Question 1027759
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Rule 1:  *[tex \LARGE \ \ \ \ \ \ \ \ \ \ a^{-n}\ =\ \frac{1}{a^n}]


Rule 2:  *[tex \LARGE \ \ \ \ \ \ \ \ \ \ a^{\frac{m}{n}}\ =\ \sqrt[n]{a^m}\ =\ \left(\sqrt{a}\right)^m]


Combining Rule 1 and Rule 2:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \left(1\ +\ 3x\right)^{-\frac{1}{2}}\ =\ \frac{1}{\sqrt{1\ +\ 3x}}]


Rationalizing the denominator:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{\sqrt{1\ +\ 3x}}{1\ +\ 3x}]


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

*[tex \Large \ \
*[tex \LARGE \ \ \ \ \ \ \ \ \ \  

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