1. **State the problem:** We want to find how doubling the radius of a right cone affects its surface area. 2. **Formula for surface area of a right cone:** $$S = \pi r^2 + \pi r l$$ where $r$ is the radius and $l$ is the slant height. 3. **Given values:** - Original radius $r = 3$ in. - Slant height $l = 7.6$ in. 4. **Calculate original surface area:** $$S_1 = \pi (3)^2 + \pi (3)(7.6) = 9\pi + 22.8\pi = 31.8\pi$$ 5. **Double the radius:** $$r_2 = 2 \times 3 = 6$$ 6. **Calculate new surface area:** $$S_2 = \pi (6)^2 + \pi (6)(7.6) = 36\pi + 45.6\pi = 81.6\pi$$ 7. **Find the ratio of new to original surface area:** $$\frac{S_2}{S_1} = \frac{81.6\pi}{31.8\pi} = \frac{81.6}{31.8}$$ 8. **Simplify the ratio:** $$\frac{81.6}{31.8} \approx 2.566$$ 9. **Interpretation:** Doubling the radius increases the surface area by about 2.57 times. **Final answer:** The surface area is about 2.57 times the original surface area when the radius is doubled.