document.write( "Question 157645: this is is kinda confusing, need your help guys.\r
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document.write( "alan and dave leave from the same point driving in opposite directions, alan driving at 55 miles pero hour and dave at 65 miles per hour. alan has a one hour head start. how long will they be able to talk on their phones if the phones have a 250 mile range? \n" );
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Algebra.Com's Answer #116211 by ptaylor(2198)![]() ![]() You can put this solution on YOUR website! OK \n" ); document.write( "Distance(d) equals Rate(r) times Time(t) or d=rt; r=d/t and t=d/r\r \n" ); document.write( "\n" ); document.write( "So, before Dave gets started, Alan has already driven 55*1 or 55 miles (and they have talked an hour on their phones)\r \n" ); document.write( "\n" ); document.write( "Let t=time it takes them to get 250 mi apart after Dave leaves\r \n" ); document.write( "\n" ); document.write( "Total distance Alan drives 55+55t \n" ); document.write( "Total distance Dave drives=65t\r \n" ); document.write( "\n" ); document.write( "Now when the above two distances add up to 250 mi, they will be at the outer limits of phone reception, so: \n" ); document.write( "55+55t+65t=250 subtract 55 from each side \n" ); document.write( "55-55+65t+55t=250-55 collect like terms \n" ); document.write( "120t=195 divide each side by 120 \n" ); document.write( "t= 1 5/8 hr or 1hr 37.5 min--------time it takes to get 250 mi apart after Dave leaves\r \n" ); document.write( "\n" ); document.write( "Time that they can talk on the phone is the 1 hour that Alan was driving alone plus the 1 5/8 hour that the were both driving and this equals 2 5/8 hrs\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "Hope this helps---ptaylor \n" ); document.write( " |