Which equation can be used to find y, the year in which
both bodies of water have the same amount of mercury?
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.
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

When the problem specifies an initial value, this means that value is not with the variable, because is an initial constant. When the problem specifies a rate, this is a coefficient of the variable, which expresses time (years.)

So, 0.05 parts per billion with a rate of 0.1 means: 0.05 + 0.1y, because the ratio is rising, that means is positive.

Similarly, 0.12 parts per billion with a rate of 0.06 means: 0.12 + 0.06y, because the ratio is also increasing, is positive.

Therefore, the right answer is C.

GoddessUltimaWolf GoddessUltimaWolf · Answer 2 · 2020-11-12T15:41:02+01:00

GoddessUltimaWolf GoddessUltimaWolf

12-11-2020

Answer:

C.) 0.05 + 0.1 y = 0.12 + 0.06 y

Body 1:

Initial = 0.05

Rate =0.1

Increase = 0.05+ 0.1y

Body 2 :

Initial = 0.12

Rate =0.06

Increase = 0.12 +0.06 y

Step by Step Explanation:

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