Return Styles: Pseud0ch, Terminal, Valhalla, NES, Geocities, Blue Moon.

Pages: 1-

how do I learned math in 1½ months

Name: Anonymous 2006-04-29 14:44

I'm taking an exam in a month and a half. Here is an old copy: http://www.studyjapan.go.jp/en/toj/pdf/013e.pdf

This is a big problem because I didn't take Algebra II in high school :( How would I go about learning this kind of stuff on the internet in a month and a half (or what are my best chances)?

Name: Anonymous 2006-04-29 16:41

PROTIP: Put 100 in ``Marks.''

Name: Anonymous 2006-04-29 18:25

Ha, I can't even do the first question. Do you get to use a calculator? Is the answer really zero (and if so, why?), or just really close to zero?

Name: Anonymous 2006-04-29 18:44

That first question is straight out of a mathematics competition. They never teach that kind of stuff in actual mathematics classes.

Name: Anonymous 2006-04-29 20:02

>>4 Well they seem to expect me to know how to do it :_(

Name: Anonymous 2006-04-29 21:19

Indeed... The rest of the questions are pretty standard fare, you just need to read through lots of things. If you don't get every question right you'll at least get some. Try tutorial.math.lamar.edu/AllBrowsers/1314/1314.asp for a bunch of free maths notes.

Name: Anonymous 2006-04-29 21:44 (sage)

>>6
THANK YOU that's exactly the kind of thing I was looking for!

Name: Anonymous 2006-04-29 23:27

I just had to figure this out... Question 1 relies on two identities, apparently. They're given at:
1. http://mathworld.wolfram.com/SquareRoot.html from "A nested radical of the form"
2. http://en.wikipedia.org/wiki/Square_root from "the following identity often comes handy"

Name: Anonymous 2009-12-15 19:22

>>1
Fuck, I don't even know if I could pass this exam.

Name: Anonymous 2009-12-15 21:08

>>4
If it's for a scholarship, do you think it'd be so easy that anyone who graduate HS can pass?

Name: Anonymous 2009-12-16 3:14

>>8
You don't need to know them, though; If you start with the assumption that the two parts are equal you can prove it with grade school math.

Name: Anonymous 2009-12-16 13:33

Haskell boobs!

Name: Anonymous 2009-12-16 16:23

>>8
After playing around a little one sooner or later finds that (\sqrt{2} + \sqrt{3})^2 = \sqrt{5+2\sqrt{6}} . That's all there is to it.

Is there a nice way to to 1.3?

Name: Anonymous 2009-12-16 17:42

1.3, god damn.

There are 2 solutions.

Maybe the facts that 4^x = 2^2x, and 2^(x+1) = 2*2^x are useful.

Now I guess one must just take logs in some way...

Name: Anonymous 2009-12-16 19:45

Let y = 2^x.  The equation is

0 = y^2 + 2y - 15 = (y-3)(y+5)

Since 2^x > 0, 2^x = 3 and x = log(3) / log(2)

Name: Anonymous 2009-12-16 19:46

>>15
errr whoops, middle term should be -2y, so switch the 3 and 5.

Name: Anonymous 2009-12-17 12:48

>>15
Wrong answer, according to Maple.

Name: Anonymous 2009-12-17 15:47

>>17
Maple
didn't read >>16

idiot detected.

Name: Anonymous 2009-12-17 17:09

>>18
Still wrong, bubba.

Name: Anonymous 2009-12-17 18:03

>>19

He's right. x = log(5) / log(2)

Mathematica might have slight rounding errors due to numerical storage.

Name: Anonymous 2009-12-17 18:25

I know the front cover says "Special training college students", but is this aimed at high school or university level students?

Name: Anonymous 2009-12-17 18:29

6.4 looks like the most time-consuming/tough question (bar the last hurdle of tedium)

Name: Anonymous 2009-12-17 18:30

The programs say this is a second solution:

x = (i*Pi + Log(3)) / log(2)

Name: Anonymous 2009-12-17 18:35

>>21
High school students applying for scholarships for university, I would guess. The test could be passed with a good high school education, and a bit of preparation.

Name: Anonymous 2009-12-17 18:37

I don't suppose OP is still around, 3 years after he made this thread?

Name: Anonymous 2009-12-17 18:38

>>23

Yeah, that's if solutions are allowed to be complex rather than only real. However there's only one box given for the solution, and given the level of the other questions I think it's safe to assume only real solutions are to be considered.

Name: Anonymous 2009-12-18 3:58

the last question is a joke
Don't change these.
Name: Email:
Entire Thread Thread List
All trademarks and copyrights on this site are owned by their respective parties. All comments and posts are the responsibility of their own posters.
- Tinychan + 2ch Mode -