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anonymous

  • one year ago

can some one help me?

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  1. anonymous
    • one year ago
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    https://media.glynlyon.com/g_geo_2012/9/q415a.gif

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  2. anonymous
    • one year ago
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    Prove: In an equilateral triangle the three medians are equal.

  3. jim_thompson5910
    • one year ago
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    what do you have so far?

  4. anonymous
    • one year ago
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    absolutely nothing ive been staring at the page for an hour

  5. jim_thompson5910
    • one year ago
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    |dw:1439083430829:dw|

  6. jim_thompson5910
    • one year ago
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    |dw:1439083442888:dw|

  7. jim_thompson5910
    • one year ago
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    |dw:1439083487262:dw|

  8. jim_thompson5910
    • one year ago
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    do you see how P is the midpoint of AB?

  9. anonymous
    • one year ago
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    yes

  10. jim_thompson5910
    • one year ago
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    they use the midpoint formula to find point P

  11. jim_thompson5910
    • one year ago
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    that means you add up the x coordinates and divide by 2 |dw:1439083618310:dw|

  12. jim_thompson5910
    • one year ago
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    make sense?

  13. anonymous
    • one year ago
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    somewhat

  14. jim_thompson5910
    • one year ago
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    I added the x coordinates 0 and 2a to get 2a then I divided 2a by 2 to get 'a'

  15. jim_thompson5910
    • one year ago
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    |dw:1439083746502:dw|

  16. anonymous
    • one year ago
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    ok got it

  17. jim_thompson5910
    • one year ago
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    the same happens with the y coordinates |dw:1439083780104:dw|

  18. anonymous
    • one year ago
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    ok

  19. jim_thompson5910
    • one year ago
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    |dw:1439083807218:dw|

  20. jim_thompson5910
    • one year ago
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    so hopefully you see how (a,0) was found

  21. anonymous
    • one year ago
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    ya think so

  22. jim_thompson5910
    • one year ago
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    the same idea is applied to find Q and R

  23. UsukiDoll
    • one year ago
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    feels like a proof... like verify (a,0) using A and B

  24. anonymous
    • one year ago
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    okay

  25. anonymous
    • one year ago
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    i think i understand.. thanks

  26. UsukiDoll
    • one year ago
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    |dw:1439083816461:dw|

  27. jim_thompson5910
    • one year ago
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    for the other stuff, you use the distance formula \[\Large d = \sqrt{\left(x_{2}-x_{1}\right)^2+\left(y_{2}-y_{1}\right)^2}\]

  28. anonymous
    • one year ago
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    kk

  29. anonymous
    • one year ago
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    |dw:1439083949599:dw|

  30. UsukiDoll
    • one year ago
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    ???

  31. UsukiDoll
    • one year ago
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    this assignment looks painful D: but I sort of see where they are going.

  32. anonymous
    • one year ago
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    it is very painful i have 7 more

  33. UsukiDoll
    • one year ago
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    clicked close by accident.

  34. UsukiDoll
    • one year ago
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    so we need Q and R hmmm Q is (3a/2,b/2) R is (a/2, b/2)

  35. anonymous
    • one year ago
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    lol i wish. It only lets me do that for quixs and test

  36. UsukiDoll
    • one year ago
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    no I said I accidentally closed the tab on my end XD

  37. anonymous
    • one year ago
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    ohhhhh lol

  38. UsukiDoll
    • one year ago
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    so I couldn't reach you for a bit XD

  39. UsukiDoll
    • one year ago
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    if Q ( 3a/2, b/2) then there are points that can lead us to this. if R ( a/2,b/2) then there are points that lead to this result points A and C may lead us to R

  40. UsukiDoll
    • one year ago
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    A (0,0) C (a,b) |dw:1439084550670:dw|

  41. anonymous
    • one year ago
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    okay

  42. UsukiDoll
    • one year ago
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    |dw:1439084709293:dw|

  43. UsukiDoll
    • one year ago
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    a HA! B AND C can make Q!

  44. UsukiDoll
    • one year ago
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    B(2a,0) C( a,b) (2a+a/2, 0+b/2) Q(3a/2,b/2)

  45. UsukiDoll
    • one year ago
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    |dw:1439084868914:dw|

  46. UsukiDoll
    • one year ago
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    |dw:1439084931274:dw|

  47. anonymous
    • one year ago
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    ohh!!! i understand now thank you sooo much!!!!

  48. UsukiDoll
    • one year ago
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    I think midpoint formula and distance formula is bring used here.

  49. UsukiDoll
    • one year ago
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    QA hmm Q(3a/2, b/2) A (0,0)

  50. UsukiDoll
    • one year ago
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    |dw:1439085179111:dw|

  51. UsukiDoll
    • one year ago
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    \[\Large d = \sqrt{\left(x_{2}-x_{1}\right)^2+\left(y_{2}-y_{1}\right)^2} \] \[\Large d = \sqrt{(0-\frac{3a}{2})^2+(0-\frac{b}{2})^2} \] \[\Large d = \sqrt{(-\frac{3a}{2})^2+(-\frac{b}{2})^2} \] \[\Large d = \sqrt{(\frac{9a^2}{4})+\frac{b^2}{4}} \] if we let \[\large b = a \sqrt{3} \] \[\Large d = \sqrt{(\frac{9a^2}{4})+\frac{(a \sqrt{3})^2}{4}} \] \[\Large d = \sqrt{(\frac{9a^2}{4})+\frac{(3a^2)}{4}} \] \[\Large d = \sqrt{\frac{12a^2}{4}}\] \[\Large d = \sqrt{3a^2} \] \[\Large d = a\sqrt{3} \] Now do the same process for RB for R ( a/2,b/2) and B (2a,0)

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