Question 242727
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The amount of pure copper in 150 lbs. of Babbitt is:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 0.03\,\cdot\,150] lbs.


The amount of pure copper in *[tex \LARGE x] lbs. of pure copper is *[tex \LARGE x] lbs.


The sum of the two quantities above is the amount of pure copper in the 17% alloy which can be represented by:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 0.17(150\ +\ x)]


So we can say:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 0.03(150)\ +\ x\ =\ 0.17(150\ +\ x)]


Just solve for *[tex \LARGE x]



John
*[tex \LARGE e^{i\pi} + 1 = 0]
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