MATHEMATICS COLLEGE

A stop sign is in the shape of a regular octagon. Each side measures 12.4 inches and the apothem of the octagon measures 15 inches. A stop sign is shown. What is the area of the stop sign? 93 square inches 186 square inches 744 square inches 1,488 square inches

Answers

Answer 1

Answer:

The area of the stop sign which is an octagon is 744 square inches.

Option C is the correct answer.

We have,

To find the area of the stop sign, we can use the formula for the area of a regular polygon:

Area = (1/2) × perimeter × apothem

The perimeter of the stop sign is 8 × 12.4 = 99.2 inches.

Using the given apothem of 15 inches, we can plug in the values and calculate:

Area = (1/2) × 99.2 × 15

Area = 744 square inches

Therefore,

The area of the stop sign is 744 square inches.

Learn more about octagons here:

brainly.com/question/30327829

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Answer 2

Answer:

Answer:744 square inches

Step-by-step explanation:

Just did it it correct trust me

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Answers

Answer:

I belive it would be 1/5. ;;;

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Answers

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$(r^2+2rt) (-2st-rs)$

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(9.4 x 10^6)(3.2 x 10^5)
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Answers

Answer:3.008x10^12

Step-by-step explanation:

Answer:

3008000000000 or 3.008×10^12

Step-by-step explanation:

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100 is 10 times much as

Answers

Answer:

?? If you answer is how does 10 become 100, you multiply 10 by 10?? Sorry

Step-by-step explanation:

Answer:

100 is 10 times as much as 10

Step-by-step explanation:

10 times 10 is 100.

How do you do this 6-2x=10

Answers

-2x=10-6

The answer: x= -2

Answer:

x= -2

Step-by-step explanation: (remember the 6 in this equation is positive so you'll subtract it and the 2 is actually -2x. First, you'll subtract 6 from itself and 10, the two 6's cancel out and 10-6=4. Then, you'll divide -2 and 4 by -2, the two -2's cancel out and 4 divided by -2 is -2. Finally, you'll see that x = -2. This is all done by the order of operations also known as PEMDAS. )

6 - 2x =10

-6 -6

-----------------

-2x = 4

------------

-2 = -2

x = -2

Write the formula for Newton's method and use the given initial approximation to compute the approximations x_1 and x_2. f(x) = x^2 + 21, x_0 = -21 x_n + 1 = x_n - (x_n)^2 + 21/2(x_n) x_n + 1 = x_n - (x_n)^2 + 21 x_n + 1 = x_n - 2(x_n)/(x_n)^2 + 21 Use the given initial approximation to compute the approximations x_1 and x_2. x_1 = (Do not round until the final answer. Then round to six decimal places as needed.)