one square yard would take 9 blocks
and one cubic yard would take 27 blocks,
that is if the question is written correctly
the sides of the cubical box would be 7 meters
253.125 blocks.
When you say 1 foot block, are they square blocks with side equal to 1 foot, in that case you would need 240 blocks
I believe it would have nine edges.
There are nine edges.
9 cause the volume is 9
1 ft = 12 in → question is how many inch cubes fit in a 12 in by 12 in by 12 in cube. 12 small blocks fit across the front of the box in a row 12 rows of these 12 small blocks fit back from the back in a layer of blocks 12 layers of these blocks fit in the box Thus you will need 12 × 12 × 12 = 1,728 blocks.
the sides of the cubical box would be 7 meters
96 g
253.125 blocks.
210
When you say 1 foot block, are they square blocks with side equal to 1 foot, in that case you would need 240 blocks
The modifications needed to make a 1982 'vette as fast as an Enzo would go beyond a couple of 'blocks', unfortunately.
The only information you've given is that there is a surface with an area of 50,000 ft2. We need to know the size of the blocks to know how many are going to fit. If we assume for the blocks length L and width W, then all we can do for now is assume this formula: 50,000 ft2/L*W = number of the blocks needed
Approximately 1667 cubical grains of NaCl would equal one millimeter, assuming each grain has a side length of 1 mm.
The coefficient of cubical expansivity is a measure of how the volume of a substance changes with temperature. It is defined as three times the linear coefficient of thermal expansion. It is denoted by the symbol β and has units of K^-1.
Yes. All prisms have edges. Edges are defined as where two faces meet. The base of the prism would have 6 edges since the base would be a hexagon. The second base (or the top) would also have 6 edges since it would have to be another hexagon. lastly, there would be 6 edges connecting the 6 vertices of the bases for a total of 18 edges.