Question

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.

Answer 1

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
0.05y + 0.1 = 0.12y + 0.06
0.05 + 0.1y = 0.12 + 0.06y
0.05y – 0.1 = 0.12y – 0.06

Answer 2

take note that it is rising. then addition is involved

Answer 3

alright, let see, like he said, its rising so it'll be addition, next..

Answer 4

the one that's constant with no variables are those initial values 0.05 and 0.12

Answer 5

since the values of both bodies are increasing at rates 0.1 and 0.06 each year then the variables should be in that respective rates as to indicate the amount for a certain period of time

Answer 6

C

Answer 7

what do u think @shinalcantara

Answer 8

there's no c though..XD the third choice

Answer 9

c, third choice, same thing

Answer 10

afdafdffgadkagjg @justsmile531 did you get the idea?

Answer 11

chill i'm just kidding @alex1248191

Answer 12

Yeah , lol thanks