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Thread: helllp

  1. #1
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    Exclamation helllp

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  2. #2
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    Re: helllp

    The points described are points with either $\displaystyle y = 2x$ or $\displaystyle x = 2y$. Let's solve for $\displaystyle y$ in the second equation: $\displaystyle y = \dfrac{1}{2}x$. Graph these two equations and see what you wind up with.
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  1. helllp
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