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ladiesman217

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  • 11 months ago
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  1. ladiesman217
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    \[3x^2=33x+24\]

    • 11 months ago
  2. ladiesman217
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    @ajprincess @nubeer @.Sam. can you help me please

    • 11 months ago
  3. ajprincess
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    \(3x^2=33x+24\) \(3x^2-33x+24=0\) Use quadratic formula. \[x=\frac{ -b\pm\sqrt{b^2-4ac} }{ 2a }\] the values of a, b and c can be found by comparing the equation \(3x^2-33x+24=0\) with the general equation \(ax^2+bx+c=0\). Does that help? @ladiesman217

    • 11 months ago
  4. ladiesman217
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    ya

    • 11 months ago
  5. gohangoku58
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    **-24 dividing the entire equation by 3 would simplify calculations

    • 11 months ago
  6. ajprincess
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    it is actually \(3x^2-33x-24=0\). am sorry abt the mistake

    • 11 months ago
  7. ajprincess
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    ya. first factor out 3 and then divide by 3 both sides. After that use quadratic formula:)

    • 11 months ago
  8. ladiesman217
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    the equation i got was\[\frac{ -(-3)\sqrt{3^{2}-4(3)(-24)} }{ 2*3 }\]

    • 11 months ago
  9. ajprincess
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    b=-33

    • 11 months ago
  10. ladiesman217
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    this is the correct equation right \[\frac{ -(-33)\sqrt{-33^{2}-4(3)(-24)} }{ 2*3 }\]

    • 11 months ago
  11. ajprincess
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    it will be \[\frac{ -(-33)\sqrt{(-33)^2-4*3*(-24)} }{ 2*3}\]

    • 11 months ago
  12. ladiesman217
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    i got\[\frac{ 11\pm \sqrt{153} }{ 2 }\]

    • 11 months ago
  13. ajprincess
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    ya right:)

    • 11 months ago
  14. ajprincess
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    good work:)

    • 11 months ago
  15. ladiesman217
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    thnx

    • 11 months ago
  16. ajprincess
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    welcome:)

    • 11 months ago
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