document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1183767>Question 1183767</A>: <I><SPAN STYLE=\"word-break: break-all;\"> the intensity at which an object is illuminated varies inversely as the square of its distance from the light source. At 1.5 ft from a light bulb, the intensity of the light is 60 units. What is the intensity at 1 ft?<BR>\n" );
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 #814225 by </B><A HREF=/tutors/aboutme.mpl?userid=greenestamps><B>greenestamps(13198)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=greenestamps><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=greenestamps><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=814225\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> <br><BR>\n" );
document.write( "The ratio of the new to the old distance is 1/1.5 = 2/3, so the intensity is changed by a factor of <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=%281%2F%282%2F3%29%29%5E2+=+%283%2F2%29%5E2+=+9%2F4 ALIGN=MIDDLE  ALT=\"%281%2F%282%2F3%29%29%5E2+=+%283%2F2%29%5E2+=+9%2F4\"   DISABLED_style= \"max-width:90%; height:auto;\" ><br><BR>\n" );
document.write( "ANSWER: <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=60%2A%289%2F4%29+=+9%2A15+=+135 ALIGN=MIDDLE  ALT=\"60%2A%289%2F4%29+=+9%2A15+=+135\"   DISABLED_style= \"max-width:90%; height:auto;\" ><br><BR>\n" );
document.write( " </SPAN>\n" );
document.write( "</TD></TR></TABLE>\n" );