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Math Help - class help please!

  1. #1
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    class help please!

    I've been needing some help with my algebra class.. thank you!


    The ordered pair (2, 16) is a solution for which equation(s)? Check all that apply.
    A. y = 2x2
    B. y = (2x)2
    C. y = 2x + 2
    D. y = (x + 2)2


    Here is my final problem I'm stuck at.. I read through my book several times but it does not mention what I do... My teacher said we whent over it in class.. But being absent has it's downs. =/
    Last edited by mr fantastic; January 12th 2010 at 05:45 PM. Reason: Removed various requests etc.
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  2. #2
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    Quote Originally Posted by daniking View Post


    The ordered pair (2, 16) is a solution for which equation(s)? Check all that apply.
    A. y = 2x2
    B. y = (2x)2
    C. y = 2x + 2
    D. y = (x + 2)2

    For each equation substitute in x=2, if the result gives 16 then it is a solution, i'll do the first one

    y=2x^2

    x=2

    y=2(2)^2= 2\times 4 = 8 \neq 16
    Last edited by mr fantastic; January 12th 2010 at 05:50 PM.
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  3. #3
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    Quote Originally Posted by pickslides View Post
    For each equation substitute in x=2, if the result gives 16 then it is a solution, i'll do the first one

    y=2x^2

    x=2

    y=2(2)^2= 2\times 4 = 8 \neq 16

    Ummmm Yeah that answer you gave me was wrong lol.

    But it still helped me out so ty (:
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  4. #4
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    Quote Originally Posted by daniking View Post
    Ummmm Yeah that answer you gave me was wrong lol.

    But it still helped me out so ty (:
    pickslides knows that this option is wrong. In fact, he clearly indicated that it was wrong. He started with option A and showed it was wrong as an example of how to follow his good advice:
    Quote Originally Posted by pickslides
    For each equation substitute in , if the result gives 16 then it is a solution
    The expectation was that you would test the remaining options in the same way until you found the option that worked.

    Please try to focus on understanding the help you are given instead of trying to score a cheap point.

    Thread closed.
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