Respuesta :
Answer:
Option A) [tex]\frac{1}{x^2y^2}[/tex] is correct.
Therefore the simplified given expression [tex]\frac{x^0y^{-3}}{x^2y^{-1}}=\frac{1}{x^2y^2}[/tex]
Step-by-step explanation:
Given expression is [tex]\frac{x^0y^{-3}}{x^2y^{-1}}[/tex]
To simplify the given expression:
[tex]\frac{x^0y^{-3}}{x^2y^{-1}}[/tex]
Above expression can be written as
[tex]\frac{x^0y^{-3}}{x^2y^{-1}}=\frac{(1)y^{-3}}{x^2y^{-1}}[/tex]
(since [tex]x^0=1[/tex] ,anything variable to the power "0' is 1)
[tex]\frac{x^0y^{-3}}{x^2y^{-1}}=\frac{y^{-3}}{x^2y^{-1}}[/tex]
[tex]\frac{x^0y^{-3}}{x^2y^{-1}}=\frac{1}{x^2y^{-1}y^3}[/tex] (since [tex]a^{-m}=\frac{1}{a^m}[/tex] )
[tex]\frac{x^0y^{-3}}{x^2y^{-1}}=\frac{1}{x^2y^{-1+3}}[/tex] (using the property [tex]a^m+a^n=a^{m+n}[/tex])
[tex]\frac{x^0y^{-3}}{x^2y^{-1}}=\frac{1}{x^2y^2}[/tex]
Therefore [tex]\frac{x^0y^{-3}}{x^2y^{-1}}=\frac{1}{x^2y^2}[/tex]
Therefore Option A) [tex]\frac{1}{x^2y^2}[/tex] is correct.
Therefore the simplified given expression [tex]\frac{x^0y^{-3}}{x^2y^{-1}}=\frac{1}{x^2y^2}[/tex]
