in centimeters find the dimensions for a can that holds between 280cm3 and 285cm3.?
410603
Well, one possibility is radius=3.3cm and height=8.3cm., this measure is proportional with a normal can. You have to remember that a can is a cylinder, so the dimensions must match with 3.141592*r^2 * h=280 - 285
594911
Radius of the cylinder = 225 mm / 2 = 112.5 mm\n\nThe length is given in metres, so work in metres \n112.5 mm = 0.1125 m \n\nCross sectional area of cylinder = pi * r^2 = 0.03976078... sq m\n\nVolume = length * cross sectional area \n= 1 m * 0.03976078 sq m = 0.03976078 cu m\n1 cu m = 1000 litres\ntherefore volume = 39.8 litres to 3 s f
296070
36.25 cans, not figuring for waste or spillage, so answer would be 35-45. More than likely you will have to put two coats to get a good  even coverage., so you may want to double this figure.***
471793
Well here's some extra info you need to solve the problem:\n\n1 cubic inch = 0.00432900431 US gallons\n\nSo one can has a volume of:\n\nV=(pi)(5)(3/2)^2=35.34291 cubic inches\n\nThey filled 86400 cans which has a volume of:\n\nVtotal = V*86400 = 35.34291*86400 = 3053627.424 cubic inches\n\nThen you need to convert this to gallons so:\n\nVgallons = 0.00432900431 * 3053627.424 = 13219.1662 Gallons
763442
(m)\n\n48 inches = 121.92 centimeters\n48 inches = 1 219.2 millimeters\n48 inches = 1.2192 meters\n48 inches = 0.0012192 kilometers
886896
0.696850393"\n1"=25.4mm=2.54cm
975485
H*W= area\nheight times width\n\none side has a height of 8 and its width is 9 thus the area would be 72
705154
9cm3\n\n9x9x9=729!!!!!
419905
volume = 3x2x2 = 12 cubic feet\n\n1 litre = 100 cubic centimetres\n\n30 centremetres = 1 ft\n\nwork the rest out, come you can do it
84844
Height = H = 10"\nRadius = r = 4"\n\nVOLUME (not area, volume is 3-dimensional) = (π/3)(r^2*H) = (160/3)*π\n\nSURFACE AREA = (πr)(sqrt(r^2+H^2)) = 4(sqrt(116))*pi
865942
b = 70sin55\n\nb = 57.34
403093
Volume = L*W*H\nV = (15cm)(35cm)(H)\n(5.4g)/(2.7g/cm^3)=525cm^2*H\n2cm^3 = 525cm^2*H\n\nThickness or height = (2/525)cm\nT = 0.0038095238095238... the 380952 repeats
762991
Volume = pi * r^2 * h\n\nwhere\nr is the radius\nand\nh is the height\n\nFor example:\nr = 0.9m\nh = 1.2m\nvolume = 3.14159 * 0.9 * 0.9 * 1.2 = 3.0536 cubic metres or 3053.6 litres\n\n_
175063
2.54 centimeters in an inch\n25.4 milimeters in an inch
198865
Volume = (area of the base) x (height)
760371
1. The ratio of h to r is not dependent on the can being made of aluminium. The surface area is just a concept - it cannot be aluminium.\n\n2. Is a cube (assuming that no aluminium is used in forming the edges) \n\n3. You need to do this bit.\n\nFurther Investigation  - you need to do this bit , it looks as though if you limit your investigation to regular prisms that will be O.K., investigating cones would be impressive. \n\nIn so far as the wastage is concerned. If you find out what the area+wastage is for the optimally sized circular cylinders and compare with optimally sized hexagonal prisms then this may be sufficient?
732533
D = M/V\n\nDensity= Mass/Volume