Question #23ae3

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Question #23ae3

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Answer 1 · 2017-11-06T02:30:22

about 42.7903455

Explanation:

$A = (- 9, - 6)$ $A=(-9, -6)$
$B = (- 5, 0)$ $B=(-5, 0)$
$C = (10, - 7)$ $C=(10,-7)$

Find the length of line segments $¯ ¯¯¯¯ ¯ A B$ $bar(AB)$ , $¯ ¯¯¯¯ ¯ B C$ $bar(BC)$ , and $¯ ¯¯¯¯ ¯ A C$ $bar(AC)$

$d_{A B}$ $d_(AB)$
$= \sqrt{{(x_{2} - x_{1})}_{2}^{2} + {(y_{2} - y_{1})}_{2}^{2}}$ $=sqrt((x_2-x_1)^2+(y_2-y_1)^2)$
$= \sqrt{{((- 5) - (- 9))}^{2} + {((0) - (- 6))}^{2}}$ $=sqrt(((-5)-(-9))^2+((0)-(-6))^2)$
$= \sqrt{{(4)}^{2} + {(6)}^{2}}$ $=sqrt((4)^2+(6)^2)$
$= \sqrt{16 + 36}$ $=sqrt(16+36)$
$= \sqrt{52}$ $=sqrt(52)$
etc.

$d_{A B} = \sqrt{52}$ $d_(AB)=sqrt(52)$
$d_{B C} = \sqrt{274}$ $d_(BC)=sqrt(274)$
$d_{A C} = \sqrt{362}$ $d_(AC)=sqrt(362)$
Total $= \sqrt{52} + \sqrt{274} + \sqrt{362} \approx 42.7903455$ $=sqrt(52)+sqrt(274)+sqrt(362)~~42.7903455$

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Question #23ae3

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