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Alex_Xolod [135]
2 years ago
7

Help me im stuck answer a b c and d please I don't know if they right

Mathematics
2 answers:
TEA [102]2 years ago
6 0

Answer:

It’s b

Step-by-step explanation:

dedylja [7]2 years ago
4 0

Answer:

Step-by-step explanation:

1   -   B

2  -  C

3  -  A

4  -  D

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Lim (n/3n-1)^(n-1)<br> n<br> →<br> ∞
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, <em>e</em> :

\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/(<em>n</em> - 9) with 1/<em>m</em>, so that

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

From the relation 9<em>m</em> = <em>n</em> - 9, we see that <em>m</em> also approaches infinity as <em>n</em> 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
9. Find (f.g)(a) if f(a) = . and g(a) = 2². Fully simplify your answer.
Aleks04 [339]

D is your answer ....

8 0
3 years ago
Order these decimals from smallest to largest 1.89 -0.81 -2.6 0.24 -1.7
tiny-mole [99]

Answer:

-2.6, -1.7, -0.81, 0.24, 1.89

Step-by-step explanation:

5 0
2 years ago
Read 2 more answers
Which expression shows that the quotient
Harrizon [31]

Answer:

Option (2)

Step-by-step explanation:

Given expression is \frac{2}{(3x-1)} ÷ \frac{6}{6x-1}

We further simplify this expression,

\frac{2}{(3x-1)} ÷ \frac{6}{6x-1}

= \frac{2}{(3x-1)}\times \frac{6x-1}{6}

= \frac{6x-1}{3(3x-1)}

= \frac{6x-1}{(9x-3)}

Therefore, \frac{6x-1}{(9x-3)} will be the quotient of the given expression.

x\neq \frac{1}{3} and x\neq \frac{1}{6} for which the given expression is not defined.

Option (2) will be the answer.

6 0
3 years ago
. 6.8 • 102<br> Help me c:
AnnZ [28]

Answer:

693.6

Step-by-step explanation:

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3 years ago
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