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3 years ago

9

What is the domain of the function?

Formula1" title="y = \sqrt[3]{x - 1} " alt="y = \sqrt[3]{x - 1} " align="absmiddle" class="latex-formula">

1 answer:

BabaBlast [244]3 years ago

3 0

Answer: The first option.

Step-by-step explanation:

This domain has no domain restrictions. Therefore, it covers all real numbers. x∈R

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Read 2 more answers

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fgiga [73]

Answer:

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Step-by-step explanation:

Given

$((\sqrt 3)(cos\frac{\pi}{18} + i\ sin\frac{\pi}{18}))^6$

Required

Determine the modulus and the argument

We have that:

$z = ((\sqrt 3)(cos\frac{\pi}{18} + i\ sin\frac{\pi}{18}))^6$

Expand:

$z = (\sqrt 3)^6(cos\frac{\pi}{18} + i\ sin\frac{\pi}{18})^6$

$z = 27(cos\frac{\pi}{18} + i\ sin\frac{\pi}{18})^6$

A complex equation can be expressed as:

$z = |z| e^{i\theta}$

Where

$|z| = modulus$

$\theta = argument$

Where

$e^{i\theta} = (cos\frac{\pi}{18} + i\ sin\frac{\pi}{18})$

So:

$z = 27(cos\frac{\pi}{18} + i\ sin\frac{\pi}{18})^6$ becomes

$z = 27(e^{i\frac{\pi}{18}})^6$

By comparison:

$e^{i\theta} = (e^{i\frac{\pi}{18}})^6$

This gives:

${i\theta} = i\frac{\pi}{18}}*6$

${i\theta} = i\frac{6\pi}{18}}$

${i\theta} = i\frac{\pi}{3}}$

Divide through by i

$\theta = \frac{\pi}{3}$

Hence, the modulus, z is:

$|z| = 27$

And the argument $\theta$ is

$\theta = \frac{\pi}{3}$

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