3 years ago

Jim is running on a trail that is 5/4 of a mile long. How many miles has he run so far?

Mathematics

1 answer:

vekshin13 years ago

6 0

1 mile and 1/4 of a mile

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A triangle ABC has its vertices at A(-2, -3), B(2, 1), and C(5,-1).

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$~\hfill \stackrel{\textit{\large distance between 2 points}}{d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2}}~\hfill ~ \\\\[-0.35em] ~\dotfill\\\\ A(\stackrel{x_1}{-2}~,~\stackrel{y_1}{-3})\qquad B(\stackrel{x_2}{2}~,~\stackrel{y_2}{1}) ~\hfill AB=\sqrt{( 2- (-2))^2 + ( 1- (-3))^2} \\\\\\ AB=\sqrt{(2+2)^2+(1+3)^2}\implies AB=\sqrt{32}\implies \boxed{AB=4\sqrt{2}} \\\\[-0.35em] ~\dotfill$

$B(\stackrel{x_1}{2}~,~\stackrel{y_1}{1})\qquad C(\stackrel{x_2}{5}~,~\stackrel{y_2}{-1}) ~\hfill BC=\sqrt{( 5- 2)^2 + ( -1- 1)^2} \\\\\\ BC=\sqrt{3^2+(-2)^2}\implies BC=\sqrt{9+4}\implies \boxed{BC=\sqrt{13}} \\\\[-0.35em] ~\dotfill\\\\ C(\stackrel{x_1}{5}~,~\stackrel{y_1}{-1})\qquad A(\stackrel{x_2}{-2}~,~\stackrel{y_2}{-3}) ~\hfill CA=\sqrt{( -2- 5)^2 + ( -3- (-1))^2} \\\\\\ CA=\sqrt{(-7)^2+(-3+1)^2}\implies CA=\sqrt{49+(-2)^2}\implies \boxed{CA=\sqrt{53}}$