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A parallelogram has an area of x^2 + 9x - 36. A. What are expressions for the length and width of the parallelogram?B. If x is an integer, what is the least possible value of x for a parallelogram to exist? Explain.Please help! ​

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Hrishii

Step-by-step explanation:

[tex]Area \: of \: parallelogram \\ = {x}^{2} + 9x - 36 \\ = {x}^{2} + 12x - 3x - 36 \\ = x(x + 12) - 3(x + 12) \\ = (x + 12)(x - 3) \\ \implies \\ length = (x + 12) \: units \\ width = (x - 3) \: units \\ [/tex]

For this parallelogram to exist least value of x should be 4.

Because if x = 3

(x - 3)= 3 - 3 =0

Thus width of parallelogram would become zero as a result area of parallelogram would become zero i. e. Parallelogram won't exist.

Cetacea

Answer:

Length = x + 12.

Width = x - 3.

The least possible value of x for a parallelogram to exist is = 3.

Given that the area of the parallelogram is: [tex]x^2 + 9x - 36[/tex].

We factor it.

[tex]x^2 + 9x - 36\\=(x+12)(x-3)[/tex]

We know that area = length*width.

Comparing we get:

Length = x + 12.

Width = x - 3.

We know that length can not be negative.

So, x - 3 ≥ 0

or, x ≥ 3.

So the least possible value of x for a parallelogram to exist is = 3.

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