Is there a tetrahedron volume 216 subdividable into tetrahedra of volumes 125, 64, 27?
953022
There sure is!  \n\nFirst, recall this: given a tetrahedron, choose any of its faces to be its base.  Now find the area of the base (remember all the faces are triangles).  Next, find the perpendicular distance from the fourth vertex of the tetrahedron to the plane that contains the base.  This distance is the height of the tetrahedron.  Now the volume of the tetrahedron is 1/3 X area of its base X its height.  \n\nNow, using three-dimensional Cartesian co-ordinates, consider the tetrahedron ABCD where A=(0,0,0), B=(27,0,0), C=(0,16,0), and D=(0,0,3).  Face ABC is a right triangle in the x-y plane.  The two perpendicular legs of ABC have lengths 16 and 27, so the area of face ABC is 1/2 X 16 X 27 = 216.  The distance of D from the x-y plane is 3, so the volume of ABCD is 1/3 X 3 X 216 = 216.  \n\nLet E=(8,2,0).  E lies in the x-y plane in the interior of ABC.  Join E to A, B, C, and D with line segments.  Triangle ABC is subdivided into triangles ABE, BCE, and ACE, and tetrahedron ABCD is subdivided into tetrahedra ABED, BCED, and ACED, and the heights of these three sub-tetrahedra are also 3.\n\nThe distance from E to the x-axis is 2, so triangle ABE has \narea = 1/2 X 2 X 27 = 27 and tetrahedron ABED has \nvolume = 1/3 X 3 X 27 = 27.  \n\nThe distance from e to the y-axis is 8, so triangle BCE has \narea = 1/2 X 8 X 16 = 64 and tetrahedron BCED has \nvolume = 1/3 X 3 X 64 = 64.  \n\nArea of ABC = 216, area of ABE = 27, and area of BCE = 64, so triangle ACE has area = 216 - 27 - 64 = 125 and tetrahedron ACED has area 1/3 X 3 X 125 = 125.  \n\nThat does it!
825337
NO! you have to replace V2 with given volumes:\nP = (V1P1)/V2\nP = (4 x 1)/3 \nP = 4/3 \n\nNow do the same thing for V2= 2. V2= 1 , V2= 5 and V2= 25
691096
the answer is 56 cu.in i.e. choice (D)\nVol of the cube is side*side*side and so is 6*6*6 cu.in i.e.216 cu.in..........................(1)\nVol of the rectangular solid is length*breadth*height and so is 10*4*4 cu.in i.e.160 cu.in................................(2)\nComparing (1) and (2) it can be seen that the cube-shaped box is bigger than the rectangular box by\n216 less 160 cu.in i.e.50 cu.in\nHence the choice (D)
65323
eBay turns up several hits on this product. See the listings I located below. You may need to get an incomplete set and watch eBay or similar sites to fill in the missing volumes. \n\nGood luck!
881616
The sum of the first n tetrahedral numbers is \n\nn(n+1)(n+2)(n+3)/4!\n\n(The sum of first n of these is in turn n(n+1)(n+2)(n+3)(n+4)/5!, and so on.)
814455
0 with 4 faces, 24 with 3 faces, 60 with 2 faces, 48 with 1 face and 12 with 0 faces.  \nNote, this is if the 3 cubes were not touching each other.  I made an assumption when you said that they were originally 3 separate cubes.  I believe, after reading the second answer, that I made an incorrect assumption since you said "total volume".  If the 3 cubes were touching each other when painted, I am wrong.
781998
27 cubes, then 81, then 243, then 729...
656895
It had been conjectured that two equal partial spheres sharing a boundary of a flat disk separate two volumes of air using a total surface area that is less than any other boundary. This equal-volume case was proved by Hass in 1995, who reduced the problem to a set of 200260 integrals which they carried out on a PC. Frank Morgan, Michael Hutchings, Manuel Ritoré, and Antonio Ros finally proved the conjecture for arbitrary double bubbles in early 2000. \n\nIn this case of two unequal partial spheres, Morgan et al. showed that the separating boundary which minimizes total surface area is a portion of a sphere which meets the outer spherical surfaces at dihedral angles of . Furthermore, the curvature of the partition is simply the difference of the curvatures of the two bubbles.\n\nProbably not any huge influence on society, however.
189752
sphere, cube, pyramid, cilinder
125798
2p, 5d, 4d, 3p, 3d exist.
63817
tsniperx\n2 months ago\n\nVolume of one block cube = 10cm*10cm*10cm = 1000cm^3\nVolume of two block cube = 2000 cm^3\nVolume of one sphere = (4/3)*pi*r^3 = (4/3)*pi*(5cm)^3\nVolume of one sphere= 523.59877 cm^3\n\nno spheres= Volume of clay / Volume of sphere= 2000cm^3 / 523.59877cm^3\nNo spheres must be integer = 3 spheres
241816
It is certainly possible. I'll give you an example for H-O-Y\n\n\nFirst, let's find the index of the last group starting with H, which is H-Y-Z. Every group before H-Y-Z and H-Y-Z itself always includes a letter that comes before I, and every group after H-Y-Z are the only ones not containing the first 8 letters. Therefore, there are C(26-8,3)=816 groups. This makes H-Y-Z 2600-816 = 1784th group.\n\nNow, let's only consider the groups starting with H. Let's find the index of H-O-Z. Every group after H-O-Z are does not contain letters before P if we do not consider the first letter. There are C(26-15,2) = 55 such groups. Therefore, H-O-Z is 1784-55 = 1729th group.\n\nH-O-Y comes right before H-O-Z. Therefore, the index of H-O-Y is 1728.
319154
http://home22.inet.tele.dk/hightower/spheres.htm
903397
You might try asking this question under "Mathematics."
402991
if you are talking about lattice structures then all the basic ones can be found on this website:\nhttp://members.iworld.net/joo/physics/curri-sub/crystal/lattice.html
544364
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.
911884
"Crystal Field Splitting can be defined only for tetrahedral and octahedral complexes."  Raymond Chang; Physical Chemistry for the chemical and biological sciences; Univ. Science Books, Sausalito CA; 2000; P.652