document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=155653>Question 155653</A>: <I><SPAN STYLE=\"word-break: break-all;\"> write an equation that expresses the relationship.  Use k as the constant of variation.<BR>\n" );
document.write( "d varies directly as the square of y\r<BR>\n" );
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document.write( "a) d= k sqrt y\r<BR>\n" );
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document.write( "b) d= k/sqrt y\r<BR>\n" );
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document.write( "c) d= ky^2\r<BR>\n" );
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document.write( "d) d= k/ y^2 </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #114679 by </B><A HREF=/tutors/aboutme.mpl?userid=Alan3354><B>Alan3354(69443)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=Alan3354><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=Alan3354><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=114679\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> write an equation that expresses the relationship. Use k as the constant of variation.<BR>\n" );
document.write( "d varies directly as the square of y <BR>\n" );
document.write( "a) d= k sqrt y <BR>\n" );
document.write( "b) d= k/sqrt y <BR>\n" );
document.write( "c) d= ky^2 <BR>\n" );
document.write( "d) d= k/ y^2<BR>\n" );
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document.write( "It's c)  d = ky^2<BR>\n" );
document.write( "That's the only one that is a function of y^2.  d  has y^2, but in the denominator it is y^(-2), so that's eliminated, or that would be called \"varies inversely as the square of y\" </SPAN>\n" );
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