Evaluate. [9/4]^-1/2 I'm not sure how to solve this one because its' a fraction.

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Evaluate. [9/4]^-1/2 I'm not sure how to solve this one because its' a fraction.

Mathematics
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\[\LARGE (\frac{9}{4})^{-\frac{1}{2}}\] is this your question? can you see the latex?
43. B
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\[\huge \left( \frac{ a }{ b } \right)^{-m} \implies \left( \frac{ b }{ a } \right)^m\] use this

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So you should be able to evaluate, another thing \[\huge \left( \frac{ b }{ a } \right)^m = \frac{ b^m }{ a^m }\]
4^1/2 / 9^1/2
I would've just use the negative exponent rule \[\LARGE (\frac{9}{4})^{-\frac{1}{2}} \] \[\LARGE \frac{1}{(\frac{9}{4})^{\frac{1}{2}}}\] then change the bottom to radical form
\[\LARGE \frac{1}{\sqrt{\frac{9}{4}}}\]
what is the square root of 9 what is the square root of 4
3 and 2
But what do I do with those
Ok so you have \[\left( \frac{ 3 }{ 2 } \right)^{-1} \implies \left( \frac{ 2 }{ 3 } \right)^1 = \frac{ 2 }{ 3 }\] as shown by the property above
Or if you are doing it usuki's way \[\huge \frac{ 1 }{ \frac{ a }{ b } } \implies \frac{ b }{ a }\] which is the same as above haha.
Thank you both so much, can y'all medal each other? I can only medal one person but I'm grateful for both of your help.
thanks @Astrophysics for finishing this problem while I was away.
Np, and don't worry about medals I always got Usuki's back on those xD

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