Equate the volume of the hemisphere to the volume of the water displaced in the cylinder.

You're on a roll — here are the solutions to the problems. 15. A cylinder has a radius of 10.5cm. A hemispherical solid is fully immersed in the cylinder containing water and the level water rose by a height of 6.2cm. Calculate the radius of the hemisphere to 1 d.p. Step 1: Equate the volume of the hemisphere to the volume of the water displaced in the cylinder. The volume of water displaced is equal to the volume of the hemisphere. Volume of hemisphere (V_h) = (2)/(3) r_h^3, where r_h is the radius of the hemisphere. Volume of water displaced in cylinder (V_c) = R_c^2 h_w, where R_c is the radius of the cylinder and h_w is the height the water level rose. V_h = V_c (2)/(3) r_h^3 = R_c^2 h_w Step 2: Substitute the given values into the equation. Given R_c = 10.5 cm and h_w = 6.2 cm. (2)/(3) r_h^3 = (10.5)^2 (6.2) Step 3: Solve for r_h. Cancel from both sides: (2)/(3) r_h^3 = (10.5)^2 (6.2) (2)/(3) r_h^3 = 110.25 × 6.2 (2)/(3) r_h^3 = 683.55 Multiply both sides by (3)/(2): r_h^3 = (683.55 × 3)/(2) r_h^3 = (2050.65)/(2) r_h^3 = 1025.325 Take the cube root of both sides: r_h = [3]1025.325 r_h ≈ 10.0838 cm Step 4: Round the radius to 1 decimal place. r_h ≈ 10.1 cm The radius of the hemisphere is 10.1 cm. 16. Complete the figure below to show rotational order 3. To complete the figure to show rotational order 3, the final shape must look identical after a rotation of 360^ / 3 = 120^ about its center. The small dot in the figure indicates the center of rotation. The given figure shows one "point" or "arm" of a three-pointed star. To complete it, you need to draw two more identical points, each rotated by 120^ from the previous one around the central dot. The completed figure will be a three-pointed star, where each point is identical to the one shown and is positioned at 120^ intervals around the center. What's next? 📸