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Mathematics

1 answer:

Dmitrij [34]2 years ago

4 0

The volumes of the given figures are:

Half-sphere: 2,093.3 m³
Cylinder: 3,140 m³
Cone: 1,046.67 m³

<h3>How to find the volume of the 3 figures?</h3>

First, let's take the half-sphere.

The volume of half a sphere of radius R is:

$V = \frac{2}{3}*pi*R^3$

Where pi = 3.14

In this case, R = 10m, then the volume is:

$V = \frac{2}{3}*3.14*(10m)^3 = 2,093.3 m^3$

For a cylinder of radius R and height H, the volume is:

$V = pi*R^2*H$

Here w ehave:

R = 10m and H = 10m, then:

$V = 3.14*(10m)^2*10m = 3,140 m^3$

For a cone of radius R and height H, the volume is:

$V = pi*R^2*H/3$

Here we have:

R = 10m

H = 10m

So the volume is:

$V = 3.14*(10m)^2*10m/3 =1,046.67 m^3$

If you want to learn more about volumes:

brainly.com/question/1972490

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WHAT IS THE REMAINDER WHEN <img src="https://tex.z-dn.net/?f=32%5E%7B37%5E%7B32%7D%20%7D" id="TexFormula1" title="32^{37^{32} }"

Feliz [49]

Recall Euler's theorem: if $\gcd(a,n) = 1$ , then

$a^{\phi(n)} \equiv 1 \pmod n$

where $\phi$ is Euler's totient function.

We have $\gcd(9,32) = 1$ - in fact, $\gcd(9,32^k)=1$ for any $k\in\Bbb N$ since $9=3^2$ and $32=2^5$ share no common divisors - as well as $\phi(9) = 6$ .

Now,

$37^{32} = (1 + 36)^{32} \\\\ ~~~~~~~~ = 1 + 36c_1 + 36^2c_2 + 36^3c_3+\cdots+36^{32}c_{32} \\\\ ~~~~~~~~ = 1 + 6 \left(6c_1 + 6^3c_2 + 6^5c_3 + \cdots + 6^{63}c_{32}\right) \\\\ \implies 32^{37^{32}} = 32^{1 + 6(\cdots)} = 32\cdot\left(32^{(\cdots)}\right)^6$