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kaisan

how do you graph the slope 5/4 using rise over run?

  • one year ago
  • one year ago

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  1. jim_thompson5910
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    start at any point that is on the line then go up 5 units, then go 4 units to the right to get to the next point

    • one year ago
  2. kaisan
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    can you start at the origin?

    • one year ago
  3. jim_thompson5910
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    say we have this starting point |dw:1359348640310:dw|

    • one year ago
  4. jim_thompson5910
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    if you go up 5 units you land here |dw:1359348680876:dw|

    • one year ago
  5. jim_thompson5910
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    if you go 4 units to the right, you will land here |dw:1359348711180:dw|

    • one year ago
  6. jim_thompson5910
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    |dw:1359348750015:dw|

    • one year ago
  7. jim_thompson5910
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    then you would draw a line through those two points |dw:1359348773195:dw|

    • one year ago
  8. jim_thompson5910
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    now you can move that "starting point" to wherever you want

    • one year ago
  9. kaisan
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    i still don't understand about who to find the starting point

    • one year ago
  10. julianahe
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    You need a Y- intercept

    • one year ago
  11. julianahe
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    the question might give you it

    • one year ago
  12. kaisan
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    the y-int 5/4 also

    • one year ago
  13. julianahe
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    |dw:1359350257526:dw|

    • one year ago
  14. julianahe
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    are you sure? well if you dont have you you can start anywhere from the graph...the graph i drew is negative slope tho

    • one year ago
  15. bettyboop8904
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    yeah that last graph probably confused them lol ; D

    • one year ago
  16. bettyboop8904
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    Can you write out the entire question as its seen?

    • one year ago
  17. bettyboop8904
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    @kaisan

    • one year ago
  18. julianahe
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    which graphj

    • one year ago
  19. kaisan
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    ok its y=5/4x-5/4

    • one year ago
  20. bettyboop8904
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    the one with the negative slope

    • one year ago
  21. bettyboop8904
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    ok if you draw the graph the y-intercept with be 5/4 or 1 1/4

    • one year ago
  22. bettyboop8904
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    |dw:1359350723423:dw|

    • one year ago
  23. bettyboop8904
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    then the slope is 5/4 rise over run: so you rise 5 and you run 4

    • one year ago
  24. bettyboop8904
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    |dw:1359350933733:dw|

    • one year ago
  25. kaisan
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    so its best to start with the y-intercept?

    • one year ago
  26. bettyboop8904
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    yes that's what I always do. I plug the y-intercept in first and then you do the slope to find the next point so you can draw the line = )

    • one year ago
  27. kaisan
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    so if the y-int was -5/4 i would go 5 units first then to the right 4 units? but i still don't understand to start when i'm graphing, i tried starting on the origin but my online hw says its wrong

    • one year ago
  28. kaisan
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    @bettyboop8904

    • one year ago
  29. bettyboop8904
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    what do you mean starting on the origin? you should only be putting the y-intercept in. In the graph you have x and y values. With the y-intercept being 5/4 the point is written as \[(0,\frac{ 5 }{ 4 })\]

    • one year ago
  30. bettyboop8904
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    If you were to start at the origin the starting equation would have to read y=(5/4)x+0 or y=(5/4)x

    • one year ago
  31. bettyboop8904
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    @kaisan

    • one year ago
  32. kaisan
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    oh, ok thanks

    • one year ago
  33. bettyboop8904
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    but if your slope was -(5/4) then the equation would read \[y=-\frac{ 5 }{ 4 }x +\frac{ 5 }{ 4 }\] then you would fall down 5 and then go to the right 4 or you could rise 5 and then go to the left 4; Just as long as when you do the rise over run you are only using one negative motion (fall down or go left bc those are negative x- and y-values) for only one of the numbers. You can place the negative either on the top or on the bottom but never both. @kaisan

    • one year ago
  34. bettyboop8904
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    |dw:1359353149250:dw|

    • one year ago
  35. bettyboop8904
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    this shows both ways on the graph

    • one year ago
  36. kaisan
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    thx

    • one year ago
  37. bettyboop8904
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    your welcome = )

    • one year ago
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