What is the square root of 108 divided by the square root of 2q^6?? I really REALLY need the answer fast!

Mathematics

2 answers:

dedylja [7]3 years ago

8 0

$\sqrt{108}:\sqrt{2q^6}=\sqrt{\frac{108}{2q^{3\times2}}}=\sqrt{\frac{54}{(q^3)^2}}=\sqrt{\frac{9\times6}{(q^3)^2}}=\frac{\sqrt{9\times6}}{\sqrt{(q^3)^2}}=\frac{\sqrt9\times\sqrt6}{q^3}=\frac{3\sqrt6}{q^3}$

Kruka [31]3 years ago

4 0

Answer in attachment below.... Open it up in a new window to see it in full.

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Step-by-step explanation:

$-3\left[\cos \left(\frac{-\pi }{4}\right)+i\sin \left(\frac{-\pi \:}{4}\right)\right]\:\div \:2\sqrt{2}\left[\cos \left(\frac{-\pi \:\:}{2}\right)+i\sin \left(\frac{-\pi \:\:\:}{2}\right)\right]$

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$\frac{-3\left(\frac{\sqrt{2}}{2}-\frac{\sqrt{2}}{2}i\right)}{2\sqrt{2}\left(0-1\right)i}$