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Math Help - College Algebra problems..

  1. #1
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    College Algebra problems..

    Hey, I have some question on college algebra
    1. Is the line y=|x| symetric about the line y=x

    2. How do you prove points A(-1,2) B(3,-2) C(-6,-3) are vertices of a right triangle?

    3. given f(x)= 2x if 0<x<1, 1-x if 1=<x<2 , 0 if 1=<x=<3: find the domain,f(1), f(2), f(3), and f(0.1).

    4. object thrown moves at a height in meters per second is represented With h(t)=20t-4.9t^2: find the height in x seconds, if time increases by 2 how migher is the object, how high does it move per second, if time increased by h how much higher is it?
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  2. #2
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    Quote Originally Posted by daniel4616
    Hey, I have some question on college algebra
    1. Is the line y=|x| symetric about the line y=x
    No, it's symmetric about x=0.

    For x>=0 it is the same as the line y=x
    For x<0 it is the same as y=-x.

    RonL
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  3. #3
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    Quote Originally Posted by daniel4616
    Hey, I have some question on college algebra


    2. How do you prove points A(-1,2) B(3,-2) C(-6,-3) are vertices of a right triangle?
    You compute the lengths of the sides AB, BC and AC, then show that
    they satisfy Pythagoras' theorem (AB^2+AC^2=BC^2).

    RonL
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  4. #4
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    Quote Originally Posted by daniel4616
    3. given f(x)= 2x if 0<x<1, 1-x if 1=<x<2 , 0 if 1=<x=<3: find the domain,f(1), f(2), f(3), and f(0.1).
    A. You appear to have an error in the function definition, the way you have
    defined it f(x) has two values for 1=<x<2. I will assume that the last
    part of the definition should be: f(x)=0 2=<x=<3.

    B. The domain is the set on which the function is defined. So in this case
    f(x) is defined for all x such that 0<x=<3.

    C. 1 lies in the second specified range so f(1)=1-1=0.
    2 lies in the third specified range so f(2)=0, same for 3.
    0.1 lies in the first specified range so f(0.1)=2x0.1=0.2.

    RonL
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  5. #5
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    Quote Originally Posted by daniel4616

    4. object thrown moves at a height in meters per second is represented With h(t)=20t-4.9t^2: find the height in x seconds, if time increases by 2 how migher is the object, how high does it move per second, if time increased by h how much higher is it?
    Try to reword this more closely to the original. As it is I would be guessing
    at what you actually want to ask.


    RonL
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  6. #6
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    Quote Originally Posted by CaptainBlack
    You compute the lengths of the sides AB, BC and AC, then show that
    they satisfy Pythagoras' theorem (AB^2+AC^2=BC^2).

    RonL
    I know how to do it, the problem is that when I find the lengths, I dont get an euqation equal to a2+b2=c2
    I get AB= sqrt32
    BC= sqrt 82
    AC=sqrt 50

    The height of an object thrown in the air depends on the time it has been thrown. One situation the hieght in meters of an object after t seconds can be represented by h(t)=20t-4.09t^2. Whats the hieghts after x seconds, if time nicreases by 2 how much higher is the object, how much higher is the object per second nicrease? how much higher is the object per unit increase?
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  7. #7
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    Hello, daniel4616!

    1. Is the line y=|x| symetric about the line y=x ?

    Did you make a sketch?
    Code:
                  |
          *       |       *
            *     |     *
              *   |   *
                * | *
          - - - - * - - - -
                  |



    2. How do you prove points A(-1,2) B(3,-2) C(-6,-3) are vertices of a right triangle?

    An alternate method . . .


    Find the slopes of AB, BC, and AC.

    m_{AB} = \frac{-2 - 2}{3-(-1)} = \frac{-4}{4} = -1

    m_{BC} = \frac{-3-(-2)}{-6-3} = \frac{-1}{-9} = \frac{1}{9}

    m_{AC} = \frac{-3 - 2}{-6-(-1)} = \frac{-5}{-5} = 1


    Since m_{AB} and m_{AC} are negative reciprocals: AB \perp AC


    Therefore: \angle BAC = 90^o and \Delta ABC is a right triangle.



    #4 is very badly worded . . .
    4. An object thrown is a height represented by: h(t)\:=\:20t-4.9t^2
    (a) Find the height in x seconds
    (b) If time increases by 2 seconds, how much higher is the object? . . . should be asked last
    (c) How high does it move per second? . . . How high? or "how far"?
    (d) If time increased by h, how much higher is it? . . . but h represents height

    (a) In x seconds, the height is: h(x)\:=\:20x - 4.9x^2


    (d) Suppose time is increased from t to t + k.

    Its height at time t is: h(t)\:=\;20t - 4.9t^2 meters.

    Its height at time t+k is: h(t+k)\;=\;20(t+k) - 4.9(t+k)^2
    . . . = \;20t + 20k - 4.9t^2 - 9.8kt - 4.9k^2

    The difference is: (20t + 20k - 4.9t^2 - 9.8kt - 4.9k^2) - (20t - 4.9t^2)

    Therefore, it is: 20k - 9.8kt - 4.9k^2 meters higher.


    (b) If k = 2, we have: h(t + 2)\;=\;20(2) - 9.8(2)t - 4.9(2^2)\;=\;20.4 - 19.6t meters higher.

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  8. #8
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    [QUOTE=daniel4616]I know how to do it, the problem is that when I find the lengths, I dont get an euqation equal to a2+b2=c2
    I get AB= sqrt32
    BC= sqrt 82
    AC=sqrt 50
    QUOTE]

    AB^2+AC^2=32+50=82=BC^2

    RonL
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