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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 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 area of the decorative border can be modeled by the following equation, where x represents the width of the decorative border.


( )x2 + ( )x =


Is it reasonable for the border to be 2.5 inches wide ( ) 

(yes/no)

Respuesta :

Refer to the attached image.

Let the width of the border be 'x' inch.

Area of border = Area of rectangle ABCD - Area of rectangle PQRS

Area of rectangle ABCD = [tex]length \times width[/tex]

= [tex]AB \times BC[/tex]

= [tex](13+2x) \times (9+2x)[/tex] square units.

Area of rectangle PQRS = [tex]PQ \times RQ[/tex]

= [tex]13 \times 9[/tex]

= 117 square units.

So, area of border = [tex](13+2x)(9+2x)-117 = 48[/tex]

As it is given that the area of border is 48 square units.

So, [tex]117+26x+18x+4x^2-117 = 48[/tex]

[tex]44x + 4x^2-48=0[/tex]

[tex]4x^2 + 44x = 48[/tex]

So, the complete equation is [tex]4x^2 + 44x = 48[/tex].

Now, we will find 'x' in the equation [tex]x^2+11x-12=0[/tex],

[tex]x^2 + 12x - x -12 =0[/tex]

[tex]x(x+12)-1(x+12)=0[/tex]

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

So, x = 1 or x = -12

Width can'not be negative.

So, x = 1 .

Therefore, the width of the border can'not be 2.5 inch.

Therefore, the width of the border is 1 inch.

Ver imagen pinquancaro

Answer:yes

Step-by-step explanation:

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