document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=557553>Question 557553</A>: <I><SPAN STYLE=\"word-break: break-all;\"> the electric power generated by a windmill varies jointly with the square of the diameter of the area swept out by the blades and the cube of the wind velocity. If a windmill with 6-foot-diameter blades and a 20-mile-per-hour wind generates 10,823 watts, how much power would be generated if the blades were 10 feet in diameter and the wind speed were 12 miles per hour?\r<BR>\n" );
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document.write( "i thought that the solution was P(power) = d^2 (diameter)+ v^3 (wind speed), when i substitute in the values given, 10823= (6*2)^2 + 20^3, i came up with 10823=8144, so not the right equation.  can anyone offer any help?  Thank you! </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #362653 by </B><A HREF=/tutors/aboutme.mpl?userid=jim_thompson5910><B>jim_thompson5910(35256)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=jim_thompson5910><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=362653\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Hint: If the \"power generated by a windmill varies jointly with the square of the diameter of the area swept out by the blades and the cube of the wind velocity\", then the equation is <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=P=kd%5E2v%5E3 ALIGN=MIDDLE  ALT=\"P=kd%5E2v%5E3\"   DISABLED_style= \"max-width:90%; height:auto;\" > (note: everything is multiplied together) </SPAN>\n" );
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