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TeemoTheTerific

  • one year ago

evaluate (a^4)^3 when a = (2r3)^1r6 Ill rewrite it neater

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  1. TeemoTheTerific
    • one year ago
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    \[\huge(a ^{4})^{3} \]

  2. TeemoTheTerific
    • one year ago
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    \[ \huge a= (\frac{ 2 }{ 3 })^{\frac{ 1 }{ 6 }}\]

  3. TeemoTheTerific
    • one year ago
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    evaluate

  4. TeemoTheTerific
    • one year ago
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    i just sub a equal whatever in a^12 but i can never get it right

  5. jim_thompson5910
    • one year ago
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    (a^4)^3 = a^(4*3) (a^4)^3 = a^(12) So if you want to find (a^4)^3, then you can find a^(12) instead since they are the same

  6. jim_thompson5910
    • one year ago
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    \[ \Large a= (\frac{ 2 }{ 3 })^{\frac{ 1 }{ 6 }}\] \[ \Large a^{12}\] \[ \Large \left((\frac{ 2 }{ 3 })^{\frac{ 1 }{ 6 }}\right)^{12}\] \[ \Large (\frac{ 2 }{ 3 })^{\frac{ 1 }{ 6 } * 12}\] \[ \Large (\frac{ 2 }{ 3 })^{\frac{ 12 }{ 6 }}\] \[ \Large (\frac{ 2 }{ 3 })^{2}\] I'll let you finish

  7. TeemoTheTerific
    • one year ago
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    so the final answer is 4r9?

  8. TeemoTheTerific
    • one year ago
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    wow thanks so much anyways

  9. jim_thompson5910
    • one year ago
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    you mean 4/9 right?

  10. jim_thompson5910
    • one year ago
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    not sure how/where the r comes into play

  11. jim_thompson5910
    • one year ago
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    btw \[\Large \frac{4}{9} = 4/9\]

  12. TeemoTheTerific
    • one year ago
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    oh lol in my calc to write something over something its something r something :P

  13. jim_thompson5910
    • one year ago
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    oh interesting, never seen or heard of that before

  14. TeemoTheTerific
    • one year ago
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    actaully some question written in australia are written 8r9 which means 8/9

  15. jim_thompson5910
    • one year ago
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    i gotcha, that makes more sense now

  16. jim_thompson5910
    • one year ago
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    never ever seen that notation before

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