How much water should be added to 6 gallons of pure acid in order to obtain the 24% acid solution?

19 gallons

Step-by-step explanation:

Let x gallons be the amount of water which should be added to 6 gallons of pure acid in order to obtain the 24% acid solution.

Amount of water = x gallons

Amount of pure acid = 6 gallons

Amount of acid solution = x + 6 gallons

Amount of acid in acid solution = 24%

So, there are gallons of acid in acid solution.

Hence,

Solve this equation. Multiply it by 100:

8(-10x+3.5y-7)

8×-10x + 8×3.5y + 8×-7

-80x + 28y - 56

therefore

-80x+28y-56 is the correct answer

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.