plot enough points so that the resulting  figure forms  a net for a cube . how many points do you plot?
551111
8. One for each corner of the cube. Think about how many corners are on a die or a box. Each of these is a point.
659736
Not a site, but I suggest you invest in something called Mathcad.\nTeach, please note, it is a site not a sight! Unless you were 'looking' for the answer... sorry!
557771
Draw a straight line.\nDraw a good cuve (mirror), using a compass or a protractor(for later mark a point as centre and place the protractor on this point coinciding with the axis.)\nThe curve should be of equal length on each side of principle axis.\nNow mark the pole, radius and focal point (twice the radius).\nIn case u use protractor, take focal lenght as twice the distance between the marked pont and pole.\n\nRemember how the rays reflect/refract, and accordingly draw two incident and two reflected/refracted rays.\nGive the standard sign convention...I hope plane mirror and prism are easier.
724460
Yes, you'll need to find the area of each of the faces of the cube.\n\nReminder a cube is a 6 sided figure with 6 square faces.  It is the same shape as a dice that you throw in board games.\n\nEach edge is 10 feet,\n\nso the area of each face is 10 x 10 feet squared\n\n=100feet squared.\n\nFour of these sides will have a total area of 4 x 100 feet squared\n\n=400 feet squared.\n\nThe paint used costs $5 per square foot,\n\nso total cost for the 4 faces with ceaper paint = $400 x 5\n\n=$2000\n\nNow consider the other 2 faces,\n\nThey have a total area of 2 x 100 square feet\n\n= 200 sq. ft.\n\nCost of painting these is $10 per sq.ft.\n\n=$10 x 200 \n\n= $2000\n\nAs each lot of painting costs $2000\n\nThe whole cube (6 faces) costs $2000+ 2000\n\n=$4000
1016603
If you use degrees, there are 360 points.\n\nIf in mills, there are 6400 points.
223867
It depends on the object.  You find the area of all the sides and then add them together.
747559
Simply draw a chart that has each of the 6 variables at right angles to each other, draw lines between the linked variables such coke-greenbike (#3), then use process of elimination.\n\nYeah, right... draw a 6-dimensional hypercube....\n\n Jeez, I've spent about 3 hours on this, and so far it looks like Carla, who was on the yellow bike which was 1st in line, was Bigbird and brought rootbeer and jelly beans.  Either Emily, who was dressed as either dracula or the clown, or Daniel dressed as the other, brought the candycorn.  But at the moment I have a contradiction, which means that somewhere I made a logical error.  I'll try to get back to this sometime later - good luck
94604
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). "
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?
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
313944
You are usually given a point to plot.  The point is written as (-4, 2).  The first number represents x and the second represents y such as (x,y)\n\nTo plot the point, you start at the origin (middle) which is (0,0)\n\nThe x-axis is the horizontal line that goes left and right.  You put your pencil on the origin *but don't make any marks*\n\nLook at the first number in the ordered pair (-4,2)\nIt's -4.   You move across to -4 (on the x-axis...the horizontal line) and stop your pencil there but don't make a mark.\n\nThen look at the second number (-4,2) and you see it's 2.  Since 2 is positive you move up to spaces and then plot your point.  If it was negative, you would move down that number of spaces on the y-axis (vertical line).\n\nHope this helps.  Plotting points is fun.
645334
for a cube with 6 faces, the edges would be 8
971807
(0,0)\n(3,0)\ndraw the graphs and then find them out!\nif u can't, then mail me!
838358
i can't give you the answer but i used my ti-83 and i can tell you the points of the graph...\n(1,0) (2,4) (3,8) (4,0)\n\nHope this helps
647980
it would be N+1\n\nBecause a line splits the cirlce in two, two lines in three, etc.
826187
I am guessing you want the vector projection onto one of the 6 sides rather than one of the 12 edges.\n\nThe diagonal vector is (1, 1, 1).  Projecting it onto any side simply removes the coordinate in the direction of projection.  Projecting it onto the side in the xy plane gives (1, 1, 0). \n\nIf you did mean an edge (the line where 2 sides meet), then the projection is always two 0s and a 1.  The one is in the direction of the edge.
372589
How about making yourself a model, finding its point group then working out the reducible representation?