document.write( "Question 166418: 84.)Gone Fishing. Debbie traveled by boat 5 miles upstream to fish in her favorite spot. Because of the 4 - mph current, it took her 20 minutes longer to get there than to return. how fast will her boat go in still water? \n" ); document.write( "
Algebra.Com's Answer #122703 by Alan3354(69443)\"\" \"About 
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84.)Gone Fishing. Debbie traveled by boat 5 miles upstream to fish in her favorite spot. Because of the 4 - mph current, it took her 20 minutes longer to get there than to return. how fast will her boat go in still water?
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\n" ); document.write( "time = distance/speed
\n" ); document.write( "time upstream = 5/(b - 4) b is the boat's speed
\n" ); document.write( "time downstream = 5/(b+4) = time upstream - 1/3 (20 mins is 1/3 hours)
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\n" ); document.write( "5/(b+4) = 5/(b-4) - 1/3
\n" ); document.write( "Multiply by 3*(b+4)*(b-4)
\n" ); document.write( "15*(b-4) = 15*(b+4) - b^2 + 16
\n" ); document.write( "15b-60 = 15b+ 60 -b^2 + 16
\n" ); document.write( "-60 = 76 - b^2
\n" ); document.write( "b^2 = 136
\n" ); document.write( "b = 11.66 mph\r
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