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Simplify (8y^)-2
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\[(8y)^{-2}\] is that what it looks like?
yes
okay well when you have negative exponents, to make the exponents positive, you put them in a denominator over 1, so: \[(8y)^{-2} = \frac{ 1 }{ (8y)^{2} }\]
now just square the bottom
so the answer would be 64y2?
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\[\frac{ 1 }{ (8y)^{2} } = \frac{ 1 }{ 8^{2}y^{2} }=\frac{ 1 }{ 64y^{2} }\]
so write the answer as 1 over 64y2?
thanks
yes, it is a fraction
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