document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=504519>Question 504519</A>: <I><SPAN STYLE=\"word-break: break-all;\"> in studying the frictional effects on a flywheel, the revolutions per minute R that it makes as a function of the time ( in minutes) is given by R = 4520(0.750)^2.50t. Find t for R = 1950 r/min. </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #339694 by </B><A HREF=/tutors/aboutme.mpl?userid=lwsshak3><B>lwsshak3(11628)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=lwsshak3><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=339694\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> n studying the frictional effects on a flywheel, the revolutions per minute R that it makes as a function of the time ( in minutes) is given by R = 4520(0.750)^2.50t. Find t for R = 1950 r/min<BR>\n" );
document.write( "**<BR>\n" );
document.write( " R = 4520(0.750)^2.50t<BR>\n" );
document.write( "1950= 4520(0.750)^2.50t<BR>\n" );
document.write( "(0.750)^2.50t=1950/4520=0.4314<BR>\n" );
document.write( "take log of both sides<BR>\n" );
document.write( "2.5t log(0.750)=log(0.4314)<BR>\n" );
document.write( "2.5t=log(0.4314)/ log(0.750)=2.9224<BR>\n" );
document.write( "t=2.9224/2.5=1.1690 minutes </SPAN>\n" );
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