document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=238782>Question 238782</A>: <I><SPAN STYLE=\"word-break: break-all;\"> The surface area \"S\" of a right prism is given by s=2b+ph where \"b\" is the area of the base. P is the perimeter of base and \"h\" is the height of the prism. Solve for P </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #175456 by </B><A HREF=/tutors/aboutme.mpl?userid=stanbon><B>stanbon(75887)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=stanbon><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=175456\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> The surface area \"S\" of a right prism is given by s=2b+ph where \"b\" is the area of the base. P is the perimeter of base and \"h\" is the height of the prism. Solve for P<BR>\n" );
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document.write( "s = 2b + ph<BR>\n" );
document.write( "ph = s-2b<BR>\n" );
document.write( "p = (s-2b)/h<BR>\n" );
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document.write( "Cheers,<BR>\n" );
document.write( "Stan H. </SPAN>\n" );
document.write( "</TD></TR></TABLE>\n" );