Question

3root9x-1 =2

Answer 1

Dont understand.. do you mean \[\sqrt[3]{9x-1} =2\]

Answer 2

yes, i just don't know how you enter it that way

Answer 3

Equation, ... if you look down..

Answer 4

It become \[9x-1=2^{3}\]

Answer 5

I see that

Answer 6

\[2^{3}= 8 .\]
\[8=9x-1\]

Answer 7

Solve X

Answer 8

ok, so x=1?

Answer 9

yes

Answer 10

ok, that was easy!!

Answer 11

yepp.. :)

Answer 12

Do you know how to rewrite using radical notation?

Answer 13

i can give it a try

Answer 14

I have (16 little 3/2)little 2/6 hard to write the little numbers!!

Answer 15

aha..

Answer 16

if you mean \[(3\div2)^{16}\]

Answer 17

no 16 is in parenthesis with a little 3/2 off to the right then there is a little 2/6 to the right of that

Answer 18

aha..

Answer 19

Would you like me to caluc it?

Answer 20

the directions say to write using radical notation

Answer 21

You just mutilpy, 3\2 with 2\6

Answer 22

and then you get a notation 1\2 over 16 = 4

Answer 23

if their is a notation over another notation you can mutiply with each other..

Answer 24

ok, so it will be 2root4 then??

Answer 25

1\2 root 4

Answer 26

3\2 x 2\6 = 6\12 rewrite to 1\2 by dividing with 6 on both side

Answer 27

Ok, I think I got it!! Thanks :)

Answer 28

Your welcome!