<PRE><B>Question 194734</B>
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If the boat goes twice as far, it must be going twice as fast.  So if the speed of the boat in still water is &lt;b&gt;&lt;i&gt;r&lt;/i&gt;&lt;/b&gt;, then it must be going &lt;b&gt;&lt;i&gt;r&lt;/i&gt; - 2&lt;/b&gt; against the 2 mph current, and &lt;b&gt;&lt;i&gt;r&lt;/i&gt; + 2&lt;/b&gt; with the current.


And since it is going twice as fast with the current as against the current:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ r + 2 = 2(r - 2)]


Just solve for &lt;b&gt;&lt;i&gt;r&lt;/i&gt;&lt;/b&gt;.



John
*[tex \LARGE e^{i\pi} + 1 = 0]
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