Given r(x) = StartFraction 11 Over (x minus 4) squared EndFraction , which represents a domain restriction on r(x) and the corresponding inverse function? X > 4; r–1(x) = 4 minus StartRoot StartFraction 11 Over x EndFraction EndRoot x > 4; r–1(x) = 4 + StartRoot StartFraction 11 Over x EndFraction EndRoot x > –4; r–1(x) = 4 minus StartRoot StartFraction 11 Over x EndFraction EndRoot x > –4; r–1(x) = 4 + StartRoot StartFraction 11 Over x EndFraction EndRoot

x > 4; r–1(x) = 4 + StartRoot StartFraction 11 Over x EndFraction EndRoot

Step-by-step explanation:

We are given that

We have to find the domain of restriction on r(x) and corresponding inverse function.

Let

Now, replace x by y and y by x then, we get

The function is not defined at x=4

But,

The inverse function defines for all positive real values.

Therefore, domain of r(x)=x>4 and inverse function

Option:

x > 4; r–1(x) = 4 + StartRoot StartFraction 11 Over x EndFraction EndRoot

lo98765432wqasdfghjkloi87654567890

step-by-step explanation:

the answer is 150.6

step-by-step explanation:

you multiply .02 by 30 and get .6.

and then you multiply 5.00 by 30 and get 150.

hope this !