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?

0.05 – 0.1y = 0.12 – 0.06y
0.05y + 0.1 = 0.12y + 0.06
0.05 + 0.1y = 0.12 + 0.06y
0.05y – 0.1 = 0.12y – 0.06

Explanation : As per the given set of information,one can easily arrive at the algebraic expression with one variable and use this equation to find y as the year in which both the bodies of water have the same amount of mercury.

Answer Link

jainveenamrata jainveenamrata · Answer 2 · 2022-03-23T06:45:49+01:00

The year in which both bodies of water have the same amount of mercury is 0.05 + 0.1 y = 0.12 + 0.06y. Then the correct option is C.

What is the linear system?

A Linear system is a system in which the degree of the variable in the equation is one. It may contain one, two, or more than two variables.

The levels of mercury in two different bodies of water are rising.

Let y be the year.

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. That is

0.05 + 0.1 y

In the second body of water, the initial measure is 0.12 ppb and the rate of increase is 0.06 ppb each year. That is

0.12 + 0.06y

The year in which both bodies of water have the same amount of mercury. Then the equation will be

0.05 + 0.1 y = 0.12 + 0.06y

More about the linear system link is given below.

https://brainly.com/question/20379472