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dddan

  • 2 years ago

how do you simplify (18u^6v^5+24u^3v^3)/42u^2v^5

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  1. JakeV8
    • 2 years ago
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    Try factoring the top first. Factor out all the numbers, u, and v terms you can. Then look for opportunities to cancel with the terms on the bottom.

  2. dddan
    • 2 years ago
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    ok i factored out the top but i still dont know what to do next..

  3. JakeV8
    • 2 years ago
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    what did you get?

  4. dddan
    • 2 years ago
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    3u^3v^3(6u^3v^2+8) / 42u^2v^5

  5. JakeV8
    • 2 years ago
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    ok, great Now you can cancel the 3 with the 42... 3 "goes into" 42 by 14

  6. JakeV8
    • 2 years ago
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    so 14 is left on bottom and you can cancel the u^2 on the bottom with the u^3 in the front on top, leaving just u out front and cancel the v^3 out in front on top with the v^5 on the bottom, leaving v^2 on the bottom

  7. dddan
    • 2 years ago
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    alright so what to i do with (6u^3v^2+8)?

  8. JakeV8
    • 2 years ago
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    Hold on... Is (18u^6v^5+24u^3v^3)/42u^2v^5 written like: \[\frac{ 18u ^{6}6v ^{5}+24u ^{3}v ^{3} }{ 42u ^{2}v ^{5} }\]

  9. dddan
    • 2 years ago
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    yup

  10. JakeV8
    • 2 years ago
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    You can simplify that part in your last post slightly by multiplying the 6 and 3 together to get (18u^3v^2 + 8) and you could then factor out another 2, to get 2(9u^3v^2 + 4), but you can't cancel that 2 with anything on the bottom because you only had 13 left on the bottom.

  11. dddan
    • 2 years ago
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    you mean 14 on the bottom..

  12. JakeV8
    • 2 years ago
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    so like you had, 3u^3v^3(6u^3v^2+8) / 42u^2v^5 simplify parenthesis and factor out 2 to get: (3u^3v^3)(2)(9u^3v^2 + 4) / 42u^2v^5 and cancel whatever you can to get: 2u(9u^3v2 + 4) / 13v^3

  13. JakeV8
    • 2 years ago
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    yes, sorry, 14 on bottom

  14. JakeV8
    • 2 years ago
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    so you CAN cancel that extra 2 on top, leaving 7 on the bottom

  15. JakeV8
    • 2 years ago
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    good catch!

  16. dddan
    • 2 years ago
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    alright so now we got rid of the 2 and then put the 7 on the bottom, i then factored the u(9u^3v^2+4) and simplified the v's and got (9u^3+4u)/7....

  17. JakeV8
    • 2 years ago
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    you lost me... but I also messed up... I multiplied the 6 and 3 inside the parenthesis to get 18, but then when I factored out the 2 (to leave 9), I left the original 3 in the term :( Sorry

  18. JakeV8
    • 2 years ago
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    wait, hold on, that 3 was the exponent on u. typing this stuff is killing me... sorry for the confusion.

  19. JakeV8
    • 2 years ago
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    Should be: 2u(9u^3v^2 + 4) / 14v^3 Then cancel the 2: u(9u^3v^2 + 4) / 7v^3 An you are done at that point... all simplified.

  20. JakeV8
    • 2 years ago
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    Ugly, yes, but simplified

  21. dddan
    • 2 years ago
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    ya i tried to go a step further but i will leave at that this is complicated enough. thank u so much!

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