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SashulF [63]
3 years ago
11

Angles E and G are

Mathematics
1 answer:
mash [69]3 years ago
4 0
A and b as I believe
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30 points for the correct answer
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Answer:

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Step-by-step explanation:

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The sector COB is cut from the circle with center O. The ratio of the area of the sector removed from the whole circle to the ar
mafiozo [28]

Answer:

Ratio = \frac{R^2 - r^2 }{ r^2}

Step-by-step explanation:

Given

See attachment for circles

Required

Ratio of the outer sector to inner sector

The area of a sector is:

Area = \frac{\theta}{360}\pi r^2

For the inner circle

r \to radius

The sector of the inner circle has the following area

A_1 = \frac{\theta}{360}\pi r^2

For the whole circle

R \to Radius

The sector of the outer sector has the following area

A_2 = \frac{\theta}{360}\pi (R^2 - r^2)

So, the ratio of the outer sector to the inner sector is:

Ratio = A_2 : A_1

Ratio = \frac{\theta}{360}\pi (R^2 - r^2) : \frac{\theta}{360}\pi r^2

Cancel out common factor

Ratio = R^2 - r^2 : r^2

Express as fraction

Ratio = \frac{R^2 - r^2 }{ r^2}

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How can you prove a triangle is an equilateral triangle?
Jlenok [28]
Each side is equal  its in the name.  hope this help!!
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