Chapter 8 - Triangle Trigonometry and Polar Coordinates - 8.3 Polar Coordinates - Exercises and Problems for Section 8.3 - Exercises and Problems - Page 347: 18

$$(\sqrt{109},1.28)$$

Converting the Cartesian coordinates to polar coordinates, we find: $$r=\sqrt{\left(-3\right)^2+10^2} = \sqrt{109} \\ θ=\arctan \left(\frac{10}{-3}\right) = 1.28$$ Thus, it follows: $$(\sqrt{109},1.28)$$