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DRAGONMOUNT

A WHEEL OF TIME COMMUNITY

Zardi

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About Zardi

  • Birthday 09/12/1984
  1. I'd like to be a mentor. I was signed up at one point and got a minty, but she disappeared within a week or so. I dunno if you still have me on the list or not, but if not, please add me.
  2. Out of the country and away from internet from March 13th - March 23rd.
  3. Well...apparently not. I wouldn't post in any thread without reading the first post.
  4. *is totally confused* I guess I don't come in here often enough.
  5. I don't usually think of baby birds as being cute, but these are. :D
  6. Whoops, sorry nephi, didn't see that. It's up to you :) You're the one with the quilt, though, so if they want a picture of it, maybe it should be you :-\ Or you can take a picture and I'll write a bit about it, doesn't matter :) I haven't had time to do much of anything except school work lately, but I hope to get a lot more patches finished. I'll try my best, I promise :( If we don't get enough done, though, we could still make a nice pillow case or something similar. I'm trying to picture 100 hexagons, and it isn't working too well :P
  7. Nyn, are you allowed to use technology to solve this (like everyone has done so far), or do you have to explain how to do it by hand? Here's a way to do it by hand: x^2 + y = 41 y^2 + x = 31 Write the second one as x = 31 - y^2 Then plug that into the first one to get: (31-y^2)^2 + y = 41 ...which simplifies to: y^4 - 62y^2 + 31^2 + y - 41 = y^4 - 62y^2 + y + 920 = 0 Then, you look at the factors of 920: 2, 5, and 23, and you try dividing the polynomial by (y - factor) or (y + factor). In this case, you can divide by (y - 5), so then you get: (y-5)(y^3 + 5y^2 - 37y - 184) = y^4 - 62y^2 + y + 920 = 0 This shows you that the two graphs intersect at y = 5. Plug that into one of your original equations to get x = 6, so the first solution you find is (6,5). Now, you can use any method of solving cubic equations to solve y^3 + 5y^2 - 37y - 184 = 0. You may have learned some in your class, but if not, try here: http://en.wikipedia.org/wiki/Cubic_equation#Cardano.27s_method You might still need a basic calculator to approximate the fractions (they will get very messy otherwise), but at least you won't need a graphing calculator to do all the work for you.
  8. Ironcross, that looks fantastic!! Thank you ;D
  9. No set deadline right now. I have too many to keep track of this month in real life, so I'll start collecting things in earnest for the card sometime in November.
  10. Tk, send it to bob_queen_of_dragons at hotmail dot com. Please put something like "Harriet's card" in the subject line so I don't think it's junk mail. Images on the signature pages will be a few inches long on each side, so please make sure your image is at least 900x900 pixels when you send it to me :)
  11. I'll be taking care of signatures, but not for a couple of weeks. I have too many deadlines to keep track of this month. I'll announce something here when I'm ready to take people's signatures :) Someone else was going to do a collage as well...not sure what's happening with that just yet. We will post all that info when it's ready :) And of course you can be added to the list of crocheters ;D
  12. I'm putting the card/book together with all the signatures and stuff, so you can just type your submission here and let me know that it's the final copy, and I'll include it. If you need special formatting of some sort, then you can send a Word document to bob_queen_of_dragons at hotmail dot com Right now we have several people who say they're interested in writing and I didn't want to leave anyone out, so let's say just one per person for now.
  13. We want to have the whole thing finished in time for Christmas, but of course, neph has to have time to assemble the whole thing. There isn't a definite deadline yet. You can use any colors you want :)
  14. Gondring, I'd love to see it! Faelene, your hexagons look fine to me. Your stitches are very even. When you said they looked crappy I was picturing a huge knotted mess :P
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