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mathmath333

  • one year ago

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  1. anonymous
    • one year ago
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    @ganeshie8

  2. mathmath333
    • one year ago
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    |dw:1440718821577:dw|

  3. mathmath333
    • one year ago
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    \(\large \color{black}{\begin{align} & \normalsize \text{Find the number of rectangles.}\hspace{.33em}\\~\\ \end{align}}\)

  4. mathmath333
    • one year ago
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    the lines are perpendicular to each other.

  5. ChillOut
    • one year ago
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    Begin by choosing two pairs of horizontal lines and two pairs of vertical lines so you can have a rectangle.

  6. anonymous
    • one year ago
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    I just know theres a ton

  7. ChillOut
    • one year ago
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    Let \(\left(\begin{matrix}a \\ 2\end{matrix}\right)\) be the number of vertical lines that can be picked, and \(\left(\begin{matrix}b \\ 2\end{matrix}\right)\) the number of horizontal lines that can be picked. For each pair of horizontal lines, there's a pair of vertical lines that will form a rectangle with them. Therefore you have \(\left(\begin{matrix}n \\ 2\end{matrix}\right)\)*\(\left(\begin{matrix}b \\ 2\end{matrix}\right)\) as the number of rectangles.

  8. ChillOut
    • one year ago
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    That would be my guess :)

  9. mathmath333
    • one year ago
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    n=4,b=6 ?

  10. ChillOut
    • one year ago
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    Yes. And sorry, meant to be "a" in the last line.

  11. mathmath333
    • one year ago
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    ok thanks

  12. ChillOut
    • one year ago
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    No problem, do you have an answer to the problem?

  13. mathmath333
    • one year ago
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    no

  14. mathmath333
    • one year ago
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    i m relying on u

  15. ChillOut
    • one year ago
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    All right then!

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