
Topic review (newest first)
 Jenilia
 20060326 17:20:25
 RickyOswaldIOW
 20060325 16:19:11
I'm 19 and studying alevel mathematics. You're 12 you say?
 George,Y
 20060324 02:09:05
By the way, is SAT that HaaaaarD?
 George,Y
 20060324 02:07:34
Yeah, as Mr Hollis put in his Caculus book, we are using sum of infinite series IMPLICITLY.
Numbers are a theory, "theory" means approximation
0.99... is accurate to leave perhaps only one particle uncollected from recursively cutting a cake into equally 10 pieces, then collecting 9 and leaving the other one to be next round cutee.
 ganesh
 20060323 21:56:52
How else can 0.9999......recurring indefinitely be expressed as a fraction? It may appear inaccurate, but think of it, you may never encounter the number 0.999999......... in any area of mathematics!
 Jenilia
 20060323 21:53:01
But then it ist accurate right?
 ganesh
 20060323 21:24:35
All recurring decimals, that is decimal numbers where numbers are repeated without ending after the decimal, can be converted into fractions. For example, 0.21212121..... can be coverted into a fraction this way. Let x = 0.212121.... 100x = 21.212121.... Finding the difference of the two, 99x=21, therefore, x=21/99 or 7/33.
When we try to convert 0.99999.. this way, this is what happens: Let x = 0.99999999.... 10x = 9.99999999.... Finding the difference of the two, 9x=9 or x=1. Yes, 0.999999......... cannot be expressed as a fraction and can only be written as 1!
 ganesh
 20060323 21:17:32
Jenilia, To find the sum of 1/5 + 1/50 + 1/500 + 1/5000..., rewrite it as 1/5(1+1/10+1/100+1/1000....) The series of numbers inside the bracket forms a Geometric Progression. The sum would be 1/(11/10)=1/(9/10)=10/9. Therefore, the sum 1/5 + 1/50 + 1/500 + 1/5000... = 1/5(10/9)=2/9=0.2222......
This is the answer you got!
 Jenilia
 20060323 20:39:44
Another problem: Find the fraction in its simplest form of 1/5 + 1/50 + 1/500 + 1/5000... I do know how to work this out if it is correct, I can simplify it as 0.2+0.02+0.002...=0.2222... I'm using Algebra, Let x be 0.222... 10x=2.222 9x (10xx)=2 x=2/9 Now, If I have a No. like 0.999999...,is there anyway to simplify it?
 Jenilia
 20060319 23:08:13
Thank you sooo much ganesh!
 ganesh
 20060318 17:33:02
Q 30: Team D wins against A and B, draws with C and has a total of 7 points. Team C wins against B, draws with D and A, and has a total of 5 points. Team B wins against A, losses to both D and C, and has a total of 3 points. Team A draws with C, losses to D and B, and has a total of 1 points.
Points Tally:
Team Played Won Lost Drawn Points D 3 2 0 1 7 C 3 1 0 2 5 B 3 1 2 0 3 A 3 0 2 1 1
 ganesh
 20060318 17:18:22
Jenilia, I shall try Q22 first. The question is to find the total of the first 100 numbers of the series. 1 to 9 is 9 numbers. It can be seen that thereafter, 10, 11, 12 etc have been given as two separate digits. 10 to 99 would be 2 digits each, therefore, 180 numbers. But we require only 99 more of these. Hence, 10 to 54 is 45 numbers and the 5 in 55 is to be taken. 1 occurs 16 times, 2 occurs 16 times, 3 occurs 16 times, 4 occurs 16 times, 5 occurs 11 times, 6, 7, 8, and 9 occur 5 times each. Therefore, the total would be 16(1+2+3+4) + 11(5)+5(6+7+8+9) =16(10) + 55 + 5(30) = 160 + 55 + 150 = 365.
 Jenilia
 20060318 13:51:33
Q30, Four football teams A,B,C and D are in the same group. Each team plays 3 matches, one with each of the other 3 teams. The winner of each match scores 3 points; the loser scores 0 points; and if a match is a draw, each team scores 1 point. After all the matches, the results are as follows: (1) The total score of 3 matches for the 4 teams are consecutive odd numbers. (2) D has the highest total score. (3) A has exactly 2 draws, one of which the match with C. Find the total score for each team.
 Jenilia
 20060318 13:43:05
It looks like I have another problem here. It is taken from the Singapore Mathematical Olympiad 2002. Q22, Find the sum of the first 100 No. in the following sequence. 1,2,3,4,5,6,7,8,9,1,0,1,1,1,2,1,3,1,4,1,5,1,6,1,7,1,8,1,9,2,0,...
 Jenilia
 20060317 23:53:20
Thanks so much for helping me. I definitely understand better now!
