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Kuoministers

  • one year ago

Simplify 2^3 * 4n^2 * 2^-n HELP PLS :D

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  1. Kuoministers
    • one year ago
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    \[2^{3} \times 4n ^{2} \times 2^{-n}\]

  2. Kuoministers
    • one year ago
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    \[2^{3} \times 2^{2}n ^{2} \times 2^{-n}\]

  3. Kuoministers
    • one year ago
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    @Dido525

  4. Kuoministers
    • one year ago
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    @RolyPoly

  5. Kuoministers
    • one year ago
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    pls help

  6. Kuoministers
    • one year ago
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    help @exraven

  7. Kuoministers
    • one year ago
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    pls help anyone? @AravindG

  8. AravindG
    • one year ago
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    |dw:1358401518683:dw|

  9. AravindG
    • one year ago
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    is that the qn ?

  10. Kuoministers
    • one year ago
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    yes!!

  11. AravindG
    • one year ago
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    |dw:1358401570163:dw|

  12. AravindG
    • one year ago
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    |dw:1358401640822:dw|

  13. Kuoministers
    • one year ago
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    im still unsure of what to do?

  14. AravindG
    • one year ago
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    why??

  15. Kuoministers
    • one year ago
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    would it be 16n?

  16. AravindG
    • one year ago
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    no !

  17. AravindG
    • one year ago
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    |dw:1358401736501:dw|

  18. Kuoministers
    • one year ago
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    no?

  19. AravindG
    • one year ago
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    |dw:1358401756772:dw|

  20. AravindG
    • one year ago
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    |dw:1358401776612:dw|

  21. AravindG
    • one year ago
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    that is the simplified form

  22. Kuoministers
    • one year ago
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    would that be final answer?

  23. AravindG
    • one year ago
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    yep :)

  24. Kuoministers
    • one year ago
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    answer in book is 2^ n-3?

  25. AravindG
    • one year ago
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    but i am sure of my answer ! check the question again

  26. Kuoministers
    • one year ago
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    2^3 times (2n)^2 times 2^ -n opps one difference

  27. AravindG
    • one year ago
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    :O

  28. Kuoministers
    • one year ago
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    is there any difference?

  29. Kuoministers
    • one year ago
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    can you pls work it out now lol?

  30. AravindG
    • one year ago
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    i dont see any diff in question

  31. Kuoministers
    • one year ago
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    anyways thanks

  32. RolyPoly
    • one year ago
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    \[2^3 \times (2n)^2 \times 2^{-n}\]\[= 2^{3-n} \times 2^2n^2\]\[=2^{3-n+2}n^2\]\[=...\]

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