Question 613678
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You don't know how badly I wanted to answer the EXACT question you asked by simply saying "yes".  BTW, there is no such thing as a "reciprorcal"  The word is "reciprocal".


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 2(x\ -\ 3)\ >\ \frac{1}{5}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 2x\ -\ 6\ >\ \frac{1}{5}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 2x\ >\ \frac{31}{5}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ >\ \frac{31}{10}]


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