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dpflan

  • 3 years ago

d=square root of (square root 2 -5 sr 2)^2 +(- sr 3 - seven sr 3)^2

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  1. dpflan
    • 3 years ago
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    all right. the equations is. \[d = \sqrt{2 - 5 \sqrt{2}} + (-\sqrt{3} - 7 * \sqrt{3})^2\]

  2. dpflan
    • 3 years ago
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    is that right?

  3. mspicer
    • 3 years ago
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    well, the 1st sr is over the whole problem....

  4. mspicer
    • 3 years ago
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    nm.....duh

  5. mspicer
    • 3 years ago
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    I've been doing this hw since yesterday, I'm in a dark zone....I am just feed up with gettin to a prob. then stuck, which thats been all 87 prob.

  6. dpflan
    • 3 years ago
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    sorry can you try rewriting the problem

  7. dpflan
    • 3 years ago
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    there are a lot of sqrt's and I can't tell which numbers of grouped under square roots and not, o_0...

  8. mspicer
    • 3 years ago
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    \[ d=\sqrt (\sqrt 2-5\sqrt2)^2+(-\sqrt 3 -7\sqrt3)^2\]

  9. mspicer
    • 3 years ago
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    It sqrt's the whole equation.... make sence? I hope

  10. dpflan
    • 3 years ago
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    ok

  11. dpflan
    • 3 years ago
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    so, let's start with each paranthetical group, focusing on the left one with root(2)

  12. dpflan
    • 3 years ago
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    can you evaluate \[\sqrt{2} - 5\sqrt{2}\]

  13. mspicer
    • 3 years ago
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    got lost...yep

  14. mspicer
    • 3 years ago
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    -4\[-4\sqrt 2\]

  15. dpflan
    • 3 years ago
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    now square it

  16. mspicer
    • 3 years ago
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    \[-8\sqrt3\]

  17. mspicer
    • 3 years ago
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    thats where I'm lost....how do you ^2 a sqrt?

  18. dpflan
    • 3 years ago
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    ok, so a square root is just this: you have a number, x, that you multiply by itself, x * x, to yield another value, x*x = y, so the square root of y is x.

  19. mspicer
    • 3 years ago
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    \[do you 16*2 then ^2 or do you ^ the 4 and the 2 separate?\]

  20. dpflan
    • 3 years ago
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    square root of 2 is actually a value less than two. if you have multiply \[\sqrt{2} * \sqrt{2}\] you'll get 2

  21. dpflan
    • 3 years ago
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    so in that case you had, y = x*x, so the square root of y is x, here you could say that y =2, therefore x = \[\sqrt{2}\]

  22. mspicer
    • 3 years ago
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    so you get 16 * 2 then square?

  23. dpflan
    • 3 years ago
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    16 *2 is the result of \[-4\sqrt{?}\] sqaured

  24. dpflan
    • 3 years ago
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    sorry, that ? is a 2

  25. dpflan
    • 3 years ago
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    i'm confused as to why you square it again?

  26. dpflan
    • 3 years ago
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    you've solved the left paranthetical group, so solve the right

  27. mspicer
    • 3 years ago
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    \[d=\sqrt (x2 -x1)^2 +(y2-y1)^2\] right?

  28. dpflan
    • 3 years ago
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    this is the distance equation, yes

  29. dpflan
    • 3 years ago
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    did you solve the problem?

  30. mspicer
    • 3 years ago
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    1024+36864 right?

  31. mspicer
    • 3 years ago
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    sqrt? right?

  32. dpflan
    • 3 years ago
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    so you have 16*2 on the left and what on the right? i get 64 * 3 on the right

  33. dpflan
    • 3 years ago
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    so that is 16*2 + 64*3

  34. dpflan
    • 3 years ago
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    squareroot( 16 *2 + 64*3 ), unless there are other operations missing in the original equation...

  35. mspicer
    • 3 years ago
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    Sorry! I keep gettin stuck with trin to post my comments to you!! I got 1024 + 36864 then sqrt 37288? am I right so far?

  36. dpflan
    • 3 years ago
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    i think the numbers you have are way too big, how did you get 1024?

  37. mspicer
    • 3 years ago
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    I squared the 16*2 and 64*3....?

  38. mspicer
    • 3 years ago
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    Are you still with me? If not then thanx for all your help...if you are then am I right or wher am I going wrong?

  39. dpflan
    • 3 years ago
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    why did you square them?

  40. dpflan
    • 3 years ago
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    you already did, that's how you have 16*2 and not -4*sqrt(2)

  41. mspicer
    • 3 years ago
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    look at the prob again.....in th () you have to get the sqrt down to be squared...

  42. mspicer
    • 3 years ago
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    If not then how do you ^ 2 a sqrt?

  43. dpflan
    • 3 years ago
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    a sqrt is just a number that if you square that number, you get another number...so if you have \[\sqrt{2}\], if you square that you get 2. if you "^2 a sqrt" those two operations are cancelling eachother out...so you did this when you found that

  44. dpflan
    • 3 years ago
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    \[(-4*\sqrt{2})^2 = 16 * 2\]

  45. dpflan
    • 3 years ago
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    right there you squared a square root

  46. mspicer
    • 3 years ago
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    oh, ok....

  47. dpflan
    • 3 years ago
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    rt(2) squared is 2

  48. mspicer
    • 3 years ago
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    so how do I do the -8 sqrt 3 ^2?

  49. dpflan
    • 3 years ago
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    the same way as you did with the left, ;) so what is -8 squared?

  50. mspicer
    • 3 years ago
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    64

  51. dpflan
    • 3 years ago
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    what is \[\sqrt{3}\] squared?

  52. mspicer
    • 3 years ago
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    3?

  53. dpflan
    • 3 years ago
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    right

  54. dpflan
    • 3 years ago
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    so you have 16*2 + 64*3

  55. mspicer
    • 3 years ago
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    so i sqrt 160?

  56. dpflan
    • 3 years ago
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    you have 32 + 192

  57. dpflan
    • 3 years ago
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    that is the value you square root

  58. mspicer
    • 3 years ago
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    the 32 is negative?

  59. mspicer
    • 3 years ago
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    no.....

  60. dpflan
    • 3 years ago
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    both value are positive, they were calculated from negative values that you then squared, and when you multiply two negative numbers you get a positive

  61. mspicer
    • 3 years ago
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    So d=14.96?

  62. dpflan
    • 3 years ago
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    right, \[\sqrt{204}\]

  63. mspicer
    • 3 years ago
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    So you are in Georgia IT?

  64. dpflan
    • 3 years ago
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    sorry, that wwas poor adding

  65. dpflan
    • 3 years ago
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    \[\sqrt{224}\]

  66. dpflan
    • 3 years ago
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    yes, i am

  67. dpflan
    • 3 years ago
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    was this helpful?

  68. dpflan
    • 3 years ago
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    where are you from?

  69. mspicer
    • 3 years ago
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    Great Help!! Thanx more than you'll ever know!! Good ol' Tennessee

  70. dpflan
    • 3 years ago
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    right on, no problem, :)

  71. mspicer
    • 3 years ago
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    Maybe you can help me again soon!! God Bless!

  72. dpflan
    • 3 years ago
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    of course, would be glad too

  73. dpflan
    • 3 years ago
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    thanks

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