How do you write an equation in the form f(x) = kx^n for the direct variation functions given f(0.3) = 45 and n = 2?

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Answer 1 · 2018-06-26T13:29:06

Function is $f(x)=500x^2$

As $f(0.3)=45$ and function is of the form $f(x)=kx^n$ and $n=2$

Hence $f(0.3)=k(0.3)^2=45$

or $0.09k=45$

and $k=45/0.09=45xx100/9=500$

Hence function is $f(x)=500x^2$

Answer 2 · 2018-06-26T13:35:44

$color(blue)(f(x)=500x^2)$

$y prop kx^n$

Where $bbk$ is the constant of variation.

To find $bbk$ :

From $f(0.3)=45$ , $n=2$

$:.$

$45=k(0.3)^2$

$k=45/(0.3)^2=500$

So we have:

$f(x)=500x^2$