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Leto [7]
3 years ago
14

A regular polygon has 15 sides. Which is a possible angle of rotational symmetry for the figure?

Mathematics
1 answer:
sasho [114]3 years ago
5 0

Answer with explanation:

Number of sides of Regular Polygon = 15

Sum of Interior angles of Regular Polygon = (n-2)×180°

Sum of Interior angles of Regular Polygon having 15 Sides= (15-2)×180°

          =13 × 180°

           = 2340°

As, Each interior angle of Regular Polygon is equal.

Measure of each Interior angle

      =\frac{2340}{15}\\\\=156

     =156°

Measure of each Exterior angle = 180° -156°=24°

Angle of Rotational Symmetry of Polygon having 15 Sides=24°,48°,....=24°×n, Where ,n=1,2,3,....

→→You can find exterior angle by this method also.

Sum of exterior angles of any Polygon =360°

Measure of each interior angle of Regular polygon having 15 Sides

         =\frac{360}{15}\\\\=24{\text{Degree}}

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Explain why the slope of line a is 2/6.
kifflom [539]

Answer:

The slope of a line is 2/6 because slope also means rise over run (y/x). So the line's slope is 2/6 because it goes up (rises) in the y-direction 2 units and moves to the right (runs) 6 units, from one point to another.

Step-by-step explanation:

5 0
2 years ago
Samuel made a list of his test scores: 88, 100, 92, 80, 85, 94, and 90. What is the lowest test score he can get on his next tes
yulyashka [42]

Answer:

he needs to get a 91 to have a mean score of 90

Step-by-step explanation:

add up all the known scores to get 629

let 's' = lowest test score

(629 + s) ÷ 8 = 90   [we divide by 8 because there are 7 known and 1 unknown score)

cross-multiply to get:

629 + s = 720

s = 720-629

s = 91

4 0
2 years ago
15 round to the nearest ten
babymother [125]
15 round by the nearest 10 is 20.
4 0
3 years ago
Prove that tan 10+tan 70'+taniou = tanio tanfo'tam 1oo.​
Galina-37 [17]

Question:

Prove that:

tan(10) + tan(70) + tan(100) = tan(10). tan(70). tan(100)

Answer:

Proved

Step-by-step explanation:

Given

tan(10) + tan(70) + tan(100) = tan(10). tan(70). tan(100)

Required

Prove

tan(10) + tan(70) + tan(100) = tan(10). tan(70). tan(100)

Subtract tan(10) from both sides

- tan(10)+tan(10) + tan(70) + tan(100) = tan(10). tan(70). tan(100) - tan(10)

tan(70) + tan(100) = tan(10). tan(70). tan(100) - tan(10)

Factorize the right hand size

tan(70) + tan(100) = -tan(10)(-tan(70). tan(100) + 1)

Rewrite as:

tan(70) + tan(100) = -tan(10)(1-tan(70). tan(100))

Divide both sides by 1-tan(70). tan(100)

\frac{tan(70) + tan(100)}{1-tan(70). tan(100)} = \frac{-tan(10)(1-tan(70). tan(100))}{1-tan(70). tan(100))}

\frac{tan(70) + tan(100)}{1-tan(70). tan(100)} = -tan(10)

In trigonometry:

tan(A + B) = \frac{tan(A) + tan(B)}{1 - tan(A)tan(B)}

So:

\frac{tan(70) + tan(100)}{1 - tan(70)tan(100)} can be expressed as: tan(70 + 100)

\frac{tan(70) + tan(100)}{1-tan(70). tan(100)} = -tan(10) gives

tan(70 + 100) = -tan(10)

tan(170) = -tan(10)

In trigonometry:

tan(180 - \theta) = -tan(\theta)

So:

tan(180 - 10) = -tan(10)

Because RHS = LHS

Then:

tan(10) + tan(70) + tan(100) = tan(10). tan(70). tan(100) has been proven

6 0
3 years ago
#2
ss7ja [257]
C. 15a+100
This is a binomial
4 0
3 years ago
Read 2 more answers
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