The width of maya’s poster is 2 inches shorter than the length. the graph models the possible area (y) of maya’s poster determined by its length (x).
which value of x makes sense for the length of maya’s poster?
x < 0
x < 2
x > 0
x > 2

D is the answer

Step-by-step explanation:

The width of Maya’s poster is 2 inches shorter than the length. The graph models the possible area (y) of Maya’s poster determined by its length (x).

Which value of x makes sense for the length of Maya’s poster?

x < 0  

x < 2  

x > 0

x > 2 -- the answer is D

x > 2

Step-by-step explanation:

Here, x represents the length of the poster,

Since, width of the poster is 2 inches shorter than its length,

⇒ Width of the poster = ( x - 2 ) inches,

Thus, the area of the poster = length × width,

⇒ A = x (x-2)

We know that,

Area of a poster must be greater than zero,

⇒ A > 0,

⇒ x(x-2) > 0

⇒ x > 0 and x-2 > 0

If x - 2 > 0 ⇒ x > 2,

That is, the length of the poster must be greater than 2 inches.

c) yes, because of asa or aas

idk

step-by-step explanation: