Find the area of a regular octagon with apothem K and side of 10.40K
60K
80K

????

Answers

Answer 1
Answer:

Answer: 40K

Step-by-step explanation:

We know that the area of a regular polygon is given by :-

A=(1)/(2)*Apothem*Perimeter

Given: The side of regular octagon = 10

Then, the perimeter of the regular octagon =8(side)=8(10)=80

Now, the area of given regular octagon is given by :-

A=(1)/(2)*K*80\n\Rightarrow\ A=40K


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A.) What is the exact surface area of the container with a radius of 9 in and a height of 36 in?

Simplify 5(MN + 1) - 5.

Answers

1. Distribute the 5 so that you get 5MN + 5 - 5.
Then you subtract the 5's and you're left with 5MN.

Solve the following system of equations algebraically:
y = x^2+7x−4
y = −x−4

Answers

y = x² + 7x - 4
y = -x - 4

x² + 7x - 4 = -x - 4
      + x       + x
x² + 8x - 4 = -4
           + 4  + 4
     x² + 8x = 0
x(x) + x(8) = 0
    x(x + 8) = 0
x = 0  or  x + 8 = 0
                   - 8  - 8
                     x = -8

      y = -x - 4
      y = -0 - 4
      y = 0 - 4
      y = -4
(x, y) = (0, -4)
         or
       y = -x - 4
       y = -(-8) - 4
       y = 8 - 4
       y = 4
 (x, y) = (-8, 4)

The solutions are (0, -4) and (8, -4).
y=x²+7x-4
y=-x-4

We can solve this system of equation by equalization method.
x²+7x-4=-x-4
x²+7x+x-4+4=0
x²+8x=0
x(x+8)=0
we have two linear equations :
x=0                                ⇒  x=0        ⇒  y=-x-4=-0-4=-4
(x+8)=0                         ⇒  x=-8      ⇒  y=-x-4=-(-8)-4=8-4=4
 
We have two solutions:
solution₁:  x=0;  y=-4
solution₂:  x=-8; y=4

∆STV ~ ∆PQR. Find m∠P.129


9


86


6

Answers

Answer: 86^(\circ)

Step-by-step explanation:

Given :\triangle{STV}\sim\triangle{PQR}

We know that the corresponding angles of two similar triangles are congruent.

i.e. \angle{S}\cong\angle{P}                  (1)

\angle{T}\cong\angle{Q}

\angle{V}\cong\angle{R}

From the given picture , we have \angle{S}=86^(\circ)

Then from (1), \angle{P}=86^(\circ)

Hence, \angle{P}=86^(\circ)

m < P = m < S becuase the triangles are similar.

m < P = 86 degrees

Why are circles so important to constructions?

Answers

Its a way to find a range of length without using exact measuments (if your using a compass)
Also you need to know constructions for the weekly tests so your school can get more money

Give one equation that could generate the ordered pair (2,8)

Answers

1).   x = 2

2).   y = 8

3).  y = 4x

4).  y = 10x - 12

5).  y = 2 x²

which are the roots of the quadratic function f(b) = b2 – 75? check all that apply. b = 5 square root of 3 b = -5 square root of 3 b = 3 square root of 5 b = -3 square root of 5 b = 25 square root of 3 b = -25 square root of 3

Answers

The answers are b = 5 square root of 3; b = -5 square root of 3. f(b) = b^2 – 75. If f(b) = 0, then b^2 – 75 ) 0. b^2 = 75. b = √75. b = √(25 * 3). b = √25 * √3. b = √(5^2) * √3. Since √x is either -x or x, then √25 = √(5^2) is either -5 or 5. Therefore. b = -5√3 or b = 5√3.

Answer:

b=5√(3) and b=-5√(3) are the roots of given quadratic equation.

Step-by-step explanation:

Given quadratic equation is f(b)=b^2-75

We have to check all the given options.

If the value of f(b) gives 0 when put the value of b in above equation then only that b value is the root of quadratic equation.

b=5√(3): (5√(3))^(2)-75=75-75=0

b=-5√(3): (-5√(3))^(2)-75=75-75=0

b=3√(5): (3√(5))^(2)-75=45-75=30\neq 0

b=-3√(5): (-3√(5))^(2)-75=45-75=30\neq 0

b=25√(3): (25√(3))^(2)-75=1875-75=1800\neq 0

b=-25√(3): (-25√(3))^(2)-75=1875-75=1800\neq 0

hence, only first two values b=5√(3),-5√(3) gives the value of f(b)=0 .

b=5√(3) and b=-5√(3) are the roots of given quadratic equation.