## College Algebra (11th Edition)

$3\sqrt{10}$
We find the distance between the first and fourth points, $(0,-6)$ and $(3,3)$ by using the distance formula: $d=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2)}$ $d=\sqrt{(3-0)^{2}+(3-(-6))^{2}}$ $=\sqrt{3^{2}+9^{2}}$ $=\sqrt{9+81}$ $=\sqrt{90}$ $=\sqrt{9*10}$ $=3\sqrt{10}$