document.write( "Question 834303: John and Brian leave Williston at the same time. JOhn drives north and Brian drives east. John's average speed is 10 miles per hour slower than Brian's. At the end of one hour they are 50 miiles apart. Find Brian's average speed.\r
\n" ); document.write( "\n" ); document.write( "50+10 =60
\n" ); document.write( "Brian was driving 50 and John 60 \n" ); document.write( "
Algebra.Com's Answer #502965 by richwmiller(17219)\"\" \"About 
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Your solution has many problems
\n" ); document.write( "John's average speed is 10 miles per hour slower than Brian's
\n" ); document.write( "but you have
\n" ); document.write( "Brian was driving 50 and John 60
\n" ); document.write( "50^2+60^2=x^2
\n" ); document.write( "that doesn't put them 50 miles apart.
\n" ); document.write( "x^2=6100
\n" ); document.write( "x=10*sqrt(61)
\n" ); document.write( "x=78.102 miles apart\r
\n" ); document.write( "\n" ); document.write( "(x+10)^2+(x)^2=50^2
\n" ); document.write( "x=30 x+10=40\r
\n" ); document.write( "\n" ); document.write( "check\r
\n" ); document.write( "\n" ); document.write( "40^2+30^2=50^2
\n" ); document.write( "1600+900=2500
\n" ); document.write( "ok\r
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