Thread: Math Question, hard

  1. #16
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    yeah.

    I was trying with logarithms

    Oskilian

  2. #17
    & the hat of GPL slaying Thantos's Avatar
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    2 and 4 are correct. Now whats the second set of numbers?

  3. #18
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    y^2 = x is a function

    >>no it isn´t, a function must have only one X value for each Y >>value, and this one doesn´t (for example: if X=4, y can be 2 >>or -2)

    Osk, you're trying to writ some code where you need to understand 3D. OK, that's fair. But, then admit that you can have more than one value of each X and Y.

    For example, let's say I want to define a circle in 3D (even the circle has only 2D, it's in a space with 3D):

    1) x^2 + y^2 = 1
    2) z = 0

    The ecuation 1) defines the limits (I don't know how to express this in Enlgish) of the function in the X-Y plane, and the function 2) express the position of the circle in the Z axis.

    And the function Y^2 = x is used many times when you need a 3D boundary to calculate for example triple integrals, which give you the volume of a body.

    By the way, I took this examples from my first year at uni (5 or 6 years ago I think).

    And these functions are quite simple, because a value of x has only two values of y; there are functions that have more y values for each x value. Like the Rosae Function, this function draws like a rose, but I don't know if you call it like this in English as well. If I find something in my notes I'll post it.
    Last edited by Gades; 12-04-2001 at 03:26 AM.

  4. #19
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    Originally posted by Thantos
    2 and 4 are correct. Now whats the second set of numbers?
    x = 4
    y = 2

    really, thats trivial

  5. #20
    A Banana Yoshi's Avatar
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    That the same one as mine first answer

    x=2
    y=4
    --------------------------
    engineer223
    Yoshi

  6. #21
    & the hat of GPL slaying Thantos's Avatar
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    No re-read my comment. I said 2 and 4 were correct. But there is another correct set of numbers. Meaning that 2 and 4 isn't the only set of correct numbers.

  7. #22
    Is it -2 and -4

  8. #23
    & the hat of GPL slaying Thantos's Avatar
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    Yeppers

  9. #24
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    Re: y^2 = x is a function

    a circle is not a function, it's an equation, but not a function

    Oskilian

  10. #25
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    y = sqrt(x)
    It passed the vertical line test so it is a function

    Next time, make "y" alone before calculating.
    -------------------
    Engineer223
    Last edited by Yoshi; 12-06-2001 at 07:10 PM.
    Yoshi

  11. #26
    Back! ^.^ *Michelle*'s Avatar
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    GEEZ! Engine, you don't get my point...

    > y˛ = x is a function
    sqrt both sides

    y = sqrt(x)
    It passed the vertical line test so it is a function<

    That was exactly what I said, instead I said it the other way around... You just proved my point, y sqrt(x) is a function while y^2=x is exact same equation but not a function. ONLY y = sqrt(x) is a function the other one is not. The definition for a function, is that it's linear graph thing is not touched twice by a vertical line. And we are talking about 1D here ( is there such a thing as 1D?...) ...
    Michelle (o^.^o)
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  12. #27
    Linguistic Engineer... doubleanti's Avatar
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    the thing with your examples is whether to decide to take the negative portion of a sqrt... and it's just a matter of orientation...
    hasafraggin shizigishin oppashigger...

  13. #28
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    you must undertand that you cannot do that!

    the inverse function of the second power IS NOT int sqare root!!!

    graphing y^2=x will give you a different graph than y=sqrt(x)

    now this is true:

    y^2=x
    y=(+/-)sqrt(x)

    but neither of them are functions, let me copy the "function" definition from my Calculus book

    Originally written by Earl W. Swokowsky
    An f function of a D group over an E group is a correspondence which assigns each element x of D one unique element y of E
    one UNIQUE element; in the equation y^2=x, y=2 and y=-2 are assigned to x=4. There are two solutions for the eqation, and THAT makes y^2 a non-function

  14. #29
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    YES!!! urrrggghh....

    you see if you start with

    y=ROOT(x)

    that IS a function.

    but if you move the root sign over, you will get y^2=X

    which IS NOT a function. (because x will be both -2 and 2)

    YOU DON'T START WITH y^2=x, ok?
    Michelle (o^.^o)
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  15. #30
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    Do I have to graph it?

    Proof
    y=sqrt(x)
    ---------------------
    Engineer223
    Last edited by Yoshi; 12-05-2001 at 10:57 PM.
    Yoshi

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