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anonymous

  • 5 years ago

in the problem y= 2/3x+3 2/3 is my slope right if that is so 3 is my first cordinate point to start graphing from

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  1. anonymous
    • 5 years ago
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    to define the line you need only 2 points. if you let x=0, so y=3 2/3 (not 3)... if you let x=3, y = 2/3 *3 +3 2/3 = 2+ 3 2/3

  2. anonymous
    • 5 years ago
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    ok that clears up my confusion ty i should be able to graph it now

  3. anonymous
    • 5 years ago
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    Terry...should the original equation be\[y=2/3x+3\] or \[y=2/3x + 3(and)2/3\]? That is, I read your initial question to say that you wanted to check that the slope of the equation \[y=2/3x+3\] is 2/3.

  4. anonymous
    • 5 years ago
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    the problem uis set up in fraction form so its 2/3x +3

  5. anonymous
    • 5 years ago
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    Your original question was misleading, I believe, and inik's answer was to a different problem. The slope of the line given by the equation y=2/3x+3 is 2/3, as I believe you indicated. Equations of the form y=mx+b have slope m and y intercept b, so your first point to start graphing from would be the y intercept, b, 3, as you had in your first post. At x=0, y=2/3*0+3 such that the point at x=0 is (0,3).

  6. anonymous
    • 5 years ago
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    ok ill try graphing the problem that way and see if that looks better than the previous way i graphed the equation

  7. anonymous
    • 5 years ago
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    i get mislead when it comes to graphing equations that include fractions a common mistake alot of people make

  8. anonymous
    • 5 years ago
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    It was just the phrasing. Your problem read like this: "in the problem y= 2/3x+3 2/3 is my slope right" It would have been more clear had you said "in the problem y=2/3x+3, the slope is 2/3, right?" Commas can make all the difference; I just wanted to make sure you did not get misled with your work. Good luck! :)

  9. anonymous
    • 5 years ago
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    terry is right. you need just two points to plot a graph of a straight line. but its better to start with a y intercept first cause you don't have a calculate the abscissa (x cordinate) in that case. so your first point be 3 in the y axis. now either you can find one more point using the method told by terry or you can use the slope (may get hard sometimes) to plot the graph. How to use the slope: \[\tan \theta = m\] so \[\theta = \tan^{-1} m\] find the angle from the above method and draw a line making that angle with the horizontal and passing throug the y -intercept

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