document.write( "Question 887905: a speed boat takes 3 hours longer to go 45 miles up a river than to return. if the boat cruises at 20 mph in still water, what is the rate of the current \n" ); document.write( "

Algebra.Com's Answer #537193 by lwsshak3(11628) $\"\"$ $\"About$

You can put this solution on YOUR website!
a speed boat takes 3 hours longer to go 45 miles up a river than to return. if the boat cruises at 20 mph in still water, what is the rate of the current
\n" ); document.write( "***
\n" ); document.write( "let c=rate of current
\n" ); document.write( "20+c=speed of boat down river
\n" ); document.write( "20-c=speed of boat up river
\n" ); document.write( "travel time=distance/speed
\n" ); document.write( " $\"45%2F%2820-c%29-45%2F%2820%2Bc%29=3\"$
\n" ); document.write( "lcd: (20+c)(20-c)
\n" ); document.write( " $\"45%2820%2Bc%29-45%2820-c%29=3+%2820%2Bc%29%2820-c%29\"$
\n" ); document.write( "900+45c-900+45c=3(20^2-c^2)=3(400-c^2)=1200-3c^2
\n" ); document.write( "90c=1200-3c^2
\n" ); document.write( "3c^2+90c-1200=0
\n" ); document.write( "c^2+30c-400=0
\n" ); document.write( "(c+40)(c-10)=0
\n" ); document.write( "c=-40 reject
\n" ); document.write( "or
\n" ); document.write( "c=10
\n" ); document.write( "rate of the current=10 mph \n" ); document.write( "

\n" );