document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1123098>Question 1123098</A>: <I><SPAN STYLE=\"word-break: break-all;\"> the function f(r)=&#960;r^2 determines the area of a circle in cm2, A, in terms of the radius of the circle in cm, r. \r<BR>\n" );
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document.write( "a. Write a function g that determines the radius of a circle in cm, r, in terms of the area of the cirlce in cm2, A.<BR>\n" );
document.write( "g(A)= \r<BR>\n" );
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document.write( "b. Use function notation to answer the following questions:<BR>\n" );
document.write( "-What is the area (in cm2) of a circle whose radius is 2.21 cm?\r<BR>\n" );
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document.write( "-What is the radius (in cm) of a circle whose area is 1798 cm2?\r<BR>\n" );
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document.write( "-How much greater is the area of a circle whose radius is 15.3 cm than the area of a circle whose radius is 15.2 cm?<BR>\n" );
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document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #739340 by </B><A HREF=/tutors/aboutme.mpl?userid=Boreal><B>Boreal(15235)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=Boreal><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=Boreal><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=739340\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> area=pi*r^2 so sqrt (area/pi)=r<BR>\n" );
document.write( "g(r)=sqrt)area/pi<BR>\n" );
document.write( "A=pi*2.21^2=15.34 cm^2<BR>\n" );
document.write( "r=sqrt(1798/pi)=23.92 cm<BR>\n" );
document.write( "it is the ratio of (15.3/15.2)^2=1.013 times greater<BR>\n" );
document.write( "15.3^2*pi=735.42<BR>\n" );
document.write( "15.2^*pi=725.83<BR>\n" );
document.write( "That ratio is 1.013 </SPAN>\n" );
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