Respuesta :

agamal91
the perimeter of any shape is equal to the sum of its sides.
then [tex]the perimeter=31+(4x^{2}+8x)+(7x+30)+(3x^{2}-5x+20)[/tex]
[tex] the perimeter= (4x^{2}+3x^{2})+(7x+8x-5x)+20+30+31=7x^{2}+10x+81[/tex]
for x=2
[tex] the perimeter= 7*(2)^{2}+10*2+81=7*4+20+81=28+101=129[/tex] 

Note:
Please double check the substitution but the procedure is 100% right.

Answer:

Step-by-step explanation:

Since perimeter of any polygon is represented by sum of its all sides.

Therefore, perimeter of the given quadrilateral = (4x² + 8x) + (3x² - 5x + 20) + (7x + 30) + 31

4x

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