 a solution with a salt concentration of 0.8kg/L is added at a rate of 5L/min.

. The solution is kept thoroughly mixed and is drained from the tank at a rate of 3L/min.

As 5L/min is entering and 3L/min is drained out, there is a 2L increase per minute. Therefore, the amount of water at any given time t = (100 +2t) L

t = (50 + t ) L

Since it is given that we should  consider y(t) to be the  amount of salt (in kilograms) in the tank after t minutes.

Then , the differential equation that  y satisfies can be computed as follows:

How much salt is in the tank after 40 minutes?

So,

suppose :

Then ,

Taking the integral on both sides; we have:

sin3a=12/15

step-by-step explanation:

since one side was multiplied by 3 the other side must also be multiplied by 3.

18.36. give me brainliest!

step-by-step explanation:

you have to multiply 1.53 by 12 so the equation is 12x1.53 and then that is your answer.