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Mathematics 7 Online
OpenStudy (anonymous):

If y varies directly as x^2 and y=12 when x=2, find y when x=5. Help please!

OpenStudy (whpalmer4):

okay, if \(u\) varies directly with \(v\), then we can write \[u = kv\] Here you have \(y\) varying directly with \(x^2\). What will the relationship look like, given what I just stated?

OpenStudy (whpalmer4):

\(k\) is the constant of variation, btw.

OpenStudy (anonymous):

You wrote u=kv...

OpenStudy (anonymous):

@whpalmer4

OpenStudy (whpalmer4):

Yes, I'm aware of that :-) If you have "thing 1" varying directly with "thing 2", "thing 1" = k* "thing 2"

OpenStudy (anonymous):

Yes, but it's the x^2 that's tripping me up. Would direct variation work the same way?

OpenStudy (whpalmer4):

"thing 1" is y "thing 2" is x^2

OpenStudy (anonymous):

So I'd get y=k(x^2)?

OpenStudy (whpalmer4):

it can be a considerably more complicated expression than \(x^2\), but it still works the same way: \(u = k * v\) if \(u\) varies directly with \(v\) yes! \[y = kx^2\]Now, you know a pair of \((x,y)\) that go together. Use them to find the value of \(k\)...

OpenStudy (anonymous):

I just don't know how... I don't know the step that comes after y=kx^2.

OpenStudy (anonymous):

@whpalmer4

OpenStudy (whpalmer4):

" y=12 when x=2" plug those numbers into your new formula and solve for \(k\)

OpenStudy (anonymous):

And what I get will be my correct answer?

OpenStudy (whpalmer4):

no, but it will be a necessary step to finding the final answer.

OpenStudy (whpalmer4):

\[12 = k(2)^2\]\[12 = k*4\]\[k =\]

OpenStudy (anonymous):

When I come up with it I'll put it on here.

OpenStudy (anonymous):

I got the answer of y=30. Is this correct?

OpenStudy (whpalmer4):

Uh, no. Can you show me what you got for \(k\)?

OpenStudy (anonymous):

I got 6=k :/

OpenStudy (whpalmer4):

\[12 = k(2)^2\]\[12 = k*4\]\[k=\]

OpenStudy (whpalmer4):

it looks like you're forgetting to square \(x\)...

OpenStudy (whpalmer4):

both in finding \(k\) and then in finding the value of \(y\) that goes with the new value of \(x\)

OpenStudy (whpalmer4):

at least you're consistent — we just have to fix one problem and then you'll be doing it right :-)

OpenStudy (anonymous):

Oh, k=3?

OpenStudy (whpalmer4):

yes! so the new, improved value of \(y\) when \(x=5\) is?

OpenStudy (anonymous):

So the second answer would be y=15?

OpenStudy (whpalmer4):

I was afraid you would answer that. the equation is \[y = kx^2 = k*x*x\]

OpenStudy (whpalmer4):

\[y = 3*5*5 = \]

OpenStudy (anonymous):

75!

OpenStudy (whpalmer4):

yes

OpenStudy (anonymous):

Thank you so much for your help and patience, I appreciate it so much!

OpenStudy (whpalmer4):

what would the value of \(y\) be if \(x = -1\)?

OpenStudy (anonymous):

3?

OpenStudy (whpalmer4):

yes, very good! I made a graph of the curve that you have here, and put on crossing lines at the various points we visited.

OpenStudy (anonymous):

You are unbelievable helpful. Thank you!!!

OpenStudy (whpalmer4):

You're welcome!

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