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anonymous

  • one year ago

(3x^2/4y^2)^-2 SIMPLIFY Help?

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  1. jim_thompson5910
    • one year ago
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    |dw:1434068972553:dw|

  2. jim_thompson5910
    • one year ago
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    |dw:1434069000316:dw|

  3. jim_thompson5910
    • one year ago
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    after doing the flip, we will have |dw:1434069045031:dw|

  4. jim_thompson5910
    • one year ago
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    with me so far?

  5. anonymous
    • one year ago
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    yep

  6. jim_thompson5910
    • one year ago
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    now we square both top and bottom |dw:1434069090790:dw|

  7. jim_thompson5910
    • one year ago
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    Since \[\Large \left(\frac{A}{B}\right)^2 = \frac{A^2}{B^2}\]

  8. jim_thompson5910
    • one year ago
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    do you see how to finish up?

  9. anonymous
    • one year ago
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    no, im still confused. do you jus multiply the eponents together or add?

  10. jim_thompson5910
    • one year ago
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    you can think of \[\Large (4y^2)^2\] as \[\Large (4y^2)*(4y^2)\]

  11. lulubj
    • one year ago
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    16y^4?

  12. lulubj
    • one year ago
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    over 9x^4 right

  13. jim_thompson5910
    • one year ago
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    lulubj, let the original asker answer the question I'm asking

  14. lulubj
    • one year ago
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    i need it too

  15. anonymous
    • one year ago
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    ohh i get what your doing though.

  16. lulubj
    • one year ago
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    dont know if thats right though

  17. jim_thompson5910
    • one year ago
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    it's correct

  18. jim_thompson5910
    • one year ago
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    |dw:1434069298881:dw|

  19. anonymous
    • one year ago
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    what if it was 4y^3 is it the same proccess?

  20. jim_thompson5910
    • one year ago
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    yes same idea but the y exponent in the answer will be different

  21. anonymous
    • one year ago
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    so it would be 4y^9?

  22. anonymous
    • one year ago
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    16y^9 ?

  23. jim_thompson5910
    • one year ago
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    |dw:1434069408797:dw|

  24. jim_thompson5910
    • one year ago
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    the exponents 3 and 2 multiply to get 6

  25. jim_thompson5910
    • one year ago
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    or you can think of it as y^3*y^3 = y^(3+3) = y^6

  26. anonymous
    • one year ago
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    oh okay thanks! i get it :)

  27. jim_thompson5910
    • one year ago
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    you're welcome

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