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Here_to_Help15

  • one year ago

H!E!L!P! M!E! P!L!E!A!S!E!

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  1. Here_to_Help15
    • one year ago
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    @Hero

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  2. Here_to_Help15
    • one year ago
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    @amistre64 now can you help me :D

  3. Here_to_Help15
    • one year ago
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    @Hero

  4. Here_to_Help15
    • one year ago
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    @campbell_st

  5. amistre64
    • one year ago
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    tell me your thoughts, how do we define an origin?

  6. Here_to_Help15
    • one year ago
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    Thats where it went gmergme4gtm5tmpt5p4 < Confusing i dont understand it

  7. amistre64
    • one year ago
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    i dont know what that means

  8. Here_to_Help15
    • one year ago
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    Exactly....

  9. Here_to_Help15
    • one year ago
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    I dont know what the question means either

  10. amistre64
    • one year ago
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    then you might need to google what an origin is ... otherwise we will not be speaking on the same terms

  11. Here_to_Help15
    • one year ago
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    wait i know what it means now please continue :)

  12. Here_to_Help15
    • one year ago
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    Its the starting point :)

  13. amistre64
    • one year ago
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    we want to find the origin of the signal, how do we define an origin? what is the origin on our conventional graphing system? we start at 0,0 right?

  14. Here_to_Help15
    • one year ago
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    Wow so many questions and yes right!

  15. amistre64
    • one year ago
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    this is a study session, how can we study if im the only one that does the thinking?

  16. Here_to_Help15
    • one year ago
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    Sorry its just this is our final unit and im just coming to terms with these problems :(

  17. amistre64
    • one year ago
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    consider the form: u^2 + v^2 = 36 the origin is at u=0, v=0 now compare that to your form where u = x+6, and v=y+4 how do we relate the origin to your problem?

  18. amistre64
    • one year ago
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    this is our study session, when you decide to study is not my concern :) but we will still study it regardless

  19. Here_to_Help15
    • one year ago
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    :)

  20. amistre64
    • one year ago
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    u = x+6, and v=y+4 our origin is u=0, v=0 therefore, in relation to x and y 0 = x+6 and 0 = y+4 what are our values of x and y in relation to the origin?

  21. amistre64
    • one year ago
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    you shold already know that, im not here to teach you the entirety of algebra. they cover solving for a single variable way before they cover this stuff.

  22. Here_to_Help15
    • one year ago
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    Would i get y by it self? and x?

  23. amistre64
    • one year ago
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    x is the name of a variable, there is nothing special in a name

  24. amistre64
    • one year ago
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    0 = x+6 , solve for x 0 = y+4 , solve for y

  25. Here_to_Help15
    • one year ago
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    -6=x -4=y ?

  26. amistre64
    • one year ago
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    good, so in relation to our xy plane, the origin is at: -6,-4 right?

  27. Here_to_Help15
    • one year ago
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    Yes

  28. amistre64
    • one year ago
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    |dw:1433281584913:dw|

  29. amistre64
    • one year ago
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    so in relation to the xy plane, we now know where it originates at ... our range is ... hmm, would you agree its a circle?

  30. Here_to_Help15
    • one year ago
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    Hmm i believe so

  31. amistre64
    • one year ago
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    well, the circle equation is just the distance from a central points, the radius measures the distance from the center of a circle what is our distance formula?

  32. Here_to_Help15
    • one year ago
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    \[2\sqrt{13} ?\]

  33. amistre64
    • one year ago
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    thats not a distance formula, thats a number

  34. Here_to_Help15
    • one year ago
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    Wait do we have 2 points or is it that one that i solved for ?

  35. amistre64
    • one year ago
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    we have 2 points, an origin (-6,-4), and all (or any) points (x,y) from it

  36. amistre64
    • one year ago
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    what is the distance between the points (-6,-4) and (x,y)?

  37. Here_to_Help15
    • one year ago
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    If i told you im a little lost would you leave me ?

  38. amistre64
    • one year ago
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    at this point, yes :) what is our distance formula?

  39. Here_to_Help15
    • one year ago
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    lol wait are you serious and do you want me to write the formula?

  40. amistre64
    • one year ago
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    i want you to write the formula for distance, this will help us determine the range

  41. Here_to_Help15
    • one year ago
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    Ok

  42. Here_to_Help15
    • one year ago
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    \[d=\sqrt{(x _{2}}-x _{1})^{2} + (y _{2}-y _{1})^{2}\] = ?

  43. Here_to_Help15
    • one year ago
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    Sorry im not so good with that equation button but everything is squared

  44. freeG13
    • one year ago
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    what is the question?

  45. amistre64
    • one year ago
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    good, but lets square each side (x2-x1)^2 + (y2-y1)^2 = d^2 let -6,-4 be the point x1,y1 and let x,y be some other point (x-(-6))^2 + (y-(-4))^2 = d^2 how does this compare to (x+6)^2 + (y+4)^2 = 36

  46. Here_to_Help15
    • one year ago
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    Thanks but @freeG13 i will show you if @amistre64 gives up on me :) but i doubt he will

  47. Here_to_Help15
    • one year ago
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    Are the exact thing its just the bottom is the simplified version

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  48. amistre64
    • one year ago
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    so, d^2 = 36 then, what is d?

  49. amistre64
    • one year ago
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    |dw:1433282849596:dw|

  50. Here_to_Help15
    • one year ago
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    \(\huge\tt\color{#f9bec7}{6!}\)

  51. Here_to_Help15
    • one year ago
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    Sorry openstudy is super laggy

  52. Here_to_Help15
    • one year ago
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    I knew it was 6 just it wouldnt let me type it

  53. Here_to_Help15
    • one year ago
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    Ok so its 6 now what?

  54. amistre64
    • one year ago
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    we have a range of 6 (miles kilometers feet?) in all directions from the origin

  55. amistre64
    • one year ago
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    what is our questions asking for?

  56. Here_to_Help15
    • one year ago
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    Thats the centre of the the signal

  57. Here_to_Help15
    • one year ago
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    \[(x-h)^{2} + (y-k)^{2} =r ^{2}\]

  58. Here_to_Help15
    • one year ago
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    I could have used this formula as well :)

  59. Here_to_Help15
    • one year ago
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    Basically go left 6 and down 4 from the origin and the signal would be 6 units

  60. amistre64
    • one year ago
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    of course you could have, but then we would not have had much to study ;)

  61. Here_to_Help15
    • one year ago
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    lol :) ok i know i am the biggest pain in the behind but can we do a couple more? Not like this (multiple choice)

  62. amistre64
    • one year ago
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    not at the moment, i have a nother question that needs my attention at the moment.

  63. Here_to_Help15
    • one year ago
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  64. Here_to_Help15
    • one year ago
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    I will attend it

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