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jchavez96 Group Title

Simplify: (40a/b^2-5b+4) / (5a^2/2b-2)

  • one year ago
  • one year ago

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  1. ashna Group Title
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    40a / (b-1)*(b-4) / 5a^2 / 2(b-1) 40 cancels with 5 leaving 8 a^2 cancels with a leaving a in denominator b-1 cancels with b-1 so left with 8/b-4 / a/2 invert and multiply 16/ab--4a

    • one year ago
  2. ashna Group Title
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    can you understand ? @jchavez96

    • one year ago
  3. jchavez96 Group Title
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    Sorta , I'm working on it right now . @ashna

    • one year ago
  4. ashna Group Title
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    okay :)

    • one year ago
  5. jchavez96 Group Title
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    These are my options ashna , I'm having trouble figuring it out .

    • one year ago
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  6. jchavez96 Group Title
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    @ashna

    • one year ago
  7. ashna Group Title
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    4th option is right , you want to know how , or do you know ?

    • one year ago
  8. ashna Group Title
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    @jchavez96

    • one year ago
  9. jchavez96 Group Title
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    Can you explain it to me please ?

    • one year ago
  10. ashna Group Title
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    now take b^2-5b+4 and simplify it like u do spliting the middle term so....b^2-5b+4 b(b-4)-1(b-4)

    • one year ago
  11. ashna Group Title
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    (b-1) (b-4)

    • one year ago
  12. ashna Group Title
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    now {40a/(b-1)(b-4)}/{5a^2/2b-2}

    • one year ago
  13. ashna Group Title
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    take the main denominater in reciprocal form then cancel 40a with 5a^2

    • one year ago
  14. ashna Group Title
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    now 16b-16/(b-1)(b-4) = 16(b-1)/(b-1)(b-4)

    • one year ago
  15. ashna Group Title
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    cancel b-1 u'll get opt4

    • one year ago
  16. ashna Group Title
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    (i.e) = 16/b-4

    • one year ago
  17. ashna Group Title
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    @jchavez96 understood now ?

    • one year ago
  18. jchavez96 Group Title
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    I see . . . Thank you !! What is the b=4 for ? The one with the equal sign crossed out ?

    • one year ago
  19. jchavez96 Group Title
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    @ashna

    • one year ago
  20. ashna Group Title
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    dat's b not equal to 4 just substute b as 4 u wont get correct answer

    • one year ago
  21. ashna Group Title
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    understood @jchavez96

    • one year ago
  22. jchavez96 Group Title
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    Yes ! Thank you @ashna ! I really appreciate it !

    • one year ago
  23. ashna Group Title
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    no mention :)

    • one year ago
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