A chemical company mixes pure water with their premium antifreeze solution to create an inexpensive antifreeze mixture. The premium antifreeze solution contains 95% pure antifreeze. The company wants to obtain 285 gallons of a mixture that contains 55% pure antifreeze. How many gallons of water and how many gallons of the premium antifreeze solution must be mixed?

Step-by-step explanation:

Let 'a' represent the amount of 95% antifreeze in the mix. Then the amount of water is (285 -a). The mix has this amount of antifreeze in it:

  0.95a +0(285 -a) = 0.55(285)

  a = 0.55/0.95(285) = 165 . . . gallons of 95%

  285 -a = 285 -165 = 120 . . . . gallons of water

There are 120 gallons of water and 165 gallons of premium antifreeze in the mixture.

.5 lb

step-by-step explanation:

12 in / 6 in =2 in

.25 x 2 = .5 lb

the answer is (4,3)

since the rule for r0, 90º is (-y,x) and it was already applied to form the image, all you have to do is reverse the answer using this formula.