document.write( "Question 56574: Plese help me out on this.I dont know what equation to use. Here is the problem-A cheeta is 300 ft. from its prey.It starts to sprint towards its prey at 90 ft. per second. At the same time, thee prey starts to sprint at 70 ft. per second. When will the cheeta catch its prey? \n" ); document.write( "
Algebra.Com's Answer #38403 by ankor@dixie-net.com(22740)![]() ![]() You can put this solution on YOUR website! A cheeta is 300 ft. from its prey.It starts to sprint towards its prey at 90 ft. per second. At the same time, thee prey starts to sprint at 70 ft. per second. When will the cheeta catch its prey?\r \n" ); document.write( "\n" ); document.write( "Let d = distance the prey travel when it is caught \n" ); document.write( "Then (d + 300) = distance the cheetah travels when it catches \n" ); document.write( ": \n" ); document.write( "Both animals will be traveling for same number of seconds \n" ); document.write( "Write time equation: t = dist/speed \n" ); document.write( ": \n" ); document.write( "Cheetah's time = Prey's time \n" ); document.write( " \n" ); document.write( ": \n" ); document.write( "Cross mult: \n" ); document.write( "70(d+300) = 90d \n" ); document.write( "70d + 21000 = 90d \n" ); document.write( "21000 = 90d - 70d \n" ); document.write( " 20d = 21000 \n" ); document.write( " d = 21000/20 \n" ); document.write( " d = 1050 ft traveled by the unfortunate one \n" ); document.write( ": \n" ); document.write( "and 1350 ft traveled by the hungry one: \n" ); document.write( ": \n" ); document.write( "Time = 1350/90 = 15 sec \n" ); document.write( ": \n" ); document.write( "Check: 1050/70 = 15 sec also\r \n" ); document.write( "\n" ); document.write( " \n" ); document.write( " |