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Anna11 [10]
3 years ago
13

In ∆ABC the angle bisectors drawn from vertices A and B intersect at point D. Find m∠ADB if: m∠C=γ

Mathematics
1 answer:
mylen [45]3 years ago
5 0

Answer:

  ∠ADB = γ/2 +90°

Step-by-step explanation:

Here's one way to show the measure of ∠ADB.

  ∠ADB = 180° - (α + β) . . . . . sum of angles in ΔABD

  ∠ADB + (2α +β) + γ + (2β +α) = 360° . . . . . sum of angles in DXCY

Substituting for (α + β) in the second equation, we get ...

  ∠ADB + 3(180° - ∠ADB) + γ = 360°

  180° + γ = 2(∠ADB) . . . . . . add 2(∠ADB)-360°

  ∠ADB = γ/2 + 90° . . . . . . . divide by 2

_____

To find angles CXD and CYD, we observe that these are exterior angles to triangles AXB and AYB, respectively. As such, those angles are equal to the sum of the remote interior angles, taking into account that AY and BX are angle bisectors.

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satela [25.4K]

Answer:

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at n = 9: number of action figures = 51

Step-by-step explanation:

first row: 3

second row: 6 more than the row before it (3) = 6 + 3 = 9

third row: 6 + 9 = 15

arithmetic series: a_n = a_1 + (n-1)d, where

a_n is the nth term in the output

a_1 is the first output

n is the input

d is the difference between terms

here, we are given the row, and we want to figure out the number of action figures. thus, row = input and number of action figures = output.

the first output, in the first row, is 3

the difference between the number of action figures in each row is 6

thus, our formula is

a_n = 3 + (n-1)6 = 3 + 6(n-1)

when the row is 9, the number of action figures is equal to

3 + 6(9-1) = 3 + 6 * 8 = 51

8 0
2 years ago
(2^(3)*(-2)^(3)*3^(3))/(2^(3*2)*4^(4(-2))*8^(3))=?
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Answer:

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Step-by-step explanation

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3 years ago
-5(x+1)+3x=6-(4x-3) solve for x
PilotLPTM [1.2K]

Answer: x = -3

Step-by-step explanation:Step by step solution :

Step  1  :

Pulling out like terms :

1.1     Pull out like factors :

  -x - 3  =   -1 • (x + 3)  

Equation at the end of step  1  :

Step  2  :

Solving a Single Variable Equation :

2.1      Solve  :    -x-3 = 0  

Add  3  to both sides of the equation :  

                     -x = 3  

Multiply both sides of the equation by (-1) :  x = -3  

One solution was found :

                  x = -3

7 0
3 years ago
3/6 -1/6= <br> HELP ITS FOR A TEST
DiKsa [7]
Answer:
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8 0
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Read 2 more answers
a 15-foot telephone pole has a wire that extends from the top of the pole to the ground. The wire and the ground form a 42 degre
scoundrel [369]

Answer:

The length of the wire is 22.42 feet

The distance from the base of the pole to the spot where the wire touches the ground is 16.66 feet

Step-by-step explanation:

* Lets explain the situation in the problem

- The telephone pole , the wire and the ground formed a right triangle

- The wire is the hypotenuse of the triangle

- The height of the telephone pole and the distance from the base of

 the pole to the spot where the wire touches the ground are the legs

 of the triangle

- The angle between the wire and the ground is 42°

- The angle 42° is opposite to the height of the telephone pole

- The height of the telephone pole is 15 feet

* Lets use the trigonometry functions to find the length of the wire

 (hypotenuse) and the distance from the base of the pole to the spot

 where the wire touches the ground

∵ sin Ф = opposite/hypotenuse

∵ Ф = 42° and its opposite side = 15 feet

∴ sin 42 = 15/hypotenuse ⇒ by using cross multiplication

∴ sin 42° (hypotenuse) = 15 ⇒ divide both sides by sin 42

∴ hypotenuse = 15/sin 42° = 22.42 feet

∵ The length of the wire is the hypotenuse

∴ The length of the wire is 22.42 feet

∵ The distance from the base of the pole to the spot where the wire

   touches the ground is the adjacent side to the angle 42°

∵ tan Ф = opposite/adjacent

∴ tan 42° = 15/adjacent ⇒ by using cross multiplication

∴ tan 42° (adjacent) = 15 ⇒ divide both sides by sin 42

∴ adjacent = 15/tan 42° = 16.66 feet

∵ The adjacent side is the distance from the base of the pole to the

  spot where the wire touches the ground

∴ The distance from the base of the pole to the spot where the wire

   touches the ground is 16.66 feet

7 0
4 years ago
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