Question

Will give fan and medal! Simplify. square root of 9 over sixth root of 9

9 to the power of negative 2 over 3
9 to the power of negative 1 over 3
8 to the power of 3 over 10
9 to the power of 1 over 3

Answer 1

@campbell_st

Answer 2

@cool41
@rydertheepic @lovepurple

Answer 3

D

Answer 4

Can you tell me how?

Answer 5

You know what a square root is right?

Answer 6

yup

Answer 7

Well whats the square root of 9

Answer 8

3

Answer 9

Good now whats the square root of 9 over 6th root of 9

Answer 10

i have no idea

Answer 11

Well try to figure it out

Answer 12

Look back in the book if you have one

Answer 13

I dont have one its online

Answer 14

k12?

Answer 15

its gonna be 3/something

Answer 16

campbell how did you do it you showed me yesterday

Answer 17

Well look it up or Ill ask ab00t @abb0t GET IN HERE

Answer 18

well start by writing them using index notation

\[\sqrt[n]{x} = x ^{\frac{1}{n}}\]

so

\[\sqrt{9} = 9^{\frac{1}{2}}\]

and the other one is

\[\sqrt[6]{9} = 9^{\frac{1}{6}}\]

now the index law for division of the same base is subtract the powers

\[9^{\frac{1}{2}} \div 9^{\frac{1}{6}} = 9^{\frac{1}{2} - \frac{1}{6}}\]

and just finish it off for the answer

Answer 19

9 1/4?

Answer 20

well negative

Answer 21

But thats not an answer choice.