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document.write( "if a child walks at 5 km/h from his house and reaches his school 3 minutes late.however,if he walks at 6km/h,he reaches 5 minutes early.how far is the school from his house? \n" );
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Algebra.Com's Answer #430181 by checkley79(3341)![]() ![]() ![]() You can put this solution on YOUR website! D=RT \n" ); document.write( "D=5(T+3) \n" ); document.write( "D=6(T-5) \n" ); document.write( "5(T+3)=6(T-5) \n" ); document.write( "5T+15=6T-30 \n" ); document.write( "5T-6T=-30-15 \n" ); document.write( "-T=-45 \n" ); document.write( "T=45 MINUTES IS THE TIME. \n" ); document.write( "PROOF: \n" ); document.write( "5(45+3)=6(45-5) \n" ); document.write( "5*48=6*40 \n" ); document.write( "240=240 \n" ); document.write( "D=5(45+3) 5 KM/H* 48 MIN. \n" ); document.write( "D=5*.8 \n" ); document.write( "D=4 KM. DISTANCE TO THE SCHOOL. \n" ); document.write( "D=6(45-5) 6KM/H* 40 MIN. \n" ); document.write( "D=6*2/3 \n" ); document.write( "D=12/3=4 KM. DITTO. \n" ); document.write( " |