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dedylja [7]
3 years ago
11

Find the area of a regular hexagon with side length 10 m. Round to the nearest tenth.​

Mathematics
1 answer:
Maksim231197 [3]3 years ago
4 0

Answer:

A=259.8\ m^2

Step-by-step explanation:

we know that

The area of a regular hexagon is the same that the area of 6 equilateral triangles

The area of 6 equilateral triangles applying the law of sines is equal to

A=6[\frac{1}{2}b^2sin(60^o)]

where

b is the length side of the regular hexagon

we have

b=10\ m

substitute

A=6[\frac{1}{2}(10)^2sin(60^o)]

A=259.8\ m^2

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You would have to find out how many square feet the lawn is first:
70 x 50 = 3500 square feet
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5 0
2 years ago
What is <2 if <4 is 35
Rudiy27

Answer:

2 is 17.5

Step-by-step explanation:

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2 years ago
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Answer:

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Step-by-step explanation:

Remember, a is congruent to b modulo d if d divides a-b.

Now, the problem says that b=4 and d=14.

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7 0
3 years ago
Eight more than seven times a number is equal to four more than five times the number. What is the number?
Alex17521 [72]

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