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3 years ago

What is the diameter of a hemisphere with a volume of 257 m^3

Mathematics

1 answer:

Alex73 [517]3 years ago

4 0

Answer:

9.9

Step-by-step explanation:

\text{Volume of Hemisphere}\text{:}

Volume of Hemisphere:

\,\,257

257

\text{Volume of Sphere}\text{:}

Volume of Sphere:

\,\,514

514

Double volume of hemisphere to get volume of the entire sphere

\text{Volume of a Sphere:}

Volume of a Sphere:

V=\frac{4}{3}\pi r^3

πr

514=

\,\,\left(\frac{4}{3}\pi\right) r^3

(

π)r

514=

\,\,(4.1887902)r^3

(4.1887902)r

Evaluate 4/3pi in calc

\frac{514}{4.1887902}=

4.1887902

514

\,\,\frac{(4.1887902)r^3}{4.1887902}

4.1887902

(4.1887902)r

Evaluate \frac{4}{3}\pi

π in calc

122.7084611=

\,\,r^3

\sqrt[3]{122.7084611}=

122.7084611

\,\,\sqrt[3]{r^3}

Cube root both sides

4.9692575=

\,\,r

\text{Then the diameter equals }9.938515

Then the diameter equals 9.938515

diameter is radius times 2

\text{Final Answer:}

Final Answer:

d\approx 9.9\text{ m}

d≈9.9 m

Round to nearest tenth

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3 years ago

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Answer:

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Step-by-step explanation:

2t = 1 / 2

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3 years ago

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Which equation represents the magnitude of an

Basile [38]

Answer:

$M=\log{(100S)}$

Step-by-step explanation:

The <em>Richter scale</em>, the standard measure of earthquake intensity, is a <em>logarithmic scale</em>, specifically logarithmic <em>base 10</em>. This means that every time you go up 1 on the Richter scale, you get an earthquake that's 10 times as powerful (a 2.0 is 10x stronger than a 1.0, a 3.0 is 10x stronger than a 2.0, etc.).

How do we compare two earthquake's intensities then? As a measure of raw intensity, let's call a "standard earthquake" S. What's the magnitude of this earthquake? The magnitude is whatever <em>power of 10</em> S corresponds to; to write this relationship as an equation, we can say $10^m=S$ , which we can rewrite in logarithmic form as $m=\log{S}$ .

We're looking for the magnitude M of an earthquake 100 times larger than S, so reflect this, we can simply replace S with 100S, giving us the equation

$M=\log{(100S)}$ .

To check to see if this equation is right, let's say we have an earthquake measuring a 3.0 on the Richter scale, so $3=\log{S}$ . Since taking 100 times some intensity is the same as taking 10 times that intensity twice, we'd expect that more intense earthquake to be a 5.0. We can expand the equation $M=\log{(100S)}$ using the product rule for logarithms to get the equation

$M=\log{(100S)}=\log{100}+\log{S}$

And using the fact that $\log{100}=2$ and our assumption that $\log{S}=3$ , we see that $M=2+3=5$ as we wanted.

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4 years ago