document.write( "Question 735649: \r
\n" ); document.write( "\n" ); document.write( "A tugboat goes 180 miles upstream in 10 hours. The return trip downstream takes 5 hours. Find the speed of the tugboat without a current and the speed of the current.
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Algebra.Com's Answer #449508 by checkley79(3341)\"\" \"About 
You can put this solution on YOUR website!
D=RT
\n" ); document.write( "180=(R-C)10
\n" ); document.write( "180=10R-10C DIVIDE BY 2 & ADD.
\n" ); document.write( "180=(R+C)5
\n" ); document.write( "180=5R+5C
\n" ); document.write( "90=5R-5C
\n" ); document.write( "---------------
\n" ); document.write( "270=10R
\n" ); document.write( "R=270/10
\n" ); document.write( "R=27 MPH. ANS. FOR THE RATE OF THE BOAT.
\n" ); document.write( "180=(27-C)10
\n" ); document.write( "180=270-10C
\n" ); document.write( "10C=270-180
\n" ); document.write( "10C=90
\n" ); document.write( "C=90/10
\n" ); document.write( "C=9 MPH. ANS.FOR THE CURRENT.
\n" ); document.write( "PROOF:
\n" ); document.write( "180(27+9)5
\n" ); document.write( "180=36*5
\n" ); document.write( "180=180
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