kaciebensberger13
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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 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




(answer needed asap)

Respuesta :

MrRoyal

Answer:

[tex]0.05 + 0.1y = 0.12 - 0.06y[/tex]

Step-by-step explanation:

Given

(A)

[tex]Initial = 0.05[/tex]

[tex]Rate = 0.1[/tex]

(B)

[tex]Initial = 0.12[/tex]

[tex]Rate = -0.06[/tex] -- it is negative, because it decreases

Required

When both will be equal

The current amount is calculated using:

[tex]Amount = Initial + Rate * y[/tex]

Where:

[tex]y \to years[/tex]

For A, we have:

[tex]Amount = 0.05 + 0.1 * y[/tex]

[tex]Amount = 0.05 + 0.1y[/tex]

For B, we have:

[tex]Amount = 0.12 - 0.06*y[/tex]

[tex]Amount = 0.12 - 0.06y[/tex]

Both are equal when:

[tex]0.05 + 0.1y = 0.12 - 0.06y[/tex]

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