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anonymous

  • 5 years ago

3root9x-1 =2

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  1. anonymous
    • 5 years ago
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    Dont understand.. do you mean \[\sqrt[3]{9x-1} =2\]

  2. anonymous
    • 5 years ago
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    yes, i just don't know how you enter it that way

  3. anonymous
    • 5 years ago
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    Equation, ... if you look down..

  4. anonymous
    • 5 years ago
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    It become \[9x-1=2^{3}\]

  5. anonymous
    • 5 years ago
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    I see that

  6. anonymous
    • 5 years ago
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    \[2^{3}= 8 .\] \[8=9x-1\]

  7. anonymous
    • 5 years ago
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    Solve X

  8. anonymous
    • 5 years ago
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    ok, so x=1?

  9. anonymous
    • 5 years ago
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    yes

  10. anonymous
    • 5 years ago
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    ok, that was easy!!

  11. anonymous
    • 5 years ago
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    yepp.. :)

  12. anonymous
    • 5 years ago
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    Do you know how to rewrite using radical notation?

  13. anonymous
    • 5 years ago
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    i can give it a try

  14. anonymous
    • 5 years ago
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    I have (16 little 3/2)little 2/6 hard to write the little numbers!!

  15. anonymous
    • 5 years ago
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    aha..

  16. anonymous
    • 5 years ago
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    if you mean \[(3\div2)^{16}\]

  17. anonymous
    • 5 years ago
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    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

  18. anonymous
    • 5 years ago
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    aha..

  19. anonymous
    • 5 years ago
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    Would you like me to caluc it?

  20. anonymous
    • 5 years ago
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    the directions say to write using radical notation

  21. anonymous
    • 5 years ago
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    You just mutilpy, 3\2 with 2\6

  22. anonymous
    • 5 years ago
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    and then you get a notation 1\2 over 16 = 4

  23. anonymous
    • 5 years ago
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    if their is a notation over another notation you can mutiply with each other..

  24. anonymous
    • 5 years ago
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    ok, so it will be 2root4 then??

  25. anonymous
    • 5 years ago
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    1\2 root 4

  26. anonymous
    • 5 years ago
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    3\2 x 2\6 = 6\12 rewrite to 1\2 by dividing with 6 on both side

  27. anonymous
    • 5 years ago
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    Ok, I think I got it!! Thanks :)

  28. anonymous
    • 5 years ago
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    Your welcome!

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spraguer (Moderator)
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