answer:- The area of each small square in the grid is 3 times 3 = 9 square cm, and the total area of the grid is 16 times 9 = 144 square cm. - The diameter of each of the four circles is equal to a side of a small square, or 3 cm, hence the radius of each circle is 1.5 cm. The area of one circle is pi cdot 1.5^2 = 2.25pi. The total area of the four circles is 4 times 2.25pi = 9pi. - The visible shaded area is 144 - 9pi square cm. Thus, C = 144 and D = 9, making C+D = 144 + 9 = boxed{153}.

answer:Sarah runs around a block which has a side length of 450 feet. Tim runs on a path that is 25 feet wider along each edge due to the width of the street. 1. Calculate the side length of Tim's path: - Each side of Tim's path is (450 + 2 times 25 = 500) feet (as he has to run on both the sides of the block because of the street width). 2. Calculate the perimeter of Sarah's path: - Perimeter for Sarah = (4 times 450 = 1800 ) feet. 3. Calculate the perimeter of Tim's path: - Perimeter for Tim = (4 times 500 = 2000 ) feet. 4. Find the difference: - Difference = (2000 - 1800 = 200 ) feet. So, Tim runs boxed{200} more feet than Sarah.

answer:Let's solve the problem step by step by examining both cases where the circular cap's area is included and not included in the surface area calculation. Case 1: Not including the circular cap in the surface area 1. **Define the variables:** - Radius of the sphere: (R) - Radius of the circular cap: (varrho) - Height of the spherical segment: (m) 2. **Surface area of the spherical cap:** The surface area of the spherical cap without including the base circular cap is given by: [ 2 pi R m ] 3. **Given condition:** According to the problem, the surface area of the spherical cap is (c) times the area of the circular cap: [ 2 pi R m = c pi varrho^2 ] 4. **Solve for (m):** Rearrange the equation to solve for (m): [ 2 pi R m = c pi varrho^2 ] [ m = frac{c varrho^2}{2 R} ] 5. **Use the Pythagorean theorem:** In the right triangle formed by the center of the sphere (O), the midpoint of the circular base (O_1), and the edge of the circular base (A): [ R^2 = varrho^2 + (R - m)^2 ] 6. **Substitute (varrho^2):** [ varrho^2 = 2 R m - m^2 ] 7. **Plug in (varrho^2) into the equation for (m):** [ m = frac{c (2 R m - m^2)}{2 R} ] 8. **Simplify:** [ m = frac{2 c R m - c m^2}{2 R} ] [ m = frac{2 c R m - c m^2}{2 R} ] [ 2 R m = 2 c R m - c m^2 ] [ c m^2 = 2 c R m - 2 R m ] [ c m^2 = 2 R (c - 1) m ] [ m = frac{2 R (c - 1)}{c} ] Hence, the height (m) of the spherical segment in case the circular cap's area is not included is: [ boxed{frac{2 R (c - 1)}{c}} ] Case 2: Including the circular cap in the surface area 1. **Surface area of the spherical cap:** When the base circular cap is included, the surface area becomes: [ 2 pi R m + pi varrho^2 ] 2. **Given condition:** According to the problem, the surface area of the spherical cap is (c) times the area of the circular cap: [ 2 pi R m + pi varrho^2 = c pi varrho^2 ] 3. **Solve for (m):** Rearrange the equation to solve for (m): [ 2 pi R m + pi varrho^2 = c pi varrho^2 ] [ 2 pi R m = (c - 1) pi varrho^2 ] [ m = frac{(c - 1) varrho^2}{2 R} ] 4. **Substitute (varrho^2):** [ varrho^2 = 2 R m - m^2 ] Substitute (varrho^2) into the equation for (m): [ m = frac{(c - 1) }{2 R} (2 R m - m^2) ] 5. **Simplify:** [ m = frac{(c - 1) (2 R m - m^2)}{2 R} ] [ 2 R m = (2 R m - m^2)(c - 1) ] [ 2 R m = 2 R m c - (c - 1) m^2 ] [ (c - 1) m^2 = 2 R (c - 2) m ] [ m = frac{2 R (c - 2)}{c - 1} ] Hence, the height (m) of the spherical segment in case the circular cap's area is included is: [ boxed{frac{2 R (c - 2)}{c - 1}} ] Conclusion: The height of the spherical segment could be: [ boxed{frac{2 R (c - 1)}{c}} quad text{or} quad boxed{frac{2 R (c - 2)}{c - 1}} ] depending on the definition used for the surface area of the spherical cap.

answer:(I) Solution: Let the common ratio of the geometric sequence {a_n} be q. Given a_1+a_2+a_3=26 and S_6=728, we have begin{cases} a_1(1+q+q^2)=26 dfrac{a_1(1-q^6)}{1-q}=728 end{cases}, solving these equations yields: begin{cases} a_1=2 q=3 end{cases}. Therefore, a_n=2cdot 3^{n-1}; (II) Proof: From (I), we have a_1=2, q=3, Therefore, S_n= dfrac{2(1-3^n)}{1-3}=3^n-1, Therefore, S_{n+1}=3^{n+1}-1, S_{n+2}=3^{n+2}-1, Therefore, S_{n+1}^2-S_nS_{n+2}=(3^{n+1}-1)^2-(3^n-1)(3^{n+2}-1) =3^{2n+2}-2cdot3^{n+1}+1-3^{2n+2}+3^n+3^{n+2}-1=boxed{4times3^n}.