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kifflom [539]
2 years ago
7

Find the volume of the cone.

Mathematics
1 answer:
juin [17]2 years ago
5 0

The answer is 48π units³ or 150.72 units³.

To find the volume of the cone, use the formula : <u>1/3 × πr²h</u>

We are given that r = 6 and h = 4.

Solving :

  • V = 1/3 × π × 6² × 4
  • V = 12 × 4 × π
  • V = 48π units³ (in terms of π)
  • V = 150.72 units³
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Pls help i will give brainly
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Answer:

A

Step-by-step explanation:

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2 years ago
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What is the perimeter and area of the figure
aleksandrvk [35]

the perimeter is 35 and are is 40

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2 years ago
How would you calculate the area and perimeter of 3yd 6yd 5yd 10yds
spayn [35]
Okay, start with the smaller square, you know the side length is 3 and that the width is 5, multiply this together:
3 * 5 = 15

That's the smaller left side of this shape, now the bigger side, divide the 10 yards at the bottom by 2 to properly represent the other side then multiply this by the 6 over there.
5 * 6 = 30

Conclusion for Area: After adding the results the answer should be 45.

The perimeter is solely adding all the sides together. You have a 10, two 5's, one 6, and two 3's. Add this together:

10 + 5 + 5 + 6 + 3 + 3 = 32

Conclusion for Perimeter: So the perimeter would be 32.

I hope this helps in someway, have a great rest of your day! ^ ^
| | Ghostgate | |
3 0
3 years ago
Find all solutions of each equation on the interval 0 ≤ x &lt; 2π.
Korvikt [17]

Answer:

x = 0 or x = \pi.

Step-by-step explanation:

How are tangents and secants related to sines and cosines?

\displaystyle \tan{x} = \frac{\sin{x}}{\cos{x}}.

\displaystyle \sec{x} = \frac{1}{\cos{x}}.

Sticking to either cosine or sine might help simplify the calculation. By the Pythagorean Theorem, \sin^{2}{x} = 1 - \cos^{2}{x}. Therefore, for the square of tangents,

\displaystyle \tan^{2}{x} = \frac{\sin^{2}{x}}{\cos^{2}{x}} = \frac{1 - \cos^{2}{x}}{\cos^{2}{x}}.

This equation will thus become:

\displaystyle \frac{1 - \cos^{2}{x}}{\cos^{2}{x}} \cdot \frac{1}{\cos^{2}{x}} + \frac{2}{\cos^{2}{x}} - \frac{1 - \cos^{2}{x}}{\cos^{2}{x}} = 2.

To simplify the calculations, replace all \cos^{2}{x} with another variable. For example, let u = \cos^{2}{x}. Keep in mind that 0 \le \cos^{2}{x} \le 1 \implies 0 \le u \le 1.

\displaystyle \frac{1 - u}{u^{2}} + \frac{2}{u} - \frac{1 - u}{u} = 2.

\displaystyle \frac{(1 - u) + u - u \cdot (1- u)}{u^{2}} = 2.

Solve this equation for u:

\displaystyle \frac{u^{2} + 1}{u^{2}} = 2.

u^{2} + 1 = 2 u^{2}.

u^{2} = 1.

Given that 0 \le u \le 1, u = 1 is the only possible solution.

\cos^{2}{x} = 1,

x = k \pi, where k\in \mathbb{Z} (i.e., k is an integer.)

Given that 0 \le x < 2\pi,

0 \le k.

k = 0 or k = 1. Accordingly,

x = 0 or x = \pi.

8 0
2 years ago
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marishachu [46]

Answer:

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Step-by-step explanation:

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½×¼= 1/8 ( Ms Laura)

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1/7 × 1/16 = 1/ 112 (the last turn)

so, Ms. Sarah eats 1/ 112 of a piece.

3 0
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