Mathematics High School

PLEASE HELP - Find the area of a regular octagon with a side length of 5 cm. Round to the nearest tenth. (Please explain in full detail the process, im trying to learn how to do these) thank you!!

Answers

Answer 1

Answer:

120.7

Step-by-step explanation:

The formula for finding the area of a regular octagon is: A = 2(1+[tex]\sqrt{2}[/tex])[tex]a^{2}[/tex].

Using the formula I got 120.7(rounded).

Answer 2

The area of the regular octagon with a side length of 5 cm is approximately 120.7cm².

We have,

To find the area of a regular octagon with a side length of 5 cm, you can use the following formula:

Area = 2 * (1 + √2) * (Side Length)²

Plugging in the values:

Area = 2 * (1 + √2) * (5 cm)² ≈ 120.7 cm²

Therefore,

The area of the regular octagon with a side length of 5 cm is approximately 120.7cm².

Learn more about octagons here:

https://brainly.com/question/30182906

#SPJ4

Related Questions

Consider the following expression and the simplified expression. Expression Simplified Expression 3 x squared + 5 y squared box + 3 box + 4 y squared + 6 9 x squared minus y squared + 9 Which terms could be in the boxes to make the expressions equivalent? Positive 6 x squared and Negative 6 y squared Positive 6 x squared and Negative 10 y squared Positive 9 x squared and Negative 10 y squared Positive 9 x squared and Negative 6 y squared

Answers

Answer:

The correct answer is:

[tex]+6x^{2}\\-9y^2[/tex]

Step-by-step explanation:

We are given the term:

[tex]3x^{2} +5y^{2} [\text{ \ }] +3 [\text{ \ }] +4y^{2} +6 = 9x^{2} -y^{2} +9[/tex]

We have to fill in to the empty spaces such that the above equation gets satisfied.

First of all, let us simplify the LHS (Left Hand Side):

[tex]3x^{2} +5y^{2} [\text{ \ }] +3 [\text{ \ }] +4y^{2} +6\\\Rightarrow 3x^{2} +5y^{2} +4y^{2} [\text{ \ }] [\text{ \ }] +6 +3\\\Rightarrow 3x^{2} +9y^{2} [\text{ \ }] [\text{ \ }] +9[/tex]

Now, let us equate the LHS and RHS (Right Hand Side):

[tex]\Rightarrow 3x^{2} +9y^{2} [\text{ \ }] [\text{ \ }] +9 = 9x^{2} -y^{2} +9[/tex]

Equating the coefficients of [tex]x^{2}\ and\ y^{2}[/tex] in LHS and RHS:

One box will have value = [tex]9x^{2} -3x^{2} =+6x^{2}[/tex]

Other box will have value = [tex]-y^{2} -9y^{2} =-10y^{2}[/tex]

The correct answer is:

[tex]+6x^{2}\\-9y^2[/tex]

So, if we fill the boxes with above values, the expression will be simplified as given.

Answer:

The correct answer is B. Positive 6 x squared and Negative 10 y squared

Step-by-step explanation:

intro to geometric sequences (help pls)

Answers

Answer:

Option B

Step-by-step explanation:

The formula for geometric sequence is given by:

[tex]a_{n} = a_{1}r^{n-1}[/tex]

Where,

[tex]a_{n}[/tex] = nth term of sequence

[tex]a_{1}[/tex] = 1st term of the sequence

[tex]r[/tex] = common ratio (ration of the second term to the first term)

So,

Here:

[tex]a_{1}[/tex] = 12

[tex]r[/tex] = 6/12 = 1/2

Plugging in the values of [tex]a_{1}[/tex] and r in the above formula:

=> [tex]a_{n} = 12 * (\frac{1}{2}) ^ {n-1}[/tex]

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