Question

express the following fraction in simplest form, only using positive exponents.
$\frac{20y^{2}h^{-1}}{(5y^{5}h^{2})^{2}}$

Explanation:

Step1: Expand denominator using exponent rule

$$(5y^5h^2)^2 = 5^2 \cdot (y^5)^2 \cdot (h^2)^2 = 25y^{10}h^4$$

Step2: Rewrite negative exponent in numerator

$$20y^2h^{-1} = \frac{20y^2}{h}$$

Step3: Rewrite fraction as multiplication

$$\frac{20y^2}{h} \cdot \frac{1}{25y^{10}h^4} = \frac{20y^2}{25y^{10}h^5}$$

Step4: Simplify coefficients and variables

Simplify $\frac{20}{25}=\frac{4}{5}$; use $y^{2-10}=y^{-8}=\frac{1}{y^8}$
$$\frac{4}{5y^8h^5}$$

Answer:

$\frac{4}{5y^8h^5}$