document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=337662>Question 337662</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Find the measures of the supplementary angles whose measures are 2a-10 and 4a-10.<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 #242066 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=242066\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Find the measures of the supplementary angles <BR>\n" );
document.write( "whose measures are 2a-10 and 4a-10.<BR>\n" );
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document.write( "If the angles are supplementary,<BR>\n" );
document.write( "2a-10 + 4a-10 = 180 degrees<BR>\n" );
document.write( "6a-20 = 180<BR>\n" );
document.write( "6a = 200<BR>\n" );
document.write( "a = 33 1/3<BR>\n" );
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document.write( "Then 2a-10 = 66.66-10 = 56.66 degrees<BR>\n" );
document.write( "And 4a-10 = 123.33 degrees<BR>\n" );
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document.write( "Cheers,<BR>\n" );
document.write( "Stan H. </SPAN>\n" );
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