1. **State the problem:** We need to find the perimeter of a rectangle with vertices at approximately $(-3,0)$, $(-1,6)$, $(2,5)$, and $(0,-1)$. The perimeter is the total distance around the rectangle. 2. **Recall the formula:** The perimeter $P$ of a rectangle is given by $$P = 2(\text{length} + \text{width})$$ where length and width are the lengths of adjacent sides. 3. **Find the lengths of the sides:** We use the distance formula between two points $A(x_1,y_1)$ and $B(x_2,y_2)$: $$d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$$ 4. **Calculate side lengths:** - Side 1 between $(-3,0)$ and $(0,-1)$: $$d_1 = \sqrt{(0 - (-3))^2 + (-1 - 0)^2} = \sqrt{3^2 + (-1)^2} = \sqrt{9 + 1} = \sqrt{10} \approx 3.162$$ - Side 2 between $(0,-1)$ and $(2,5)$: $$d_2 = \sqrt{(2 - 0)^2 + (5 - (-1))^2} = \sqrt{2^2 + 6^2} = \sqrt{4 + 36} = \sqrt{40} \approx 6.325$$ - Side 3 between $(2,5)$ and $(-1,6)$: $$d_3 = \sqrt{(-1 - 2)^2 + (6 - 5)^2} = \sqrt{(-3)^2 + 1^2} = \sqrt{9 + 1} = \sqrt{10} \approx 3.162$$ - Side 4 between $(-1,6)$ and $(-3,0)$: $$d_4 = \sqrt{(-3 - (-1))^2 + (0 - 6)^2} = \sqrt{(-2)^2 + (-6)^2} = \sqrt{4 + 36} = \sqrt{40} \approx 6.325$$ 5. **Identify length and width:** Opposite sides are equal, so length $= 6.325$ and width $= 3.162$. 6. **Calculate perimeter:** $$P = 2(6.325 + 3.162) = 2(9.487) = 18.974$$ 7. **Round to 1 decimal place:** $$P \approx 19.0$$ **Final answer:** The perimeter of the rectangle is $19.0$ units.