Find the dimensions of a rectangle with a perimeter of 52cm if it’s length is 4cm more than its width

So the perimeter(P) of a rectangle would be:
P= 2L+2W
L being the length and W being the width.
The problem says the length is 4cm more than the width, so L= 4+W.
So if we substitute L with 4+W, we get:
P= 2(4+W) + 2W
Use the Distributive Property
P= 8+2W+2W
Combine like terms
P=8+4W
Since we're given the perimeter, we could replace P with 52. So:
52=8+4W
Subtract 8 to both sides
44=4W
Divide 4 to both sides
11=W
Therefore, the width is 11cm
And since the length is 4cm more than the width, we could add 4cm to 11cm to find that the length is 15cm
Thus, the dimensions of the rectangle are 15cm by 11cm

D) the x-intercept represents adam's commission pay when he had zero total sales

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