Question 467328
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Let *[tex \Large x] represent the amount drained (and the amount of 100% anti-freeze).  Then the amount of anti-freeze that remains after draining *[tex \Large x] gallons is *[tex \Large 40\ -\ x].  20% of that which remains is antifreeze, to which you want to add *[tex \Large x] gallons of 100% antifreeze which is, by definition, *[tex \Large x] gallons of antifreeze.  The two add up to 40 gallons of 40% antifreeze which is 16 gallons of antifreeze, so:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 0.2(40\ -\ x)\ +\ x\ =\ 0.4(40)]


Solve for *[tex \Large x]


John
*[tex \LARGE e^{i\pi} + 1 = 0]
My calculator said it, I believe it, that settles it
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