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adajiamcneal

  • one year ago

please help

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  1. adajiamcneal
    • one year ago
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  2. jdoe0001
    • one year ago
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    \(\bf \begin{array}{lllll} &x_1&y_1&x_2&y_2\\ % (a,b) &({\color{red}{ -2}}\quad ,&{\color{blue}{ 15}})\quad % (c,d) &({\color{red}{ 9}}\quad ,&{\color{blue}{ -18}}) \end{array} \\\quad \\ % slope = m slope = {\color{green}{ m}}= \cfrac{rise}{run} \implies \cfrac{{\color{blue}{ y_2}}-{\color{blue}{ y_1}}}{{\color{red}{ x_2}}-{\color{red}{ x_1}}} \\ \quad \\ % point-slope intercept y-{\color{blue}{ y_1}}={\color{green}{ m}}(x-{\color{red}{ x_1}})\qquad \textit{plug in the values and solve for "y"}\\ \qquad \uparrow\\ \textit{point-slope form}\)

  3. adajiamcneal
    • one year ago
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    . . . . so i plug in the x and ys?

  4. jdoe0001
    • one year ago
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    actually, you don't need to solve for "y: since you're asked to leave it in point-slope form anyhow

  5. jdoe0001
    • one year ago
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    plug in \(x_1\ and\ y_1\) and the found slope

  6. jdoe0001
    • one year ago
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    so.. find the slope first

  7. adajiamcneal
    • one year ago
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    . . . and how do i do that

  8. jdoe0001
    • one year ago
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    well... see above

  9. jdoe0001
    • one year ago
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    \(\bf slope = {\color{green}{ m}}= \cfrac{rise}{run} \implies \cfrac{{\color{blue}{ y_2}}-{\color{blue}{ y_1}}}{{\color{red}{ x_2}}-{\color{red}{ x_1}}}\)

  10. adajiamcneal
    • one year ago
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    so i plug it in like i said or just x 1 and y 1

  11. jdoe0001
    • one year ago
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    they do have a value, it's given in coordinates

  12. jdoe0001
    • one year ago
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    \(\begin{array}{lllll} &x_1&y_1&x_2&y_2\\ % (a,b) &({\color{red}{ -2}}\quad ,&{\color{blue}{ 15}})\quad % (c,d) &({\color{red}{ 9}}\quad ,&{\color{blue}{ -18}}) \end{array}\)

  13. adajiamcneal
    • one year ago
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    and then i just divide them?

  14. jdoe0001
    • one year ago
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    or leave them as fraction, yes

  15. adajiamcneal
    • one year ago
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    |dw:1442534647425:dw| like this

  16. jdoe0001
    • one year ago
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    hmmm one sec

  17. jdoe0001
    • one year ago
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    \(\bf \begin{array}{lllll} &x_1&y_1&x_2&y_2\\ % (a,b) &({\color{red}{ -2}}\quad ,&{\color{blue}{ 15}})\quad % (c,d) &({\color{red}{ 9}}\quad ,&{\color{blue}{ -18}}) \end{array} \\\quad \\ % slope = m slope = {\color{green}{ m}}= \cfrac{rise}{run} \implies \cfrac{{\color{blue}{ -18}}-{\color{blue}{ 15}}}{{\color{red}{ 9}}-{\color{red}{ (-2)}}}\implies \cfrac{-33}{9+2}\implies \cfrac{-\cancel{33}}{\cancel{11}}\implies ?\)

  18. adajiamcneal
    • one year ago
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    3?

  19. jdoe0001
    • one year ago
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    well... don't miss the " - " in front of the 33 so is -3 so, that's the slope, so plug that in, along with \(x_1\ and\ y_1\)

  20. adajiamcneal
    • one year ago
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    wait so what do i do with the -3?

  21. jdoe0001
    • one year ago
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    \(\bf \begin{array}{lllll} &x_1&y_1&x_2&y_2\\ % (a,b) &({\color{red}{ -2}}\quad ,&{\color{blue}{ 15}})\quad % (c,d) &({\color{red}{ 9}}\quad ,&{\color{blue}{ -18}}) \end{array} \\\quad \\ % slope = m slope = {\color{green}{ m}}= -3 \\ \quad \\ % point-slope intercept y-{\color{blue}{ 15}}={\color{green}{ -3}}(x-{\color{red}{ (-2)}})\\ \qquad \uparrow\\ \textit{point-slope form}\)

  22. jdoe0001
    • one year ago
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    x-(-2) => x+2 so... bear in mind that

  23. adajiamcneal
    • one year ago
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    so do i add 15 to the 3? to get y

  24. jdoe0001
    • one year ago
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    well... if you need it in "slope-intercept" form, yes, you could solve for "y:, yes BUT you're asked to leave it in "point-slope" form :)

  25. adajiamcneal
    • one year ago
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    so then what do i do?

  26. jdoe0001
    • one year ago
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    nothing, be happy, eat ice-cream

  27. adajiamcneal
    • one year ago
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    oh..im done ?

  28. jdoe0001
    • one year ago
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    yeap

  29. adajiamcneal
    • one year ago
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    thank you so much :)

  30. jdoe0001
    • one year ago
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    yw

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