http://wiki.sommarmatte.se/wikis/2009/bridgecoursemechanics/index.php?action=history&feed=atom&title=Solution_18.3b
Solution 18.3b - Revision history
2026-04-08T01:44:41Z
Revision history for this page on the wiki
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http://wiki.sommarmatte.se/wikis/2009/bridgecoursemechanics/index.php?title=Solution_18.3b&diff=1657&oldid=prev
Ian at 17:05, 27 March 2011
2011-03-27T17:05:51Z
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<td colspan='2' style="background-color: white; color:black;">Revision as of 17:05, 27 March 2011</td>
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<tr><td class='diff-marker'>-</td><td style="background: #ffa; color:black; font-size: smaller;"><div><math>s=\frac{3\times {{10}^{2}}}{<del style="color: red; font-weight: bold; text-decoration: none;">25</del>}-\frac{{{10}^{3}}}{250}=<del style="color: red; font-weight: bold; text-decoration: none;">12</del>-4=<del style="color: red; font-weight: bold; text-decoration: none;">8</del>\text{ m}</math></div></td><td class='diff-marker'>+</td><td style="background: #cfc; color:black; font-size: smaller;"><div><math>s=\frac{3\times {{10}^{2}}}{<ins style="color: red; font-weight: bold; text-decoration: none;">50</ins>}-\frac{{{10}^{3}}}{250}=<ins style="color: red; font-weight: bold; text-decoration: none;">6</ins>-4=<ins style="color: red; font-weight: bold; text-decoration: none;">2</ins>\text{ m}</math></div></td></tr>
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Ian
http://wiki.sommarmatte.se/wikis/2009/bridgecoursemechanics/index.php?title=Solution_18.3b&diff=1386&oldid=prev
Ian: New page: <math>s=\frac{3\times {{10}^{2}}}{25}-\frac{{{10}^{3}}}{250}=12-4=8\text{ m}</math>
2010-10-06T17:33:22Z
<p>New page: <math>s=\frac{3\times {{10}^{2}}}{25}-\frac{{{10}^{3}}}{250}=12-4=8\text{ m}</math></p>
<p><b>New page</b></p><div><math>s=\frac{3\times {{10}^{2}}}{25}-\frac{{{10}^{3}}}{250}=12-4=8\text{ m}</math></div>
Ian