Question
A rectangular classroom has a length that is six feet shorter than its width, w
. Mrs. Lordes wants to increase both dimensions by 2 feet.
The equation that represents the new area, N, is
A.N=w^2-4w
B.N=w^2+2w
C.N=w^2-8w+12
D.N=w^2-2w-8
. Mrs. Lordes wants to increase both dimensions by 2 feet.
The equation that represents the new area, N, is
A.N=w^2-4w
B.N=w^2+2w
C.N=w^2-8w+12
D.N=w^2-2w-8
Answers
Answered by
GPT-4o mini
Let the width of the classroom be \( w \) feet. The length of the classroom, being six feet shorter than the width, can be expressed as \( w - 6 \) feet.
The original area, \( A \), of the classroom can be calculated using the formula for area of a rectangle:
\[
A = \text{length} \times \text{width} = (w - 6) \times w = w^2 - 6w
\]
Now, if Mrs. Lordes increases both dimensions by 2 feet, the new width becomes \( w + 2 \) and the new length becomes \( (w - 6) + 2 = w - 4 \).
The new area \( N \) is then:
\[
N = (w - 4)(w + 2)
\]
We can expand this equation:
\[
N = w^2 + 2w - 4w - 8 = w^2 - 2w - 8
\]
So the new area \( N \) can be represented by the equation:
\[
N = w^2 - 2w - 8
\]
Thus, the correct answer is:
**D. \( N = w^2 - 2w - 8 \)**
The original area, \( A \), of the classroom can be calculated using the formula for area of a rectangle:
\[
A = \text{length} \times \text{width} = (w - 6) \times w = w^2 - 6w
\]
Now, if Mrs. Lordes increases both dimensions by 2 feet, the new width becomes \( w + 2 \) and the new length becomes \( (w - 6) + 2 = w - 4 \).
The new area \( N \) is then:
\[
N = (w - 4)(w + 2)
\]
We can expand this equation:
\[
N = w^2 + 2w - 4w - 8 = w^2 - 2w - 8
\]
So the new area \( N \) can be represented by the equation:
\[
N = w^2 - 2w - 8
\]
Thus, the correct answer is:
**D. \( N = w^2 - 2w - 8 \)**