How many dimensions would you need for the cube of a three-dimensional object?
340934
I think there's several ways to conceptualize this stuff.  First of all, a number raised to a power doesn't necessarily involve any extra dimensions at all.  3^3 = 9, and 9 is a number that can be graphed on a one dimensional line.  Consider, what dimension does 4^1.5 exist on, the 1.5th dimension?  That doesn't make much sense.  The answer is simply 4^1.5 = 8, and 8 can also be graphed on a one dimensional line.  Or how about 3^pi?  Do we get the pi'th dimension?  Mmm, a dimension of pie...  Er, what was I saying?\n\nAnyway if we want to have a real cube in three dimensional space, we need at least three points in 3D space to represent it (actually we can represent it with 6 pieces of info instead of the 9 x,y,z's that make up the 3 points).  So what happens when we "cube" those three points, e.i., raise the points of that cube to the 3rd power?  We end up with a bigger cube in three dimensional space, just like we ended up with a bigger number in the one dimensional space example before.\n\nHowever, the answer to your question might possibly be 5 dimensions.  If you think of a number to the first power as being in the first dimension (on a line graph), and you cube it, it only moves up two dimensions, from the 1st to the 3rd (a cubic graph of space).  So taking a 3 dimensional number and cubing it would also move it up just two more dimensions from the 3rd to the 5th.  I can't think of any way to write that with a conceptually valid math expression, though.\n\nI'm not exactly sure how we'd get to 6 dimensions, unless you're thinking something like x^3*x^3 = x^(3+3) = x^6, but that doesn't seem reasonable to me, either.\n\nInstead, let's try this:\n3^1 = 3\n3^2 = 9, our previous number times our base number, 3.\n3^3 = 27, our previous number times our base number, 3.  So the cube of 3, our starting example, is 27.\n\nWhat's the cube of our cube (27)?  I.e., what's 27^3?\n27^1 = 27\n27^2 = 729 (27 times 27)\n27^3 = 19683 (729 times 27)\n\nSo for our final number, we want to find out how many "dimensions" it took to reach that, right?  So, what exponent of 3 does 19683 represent?  In other words, 3^x = 19683, what's x?  To find that out, we need to convert it to logarithmic form.  log_base3(19683) = x.  x = 9.  On a standard calculator, enter log(19683)/log(3) to find that out.  And you can check the answer by doing 3^9, which comes out to be 19683.  You can also follow that whole process using starting numbers other than 3.\n\nTo write it out more formally, (x^3)^3 = x^(3*3), by one of the properties of exponents, which = x^9.\n\nOr what if we did x^(3^3), which would be x^27.  27 "dimensions" seems far too excessive for the cube of a cube, to me.\n\nSo, my final answer is 9 "dimensions", though with the caveat that I think "dimensions" is misleading.
735050
The total number of cubes will be:\nn * (n + 3) * (n + 9) cubes.\n\nTo get the number of cubes inside, subtract 2 from each dimension:\nInside cubes = (n - 2) * (n + 1) * (n + 7)\n\nNow expand these out...\nn * (n + 3) * (n + 9)\n= n (n² + 12n + 27)\n= n^3 + 12n² + 27n\n\nInside cubes:\n(n - 2) * (n + 1) * (n + 7)\n= (n-2) * (n² + 8n + 7)\n= n^3 + 8n² + 7n - 2n² - 16n - 14\nGroup like terms:\n= n^3 + 6n² - 9n - 14\n\nRemember that the inside cubes are exactly half of the total cubes, so double it:\n= 2n^3 + 12n² - 18n - 28\n\nNow equate this to the total cubes: \nn^3 + 12n² + 27n = 2n^3 + 12n² - 18n - 28\n\nPull everything to one side (so subtract n^3 + 12n² + 27n):\nn^3 - 45n - 28 = 0\n\nPossible answers for n will be the factors of 28:\n±1, ±2, ±4, ±7, ±14, ±28\n\nBy inspection, 7 works:\n7^3 = 343\n-45 * 7 = -315\n343 - 315 - 28 = 0\n\nSo the answer is:\n7 x 10 x 16 = 1120 blocks\n\n(Note: inside you have 5 x 8 x 14 = 560 blocks)
359632
I'm guessing that you're asking what does it mean to say something is 2.5 dimensional, as opposed to 2 dimensional or 3 dimensional.  \n\nAnytime you have a fractional dimension like that, you're dealing with a fractal.  That's where the word "fractal" comes from.  The dimension for an object that is self-similar at many size scales in a complex way can be thought of as an extension of the obvious fact that a uniform object of unit length d in two dimensions has an area of d^2.  In three dimensions, a uniform object of unit  d (a cube with sides d units long) has a volume of d^3, and so on.  \n\nSo, you can generalize this concept and say that something has the dimensionality of whatever power the scale along one axis has to be raised to in order to give the area/volume/whatever.   Since a fractal object has repetition at all scales of some sort of complex pattern, the relationship between scale and total size, the dimensionality, is a non-integer.\n\nSo, for example, if you had a volume which had been "Swiss cheesed" by some sort of algorithmic removal of sub-volumes, and sub-volumes of the remaining volumes, and so on, and so on, to make a fractal sponge, it would have a dimensionality between 2 and 3.
305696
5\nIt is helpful if you know the first 5 or 6 cubes.\n1^3 = 1\n2^3 = 8\n3^3 = 27\n4^3 = 64\n5^3 = 125\n6^3 = 216
590580
Volume = length x width x height = 3*3*3 = 27 cubic inches
752121
If you're familiar with 2 D coordinates, 3 D coordinates should be no problem.  2 dimensions (with 2 (x,y) coordinates needed to describe a point) describe a flat surface (area) with length and width.  3 dimensions describe a space (volume) length, width and depth.  To describe a point 3 (x,y,z) values are needed.  4D adds time so that 4 values are needed to describe a point (in time).  Trigonometry and linear algebra become very important and helpful to solve problems.
1103163
no, two- dimensional objects aren't real in that sense... even a sheet of paper (as close to 2-d as you can get in the physical world) is really 3 dimensional and does have wieght and mass.
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The Cube has 43,252,003,274,489,856,000 different possible configurations.\n\nI cheated and looked up the answer, but to figure it out you'll want to use combinatorics.  I'll have to work on that part of it...\n\nIf you're taking the cube apart and counting how many different ways you can put it back together, this is a relatively simple problem (although it doesn't match the above answer).\n\nThe center cubes don't move at all, so there is only one possible configuration of them.\n\nThere are 12 edge cubes, which each have 2 outside edges and therefore two orientations for each location.  The number of possible combinations of these edges is (2*12)*(2*11)*...*(2*2)*(2*1) = (2^12)*12!\n\nThere are 8 corner cubes, which each have 3 outside edges and therefore three orientations for each location.  The number of possible combinations of these corners is (3*8)*(3*7)*...*(3*2)*(3*1) = (3^8)*8!\n\nMultiply the number of possible combinations of edge cubes by the number of possible combinations of corner cubes and you end up with 519,024,039,293,878,300,000 combinations.  \n\nThis is different than the answer I looked up because there are possible combinations when you take the cube apart that cause it to no longer be solvable.  I can't think of an easy way to find the number of combinations that keep solvability.
77442
It implies that the box is 10 square so each dimension is 10.  Volume of a cube = h x d x w , or in this case 10 x 10 x 10 = 1000
805518
These numbers are cubes. 1 cubed (1 x 1 x 1) is 1, 2 cubed (2 x 2 x 2) is 8, 3 cubed (3 x 3 x 3) is 27, and 4 (4 x 4 x 4) cubed is 64. Therefore, the next number is 125 or 5 cubed (5 x 5 x 5). This pattern continues with 6 cubed, 7 cubed, etc.
600049
The three dimensions they are most likely talking about are the x,y, and z planes in space. Time is the fourth dimension, and there could be more after that. read stephen hawkings book A Brief History of Time. He explains how to conceptualize dimensions in your head very well.
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You can divide the cube into 9 1x3x3 slices.\n\nEach one contains the following prisms:\n1x1x2: 12\n1x1x3: 6\n1x2x2: 4\n1x2x3: 4\n1x3x3: 1\nSubtotal 27 x 9 = 81\n\nYou can divide the cube into 6 2x3x3 slices.\nEach one contains the following prisms that weren't counted above:\n\n2x2x2: 4\n2x2x3: 4\n2x3x3: 1\n\nSubtotal 9 x 6 = 54\n\nThat's 135.\n\nIf you count the 3x3x3, you have 136.
445986
Just multiply the three numbers together - length * width * height.
643342
A cube is like a dice/ it has 6 sides. Also part of the definition of a cube is each side is a square. So it is also 6 squares. Each square is 7 x7 or 7 "squared" so each of the 6 sides is 49cm squared. cm x cm is cm squared, just like 7 x 7 is 7 squared. Don't forget there are 6 of those, multiply 49 by 6 for your total answer in cm squared.
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Mathematically n dimensions can be described but our experience is only 3 dimensions So the properties of fourth dimensonal shapes can be mathematically described and a fourth dimensional quantity that is in units^4 can also be calculated\n\nThis is an argument that the 4th dimension is time. That cannot be the case when one considers the properties in 3d are L, L, and L One would immediately envisage that the 4th, 5th .... nth dimensions would also all be L\n\nA tesseract, the 4 dimensional equivalent to the 2 dimensional square and 3 dimensional cube, is not a time related object at all. Its 3D projection can be build and the objects equivalent of a 2D shadow seen easily. In 4-space it occupies L^4 units^4. Wikepedia has some interesting information and projections of a tesseract
334077
You need to give more information about the shape to get a definitive answer.  If the cell were cubical, then the volume increases 1000 times if all three sides increase 10 times in length.
158992
sorry i must really correct the first answer before i continue;living in a 10 dimensional world is just a theory,known as string theory,and that is all it is,there is no proof\n\n(and the theory ACTULLY states that there are 11 dimensions)\n\na 2d object could not even physically exist because even if you had the smallest possible object that could exist indepdantly(an atom) it still has dimensions of length width and height,however small it may be it is not 2d\n\nIn writing the answer im kinda curious as to Could you theoretically stack and infinate number of 2d objects on top of each other and it still remain 2d?? anyone know?