The answer to the second semi-mathematical puzzle Mike Reed Unregistered Posts: 82 Threads: 15 Joined: Jan 2008 02-14-2013, 10:05 AM " int_(1)^(3^(1/3)) z^2 dz cos(3pi/9) = ln e^(1/3) (Note: Readers outside the US, this works better if you allow yourself the agony of pronouncing the letter “z” as “zee”.) I will post this answer one week from today" OK, the week has passed; here's the answer: The integral of z squared dz from 1 to the cube root of 3 times the cosine of 3 pi over 9 equals log of the cube root of e How did you do in solving it? mike Les Koller Unregistered Posts: 253 Threads: 20 Joined: Jun 2012 02-16-2013, 06:30 PM Not sure how to read it in Lymerick meter...is this close? The integral of z squared dz from one to cube root of 3 times the cos of 3pi divided by nine = the log of the cube root of three Marcus von Cube, Germany Unregistered Posts: 3,283 Threads: 104 Joined: Jul 2005 02-17-2013, 06:40 AM ```The integral of z squared dz from one to cube root of 3 times the cos of 3pi divided by nine = the log of the cube root of three``` The formum software has wrapped the text. What you see here is enclosed in [pre]...[/pre] tags. As an alternative use [nl] at each line end. Derek Walker (UK) Unregistered Posts: 50 Threads: 1 Joined: Sep 2011 02-18-2013, 02:40 AM Shouldn't 'cosine' and 'nine' be at the ends of lines, to rhyme: The integral of z squared dzfrom 1 to the cube root of 3 times the cosine of 3 pi over 9 equals log of the cube root of e « Next Oldest | Next Newest »

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