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Kobotan [32]
2 years ago
15

Each exterior angle of a regular decagon has a measure of (3x + 6)". What is the value of x?

Mathematics
2 answers:
Oxana [17]2 years ago
7 0

Answer:

B = 10

Step-by-step explanation:

Just took test edg 2020

dedylja [7]2 years ago
4 0

Answer:10

Step-by-step explanation:

Each exterior angle is (3x + 6)°

A decagon has 10 sides

Each exterior angle=360/n

n=number of sides

3x+6=360/10

Cross multiply

10(3x + 6)=360

30x+60=360

Collect like terms

30x=360-60

30x=300

Divide both sides by 30

30x/30=300/30

x=10

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Which statement must be true about the diagram?
Tcecarenko [31]

Answer:

It's the third choice.

Step-by-step explanation:

M < NKM and m < MKL both equal  61 degrees

so KM is a bisector of < NKL.

5 0
3 years ago
Read 2 more answers
I’m confused?????????
QveST [7]

Answer:

The equation of the line is y - 3 = 2.5(x - 2) ⇒ D

Step-by-step explanation:

The rule of the slope of a line is m = \frac{y2-y1}{x2-x1} , where

  • (x1, y1) and (x2, y2) are two points on the line

The point-slope form of a line is y - y1 = m(x - x1), where

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From the given figure

∵ The line passes through points (2, 3) and (0, -2)

∴ x1 = 2 and y1 = 3

∴ x2 = 0 and y2 = -2

→ Substitute them in the rule of the slope to find it

∵ m = \frac{-2-3}{0-2}=\frac{-5}{-2}=2.5

∴ m = 2.5

→ Substitute the values of m, x1, y1 in the form of the equation above

∵ m = 2.5, x1 = 2, y1 = 3

∵ y - 3 = 2.5(x - 2)

∴ The equation of the line is y - 3 = 2.5(x - 2)

6 0
2 years ago
12 less the product of an unknown number and three equals -9.
chubhunter [2.5K]
In standard form its written 
3x-12=-9
if you need help solving then here you go as well
3x-12=-9
add 12 to both sides
3x=3
divide both sides by 3 
x=1
4 0
2 years ago
PLEASE HELP ITS AN EMERGENCY
kirill [66]

Answer:

the answer is d

Step-by-step explanation:

3 0
2 years ago
Does anyone know how to do this ?!?
KonstantinChe [14]
Download( photo math). you can scan the question or write it. it answers as well as explains how to get the answer.
7 0
2 years ago
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