A poster company has decided that all of its posters will be shipped in cylindrical containers with a height of 17 inches and a circumference of 2 inches. The curved surface area of the container will be wrapped with a paper label containing the company logo. Calculate the curved surface area wrapping of the container. (this is all they gave me as a source I need help asap!)
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Calculate the volume of a cylindrical can whose height is 15.4 cm and whose base diameter is 10.2 cm. (round to one decimal place)
2. A poster company has decided that all of its posters will be shipped in cylindrical containers with a height of 17 inches and a circumference of 2 inches. The curved surface area of the container will be wrapped with a paper label containing the company logo. Calculate to the nearest square inch, the curved surface area wrapping of the container.
3. The Great Pyramid of Giza is considered to be one of the Seven Wonders of the Ancient World. It is approximately 147 meters tall, and has a square base with sides measuring approximately 440 meters. Calculate the volume to the nearest cubic meter.
4. Cups Inc. is a firm which makes conical paper cups for water dispensers. One of the paper cups it makes has a diameter of 6 cm, a height of 7 cm and no base. Determine the slant height of the paper cup to the nearest 0.1 cm.
5. To the nearest whole number, how many ml of water will a paper cup hold?
6. Determine how much paper is required, to the nearest square cm, to make a cup.
7. Calculate the surface area of a soccer ball with a diameter of 21 cm.
8.Calculate the volume of a spherical ornament with a radius of 4 cm.
Battleaxe
9 years ago
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1) V = (pi)(r^2)(h) = (3.14)(5.1^2)(15.4) = 1257.7 cm^3
2) Surface area = circumference times height, (2)(17) = 34 in^2
3) For a square, isosceles pyramid; V = (1/3)(L^2)(H) = (1/3)(440^2)(147) = 9,486,400 m^3
4) Assume a right triangle with legs of 7cm and 3 cm (the radius). Slant height is L:
L^2 = 7^2 + 3^2 so L = SQR(58) = 7.6 cm
5) V = (1/3)(pi)(r^2)(h) = (1/3)(3.14)(9)(7) = 65.9 cm^3
6) SA(cone, neglecting the base) = (pi)(r)(L) we determined L in problem (4):
SA = (3.14)(3)(7.6) = 71.6 cm^2 not including the overlap needed for gluing the cup together.
7) Assuming that the soccer ball is a smooth sphere, A = (4)(pi)(r^2) = (4)(pi)(10.5^2) = 1384.7 cm^3
8) V = (4/