Model With Mathematics A photographer is placing photos in a mat for a gallery show.
Each mat she uses is x in, wide on each side.
The total area of each photo and mat is
shown

a. The possible dimensions of the total area are;

The width of the total area is (2·x + 8) inches

The length of the total area is (2·x + 10) inches

b. The dimensions of the photos are as follows;

The length of the photos are 10 inches

The width of the photos are 8 inches

Step-by-step explanation:

a. Given that the area, A = 4·x² + 36·x + 80

We get;

A = 4 × (x² + 9·x + 20) = 4 × (x + 4) × (x + 5) = (2·x + 8)·(2·x + 10)

Therefore, the possible dimensions of the total area (photo + mat) are;

The width of the total area (photo + mat) = (2·x + 8) in.

The length of the total area (photo + mat) = (2·x + 10) in.

b. The dimensions of the photos alone are shorter than the dimensions of the photo and mat combined by 2·x each

Therefore, we have the dimensions of the photos are as follows;

The length of the photo = (2·x + 10) in. - 2·x in. = 10 in.

The width of the photos = (2·x + 8) in. - 2·x in. = 8 in.

sorry, i can't understand your question. will later if you describe more clearly!

: )

answer is the third choice:

3x^5+4x^4+x^3+2x^2-10x-7=0

step-by-step explanation: