I NEED THIS ANSWER BY TONIGHT How many gallons of a 90% antifreeze solution must be mixed with 100 gallons of 20% antifreeze to get a mixture that is 80% antifreeze? Use the six-step method.

Solution : 600 gallons

Step-by-step explanation:

Let's say that x is the gallons of 90% antifreeze solution. Let's start by converting each percent to decimal form.

90% = 0.9

20% = 0.2

80% = 0.8

Knowing that we have the following set up,

0.9(x) + 0.2(100) = 0.8(x + 100) ⇒ 0.8 is associated with " x + 100 " as it is a mixture of the gallons of the 90% antifreeze solution, and the gallons of the 20% antifreeze solution

We can solve for x in this equation, and receive our solution. This is represented by the following attachment. It proves that there are 600 gallons of the 90% antifreeze solution.

the answer is b: 18

step-by-step explanation:

use the pythagorean theorem to .

a^2+b^2=c^2

plug 18 for "r"

the 2 legs are 18 and 24

18^2+24^2=c^2

324+576=c^2

900=c^2

c=30

30-12=18

18 is the radius

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