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Saylilbaby

  • one year ago

find the excluded value pls...medal

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  1. saylilbaby
    • one year ago
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    (3x^2+5x-8)/(6x+16)*(4x^2+x)/(3x^2-3x)

  2. saylilbaby
    • one year ago
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    @Australopithecus @jdoe0001 @pooja195 @TheSmartOne @Awolflover1

  3. saylilbaby
    • one year ago
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    2/(x^2-1)+x/(x^2-2x+1)

  4. saylilbaby
    • one year ago
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    @Astrophysics

  5. xapproachesinfinity
    • one year ago
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    let me rewrite that \(\Large \frac{(3x^2+5x-8)}{(6x+16)} \times \frac{(4x^2+x)}{(3x^2-3x)}\)

  6. xapproachesinfinity
    • one year ago
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    that's what you have?

  7. xapproachesinfinity
    • one year ago
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    for excluded values you look at what makes the denominator equal to zero

  8. xapproachesinfinity
    • one year ago
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    but before doing so we would like to simplify first factoring if possible

  9. saylilbaby
    • one year ago
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    the answer was (3x^33-8)(4x+1)/6(3x+8)(x-1) @xapproachesinfinity

  10. xapproachesinfinity
    • one year ago
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    can we factor \(3x^2+5x-8\)

  11. xapproachesinfinity
    • one year ago
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    answer for what?

  12. saylilbaby
    • one year ago
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    yes we can factor that @xapproachesinfinity

  13. xapproachesinfinity
    • one year ago
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    so what are the factors

  14. saylilbaby
    • one year ago
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    (3x+8)(x-1) @xapproachesinfinity

  15. xapproachesinfinity
    • one year ago
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    yes!

  16. xapproachesinfinity
    • one year ago
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    so we have \(\Large \frac{(3x+8)(x-1)}{(6x+16)}\times \frac{(4x+1)}{3(x-1)}\) \(\Large \frac{(3x+8)(4x+1)}{3(6x+16)}\)

  17. xapproachesinfinity
    • one year ago
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    so what make denominator now equal to 0

  18. xapproachesinfinity
    • one year ago
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    6x+16=0 ===> x=-16/6 x=-8/3

  19. xapproachesinfinity
    • one year ago
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    excluded value -8/3

  20. saylilbaby
    • one year ago
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    ok thanks can you help with the secondequation @xapproachesinfinity

  21. xapproachesinfinity
    • one year ago
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    actually i have to go for some time i fasting i need to go eat :)

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