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expeople1 [14]
3 years ago
7

Can a polygon be a pentagon yes or no please answer!

Mathematics
2 answers:
Naddika [18.5K]3 years ago
5 0
A polygon is plane<span> shape with <em><u>straight</u></em> sides. Therefore, a pentagon is a polygon!</span>
alexdok [17]3 years ago
3 0

<em></em>
Yes, a pentagon can be a polygon.

Any shape with straight sides can be a polygon, including hexagons, pentagons, triangles, squares, rectangles, etc. But this does not include shapes such as a circle or oval since they do not have straight sides.

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Aleks04 [339]
B is correct answer.
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If a diameter intersects a chord of a circle at a right angle, what conclusion can be made?
drek231 [11]

Answer:

The chord is bisected.

Step-by-step explanation:

see the attached figure to better understand the problem

In the circle of the figure

The diameter is the segment DE

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Linear

Step-by-step explanation:

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What is the volume of the sphere that has a diameter of 3? use 3.14 for pie​
aev [14]

Hey ! there

Answer:

  • <u>1</u><u>1</u><u>3</u><u>.</u><u>0</u><u>4</u><u> </u><u>unit </u><u>cube</u>

Step-by-step explanation:

In this question we are provided with a sphere <u>having</u><u> </u><u>radius </u><u>3 </u><u>units </u>and <u>value </u><u>of </u><u>π </u><u>is </u><u>3.</u><u>1</u><u>4</u><u> </u><u>.</u><u> </u>And we're asked to find the<u> </u><u>volume</u><u> of</u><u> </u><u>sphere</u><u> </u><u>.</u>

For finding volume of sphere , we need to know its formula . So ,

\qquad \qquad \: \underline{\boxed{ \frak{Volume_{(Sphere)} =  \dfrac{4}{3} \pi r {}^{3} }}}

<u>Where</u><u> </u><u>,</u>

  • π refers to <u>3.</u><u>1</u><u>4</u>

  • r refers to <u>radius</u><u> of</u><u> sphere</u>

<u>Sol</u><u>u</u><u>tion </u><u>:</u><u> </u><u>-</u>

Now , we are substituting value of π and radius in the formula ,

\quad \longrightarrow \qquad \: \dfrac{4}{3}   \times 3.14 \times (3) {}^{3}

Simplifying it ,

\quad \longrightarrow \qquad \: \dfrac{4}{3}  \times 3.14 \times 3 \times 3 \times 3

Cancelling 3 with 3 :

\quad \longrightarrow \qquad \: \dfrac{4}{ \cancel{3}}  \times 3.14 \times 3 \times 3 \times  \cancel{3}

We get ,

\quad \longrightarrow \qquad \:4 \times 3.14 \times 9

Multiplying 4 and 3.14 :

\quad \longrightarrow \qquad \:12.56 \times 9

Multiplying 12.56 and 9 :

\quad \longrightarrow \qquad \:    \pink{\underline{\boxed{\frak{113.04  \: unit \: cube}}}} \quad \bigstar

  • <u>Henceforth</u><u> </u><u>,</u><u> </u><u>volume</u><u> </u><u>of</u><u> </u><u>sphere</u><u> </u><u>having </u><u>radius </u><u>3 </u><u>units </u><u>is </u><em><u>1</u></em><em><u>1</u></em><em><u>3</u></em><em><u> </u></em><em><u>.</u></em><em><u>0</u></em><em><u>4</u></em><em><u> </u></em><em><u>units </u></em><em><u>cube </u></em><em><u>.</u></em>

<h2><u>#</u><u>K</u><u>e</u><u>e</u><u>p</u><u> </u><u>Learning</u></h2>

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2 years ago
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