All categories

3 years ago

4. About how long ago did the first people migrate to North America?

Social Studies

2 answers:

Strike441 [17]3 years ago

7 0

The first people to migrate to North America was about 100,000 years ago.

lianna [129]3 years ago

6 0

Hello there army , I would say about 100,000 years ago

You might be interested in

What do communism and fascism have in common ?

enot [183]

Answer:

Communism and fascism leave little to no power to the citizens and are heavily dictated by the government or dictator.

5 0

3 years ago

What were benjamin Banneker's main childhood interests

lesantik [10]

Reading books and writting.

8 0

3 years ago

hRorey was just diagnosed with a personality disorder that fits into the detachment category of the DSM-5. What behavior is he m

Degger [83]

DSM-5 lists ten specific personality disorders: paranoid, schizoid, schizotypal, antisocial, borderline, histrionic, narcissistic, avoidant, dependent and obsessive–compulsive personality disorder.

<h3>What are the Cluster B DSM-5 personality disorders?</h3><h3>Cluster B personality disorders</h3>

They include antisocial personality disorder, borderline personality disorder, histrionic personality disorder and narcissistic personality disorder.

<h3>What does the DSM-5 stand for?</h3>

Diagnostic and Statistical Manual of Mental Disorders (DSM-5-TR).

Learn more about DSM-5 here:

<h3>brainly.com/question/16032005</h3><h3 /><h3>#SPJ4</h3>

8 0

1 year ago

Blazers = confidence to go up !! :)

juin [17]

Answer:

your good at posing I like the photo and the background too it looks good

your really pretty

8 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