how many vertices does a cube have?a convex lens curves _____.a pluvial day is one with _____>?
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What on earth are you posting this here for  ? It is NOT earth science for sure. And it is definitely a homework problem.\n\n------------\n\nI guess it isnt. I found it as part of some science trivia quiz.
932841
This might help --> http://www.uwm.edu/~ericskey/TANOTES/Geometry/node15.html
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what are you supposed to prove?\n\nedit:\nyou mean why the three vertices of an equilateral triangle are concyclic? i am unable to make out theproblem
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A convex lense is made up of individual prisms arranged in such a way that the pointed axis, which is suppossed to be the top, of one is alined to the base of the other. \nWhen ligh hits the lense, refraction occurs due to the arrangement of the prisms.\nAt the principal axis, two of these prisims are facing one another tip-to-tip in which case the light does not neccessarily pass through any of the prisms itself but through the interface of the apex of the two adjoining prisms.\nThe reverse is true for concave lesnses.
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a) isoceles\nb) hexagon\nc) hemisphere\nD) equilateral\ne) \nf)\nG) sphere or spherical\nh)\ni)\nj) scalene\n\nthese are the ones i know. next time, do your own homework!
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They're polyhedrons.  The 5 types are:  Tetrahedron (a 3-d figure with 4 equilateral triangle faces, 4 vertices, & 6 edges); Cube (a 3-d figure with 6 square faces, 4 vertices, & 12 edges); Octahedron (a 3-d figure with 8 equilateral triangle faces, 6 vertices, & 12 edges)); Dodecahedron (a 3-d figure with 12 regular pentagonal faces, 20 vertices, & 30 edges); and Icosahedron (a 3-d figure with 20 equilateral triangle faces, 12 vertices, & 30 edges).
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Spiral, lenticular, elliptical and irregular\n\nSee the link
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the radius of curvature of a plane mirror is infinity( iam dead sure)\n\n\nThe term radius of curvature has specific meaning and sign convention in optical design. A spherical lens or mirror surface has a center of curvature located in (x,y,z) either along or decentered from the system local optical axis. The vertex of the lens surface is located on the local optical axis at a distance equal to the radius of curvature from the center of curvature. The sign convention for the optical radius of curvature is as follows:\n\nIf the vertex lies to the left of the center of curvature, the radius of curvature is positive. \nIf the vertex lies to the right of the center of curvature, the radius of curvature is negative. \nThus when viewing a biconvex lens from the side, the left surface radius of curvature is positive, and the right surface has a negative radius of curvature.\n\nNote however that in some areas of optics other than design, other sign conventions are sometimes used.
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Magnification is dimensionless\n\nWhat is the focal length of the concave mirror in b\n\nConvex mirror has wider field of view.
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im a geo student, but i havent done proofs for a long time sorry. ok\n2. false, because squares are parallelograms with 4 congruent sides and 4 right angles, while kites have 2 pairs of congruent sides and only 1 pair of congruent angles that are NOT right angles\n3. false, because a rhombus is a parallelogram with 4 congruent sides, while a kite is any thing BUT a parallelogram. enuff said\n4. false, because kites are quadrilaterals with only two pairs of congruent sides, while the quadrilateral mentioned has 3 congruent sides\n5. false, for example- do trapezoids have 4 congruent sides, and 4 right angles, and is a parallelogram (characteristic of squares)? i think NOT\n6. read above- trapezoids are not parallelograms.
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A quotation from\nhttp://www.ams.org/featurecolumn/archive/sylvester3.html\n\n"In retrospect, it is clear that one could easily see that there had to be at least three ordinary lines for any point-line configuration. This follows from Melchior's work. Later it was shown that the number of ordinary points had to grow larger as the number of points in a plane finite configuration grew. Both G. Dirac and T. Motzkin were among those who explicitly conjectured that there would be at least n/2 ordinary points for a set of n plane points. However, it should again be noted that for n =7 and 13, there are configurations for which this bound does not hold. The additional requirement that n be large enough has to be added. It was announced by Hansen that he had a proof that there were always at least n/2 (n > 13), but the proof, unfortunately, had an error. Despite this, there were many correct new results in Hansen's investigations related to Sylvester-Gallai results. The number of ordinary points is known to be at least 3n/7 (n > 2), as was shown by L. Kelly and W. Moser, and the best known result (due to J. Csima and E. Sawyer) is that the number of ordinary lines is at least 6n/13 (n > 6). "
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sphere, elliptic, plane, euclidean, pseudosphere, hyperbolic
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Suppose the shape is represented by a graph ( http://mathworld.wolfram.com/Graph.html ). You want to find a path through the graph that passes through every edge without travelling on an edge more than once. This is an easy problem in graph theory, and such a path is called an Euler path. An Euler path through a connected graph exists if and only if the graph contains at most two vertices with odd degree (the number of edges the vertex has). This is a trivial theorem.\nSuppose a graph is connected, which means there exists some path between any pair of vertices. For each vertex, there is some number of edges connected to it. Suppose the number of edges is even for all the vertices. Then there is no problem drawing an Euler path through these edges. Explain why (Examine the problem at each vertex).\nFrom the above, the only possible problems must come from vertices with odd degree. Without loss of generality, we proceed by adding vertices of odd degree to the nice graph described above. Adding one vertex is no problem, we start at that vertex, travel along the one additional edge and then we're in a nice graph that already contains an Euler path, since the vertex we left at the start now has even degree, for all intents and purposes. \nIf we add two vertices of odd degree, we are alright as well, with the exception that any path must start at one of these vertices and end at the other. Show this by labelling each odd degree vertex with little "in" and "out" signs on the edges.\nIt should now be obvious that we can't add any more vertices with odd degree for the simple reason that a path that doesn't double back on itself can only have 2 endpoints. Adding more vertices doesn't change the fact that you put your pencil down at one odd vertex and you're required to lift it back up at one other odd vertex. After that, the game is up.
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vertex (h,k)\nfocus (h,k + p)\n\n(x - h)^2 = 4p(y - k)\n\nvertex : (2,-6)\nfocus : (2,-6 + p)\n\nk + p = -8\n-6 + p = -8\np = -2\n\n(x - h)^2 = 4p(y - k)\n(x - 2)^2 = 4 * -2(y - (-6))\nx^2 - 2x - 2x + 4 = 4 * -2(y + 6)\nx^2 - 4x + 4 = -8(y + 6)\nx^2 - 4x + 4 = -8y - 48\nx^2 - 4x + 8y + 52 = 0\n\nANS : d.) x^2 - 4x + 8y + 52 = 0
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polygons are 2 dimensional shapes. they must have all straight sides and they must be closed. some examples are squares triangles, etc.  circles aren't because they  have 1 curved side.\n\npolyhedrons are 3-d shapes that have all flat sides and have no openings. ex: cube. spheres, cones, and cylinders are not because they all have curved sides.
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maybe a hexagonal prism http://en.wikipedia.org/wiki/Hexagonal_prism http://en.wikipedia.org/wiki/Prism_%28geometry%29
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The correct answer was given above; the sum of the interior angles is 360, which is a necessary property for a shape to be able to tessellate. That this property is sufficient for quadrilaterals is shown in the source.