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3 years ago

If a shape is a regular pentagon with five sides what must be true

Mathematics

2 answers:

Dimas [21]3 years ago

7 0

A pentagon is a closed 5 sided polygon.

A regular polygon has all sides equal as well as all angles equal.

°The formula to find the sum of all angles of a polygon is = (n-2) ×180

So for a pentagon, n= 5

So the sum of all the angles must be (5-2)× 180= 3 × 180 = 540.

Now as it is a pentagon, it shall have 5 vertices.

So there are 5 angles in a pentagon and all must be equal.

Hence each interior angle must be 540/5 = 108

So for a regular polygon with 5 sides must have each interior angle = 108

Effectus [21]3 years ago

4 0

Answer:

Step-by-step explanation:

A) it is symmetrical

B) it has five lines of symmetry

C) it has reflectional symmetry

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4 years ago

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garri49 [273]

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3 years ago

Y=3x+1 on a table graph

fgiga [73]

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3 0

3 years ago

If the product of any two rational numbers is 2 and one of them is 1/7, find the other number.

ozzi

Answer:

The other number is 14.

Step-by-step explanation:

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$\frac{2}{1/7}$

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Hope this helped!

5 0

3 years ago

Need help pleaseI was bad at math in school so lwant to learn

aleksklad [387]

The probability of an event is expressed as

$Pr(\text{event) =}\frac{Total\text{ number of favourable/desired outcome}}{Tota\text{l number of possible outcome}}$

Given:

$\begin{gathered} \text{Red}\Rightarrow2 \\ \text{Green}\Rightarrow3 \\ \text{Blue}\Rightarrow2 \\ \Rightarrow Total\text{ number of balls = 2+3+2=7 balls} \end{gathered}$

The probability of drwing two blue balls one after the other is expressed as

$Pr(\text{blue)}\times Pr(blue)$

For the first draw:

$\begin{gathered} Pr(\text{blue) = }\frac{number\text{ of blue balls}}{total\text{ number of balls}} \\ =\frac{2}{7} \end{gathered}$

For the second draw, we have only 1 blue ball left out of a total of 6 balls (since a blue ball with drawn earlier).

Thus,

$\begin{gathered} Pr(\text{blue)}=\frac{number\text{ of blue balls left}}{total\text{ number of balls left}} \\ =\frac{1}{6} \end{gathered}$

The probability of drawing two blue balls one after the other is evaluted as

$\begin{gathered} \frac{1}{6}\times\frac{2}{7} \\ =\frac{1}{21} \end{gathered}$

The probablity that none of the balls drawn is blue is evaluted as

$\begin{gathered} 1-\frac{1}{21} \\ =\frac{20}{21} \end{gathered}$

Hence, the probablity that none of the balls drawn is blue is evaluted as

$\frac{20}{21}$

8 0

1 year ago