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anonymous

  • one year ago

The sides of a square are 3 cm long. One vertex of the square is at (2,0) on a coordinate grid marked in centimeter units. What other point could be a vertex of the square?

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  1. anonymous
    • one year ago
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    try plotting it

  2. anonymous
    • one year ago
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    Okay

  3. anonymous
    • one year ago
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    Done

  4. anonymous
    • one year ago
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    I forgot what equation we used to finde other points

  5. anonymous
    • one year ago
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    well you dont really need an equation if you have plotted it

  6. anonymous
    • one year ago
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    what does your graph look like?

  7. anonymous
    • one year ago
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    I mean to find another point, I think I'm confusing it with a parabola. is it supposed to look like a parabola?

  8. anonymous
    • one year ago
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    or a straight line?

  9. anonymous
    • one year ago
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    I'm thinking since it has a vertex it's supposed to look like a parabola

  10. anonymous
    • one year ago
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    a vertex of a square is a corner of the square

  11. anonymous
    • one year ago
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    Ohhh I did not know that I'm sorry now I'm thinking of vertices ughhh okay so then it's just a straight line

  12. UnkleRhaukus
    • one year ago
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    |dw:1434176148900:dw|

  13. UnkleRhaukus
    • one year ago
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    |dw:1434176189512:dw|

  14. anonymous
    • one year ago
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    Yes!

  15. anonymous
    • one year ago
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    Thats what mine currently resembles

  16. UnkleRhaukus
    • one year ago
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    can you draw a square with this point as a corner (vertex), and with sides all equal to three?

  17. UnkleRhaukus
    • one year ago
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    (there are lots of ways you could do this )

  18. anonymous
    • one year ago
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    I'll try but I don't think mine will be accurate

  19. UnkleRhaukus
    • one year ago
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    let me see

  20. UnkleRhaukus
    • one year ago
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    are you going to choose the above point as: the top-left, top-right, bottom-left, or bottom right, corner of your square?

  21. anonymous
    • one year ago
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    |dw:1434176691232:dw|

  22. anonymous
    • one year ago
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    That's how I drew it

  23. UnkleRhaukus
    • one year ago
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    but that square has sides of length two, we want sides of length three!

  24. anonymous
    • one year ago
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    oh! my bad

  25. anonymous
    • one year ago
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    |dw:1434176939212:dw|

  26. anonymous
    • one year ago
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    I think i fixed it lol

  27. UnkleRhaukus
    • one year ago
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    Good! So, what are the coordinates of the other vertices of the square now?

  28. UnkleRhaukus
    • one year ago
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    |dw:1434177028303:dw|

  29. anonymous
    • one year ago
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    (-3,3)

  30. anonymous
    • one year ago
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    is one of them

  31. anonymous
    • one year ago
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    (0,-1)

  32. anonymous
    • one year ago
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    (-1, -3)

  33. UnkleRhaukus
    • one year ago
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    (-3,3) is not |dw:1434177119461:dw|

  34. anonymous
    • one year ago
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    oh never mind i meant -3, and 2

  35. UnkleRhaukus
    • one year ago
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    |dw:1434177260884:dw|

  36. anonymous
    • one year ago
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    are these my possible answers?

  37. anonymous
    • one year ago
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    those coordinates

  38. UnkleRhaukus
    • one year ago
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    yea, these are the three options for the square you chose

  39. anonymous
    • one year ago
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    I think I did the wrong square DX

  40. anonymous
    • one year ago
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    gosh darn it

  41. anonymous
    • one year ago
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    I had to get coordinates (-4,0) (0,-1) (1,-1) (4,1) or (5,0)

  42. anonymous
    • one year ago
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    but thank you you taught me how to do the problem :) I'm gonna try that graphing again

  43. UnkleRhaukus
    • one year ago
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    remember the coordinate pairs are (x,y), [right/left , then up/down]

  44. anonymous
    • one year ago
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    |dw:1434177584869:dw|

  45. anonymous
    • one year ago
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    I DID ITTTT!!!!!!!!!!

  46. anonymous
    • one year ago
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    HOORAHHHHHH

  47. anonymous
    • one year ago
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    but I couldn'thave done it without thee so thank you lol @UnkleRhaukus

  48. UnkleRhaukus
    • one year ago
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    Nice work!!!

  49. anonymous
    • one year ago
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    :)

  50. anonymous
    • one year ago
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    Also, if it's not too much of a problem, I've got another graphing question I could use some help on ! If not, it's totally cool! Thank you !

  51. anonymous
    • one year ago
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    Nm false alarm!

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