document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=121824>Question 121824</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Hi may I have some help with the following word problem.\r<BR>\n" );
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document.write( "A jeweler has three small solid spheres made of gold, of radius 2mm, 3mm, and 4mm. He decides to melt these down and make just one sphere out of them. What will the radius of this larger sphere be? </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #89432 by </B><A HREF=/tutors/aboutme.mpl?userid=checkley71><B>checkley71(8403)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=checkley71><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=89432\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> ASSUMING ALL SPHERES ARE THE SAME THICKNESS THEN:<BR>\n" );
document.write( "AREA=PIR^2=3.14*2^2=3.14*4=12.56 MM^2<BR>\n" );
document.write( "AREA=3.14*3^2=3.14*9=28.26 MM^2<BR>\n" );
document.write( "AREA=3.14*4^2=3.14*16=50.24 MM^2<BR>\n" );
document.write( "TOTAL AREA=12.56+28.26+50.24=91.06 MM^2<BR>\n" );
document.write( "91.06=3.14R^2<BR>\n" );
document.write( "R^2=91.06/3.14<BR>\n" );
document.write( "R^2=29<BR>\n" );
document.write( "R=SQRT29 <BR>\n" );
document.write( "R=5.385 MM IS THE NEW RADIUS. </SPAN>\n" );
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