Verified Answer :

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\(P(7,2) \\ x_1=7 \\ y_1=2 \\ \\ Q(2,5) \\ x_2=2 \\ y_2=5 \\ \\ \hbox{the distance between the points:} \\ d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}=\sqrt{(2-7)^2+(5-2)^2}= \\ \sqrt{(-5)^2+3^2}=\sqrt{25+9}=\sqrt{34} \approx 5.83\)

The distance is 5.83 units.

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