A chemical company mixes pure water with their premium antifreeze solution to create an inexpensive antifreeze mixture. The premium antifreeze solution contains
80
%

pure antifreeze. The company wants to obtain
320

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?

let w = amt of water required

:

An equation based on the the percent water, an 80% solution is 20% water.

.20(320-w) + w = .50(320)

64 - .2w + w = 160

.8w = 160 - 64

w = 96%2F.8

w = 120 gal of water to produce 320 gal of 50% antifreeze

:

320 - 120 = 200 gal of 80% solution

Water: _120___ gallons

Premium Antifreeze: __200___ gallons

Step-by-step explanation:

7! = 7×6×5×4×3×2

3! = 3× 2

as 7! is multiple of 3! because 7! contains 3! that is 3× 2 as factor

so 7! /(3! ) = 7× 6× 5× 4

= 42 × 20

= 840

its b) 840✌✌✌

dhruv⭐

step-by-step explanation:

the true ones are a b and d