document.write( "Question 555102: With a strong wind at its back, a ship can make a 425 mile journey in 90 minutes less than it can without the breeze. If its average speed is increased by 15 mph with the aid of the wind, about how long did the joutney take with the breeze? \n" ); document.write( "
Algebra.Com's Answer #361702 by ankor@dixie-net.com(22740) You can put this solution on YOUR website! With a strong wind at its back, a ship can make a 425 mile journey in 90 minutes less than it can without the breeze. \n" ); document.write( " If its average speed is increased by 15 mph with the aid of the wind, about how long did the journey take with the breeze? \n" ); document.write( ": \n" ); document.write( "From the information given, we can find the speed of the boat in still water \n" ); document.write( "Let s = boat speed in still water \n" ); document.write( ": \n" ); document.write( "write a time equation, change 90 min to 1.5 hrs \n" ); document.write( ": \n" ); document.write( "no breeze time - with breeze time = 1.5 hrs \n" ); document.write( " \n" ); document.write( "Multiply by s(s+15) \n" ); document.write( "425(s+15) - 425s = 1.5s(s+15) \n" ); document.write( "425s + 6375 - 425s = 1.5s^2 + 22.5s \n" ); document.write( "6375 = 1.5s^2 + 22.5s \n" ); document.write( "A quadratic equation \n" ); document.write( "1.5s^2 + 22.5s - 6375 = 0 \n" ); document.write( "Simplify divide by 1.5 \n" ); document.write( "s^2 + 15s - 4250 = 0 \n" ); document.write( "Use the quadratic formula to find the s \n" ); document.write( "I got a positive solution of s = 58.12 mph \n" ); document.write( ": \n" ); document.write( "\" If its average speed is increased by 15 mph with the aid of the wind, about how long did the journey take with the breeze?\" \n" ); document.write( " \n" ); document.write( ": \n" ); document.write( "Check this by finding the time without the wind \n" ); document.write( " \n" ); document.write( ": \n" ); document.write( "7.31 - 5.81 = 1.5 hrs \n" ); document.write( " |