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anonymous

  • one year ago

Choose the point-slope form of the equation below that represents the line that passes through the point (6, -3) and has a slope of one half.

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  1. anonymous
    • one year ago
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    Help?

  2. anonymous
    • one year ago
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    @iambatman ?

  3. anonymous
    • one year ago
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    Y2-y1 over x2-x1

  4. anonymous
    • one year ago
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    Sorry y=mx+b

  5. anonymous
    • one year ago
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    So This Is The Equation so far 6=1/2 (-3)+b Now You Have To Solve for b

  6. pooja195
    • one year ago
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    \[\huge~\rm~y-y_1=m(x-x_1)\]

  7. pooja195
    • one year ago
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    (x1,y1) m=slope

  8. pooja195
    • one year ago
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    And where are your options??

  9. anonymous
    • one year ago
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    Either equation will give you the same answer

  10. anonymous
    • one year ago
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    The choices are y - 6 = 1/2 (3 + x) y = 1/2x - 6 y - 3 = 1/2 (x - 6) x - 2y = 12

  11. anonymous
    • one year ago
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    @pooja195 @Niaa08

  12. pooja195
    • one year ago
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    Ok so we dont need to solve \[ \huge~\rm~y−y_1=m(x−x_1)\] (x1,y1) m=slope Just plug in the given info

  13. anonymous
    • one year ago
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    Im not so sure how to, im kinda new at this...

  14. abb0t
    • one year ago
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    Answer is \(\sf \color{red}{C}\) in other words, choice number 3.

  15. pooja195
    • one year ago
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    (6, -3) (x1,y1) x1=6 y1=-3 And the slope is given slope =1/2

  16. anonymous
    • one year ago
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    Then what?

  17. pooja195
    • one year ago
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    plug the info into the formula

  18. abb0t
    • one year ago
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    Although, it should be \(\sf \ y+3 = \frac{1}{2} x-6\)

  19. abb0t
    • one year ago
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    not -3

  20. abb0t
    • one year ago
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    But choice 3 looks like the most correct.

  21. pooja195
    • one year ago
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    @abb0t please stop giving direct answers and let the user learn

  22. anonymous
    • one year ago
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    oops yeah it's +3 not -3

  23. abb0t
    • one year ago
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    Yep, then that's the answer @TheDragon :)

  24. pooja195
    • one year ago
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    Wow.

  25. anonymous
    • one year ago
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    Thanks @abb0t @pooja195 for the help

  26. abb0t
    • one year ago
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    You're very welcome. If you need anymore just tag me <3

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