How do you simplify $\frac{f^{- 3} g^{2}}{h^{- 4}}$ $(f^-3g^2)/(h^-4)$ ?

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Answer 1 · 2018-07-15T06:53:59

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$\frac{f^{- 3} g^{2}}{h^{- 4}} = \frac{1}{f^{3}} \cdot h^{4} \cdot g^{2}$ $color(red)((f^-3g^2)/(h^-4)=1/(f^3) * h^4*g^2$

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We are given the expression:

$\frac{f^{- 3} g^{2}}{h^{- 4}}$ $color(red)((f^-3g^2)/(h^-4)$

Formula used:

$a^{- b} = \frac{1}{a^{b}}$ $color(blue)(a^(-b)=1/a^b$

$\frac{1}{\frac{1}{a}} = a$ $color(blue)(1/(1/a) = a$

Using this formula, we can write

$f^{- 3} = \frac{1}{f^{3}}$ $f^(-3) = 1/f^3$

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

Hence, we get

$\frac{f^{- 3} g^{2}}{h^{- 4}} = \frac{1}{f^{3}} \cdot h^{4} \cdot g^{2}$ $color(red)((f^-3g^2)/(h^-4)=1/(f^3) * h^4*g^2$

Hope it helps.

Answer 2 · 2018-07-15T07:27:15

nothing cancels down as they are three different variables

But $a^{- b} = \frac{1}{a^{b}}$ $a^-b=1/a^b$ so we can rewrite the expression with no negative powers

$\frac{f^{- 3} g^{2}}{h^{- 4}} = \frac{g^{2} h^{4}}{f^{3}}$ $(f^-3g^2)/(h^-4)=(g^2h^4)/f^3$

How do you simplify (f^-3g^2)/(h^-4)f−3g2h−4(f^-3g^2)/(h^-4)?