Aregular octagon has side length 10.9 in. the perimeter of the octagon is 87.2 in and the area is 392.4 in2. a second octagon has side lengths equal to 16.35 in. find the area of the second octagon.

To solve this problem, let us first calculate for the Perimeter of the other octagon. The formula for Perimeter is:

Perimeter = n * l

Where n is the number of sides (8) and l is the length of one side. Let us say that first octagon is 1 and the second octagon is 2 so that:

Perimeter 2 = 8 * 16.35 in = 130.8 inch

We know that Area is directly proportional to the square of Perimeter for a regular polygon:

Area = k * Perimeter^2

Where k is the constant of proportionality. Therefore we can equate 1 and 2 since k is constant:

Area 1 / Perimeter 1^2 = Area 2 / Perimeter 2^2

Substituting the known values:

392.4 inches^2 / (87.2 inch)^2 = Area 2 / (130.8 inch)^2

Area 2 = 882.9 inches^2

Therefore the area of the larger octagon is about 882.9 square inches.

3. (5x + 1) - (-10x + 6)

simplify the answer choice. combine like terms. remember to distribute the negative to the terms inside the second parenthesis. also remember that two negatives = one positive, and one of each sign = negative.

- (-10x + 6) = + 10x - 6

simplify. combine like terms:

5x + 1 + 10x - 6

5x + 10x + 1 - 6

15x - 5

∴ it is your answer

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