Question

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

Answer 1

Given : 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.

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

Solution:

initial measure of mercury is 0.05 parts per billion (ppb) and is rising at a rate of 0.1 ppb each year.

So after y years  = 0.1y + 0.05

initial measure is 0.12 ppb and the rate of increase is 0.06 ppb each year.

So after y years  = 0.06y + 0.12

0.1y + 0.05  =  0.06y + 0.12

=> 0.05 + 0.1y = 0.12 + 0.06y

0.05 + 0.1y = 0.12 + 0.06y  

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Answer 2

Answer:

its C

0.05 + 0.1y = 0.12 + 0.06y  

Step-by-step explanation: