The levels of mercury in two different bodies of water are rising. In one body of water the initial measure of mercury is 0.05 parts per billion (ppb) and is rising at a rate of 0.1 ppb each year. In the second body of water the initial measure is 0.12 ppb and the rate of increase is 0.06 ppb each year.

The levels of mercury in two different bodies of water are rising. In one body of water, the initial measure of mercury is 0.05 parts per billion (ppb) and is rising at a rate of 0.1 ppb each year. In the second body of water, the initial measure is 0.12 ppb and the rate of increase is 0.06 ppb each year. Which equation can be used to find y, the year in which both bodies of water have the same amount of mercury?

An equation can be used to find y, the year in which both bodies of water have the same amount of mercury will be given as below:-

0.05 + 0.1y = 0.12 + 0.06y

What is an equation?

It is defined as the relation between two variables, if we plot the graph of the linear equation we will get a straight line.

Here, we want to find y which is the heat in which the amount of mercury in each of the water bodies is the same.

Now for the first water body:

Initial is 0.05 ppb and after y years, we have a rise of 0.1 x y = 0.1y ppb

So the amount of mercury in ppb after y years would be;

0.05 + 0.1y

For the second water body;

Initial is 0.12 and a rise of 0.06 ppb per year for y years. The rise would be

0.06 x y = 0.06y

So the total amount of mercury here is

0.12 + 0.06y

So to find y which is the year the amount of mercury in each water body is the same, we simple equate both and that would be;

0.05 + 0.1y = 0.12 + 0.06y

Therefore an equation can be used to find y, the year in which both bodies of water have the same amount of mercury is 0.05 + 0.1y = 0.12 + 0.06y

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