Inverse Variation: Solve for y when x=5

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The problem involves finding the value of y when x = 5, given that y varies inversely as the square of x. The correct relationship is expressed as y = k/x². Using the provided condition y = 1/8 when x = 1, the constant k is determined to be 1/8. Substituting x = 5 into the equation yields y = 1/8 * (1/5²), resulting in y = 8/25. The incorrect options presented were y = 8/25 and y = 1/200, with the correct answer being y = 8/25.

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Homework Statement



If y varies inversely as the square of x, and y = 1/8 when x = 1, find y when x = 5.

y = 8/25
y = 1/200

The Attempt at a Solution


The equation to find this is y=k/x, I know that. I've tried to plug in both given answers to see which ones matched but neither of them did. I'm not sure what to plug in for k or x. I just need someone to correctly set this up for me and I'll be able to solve it.
 
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amanda_ said:

Homework Statement



If y varies inversely as the square of x, and y = 1/8 when x = 1, find y when x = 5.

y = 8/25
y = 1/200


The Attempt at a Solution


The equation to find this is y=k/x, I know that.
No, it says that y varies inversely as the square of x.
amanda_ said:
I've tried to plug in both given answers to see which ones matched but neither of them did. I'm not sure what to plug in for k or x. I just need someone to correctly set this up for me and I'll be able to solve it.
 
amanda_ said:
If y varies inversely as the square of x


The Attempt at a Solution


The equation to find this is y=k/x[/QUOTE]

You'll need to try again, if the cube of x is x3, the square of x is?
 
rock.freak667 said:
You'll need to try again, if the cube of x is x3, the square of x is?

x2

So if I square 5 and cross multiply it's 8/25. Is that right?
 
Last edited:
amanda_ said:
x2

So if I square 5 and cross multiply it's 8/25. Is that right?

No, it's not. You skipped some steps. First, what is the equation? And you know that when x = 1, y = 1/8.
 

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