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anonymous

  • one year ago

please help, question in comments

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  1. anonymous
    • one year ago
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    |dw:1436942415424:dw|

  2. anonymous
    • one year ago
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    I'm dealing with rational exponents, and it's asking me to simplify this

  3. anonymous
    • one year ago
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    |dw:1436942481286:dw|

  4. anonymous
    • one year ago
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    that simple? the answer is 1?

  5. anonymous
    • one year ago
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    can you explain how you did it so I can try myself?

  6. anonymous
    • one year ago
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    |dw:1436942663435:dw|

  7. rishavraj
    • one year ago
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    @Kaelyn78 see \[a^m \times a^n = a^{m + n} ~~~ and ~~~~(a^m)^{n} = a^{mn} \]

  8. rishavraj
    • one year ago
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    so \[a^8 \times a^6 = a^{14} ~~~and (a^{14})^{\frac{ 1 }{ 7 }} = a^{\frac{ 14 }{ 7 }}\]

  9. anonymous
    • one year ago
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    so the equation simplified is \[a \frac{ 14 }{ 7 }\]

  10. rishavraj
    • one year ago
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    the equation u get is \[\frac{ a^{\frac{ 14 }{ 7 }} }{ a^2 } = \frac{ a^2 }{ a^2 } = 1\]

  11. anonymous
    • one year ago
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    thanks for your help

  12. rishavraj
    • one year ago
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    u r most welcome :))

  13. UsukiDoll
    • one year ago
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    \[\large \frac{ a^{\frac{ 14 }{ 7 }} }{ a^2 } = \frac{ a^2 }{ a^2 } \rightarrow a^{2-2} \rightarrow a^0 = 1\] a step was skipped due to the zero exponent, our result will always be 1

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