document.write( "Question 159793: a man at A walks to a neigborhoring town, B, at the rate of 5mph and returns at a alower rate of 4mph. if it took him 3 hours and 9 minutes for the entire trip, how far is town from A. \n" ); document.write( "
Algebra.Com's Answer #117874 by checkley77(12844)![]() ![]() ![]() You can put this solution on YOUR website! D=RT \n" ); document.write( "T=D/R \n" ); document.write( "3.15=D/5+D/4 \n" ); document.write( "3.15=(4D+5D)/20 \n" ); document.write( "3.15=9D/20 CROSS MULTIPLY. \n" ); document.write( "9D=20*3.15 \n" ); document.write( "9D=63 \n" ); document.write( "D=63/9 \n" ); document.write( "D=7 MILES BETWEEN A & B. \n" ); document.write( "PROOF: \n" ); document.write( "3.15=7/5+7/4 \n" ); document.write( "3.15=(4*7+5*7)/20 \n" ); document.write( "3.15=(28+35)/20 \n" ); document.write( "3.15=63/20 \n" ); document.write( "3.15=3.15 \n" ); document.write( " \n" ); document.write( " |