Question 513671
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A number:  *[tex \Large x]


Twice a number:  *[tex \Large 2x]


Half a number:  *[tex \Large \frac{x}{2}]


Thirteen less than twice a number:  *[tex \Large 2x\ -\ 13]


Seventeen more than half a number:  *[tex \Large \frac{x}{2}\ +\ 17]


Is: *[tex \Large =]


So:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 2x\ -\ 13\ =\ \frac{x}{2}\ +\ 17]


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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