Question

Evaluate. fourth root of 9 multiplied by square root of 9 over the fourth root of 9 to the power of 5

5 to the power of negative 1 over 6
5 to the power of 3 over 2
9
92

Answer 1

@studygurl14

Answer 2

Im so lost

Answer 3

This?
\(\huge\frac{\sqrt[4]{9}\sqrt{9}}{(\sqrt[4]{9})^5}\)

Answer 4

Yes

Answer 5

\(\large\sqrt[4]{9}=9^{\frac{1}{4}}\)
And...
\(\large(\sqrt[4]{9})^5=9^{\frac{5}{4}}\)

Answer 6

does that help?

Answer 7

im still really confused now would i subtract the powers?

Answer 8

@StudyGurl14

Answer 9

oh, and \(\large\sqrt{9}=9^{\frac{1}{2}}\)

Answer 10

I think the answer is D

Answer 11

So you're left with...
\(\huge\frac{9^{\frac{1}{4}}9^{\frac{1}{2}}}{9^{\frac{5}{4}}}\)

Answer 12

nvm

Answer 13

simplfy that

Answer 14

How?

Answer 15

Do this:
\(\huge 9^{\frac{1}{4}+\frac{1}{2}-\frac{5}{4}}\)

Answer 16

so the base will stay 9 and in the answers the base is always 9 it just copied it wrong.

and it would be 9 -1/-2

Answer 17

as @StudyGurl14 explained 9^(1/4+1/2-5/4)=9^-2/4=9-1/2

Answer 18

It would be
\(\large 9^{-\frac{1}{2}}\)

Answer 19

thats not an answer choice tho

Answer 20

Oh yes it is thank you

Answer 21

:)

Answer 22

can i have a medal now @Helppme6

Answer 23

\[\frac{ 1 }{ \sqrt{9} }=\frac{ 1 }{ 3}\]