document.write( "Question 222063: Debbie traveled by Boat 5 miles upstream to fish. Because of the 4mph 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 #166436 by ankor@dixie-net.com(22740)\"\" \"About 
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Debbie traveled by boat 5 miles upstream to fish.
\n" ); document.write( " Because of the 4 mph current, it took her 20 minutes longer to get there
\n" ); document.write( " than to return.
\n" ); document.write( "How fast will her boat go in still water?
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\n" ); document.write( "Let s = boat speed in still water
\n" ); document.write( "then
\n" ); document.write( "(s-4) = speed upstream
\n" ); document.write( "and
\n" ); document.write( "(s+4) = speed down stream
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\n" ); document.write( "Convert 20 min to hrs: 20/60 = \"1%2F3\"hr
\n" ); document.write( ":
\n" ); document.write( "Write a time equation: time = \"distance%2Fspeed\"\r
\n" ); document.write( "\n" ); document.write( "Upstream time = downstream time + 20 minutes (1/3 hr)
\n" ); document.write( "\"5%2F%28%28s-4%29%29\" = \"5%2F%28%28s%2B4%29%29\" + \"1%2F3\"
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\n" ); document.write( "To clear out these denominators, multiply equation by 3(s+4)(s-4)
\n" ); document.write( "3(s+4)(s-4)*\"5%2F%28%28s-4%29%29\" = 3(s+4)(s-4)*\"5%2F%28%28s%2B4%29%29\" + \"1%2F3\"*3(s+4)(s-4)
\n" ); document.write( "Results in
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\n" ); document.write( "15(s+4) = 15(s-4) + (s+4)(s-4)
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\n" ); document.write( "15s + 60 = 15s - 60 + s^2 - 16
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\n" ); document.write( "0 = 15s - 15s - 60 - 60 + s^2 - 16
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\n" ); document.write( "0 = s^2 - 136
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\n" ); document.write( "s^2 = 136
\n" ); document.write( "s = \"sqrt%28136%29\"
\n" ); document.write( "s = 11.66 mph speed in still water
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\n" ); document.write( "Check solution on calc: (speed up = 7.66 and speed down = 15.66)
\n" ); document.write( "5/7.66 - 5/15.66 = .3333 which is 1/3 of an hr
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