All categories

3 years ago

Find the area of a regular octagon with side length 6 cm

Mathematics

2 answers:

OverLord2011 [107]3 years ago

7 0

Answer: 173.82cm²

Step-by-step explanation:

prohojiy [21]3 years ago

3 0

Answer:

173.82cm²

Step-by-step explanation:

A=2(1+2)a2=2·(1+2)·62≈173.82338cm²

You might be interested in

Find the solutions to the system of equations. Select all that apply. Y=x^2-1 y=2x-2

Sliva [168]

Answer:

Option C (1, 0)

Step-by-step explanation:

We have a system with the following equations:

$y = x ^ 2-1\\\\y = 2x-2$

The first equation is a parabola.

The second equation is a straight line

To solve the system, substitute the second equation in the first and solve for x.

$2x-2 = x ^ 2 -1$

Simplify

$x ^ 2-2x + 1 = 0$

You must search for two numbers that when you add them, obtain as a result -2 and multiplying both results in 1.

These numbers are -1 and -1

Therefore

$x ^ 2-2x + 1= (x-1)(x-1)\\\\x ^ 2-2x + 1= (x-1)^2=0$

Finally the solutions are

$x = 1\\y =0$

8 0

3 years ago

Read 2 more answers

True or False, In Euclidean geometry, there is only one equilateral triangle

docker41 [41]

True, In Euclidean geometry an equilateral triangle must be a 60-60-60 triangle.

Hope This Helps!!!

5 0

3 years ago

Is it C ? am I right

sveta [45]

Answer:

You are right!!

Step-by-step explanation:

- 8 = 4 x (- 2)

- 16 = 4 x (- 4)

- 24 = 4 x (- 6)

8 0

3 years ago

There are 12 triangles and 3 squares. What is the simplest ratio of<br> squares to triangles?

grigory [225]

Squares : Triangles
3 : 12
Divide both sides by 3.
The simplest ratio is

1 : 4

4 0

3 years ago

Read 2 more answers

Find k so that the distance from (–1, 1) to (2, k) is 5 units. k= k= *there are two solutions for 2*

dalvyx [7]

Answer:

k = -3

k =5

Step-by-step explanation:

$d = \sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\\d = 5\\(-1,1) =(x_1,y_1)\\(2,k)=(x_2,y_2)\\$

$5=\sqrt{\left(2-\left(-1\right)\right)^2+\left(k-1\right)^2}\\\\\mathrm{Square\:both\:sides}:\quad 25=k^2-2k+10\\25=k^2-2k+10\\\\\mathrm{Solve\:}\:25=k^2-2k+10:\\k^2-2k+10=25\\\\\mathrm{Subtract\:}25\mathrm{\:from\:both\:sides}\\k^2-2k+10-25=25-25\\k^2-2k-15=0\\\\\mathrm{Solve\:by\:factoring}\\\\\mathrm{Factor\:}k^2-2k-15:\quad \left(k+3\right)\left(k-5\right)\\\mathrm{Solve\:}\:k+3=0:\quad k=-3\\$

$\mathrm{Solve\:}\:k-5=0:\quad k=5\\\\k =5 , k=-3$

7 0

3 years ago

Read 2 more answers