6x10x4. What is the volume of a box with these deminsions? Explain you steps.?
102106
Volume is equal to legnth times width times height. The dimensions you gave are the length, width, and height. So all you do is multiply the dimensions together.\n\nVolume=6*10*4=60*4=240
790039
Units help. \nWhat are the dimensions of a standard pallet\nWhat are the dimensions of a standard 40' container (2 what high?)\n18 per layer of what?\n20x2 of what?\nTherefore 40 what? Pallets?\n\nYou'll get some answers if you answer those questions.
738861
If you are talking about a 3-D rectangle,\n\nThe volume of a rectangle is height times width times depth.  \n\nYou can also use base times height if its easier.\n\nhope this helps!
934849
291.13 L is the correct volume.\n\nCheck your algebra when you are solving for V1.
374444
100\n\n(10mm=1cm; 10x10=100)
460515
i think the volume must be, ((12*5)/2)*15 \n\nin which (12*5)/2 is the area of the one of the bases and 15 is the height since the volume is the base area times the height
508477
Not True.\n\nSolid I,\nL=1\nB=4\nH=4\nSurface Area = 2x(1x4+1x4+4x4) = 48 sq units\nVolume = 1x4x4 = 16 cu units\n\nSolid II,\nL=2\nB=2\nH=5\nSurface Area = 2x(2x2+2x5+2x5) = 48 sq units\nVolume 2x2x5 = 20 cu units\n\nSurface Area the same [48 sq units, 48 sq units],\nbut Volume is different [16 cu units, 20 cu units]
425258
volume = height X width x depth\nv = h x w x d\nv= 11 x 16 x 24\nv= 4224 inches^3 ( inches cubed )
512055
2,800\n\nYour original solution is one tenth of a mL, so multiply that by ten to get the end volume.  Do the same thing with the number of particles: 280x10=2,800.
67957
First you multiply 10x264. Then multiply the answer by 9.658. This is a math problem, not a number.
937032
I'm assuming:\nThe box has a square top and bottom.\nThe volume v is fixed.\nr is the length of a side of the top.\nh is the height of the box.\n\nv = r^2*h\nh = v/r^2\n\nThe surface area is a top and bottom both r x r and\nthe sides which are r x h.\n\nArea = 2*r^2 + 4*r*h = 2^r^2 + 4*r(v/r^2) = 2r^2 +4v/r\n\ndA/dr = 4r -4v/r^2\n\nSet to 0 and solve\n0 = 4r -4v/r^2\n\n0 = 4r^3 -4v\n0 = r^3 - v\nv = r^3, or  r = v^(1/3)\nThat means that the container is a cube.
796040
material needed=2pir(h+r)\nvolume=pir^2h\npir^2h=1000cc (given)\nh=1000/(pir^2)\nsubstituing h in the equation 2pir(h+r)\nmaterial=2pir(1000/pir^2+r)\n=2pir*1000/pir^2+2pir^2\ndA/dt=-2000/r^2+4pir\nsetting this to zero\n-2000+4pir^3=0\n4pir^3=2000\nr^3=2000/4pi\nr=cube root of 500/3.14\n=5.4 approx\nsubstituting\nh=1000/(3.14*5.4*5.4)\n=10.9 aapprox\nso the radius is 5.4 cm approx\nand the height 10.9 approx
125346
Use Heron's formula to find the area of the base: 6 times the square root of 5.  Then multiply that by the height.\n\nThe volume is 30 times the square root of 5.
492961
Yes 152 cm is correct.\n\nTo solve the problem factor 2002 = 2x7x11x13\n\nThe closer the box is to a perfect cube the lower the ratio of the surface area to the volume.\n\nThis occurs when the sides are  11, 13 and 2x7=14\n\nIn the cube, each edge appears 4 times and so the sum of the edges will be 4 times (11+13+14) = 4x38 = 152 cm
1030347
Density = Mass / Volume\n\n4.3 = 43 / Volume\n\nVolume = 43/4.3\n\nVolume = 10 cu cm\n\n* Assume your density's dimension is 4.3 g/(cu cm)
625809
convert everything to ft. (its easier). now, find the volume of the deep area (length times width times height). then divide by the volume by the bags of mulch
373349
Volume = 1/48 * 1*1 = 1/48 cu ft\ndensity = 480 pounds per cu ft\n\nmass = density*volume = 480* 1/48 lb = 10 lb