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barxatty [35]
3 years ago
15

The volume of a sphere is 381.51 in. If v-(4/3)π=r³ what is the radius

Mathematics
1 answer:
Fudgin [204]3 years ago
7 0

\bf \textit{volume of a sphere}\\\\ V=\cfrac{4\pi r^3}{3}~~ \begin{cases} r=radius\\[-0.5em] \hrulefill\\ V=381.51 \end{cases}\implies 381.51=\cfrac{4\pi r^3}{3}\implies 3(381.51)=4\pi r^3 \\\\\\ \cfrac{3(381.51)}{4\pi }=r^3\implies \sqrt[3]{\cfrac{3(381.51)}{4\pi }}=r\implies 4.49923943538655585015\approx r

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Given the complex number z_1=3\big(\cos \frac{14\pi}{15} +i\sin \frac{14\pi}{15}\big)z 1 ​ =3(cos 15 14π ​ +isin 15 14π ​ ) and
mars1129 [50]

The product of <em>z₁ =</em> 3 · (cos 14π/15 + i · sin 14π/15) and <em>z₂ =</em> 3 √3 · (cos 11π/15 + i · sin 11π/15) in <em>rectangular</em> form with fully simplified expressions is <em>z₁ · z₂ =</em> 7.794 - i · 13.5.

<h3>How to determine the product of two complex numbers</h3>

Let be two numbers of the form <em>z = a + i · b</em>, where <em>i =</em> √-1, the product of two of these numbers in <em>rectangular</em> form is described by the following formula:

<em>z₁ · z₂ = (a + i · b) · (c + i · d) = (a · c - b · d) + i · (a · d + b · c)</em>   (1)

If we know that a = 3 · cos 14π/15, b = 3 · sin 14π/15, c = 3√3 · cos 11π/15, d = 3√3 · sin 11π/15, then the result in rectangular form is:

<em>z₁ · z₂ =</em> 7.794 - i · 13.5

The product of <em>z₁ =</em> 3 · (cos 14π/15 + i · sin 14π/15) and <em>z₂ =</em> 3 √3 · (cos 11π/15 + i · sin 11π/15) in <em>rectangular</em> form with fully simplified expressions is <em>z₁ · z₂ =</em> 7.794 - i · 13.5. \blacksquare

<h3>Remark</h3>

The statement presents typing mistakes and is poorly formatted, the correct form is introduced below:

<em>Given the complex number z₁ = 3 · (cos 14π/15 + i · sin 14π/15) and z₂ = 3 √3 · (cos 11π/15 + i · sin 11π/15), express the result of z₁ · z₂ in rectangular form with fully simplified fractions and radicals.</em>

<em />

To learn more on complex numbers, we kindly invite to check this verified question: brainly.com/question/10251853

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