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jasenka [17]
3 years ago
10

The figure below shows circle with center O and regular hexagon ABCDEF constructed inscribed in a circle.Which of the following

steps will help prove that triangle FBD is equilateral?

Mathematics
2 answers:
Vikki [24]3 years ago
6 0

Answer:

As mentions in the explanation steps

Step-by-step explanation:

In the given figure hexagonal ABCDEF is a regular hexagonal, therefore each interior angles and sides of the hexagonal are same.

Now, the following steps are  helpful to prove that the triangle FBD is equilateral.

(1). Join the the three pints (pint F, pint B, pint D ) with the help of line segments and formed a triangle FBD.

(2). In ΔFAB, ΔBCD, ΔDEF two sides of each triangle and included angle are same( because two sides of each triangle and included angle must be the sides and interior angle of the given regular hexagon ABCDEF).  So ΔFAB, ΔBCD and  ΔDEF are congruence to each other by SAS postulate.

(3). By corresponding part of congruence triangle we say that line segment FB, line segment BD and line segment DF are same.

(4). Hence ΔFBD is an equilateral triangle, because all sides are same (from step -3).

mote1985 [20]3 years ago
4 0
I hope this helps you

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Which ordered pair is a solution to the system of linear equations 1/2x-3/4y=11/60 and 2/5x+1/6y=3/10
natka813 [3]

ANSWER

( \frac{2}{3} , \frac{1}{5} )

EXPLANATION

The first equation is

\frac{1}{2} x -  \frac{3}{4} y =  \frac{11}{60} ...(1)

The second equation is

\frac{2}{5} x  +  \frac{1}{6} y =  \frac{3}{10} ...(2)

We want to eliminate y, so we multiply the first equation by

\frac{4}{5}

\frac{4}{5}  \times \frac{1}{2} x - \frac{4}{5}    \times \frac{3}{4} y =  \frac{11}{60}  \times  \frac{4}{5}

\frac{2}{5} x - \frac{3}{5} y =  \frac{11}{75} ...(3)

We now subtract equation (3) from (2)

(\frac{2}{3} x  -  \frac{2}{3} x )+ ( \frac{1}{6} y -  -  \frac{3}{5}y ) =(  \frac{3}{10}  -  \frac{11}{75} )

\frac{1}{6} y  +  \frac{3}{5}y  =\frac{3}{10}  -  \frac{11}{75}

\frac{23}{30}y =  \frac{23}{150}

Multiply both sides by

\frac{30}{23}

\implies \:  \frac{30}{23} \times  \frac{23}{30}y=  \frac{23}{150}  \times  \frac{30}{23}

\implies \: y =  \frac{1}{5}

Substitute into the first equation to solve for x .

\frac{1}{2} x -  \frac{3}{4}  \times \frac{1}{5} =  \frac{11}{60}

Multiply to obtain

\frac{1}{2} x -  \frac{3}{20} =  \frac{11}{60}

\frac{1}{2} x = \frac{11}{60} + \frac{3}{20}

\frac{1}{2} x = \frac{1}{3}

Multiply both sides by 2.

2 \times \frac{1}{2} x =2 \times  \frac{1}{3}

x = \frac{2}{3}

The solution is

( \frac{2}{3} , \frac{1}{5} )

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3 years ago
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Write each parabola in the form y-k=a(x-h)^2 aand determine its vertex.
Viefleur [7K]

Answer:

Step-by-step explanation:

a

5 0
3 years ago
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What is the side length of a 36 inch cube
Vladimir [108]

The side length is 3 inches

<h3>What is the perimeter of a cube?</h3>

The formula for perimeter of a cube is given as:

Perimeter = 12 × side length

We have that;

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  • Side length be x

Substitute into the formula

12x = 36

Divide both sides by 12

12x/12 = 36 / 12

x = 3 inches

The side length is x and equal to 3 inches.

Hence, the side length is 3 inches

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8 0
2 years ago
In triangle ABC measure of angle A equals 45°. The altitude divides the side AB into two parts of 20 and 21 units. Find BC.
Nat2105 [25]

In triangle  ABC, BC = 29 units.

Given,

In Δ ABC, ∠A = 45°.

A perpendicular line segment traced from a triangle's vertex to its opposite side is said to be the triangle's altitude.

Let altitude meet AB on D.

So, AD = 20 units and DB = 21 units and ∠ADC = ∠BDC = 90°

In ΔADC,

tan A = DC/AD

tan 45° = DC/20

          1 = DC/20

      DC = 20 units

In ΔBDC,

BC² = BD² + DC²     (Pythagorean Theorem)

      = 21²  + 20²

      = 441 + 400

      = 841

BC  = √841

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Hence, BC = 29 units.

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4 0
1 year ago
A bouquet of 8
iren2701 [21]

A rose costs $2.5 and a carnation costs $3.

Step-by-step explanation:

Let,

Roses = x

Carnations = y

According to given statement;

8x+10y=50  Eqn 1

6x+12y=51    Eqn 2

Multiplying Eqn 1 by 6;

6(8x+10y=50)\\48x+60y=300\ \ \ Eqn\ 3\\

Multiplying Eqn 2 by 8;

8(6x+12y=51)\\48x+96y=408\ \ \ Eqn 4

Subtracting Eqn 3 from Eqn 4;

(48x+96y)-(48x+60y)=408-300\\48x+96y-48x-60y=108\\36y=108

Dividing both sides by 36;

\frac{36y}{36}=\frac{108}{36}\\y=3

Putting y=3 in Eqn 3;

48x+60(3)=300\\48x+180=300\\48x=300-180\\48x=120

Dividing both sides by 48;

\frac{48x}{48}=\frac{120}{48}\\x=2.5

A rose costs $2.5 and a carnation costs $3.

Keywords: linear equations, subtraction

Learn more about linear equations at:

  • brainly.com/question/11007026
  • brainly.com/question/11207748

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8 0
3 years ago
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