Need help making quilt patterns using Trig Functions (Pre Calculas)?
956464
There are any number of geometric curves that can make interesting quilt patterns.  The website:\n\nmathworld.wolfram.com\n\ngo to: /topics/AlgebraicCurves.html\n\nIt has a number of them.  Many are expressed both in polar and cartesian coordinates.  Check out Cassini Ovals, Astroid, Bicorn, Butterfly Curve, Conchoid of de Sluze, Scarabaeus, and especially Trifolium and Quadrafolium
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Check out this web page: http://mathworld.wolfram.com/Net.html
878923
Make a program on your calculator.\nView the snippet in the link for code.
337947
You need a stringing machine, then it depends on what kind of machine you have. Would really need more details.\nStringing pattern can normally be figured out or taken from the previous set of strings.
966909
I am assuming that all you want is that simple pattern drawn.\n\nFor a predefined size of the triangle LINELIMIT:\n\nint main() {\n  int head = 1. prev; line = 0, col = 0;\n\n  for (line = 0; line < LINELIMIT; line++) {\n    prev = head;\n    for (col = 0; col < line; col++) {\n      printf("%d ", prev);\n      prev *= 2;\n    }\n\n    printf("&#92;n");\n    head *= 2;\n  }\n\n  return 0;\n};
771697
Try Google or Yahoo.  You'll find worked examples there.  I think it needs diagrams to understand; they and the accompanying explanations are too complex to attempt to reproduce here.
596750
The Pythagorean theorem only works with right triangles.  Try using SIN, COS or TAN.
775797
http://www.coolmath.com/graphit/
34052
I do not know a preprogrammed utility to do this, but can point you in the direction that you might need to go.  See the reference below for the math and an on-line converter:
953641
I can only say so much because those equations are too long for two in the morning. Not that it matters; I just won't be retyping them. But you certainly need a bunch of random things for this. First, you need a variable -- a char, probably, since you don't care about any input beyond that. I don't know what the compare thing is going to output, but that'll come.\n\nI suggest an array for the old and new charts.\n\nGeez, what's the syntax? int oldchart[11], or something? I'm getting too used to Java. Anyway. Since you're going in increments of 5 from 0 to 50, you'll have a 0, 5, 10, ... 45, 50 -- that is, 11 different speeds, right? So you'll need an array for the old chart and an array for the new chart, both with a size of 11. Do you get where I'm going?\n\nCalculating the chart takes a bit more thinking. We'll ignore the actual process of getting input for the moment and just deal with filling it in. I'll suggest a method, and explain afterwards:\n\nint speed;\nfor(int i = 0; i < 11; i++)\n{\n   speed = i*5;\n   oldchart[i] = <calculate wind chill>;\n}\n\nThe same thing could be applies to the newchart array. The index, i, will go from 0 to 10, matching the indexes of the array, right? Since you're going by increments of 5, the speed for slot i will be i*5. That takes you from 0 to 50.\n\nSo after you pull that off for the old and new, you've got charts saved in the variables. Printing them is kind of similar to what happened above -- you use a loop to output what's in the array, using i*5 for the speed.\n\nNow, granted, you could combine this and the above step so you wouldn't need the arrays at all--you'd just recalculate and print every time. But I find it easier to organize my thoughts by going a way that deals with one issue at a time... It depends on what you prefer at this point.\n\nfor(int i = 0; i < 11; i++)\n{\n   speed = i*5;\n   cout << "Speed: "<< speed << " Wind Chill: " << oldarray[i] << endl;\n}\n\nLet's see...\n\nDisplaying the menu is easy--just use cout << "your menu stuff" << endl;\nInput is a bit stranger, I think. For something like this, just use a cin and check afterwards if it's right. Um, let's see... I'll give this to you in pseudocode--you should be able to work out the actual stuff.\n\nwhile ( the user doesn't want to exit )\n//this is just a logic thing -- if the user is done, then what?\n{\n     take the user's input;\n\n     if the input is 'o' or 'O'\n     {\n          calculate and print the old chart\n     }\n\n     else if the input is 'n' or 'N'\n     {\n          calculate and print the new chart\n     }\n\n     else if the input is 'c' or 'C'\n     {\n          calculate and print both the old and new charts (?) (whatever compare does)\n     }\n\n     else if the user wants to exit\n     {\n          end the program (usu. with return 0 or something of that nature)\n     }\n\n     else the user's input was bad\n}\n\nMind that for this to go well, you'd have to mind a few issues -- as I wrote above, if the user doesn't enter what you expect them to. Also, you'd have to deal with a case where the user decided to print charts before entering data into the old or new charts. For that, you should give the temperature a default value, like 0, so the charts can still be calculated.\n\nYou can see that I wrote in a calculation for every print, which is what I was saying before with it not being necessary. You could technically use the for loop I gave for printing but just replace the reference to the array with the calculation for the wind chill.\n\n...Don't know if I covered all of it, but I should be getting to bed. ^^;
895625
This may be unexpected, but a good site with the formulas you need.\n\nhttp://en.wikipedia.org/wiki/Trig
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cot = o/a\ncsc = r/o\n\no = opposite side\na = adjacent side\nr = radius\n\ncalculate csc(2x) in terms of o and r.\nSubstitute o/a for cot(x) and substitue the value you determine for csc(2x) \nmultiply these, then convert to a trig function in terms of x.\n\nIf you know trig identities better than I do, you can reduce this expression using identities, without resorting to o, a, and r.  However, either route will work.
459338
Go to http://mathworld.wolfram.com/Cosine.html
150119
Have you tried the ezplot routine?  You can get help from the context help window.  Mathlab is the bomb.
796870
You can read about it here\n\nhttp://www.quilt.com/HowTo/AppliqueHowToPage.html
877622
ok...is what id imaigne you want is a ruler and compass construction of a line segment and you want to divide it into 4 equal parts...this isnt so bad\n\nis what you need is the bisector construction...and you do this 3 times...so you have your segment...go get yoru compass!\n\nok, now that you have your compass...put the pointy part of the compass on one of the segment endpoints...open teh compass so its open to a little more than half the length of the segment...draw an arc above and below the segment, then do the same thing with the other end point...\n\nyou now notice that those two arcs you drew intersect at 2 points...and if you connect those two points of intersectoin you would bisect the line....ok, so we have divided the line in half...well we need to do that again...\n\nso now go back to the endpoint you started on, and put the pointy end of the compass on the endpoint and open the compass to slightly more than half the distance from the endpoint to the bisector...arc above and below the line, repeat teh steps at the place where the bisector cuts the line...you would noice that these two arcs will intersect above and below the line...and you connect those points, and you have now split one half of the line into quarters...do the same thing for the other half\n\nhope this helps.\n\nmatttlocke
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resource directory for pattern making below: