Answer :
To determine the area of the rectangle, we need to multiply its length by its width. The length of the rectangle is given as [tex]\(x - 8\)[/tex] units, and the width is given as [tex]\(x + 11\)[/tex] units.
So, the area [tex]\(A\)[/tex] can be represented as:
[tex]\[ A = \text{length} \times \text{width} = (x - 8)(x + 11) \][/tex]
We'll expand this expression step-by-step to find the area in its simplified form.
First, we'll use the distributive property (also known as the FOIL method for binomials) to expand the expression:
[tex]\[ (x - 8)(x + 11) = x(x + 11) - 8(x + 11) \][/tex]
Distribute [tex]\(x\)[/tex] in the first term:
[tex]\[ x(x + 11) = x^2 + 11x \][/tex]
Distribute [tex]\(-8\)[/tex] in the second term:
[tex]\[ -8(x + 11) = -8x - 88 \][/tex]
Now, combine these two results:
[tex]\[ x^2 + 11x - 8x - 88 \][/tex]
Combine like terms:
[tex]\[ x^2 + (11x - 8x) - 88 \][/tex]
Simplify:
[tex]\[ x^2 + 3x - 88 \][/tex]
Thus, the expression that represents the area of the rectangle is:
[tex]\[ x^2 + 3x - 88 \][/tex]
So, the correct answer is:
[tex]\[ \boxed{x^2 + 3x - 88} \][/tex]
This corresponds to option C.
So, the area [tex]\(A\)[/tex] can be represented as:
[tex]\[ A = \text{length} \times \text{width} = (x - 8)(x + 11) \][/tex]
We'll expand this expression step-by-step to find the area in its simplified form.
First, we'll use the distributive property (also known as the FOIL method for binomials) to expand the expression:
[tex]\[ (x - 8)(x + 11) = x(x + 11) - 8(x + 11) \][/tex]
Distribute [tex]\(x\)[/tex] in the first term:
[tex]\[ x(x + 11) = x^2 + 11x \][/tex]
Distribute [tex]\(-8\)[/tex] in the second term:
[tex]\[ -8(x + 11) = -8x - 88 \][/tex]
Now, combine these two results:
[tex]\[ x^2 + 11x - 8x - 88 \][/tex]
Combine like terms:
[tex]\[ x^2 + (11x - 8x) - 88 \][/tex]
Simplify:
[tex]\[ x^2 + 3x - 88 \][/tex]
Thus, the expression that represents the area of the rectangle is:
[tex]\[ x^2 + 3x - 88 \][/tex]
So, the correct answer is:
[tex]\[ \boxed{x^2 + 3x - 88} \][/tex]
This corresponds to option C.