How do you simplify $\frac{{(- 3)}^{0} x^{4} y^{- 1}}{x^{- 1} y z^{- 2}}$ $((-3)^0 x^4 y^-1) /( x^-1 yz ^-2)$ ?

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How do you simplify $\frac{{(- 3)}^{0} x^{4} y^{- 1}}{x^{- 1} y z^{- 2}}$ $((-3)^0 x^4 y^-1) /( x^-1 yz ^-2)$ ?

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Answer 1 · 2017-01-26T22:41:11

$\frac{{(- 3)}^{0} x^{4} y^{- 1}}{x^{- 1} y z^{- 2}} = \frac{x^{5} z^{2}}{y^{2}}$ $((-3)^0 x^4 y^-1) /( x^-1 yz ^-2)=(x^5z^2)/y^2$

Explanation:

Some rules which apply to indices:

$a^{0} \equiv 1,$ $a^0-=1,$ $given$ $"given"$ $a \neq 0$ $a!=0$

$\frac{a^{m}}{a^{n}} \equiv a^{m - n},$ $a^m/a^n-=a^(m-n),$ $given$ $"given"$ $m - n > 0$ $m-n>0$ $where$ $"where"$ $a = 0$ $a=0$

$a^{- n} \equiv \frac{1}{a^{n}},$ $a^(-n)-=1/(a^n),$ $given$ $"given"$ $a \neq 0$ $a!=0$

$- 3^{0} = 1$ $-3^0=1$

$\frac{x^{4}}{x^{- 1}} = x^{4 - - 1} = x^{5}$ $x^4/x^-1=x^(4--1)=x^5$

$\frac{y^{- 1}}{y} = y^{- 1 - 1} = y^{- 2}$ $y^-1/y=y^(-1-1)=y^-2$

$\frac{1 (z^{0})}{z^{- 2}} = z^{0 - - 2} = z^{2}$ $(1(z^0))/z^-2=z^(0--2)=z^2$

$\frac{{(- 3)}^{0} x^{4} y^{- 1}}{x^{- 1} y z^{- 2}} = x^{5} y^{- 2} z^{2} = x^{5} \frac{1}{y^{2}} z^{2} = \frac{x^{5} z^{2}}{y^{2}}$ $((-3)^0 x^4 y^-1) /( x^-1 yz ^-2)=x^5y^-2z^2=x^5 1/y^2 z^2=(x^5z^2)/y^2$

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How do you simplify $\frac{{(- 3)}^{0} x^{4} y^{- 1}}{x^{- 1} y z^{- 2}}$ $((-3)^0 x^4 y^-1) /( x^-1 yz ^-2)$ ?

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Explanation:

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How do you simplify $\frac{{(- 3)}^{0} x^{4} y^{- 1}}{x^{- 1} y z^{- 2}}$ $((-3)^0 x^4 y^-1) /( x^-1 yz ^-2)$ ?