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calculusxy

  • one year ago

MEDAL!!!

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  1. calculusxy
    • one year ago
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    \[\huge 3n^2 - 11 = 8n \]

  2. AlexandervonHumboldt2
    • one year ago
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    solve for n?

  3. calculusxy
    • one year ago
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    yes using quadratic formula

  4. calculusxy
    • one year ago
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    \[\large 3n^2 - 8n - 11 = 0\] would that equation be correct?

  5. anonymous
    • one year ago
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    yes

  6. AlexandervonHumboldt2
    • one year ago
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    use quadratic formula \[\frac{ -b \pm \sqrt{b^2-4ac }}{ 2a }\] where ax^2+bx+c=0 yes this equation is correct substitute values:

  7. calculusxy
    • one year ago
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    \[n = \frac{ 8 \pm \sqrt{64 - 4(-33)} }{ 6 }\]

  8. anonymous
    • one year ago
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    looks good

  9. AlexandervonHumboldt2
    • one year ago
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    yeah correct now simplify this

  10. calculusxy
    • one year ago
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    \[n = \frac{ 8 \pm \sqrt{196} }{ 8 } = \frac{ 8 + 14 }{ 8 }\]

  11. AlexandervonHumboldt2
    • one year ago
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    how denominator became 8 from 6?

  12. anonymous
    • one year ago
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    how did your 6 become an 8 in the denominator

  13. AlexandervonHumboldt2
    • one year ago
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    everything is correct exept the denominator

  14. calculusxy
    • one year ago
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    sorry typo :\

  15. AlexandervonHumboldt2
    • one year ago
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    ok

  16. anonymous
    • one year ago
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    ah ;)

  17. AlexandervonHumboldt2
    • one year ago
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    it would be \[\frac{ 8 \pm 14 }{ ? }\]

  18. AlexandervonHumboldt2
    • one year ago
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    woops 6 instead of ?

  19. calculusxy
    • one year ago
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    (3.66, -1)

  20. AlexandervonHumboldt2
    • one year ago
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    correct

  21. calculusxy
    • one year ago
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    thank you @Lily2913 @AlexandervonHumboldt2

  22. AlexandervonHumboldt2
    • one year ago
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    np :)

  23. anonymous
    • one year ago
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    np :)

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