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kotegsom [21]
3 years ago
13

A circular cake with a radius of 8 inches is cut from the center into 6 equal pieces. How many inches wide, to the nearest tenth

of an inch, is the outer edge of each piece of cake?
Mathematics
2 answers:
aalyn [17]3 years ago
5 0

Answer: 8.4 inches.

Step-by-step explanation:

We know that the circumference of a circle can be calculated with this formula:

C=2\pi r

Where "r" is the radius of the circle.

We know that the circular cake (whose radius is 8 inches) is cut from the center into 6 equal pieces. Then, you need to divide the circumference of this circular cake by 6 to find the width of the outer edge of each piece of cake.

Therefore, this is (to the nearest tenth of an inch):

C=\frac{2\pi (8in)}{6}\\\\C=8.4in

Aliun [14]3 years ago
5 0

The width of the outer edge of each piece of cake  8.4 inches.

<u>SOLUTION: </u>

Given, a circular cake with a radius of 8 inches is cut from the center into 6 equal pieces. We have to find the width of the outer edge of each piece of cake.

We know that the circumference of a circle can be calculated with this formula: 2\pi r where "r" is the radius of the circle.

We know that the circular cake (whose radius is 8 inches) is cut from the center into 6 equal pieces.

Then, you need to divide the circumference of this circular cake by 6 to find the width of the outer edge of each piece of cake.

Therefore, this is (to the nearest tenth of an inch):

\text { Width }=\frac{2 \pi \times 8}{6}=\frac{\pi \times 8}{3}=\frac{8 \pi}{3}=8.377

Hence, the cake piece is 8.4 inches wide.

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