Question

below. write your answer using only positive exponents. $9x^0y^{-3}$ $\frac{9x}{y^3}$ $\frac{y^3}{9x}$ $\frac{y^3}{9}$ $\frac{9}{y^3}$ clear my selection

Explanation:

Step1: Simplify \( x^0 \)

Any non - zero number to the power of 0 is 1, so \( x^0 = 1\) (assuming \( x
eq0\)). Then the expression \( 9x^{0}y^{-3}\) becomes \( 9\times1\times y^{-3}=9y^{-3}\).

Step2: Convert negative exponent to positive

Recall the rule \( a^{-n}=\frac{1}{a^{n}}\) for \( a
eq0\) and \( n>0\). So \( y^{-3}=\frac{1}{y^{3}}\), and \( 9y^{-3}=9\times\frac{1}{y^{3}}=\frac{9}{y^{3}}\).

Answer:

\(\frac{9}{y^{3}}\) (corresponding to the option \(\frac{9}{y^{3}}\))