document.write( "Question 1109935: AN AIRPLANE FLYING EAST AT THE RATE OF 6 MILES/MINUTES ,PASSED OVER A COURT-HOUSE AT 2 PM. A SECOND PLANE FLYING NORTH AT THE RATE OF 6 MILES/MINUTES,PASSED OVER THE COURT-HOUSE AT 2:04 PM.IV THE PLANES ARE FLYING AT THE SAME ALTITUDE ,IN HOW MANY MINUTES AFTER 2 PM WILL THEY BE 36 MILES APART? \n" ); document.write( "
Algebra.Com's Answer #724915 by Alan3354(69443) You can put this solution on YOUR website! AN AIRPLANE FLYING EAST AT THE RATE OF 6 MILES/MINUTES ,PASSED OVER A COURT-HOUSE AT 2 PM. A SECOND PLANE FLYING NORTH AT THE RATE OF 6 MILES/MINUTES,PASSED OVER THE COURT-HOUSE AT 2:04 PM. IV THE PLANES ARE FLYING AT THE SAME ALTITUDE,IN HOW MANY MINUTES AFTER 2 PM WILL THEY BE 36 MILES APART? \n" ); document.write( "--------------- \n" ); document.write( "p = distance of the 1st plane from the courthouse at t minutes past 2 \n" ); document.write( "q = distance of the 2nd plane from the courthouse at t minutes past 2 \n" ); document.write( "----- \n" ); document.write( "t=0 at 2:04 \n" ); document.write( "p = 6t + 4*6 = 6t+24 \n" ); document.write( "q = 6t \n" ); document.write( "--- \n" ); document.write( "(6t+24)^2 + (6t)^2 = 36^2 \n" ); document.write( "(t+4)^2 + (t)^2 = 36 \n" ); document.write( "2t^2 + 8t -20 = 0 \n" ); document.write( "
\n" ); document.write( "\n" ); document.write( "======================== \n" ); document.write( "t =~ 1.74 minutes past 2:04 \n" ); document.write( "--> 5.74 minutes past 2:00\r \n" ); document.write( "\n" ); document.write( " \n" ); document.write( " |