Yahoo Answers: Answers and Comments for Maths Problem help needed? Urgent, please help!? [Mathematics]
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enUS
Fri, 10 Sep 2010 13:27:02 +0000
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Yahoo Answers: Answers and Comments for Maths Problem help needed? Urgent, please help!? [Mathematics]
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https://answers.yahoo.com/question/index?qid=20100910132702AABpTSu
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From Anonymous: Write as decimal:  using known sequences; t...
https://answers.yahoo.com/question/index?qid=20100910132702AABpTSu
https://answers.yahoo.com/question/index?qid=20100910132702AABpTSu
Fri, 10 Sep 2010 15:33:31 +0000
Write as decimal:  using known sequences; this method might not be the best for complicated numbers
1) 4/27
27 is divisible by 9 and 3
You know 4/9 equals .444...
4/9 divided by 3 equals 4/27, so .444... divided by 3 equals 4/27; alternately do 4/3 divided by 9
2) 11/111
111 is divisible by 3 and 37
do the same thing as a above with 11/3 or 11/37
Write as a fraction in its simplest form:
1) 0.015(05)... (that's less ambiguous notation)
0.015 alone is 15/1000 or 3/200
let variable d equal 0.000(05)...
Understand that 0.015(05)... = d + (3/200) because 0.015 + 0.000(05)... = 0.015(05)...
100,000d = 00005.0505... (this is equal to 100,000 times the original number)
5.0505...  0.0505... = 5
0.0505... is equal to 1000d so this equation can be
rewritten 100,000d  1000d = 5, which simplifies to 99,000d = 5; divide both sides of the equation by 99,000 and you get 5/99,000 = d
Since 0.015(05)... = d + (3/200) it can be represented as 5/99,000 + 3/200
The least common denominator is 495, so multiply 3 and 200 by that to get 1,485/99,000
The sum is thus 1,490/99,000 or: 149/9,900 (149 is prime)
Try to apply this to the other problems. I'm done.

From CHRIS: I can't be bothered writing out the workin...
https://answers.yahoo.com/question/index?qid=20100910132702AABpTSu
https://answers.yahoo.com/question/index?qid=20100910132702AABpTSu
Fri, 10 Sep 2010 14:36:20 +0000
I can't be bothered writing out the workings for these problems, but the first recurring decimal should be written in longhand as 0.015015015 etc, as the recurring part is that between and including the recurring symbols.
Good luck.