plotting shapes maths?
969111
what ever you want to plot-just plot it\nit may be a circle or a triangle or just an\nold square-what ever you do-just plot it\n\ni hope that this helps
70364
Well, it would depend what graphics toolkit you're using, what operating system, etc. It would probably involve:\n\n1. creating a window\n2. creating a canvas or drawing area\n3. creating all the elements of a clock on the canvas.\n\nThe "hard" part is 3, where you probably have to use trigonometry to turn your circular clock coordinates into the usual cartesian coordinates.
805339
These are quite famous problems. As you mentioned, you can do it with integration; thats probably the simplest way.\n \nIn order to see why that works, lets call f(x) = 1/x or 1/x^2 depending on which one you want. \nDraw a 'histogram', where each of the bars is 1 wide, and has height 1/(left x value). So you have a bar from x=1 to x=2 with height 1, x=2 to x=3 with height 1/2, etc.\n\nNow if you draw the curve f(x), it never goes below any of these bars. So the integral of f(x), which is the area under the curve, must be less than the total area of each bar, which is that infinite sum. So if we get a finite integral, we have must a finite sum.\n\nSimilarly, you could draw a histogram where each bar has height 1/(right x value). Then the curve is always *below* the bars, so you get a lower limit. Then if that integral is infinite, the area must be infinite.\n\nAnyway, now that we've got the official answer sorted, there are many other proofs. Heres a nice one for the 1/k case:\n= 1 + 1/2 + (1/3 + 1/4) + (1/5 + 1/6 + 1/7 + 1/8) + ...\n>= 1 + 1/2 + (1/4 + 1/4) + (1/8 + 1/8 + 1/8 + 1/8) + ...\n= 1 + 1/2 + 1/2 + 1/2 + ...\nwhich goes to infinity.\n\nAs a side note, the actual sum of 1/k^2 is pi^2/6. That always interests me - if you draw a whole bunch of circles with radius 1, 1/2, 1/3, .., then their total perimeter turns into the 1/k-type sum, and is infinite. However, their total area turns into the 1/k^2-type sum, and is finite. So you can't draw the outline of the circles, but you can colour them in.. ;)
778580
Go to this link:\n\nhttp://www.coolmath.com/graphit/index.html\n\nYou can enter your equation and graph it.
793059
give values of -3,-2,-1,0,1,2,3 to x and find the corresponding values of y\nyou get the ordered pairs\nnow plot
12336
Try the BBC's Maths series and MathsFile as well as Math.com (see links below)
556436
If you can plot (X, Y) points on a graph-there is an easy way to show the shape of a very basic parabola. \n\nUse (0, 0) as the vertex, or the point where it bends.\n\nNow Use the process called an X, Y table (its how you will plot out the basic equation (or function) of your parabola. Use easy numbers such as 3, 2, 1, 0 -1, -2 -3. That is all you need.  Often 1, 0 and -1 will work fine. Simply put any of the "easy" numbers go in for X. Solve through for Y (if X is 2 in Y=Xsquared+1 than Y will be 5) (2,5)\n\nThe easiest is Y=X squared, but just rememeber that since  both positives AND negatives square as positives. You will have (-1, 1) (-2, 4) (-3, 9)\nand also (1,1) and (2,4) and (3,9)\n\n Say for Y=x squared +2\n\nYou will have a vertical shift upward. Remember (0,0) as you starting point, now go up the Y axis two points and use the same process you did in y=x sqaured\n\nRemember that what you figure out as a positive with a negative its always backwards. So Y=x squared-2 will start at (0, -2)\n\nIf you have y=-x squared instead of facing up like in y=x squared, it will face down. \n\nThere is this rule to remember as well:\n\ny=A(X +OR- B)+OR- C \nSorry the computer doesn't let me say "plus or minus" any better way.\n\n1. +/- A amplitude \n\nY=3x sqaured will fatter than Y=Xsquraed, Y= 1/3x squared will look skinnier. \n\nNegatives Open downward, positives open upward.\n\n2.+ or - B : phase shift\nnegatives addded to X in parenthesis  move to the right, positives added to X in parenthesis move to the left. \nDONT WORRY ABOUT THIS ONE AS MUCH. \n\n3. = or - C: Vertical shift \nIF a negative number is added after X, move it down\n\nIF a positive number is added after X, move it up\n\n\nI said it both geometrically and algeabraically. Email me at josh_rittinger@yahoo.com if you still need help.
896595
http://local.wasp.uwa.edu.au/~pbourke/geometry/2circle/\n\nit tells you how
223761
if your equation is y=x^2, then the graph that you draw is going to be a parabola. You can see this by having a table first. The table should consist of 2 values...x and y. So you say when \nx=-3, y=9,\nx=-2, y=4,\nx=-1, y=1\nx=0,  y=0\nx=1,  y=1\nx=2,  y=4\nx=3,  y=9\nso now if you have a square line paper, just plot it and you will see that it will give you a parabola. Another thing is that when you draw it, first just plot the points and then when you connect the point connect them using curves and not a line,...that's why u get a u shaped graph and not just a combo of lines.
695770
plot the line for y=4x-4 and shade the portion which would correspond to less than
351583
how about instead of (a+b)^4 you plot (x+y)^4?\n\nit's quite a funny 3D plot.\n\nLooks like a conehead balancing on his head.\n\nSeriously... draw a cone rounded off on the top and then flip the paper upside down and you'll get an approximation of (a+b)^4.
736099
plot the coordinates then do the vertical line test.  if the vertical line intersects more than two points on the graph then it is not a function.  each set of coordinates you have listed are functions (i free hand it with no graphing paper or calc) so forgive me if im wrong.  also refer to www.coolmath.com. it is a good site for breaking down math
977003
Draw a straight line .now plot three points S,T and R on it ,taking ST=3.8 cm and TR=1.4 cm.\nWrite the distances 38 in betweenS and T and 14 between Tand R.your sketch is complete.\nNow ,rite the followin unde the sketch:-\nSR=ST+TR=38+14=52.
168156
get a TI 82 or above graphing calculator.  MUCH easier. program already installed, just build the type of matrix,, and plug in the numbers
168710
http://www.graphcalc.com/
150119
Have you tried the ezplot routine?  You can get help from the context help window.  Mathlab is the bomb.
774030
There are in infinate number of 2d shapes, but here are diagrams of some of them --> http://www.mathleague.com/help/geometry/polygons.htm