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)![]() ![]() You can put this solution on YOUR website! 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( "----------------------------- \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) \n" ); document.write( "--------------- \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 \n" ); document.write( "\n" ); document.write( " \n" ); document.write( " |