Afirework rocket consists of a cone stacked on top of a cylinder, where the radii of the cone and the cylinder are equal. the diameter of the cylindrical base of the rocket is 8 in and the height of the cylinder is 5 in, while the height of the cone is 3 in. calculate the surface area of the rocket. leave your answer in terms of π.
184π sq. in.
76π sq. in.
168π sq. in.
88π sq. in.

The surface area of the rocket is 76π inches² ⇒ 2nd answer

Step-by-step explanation:

* Lets explain how to solve the problem

- A firework rocket consists of a cone stacked on top of a cylinder

- The radii of the cone and the cylinder are equal

∴ Their bases are equal

- The diameter of the cylindrical base of the rocket is 8 inches

∴ The radius of the cylinder = 8 ÷ 2 = 4 inches

- The height of the cylinder is 5 inches

- The height of the cone is 3 inches

- The surface area of the rocket is the sum of the lateral area of the

 cylinder , area of one base of the cylinder and the lateral area of

 the cone

∵ The lateral are of the cylinder = perimeter of its base × its height

∵ The base of the cylinder is circle

∵ The perimeter of the circle (circumference) = 2πr

∵ r = 4 and h = 5

∴ The lateral area of the cylinder = 2π(4)(5) = 40π inches²

- Lets calculate the area of the base of the cylinder

∵ The area of the circle = πr²

∴ The area of the circle = π(4)² = 16π inches²

- To find the lateral area of the cone you must find the slant height l

∵ The slant height of the cone = √ (r² + h²) , where r is the radius of

  the base of the cone and h is the height of the cone

∵ r = 4 and h = 3

∴ l = √(4² + 3²) = √(16 + 9) = √25 = 5 inches

∵ The lateral area of the cone = 1/2 Cl , where C is the circumference

  of the base and l is the slant height

∵ C = 2πr

∴ C = 2π(4) = 8π

∵ l = 5

∴ The lateral area of the cone = 1/2 (8π)(5) = 20π inches²

- Lets find the surface area of the rocket

∵ The S.A of the rocket = L.A of the cylinder + A of the base of the

   cylinder + L.A of the cone

∵ L.A of the cylinder = 40π inches²

∵ A of the base of the cylinder = 16π inches²

∵ L.A of the cone = 20π inches²

∴ The S.A = 40π + 16π + 20π = 76π inches²

* The surface area of the rocket is 76π inches²

step-by-step explanation:

part a. the differences are decreasing, so we know the sequence is not arithmetic. the ratios are constant (=3/4), so we know the sequence is geometric.

part b. from part a, we know the common ratio is 3/4, so each term is 3/4 of the one before. the initial term is the number of gallons at h=1.

g[h] = (3/4)·g[h-1]

g[1] = 384 . . . . initial value required for a recursive definition

part c. the general formula for a geometric sequence is

a[n] = a[1]·r^(n-1)

we have a[1] = 384, r = 3/4, and the term number given by h. the formula that follows is

g[h] = 384·(3/4)^(h-1)

part d. filling in the value and doing the arithmetic, we have

g[12] = 384·(3/4)^11 ≈ 384·0.042235136 ≈ 16.2183 . . gallons

part e. each gallon is 4 quarts, so

16.2183 gal × 4 qt/gal = 64.8732 qt

so mark-up is 0.4 multiplied by 180 = 72