Respuesta :

Answer:

you are wrong

Step-by-step explanation:

[tex]\geq x^{2} \lim_{n \to \infty} a_n \left[\begin{array}{ccc}1&2&3\\4&5&6\\7&8&9\end{array}\right] \pi \sqrt[n]{x} \alpha \beta[/tex]∛Ξthats the answer

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