document.write( "Question 628664: Donna is late for a sales meeting after traveling from one town to another at a speed of 32 mph. If she had traveled 4 mph faster, she could have made the trip in 1/2 hr less time. How far apart are the towns? \n" ); document.write( "
Algebra.Com's Answer #395776 by dfrazzetto(283)![]() ![]() You can put this solution on YOUR website! d = v*t\r \n" ); document.write( "\n" ); document.write( "V1 = 32 mph\r \n" ); document.write( "\n" ); document.write( "V2 = 32+4 = 36 mph\r \n" ); document.write( "\n" ); document.write( "t1 = t\r \n" ); document.write( "\n" ); document.write( "t2 = t - 1/2\r \n" ); document.write( "\n" ); document.write( "distance is from one town to the other, so d1 and d2 are the same, set them equal\r \n" ); document.write( "\n" ); document.write( "v1*t1 = v2*t2\r \n" ); document.write( "\n" ); document.write( "32t = 36(t-1/2)\r \n" ); document.write( "\n" ); document.write( "32t = 36t - 18\r \n" ); document.write( "\n" ); document.write( "4t = 18 \r \n" ); document.write( "\n" ); document.write( "t = 4.5 hrs\r \n" ); document.write( "\n" ); document.write( "d = v1*t1 = v2*t2 = 32(4.5) = 36(4) = 144 miles \n" ); document.write( " |