Solution 18.3b - Revision history http://wiki.sommarmatte.se/wikis/2009/bridgecoursemechanics/index.php?title=Solution_18.3b&action=history Revision history for this page on the wiki en MediaWiki 1.11.1 Wed, 08 Apr 2026 01:43:26 GMT Ian at 17:05, 27 March 2011 http://wiki.sommarmatte.se/wikis/2009/bridgecoursemechanics/index.php?title=Solution_18.3b&diff=1657&oldid=prev <p></p> <table style="background-color: white; color:black;"> <col class='diff-marker' /> <col class='diff-content' /> <col class='diff-marker' /> <col class='diff-content' /> <tr> <td colspan='2' style="background-color: white; color:black;">←Older revision</td> <td colspan='2' style="background-color: white; color:black;">Revision as of 17:05, 27 March 2011</td> </tr> <tr><td colspan="2" class="diff-lineno">Line 1:</td> <td colspan="2" class="diff-lineno">Line 1:</td></tr> <tr><td class='diff-marker'>-</td><td style="background: #ffa; color:black; font-size: smaller;"><div>&lt;math&gt;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}&lt;/math&gt;</div></td><td class='diff-marker'>+</td><td style="background: #cfc; color:black; font-size: smaller;"><div>&lt;math&gt;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}&lt;/math&gt;</div></td></tr> </table> Sun, 27 Mar 2011 17:05:51 GMT Ian http://wiki.sommarmatte.se/wikis/2009/bridgecoursemechanics/index.php/Talk:Solution_18.3b Ian: New page: <math>s=\frac{3\times {{10}^{2}}}{25}-\frac{{{10}^{3}}}{250}=12-4=8\text{ m}</math> http://wiki.sommarmatte.se/wikis/2009/bridgecoursemechanics/index.php?title=Solution_18.3b&diff=1386&oldid=prev <p>New page: &lt;math&gt;s=\frac{3\times {{10}^{2}}}{25}-\frac{{{10}^{3}}}{250}=12-4=8\text{ m}&lt;/math&gt;</p> <p><b>New page</b></p><div>&lt;math&gt;s=\frac{3\times {{10}^{2}}}{25}-\frac{{{10}^{3}}}{250}=12-4=8\text{ m}&lt;/math&gt;</div> Wed, 06 Oct 2010 17:33:22 GMT Ian http://wiki.sommarmatte.se/wikis/2009/bridgecoursemechanics/index.php/Talk:Solution_18.3b