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Reading and interpreting data

Bevan Mcdonald

2021-01-02

odgovoreh

Skilled2021-01-03Added 107 answers

$i\frac{\sqrt{6}}{i\sqrt{3}\cdot i\sqrt{4}}=-i\frac{\sqrt{6}}{2}\sqrt{3}=-i\frac{\sqrt{\frac{6}{3}}}{2}=-i\frac{\sqrt{2}}{2}$.

So $a=0{\textstyle \phantom{\rule{1em}{0ex}}}\text{and}{\textstyle \phantom{\rule{1em}{0ex}}}b=-\frac{\sqrt{2}}{2}$ for a+ib. Therefore $a=0\text{and}b=-\frac{\sqrt{2}}{2}\text{or}-\frac{1}{\sqrt{2}}$before rationalisation.

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