document.write( "Question 937920: A 150 minute river cruise goes 14km upstream and then back again. The river has a current of 3km an hour going downstream. What is the boat's speed in still water and how long was the upstream journey in minutes? \n" ); document.write( "
Algebra.Com's Answer #571346 by mananth(16946)![]() ![]() You can put this solution on YOUR website! boat speed x km Time forward + time return = hours \n" ); document.write( "current speed 3 km \n" ); document.write( " \n" ); document.write( "against current x- 3 km/h \n" ); document.write( "with current x+ 3 km/h \n" ); document.write( " \n" ); document.write( "Distance= 14 km \n" ); document.write( " \n" ); document.write( "Time against + time with = 2.5 hours \n" ); document.write( "t=d/r \n" ); document.write( " \n" ); document.write( "14 /( x + 3 ) + 14 \n" ); document.write( " \n" ); document.write( "LCD = (x - 3 ) ( x + \n" ); document.write( "14 *( x - 3 ) + 14 \n" ); document.write( "14 x - -42 + 14 x + \n" ); document.write( "28 x = 2.5 x ^2 -22.5 \n" ); document.write( "2.5 x ^2 - -28 x - -22.5 = \n" ); document.write( " \n" ); document.write( "Find the roots of the equation by quadratic formula \n" ); document.write( " \n" ); document.write( "a= 2.5 , b= -28 , c= -22.5 \n" ); document.write( " \n" ); document.write( "b^2-4ac= 784 + 225 \n" ); document.write( "b^2-4ac= 1009 \n" ); document.write( " \n" ); document.write( " \n" ); document.write( " \n" ); document.write( "x1=( 28 + 31.76 )/ 5 \n" ); document.write( "x1= 11.95 \n" ); document.write( " \n" ); document.write( "x2=( 28 -31.76 ) / 5 \n" ); document.write( "x2= -0.75 \n" ); document.write( "Ignore negative value \n" ); document.write( "boat speed 11.953 km/h \r \n" ); document.write( "\n" ); document.write( "Time upstream = (14/8.953)*60 =94 minutes \n" ); document.write( " \n" ); document.write( "m.ananth@hotmail.ca \n" ); document.write( " \n" ); document.write( " |