document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1065016>Question 1065016</A>: <I><SPAN STYLE=\"word-break: break-all;\"> A solid metallic sphere of diameter 28 cm is melted and number of cones formed by the diameter 4 2/3 cm and height 3 cm .find the no. Of cones formed .......<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 #680152 by </B><A HREF=/tutors/aboutme.mpl?userid=addingup><B>addingup(3677)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=addingup><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=addingup><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=680152\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Radius = diameter/2 = 28/2 = 14<BR>\n" );
document.write( "Volume of the sphere:<BR>\n" );
document.write( "4/3(Pi*r^3) = Volume<BR>\n" );
document.write( "4/3(Pi*14^3) = 11,494 This is how much material you have to form cones<BR>\n" );
document.write( "-------------------------------------<BR>\n" );
document.write( "Volume of a cone:<BR>\n" );
document.write( "1/3(B*h) B: base and h: height. Since the base is Pi*r^2, we can re-write:<BR>\n" );
document.write( "1/3((Pi*r^2)h)<BR>\n" );
document.write( "1/3((Pi*14^2)3) = 615.75<BR>\n" );
document.write( "---------------------------------------<BR>\n" );
document.write( "11,494/615.75 = 18.667 <BR>\n" );
document.write( "You can make 18 cones and have some left over, 0.667 which is enough for 2/3 of a cone. </SPAN>\n" );
document.write( "</TD></TR></TABLE>\n" );