document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=347630>Question 347630</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Two surveyors with two-way radios leave the same point at 9:00 A.M., one walking due south at 2 mi/hr and the other due west at 4 mi/hr. How long can they communicate with one another if each radio has a maximum range of 2.98 miles? \r<BR>\n" );
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document.write( "Round the answer to the nearest minute. \r<BR>\n" );
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document.write( " </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #248574 by </B><A HREF=/tutors/aboutme.mpl?userid=checkley77><B>checkley77(12844)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=checkley77><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/images/nopicture.jpg><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=248574\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> (2t)^2+(4t)^2=2.98^2<BR>\n" );
document.write( "4t^2+16t^2=8.88<BR>\n" );
document.write( "20t^2=8.88<BR>\n" );
document.write( "t^2=8.88/20<BR>\n" );
document.write( "t^2=.444<BR>\n" );
document.write( "t=sqrt.444<BR>\n" );
document.write( "t=.666 hours.<BR>\n" );
document.write( "Proof:<BR>\n" );
document.write( "(2*.666)^2+(4*.666)^2=2.98^2<BR>\n" );
document.write( "1.332^2+2.664^2=8.88<BR>\n" );
document.write( "1.774+7.097=8.88<BR>\n" );
document.write( "8.87~8.88 </SPAN>\n" );
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