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mina [271]
2 years ago
10

How do you find prime factorization?

Mathematics
1 answer:
skad [1K]2 years ago
6 0
You can use a factor tree, and if you dont know how to use one, look it up on youtube, theres tons
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Who can pls follow me on rblx im trying to get 100 followers pls follow me tysm if you will :D
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I can my user is hxe_toxic
4 0
3 years ago
The polygons are regular polygons. What is the area of the shaded region to the nearest tenth? (see attachment)
Fynjy0 [20]
Area of shaded region = 2(4*4)=2*16=32 in squared 

hope helped 
8 0
3 years ago
Read 2 more answers
Solve the equation. <br><br> 2.2= z - 1.1 <br> z=?
fredd [130]

Answer:

z = 3.3

Step-by-step explanation:

Given

2.2 = z - 1.1 ( add 1.1 to both sides )

3.3 = z

7 0
2 years ago
Please calculate this limit <br>please help me​
Tasya [4]

Answer:

We want to find:

\lim_{n \to \infty} \frac{\sqrt[n]{n!} }{n}

Here we can use Stirling's approximation, which says that for large values of n, we get:

n! = \sqrt{2*\pi*n} *(\frac{n}{e} )^n

Because here we are taking the limit when n tends to infinity, we can use this approximation.

Then we get.

\lim_{n \to \infty} \frac{\sqrt[n]{n!} }{n} = \lim_{n \to \infty} \frac{\sqrt[n]{\sqrt{2*\pi*n} *(\frac{n}{e} )^n} }{n} =  \lim_{n \to \infty} \frac{n}{e*n} *\sqrt[2*n]{2*\pi*n}

Now we can just simplify this, so we get:

\lim_{n \to \infty} \frac{1}{e} *\sqrt[2*n]{2*\pi*n} \\

And we can rewrite it as:

\lim_{n \to \infty} \frac{1}{e} *(2*\pi*n)^{1/2n}

The important part here is the exponent, as n tends to infinite, the exponent tends to zero.

Thus:

\lim_{n \to \infty} \frac{1}{e} *(2*\pi*n)^{1/2n} = \frac{1}{e}*1 = \frac{1}{e}

7 0
3 years ago
Which of the followint iw a linear equation
SpyIntel [72]

Answer:

11.C

12.B

Step by step explanation:

4 0
3 years ago
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