How do you simplify ${(\frac{3 f^{4} g h^{4}}{32 f^{3} g^{4} h})}^{0}$ $((3f^4gh^4)/(32f^3g^4h))^0$ ?

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How do you simplify ${(\frac{3 f^{4} g h^{4}}{32 f^{3} g^{4} h})}^{0}$ $((3f^4gh^4)/(32f^3g^4h))^0$ ?

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Answer 1 · 2016-12-24T09:29:17

$1$ $1$

Explanation:

Using the $law of exponents$ $color(blue)"law of exponents"$

$color(orange)"Reminder " color(red)(bar(ul(|color(white)(2/2)color(black)(a^0=1)color(white)(2/2)|)))$

The 'contents' of the bracket are all raised to the power of 0

$rArr((3f^4gh^4)/(32f^3g^4h))^0=1$

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How do you simplify ${(\frac{3 f^{4} g h^{4}}{32 f^{3} g^{4} h})}^{0}$ $((3f^4gh^4)/(32f^3g^4h))^0$ ?

1 Answer

Explanation:

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How do you simplify ((3f^4gh^4)/(32f^3g^4h))^0(3f4gh432f3g4h)0((3f^4gh^4)/(32f^3g^4h))^0?

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Explanation:

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How do you simplify ${(\frac{3 f^{4} g h^{4}}{32 f^{3} g^{4} h})}^{0}$ $((3f^4gh^4)/(32f^3g^4h))^0$ ?