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Mathematics 13 Online
OpenStudy (neverendingcycle):

If f(x) varies directly with x^2, and f(x) = 75 when x = 5, find the value of f(8). (I can find the answer but I need help putting the equation together) A. 192 B. 25 C. 256 D. 40

OpenStudy (neverendingcycle):

@Compassionate @Luigi0210

OpenStudy (luigi0210):

Varies directly meaning it would look something like this: \(\ y=kx \)

OpenStudy (luigi0210):

In your case, it would be \(\large f(x)=kx^2\)

OpenStudy (neverendingcycle):

what does k stand for again?

OpenStudy (neverendingcycle):

constant something I think?

OpenStudy (luigi0210):

Yes, it's the constant you're solving for in the equation.

OpenStudy (neverendingcycle):

so its 75=k5^2?

OpenStudy (luigi0210):

Once you plug in all the information given \(\large 75 = k (5)^2 \) You solve for "k" Once you get k you can plug it back in with the new condition \(\large f(8) \)

OpenStudy (neverendingcycle):

so then it becomes f(8)=3x^2?

OpenStudy (neverendingcycle):

or is the x still 5?

OpenStudy (luigi0210):

* \(\large f(x)=3x^2 \) But yes, then you just plug in "8" for "x" as \(f(8) \) dictates. \(\large f(8)=3 (8)^2 \) and that will give you your solution (:

OpenStudy (neverendingcycle):

ohh okay thanks a bunch!

OpenStudy (luigi0210):

No prob :P

OpenStudy (neverendingcycle):

i got 192

OpenStudy (luigi0210):

You would be correct.

OpenStudy (neverendingcycle):

Yes!

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