How can I find if a shape is traceable?
460155
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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lines, polylines, blocks, circles etc... look in the help files if you have the software\n\nhttp://www.we-r-here.com/cad/tutorials/index.htm
496604
Try doing this with an asymmetric figure. The translation and reflection won't give the same result.\n\nHowever, you can get any translation by doing *two* reflections one after the other.
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whichever is closest, no particular pattern.\n\n/doc
274979
That's a tough one.  Thinking the only way would be to verify source of how/where data got input?\nHow else could this be done, plus this would be time consuming.\nDo you have more specifics?
919171
To answer this question, you need to know the shape that these points make.  Are they on a line? or are they some two-dimensional shape?
239205
GO TO AN APPRASER AND THEY CAN TELL YOU IF IT IS OR NOT
241795
here:\nhttp://www.dictionaryofeverything.com/cgi-bin/find.cgi
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YES\n\nhttp://www.flux.utah.edu/~aclement/templair/collection/Unknown%20-%20Unknown%20-%20A%20dragon%20in%20a%20cave.jpg\nOR\nhttp://graphics.stanford.edu/data/3Dscanrep/dragon.jpg\n\nJUST GO TO GOOGLE AND CLICK THE IMAGES LINK ABOVE THEN TYPE WHAT YOU WANT...
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For example, a square is inscribed if the square is inside a circle.  The four vertices of the square are touching the circle..\n\nThe square will be circumscribed if it will be outside the circle.  In this case the circle will be touching the square in four points on the sides.
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try http://www.track-trace.com/
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u could go to www.google.com or www.search.yahoo.com and enter figure scating and find out
773115
you can surely rule out some things based on the type of plate you are using.  assuming that you have covered that base, you can separately pick a colony that you think may be a contaminant and look for differences in growth patterns, color, morphology, gram stain, growth on other media, and if all else fails, you can sequence it to find out exactly what it is.
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Approach your teacher and ask him or her whether your rule is geometrically valid. This is a very elementary costruction but you may have come up with something which exploits some neglected area of geometry. Once you interact, you will come to know whether it really is a brainwave or whether it already exists or is plain obvious.
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since your ddoing coordinate geometry you should at least tell what shape you're trying to get.\nI'm assuming it's a parallelogram so u should make it (1,8)
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Well, I suppose you want a program that computes the trace of a matrix. The trace is simply the sum of the elements in the diagonol in a matrix. I won't show it to you with OOP since there's no reason to.\n\nint trace(int matrix[][],int numrows,int numcols) {\n\nint tracetotal = 0;\n\nif (numcols != numrows) {\nstd::cout<<"Error: Can't compute trace. Not square matrix.&#92;n";\nexit(1);\n}\n\nfor (int i = 0; i < numrows; i++) {\ntracetotal+=matrix[i][i];\n}\n\nreturn tracetotal;\n}
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If you have a TI-83 you can graph it and find it easily...\n\ny= button (plug in equation) ---> hit graph\n\nthen hit the second button(yellow)--->calc(same as trace button)\n\nbecause the graph was concave up you want to find a minimum...so click that...and simply enter once on the left side of the graph, once on the right, and then guess in the middle.  It will then give you the exact answer.\n\n(1,6) should be the correct coordinate.