For a regular 30-sided polygon, what is the degree of rotation? enter your answer in the box. °

Mathematics

2 answers:

Shkiper50 [21]4 years ago

7 0

Answer:

Step-by-step explanation:

Crank4 years ago

6 0

For a 30 sided regular polygon, We can divide the polygon (30 sides) into 30 inscribed triangles with central vertex angles of = $\frac{360}{30} =12^{o}$ .

This central vertex angle of <u>12 degrees</u> is the degree of rotation for a 30 sided polygon.

In other words, the 30 sided polygon has <u>12 degree</u> rotational symmetry about the center.

You might be interested in

The Candle Factory is producing a new candle. It has a radius of 3 inches and a height of 5 inches. What is the surface area of

garik1379 [7]

Answer:

87.92 square inches

Step-by-step explanation:

Surface area for cylinder is given by $2\pi r^2 + 2\pi rh$

where r is the radius and h is the height of cylinder.

Given that candle is cylindrical in shape

Thus, Surface area for candle is calculated as given below\\

$Surface / area \ for \ candle = 2\pi r^2 + 2\pi rh\\=>Surface / area \ for \ candle = 2*3.14*3^2 + 2*3.14*5\\=>Surface / area \ for \ candle = 2*3.14*3^2 + 2*3.14*5 = 56.52 + 31.4\\=>Surface / area \ for \ candle = 56.52 + 31.4 = 87.92$

Thus, total surface area for candle is 87.92 square inches.

8 0

3 years ago

Among a group of 100 people, 68 can speak English, 45 can speak French, 42 can speak Spanish. 27 can speak both English and Fren

Nat2105 [25]

Answer:

Step-by-step explanation:

Given

No of People who can speak English is $n(E)=68$

No of People who can speak French is $n(F)=45$

No of People who can speak Spanish is $n(S)=42$

No of People who can speak both English and French $n\left ( E\cap F\right )=27$

No of People who can speak both English and Spanish $n\left ( E\cap S\right )=25$

No of People who can speak both French and Spanish $n\left ( F\cap S\right )=16$

No of people who can speak all languages is $n\left ( E\cap F\cap S\right )=9$

no of People who can Speak at least one Language is

$n\left ( E\cup F\cup S\right )=n\left ( E\right )+n\left ( F\right )+n\left ( S\right )-n\left ( E\cap F\right )-n\left ( E\cap S\right )-n\left ( F\cap S\right )+n\left ( E\cap F\cap S\right )$