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Math Help - 3 questions that I need help on?

  1. #1
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    3 questions that I need help on?

    You don't need to answer all the questions one is enough. I need this by tomorrow morning if possible.

    How do I find all the asymptotes (oblique, vertical, horizontal) of this function?
    (3x-2)/(square root of (2x^2 + 1))? Am I supposed to rationalize the denominator because I can't seem to do it?


    Also this is kinda hard to explain but I need to find the point marked on this picture that uses the shortest combined length of lines. Disregard the faint stuff in the background. The measurements are 5 ft, 12 ft, and 3 ft.
    [url=http://i3.photobucket.com/albums/y98/DaAzNJRiCh/CCI00005-1.jpg[/url]
    The second part of the question is that if the line on the left costs double the price of the line of the right then where should the point be?



    Final question is a plane is going to take off. It is taxiing along at 1 ft/s before it speeds up at a constant acceleration. If it takes 900 ft to take off and takes off at a velocity of 93 ft/s, then what is the acceleration function, velocity function, and position function?

    Thanks
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  2. #2
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    Quote Originally Posted by DaAzNJRiCh View Post

    How do I find all the asymptotes (oblique, vertical, horizontal) of this function?
    (3x-2)/(square root of (2x^2 + 1))? Am I supposed to rationalize the denominator because I can't seem to do it?
    The vertical asympotes (informally) exist when numerator is non-zero and denominator is zero. Since the only way the denominator to be zero is,
    2x^2+1=0
    Which is not possible, thus it does not have.

    The horizontal asymptotes are the limits at +oo and -oo which are +2 and -2 respictively.
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  3. #3
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    Quote Originally Posted by DaAzNJRiCh View Post
    Final question is a plane is going to take off. It is taxiing along at 1 ft/s before it speeds up at a constant acceleration. If it takes 900 ft to take off and takes off at a velocity of 93 ft/s, then what is the acceleration function, velocity function, and position function?
    I will assume a positive direction to the direction the plane is taxiing in initially. The origin will be the point where the acceleration starts. We need to find the acceleration so
    v^2 = v0^2 + 2a(x - x0)
    The plane starts at the origin with a speed v0 = 1 ft/s. It then accelerates over a distance x = 900 ft and takes off at a speed of v = 93 ft/s.

    93^2 = 1^2 + 2a(900 - 0) gives a = 4.805 ft/s^2.

    a is constant so the acceleration "function" is a = 4.805 ft/s^2
    The speed function will be v = 1 + 4.805t ft/s
    The position function will be x = t + 2.4025t^2 ft.

    (The equations are: v = v0 + at and x = x0 + vo*t + (1/2)a*t^2 respectively.)

    -Dan
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