I need help with this question:
2x^3 y^-2
8x^-2 y^-3
What its asking me to do is to simplify the question. Can anyone help me??
$\displaystyle \frac{2x^3y^{-2}}{8x^{-2}y^{-3}} $
$\displaystyle \left(\frac{2x^3}{8x^{-2}}\right)\left(\frac{y^{-2}}{y^{-3}}\right) $
you should know
$\displaystyle y^{-n} = \frac{1}{y^n} $
moving the number from the denominator to nominator change the exponent sign, so
$\displaystyle \left(\frac{2x^3}{8x^{-2}}\right) = \frac{2x^3 . x^2}{8}= \frac{2x^{3+2}}{8}= \frac{2x^5}{8} $
$\displaystyle \left(\frac{y^{-2}}{y^{-3}}\right) = y^{-2}. y^{3}= y^{-2+3} = y $
$\displaystyle \frac{2x^3y^{-2}}{8x^{-2}y^{-3}} = \frac{2x^5y}{8}=\frac{x^5y}{4}$