document.write( "Question 119281: if y varies directly as the square of x, and y = 32, when x = 2, find y when x = 5 \n" ); document.write( "
Algebra.Com's Answer #87361 by bucky(2189)\"\" \"About 
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Since y varies directly as the square of x we can write the equation:
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\n" ); document.write( "\"y+=+k%2Ax%5E2\"
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\n" ); document.write( "where k is just a constant that makes the equation true. Notice in this equation that if x
\n" ); document.write( "gets bigger, then y must also get bigger. This is what is meant by \"y varies directly as\"
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\n" ); document.write( "We can find the value of k because we are told that when x = 2, then y = 32. Substitute these
\n" ); document.write( "two values into the above equation and you get:
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\n" ); document.write( "\"32+=+k%2A%282%5E2%29\"
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\n" ); document.write( "Square the 2 on the right side and the equation is then:
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\n" ); document.write( "\"32+=+k%2A4\"
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\n" ); document.write( "Solve for k by dividing both sides by 4 to get:
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\n" ); document.write( "\"8+=+k\"
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\n" ); document.write( "Now we can return to the original equation and substitute 8 for k to get:
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\n" ); document.write( "\"y+=+8%2Ax%5E2\"
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\n" ); document.write( "This the complete equation that relates x to y for this problem.
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\n" ); document.write( "To find the value of y when x is 5 all we need to do is to substitute 5 for x in the equation
\n" ); document.write( "to get:
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\n" ); document.write( "\"y+=+8%2A%285%5E2%29\"
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\n" ); document.write( "Squaring the 5 on the right side results in:
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\n" ); document.write( "\"y+=+8%2A25+\"
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\n" ); document.write( "and the right side multiplies out to:
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\n" ); document.write( "\"y+=+200\"
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\n" ); document.write( "So the answer to your problem is that when x = 5, y equals 200.
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\n" ); document.write( "Hope this helps you to understand the problem.
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