A chemical company makes two brands of antifreeze. The first brand is 40%pure antifreeze, and the second brand is 90%pure antifreeze. In order to obtain 160

Complete question:

A chemical company makes two brands of antifreeze. The first brand is 40% pure antifreeze, and the second brand is 90% pure antifreeze. In order to obtain 160 gallons of a mixture that contains 60% pure antifreeze, how many gallons of each brand of antifreeze must be used?

Amount of 40% pure antifreeze = y = 96gallons

Amount of 90% pure antifreeze = ((160-y) = (160 - 96) = 64gallons

Step-by-step explanation:

Brand A = 40% pure

Brand B = 90% pure

Therefore to obtain 160 gallons that contains 60% pure antibfreeze:

Let y = Amount of 40% pure antifreeze required

Therefore (160 - y) = Amount of 90% pure antifreeze required.

Expressing mathematically,

0.4y + (160 - y) (0.9) = 160×0.6

0.4y + 144 - 0.9y = 96

-0.5y = 96 - 144

-0.5y = - 48

y = 48 / 0.5

y = 96

Amount of 40% pure antifreeze = y = 96gallons

Amount of 90% pure antifreeze = ((160-y) = (160 - 96) = 64gallons

use an equation of ratios to determine the amount of gas needed to fill the tank.

    0.2             0.8

  =    

288.75             x

                                                              231 gal

so, 0.2x = (0.8)(288.75 gal), and x = = 1155 gal.

                                                                0.2

it would take 1155 gal. to fill the tank.

the cost would be

1155 gal         $3

* = $3465   (answer)

      1           1 gal

hope this you

step-by-step explanation:

given perimeter: 28cm

length: 9cm

atp,

2(l+w)=p of the reactangle

2(9+x)=28

18+2x=28

2x=28-18

x=10/2

x=5

therefore width is 5 cm