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anonymous

  • one year ago

Is relation t a function? Is the inverse of relations t a function? Relation t x 0 2 4 6 y -8 -7 -4 -4

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  1. anonymous
    • one year ago
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    A. Relation t is a function. The inverse of relation t is not a functions. B. Relation t is not a function. The inverse of relation t is a function. C. Relation t is not a function. The inverse of relation t is not a function.

  2. anonymous
    • one year ago
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    @UsukiDoll

  3. UsukiDoll
    • one year ago
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    ok so to determine a function we use the vertical line test for a graph. Similarly we use a horizontal line test for a graph to determine a one to one function. but this time we don't have a graph. Instead we have a chart. This may be a while back since I've last did charts though If I remembered correctly the function must have the x not repeating and the one to one function (inverse) must have the y not repeating

  4. anonymous
    • one year ago
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    So A?

  5. UsukiDoll
    • one year ago
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    yeah... I'm still trying to remember though... this was way back in 2008 when I did this.

  6. anonymous
    • one year ago
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    Any help is better than no help

  7. UsukiDoll
    • one year ago
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    ok... now I got my memory back after I read a paragraph somewhere .. we can't have duplicates for x or y

  8. anonymous
    • one year ago
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    So the answer is still A, correct?

  9. UsukiDoll
    • one year ago
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    there's an example in my book somewhere.. I have to find it before giving out an answer.

  10. anonymous
    • one year ago
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    ok, ill wait

  11. UsukiDoll
    • one year ago
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    ok that was mapping and proofs. It's too strong for this problem... but I do know that we can't have anything repeating

  12. anonymous
    • one year ago
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    Im gonna go with A than because the top row isn't repeating

  13. UsukiDoll
    • one year ago
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    the original function has a repeated value for y though..

  14. anonymous
    • one year ago
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    So c?

  15. Astrophysics
    • one year ago
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    Well for it to be a one - to - one function the x and y values are to be used only once. So if you can switch the x and y variables around to create a inverse relation, we would notice function t would not be a one - to - one function since there are two of the same y values, then we can conclude the inverse is not a function.

  16. UsukiDoll
    • one year ago
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    if we switch.. that would mean that there exists x values that are repeating.

  17. anonymous
    • one year ago
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    So it's C @Astrophysics

  18. UsukiDoll
    • one year ago
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    Ok I'm seeing this horizontally .. I'm going to draw vertical |dw:1435810073365:dw|

  19. UsukiDoll
    • one year ago
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    alright so there are repeats.. since we can't have that.. it's not a function .

  20. UsukiDoll
    • one year ago
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    so switching to grab the inverse |dw:1435810139494:dw| same problem occurs

  21. UsukiDoll
    • one year ago
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    Since we have repetitions or duplicates of x-values with different y-values, then this relation ceases to be a function.

  22. Astrophysics
    • one year ago
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    Well lets put it like this, |dw:1435810074921:dw| it is not a function since the x value of -4 now has two different y values, so it's not a function, I hope that makes sense

  23. Astrophysics
    • one year ago
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    yeah exactly

  24. anonymous
    • one year ago
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    So the answer is C correct?

  25. anonymous
    • one year ago
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    Since if it gets reversed than there would be duplicates on x and y

  26. anonymous
    • one year ago
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    @Astrophysics

  27. UsukiDoll
    • one year ago
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    if it's reversed we have duplicate x values if it's not reversed we have duplicate y values. either way it's not a function since each number has to be unique

  28. Astrophysics
    • one year ago
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    Yeah, that sounds good to me, since it's not 1 - to - 1.

  29. anonymous
    • one year ago
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    I have another question

  30. Astrophysics
    • one year ago
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    Note that for a 1 - to - 1 function it has to pass both vertical and horizontal line test.

  31. UsukiDoll
    • one year ago
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    no... 1 to 1 is horizontal line test only

  32. anonymous
    • one year ago
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    What is the inverse of the given function? y = 7x^2 - 3.

  33. UsukiDoll
    • one year ago
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    for graphs

  34. UsukiDoll
    • one year ago
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    ok first switch the x and y

  35. UsukiDoll
    • one year ago
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    for y=7x^2-3

  36. Astrophysics
    • one year ago
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    I just googled it to make sure and look |dw:1435810563144:dw|

  37. anonymous
    • one year ago
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    so make it x=7y^2-3

  38. UsukiDoll
    • one year ago
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    yes now solve for y

  39. anonymous
    • one year ago
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    I wish i was as good at math as you guys are

  40. UsukiDoll
    • one year ago
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    If no horizontal line intersects the graph of the function more than once, then the function is one-to-one.

  41. UsukiDoll
    • one year ago
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    :P

  42. UsukiDoll
    • one year ago
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    \[x=7y^2-3\] k now solve for y

  43. UsukiDoll
    • one year ago
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    so we need y by itself

  44. anonymous
    • one year ago
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    So how do i factor it to itself?

  45. anonymous
    • one year ago
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    7*7=49 49-3=46

  46. UsukiDoll
    • one year ago
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    whoa.

  47. anonymous
    • one year ago
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    Am i totally off

  48. UsukiDoll
    • one year ago
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    you want the inverse of the function. This has nothing to do with factoring

  49. anonymous
    • one year ago
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    So how do i do an inverse?

  50. Astrophysics
    • one year ago
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    Oh okkkk yes, I understand @UsukiDoll I get what you mean now haha, I kept saying it has to be 1 to 1 for it to pass both, but it just needs to pass horizontal for it to be 1 to 1 and if it passes a vertical line test it is just a function! :) Ok now I'll let you do your thing!

  51. UsukiDoll
    • one year ago
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    we were on the right track so our original equation y=7x^2-3 we switch the x and y x=7y^2-3 now we need to solve for y

  52. UsukiDoll
    • one year ago
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    to have y by itself we need to ______ 3 to both sides

  53. anonymous
    • one year ago
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    add

  54. anonymous
    • one year ago
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    \[y \pm \sqrt{\frac{ 3+x }{ 7}}\]

  55. UsukiDoll
    • one year ago
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    x+3=7y^2 \[\frac{x+3}{7}=y^2 \rightarrow \sqrt{\frac{x+3}{7}}, -\sqrt{\frac{x+3}{7}}\] my x is in a different place but it wouldn't matter since the answer is the same

  56. anonymous
    • one year ago
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    yay we figured it out! *high five*

  57. UsukiDoll
    • one year ago
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    yay :D *high fives*

  58. anonymous
    • one year ago
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    haha

  59. anonymous
    • one year ago
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    It's a function if we assume \(\{0,2,4,6\}\) is the domain

  60. Hero
    • one year ago
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    For the original question posted, I don't agree with "C" as an answer.

  61. anonymous
    • one year ago
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    How come @Hero

  62. Astrophysics
    • one year ago
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    I don't either actually, because I kept saying 1 to 1 function, that is true it's not a 1 to 1 function, but however it is a function as the x does not repeat itself.

  63. Hero
    • one year ago
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    Plot each point, then perform a vertical line test: https://www.desmos.com/calculator/ogb1oqtoi2

  64. anonymous
    • one year ago
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    So i was right when i said it is A?

  65. Astrophysics
    • one year ago
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    That seems better

  66. anonymous
    • one year ago
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    It's A

  67. Hero
    • one year ago
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    A is correct.

  68. Astrophysics
    • one year ago
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    Sorry for the confusion @EllenJaz17 :P

  69. anonymous
    • one year ago
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    Thanks guys!

  70. anonymous
    • one year ago
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    No problem, at least im learning

  71. anonymous
    • one year ago
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    A function is: Left total - all elements in the domain used at least once. Many to One - all elements in the domain used no more than once.

  72. UsukiDoll
    • one year ago
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    thanks for the memory refresher x never had repeats for the function , but switch x and y and now there are repeats

  73. anonymous
    • one year ago
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    Mono, did you draw that avatar?

  74. UsukiDoll
    • one year ago
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    no...

  75. Astrophysics
    • one year ago
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    Haha

  76. UsukiDoll
    • one year ago
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    yeah it was A for that question

  77. UsukiDoll
    • one year ago
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    but like I said I've done this way back . so it's not 100% until @wio showed up and I saw what he was talking about.

  78. UsukiDoll
    • one year ago
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    then it popped back into my mind... there's a bunch of Math topics that don't appear any more when you study further

  79. Astrophysics
    • one year ago
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    Yeah it was a good review, I think last time I did a problem like this was in middle school

  80. UsukiDoll
    • one year ago
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    the graph versions.. those were the ones I knew and functions... well my book had a strong approach to it... it was about mapping... surjection/injection bijection so I highly doubt it would be used.

  81. anonymous
    • one year ago
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    For the function f(x) = (8-2x)2 ,find f-1 . Determine whether f-1 is a function.

  82. anonymous
    • one year ago
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