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

Step-by-step explanation:

According to the problem, both expression are equal because they have the same amount of mercury.

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.

6mm^3 or .6cm^3

      216

      /   \

  36     6

  /   \     /

6     6   6

factor trees are nice.

it would not be increasing by nothing that i can see probably by 25 percent that is all i can sat\y for now if i have any thing else to say i will edit this.

step-by-step explanation: