Area of rectangular post card = length x width
For the given post card:
length = 4 in
width = (3+b) in
area = 24 in^2
So, substituting in the equation of the area:
24 = 4 x (3+b)
24 = 12 + 4b
24 - 12 = 4b
12 = 4b
b = 3 in
Therefore:
the length of the postcard = 4 inch
the width of the postcard = b+3 = 3 + 3 = 6 inch
I observe that 5 raised to any whole-number power
produces a number that ends with the digit 5 .
To find the inverse, we swap the variables y and x, then solve for the new y.
3a.

Swapping the variables:

Solving for y:

The domain of this inverse is

.
3b.

Swapping:

Solving for y:

The domain of this inverse is

.
3c.
![y=\sqrt[3]{\frac{x-7}{3}}](https://tex.z-dn.net/?f=y%3D%5Csqrt%5B3%5D%7B%5Cfrac%7Bx-7%7D%7B3%7D%7D)
Swapping:
![x=\sqrt[3]{\frac{y-7}{3}}](https://tex.z-dn.net/?f=x%3D%5Csqrt%5B3%5D%7B%5Cfrac%7By-7%7D%7B3%7D%7D)
Solving for y:

The domain of this inverse is all real numbers.
4a.

,


4c.
![y=\sqrt[3]{\frac{x-7}{3}}](https://tex.z-dn.net/?f=y%3D%5Csqrt%5B3%5D%7B%5Cfrac%7Bx-7%7D%7B3%7D%7D)
,

![y=\sqrt[3]{\frac{(3x^3+7)-7}{3}} \\ y=\sqrt[3]{\frac{3x^3}{3}} \\ y=\sqrt[3]{x^3} \\ y=x](https://tex.z-dn.net/?f=y%3D%5Csqrt%5B3%5D%7B%5Cfrac%7B%283x%5E3%2B7%29-7%7D%7B3%7D%7D%20%5C%5C%20y%3D%5Csqrt%5B3%5D%7B%5Cfrac%7B3x%5E3%7D%7B3%7D%7D%20%5C%5C%20y%3D%5Csqrt%5B3%5D%7Bx%5E3%7D%20%5C%5C%20y%3Dx)