document.write( "Question 805762: A boat travels 2 km upstream and 2 km back. The time for the round trip is 10 hours. The speed of the stream is 3km/hr. What is the speed of the boat in still water? \n" ); document.write( "
Algebra.Com's Answer #485465 by mananth(16946)![]() ![]() You can put this solution on YOUR website! boat speed x kph \n" ); document.write( "current speed 3 kph \n" ); document.write( " \n" ); document.write( "against current x- 3 kph \n" ); document.write( "with current x+ 3 kph \n" ); document.write( " \n" ); document.write( "Distance= 2 miles \n" ); document.write( " \n" ); document.write( "Time against + time with = 10 hours \n" ); document.write( "t=d/r \n" ); document.write( " \n" ); document.write( "2 /( x + 3 ) + 2 /(x - 3 ) = 10 \n" ); document.write( " \n" ); document.write( "LCD = (x - 3 ) ( x + 3 ) \n" ); document.write( "2 *( x - 3 ) + 2 (x + 3 ) = 10 (x^2 - 9 ) \n" ); document.write( "2 x - -6 + 2 x + 6 = 10 ( x ^2 - 9 ) \n" ); document.write( "4 x = 10 x ^2 -90 \n" ); document.write( "10 x ^2 - -4 x - 90 = 0 \n" ); document.write( " \n" ); document.write( "Find the roots of the equation by quadratic formula 5.12 \n" ); document.write( " iiiii \n" ); document.write( "a= 10 , b= -4 , c= -90 \n" ); document.write( " \n" ); document.write( "b^2-4ac= 16 + 3600 \n" ); document.write( "b^2-4ac= 3616 \n" ); document.write( " \n" ); document.write( " \n" ); document.write( " \n" ); document.write( "x1=( 4 + 60.13 )/ 20 \n" ); document.write( "x1= 3.21 \n" ); document.write( " \n" ); document.write( "x2=( 4 -60.13 ) / 20 \n" ); document.write( "x2= -2.81 \n" ); document.write( "Ignore negative value \n" ); document.write( "boat speed 3.207 kph \n" ); document.write( " \n" ); document.write( " |