No, they forgot to switch variable labels after solving for the independent variable...
y=-8x+4
y-4=-8x
(y-4)/-8=x
Now that you have solved for the independent variable x, you switch the variable labels...
y=(x-4)/-8
f^-1(x)=(x-4)/-8 which should really be rewritten as:
f^-1(x)=(4-x)/8 :P
Diametre = 2 x Radius
Radius = 7
Circumference = 2πr
= 2x22/7x7
= 44
The interval notation is 
Explanation:
The inequality is 
Let us complete the square, we get,

Simplifying, we have,


Adding both sides of the equation by 11, we get,

Since, we know that for
, if n is even then ![-\sqrt[n]{a}](https://tex.z-dn.net/?f=-%5Csqrt%5Bn%5D%7Ba%7D%3Cu%3C%5Csqrt%5Bn%5D%7Ba%7D)
Thus, writing the above expression in this form, we get,

Also, if
then
and 
Then ,we have,
and 
Solving the inequalities, we get,
and 
Merging the intervals, we have,

Hence, writing the solution in interval notation, we have,

Therefore, the answer is 
Square units are the units of measuring area. Therefore, this question asks for the area of the room.
The room has the shape of a rectangle with:
length = 13 units and width = 9 units
The are of the rectangle can be calculated using the following rule:
area of rectangle = length * width
area of the room = 13 * 9 = 117 square units
(2000 - 150) / 67
27
(3000 - 400) /6
434
(20 - 7) / .75
17