SOLUTION: if y varies directly as the square of x, and y = 32, when x = 2, find y when x = 5

Algebra ->  Equations -> SOLUTION: if y varies directly as the square of x, and y = 32, when x = 2, find y when x = 5      Log On


   

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