All categories

3 years ago

Can someone help me please?

Mathematics

1 answer:

Roman55 [17]3 years ago

3 0

Answer:

X-16 the first one

Step-by-step explanation:

The first one is equal to x-4

You might be interested in

Lim (n/3n-1)^(n-1) n → ∞

n200080 [17]

Looks like the given limit is

$\displaystyle \lim_{n\to\infty} \left(\frac n{3n-1}\right)^{n-1}$

With some simple algebra, we can rewrite

$\dfrac n{3n-1} = \dfrac13 \cdot \dfrac n{n-9} = \dfrac13 \cdot \dfrac{(n-9)+9}{n-9} = \dfrac13 \cdot \left(1 + \dfrac9{n-9}\right)$

then distribute the limit over the product,

$\displaystyle \lim_{n\to\infty} \left(\frac n{3n-1}\right)^{n-1} = \lim_{n\to\infty}\left(\dfrac13\right)^{n-1} \cdot \lim_{n\to\infty}\left(1+\dfrac9{n-9}\right)^{n-1}$

The first limit is 0, since 1/3ⁿ is a positive, decreasing sequence. But before claiming the overall limit is also 0, we need to show that the second limit is also finite.

For the second limit, recall the definition of the constant, e :

$\displaystyle e = \lim_{n\to\infty} \left(1+\frac1n\right)^n$

To make our limit resemble this one more closely, make a substitution; replace 9/(n - 9) with 1/m, so that

$\dfrac{9}{n-9} = \dfrac1m \implies 9m = n-9 \implies 9m+8 = n-1$

From the relation 9m = n - 9, we see that m also approaches infinity as n approaches infinity. So, the second limit is rewritten as

$\displaystyle\lim_{n\to\infty}\left(1+\dfrac9{n-9}\right)^{n-1} = \lim_{m\to\infty}\left(1+\dfrac1m\right)^{9m+8}$

Now we apply some more properties of multiplication and limits:

$\displaystyle \lim_{m\to\infty}\left(1+\dfrac1m\right)^{9m+8} = \lim_{m\to\infty}\left(1+\dfrac1m\right)^{9m} \cdot \lim_{m\to\infty}\left(1+\dfrac1m\right)^8 \\\\ = \lim_{m\to\infty}\left(\left(1+\dfrac1m\right)^m\right)^9 \cdot \left(\lim_{m\to\infty}\left(1+\dfrac1m\right)\right)^8 \\\\ = \left(\lim_{m\to\infty}\left(1+\dfrac1m\right)^m\right)^9 \cdot \left(\lim_{m\to\infty}\left(1+\dfrac1m\right)\right)^8 \\\\ = e^9 \cdot 1^8 = e^9$

So, the overall limit is indeed 0:

$\displaystyle \lim_{n\to\infty} \left(\frac n{3n-1}\right)^{n-1} = \underbrace{\lim_{n\to\infty}\left(\dfrac13\right)^{n-1}}_0 \cdot \underbrace{\lim_{n\to\infty}\left(1+\dfrac9{n-9}\right)^{n-1}}_{e^9} = \boxed{0}$

7 0

3 years ago

30 pts!!!!!!!!!

boyakko [2]

Divide the first one by the second and the answer will be B.

3 0

3 years ago

Read 2 more answers

Rewrite with only sin x and cos x.

aivan3 [116]

Answer:

cos x (2 sin x − 1)

Step-by-step explanation:

sin(2x) − cos x

Use double angle formula.

2 sin x cos x − cos x

Factor.

cos x (2 sin x − 1)

5 0

3 years ago

I need help plzzzz percent equations blank% of $30 =3

Lyrx [107]

Answer:

10%

Step-by-step explanation:

Hello I can help you with this question. Lets start.

The question is asking what percent of $30 is 3

so i think u actually meant $3 right but anyways

30 is 10 times 3

then clearly it would be 10%

gogo that's the answer!

4 0

3 years ago

A Ferris wheel can accommodate 40 people in 20 minutes. How many people

Montano1993 [528]

Answer:

360 people in 3 hours. 120 people per hour.

Step-by-step explanation:

How many times do 20 minutes go into 3 hours?

Well, 60 min. ÷ 20 min. = 3

So, 3 x 3 = 9

So 40 people went on the ride each 20 minutes. And there are (3) 20 minutes in every hour.

So 3 hours would be 9 x 40 = 360

120 people per hour because 40 x 3 = 120

and 20 minutes (3) times equals 1 hour.

6 0

3 years ago