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kaheart [24]
3 years ago
5

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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3 0
3 years ago
Convert the complex number z = -12 + 5i from rectangular form to polar form.​
Paha777 [63]

Answer:

z=13(\cos 157\degree +i\sin157\degree)

Step-by-step explanation:

The given complex number is

z=-12+5i

The polar form of this complex number is;

z=r(\cos \theta +i\sin \theta), where

r=\sqrt{(-12)^2+5^2}

r=\sqrt{144+25}=\sqrt{169}=13

and

\theta =\tan^{-1}(\frac{5}{-12})

This implies that;

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z=13(\cos 157\degree +i\sin157\degree)

7 0
4 years ago
Please help and thank you
Fantom [35]

Answer:

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Step-by-step explanation:

7 0
3 years ago
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Kipish [7]

Answer:

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Step-by-step explanation:

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4 0
3 years ago
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tatuchka [14]

Answer:

4.8

Step-by-step explanation:

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I hope that helps!


8 0
3 years ago
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