The regular can has a radius of 3 cm and a height of 12.5 cmThe party-size can has a radius of 18 cm and a height of 37.5 cmaHow many square centimeters of aluminum are needed to make the regular can?(Assume Kola Kola's cans have flat bottoms and tops)bHow many square centimeters of aluminum are needed to make the party-size can?
You might consider some alternatives to traditional insulation, such as a radiant barrierRadiant barriers are a foil like product that actually repel the heat from the sun in the summer, and bounce the heat back into the living area in the winterRelatively knew, they have in immediate effect on the comfort level in the home, and reduce energy costs.
Yes blown insulationWhen you go to use the blower lint gets everywhere! So we set up a plastic room in our living room around the machineYou could set up a tint maybe, or if the tubes are long enough you could keep it outside.
That depends on if the attic is a room with a walk-up staircase or a pull down ladder and it depends if the floor or ceiling beams are exposed or covered by sheetrock or plaster In my home, I insulated the roof of the attic because I use the attic as a room If I was not using the attic as a room then I would probably have insulated between the floor joists
Urgent help in math please.? The regular can has a radius of 3 cm and a height of 12.5 cmThe party-size can has a radius of 18 cm and a height of 37.5 cm aHow many square centimeters of aluminum are needed to make the regular can?(Assume Kola Kola's cans have flat bottoms and tops) Finding the cross-sectional area of a cylinderArea 2 area of round top + area of round sheet pulled out and laid flat Area 2 pi r^2 + circumference of can height Area 2 pi r^2 + 2 pi r height Area 2 pi (3cm)^2 + 2 pi (3cm) 12.5 cm Area 18 pi cm^2 + 75 pi cm^2 Area 93 pi cm^2 Area 292.168 cm^2 bHow many square centimeters of aluminum are needed to make the party-size can? Finding the cross-sectional area of a cylinderArea 2 area of round top + area of round sheet pulled out and laid flat Area 2 pi r^2 + circumference of can height Area 2 pi r^2 + 2 pi r height Area 2 pi (18cm)^2 + 2 pi (18cm) 37.5 cm Area 648 pi cm^2 + 1350 pi cm^2 Area 1998 pi cm^2 Area 6276.90 cm^2
