Respuesta :
[tex]~~~~~~~~~~~~\textit{distance between 2 points} \\\\ (\stackrel{x_1}{-2}~,~\stackrel{y_1}{3})\qquad (\stackrel{x_2}{-7}~,~\stackrel{y_2}{-7})\qquad \qquad d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2} \\\\\\ d=\sqrt{(~~-7 - (-2)~~)^2 + (~~-7 - 3~~)^2} \implies d=\sqrt{(-7 +2)^2 + (-7 -3)^2} \\\\\\ d=\sqrt{( -5 )^2 + ( -10 )^2} \implies d=\sqrt{ 25 + 100 } \implies d=\sqrt{ 125 }\implies d\approx 11.18[/tex]
Exact Distance = [tex]\boldsymbol{5\sqrt{5}}[/tex] units
Approximate Distance = 11.1803 units
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Work Shown:
[tex](x_1,y_1) = (-2,3) \text{ and } (x_2, y_2) = (-7,-7)\\\\d = \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2}\\\\d = \sqrt{(-2-(-7))^2 + (3-(-7))^2}\\\\d = \sqrt{(-2+7)^2 + (3+7)^2}\\\\d = \sqrt{(5)^2 + (10)^2}\\\\d = \sqrt{25 + 100}\\\\d = \sqrt{125}\\\\d = \sqrt{25*5}\\\\d = \sqrt{25}*\sqrt{5}\\\\d = 5\sqrt{5}\\\\d \approx 11.1803\\\\[/tex]
The exact distance is [tex]5\sqrt{5}[/tex] units, which approximates to about 11.1803 units
I used the distance formula. Round the approximate value however your teacher instructs.
A slight alternative is to plot the two points to form a right triangle. The hypotenuse goes from (-2,3) to (-7,-7). Then use the pythagorean theorem.
You can use tools like WolframAlpha or GeoGebra to confirm the answer.
