document.write( "Question 73149: If y varies directly as x^2, and if y=36 when x=2, find y when x=10. \n" ); document.write( "
Algebra.Com's Answer #52409 by psbhowmick(878)\"\" \"About 
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If y varies directly as \"x%5E2\", then
\n" ); document.write( "\"y+=+kx%5E2\"______(1) where k = constant of variation.\r
\n" ); document.write( "\n" ); document.write( "When y = 36 then x=2.
\n" ); document.write( "So substitute these values for x & y in in equation (1)
\n" ); document.write( "\"36+=+k%2A2%5E2\"
\n" ); document.write( "\"36+=+k%2A4\"
\n" ); document.write( "\"k=36%2F4=9\"\r
\n" ); document.write( "\n" ); document.write( "So, equation (1) becomes \"y+=+9x%5E2\"\r
\n" ); document.write( "\n" ); document.write( "Hence when x = 10, \"y+=+9x%5E2+=+9%2A10%5E2+=+900\".\r
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\n" ); document.write( "\n" ); document.write( "So, when x = 10, y = 900.
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