<PRE><B>Question 597182</B>
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You have one angle of 140 degrees and (n - 1) angles of 100 degrees each.  The sum of the interior angles of any n-gon is given by 180(n - 2)


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 180(n\ -\ 2)\ =\ 140\ +\ 100(n\ -\ 1)]


Solve for *[tex \Large n]


John
*[tex \LARGE e^{i\pi} + 1 = 0]
My calculator said it, I believe it, that settles it
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