how can i get eight peices from a circle that i can only cut three times?
116295
fold the paper.  think outside the box
905726
three sizes\nthree crusts\nten toppings\n\nLook in your math text under "combinations and permutations." You can do this!
855656
I'm assuming you're talking about cutting a circle into quadrants? This might be the long way to do it, but I would first find the center of the circle. Here's a web page for how to do that:\n\nhttp://www.makeitsolar.com/science-fair-ideas/90-find-circle-center.htm\n\nThen you draw your diameter. (there's half)\n\nFinally, find and draw a line perpendicular to your diameter that crosses through the center of the circle.\n\nLike I said - probably the long way to do it, but I'm a very visual person when it comes to geometry. I like to draw everything out.
679656
making crop circles and disecting people
971896
Try it like this.  The square can intersect the circle a maximum of eight times.  This happens when the circle is wider than the square at the center but becomes narrower before it reaches the edge of the square, thus crossing it at two points on each edge.  The triangle can intersect the circle a maximum of six times in the same way.  The triangle can intersect the square six times as well, if you draw two legs of the triangle cutting off edges of the square and the third connecting them, going through opposite sides.  It's hard to explain without drawing them.  Anyway, the point is, look at the possible intersections of two shapes at a time, then add them together to get your answer, which in this case is 20.
349917
Cut out a cardboard circle, draw colors on that or glue the color wheel on it.\nPut a needle through the centre (as the axis) and use 2 forks to hold the needle.\n\nIf you need a lot of spin, tie a rope around the needle and pull
882048
Here's the answer you want:  With a compass set to the radius of the circle, start marking off radii around the circle, dividing it into 6 equal parts.  Connect any 3 alternate points and you'll have your triangle.\n\nNO MATH REQUIRED!
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.
855176
Hold up eight fingers.  Now put down three of them.  That leaves five of them.\n\nThe answer is 5.\n\nIt is also equivalent to the number of points you lost asking this question... but I guess you are just having fun with Yahoo! Answers coming up with silly questions, aren't you.  Enjoy!
157538
1.) Right click on mouse\n2.)go down to the 4th coalom\n3.) click the frist one" Create Sortcut"\n\nHope that helps!
791300
Use the FOIL method; First, Outer, Inner, Last.\n\nFirst, n * n\n\nOuter, n * 5\n\nInner, n * 7\n\nLast 7 * 5.\n\nThen add them all together.\n\n22 NOV 06, 0420 hrs, GMT.
765113
Very low. I'm not actually going to do the computation for you but I'll show you how it's set up.\n\nThis would be easiest to figure out on a graphing calculator like the TI-89.\n\nI'm assuming that the 1/3500 is for a single hole, so note that you're playing 50*18 holes in the season, or 900 holes. Getting 3 holes in one and 897 not is going to be the odds of three holes in one((1/3500)^3) mutiplied by the odds of 897 not((3499/3500)^897) multiplied by the ways to picking any three random holes of the 900 for the holes-in-one(this is done by a function called choose, denoted 2c5 for the number of ways to pick any two of five slots.\n\nThe answer is:\n((1/3500)^3) * ((3499/3500)^897) * (3c900)\n\nIt's going to be a very small number indeed. Congratulations on your fantastic season, I hope you do as well again next year.
1152243
I think it's 16 because of the math involved, you cut 16 in half and get 8, cut that in half you get 4, cut that in half and get 2, then a final winner.  Try doing that with 15 or 18
1165642
One way to reason is this.  Let n be any even integer greater than or equal to 4.  Divide the circle into n equal wedges (by drawing n/2 diameter lines with angles between successive lines equal to 360/n degrees).  Now, cut the wedges out and make a "rectangle" with them: n/2 wedges with the points downward and the arcs on top of adjacent pairs of wedges touching each other, then the other n/2 wedges with the points upward and adjacent arcs touching, filling the spaces of the first n/2 wedges.  What you end up with is almost a rectangle with top and bottom each made of n/2 360/n-degree arcs, and the lateral sides slanting slightly, made from the long edges of the wedges on each end.\n\nYou can estimate the area of this figure (which is the same as the area of the original circle of radius r) by pretending it IS a rectangle.  The length is the circumference of the circle, or 2*pi*r/2 (since half the circumference is the top, half is the bottom) (by definition of pi).  The width is the length of a wedge, or r.  So the area is about pi*r^2.  \n\nTo turn the estimate into an exact value, consider what happens as n goes to infinity, that is, it increases without bound.\n\nAs n goes to infinity, the slanted edges aproach being verticle and the number of arcs making up the top and bottom approaches infinity, and the arcs themselves approach straight horizontal lines.  Thus, the figure approaches a rectangle of exactly area pi*r^2.  However, for each n, you have the same circle of the same area, so the area of that circle is pi*r^2.
799606
You mean three equal parts? Is that the only criterion? not say, "three equal parts by drawing 2 lines", or something?\nIf so, thats easy, three wedges, each of angle 120°.
75253
Normal straight cuts that all pass through the center point will give you six pieces.  Straight cuts that do not go through the center acn give you 7 pieces, with a triangle-shaped crust-free piece in the middle.  For straight cuts, that is a maximum.
611894
8 to 3  (56 to 21 and divide by 7)