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.
Answer: 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.
Answer:
9
Step-by-step explanation:
64.93-22.36=42.57
42.57/4.73=9
this triangle is obtuse meaning the degree it's currently at is 270°
1400 or 9,085858585 or balls
Answer:
I’m pretty sure its 27
Step-by-step explanation:
