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anonymous

  • one year ago

PLEASE SOMEONE HELP ITS AN SOS!!! WILL MEDAL AND FAN Find the distance between these points. C(0, 4), T(-6, -3) √(37) √(85) √(109)

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  1. anonymous
    • one year ago
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    A parallelogram whose vertices have coordinates R(1, -1), S(6, 1), T(8, 5), and U(3, 3) has a shorter diagonal of ___ . 5 √(13) √(97)

  2. anonymous
    • one year ago
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    Find the distance between these points. W(-6, -8), X(6, 8) 20 10 √8

  3. anonymous
    • one year ago
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    Find the other endpoint of a line segment with one endpoint at (-3, 5) and whose midpoint is (5, 2). (7, 2) (1, 7/2) (13, -1)

  4. DanJS
    • one year ago
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    The distance is figured using the pythagorean theorem

  5. anonymous
    • one year ago
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    ive never heard of that

  6. DanJS
    • one year ago
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    (0,4) and (-6,-3) |dw:1438902675608:dw|

  7. anonymous
    • one year ago
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    okay then what

  8. DanJS
    • one year ago
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    you can form a right triangle to get the components of a right triangle

  9. anonymous
    • one year ago
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    okay

  10. jdoe0001
    • one year ago
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    \(\bf \textit{distance between 2 points}\\ \quad \\ \begin{array}{lllll} &x_1&y_1&x_2&y_2\\ % (a,b) C&({\color{red}{ 0}}\quad ,&{\color{blue}{ 4}})\quad % (c,d) T&({\color{red}{ -6}}\quad ,&{\color{blue}{ -3}}) \end{array}\qquad % distance value d = \sqrt{({\color{red}{ x_2}}-{\color{red}{ x_1}})^2 + ({\color{blue}{ y_2}}-{\color{blue}{ y_1}})^2}\)

  11. DanJS
    • one year ago
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    |dw:1438902788817:dw|

  12. anonymous
    • one year ago
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    ohhh f.o.i.l right?

  13. DanJS
    • one year ago
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    Right, that is just memorizing the formula

  14. anonymous
    • one year ago
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    ya

  15. anonymous
    • one year ago
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    is that how i do it using foil?

  16. DanJS
    • one year ago
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    If you know how to get to it, more likely to remember even if you forget the formulas

  17. anonymous
    • one year ago
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    this is way to complicated

  18. DanJS
    • one year ago
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    That forms a right triangle, with the property d^2 = (side1)^2 + (side2)^2

  19. anonymous
    • one year ago
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    its going way over my head

  20. DanJS
    • one year ago
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    solving for Distance d, you get what @jdoe0001 typed above

  21. anonymous
    • one year ago
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    ya

  22. anonymous
    • one year ago
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    but still really confused

  23. DanJS
    • one year ago
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    \[distance = \sqrt{(\Delta x)^2 + (\Delta y)^2}\] Distance is square root of the sum of the difference in x coordinates and the difference in y coordinates of the two points

  24. anonymous
    • one year ago
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    um i dont get it still but thank you sooo much for your trouble

  25. DanJS
    • one year ago
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    (0,4) and (-6 , -3) \[distance = \sqrt{(-6 - 0)^2 + (-3 - 4)^2} = \sqrt{36 + 49} = \sqrt{85}\]

  26. anonymous
    • one year ago
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    oh so it was how i thought. I dont know why but for some reason its harder then it should be

  27. anonymous
    • one year ago
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    so all i do is foil pretty much right?

  28. DanJS
    • one year ago
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    i was just explaining what the distance formula comes from. ha sorry

  29. DanJS
    • one year ago
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    no, no Foil. You just put the numbers in and calculate it

  30. anonymous
    • one year ago
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    oh

  31. anonymous
    • one year ago
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    well guess i dont get it

  32. DanJS
    • one year ago
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    I guess all you have to take note of is, if you given 2 points (x , y) and (x1 , y1) The distance between them is calculated by.. \[distance = \sqrt{(x - x1)^2+(y-y1)^2}\]

  33. DanJS
    • one year ago
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    \[distance = \sqrt{(-6 - 0)^2 + (-3 - 4)^2} = \sqrt{(-6)^2+(-7)^2 } = \sqrt{85}\]

  34. anonymous
    • one year ago
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    oh isee what u did there

  35. DanJS
    • one year ago
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    coo, try the next one

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