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I'm blanking on how to do this, I learned it so long ago, any help would be greatly appreciated. More interested on how to do it than the answer.

Im blanking on how to do this I learned it so long ago any help would be greatly appreciated More interested on how to do it than the answer class=

Respuesta :

VirαtKσhli

Answer:

[tex] \dfrac{16 y^{22}}{x^{10}z^{10}} [/tex]

Step-by-step explanation:

Given expression is ,

[tex]\sf\longrightarrow \bigg(\dfrac{2x^3y^{-5}z^8}{8x^{-2}y^6z^3}\bigg)^{-2} [/tex]

This would be simplified using the law of exponents , some of which I will use here are ,

  • [tex] (an)^m = a^m n^m [/tex]
  • [tex] \dfrac{a^m}{a^n}=a^{m-n}[/tex]

  • [tex] a^m a^n = a^{m+n}[/tex]
  • [tex] a^{-n} = \dfrac{1}{a^n}[/tex]

Using the above laws ,

[tex]\sf \longrightarrow \bigg[ \dfrac{2}{8} \bigg(\dfrac{x^3}{x^{-2}}\bigg)\bigg(\dfrac{y^{-5}}{y^6}\bigg)\bigg(\dfrac{z^8}{z^3}\bigg) \bigg]^{-2}[/tex]

Using the second law mentioned above , we have,

[tex]\sf \longrightarrow \bigg[ \dfrac{1}{4}(x^{3+2})(y^{-5-6})(z^{8-3})\bigg]^{-2} [/tex]

Simplify ,

[tex]\sf \longrightarrow \bigg[\dfrac{1}{4} x^5y^{-11}z^5\bigg]^{-2} [/tex]

Using the first law mentioned above , we have,

[tex]\sf \longrightarrow \bigg[ \dfrac{1}{4^{-2}} x^{5(-2)} y^{-11(-2)} z^{5(-2)}\bigg] [/tex]

Simplify,

[tex]\sf \longrightarrow 4^2x^{-10}y^{22} z^{-10}[/tex]

Finally using the fourth law mentioned above , we have ,

[tex]\sf \longrightarrow \boxed{\bf \dfrac{16 y^{22}}{x^{10}z^{10}}} [/tex]

Option K is the correct answer.

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