document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1152741>Question 1152741</A>: <I><SPAN STYLE=\"word-break: break-all;\"> A steel plant has to fill an order from an automobile manufacturer for 6,710 steel ball bearings 0.18 ft. in diameter.  What is the total mass of the metal required to fill this order, if the steel’s density is  484 lbs per cu. ft? </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #774826 by </B><A HREF=/tutors/aboutme.mpl?userid=josgarithmetic><B>josgarithmetic(39617)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=josgarithmetic><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=774826\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Volume of steel:<BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=6710%284%2F3%29pi%280.18%2F2%29%5E3 ALIGN=MIDDLE  ALT=\"6710%284%2F3%29pi%280.18%2F2%29%5E3\"   DISABLED_style= \"max-width:90%; height:auto;\" ><IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=cubicFeet ALIGN=MIDDLE  ALT=\"cubicFeet\"   DISABLED_style= \"max-width:90%; height:auto;\" >\r<BR>\n" );
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document.write( "Multiply by 484 pounds per cubic feet. </SPAN>\n" );
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