How much pure water must be mixed with 10 liters of a 25% acid solution to reduce it to a 10% acid solution?
11l
151
25l

Suppose you add x liters of pure water to the 10 L of 25% acid solution. The new solution's volume is x + 10 L. Each L of pure water contributes no acid, while the starting solution contains 2.5 L of acid. So in the new solution, you end up with a concentration of (2.5 L)/(x + 10 L), and you want this concentration to be 10%. So we have

and so you would need to add 15 L of pure water to get the desired concentration of acid.

see explanation

step-by-step explanation:

since the sequence is geometric there is a common ratio r between consecutive terms, that is

r = 4 ÷ - 2 = - 2

r = - 8 ÷ 4 = - 2

r = 16 ÷ 8 = - 2

the next term - 13 must be incorrect as it should be - 32 for the sequence to be geometric.

the first 4 terms are therefore geometric with common ratio r = - 2

c.

step-by-step explanation:

she should of added 9^2 and 12^2. that is because it is a^2+b^2=c^2. c^2 is the hypotenuse and should always be bigger than the other two numbers. which is why she should of added 9 squared and 12 squared together instead of 15 squared.