Results 1 to 9 of 9

Math Help - Find the Equation

  1. #1
    Newbie
    Joined
    Oct 2006
    Posts
    15

    Question Find the Equation

    Okay, first I wasn't sure what math this is, but I'm pretty sure its Algebra.

    I'm looking to find the equation for this.

    as 1300 ---> 2400
    12---> 0

    it's an equation of adding to 1300, starting at adding 12, until I reach 2400. but the closer to 2400 the equation gets the smaller the addition becomes - ultimately going to an addition of 0 at exactly 2400. Also the addition cannot be a decimal, therefore it must round up.

    mm doing some math if 1300 is +12 and 2400 is +0 this would mean each number on the 1300-2400 scale would be .0109 (or .011 rounded) on the 12 to 0 scale.

    The equation I want would be something where if I said 2250 what would be added given that the addition (which shrinks the larger the number gets) loses .011 per additional number over 1300 on the larger scale. I know (doing the match) that it would add 1.55 to 2250 (rounded up to 2)
    Last edited by Lammalord; February 17th 2011 at 12:30 AM.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Member rtblue's Avatar
    Joined
    Mar 2009
    From
    Birmingham, Alabama.
    Posts
    221
    The question is very vague. At what rate are you adding to 1300? Are you adding in an exponential pattern? quadratic? cubic? logarithmic?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Oct 2006
    Posts
    15
    ah, sorry it starts at 12 and decreases to 1, ie 1300+12, 1312+11.87(rounded up), 1324+11.76(rounded up) ect, ect, till it gets to 2400.

    every 1 point on the top one = a .011 subtraction to the number on the bottom (thats adding to the total), if i got that right.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor
    Joined
    Dec 2007
    From
    Ottawa, Canada
    Posts
    3,161
    Thanks
    70
    Quote Originally Posted by Lammalord View Post
    ah, sorry it starts at 12 and decreases to 1, ie 1300+12, 1312+11.87(rounded up), 1324+11.76(rounded up) ect, ect, till it gets to 2400.
    every 1 point on the top one = a .011 subtraction to the number on the bottom (thats adding to the total), if i got that right.
    Very unclear and confusing.
    Change your problem to something shorter, so we can "see";
    make it from 100 to 200, starting with 12, decreasing by .45:
    Code:
     0                   100.00
     1    12.00  12.00   112.00
     2    11.55  12.00   124.00
     3    11.10  12.00   136.00
     4    10.65  11.00   147.00
     5    10.20  11.00   158.00
     6     9.75  10.00   168.00
     7     9.30  10.00   178.00
     8     8.85   9.00   187.00
     9     8.40   9.00   196.00
    10     7.95   8.00   204.00
    So .45 is deducted constantly from 12, the results rounded up and added...stop once >= 200.
    Is that what you mean?
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Newbie
    Joined
    Oct 2006
    Posts
    15
    not quite, the +12 would lose .011 per number on the 1300-2400 scale.

    so in other words it would add something like so:



    1 12.00 12 1312
    2 11.868 12 1324
    3 11.736 12 1336
    4 11.604 12 1348
    5 11.472 12 1360

    ect, ect - and it needs the .011 per number on the 1300-2400 scale in case a random number was given, say, 2250 - what would be added to it? in this case it would be 1.55 rounded up, that would make it +2, I can do the math one at a time, I just can't figure out how to get an equation for it.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor
    Joined
    Dec 2007
    From
    Ottawa, Canada
    Posts
    3,161
    Thanks
    70
    OK, I get it now (though I wonder what the purpose of this is!)
    To illustrate, I'll carry on with your example, adding column for ".011 per number":

    000 .000 00.000 00 1300
    001 .000 12.000 12 1312
    002 .132 11.868 12 1324 : 12(.011) = .132
    003 .132 11.736 12 1336
    004 .132 11.604 12 1348
    005 .132 11.472 12 1360
    006 .132 11.340 12 1372
    007 .132 11.208 12 1384
    008 .132 11.076 12 1396
    009 .132 10.944 11 1407
    010 .121 10.823 11 1418 : 11(.011) = .121
    011 .121 10.702 11 1429
    .....
    164 .022 01.594 02 2248
    165 .022 01.572 02 2250
    166 .022 01.550 02 2252 : your example of 1.55 resulting from 2250
    .....
    279 .011 00.021 01 2390
    280 .011 00.010 01 2391
    281 .011 -0.001 00 2391
    282 .000 -0.001 00 2391
    283 .000 -0.001 00 2391 : 2400 will not be reached
    .....
    infinity!

    Correct?
    Now I'll let RTBlue take over...
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Newbie
    Joined
    Oct 2006
    Posts
    15
    yes, perfect yeah it wouldn't be reached because of rounding, it's supposed to be .01090909(repeating)
    Follow Math Help Forum on Facebook and Google+

  8. #8
    MHF Contributor
    Joined
    Dec 2007
    From
    Ottawa, Canada
    Posts
    3,161
    Thanks
    70
    Correct:
    ......
    282 .010909... 00.02182... 01 2399
    283 .010909... 00.01090... 01 2400
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Newbie
    Joined
    Oct 2006
    Posts
    15
    mmm, So I still do need an equation for this - anyone?
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 1
    Last Post: May 18th 2011, 11:36 AM
  2. Replies: 8
    Last Post: March 22nd 2011, 05:57 PM
  3. find h'(x): has ln in equation
    Posted in the Calculus Forum
    Replies: 3
    Last Post: March 9th 2009, 08:03 PM
  4. Find the equation
    Posted in the Pre-Calculus Forum
    Replies: 1
    Last Post: November 24th 2008, 06:57 PM
  5. Replies: 5
    Last Post: October 13th 2008, 11:16 AM

Search Tags


/mathhelpforum @mathhelpforum