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Murrr4er [49]
3 years ago
7

The volume of a soap bubble is 1,203.6 mm^3. Find the radius and diameter of the soap bubble. Use 3.14 for it.​

Mathematics
1 answer:
madreJ [45]3 years ago
8 0

The radius of the soap bubble=6.6 mm

The diameter of the soap bubble=13.2 mm

<u>Step-by-step explanation:</u>

The volume of the soap bubble is given in the problem and using this information we have to calculate its radius and diameter.

volume=1203.6mm^3

The soap bubble is in the shape of a sphere and we can use the equation of the volume of the sphere in this problem.

volume=\frac{4}{3} \pi r^3\\1203.6=\frac{4}{3} \times 3.14\times r^3\\r^3=\frac{1203.6\times 3}{4\times3.14} \\r^3=287.48\\r=6.59 mm

The radius of the soap bubble is 6.6 mm.

diameter=2\times radius\\=2\times 6.6\\13.2 mm

The diameter of the soap bubble is 13.2 mm

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The <em>trigonometric</em> expression \frac{\tan^{2} \alpha}{\tan \alpha - 1} + \frac{\cot^{2} \alpha}{\cot \alpha - 1} is equivalent to the <em>trigonometric</em> expression \sec \alpha \cdot \csc \alpha + 1.

<h3>How to prove a trigonometric equivalence</h3>

In this problem we must prove that <em>one</em> side of the equality is equal to the expression of the <em>other</em> side, requiring the use of <em>algebraic</em> and <em>trigonometric</em> properties. Now we proceed to present the corresponding procedure:

\frac{\tan^{2} \alpha}{\tan \alpha - 1} + \frac{\cot^{2} \alpha}{\cot \alpha - 1}

\frac{\tan^{2}\alpha}{\tan \alpha - 1} + \frac{\frac{1}{\tan^{2}\alpha} }{\frac{1}{\tan \alpha} - 1 }

\frac{\tan^{2}\alpha}{\tan \alpha - 1} - \frac{\frac{1 }{\tan \alpha} }{\tan \alpha - 1}

\frac{\frac{\tan^{3}\alpha - 1}{\tan \alpha} }{\tan \alpha - 1}

\frac{\tan^{3}\alpha - 1}{\tan \alpha \cdot (\tan \alpha - 1)}

\frac{(\tan \alpha - 1)\cdot (\tan^{2} \alpha + \tan \alpha + 1)}{\tan \alpha\cdot (\tan \alpha - 1)}

\frac{\tan^{2}\alpha + \tan \alpha + 1}{\tan \alpha}

\tan \alpha + 1 + \cot \alpha

\frac{\sin \alpha}{\cos \alpha} + \frac{\cos \alpha}{\sin \alpha} + 1

\frac{\sin^{2}\alpha + \cos^{2}\alpha}{\cos \alpha \cdot \sin \alpha} + 1

\frac{1}{\cos \alpha \cdot \sin \alpha} + 1

\sec \alpha \cdot \csc \alpha + 1

The <em>trigonometric</em> expression \frac{\tan^{2} \alpha}{\tan \alpha - 1} + \frac{\cot^{2} \alpha}{\cot \alpha - 1} is equivalent to the <em>trigonometric</em> expression \sec \alpha \cdot \csc \alpha + 1.

To learn more on trigonometric expressions: brainly.com/question/10083069

#SPJ1

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