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anonymous

  • one year ago

Simplify 1 / (1+a^n) + 1/(1+a^-n)

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  1. freckles
    • one year ago
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    hint: multiply that second fraction by a^n/a^n

  2. anonymous
    • one year ago
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    Thanks but does it cancel out th denominater for the second fraction

  3. freckles
    • one year ago
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    \[1 \cdot a_n=? \\ (1+a^{-n}) \cdot a_n=1 \cdot a_n +a^{-n} \cdot a^{n} =?\]

  4. anonymous
    • one year ago
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    Wouldn't it be 1+a as the denimontar

  5. freckles
    • one year ago
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    those one n's were suppose to be exponents (not subscipts)

  6. freckles
    • one year ago
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    \[1 \cdot a^n=a^n \\ (1+a^{-n}) \cdot a^n=1 \cdot a^n+a^{-n} a^{n} \text{ by distributive law } \\ \text{ now do you know law of exponents? }\]

  7. freckles
    • one year ago
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    if you have the same base and you are multiplying what do you do with the exponents ?

  8. anonymous
    • one year ago
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    Add

  9. freckles
    • one year ago
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    \[(1+a^{-n})a^n=a^n+a^{-n+n}=?\]

  10. freckles
    • one year ago
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    -n+n=?

  11. anonymous
    • one year ago
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    0

  12. freckles
    • one year ago
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    right and a^0=?

  13. anonymous
    • one year ago
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    One

  14. freckles
    • one year ago
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    \[\frac{1}{1+a^{n}}+\frac{a^n}{1+a^{n}}=?\]

  15. freckles
    • one year ago
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    you see you have the same denominator

  16. freckles
    • one year ago
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    now you can write as one fraction

  17. freckles
    • one year ago
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    \[\frac{1+a^{n}}{1+a^{n}}=?\]

  18. anonymous
    • one year ago
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    1

  19. freckles
    • one year ago
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    right and this is of coursing assuming a>0

  20. freckles
    • one year ago
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    a could be less than 0 depending on n we could say a lot about the domain restrictions lol

  21. freckles
    • one year ago
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    but I'm sure they are just looking for 1

  22. anonymous
    • one year ago
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    Ok tysm

  23. freckles
    • one year ago
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    np

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