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Neporo4naja [7]
3 years ago
6

Which regular polygon has a minimum rotation of 45 degrees to carry the polygon onto itself?

Mathematics
2 answers:
lubasha [3.4K]3 years ago
7 0

Answer:

Hence, a regular polygon that has a minimum rotation of 45 degrees to carry the polygon onto itself is:

A. octagon.

Step-by-step explanation:

We know that the for a regular polygon with 'n' sides the minimum rotation that can carry the polygon onto itself is:

\dfrac{360\degree}{n}

Here we are given the minimum angle of  rotation such that polygon is transformed to itself as 45 degree.

i.e. we have:

\dfrac{360}{n}=45\\\\n=\dfrac{360}{45}\\\\n=8

Hence, the number of sides of a polygon are 8.

Hence, the polygon is:

A. octagon.

( Since a polygon with 8 sides is called a octagon)

SVEN [57.7K]3 years ago
6 0
Octagon, because 360/8 = 45
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The average rate of the change from x = 2 to x = 5 is about 20 , The domain is constrained at x =0

<h3>What is a Function ?</h3>

A function is a mathematical statement used to derive a relation between a dependent variable and an independent variable.

A) The vertical axis of the graph shows a company's profit f(x), in dollars, depending on the price of pencils x, in dollars, sold by the company , i.e is the x axis .

So the x-intercepts represent prices of the pencil at which the profit is 0

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5 0
2 years ago
In the circle below, suppose m FEH=272º and m EFG=116º. Find the following.​
adoni [48]

Answer:

m∠FEH = 44°

m∠EHG =  64°

Step-by-step explanation:

1) The given information are;

The angle of arc m∠FEH = 272°, the measured angle of ∠EFG = 116°

Given that m∠FEH = 272°, therefore, arc ∠HGF = 360 - 272 = 88°

Therefore, angle subtended by arc ∠HGF at the center = 88°

The angle subtended by arc ∠HGF at the circumference = m∠FEH

∴ m∠FEH = 88°/2 = 44° (Angle subtended at the center = 2×angle subtended at the circumference)

m∠FEH = 44°

2) Similarly, m∠HGF is subtended by arc m FEH, therefore, m∠HGF = (arc m FEH)/2 = 272°/2 = 136°

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Therefore;

m∠FEH + m∠HGF + m∠EFG + m∠EHG = 360° (The sum of angles in a quadrilateral EFGH)

m∠EHG = 360° - (m∠FEH + m∠HGF + m∠EFG) = 360 - (44 + 136 + 116) = 64°

m∠EHG =  64°.

5 0
3 years ago
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