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masya89 [10]
2 years ago
5

11. Large parts of Syria came under the control of

Social Studies
1 answer:
laiz [17]2 years ago
5 0
Answer is Separatists. The separatists was led by the CIS. They produced an army to fight in the ancient clone wars.
You might be interested in
what contributed to the Iconoclast Controversy in the Byzantine Empire during the eighth and ninth centuries?
Svetlanka [38]

Answer:

Correct answer is that The iconoclasts' fear that adoration of icons would lead to idolatry .

Explanation:

During the reign of Byzantine ruler Leo III the idolatry was prohibited and people were not allowed to worship icons. This created a large conflict inside of the church, that lasted until the period of Empress Irene that officially reestablished the usage of icons. It was conflict that shook only the church, but the whole country.

8 0
2 years ago
Which incident type requires one or two single resources with up to six personnel?
gregori [183]

Answer:

Incident Type 5

Explanation:

Incident type 5 is the type of incident that can be managed with one(1) or two(2) single resources with up to six personnel.

Command and General Staff positions (other than the Incident Commander) are not

initiated.

The incident is included within the early operational moment and frequently within an hour to

a limited hours after resources arrive on scene.

A good examples is a vehicle fire, an injured person, or a police traffic stop etc.

3 0
3 years ago
Read 2 more answers
In civil cases, no one is charged with a crime. True or False?
Anit [1.1K]

Answer:

I believe it's false.

Explanation:

Hope my answer has helped you!

6 0
2 years ago
Read 2 more answers
In hippocrates theory of personality, what bodily humor was associated with choleric temperament?
Julli [10]
<span>The body humor associated with choleric temperament is yellow bile. There are 3 other body humors including blood, phlegm, and black bile.</span>
6 0
3 years ago
Question 9 please both a and b
GarryVolchara [31]

a^k - b^k = (a - b)(a^{k-1} + a^{k-2}b + ... + b^{k-1})

a. Let's set a=3, b=1

3^k - 1^k = (3 - 1)(\textrm{some integer we'll call } m)

3^k = 1 + 2m

In other words 3^k is odd, i.e. 3^k has remainder 1 when divided by 2. \quad\checkmark

b. Let

k=2n, a=3, b=1

a^k - b^k = (a - b)(a^{k-1} + a^{k-2}b + ... + b^{k-1})

3^{2n} - 1^{2n} = (3 - 1 )(3^{2n-1} + 3^{2n-2} + ... + 1)

There are 2n terms in the last factor.  All the terms of the form

3^{2n-k}

are odd. So when we add up an even number of them, we get an even number, call it 2m.

3^{2n} - 1 = 2(2m)

3^{2n} - 1 = 4m

We've shown 3^{2n}-1 is divisible by four.  \quad \checkmark

Merry Christmas!

3 0
3 years ago
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