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Math Help - Pre-calculus Challenge!

  1. #1
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    Smile Pre-calculus Challenge!

    Who can help me do this challenge? I need urgent help please.
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  2. #2
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    mr fantastic's Avatar
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    Quote Originally Posted by rlarach View Post
    Who can help me do this challenge? I need urgent help please.
    You have posted many of these numerous problems already (problems 3, 4 and 6 immediately come to mind, there's probably others) and help has been given for them. If that help was insufficient for you, I suggest you find the appropriate threads, show what working you've done and state where you're stuck.

    Personally, I find this current post academically unethical, an insult to the intelligence of this forum and a gross abuse of this service. There may be some members who don't have a problem in doing all your assignment work for you, but from what I've seen here I doubt it. You've shown no evidence that you're willing to put any effort into your work or understanding the help you've been given - I think you'll find that such an attitude will result in very few future replies.

    Bottom line: Show the work you've done, state where you're stuck. Show some effort.
    Last edited by mr fantastic; January 5th 2008 at 02:39 PM. Reason: replaced "my intelligence" with the stuff in red - perhaps taking a liberty .....
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  3. #3
    Member OnMyWayToBeAMathProffesor's Avatar
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    agree

    I also agree with "mr fantastic". I use this forum only when I have tried to do a problem and i get stuck. I enter in the problem, what work i have done and why. This also helps other help you. They can then fully understand you, and provide a fitting explanation.
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  4. #4
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    Hello, rlarach!

    Here's some help . . .


    1) Determine the area of the region enclosed by the curve:
    . . |x-1| + |y-1| \:=\:1
    If we sketch the graph, the answer is obvious.
    Code:
            |
            |   (1,2)
            |     *
            |   *   *
            | *       *
       (0,1)*           *(2.1)
            | *       *
            |   *   *
        - - + - - * - - - - - 
            |   (1,0)
            |

    2) Let x be a positive real number not equal to 1.
    Determine the region of the x-y plane containing the solution
    to the inequality: . \log_x\left[\log_x(y^2)\right] \;>\;0

    We have: . \log_x\left[\log_x\left(y^2\right)\right] \:>\:0\quad\Rightarrow\quad \log_x\left(y^2\right) \:>\:x^0

    Then: . \log_x(y^2) \:>\:1\quad\Rightarrow\quad y^2 \:>\:x^1

    Hence: . x\:<\:y^2


    Code:
          |::::: ::::::::*
          |::::: :::*
          |:::::*
          |::*
          |*
        - + - - + - - -
          |*    1
          |::*
          |:::::*
          |::::: :::*
          |::::: ::::::::*

    This is the region to the left of the parabola: x \,=\,y^2
    . . and to the right of the y-axis, except where x = 1

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