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

Step-by-step explanation:

Here, we want to find y which is the heat in which the amount of mercury in each of the water bodies is same.

Now for the first water body:

Initial is 0.05 ppb and after y years, we have a rise of 0.1 * y = 0.1y ppb

So the amount of mercury in ppb after y years would be; 0.05 + 0.1y

For the second water body;

Initial is 0.12 and a rise of 0.06 ppb per year for y years. The rise would be 0.06 * y = 0.06y

So total amount of mercury here is 0.12 + 0.06y

So to find y which is the year the amount of mercury in each water body is same, we simple equate both and that would be;

0.05 + 0.1y = 0.12 + 0.06y

option c) graph of line going through 1, 4 and 3, 10

step-by-step explanation:

let y equal to margo's coin collection

let x equal to number of years

at the begining x=0, y=1 → point=(x,y)→point=(0,1)

she adds 3 coins per year, then y=1+3x

at x=1, y=1+3(1)=1+3→y=4→point=(x,y)→point=(1,4)

at x=3, y=1+3(3)=1+9→y=10→point=(x,y)→point=(3,10)

the graph of line going through (1, 4) and (3, 10), then the option c is the correct.