highflylex6511 highflylex6511
  • 11-10-2022
  • Mathematics
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moneyromo11734 moneyromo11734
  • 11-10-2022
[tex]27)\frac{4x^0y^{-2}z^3^{}^{}}{4x}=x^{0-1}y^{-2}z^3=x^{-1}y^{-2}z^3=\frac{z^3}{xy^2}[/tex][tex]undefined[/tex]

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