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3 years ago

What is the correct way to write three hundred nine million, fifty-eight thousand, three hundred four?

Mathematics

2 answers:

cestrela7 [59]3 years ago

7 0

Either 309,058,304
or three hundred million, fifty-eight thousand, three hundred four.

attashe74 [19]3 years ago

5 0

309,058,304 Hope this helped! :)

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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

The teacher never taught this please help it’s for a grade

34kurt

Answer: D 2/3

Step-by-step explanation:

8 0

3 years ago

0.5, 3/16, 5/48 in order from least to greatest

sukhopar [10]

1# 5/48
2# 0.5
3# 3/16

6 0

3 years ago

How is the total price including sales tax calculated?

klasskru [66]

Answer:

Sales tax is calculated by multiplying the cost of a good or service by the appropriate sales tax rate The answer is B (purchase price) (percent sales tax); then d. (purchase price) (percent sales tax). Hope it helps!:)

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3 years ago

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eremy loves movies. He likes to watch movies and someday he hopes to make one of his own. He likes to ask questions about movies

IrinaVladis [17]

One movie-related statistical question would be the length of people's favorite movies (in minutes). This would contain variability since people would have different favorite movies, which would have different lengths.
One movie-related question that does not yield variability is the number of movies released in 2018 that grossed more than $100 million. This would be a fixed fact, so there is no variability about this.

8 0

3 years ago