How much water should be added to 1 gallon of pure antifreeze to obtain a solution that is 10% ​antifreeze? To obtain a 10% antifreeze​ solution, _ ​gallon(s) of water should be added.

Let x = amount of water to be added with 0% antifreeze

1 = amount of pure antifreeze at 100%

x+1 = amount of the mixture at 60% antifreeze

Step-by-step explanation:

1= .60x + .60

Subtract .60 from each side:

1-.60 = .60x

.40 = .60x

Divide both sides by .60

Check: Total antifreeze = 1 gallon in total mixture of 1 + 2/3 gallon = 5/3 gallons. 1 gallon divided by 5/3 equals 3/5 which is 60%. It checks.

square root of 36/6 = 2.449489743

square root of 16/6 = 1.632993162

square root of 6 = 2.449489743

the answer is square root of 16/6 because the square root of 36/6 is the same answer as square root of 6 = 2.449489743

40 x 50 = 2000

step-by-step explanation: