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anonymous

  • one year ago

find the direct relationship between x and y and determine whether the parametric equations determine y as a function of x x=(1/3)t-1 y^2=t^2

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  1. IrishBoy123
    • one year ago
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    get t in terms of x from this one: x=(1/3)t-1

  2. anonymous
    • one year ago
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    can you explain some more?

  3. IrishBoy123
    • one year ago
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    you have x=(1/3)t-1, ie an equation giving x in terms of t work it so that you have t in terms of x, ie t = .... use normal algebra

  4. anonymous
    • one year ago
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    is this correct so far? \[t=((1/3)-1)/x\]

  5. whovianchick
    • one year ago
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    y=3x+3, yes it's a function

  6. IrishBoy123
    • one year ago
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    isn't it \[x=\frac{1}{3}t-1\]??? all you have to do is move the t across to the RHS and the x across to the RHS, applying usual rules of maipulation

  7. anonymous
    • one year ago
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    oh so just t= \[t=\frac{ 1 }{3 }x-1\]

  8. whovianchick
    • one year ago
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    Well, you have to find out if y is a function of x. first of all, if x=1/3t-1, then 3x=t-3, and t=3x+3. Then if y^2=t^2, y=t, so y=3x+3. Which is a function. I don't know where you got t=1/3x-1. That's not right

  9. IrishBoy123
    • one year ago
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    no, i said "applying usual rules of manipulation". you can't just lift them out(!!) and switch them around an example if \(x = \frac{2}{3} t + 4\) then \(x-4 = \frac{2}{3} t\) \(\frac{2}{3}t = x-4\) \(t = \frac{3}{2} x - 6\)

  10. IrishBoy123
    • one year ago
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    @whovianchick no, it is not a function but vinny has some work to do before we get there

  11. whovianchick
    • one year ago
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    Except that your math is wrong and you're forgetting y. Recheck your math. You can't just switch t and x, you have to solve for t. It's called 7th grade math. I'm in 11th grade math.

  12. anonymous
    • one year ago
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    sorry his wording confused me its t=3x-3 correct?

  13. IrishBoy123
    • one year ago
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    not quite

  14. IrishBoy123
    • one year ago
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    sign on the 3??

  15. anonymous
    • one year ago
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    positve

  16. IrishBoy123
    • one year ago
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    and when you have done that we are told that \(y^2 = t^2\) so stuff your new equation into that

  17. anonymous
    • one year ago
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    ugh sorry im working on little sleep and silly mistakes are getting me'

  18. whovianchick
    • one year ago
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    |dw:1442362440473:dw|

  19. IrishBoy123
    • one year ago
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    thanks whovian:-)

  20. IrishBoy123
    • one year ago
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    keep going vinny, we are nearly done!

  21. anonymous
    • one year ago
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    y=3x=3

  22. whovianchick
    • one year ago
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    You're welcome. You had him working with the wrong equation, and I'm nitpicky like that

  23. anonymous
    • one year ago
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    ah they are squared gh

  24. IrishBoy123
    • one year ago
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    yes so just square them, DO NOT multiply/expand out the x side,...!

  25. IrishBoy123
    • one year ago
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    "You had him working with the wrong equation," when?

  26. whovianchick
    • one year ago
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    if y^2 = t^2, then y=t Replace y for t in the equation: y=3x+3, which is a function

  27. anonymous
    • one year ago
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    yeah thats what i did

  28. anonymous
    • one year ago
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    thanks guys

  29. IrishBoy123
    • one year ago
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    no vinny it is NOT a function

  30. whovianchick
    • one year ago
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    No prob. Sorry @IrishBoy123 earlier he said something like 1/3x-1 = t and I thought it was you

  31. IrishBoy123
    • one year ago
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    you will see why if we finish this properly but if you are really tired, go to bed

  32. whovianchick
    • one year ago
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    @IrishBoy123 UMMMMMMMMMMMM y=3x+3 IS a function

  33. IrishBoy123
    • one year ago
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    it is a function, thatis ture, but that is not what we have here

  34. IrishBoy123
    • one year ago
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    "thatis ture" that is true

  35. whovianchick
    • one year ago
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    |dw:1442362767682:dw|

  36. whovianchick
    • one year ago
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    That's the answer, I'm confused what you're saying

  37. IrishBoy123
    • one year ago
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    @vinny0515 are you done for the day?

  38. IrishBoy123
    • one year ago
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    this is what we have here |dw:1442362770822:dw|

  39. whovianchick
    • one year ago
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    No it's not, and both are functions

  40. anonymous
    • one year ago
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    so the answer for the quesrion is y=3x+3

  41. IrishBoy123
    • one year ago
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    no it is not

  42. whovianchick
    • one year ago
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    |dw:1442362943289:dw|

  43. whovianchick
    • one year ago
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    What is the answer, then? We've replaced y for t, and we're done. What the heck are you talking about?

  44. whovianchick
    • one year ago
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    y=3x+3 is the answer

  45. anonymous
    • one year ago
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    how do i answer it then

  46. whovianchick
    • one year ago
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    The way I told you to. y=3x+3, and yes it's a function

  47. IrishBoy123
    • one year ago
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    \(y^2=(3x+3)^2\) is what you have if you want to simplify it you can say \(y= \color{red}\pm (3x+3)\) but whatever you do, x maps to 2 y values it is not a function

  48. anonymous
    • one year ago
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    irsih boy is right

  49. whovianchick
    • one year ago
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    @IrishBoy123 I understand that. But it's saying as one line, is every solution a function. The answer is yes. Both of those equations are functions

  50. IrishBoy123
    • one year ago
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    it is not a function as you can get 2 y's for every x y = 3x+3 is a function y = -(3x+3) is a function this is both and is not a function |dw:1442363265605:dw|

  51. IrishBoy123
    • one year ago
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    and that's us done thank you both:p

  52. anonymous
    • one year ago
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    so i write the answer as y is not a function

  53. whovianchick
    • one year ago
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    Y is a function. I think you're forgetting the concept of a function, @IrishBoy123 , it's ok if there are multiple y's for every x. What you don't want is multiply x's for every y. THAT is the definition of a function, and even with your skewed logic, it's STILL a function. Stop confusing @vinny0515. YOU ARE WRONG

  54. IrishBoy123
    • one year ago
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    |dw:1442432889018:dw|

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