contestada

O A. lw= (x - 5)(x - 5); 49 square feet
O B. /w = x(x - 5); 84 square feet
O c. /w = (x + 5)(x + 5); 289 square feet
D. lw= (x + 5)(x - 5); 119 square feet

O A lw x 5x 5 49 square feet O B w xx 5 84 square feet O c w x 5x 5 289 square feet D lw x 5x 5 119 square feet class=

Respuesta :

Answer:

D. [tex] lw = (x + 5)(x - 5) ; 119 ft^2 [/tex]

Step-by-step explanation:

Dimensions of the old square brick patio:

[tex] length (l) = x ft [/tex]

[tex] width (l) = x ft [/tex]

Note: a square has equal side measure

Dimensions of the new patio

[tex] length (l) = (x + 5) ft [/tex] ==> she increased length by 5 ft

[tex] width (l) = (x - 5) ft [/tex] she reduced width by 5 ft

Expression of the length and width of the new patio is: [tex] lw = (x + 5)(x - 5) [/tex]

Area of the new patio:

Dimension of original patio = x by x = 12 ft by 12 ft.

To find area of the new patio, replace x with 12 in the expression, [tex] lw = (x + 5)(x - 5) [/tex] , which gives you the area.

[tex] area = lw = (12 + 5)(12 - 5) [/tex]

[tex] area = (17)(7) [/tex]

[tex] area = 199 ft^2 [/tex]

Answer is D. [tex] lw = (x + 5)(x - 5) ; 119 ft^2 [/tex]

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