Inverse Variation: Solve for y when x=5

[amanda_]

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.

 

[Mark44]

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.

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.

 

[rock.freak667]

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?

 

[amanda_]

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?

 

[Mark44]

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.