3 years ago

PLEASE HELP IF YOU CAN! IM DESPERATE! DUE TODAY!

Mathematics

2 answers:

kodGreya [7K]3 years ago

8 0

Answer:

Step-by-step explanation:

sukhopar [10]3 years ago

6 0

Answer:

2x+6=-8(x>=-7), 2x+3<9(x<3), 2x-4<=-8(x<=-2), -x-8>=-5(x<=-3), x+-4<=-3(x<=1)

Step-by-step explanation:

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Read 2 more answers

Solve using demoivres theorem

pashok25 [27]

It's difficult to make out, but I think the task is to expand

$(\sqrt3-i)^4$

Write the number in polar form first:

$\sqrt3-i=2e^{-i\pi/6}$

By DeMoivre's theorem, you have

$(\sqrt3-i)^4=\left(2e^{-i\pi/6}\right)^4=2^4e^{-i4\pi/6}=16e^{-i2\pi/3}$

and converting back to Cartesian form, this number is equivalent to

$16\left(\cos\dfrac{2\pi}3-i\sin\dfrac{2\pi}3\right)=16\left(-\dfrac12+-\dfrac{\sqrt3}2\right)=-8(1+i\sqrt3)$

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