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Thread: 4 more math questions

  1. #1
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    4 more math questions

    in attachments

    think 17 is A. 25.6 not sure
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by dgenerationx2 View Post
    in attachments

    think 17 is A. 25.6 not sure
    you are correct (you used Pythagoras' Theorem, i suppose)

    what about the others?

    for 23 and 27 recall that the diagonals of a rectangle bisect each other (cut each other at their midpoints)

    hint for 29, note that the diagonals are the same length, so AC = BD
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  3. #3
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    Quote Originally Posted by dgenerationx2 View Post
    in attachments

    think 17 is A. 25.6 not sure
    17)
    $\displaystyle

    \overline{NL} = 2\times\overline{JL} = 2\times 12.8 = 25.6
    $

    OR

    $\displaystyle
    \overline{NL} = \overline{MK}
    $

    $\displaystyle
    \overline{MK} = \sqrt{\overline{NM}^2 + \overline{NK}^2} = \sqrt{16^2 +20^2} = \sqrt 656 \doteq 25.6
    $

    23) and 27)
    $\displaystyle
    \overline {RT} = \overline {TP} = \overline {PS}
    $

    $\displaystyle
    \sqrt{(\overline {RT}+ \overline {TP})^2 - \overline {PS}^2} = \overline {SR}
    $

    $\displaystyle
    \sqrt{(2\times\overline{RT})^2 - \overline{PS}^2} = \overline{SR}
    $

    $\displaystyle
    \sqrt{(2\times5)^2 - 5^2} = \overline {SR}
    $

    $\displaystyle
    \overline {SR} = \sqrt {75} = 5\sqrt{3} \doteq 8.7
    $

    29)
    $\displaystyle
    \overline{AC} = 3x
    $

    $\displaystyle
    \overline{BD} = 7x - 8
    $

    $\displaystyle
    \overline{AC} = \overline {BD}
    $

    $\displaystyle
    3x = 7x - 8
    $

    $\displaystyle
    4x = 8
    $

    $\displaystyle
    x = 2
    $

    $\displaystyle
    \overline{AC} = \overline {BD} = 3x = 3\times2 = 6
    $
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