document.write( "Question 377513: suppose that y varies directly as the sqaure root of x, and that y=32 when x=16. What is y when x=50? \n" ); document.write( "
Algebra.Com's Answer #268242 by mananth(16946)\"\" \"About 
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suppose that y varies directly as the sqaure root of x, and that y=32 when x=16. What is y when x=50?
\n" ); document.write( "...
\n" ); document.write( "\"+y+%28alpha%29++sqrt%28x%29\"
\n" ); document.write( "...
\n" ); document.write( "\"y+=+k%2Asqrt%28x%29\"
\n" ); document.write( "..
\n" ); document.write( "x=16,y=32
\n" ); document.write( "plug the value
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\n" ); document.write( "\"32=k%2Asqrt%2816%29\"
\n" ); document.write( "...
\n" ); document.write( "32=4k
\n" ); document.write( "/4
\n" ); document.write( "k=32/4
\n" ); document.write( "k=8
\n" ); document.write( "...
\n" ); document.write( "\"y+=+k%2Asqrt%28x%29\"
\n" ); document.write( "x=50, k= 8
\n" ); document.write( "\"y=8%2Asqrt%2850%29\"\r
\n" ); document.write( "\n" ); document.write( "y =56.56
\n" ); document.write( "...
\n" ); document.write( "m.ananth@hotmail.ca
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