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anonymous

  • one year ago

Graphing please

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  1. anonymous
    • one year ago
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    |dw:1439231312962:dw| Will medal!

  2. anonymous
    • one year ago
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    |dw:1439231406656:dw|

  3. anonymous
    • one year ago
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    @Michele_Laino

  4. Michele_Laino
    • one year ago
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    first graph: we have to consider these 2 cases: 1) \[x - 2 \geqslant 0\] and: 2)\[x - 2 < 0\]

  5. anonymous
    • one year ago
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    Alright. But for this hw assignmnet, I dont have to show work, i just have to graph it.

  6. anonymous
    • one year ago
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    So, is there a website where I can go on to type it in?

  7. Michele_Laino
    • one year ago
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    I understand, nevertheless I can not give you the graph, since I have to guide you to your solution

  8. anonymous
    • one year ago
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    Ok:)

  9. Michele_Laino
    • one year ago
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    in the first case your function is: \[\Large \begin{gathered} y = - \frac{1}{3}\left( {x - 2} \right) + 2 = - \frac{x}{3} + \frac{2}{3} + 2 \hfill \\ \hfill \\ y = - \frac{x}{3} + \frac{8}{3} \hfill \\ \end{gathered} \]

  10. anonymous
    • one year ago
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    Yes:)

  11. anonymous
    • one year ago
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    So, then i would just plug in some points to get the answer

  12. Michele_Laino
    • one year ago
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    in thesecond case, the equation of your function is: \[\Large \begin{gathered} y = - \frac{1}{3} \cdot - \left( {x - 2} \right) + 2 = \frac{x}{3} - \frac{2}{3} + 2 \hfill \\ \hfill \\ y = \frac{x}{3} + \frac{4}{3} \hfill \\ \end{gathered} \]

  13. Michele_Laino
    • one year ago
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    yes! for example, I consider the first function: \[\Large y = - \frac{x}{3} + \frac{8}{3}\] if x=0, then y=8/3, so that line passes at (0,8/3) whereas if y=0, then x=8, so that line passes at point (8,0)

  14. anonymous
    • one year ago
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    Oh! Okay, thank you:) I understand it now

  15. Michele_Laino
    • one year ago
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    |dw:1439232238778:dw|

  16. Michele_Laino
    • one year ago
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    in the second case, I consider this function: \[\Large y = \frac{x}{3} + \frac{4}{3}\] now, if x=0, then y= 4/3, so that line passes at (0, 4/3) whereas if y=0, then x=-4, so that line passes at (4, 0)

  17. anonymous
    • one year ago
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    ok and then i can just graph it?

  18. anonymous
    • one year ago
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    thanks so much for all ur help!

  19. Michele_Laino
    • one year ago
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    |dw:1439232482291:dw|

  20. Michele_Laino
    • one year ago
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    similarly for second graph

  21. anonymous
    • one year ago
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    Makes sense :)

  22. Michele_Laino
    • one year ago
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    for second graph, you have to consider these 2 subsequent cases: \[\Large \begin{gathered} x + 3 \geqslant 0 \hfill \\ x - 3 < 0 \hfill \\ \end{gathered} \] and repeat the procedure above

  23. anonymous
    • one year ago
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    Kk:) I will do that

  24. Michele_Laino
    • one year ago
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    ok! :)

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