Answer
$\frac{1}{3x^{-4}} + \frac{0.1x^{-2}}{3} = \frac{x^4}{3} + \frac{0.1}{3x^2}$
Work Step by Step
By the formula given on page eight,
$\frac{1}{3x^{-4}} + \frac{0.1x^{-2}}{3} = \frac{1}{\frac{3}{x^4}} + \frac{0.1}{3x^2} = \frac{x^4}{3} + \frac{0.1}{3x^2}$.
