Respuesta :
We want to simplify √(-48).
Note that
i² = -1 by definition.
Therefore
[tex] \sqrt{-48} = \sqrt{i^{2}.16.3} = \sqrt{i^{2}} \sqrt{16} \sqrt{3} =4i \sqrt{3} [/tex]
Answer: c. [tex]4i \sqrt{3} [/tex]
Note that
i² = -1 by definition.
Therefore
[tex] \sqrt{-48} = \sqrt{i^{2}.16.3} = \sqrt{i^{2}} \sqrt{16} \sqrt{3} =4i \sqrt{3} [/tex]
Answer: c. [tex]4i \sqrt{3} [/tex]