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dimulka [17.4K]
3 years ago
7

A chemical company makes two brands of antifreeze. The first brand is 55 % pure antifreeze, and the second brand is 80 % pure an

tifreeze. In order to obtain 90 gallons of a mixture that contains 75 % pure antifreeze, how many gallons of each brand of antifreeze must be used?
Mathematics
1 answer:
Likurg_2 [28]3 years ago
3 0

We have 55% and 80% antifreeze.  We need 90 gallons of 75%.  

F - 55% and E - 80%

A) F + E = 90

B) .55 F + .80 E = 67.50 (which is .75 * 90)  Multiplying A) by -.55

A) -.55 F -.55 E = -49.50 (which is -.55 * 90) then adding A) & B)

.25 E = 18

E = 72 gallons of 80% and

F = 18 gallons of 55%

********* DOUBLE CHECK ******************

72 * .80 = 57.60  and 18 * .55 = 9.90

57.60 + 9.90 = 67.50 gallons of antifreeze and we have 90 total gallons

(67.50 / 90) * 100 = 75% correct!

Source: 1728.com/mixture.htm


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Answer:

Step-by-step explanation:

Let's look at some other ratios that equal 8:6  We could increase them or decrease them.  For instance if we cut one in half we cut the other one in half, so 8:6 would be the same ratio as 4:3.  With the same thinking if we double one we double the other so 8:6 s the same ratio as  16:12.  Now, we want to go from 8 to 28.  What do we multiply 8 by to get to it?  Or, it might be easier to use 4:3 and see what you multiply 4 by.  Then you multiply 3 by that same number to get the other number.

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Use Lagrange multipliers to find the dimensions of the box with volume 1728 cm3 that has minimal surface area. (Enter the dimens
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Answer:

(x,y,z) = (12,12,12) cm

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The box is assumed to be a closed box.

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(∂L/∂y) = 2x + 2z - λxz = 0

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(2/z) = (2/y)

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Putting these into the constraint equation or the solution of the fourth partial derivative,

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