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Help Solving Exponents? ~Confused To The Max~ 1. m^2n^0 2. -3f^3g^-1
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\(n^0=1\) so the first one is \(m^2\)
\[-3f^3g^{-1}=-\frac{3f^3}{g}\] not sure what else they want you to do
@satellite73 I think the first one is m^2n^0. If you take 2n^0 this is equivalent to 1, so you would get m^1.
no
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I was writing how to do the second one and I just realized sat already did >_>
\[x ^{-n}=\frac{ 1 }{ x^n }\]
Thanks for the help ya'll.
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