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sammietaygreen

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  • one year ago
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  1. whpalmer4
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    Put both equations in slope-intercept form: y = mx + b where m is slope and b is y-intercept. If they are parallel, slopes will be =. If they are perpendicular, the product of the two slopes will be -1. Otherwise, they are none of the above.

    • one year ago
  2. whpalmer4
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    Slope of first line is -2/3. Second equation we need to solve for y: [\2x-3y=-3\]\[2x+3=3y\]\[y=\frac{2}{3}x+1\] so slope of second line is 2/3. \[\frac{2}{3}*-\frac{2}{3} = -\frac{4}{9} \] so they are not perpendicular.

    • one year ago
  3. sammietaygreen
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    Yes, yes. I know it can't be perpendicular.

    • one year ago
  4. whpalmer4
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    Well, that also shows that they are not parallel because the slopes are not equal.

    • one year ago
  5. whpalmer4
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    Need to check your work if you graphed them and got parallel lines :-)

    • one year ago
  6. sammietaygreen
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    Ohhh, I must have graphed wrong. That simplified it a bit, @jim_thompson5910 is helping me as well. I'm going to check what I did wrong

    • one year ago
  7. whpalmer4
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    I always use x = 0 as one of my graph points :-)

    • one year ago
  8. sammietaygreen
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    thanks! :)

    • one year ago
  9. whpalmer4
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    In this case, it doesn't help because they actually cross at x =0, but often that isn't the case, and the arithmetic is a bit easier!

    • one year ago
  10. sammietaygreen
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    could you help on one more on this thread?

    • one year ago
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spraguer (Moderator)
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is replying to Can someone tell me what button the professor is hitting...

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