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anonymous

  • one year ago

What is the simplified form of x minus 5 over x squared minus 3x minus 10 ⋅ x plus 2 over x squared plus x minus 12?

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  1. anonymous
    • one year ago
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    |dw:1437150497955:dw|

  2. anonymous
    • one year ago
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    @Michele_Laino

  3. anonymous
    • one year ago
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    I got 1/(x+3)(x-2) as an answer

  4. Michele_Laino
    • one year ago
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    is your expression like this: \[\frac{{x - 5}}{{{x^2} - 3x - 10}}:\frac{{x + 2}}{{{x^2} + x - 12}}\]

  5. anonymous
    • one year ago
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    x-5/(x-5)(x+2) timex x+2/(x+4)(x-3)

  6. Michele_Laino
    • one year ago
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    ok! sorry

  7. Michele_Laino
    • one year ago
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    we have these steps:

  8. anonymous
    • one year ago
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    ok

  9. Michele_Laino
    • one year ago
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    \[\begin{gathered} \frac{{x - 5}}{{{x^2} - 3x - 10}} \cdot \frac{{x + 2}}{{{x^2} + x - 12}} = \hfill \\ \hfill \\ = \frac{{x - 5}}{{\left( {x - 5} \right)\left( {x + 2} \right)}} \cdot \frac{{x + 2}}{{\left( {x - 3} \right)\left( {x + 4} \right)}} = ...? \hfill \\ \end{gathered} \]

  10. anonymous
    • one year ago
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    Was I wrong?

  11. anonymous
    • one year ago
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    what's going on? @Michele_Laino

  12. Michele_Laino
    • one year ago
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    after a simplification I got this: \[\frac{1}{{\left( {x - 3} \right)\left( {x + 4} \right)}}\]

  13. anonymous
    • one year ago
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    Thanks :) Can I ask another?

  14. Michele_Laino
    • one year ago
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    ok!

  15. anonymous
    • one year ago
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    What is the simplified form of the quantity x over 4 plus y over 3 all over the quantity x over 3 minus y over 4?

  16. anonymous
    • one year ago
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    |dw:1437151350249:dw|

  17. anonymous
    • one year ago
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    i got 3x+4y/ 4y-3x

  18. Michele_Laino
    • one year ago
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    do you mean this: \[\frac{{\frac{x}{4} + \frac{y}{3}}}{{\frac{x}{4} - \frac{y}{3}}}\]

  19. Michele_Laino
    • one year ago
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    \[\Large \frac{{\frac{x}{4} + \frac{y}{3}}}{{\frac{x}{4} - \frac{y}{3}}}\]

  20. anonymous
    • one year ago
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    no the bottom one is x/3 - y/4

  21. Michele_Laino
    • one year ago
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    ok! we have to compute the least common multiple at numerator and at denominator, of your main fraction, which is 12, so we get: \[\Large \frac{{\frac{x}{4} + \frac{y}{3}}}{{\frac{x}{3} - \frac{y}{4}}} = \frac{{\frac{{3x + 4y}}{{12}}}}{{\frac{{4x - 3y}}{{12}}}}\]

  22. anonymous
    • one year ago
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    So we can get rid of the denominator of 12

  23. Michele_Laino
    • one year ago
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    yes!

  24. anonymous
    • one year ago
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    :) YAY! Thanks!

  25. Michele_Laino
    • one year ago
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    :)

  26. anonymous
    • one year ago
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    Last one?

  27. anonymous
    • one year ago
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    What is the oblique asymptote of the function f(x) = the quantity x squared minus 5x plus 6 over the quantity x minus 4?

  28. Michele_Laino
    • one year ago
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    ok! :)

  29. anonymous
    • one year ago
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    I got x-1 as the oblique asymptote

  30. Michele_Laino
    • one year ago
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    your function is like this: \[\frac{{\sqrt {{x^2} - 5x + 6} }}{{x - 4}}\]

  31. anonymous
    • one year ago
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    No. It is without the radical sign

  32. Michele_Laino
    • one year ago
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    no, sorry, is like this? \[\frac{{{x^2} - 5x + 6}}{{x - 4}}\]

  33. anonymous
    • one year ago
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    Yup:)

  34. anonymous
    • one year ago
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    I got x-1.

  35. Michele_Laino
    • one year ago
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    If I write the asymptote like this: y=mx+q then we have: \[\Large \begin{gathered} m = \mathop {\lim }\limits_{x \to \infty } \frac{{f\left( x \right)}}{x} = \mathop {\lim }\limits_{x \to \infty } \frac{{{x^2} - 5x + 6}}{{x\left( {x - 4} \right)}} = \hfill \\ \hfill \\ = \mathop {\lim }\limits_{x \to \infty } \frac{{{x^2}\left( {1 - \frac{5}{x} + \frac{6}{{{x^2}}}} \right)}}{{{x^2}\left( {1 - \frac{4}{{{x^2}}}} \right)}} = \hfill \\ \end{gathered} \]

  36. anonymous
    • one year ago
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    Ohh okay:) Kinda confused but I'm getting it

  37. anonymous
    • one year ago
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    can't you just use synthetic divison?

  38. Michele_Laino
    • one year ago
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    that limit is 1, so your slope is right!

  39. anonymous
    • one year ago
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    Thanks:)

  40. anonymous
    • one year ago
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    I have to go but I need more help. When will you be back?

  41. Michele_Laino
    • one year ago
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    I stay here for at least 2 hours

  42. anonymous
    • one year ago
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    Ok:) Thank you. U gtg now!

  43. Michele_Laino
    • one year ago
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    wait I have to compute the quantity q

  44. Michele_Laino
    • one year ago
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    oops..please wait...

  45. Michele_Laino
    • one year ago
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    we have this formula: \[\large \begin{gathered} q = \mathop {\lim }\limits_{x \to \infty } \left\{ {f\left( x \right) - mx} \right\} = \mathop {\lim }\limits_{x \to \infty } \left\{ {f\left( x \right) - x} \right\} = \hfill \\ \hfill \\ = \mathop {\lim }\limits_{x \to \infty } \left\{ {\frac{{{x^2} - 5x + 6}}{{x - 4}} - x} \right\} = \mathop {\lim }\limits_{x \to \infty } \left\{ {\frac{{{x^2} - 5x + 6 - {x^2} + 4x}}{{x - 4}}} \right\} = \hfill \\ \hfill \\ = \mathop {\lim }\limits_{x \to \infty } \left\{ {\frac{{6 - x}}{{x - 4}}} \right\} = \mathop {\lim }\limits_{x \to \infty } \left\{ { - 1 + \frac{2}{{x - 4}}} \right\} = ...? \hfill \\ \end{gathered} \]

  46. Michele_Laino
    • one year ago
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    that limit is -1, so I confirm your answer, namely the equation of your asymptote is: y=x-1

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