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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 [tex]\( y \)[/tex], the year in which both bodies of water have the same amount of mercury?

A. [tex]\( 0.05 - 0.1y = 0.12 - 0.06y \)[/tex]

B. [tex]\( 0.05y + 0.1 = 0.12y + 0.06 \)[/tex]

C. [tex]\( 0.05 + 0.1y = 0.12 + 0.06y \)[/tex]

D. [tex]\( 0.05y - 0.1 = 0.12y - 0.06 \)[/tex]



Answer :

GinnyAnswer
To determine the year [tex]\( y \)[/tex] when the levels of mercury in both bodies of water will be the same, we must set up an equation representing the amount of mercury in each body of water over time and then equate them.

Let's break it down step by step:

1. Initial Measurement and Rate of Increase:
- For the first body of water:
- Initial measure: [tex]\( 0.05 \)[/tex] ppb
- Rate of increase: [tex]\( 0.1 \)[/tex] ppb per year
- For the second body of water:
- Initial measure: [tex]\( 0.12 \)[/tex] ppb
- Rate of increase: [tex]\( 0.06 \)[/tex] ppb per year

2. Formulating Expressions:
- The amount of mercury in the first body of water after [tex]\( y \)[/tex] years can be expressed as:
[tex]\[ \text{Mercury in first body} = 0.05 + 0.1y \][/tex]
- The amount of mercury in the second body of water after [tex]\( y \)[/tex] years can be expressed as:
[tex]\[ \text{Mercury in second body} = 0.12 + 0.06y \][/tex]

3. Setting up the Equation:
- To find the year [tex]\( y \)[/tex] when both bodies of water have the same amount of mercury, we set the two expressions equal to each other:
[tex]\[ 0.05 + 0.1y = 0.12 + 0.06y \][/tex]

Therefore, the correct equation to use is:
[tex]\[ 0.05 + 0.1y = 0.12 + 0.06y \][/tex]

This equation accurately represents the year in which the mercury levels in both bodies of water will be equal.

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