A chemical company makes two brands of antifreeze. The first brand is 30% pure antifreeze, and the second brand is 75% pure antifreeze. In order to obtain 30 gallons of a mixture that contains 60% pure antifreeze, how many gallons of each brand of antifreeze must be used?

Let x = amount of 45% antifreeze

Let y = amount of 70% antifreeze

    EQUATION 1:   x + y = 150    (total of 150 gallons mixed)

    EQUATION 2:  .45x + .75y = .55(x + y)

Simplify and solve the system of equations

    Multiply second equation by 100 on both sides to remove the decimals

          45x + 75y = 55(x + y)

    Combine like terms

          45x + 75y = 55x + 55y

          45x - 55x + 75y - 55y = 0

          -10x + 20y = 0

      Now we have the following system of equations:

             x  +    y = 150

        -10x + 20y =     0

    Multiply the first equation by -10 to get opposite coefficients for x;  add the equations to eliminate x

          10x + 10y = 1500

         -10x + 20y =       0

                    30y = 1500

     Solve for y

           30y = 1500

               y = 50

     Since the total mixed gallons is 150, x = 150 - 50 = 100

So we need 100 gallons of the 45% antifreeze and 50 gallons of the 70% antifreeze

0.6, 3/5, and 6% are the 3 same portion.

but, when they're all changed to a decimal they would equal to 0.6.

hope this , good luck!

0.02

step-by-step explanation:

time of 9 192 631 770 periods of the radiation corresponding to the transition between the 2 hyperfine levels of the ground state of the caesium 133 atom at 0 k (but other seconds are sometimes used in astronomy)