document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1011874>Question 1011874</A>: <I><SPAN STYLE=\"word-break: break-all;\"> A biker sees a derr crossing the street and puts on the breaks. The distance that he travels in feet can be modeled by D(s)=30+5squaroot(s) where s is the speed iof the bikes in miles per hr. if the bike travels a distance of 58 ft what was his speed? </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #627699 by </B><A HREF=/tutors/aboutme.mpl?userid=Theo><B>Theo(13342)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=Theo><IMG SRC=http://www.algebra.com/images/aboutme-small.gif BORDER=0 alt=\"About Me\" VALIGN=MIDDLE></A>&nbsp;<IMG SRC=http://www.algebra.com/cgi-bin/show-face.mpl?user=Theo><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=627699\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> the equation is d(s) = 30 + 5*sqrt(s).\r<BR>\n" );
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document.write( "you are given that d(s) = 58.\r<BR>\n" );
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document.write( "the equation becomes 58 = 30 + 5 * sqrt(s).\r<BR>\n" );
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document.write( "subtract 30 from both sides of the equation to get 28 = 5 * sqrt(s).\r<BR>\n" );
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document.write( "divide both sides of the equation by 5 rto get 28/5 = sqrt(s).\r<BR>\n" );
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document.write( "square both sides of the equation to get (28/5)^2 = s.\r<BR>\n" );
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document.write( "solve for s to get s = 31.36.\r<BR>\n" );
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document.write( "replace s in the original equation to get:\r<BR>\n" );
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document.write( "58 = 30 + 5 * sqrt(31.36)\r<BR>\n" );
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document.write( "evalute the equation to get 58 = 58.\r<BR>\n" );
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document.write( "this confirms the solution is correct.\r<BR>\n" );
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document.write( " </SPAN>\n" );
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