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

Complete question is;

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?

A) 0.05 – 0.1y = 0.12 – 0.06y

B) 0.05y + 0.1 = 0.12y + 0.06

C) 0.05 + 0.1y = 0.12 + 0.06y

D) 0.05y – 0.1 = 0.12y – 0.06

Option C: 0.05 + 0.1y = 0.12 + 0.06y

Step-by-step explanation:

In the first body, the initial measure of mercury is 0.05 parts per billion (ppb) while it's rising at a rate of 0.1 ppb each year. We are told to use y for the number of years.

Thus, amount of mercury for y years in this first body is;

A1 = 0.05 + 0.1y

Now, for the second body, we are told that;the initial measure is 0.12 ppb and the rate of increase is 0.06 ppb each year.

Thus, amount of mercury for y years in this second body is;

A2 = 0.12 + 0.06y

Since we want to find the year in which both bodies of water have the same amount of mercury. Thus, it means;

A1 = A2

Thus;

0.05 + 0.1y = 0.12 + 0.06y

This corresponds to option C

answer: huh

step-by-step explanation: