Question 551175
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A fraction to the negative 1 power is just the reciprocal, so:


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


Then move negative exponents from numerator to denominator and from denominator to numerator:


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



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