How would you draw 10 distinct different drawings of octane isomers?
956863
Gotta love wiki\n\nhttp://en.wikipedia.org/wiki/Octane\n\nclick on each one of the 18 isomers, then you can draw. Basically, you shift a methyl group from carbon 1 on down the line so you don't have a straight hydrocarbon chain.
774231
an octet configuration with eight valence electrons each (on each atom) ?
176182
Draw 2 concentric cirles 1 smaller and 1 larger then a teeth on the outer circle. Then use the circular array tool to array 6 the teeths or tooths (iuno) around the outer circle. And make it look like a gear.
978173
You need to start off with a six-pointed star (constructed with two equilateral triangles).\n\ntriangle AEC and triangle FBD\n\nNow connect A to D, then connect B to E, and finally C to F. If you \ndrew everything correctly, those three segments should meet at a point \nin the center of the star.  \n\nThe 3 line segments you just drew plus the six sides of the star make \n9 lines. There are 5 points of intersection on each line, and the \ntotal number of intersections is 19.
827293
Triacontakaihenagon\n\nHere's a list of the prefixes for them (BTW they're from Greek) : \n\n1= mono\n2= di\n3= tri\n4= tetra\n5= penta\n6= hexa\n7= hepta\n8= octa\n9= ennea\n10= deca\n11= hendeca\n12= dodeca\n13= triskaideca\n14= tetrakaideca\n19= enneakaideca\n20= icosa\n21= icosikaihena\n22= icosikaidi\n23= icosikaitri\n29= icosikaiennea,\n30= triaconta\n31= triacontakaihena\n39= triacontakaiennea\n40= tetraconta\n50= pentaconta \n60= hexaconta \n70= heptaconta \n80= octaconta\n90= enneaconta\n100= hecta\n\nHope this helps~!
661664
you can use a dot to separate two or more algarism blocks, like:\n999.999.999\n\nThis way you will have 1000 * 1000 * 1000 combinations of algarisms. Since 000.000.000 until 999.999.999.\n\n000.000.000, 000.000.001, 000.000.002, ..., 999.999.997, 999.999.998, 999.999.999
785873
M.C. Escher has already done a number of drawings based on the hyperbolic plane, based on Henri Poincaire's method of representing the entire hyperbolic plane.  Check this out: \n\nhttp://www.sciencenews.org/articles/20030830/mathtrek.asp
703879
This sounds like plane geometry where you have an n sided polygon and want to know how many diagonals you can draw.  From each vertex, you can draw a diagonal to every other vertex except to yourself and the two next to you (those two line make sides of the polygon not diagonals).  So for n sides (which means n vertices), you can draw n-3 diagonals at each vertex.  This gives (n-3)*n.  But this just counts the starting end of each line. Whenever you draw one of these lines, it goes to another vertex.  If we drew (n-3)*n lines, we would traces over each diagonal twice, once from each end so there are only half as many diagonals as that.\n\nThe answer then is: (n-3)*n/2
989768
Pure Ethyl alcohol has a hellvua (R+M)/2 octane number at 115!\n\nFirst number is RON, second number is MON\n\nmethanol                    133 : 105         \nethanol                       129 : 102 \niso propyl alcohol       118 :  98 \nmethyl tertiary butyl ether   116 : 103 (this is MTBE the stuff they just took out of gasoline) \nethyl tertiary butyl ether    118 : 102 \ntertiary amyl methyl ether    111 :  98\n\nHope this does the trick for you, and no I am not a chemical engineer, but I do know how to use Google! :^)
712605
#10.\n\n3,4 - dimethylhept - 3 - ene\n\nI cant really draw the other ones.\nJust look up your reaction mechanisms.\n\nGood Luck!
801515
the best program to use is power point or microsoft word.\ngo to tool then go down to pichures, then some options should apear click on cip art. if a poker symbol dosen't come up try search the web for more results.
403308
It is correct, but there's no way to paste diagrams here. Just draw the intersecting circles (i.e. 7 regions) and then shade the appropriate areas.
780311
You are on the right track.\nThis a geometrical problem that requires drawings.\nCheck this link http://departments.kings.edu/chemlab/chemlab_v2/packgeo.html\n\nI'll get back to you later \n\nso later it is\n\n3--------R=r(1+2/3^.5)\n7--------R=3r\n19------R=5r\n37------R=7r\n61------R=9r
764178
Trick question, multiple answers. . .\n\n1. CH3(CH2)3OH\n2. C2H5CH(OH)CH3\n3. (CH3)2CHCH2OH\n4. (CH3)3COH\n5. (C2H5)2O\n6. CH3(CH2)2OCH3\n7. (CH3)2CHOCH3\n\nOk, #4 is tert-butanol.  All of these are structures for C4H10O\n\ngo to http://www.cem.msu.edu/~reusch/VirtualText/intro3.htm and scroll down to the isomers section.  All the diagrams are listed there.
1070589
Assuming you mean as 4 shapes equivalent to the original L shape, the answer is given in the attached image.
436394
7\n\nRed, Green, Blue, Indigo, Violet, Orange, Yellow
823299
How about drawing a bunch of straight lines of differing length, from a fixed point, so that the tips create the image of an arc?