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About Petanquell

  • Rank
  • Birthday 11/27/1991

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  • Location
    Czech Republic
  • Interests
    Music, drums, frisbee and american trains made in late sixtees
  • Favorite Area of Science
    Mathematics, biology and maybe physics
  • Occupation
    Be useful :)
  1. I thing Big Nose said it perfect. [math] 8^{x^2 - 2x} = \frac{1}{2} [/math] the first step is: [math] 8^{x^2 - 2x} = 8^{-\frac{1}{3}}[/math] using fact that [math]\frac{1}{2}=8^{-\frac{1}{3}}[/math] Than using the old rule "when bases are equal than exponents must be equal too" or with "more official" way, use logarithm with base 8 on both sides we get: [math]\mathrm{log_{8}}(8^{x^2 - 2x})=\mathrm{log_{8}}(8^{-\frac{1}{3}}) [/math] [math](x^2 - 2x) \cdot \mathrm{log_{8}}(8)=-\frac{1}{3} \cdot \mathrm{log_{8}(8)} [/math] and we get to easy equation [math] x^2 - 2x= -\frac{1}{3} [
  2. I've found only this but it's barely useful... http://www.newsobserver.com/news/story/1591089.html Pq
  3. Hey guys, I've never got problem with solving quadratic, logarithmic or goniometric equations, but I don't know how to solve them when they are all together. The most simple example is: [math] \textup{log}(x)+\textup{sin}(x)+x^2+x+1=0 [/math] Is there any way to solve them without "computer help"? Thanks, pq
  4. Page 40 contains this exercise: [math](x+1)^2=-25[/math] [math]x+1=\pm-25[/math] [math]x=-1\pm-25[/math] [math]x=-1-25 \textrm{ or } x=-1+25[/math] [math]x=-26,24[/math] and note probably about complex solution..(?) What tells you that it hasn't got real solutions? Pq
  5. Well, I hate people doing this but it's well known trick: 1) go to the shop 2) buy the most expensive calculator 3) do an exam 4) complain and return your calculator to the shop Pq
  6. Of course there are some things we can't understand on the first sight, but like Feynmann said: "It's like a park with huge number of ways. If you can't find one, there are some other reaching the same place." Or other point of view, everything must be proved so folowing the prove must lead to understanding. Could this be a way to avoid incomprehension? Pq
  7. It will be grievous to have a limit on high school....
  8. Hey guys, I talked with a guy who studies mathematics and he told me that his friends from collage found their "limits" (mostly in third grade) and now the don't have a clue what's going on, and just learn mechanically like medival scholars did. He also told me that only some chosen ones still understand it. Does anyone you know of feel a similar effect? And now something completly diffrent, will you press (softly, of course) on your child to do math?
  9. I'll try it explain with my words despite my level of english. I have three very related reasons: The first reason should be clear, I'm bored and curious. In a curious-period I wonder how things work and how they were invented. In a bored-period I just sit and read, or work on a book I'm writing for my schoolmates, because they complain about the speed of our math teacher (she's really fast, I guess it's about 180 words per minute ). The second reason is a little related to the book I write. I want to motivate others to do mathematics because I consider mathematics as a basic knowled
  10. Well , we haven't learned trigonimetric functions yet so I can't see it there. But I'll take a look sometime... And that conversation, I don't remember why I did it. Maybe because Maple doesn't take degrees, really don't know. pq
  11. Ok, i'll try First of all, I coversed degrees to radians [math]T*\sin(37\pi/180)+\frac{4}{3}*T*\sin(53\pi/180)-100N = 0[/math] [math]T*\sin(37\pi/180)+\frac{4}{3}*T*\sin(53\pi/180)=100N[/math] [math]T(\sin(37\pi/180)+\frac{4}{3}*\sin(53\pi/180))=100N[/math] [math]T=\frac{100}{\sin(37\pi/180)+\frac{4}{3}*\sin(53\pi/180)}N[/math] And aproximate... pq
  12. is it [math] T*\sin{37}+1.33*T*\sin{53}-100N=0[/math] ? Eventually, can I use 4/3 exept 1,33 ?
  13. Much better jetpack have Yves Rossy here.. That's impressive... Does someone here fly with hang*glider? It must be possible to fly with something like it even if it will be grafted on your back...
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