Acafé owner is designing a new menu and wants to include a decorative border around the outside of her food listings. due to the cost of printing, the border should have an area of 48 square inches. the width of the border needs to be uniform around the entire menu. she has already determined that her food listings will fit within a 13-inch by 9-inch rectangular area.

The 13-in. by 9-in. rectangle where the food listings fit has an area of 13 in. * 9 in. = 117 in.^2

Adding 48 in.^2 for the border, the total area of the menu with the border will be 117 in.^2 + 48 in.^2 = 165 in.^2

The border has to have uniform width around the menu. We need to find the width of the border. Let the border be x inches wide. Then since you have a border at each of the 4 sides, the border will add 2x to the length of the rectangle and 2x to the width of the rectangle. The menu will have a length of 2x + 13 and a width of 2x + 9. The area of the larger rectangle must by 165 in.^2. The area of a rectangle is length times width, so we get our equation:

(2x + 13)(2x + 9) = 165

Multiply out the left side (use FOIL or any other method you know):

4x^2 + 18x + 26x + 117 = 165

4x^2 + 44x + 117 = 165

4x^2 + 44x - 48 = 0

Divide both sides by 4.

x^2 + 11x - 12 = 0

Factor the left side.

(x + 12)(x - 1) = 0

x + 12 = 0 or x - 1 = 0

x = -12 or x = 1

The solution x = -12 is not valid for our problem because the width of a border cannot be a negative number. Discard the negative solution.

The solution is x = 1.

The border is 1 inch wide.

Check. Add 2 inches to the length and width of the food listings rectangle to get 15 inches by 11 inches. A = 15 in. * 11 in.= 165 in.^2. Now subtract the area of the border, 48 in.^2, 165 in.^2 = 48 in.^2 = 117 in.^2, and you get the area of the 13-in. by 9-in. rectangle. This shows that our solution is correct.

y=-2.5x+12.5

step-by-step explanation:

5y=2x+10

y=0.4x+2

slope=-2.5

5=-2.5*3+b

b=12.5