document.write( "Question 26014: Could you please help me with this problem: The excursion boat \"Holiday\" travels 35 km upstream and then back again in 4hours and 48 minutes. If the speed on the \"Holiday\" in still water is 15 km/h, what is the speed of the current? \r
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Algebra.Com's Answer #13976 by Paul(988)\"\" \"About 
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4 hours and 48 minutes = 4.8hours
\n" ); document.write( "Still water speed = 15
\n" ); document.write( "Speed with the current = 15+x
\n" ); document.write( "Speed against the current = 15-x
\n" ); document.write( "total distance == 35
\n" ); document.write( "Equation:
\n" ); document.write( "\"35%2F15%2Bx+-+35%2F15-x=4.8\"
\n" ); document.write( "35[(15+x)+(15-x)]=4.8[(15+x)(15-x)] SIMPLIFY
\n" ); document.write( "\"35%2830%29=4.8%28-x%5E2%2B225%29\"
\n" ); document.write( "\"1050=-4.8x%5E2%2B1080\"
\n" ); document.write( "\"4.8x%5E2=1080-1050\"
\n" ); document.write( "\"4.8x%5E2=30\"
\n" ); document.write( "\"4.8x%5E2%2F4.8=30%2F4.8\"
\n" ); document.write( "\"x%5E2=6.25\"
\n" ); document.write( "\"x=sqrt%286.25%29%29\"
\n" ); document.write( "x=2.5\r
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\n" ); document.write( "\n" ); document.write( "Hence, the speed of the current is 2.5km/h.
\n" ); document.write( "Paul.
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