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A café owner is designing a new menu and wants to include a decorative border around the outside of her food listings. Due to th

kipiarov [429]

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.

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3 years ago

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Find exact values for sin θ and tan θ if cos θ = -4/9 and tan θ > 0.

julsineya [31]

Answer:

Part 1) $sin(\theta)=-\frac{\sqrt{65}}{9}$