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s2008m [1.1K]
3 years ago
11

Write an integer that represents a bank withdrawl of $50.

Mathematics
1 answer:
katen-ka-za [31]3 years ago
4 0
You're taking money away so I'm pretty sure that the answer is -50.
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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
Mike withdrew $32 from his bank account. The transaction slip said his balance was $289.14. What was his previous balance?
Temka [501]
<span>His previous balance was $321.14
289.14 + 32 = 321.14

hope this helps
</span>
4 0
3 years ago
If the sum of four consecutive numbers is 854, what is the third one?
lukranit [14]

Answer:

214

Step-by-step explanation:

The set up would be x + (x + 1) + (x + 2) + (x + 3) = 854, naturally as the numbers are consecutive. Solving for x:

x + (x + 1) + (x + 2) + (x + 3) = 854,

x + x + 1 + x + 2 + x + 3 = 854,

4x + 1 + 2 + 3 = 854,

4x + 6 = 854,

4x = 854 - 6 = 848,

x = 848/4 = 212

The third number then should be 212 + 2 = 214

7 0
3 years ago
  (5 pt) Enter a number in the box to make this a true number sentence.  5 × (6 – 4) = (5 × 6) – (5 × )
Sloan [31]
5x+1(6-4) = +1(5x6)-(5x)
this would be a hard question to answer but i think i can do it ((( nNOTTTT))))
5 0
3 years ago
What’s 80 percent in fractions
Greeley [361]
The answer to that is 80 percent as a fraction is 8/10
4 0
3 years ago
Read 2 more answers
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