pharmacist has an 18% alcohol solution and a 40% alcohol solution. How much of each should he mix together to make 10 liters of a 20% alcohol solution?

9.09 Liters of 18% solution

.91 Liters of 40% Solution

Step-by-step explanation:

You have to solve a system of equations for this problem. But first, let's figure out what we know.

We know we end up with 10L of total solution, at 20% concentration.

So, we know we have 10 * 0.2 = 2L of total alcohol in the solution.

Consider the 18% solution "x" and the 40% solution "y." Express the percentages as decimals, so .18 and .4 respectively. Putting this all together,

we create the first equation:

.18x + .4y = 2

We also know that the amount of 18% solution, or x, combined with the 40% solution, y, must equal 10L. So our second equation is

x + y = 10

We'll solve the second one first, for x. Rearranging the equation gives you

x = 10 - y

Now, put this value for x into the first equation, in place of the current x term. Doing so gives you

.18(10-y) + .4y = 2

Distributing the ".18" gives you

1.8 -.18y + .4y = 2, which is the same as

0.22y = 0.2 when you combine the like terms.

This means, y = .2/.22 = 0.91L of 40% solution

Inserting our new y-value into the second equation gives us

x + 0.91 = 10, or x = 10 - 0.91 = 9.09L of 18% solution

step-by-step explanation:

x^2 -12x + 36 + y^2 = 36

x^2 + y^2 = 12x

r^2 = 12rcos(theta)

r = 12 cos(theta)

the answer would be 484

step-by-step explanation:

just multiply 0.484 times 1,000 (formula)