Question

1. Contaminated water is subjected to a cleaning process. The concentration of the

Answer 1

Answer:

The concentration of pollutants in the water can be defined as the quotient between the mass of pollutants and the total mass of water.

Assuming that 1L of water weighs 1000 grams, we know that:

Mass of pollutants = 5mg = 0.005g.

Initial concentration = 0.005g/1000g = 0.000005

If we multiply this by 100%, we get the percentage:

0.000005*100% = 0.0005%

Now, the mass of pollutants decreases by 10% each hour.

So if initially, we have 5mg

After one hour, we will have: 5mg - 0.1*5mg = 5mg*(0.9).

After another hour, we will have: 5mg*(0.9) - 0.1*5mg*0.9 = 5mg*(0.9)^2.

And so on, then after n hours, the mass of pollutants will be:

M(n) = 5mg*(0.9)^n

Or we can write this in grams as:

M(n) = 0.005g*(0.9)^n

Then the concentration as a function of time in hours will be:

C(n) = M(n)/1000g = (0.005g/1000g)*(0.9)^n

Notice that the thing between the parentheses is the initial concentration, then if we write this in percentage form:

c(n) = 100%*(0.005g/1000g)*(0.9)^n = 0.0005%*(0.9)

The function that represents the concentration of polution in the water as a function of hours is:

c(n) = 0.0005%*(0.9)