Respuesta :

sqdancefan

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Answer:

  (-7, -11)

Step-by-step explanation:

  [tex]\dfrac{x^2+3x-28}{x+7}=\dfrac{(x+7)(x-4)}{x+7}=x-4\qquad x\ne-7[/tex]

The hole is at x=-7. The left- and right-limits at that point are -7-4 = -11. So, the coordinates of the hole are (-7, -11).

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