to draw ten rectangles and shade them in fractions?
912774
Yes, what's the question? I'll take a guess ...\n\nIf your rectangle is divided into ten equal pieces, then each one will be a tenth, so if you shade 3 for example, that will be 3 tenths ... any help?
66715
Try out Smartdraw for presentations.\nOr Gimp for designing
411393
Yes 10, think of vertices like the corners.
1111648
200 x 200\n\nor\n\n40, 000 colors
1006293
The following link  has a tutorial \n\nhttp://www.ncsu.edu/project/graphicscourse/gc/acadtut/acadtut2000/acadtut-home.html
834309
depending on how many squares (cells) and rows/columns you would be using to create it.\n\nfor example,if you wanted 4 squares across & 2 rows down \n\n1 2  3  4 \n5 6  7  8 \n\nusing red & black alternating  you would write it like:\n\n<table border="0" cellspacing="0">\n<tr>\n<td bgcolor="#FF0000">Cell 1</td>\n<td bgcolor="#000000">Cell 2</td>\n<td bgcolor="#FF0000">Cell 3</td>\n<td bgcolor="#000000">Cell 4</td>\n</tr>\n<tr>\n<td bgcolor="#FF0000">cell 5</td>\n<td bgcolor="#000000">cell 6</td>\n<td bgcolor="#FF0000">cell 7</td>\n<td bgcolor="#000000">cell 8</td>\n</tr>\n</table>\n</tr>\n</table>\n\nfor using the pattern techinque only and no text (text was used so i can demonstrate to ya) remove the cell 1,cell 2,etc
593727
I can remember upto 4 to 5 ways of doing it and not more than that.  btw you are talking about halving a square (diagram)! Right ?
394215
Divide.\n\nFor example, how many twos are in 10?\n\n10/2 = 5\n\n5 twos in 10.\n\nNow let's try to divide 15 by 5/4. Do you know how to divide by a fraction? Invert the fraction and multiply.\n\nSo we get\n\n15 * 4/5 = 12\n\nTo check, we can multiply 5/4 by 12 and see if we get 15.\n\n12 * 5/4 = 15\n\nYes!!!
957
make a small hole in piece of paper with a needle, reflect the light that goes thru the hole onto some shaded surface
454825
We can do this 2 different ways.\n\nFirst: Use the Pythagorean Theorem. Since all angles are right angles, the diagonal (measure from corner to corner) is the hypotenuse of the right triangle formed.\n\nBy the Pythagorean Theorem, 10^2 + 10^2 =x^2, where x=diagonal\n\nx^2=200\nx=sqrt(200)\nx=10 sqrt 2\n\n2nd way: use 45-45-90 Triangles\n\nWhen you draw any diagonal in a square, you get a 45-45-90 degree right triangle. This is because the 2 sides are the same, so the 2 base angles must be also. The top is 90, so the others must be 45 and 45 to equal each other.\n\nThe sides to a 45-45-90 right triangle are always in the ratio 1:1: sqrt(2) where sqrt(2) is the hypotenuse.\n\nYou know that the sides are 10 feet. Therefore the hypotenuse is 10 sqrt2 feet
397805
This problem is certainly tractable, you just need to\nfind a consistent/methodical way of counting the\nrectangles -- without missing any and without counting\nany of them multiple times.\n\nThe way I chose is to define each rectangle as being\ndefined by the square in the upper left corner and the\nsquare in the lower right corner.  Consider the following\nsimplified 2x3 case (where each square is labelled by\nits X and Y coordinates:\n\n0,0     0,1   0,2\n1,0     1,1   1,2\n\n\nThis 2X3 example has the following rectangles starting\nat 0,0.  Remember, each rectangle is defined by the\nsquare in its upper left and the square in the lower right:\n0,0   0,0\n0,0   0,1\n0,0   0,2\n0,0   1,0\n0,0   1,1\n0,0   1,2\n\nThen it also has the set anchored in the upper left by 0,1:\n0,1   0,1\n0,1   0,2\n0,1   1,1\n0,1   1,2\n\nThen it has the set anchored in the upper left by 0,2:\n0,2   0,2\n0,2   1,2\n\nThen those anchored by 1,0:\n1,0   1,0\n1,0   1,1\n1,0   1,2\n\nThose anchored by 1,1:\n1,1   1,1\n1,1   1,2\n\nAnd finally those anchored by 1,2:\n1,2   1,2\n\n\nRewriting the matrix another way, showing how many\nrectangles are anchored by each location in the matrix:\n\n2x3    2x2   2,1\n1x3    1x2   1x1\n\n\nand then doing the math:\n\n6      4     2\n3      2     1\n\nfor a total of 6+4+2+3+2+1 = 18\n\n\nDoing the same analysys for a 4x5 gives:\n\n4x5   4x4   4x3   4x2  4x1\n3x5   3x4   3x3   3x2  3x1\n2x5   2x4   2x3   2x2  2x1\n1x5   1x4   1x3   1x2  1x1\n\nor rewriting after doing the multiplication:\n\n20    16    12    8    4\n15    12     9    6    3\n10     8     6    4    2\n 5     4     3    2    1\n\nAdding up the columns gives:\n\n50    40    30   20   10\n\nFor a total of 50+40+30+20+10 = 150 total rectangles
510559
three equally spaced horizontal lines, three equally spaced vertical lines. Plant a tree on each of the nine intersections. Put a line through each of the two major diagonals. Number the trees:\n1 2 3\n4 5 6\n7 8 9\ndraw a line through 8 and 6. draw a line through 4 and 3. That's 10 lines. Put the 10th tree at the intersection of these last two lines. It can be seen that this last tree doesn't lie on any of the other 8 lines.\nQED
1112728
to draw lines u hav to select 2 points at a time (as no 3 r collinear)\n\nthus no of selections of such points is nC2= n!/ [ 2! (n-2)!]
177784
take a seat and do some fractions
455504
I will explain a method that is the most easy on this planet, and a unique one.\n\nI will illustrate with the help of a simple example.\n\nEg. 4-2.\n\nDraw no. line as follows:\n\n<-6--5--4--3--2--1--0--1--2--3--4--5--6--7-->\n\nNumbers left to zero are negative integers\nand to the right are the positive ones.\n\nLocate 4 on this line. Now move 2 steps behind 4 as the operation to be carried out is subtraction.\n\nThis will give answer as 2. \n\nSimilarly for addition move the steps ahead.
590575
USE HTML AND PAINT DUH
359093
Square and Rectangle= lenght*height\ntriangle=base*height divided by 2