Page 3 of 7 FirstFirst 12345 ... LastLast
Results 21 to 30 of 65

Thread: Math problems

  1. #21
    darkus-s6 is offline Senior Member Always Around
    Join Date
    Jul 2007
    Location
    somewhere but not here
    Posts
    1,968

    Default

    the gravitational force on a mass inside a spherical shell = 0
    when a mass m is inside a simmetric spherical distribution of mass, that part of the mass outside its radius doesn't contribute to the net force on it
    Last edited by darkus-s6; 02-04-2008 at 06:56 PM.

    "sleep with me tonight, sweety" ~sakura_hana~

  2. #22
    adonai is offline Senior Member Community Builder
    Join Date
    Aug 2006
    Posts
    5,630

    Default

    Correct, though I have no idea what you're trying to say in that second sentence.

  3. #23
    darkus-s6 is offline Senior Member Always Around
    Join Date
    Jul 2007
    Location
    somewhere but not here
    Posts
    1,968

    Default

    ^ that's the shell theorem

    Ok, so here is another one:

    13 musicians are playing a song. 6 of them are singing, 4 are playing tambourine, and 8 are playing some other instruments. How many are playing tambourine and some other instrument, if there are two musicians that only sing?

    "sleep with me tonight, sweety" ~sakura_hana~

  4. #24
    Volvogga's Avatar
    Volvogga is offline Senior Member Always Around
    Join Date
    Sep 2005
    Location
    MI, USA
    Posts
    2,076

    Default

    You mean 18 musicians are playing a song?
    If so, then 16... if not... uhh... well lets see:

    Assuming no one is extremely talented with their feet, then we would conclude that tambourine and 'other instruments' take up the 'instrument' slot for a person. Beyond that, we assume microphone stands and what not for various 'other instruments' allowing one to play and sing at the same time. With that in mind, we first deal with tambourine and other, being 4 and 8 respectively. The sum of this is 12. There are 12 musicians out of 13, so one person must exclusively sing.

    The new lineup is only 2 singers that ONLY sing, so we know the 1 must be the singer from before, and the other is most likely a dual singer/player again that has given up the instrument... making the total players one less than the previous total.

    Answer: 11.
    Vol~

    thanks to Silverwmoon!

  5. #25
    irecinius is offline Senior Member Community Builder
    Join Date
    Oct 2005
    Location
    FL
    Posts
    2,785

    Default

    There are 13 playing
    minus 2 that only sing, leaving us with 11 musicians that are playing any instrument (tambourine + others).

    This would also mean one of them is dual plyaing the tambourine and other instrument, and maybe even singing at the same time.
    Last edited by irecinius; 02-07-2008 at 01:56 AM. Reason: I wanted to say 12..kukuk
    "Chile is a thin and tall country"

  6. #26
    darkus-s6 is offline Senior Member Always Around
    Join Date
    Jul 2007
    Location
    somewhere but not here
    Posts
    1,968

    Default

    Yes there is only one musician playing tambourine, an other instrument and possibly singing too, you're right irecinius

    "sleep with me tonight, sweety" ~sakura_hana~

  7. #27
    Volvogga's Avatar
    Volvogga is offline Senior Member Always Around
    Join Date
    Sep 2005
    Location
    MI, USA
    Posts
    2,076

    Default

    Oh, hahahaaha. Damn, good one.
    I read that completely wrong. XD

    Irecinius, all you buddy.

    I like this thread.
    Vol~

    thanks to Silverwmoon!

  8. #28
    irecinius is offline Senior Member Community Builder
    Join Date
    Oct 2005
    Location
    FL
    Posts
    2,785

    Default

    lol... its my turn then? i'm not very good with questions, well lets see~

    ABCDE * 4 = EDCBA, find A, B, C, D and E where each is a unique integer from 0 to 9.
    "Chile is a thin and tall country"

  9. #29
    adonai is offline Senior Member Community Builder
    Join Date
    Aug 2006
    Posts
    5,630

    Default

    Are you sure that that's possible?

    I'm starting at the edge, so A could be 0, 1, or 2, since 4A cannot be > 9.

    It can't be 0; then E would have to be be 5 (5*4=20), but that's impossible since even 9999*4 != 5xxxx

    It can't be 1 since 4E != x1, no matter what E is.

    It can't be 2 since E must then be 8 or 9, 8*4=32, 9*4=36, neither of which ends in 2.

  10. #30
    irecinius is offline Senior Member Community Builder
    Join Date
    Oct 2005
    Location
    FL
    Posts
    2,785

    Default

    Yes it is, I'll give that you started on the right track, but derailed midway...


    [Edit] I can just give the answer if someone wants.
    Last edited by irecinius; 02-08-2008 at 04:21 PM.
    "Chile is a thin and tall country"

 

 
Page 3 of 7 FirstFirst 12345 ... LastLast

Bookmarks

Posting Permissions

  • You may not post new threads
  • You may not post replies
  • You may not post attachments
  • You may not edit your posts
  •  
vBulletin Skin by: ForumThemes.com
Powered by vBulletin® Version 4.2.0
Copyright © 2014 vBulletin Solutions, Inc. All rights reserved.
SEO by vBSEO 3.6.0 PL2

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79