Ask your own question, for FREE!
Mathematics 10 Online
b1az3:

The expression can best be rewritten as ______.

b1az3:

AZ:

\(\Large\sqrt[c]{a^b} = a^{\frac{b}{c}}\) keep in mind that a normal square root \(\sqrt{a}\) is the same thing as \(\Large\sqrt[2]{a}\)

b1az3:

so is's 3\/9

AZ:

that's not even an answer choice your question is basically, what is \(\Large \sqrt[2]{9^3}\) use what I just told you above put the numbers in the same order as I put the letters

b1az3:

9 2/3

AZ:

I have a really dumb question for you but I want to make sure before I put any more effort into this question that it isn't a wasted attempt Are you colorblind?

AZ:

I will assume the silence means no \( \LARGE\sqrt[\color{red}{c}]{\color{orange}{a}^\color{cyan}{b}} = \color{orange}{a}^{\frac{\color{cyan}{b}}{\color{red}{c}}}\)

AZ:

so what is \( \LARGE\sqrt[\color{red}{2}]{\color{orange}{9}^\color{cyan}{3}} =\)

b1az3:

27

AZ:

....

AZ:

use the colors, mate \(\Large 9^3 \) is not the same thing as 27

AZ:

ohh you solve it all out but still- that's not what your question is asking you to do

AZ:

\(\color{#0cbb34}{\text{Originally Posted by}}\) @AZ I will assume the silence means no \( \LARGE\sqrt[\color{red}{c}]{\color{orange}{a}^\color{cyan}{b}} = \color{orange}{a}^{\frac{\color{cyan}{b}}{\color{red}{c}}}\) \(\color{#0cbb34}{\text{End of Quote}}\) so what is \( \LARGE\sqrt[\color{red}{2}]{\color{orange}{9}^\color{cyan}{3}} =\)

supie:

as AZ said, what you have is \( \LARGE\sqrt[\color{red}{2}]{\color{orange}{9}^\color{cyan}{3}}\) and you have to use \( \LARGE\sqrt[\color{red}{c}]{\color{orange}{a}^\color{cyan}{b}} = \color{orange}{a}^{\frac{\color{cyan}{b}}{\color{red}{c}}}\) so basically, from you have to put the numbers from \( \LARGE\sqrt[\color{red}{2}]{\color{orange}{9}^\color{cyan}{3}}\) into \(a^{^{\frac{b}{c}}}\) (a=9, b=3, and c=2) I hope that makes sense ._.

Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!
Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!