All categories

3 years ago

The figure below is a regular octagon with a side length of 8 inches.

Mathematics

1 answer:

Masteriza [31]3 years ago

3 0

Answer:

309in²

Step-by-step explanation:

Area of an octagon is expressed as;

A=2(1+√2)a²

a is the side length of the octagon

Given

a = 8

Substitute into the formula

A = 2(1+√2)(8)²

A = 2(1+√2)(64)

Expand

A = 2(64+64√2)

A = 128+128√2

A = 128+128(1.414)

A = 128+181.01

A = 309in²

Hence the area of the octagon is 309in²

You might be interested in

Please solve my difficult question fast

kicyunya [14]

The 3 angles form the straight line AB. A straight line equals 180 degrees.

The 3 angles when added together need to equal 180:

2x + 65 + (x + 65) = 180

Simplify by combining like terms:

3x + 130 = 180

Subtract 130 from both sides

3x = 50

Divide both sides by 3

X = 50/3

X = 16 2/3 (16.66667 as a repeating decimal)

Now you have x if you need to solve all the angles replace x with its value and sole:

2x = 2(16 2/3) = 33 1/3

X + 65 = 16 2/3 + 65 = 81 2/3

7 0

3 years ago

Read 2 more answers

Which statement below is an example of the

Hitman42 [59]

Answer:

Step-by-step explanation:

Let say, a=9,b=11,c=6

Then associative properties says that every element is related to each other, here "b" associates a &c.

Therefore, a is related to b+c

c is related to a+b

3 0

3 years ago

A ferry will safely accommodate 82 tons of passenger cars. Assume that the meanweight of a passenger car is 2 tons with standard

Shalnov [3]

Answer:

The probability that the maximum safe-weight will be exceeded is <u>0.0455 or 4.55%</u>.

Step-by-step explanation:

Given:

Maximum safe-weight of 37 cars = 82 tons

∴ Maximum safe-weight of 1 car (x) = 82 ÷ 37 = 2.22 tons (Unitary method)

Mean weight of 1 car (μ) = 2 tons

Standard deviation of 37 cars = 0.8 tons

So, standard deviation of 1 car is given as:

$\sigma=\frac{0.8}{\sqrt{37}}=0.13$

Probability that maximum safe-weight is exceeded, P(x > 2.22) = ?

The sample is normally distributed (Assume)

Now, let us determine the z-score of the mean weight.

The z-score is given as:

$z=\frac{x-\mu}{\sigma}\\\\z=\frac{2.22-2}{0.13}\\\\z=\frac{0.22}{0.13}=1.69$

Now, finding P(x > 2.22) is same as finding P(z > 1.69).

From the z-score table of normal distribution curve, the value of area under the curve for z < 1.69 is 0.9545.

But we need the area under the curve for z > 1.69.

So, we subtract from the total area. Total area is 1 or 100%.

So, $P(z > 1.69) = 1 - P(z < 1.69)$

$P(z>1.69)=1-0.9545=0.0455\ or\ 4.55\%$

Therefore, the probability that the maximum safe-weight will be exceeded is 0.0455 or 4.55%.

8 0

3 years ago

In a landscape plan, a rectangular flower bed is designed to be 4 meters longer than it is wide. If 60 square meters are needed

Rudiy27

Answer:

i dk and stop callin me slime

Step-by-step explanation:

4 0

3 years ago

Yvette uses 3.5 cans of blue paint to paint her bedroom and b cans of blue paint to paint her kitchen. If she used 6.25 cans tot

liberstina [14]

Answer:3.25

Step-by-step explanation:

8 0

3 years ago