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Math Help - Help me this problem

  1. #1
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    Help me this problem

    A man walks 5 miles North, 3 miles West, 8 miles South, and 7 miles East. What is his distance from his starting point?

    sri
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  2. #2
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    Mark a point on a grid representing where he starts (0,0). Draw a line 5 units up (north), 3 units left (west), 8 units down (south), 7 units right (east).

    Now you have a starting point and an ending point. The distance between two points (a,b) and (c,d) is given by the formula \sqrt{(c-a)^2+(d-b)^2}. Can you solve this now?

    You can also do it in your head if you want. 5-8 = -3, so he walked a net distance of 3 miles south. 7-3 = 4, so he walked a net distance of 4 miles east.

    If he started at (0,0) he ended at (4,-3) is that what you arrived at?
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  3. #3
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    Quote Originally Posted by sri340 View Post
    A man walks 5 miles North, 3 miles West, 8 miles South, and 7 miles East. What is his distance from his starting point?

    sri
    immagine this on a coordinate system

    North is positive y south i negative y west is negative x and east is positive x.

    is he starts at the origin (0,0) then

    north 5 is (0,0)+(0,5)=(0,5)
    west 3 is (-3,0)+(0,5)=(-3,5)
    south 8 is (0,-8)+(-3,5)=(-3,-3)
    east 7 is (7,0)+(-3,-3)=(4,-3)

    so he is 5 miles from the origin becuase he is on a 3,4,5 right triangle

    or we could use the distance formula

    d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}

    using the ending (4,-3) and starting (0,0) points we get...

    d=\sqrt{(4-0)^2+(-3-0)^2}=\sqrt{(4)^2+(-3)^2}=\sqrt{16+9}=\sqrt{25}=5

    I hope this helps.
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