document.write( "Question 789268: A and B are riding bicycles on perpendicular roads. Suppose that A is 9km from the intersection and riding toward it at 20 kph, and B is 7km from it riding away from it at 25kph. After how many hours will they be 13km apart? \n" ); document.write( "
Algebra.Com's Answer #478940 by ankor@dixie-net.com(22740)\"\" \"About 
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A and B are riding bicycles on perpendicular roads.
\n" ); document.write( " Suppose that A is 9km from the intersection and riding toward it at 20 kph, and B is 7km from it riding away from it at 25kph.
\n" ); document.write( " After how many hours will they be 13km apart?
\n" ); document.write( ":
\n" ); document.write( "Let t = time for this to be true
\n" ); document.write( "A pythag problem a^2 + b^2 = c^2, where
\n" ); document.write( "a = (9-20t)
\n" ); document.write( "b = (7+25t)
\n" ); document.write( "c = 13
\n" ); document.write( ":
\n" ); document.write( "(9-20t)^2 + (7+25t)^2 = 13^2
\n" ); document.write( "FOIL
\n" ); document.write( "81 - 180t - 180t + 400t^2 + 49 + 175t + 175t + 625t^2 = 169
\n" ); document.write( "Combine like terms
\n" ); document.write( "400t^2 + 625t^2 - 360t + 350t + 81 + 49 - 169 = 0
\n" ); document.write( "A quadratic equation
\n" ); document.write( "1025t^2 - 10t - 39 = 0
\n" ); document.write( "Using the quadratic formula got a positive solution of
\n" ); document.write( "t = .2 hrs, they will be 13 km apart
\n" ); document.write( ":
\n" ); document.write( ":
\n" ); document.write( "See if this works
\n" ); document.write( "9-(.2*20) = 5 km for a
\n" ); document.write( "7+(.2*25) = 12 km for b
\n" ); document.write( "c = \"sqrt%285%5E2+%2B+12%5E2%29\"
\n" ); document.write( "c = 13, confirms our solution
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