Respuesta :

we have the following:

[tex]\frac{j^3k}{h^0}[/tex]

replacing, j =-1, h=8 and k = -12

[tex]\begin{gathered} \frac{(-1)^3\cdot(-12)}{8^0} \\ \frac{-1\cdot-12}{1}=12 \end{gathered}[/tex]

The answer correct is B. 12

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