Question

Evaluate.
[9/4]^-1/2

I'm not sure how to solve this one because its' a fraction.

Answer 1

\[\LARGE (\frac{9}{4})^{-\frac{1}{2}}\] is this your question?
can you see the latex?

Answer 2

43. B

Answer 3

\[\huge \left( \frac{ a }{ b } \right)^{-m} \implies \left( \frac{ b }{ a } \right)^m\] use this

Answer 4

So you should be able to evaluate, another thing \[\huge \left( \frac{ b }{ a } \right)^m = \frac{ b^m }{ a^m }\]

Answer 5

4^1/2 / 9^1/2

Answer 6

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

Answer 7

\[\LARGE \frac{1}{\sqrt{\frac{9}{4}}}\]

Answer 8

what is the square root of 9
what is the square root of 4

Answer 9

3 and 2

Answer 10

But what do I do with those

Answer 11

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

Answer 12

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.

Answer 13

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.

Answer 14

thanks @Astrophysics for finishing this problem while I was away.

Answer 15

Np, and don't worry about medals I always got Usuki's back on those xD