Tessellation question help!?
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One of the most simple requirements for tiling: if a polygon tiles a plane, then in each of polygon vertexes on the plane, the sum of adjacent angles should equal 360 degrees (one of the angles may be a side of another polygon, making up 180 degrees). \n\nFor example, imagine or draw a "house-shaped" pentagon, that is a square with a right-angled triangle on top (right angle pointing up). It has two angles of 90 degrees at the bottom, two 135 degree angles where "walls" and "roof" meet, and a 90 degree angle at the top ("roof" tip). You can tile them side-by-side and then tile the same pentagons on top, flipped upside down. At the "tip of the roof", you will see that two 135 degree angles and one 90 degree angle meet. 135+135+90=360. If it would not be the case, there would be a crack in the tiling.\n\nHowever, some polygons have such angle values that it is impossible to make a sum of angles equal to 360 or 180 degrees. For example, consider regular pentagons. They have angles of 108 degrees. However you add up 108 degree angles, you cannot make up 360 or 180.
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1st list of elements=\n\n{X u Y}n Z\n{2,4,6,8,10} n {3,6,10,12}\n= {6,10}\n\n2nd list of elements=\n\n{Y n Z}' /X\n{6,10} / X\n{6,10} / {2,4,6,8}\n={10}
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1) think of  a series of disks stacked on top of eachother, so thier centers are aligned.  Each disk has a slightly smaller radius then the one under it.  If the disks are very thin (infinately so), and you stack enough of them so that the top disk has a radius of zero (a point), then you have a  cone. \n\n2) Take a circle, and remove some portion of it- like a pizza with one or more pieces removed. Take the remaining portion of the circle, and roll it, so that the straight edges adjoining the removed section are connected together along their entire lengths, and you have a cone.
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Find the closest brick like shape lxbxh, also closest spherical shape by mean radius. Then by water displacement find the volume. Then find the surface area taking those regular shapes. Then make an avrage of the two. If the mango is long then take cylindrical shape intead of spherical.\n\nOtherwise put soft, wet tissue paper without much wrinkle like a paste/coating (you may tear some pieces and use small pcs). Then dry it. Peel off carefully. Place on a graph paper and measure. You may cut and paste.
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This might help --> http://www.uwm.edu/~ericskey/TANOTES/Geometry/node15.html
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If you have the letter names, use the Periodic Table of Elements.  If not, use PH paper, and that might help.  Go by color, soluability, and texture.
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This is  basic\nslope= m=(y2-y1)/(x2-x1)=(12-7)/(8-6)=5/2\nangle=arctan(5/2)
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This will happen for only two irregular pentagons in general:\n\nAll triangles as the angle sum of the three angles is a straight angle so the the angles so placed will always make straight lines ... and thus tessellate\n\nAll quadrilaterals as the angle sum of the four angles is a straight angle so the the angles so placed will make a complete revolution ... and thus tessellate\n\nOther particular polygons MAY tessellate BUT no more general cases will occur
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6 faces, 12 edges, 8 vertices
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Check Out This Website!! \n\nGives You Examples And Everything!! \n\nhttp://www.mathleague.com/help/geometry/geometry.htm
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(0,0)\n(3,0)\ndraw the graphs and then find them out!\nif u can't, then mail me!
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this is impossible to answer, there are many shapes that could be made this way.  do you have any other information like right angles or angles at corners?
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I am only using visual techniques and not math theorems here:\n\nThe first plane will divide space into 2 parts.\n\nIf totally random the second plane will probably not be parallel with the first.  This will divide the 2 parts into 4.  If it did happen to be parallel with the first, then the space is divided into 3 parts.\n\nThe third plane will most likely not be parallel with either of the other two.  It would divide the space into 8ths.  If it was parallel to either of the other two then the minimum parts is 4 and the maximum is 6 parts.\n\nThe fourth plane will most likely not be parallel with any of the first 3, and can divide the space into maximum of 15 parts.  The fifth will make that 26.  Parallel planes will reduce this number.\n\nThe space will be divided into a minimum of 6 and a maximum of 26 parts with 5 planes.\n\nI just found a great description at the site below:
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Fractals\n\nTake the serpenti carpet, for example. This tesselation is derived first from a square the square has its middle section taken out creating 8 new, smaller squares. This process is repeated to infinity. It is a tesselation because it is a repeated pattern. The math aspect is this:\n\nEach time we take out the middle ninth, thus leaving 8/9 of what we started with. If we continue the process infinitely, the area will be 1 square unit * (8/9)^infinity, or 0. However, the shape still exists in a 2-D plane. This explanation only scratches the surface. If you're interested, check out wikipedia, mathworld, or simply google tesselation, serpenti carpet, cantor dust, tesselation, and even the Koch Curve. If your REALLY interested you might want to check out "To Infinity and Beyond: A cultural HIstory of the infinite." You can find it on amazon, and it will give the readers digest, don't need to be a math major version of everything about tesselations.
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Hi,\nThe answer is C.\n\nYou could make a ray-diagram and see for yourself...
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a) cube...final answer
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a) 4 "things" attached to central atom and no lone pairs then: sp3-> tetrahedral e- arrangement, tetrahedral shape, 109.5°\nb) 4 "things" attached to central atom one lone pair then: sp3 -> tetrahedral e- arrangement, trigonal pyramidal shape, < 109.5° (about 107°)\nc) 3 "things" attached to central atom one lone pair then: sp2 -> trigonal planar e- arrangement, bent shape, < 120° (guess about 115°?). This is most challanging, try to figure out how many e- in total you have (18), then arrange them around the central atom using usual valances.