document.write( "Question 1148047: Hi\r
\n" ); document.write( "\n" ); document.write( "a boy on his way to his friends house just missed his bus and that the next bus would leave in 38 minutes. He decided to run to his friends house instead at an average speed of 12 mph and arrived at the same as the next bus. If the bus traveled at an average speed of 50mph, how far was he from his friends place when he missed the first bus.\r
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Algebra.Com's Answer #769423 by ankor@dixie-net.com(22740)\"\" \"About 
You can put this solution on YOUR website!
a boy on his way to his friends house just missed his bus and that the next bus would leave in 38 minutes.
\n" ); document.write( " He decided to run to his friends house instead at an average speed of 12 mph and arrived at the same as the next bus.
\n" ); document.write( " If the bus traveled at an average speed of 50mph, how far was he from his friends place when he missed the first bus.
\n" ); document.write( ":
\n" ); document.write( "let d = distance to friends place
\n" ); document.write( "A time equation, time = dist/speed; change 38 min to hrs
\n" ); document.write( "\"d%2F12\" = \"38%2F60\" + \"d%2F50\"
\n" ); document.write( "mult equation 300 to get rid of the fractions
\n" ); document.write( "25d = 5(38) + 6d
\n" ); document.write( "25d - 6d = 190
\n" ); document.write( "19d = 190
\n" ); document.write( "d = 190/19
\n" ); document.write( "d = 10 mi to friends house
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