Answer:
[tex]\frac{y^2}{5}[/tex]
Step-by-step explanation:
[tex](5x^0y^{-2})^{-1}[/tex]
All quantity raised to zero is equal to 1. Therefore x⁰ = 1
[tex](5(1)y^{-2})^{-1}[/tex]
Apply the rule [tex]a^{-b} = \frac{1}{a^b}[/tex] to solve [tex]y^{-2}[/tex]
[tex](5)(\frac{1}{y^2} )^{-1}[/tex]
[tex](\frac{5}{y^2} )^{-1}[/tex]
Apply the rule [tex]a^{-b} = \frac{1}{a^b}[/tex] again
[tex](\frac{5}{y^2} )^{-1} = \frac{1}{(\frac{5}{y^2})^1 }[/tex]
[tex](\frac{5}{y^2} )^{-1} = \frac{y^2}{5}[/tex]
#CarryOnLearning
