3root9x-1 =2

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Dont understand.. do you mean \[\sqrt[3]{9x-1} =2\]
yes, i just don't know how you enter it that way
Equation, ... if you look down..

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Other answers:

It become \[9x-1=2^{3}\]
I see that
\[2^{3}= 8 .\] \[8=9x-1\]
Solve X
ok, so x=1?
yes
ok, that was easy!!
yepp.. :)
Do you know how to rewrite using radical notation?
i can give it a try
I have (16 little 3/2)little 2/6 hard to write the little numbers!!
aha..
if you mean \[(3\div2)^{16}\]
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
aha..
Would you like me to caluc it?
the directions say to write using radical notation
You just mutilpy, 3\2 with 2\6
and then you get a notation 1\2 over 16 = 4
if their is a notation over another notation you can mutiply with each other..
ok, so it will be 2root4 then??
1\2 root 4
3\2 x 2\6 = 6\12 rewrite to 1\2 by dividing with 6 on both side
Ok, I think I got it!! Thanks :)
Your welcome!

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