1. **Problem 1: Find the perimeter of triangle ABC with vertices A(9,9), B(4,-4), and C(-4,1).** 2. Use the distance formula between two points $P(x_1,y_1)$ and $Q(x_2,y_2)$: $$d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$$ 3. Calculate each side length: - $AB = \sqrt{(4 - 9)^2 + (-4 - 9)^2} = \sqrt{(-5)^2 + (-13)^2} = \sqrt{25 + 169} = \sqrt{194}$ - $BC = \sqrt{(-4 - 4)^2 + (1 - (-4))^2} = \sqrt{(-8)^2 + (5)^2} = \sqrt{64 + 25} = \sqrt{89}$ - $CA = \sqrt{(9 - (-4))^2 + (9 - 1)^2} = \sqrt{(13)^2 + (8)^2} = \sqrt{169 + 64} = \sqrt{233}$ 4. The perimeter $P$ is the sum of the side lengths: $$P = AB + BC + CA = \sqrt{194} + \sqrt{89} + \sqrt{233}$$ 5. Approximate the values: - $\sqrt{194} \approx 13.928$ - $\sqrt{89} \approx 9.434$ - $\sqrt{233} \approx 15.264$ 6. Therefore, $$P \approx 13.928 + 9.434 + 15.264 = 38.626$$ 7. **Problem 2: Find the circumference of a circle centered at the origin passing through (-3,0).** 8. The radius $r$ is the distance from the center $(0,0)$ to the point $(-3,0)$: $$r = \sqrt{(-3 - 0)^2 + (0 - 0)^2} = \sqrt{9} = 3$$ 9. The circumference $c$ of a circle is given by: $$c = 2\pi r$$ 10. Substitute $r=3$: $$c = 2\pi \times 3 = 6\pi$$ 11. Approximate: $$c \approx 6 \times 3.1416 = 18.85$$ **Final answers:** - Perimeter of triangle ABC $\approx 38.63$ - Circumference of the circle $= 6\pi \approx 18.85$