Digital ACT Math Practice Question #2103
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Notice $\Delta x = 3, \Delta y = 4 \implies \text{hypotenuse} = 5$. Then $5$ and $\Delta z = 12 \implies 5-12-13$ triangle! $d = 13$.
The 3D distance formula is $d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2 + (z_2-z_1)^2} = \sqrt{(5-2)^2 + (3-(-1))^2 + (-8-4)^2} = \sqrt{3^2 + 4^2 + (-12)^2} = \sqrt{9 + 16 + 144} = \sqrt{169} = 13$....
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