A chemical company makes two brands of antifreeze. The first brand is 55%

pure antifreeze, and the second brand is

80%

pure antifreeze. In order to obtain

130

gallons of a mixture that contains

70%

pure antifreeze, how many gallons of each brand of antifreeze must be used?

9514 1404 393

Step-by-step explanation:

Let x represent the quantity of 80% solution. Then the quantity of 55% solution is (130-x) and the total amount of antifreeze in the mix is ...

  0.55(130 -x) +0.80(x) = 0.70(130)

  0.25x +71.5 = 91 . . . simplify

  0.25x = 19.5 . . . . . . subtract 71.5

  x = 78 . . . . . . . . . . . divide by 0.25; amount of 80%

  130-78 = 52 . . . . amount of 55%

52 gallons of the 55% brand, and 78 gallons of the 80% brand must be used.

ad = ae + ed = y + 7

similar triangles therefor

ac/ab = ad/ae

6/4 = (y + 7) / y

cross multiply

6y = 4(y + 7)

6y = 4y + 28

2y = 28

y = 14

step-by-step explanation:

europa2433

europa2433

step-by-step explanation: