Ask question

Ask question

All categories

2 years ago

10

Suppose you are doing a key exchange with John using generator 3 and prime 29. Your secret number is 4. John sends you the value

8. Determine the shared secret key

1 answer:

jolli1 [7]2 years ago

5 0

Answer:

sorry

Step-by-step explanation:

You might be interested in

A radio disc jockey has 7 songs on this up coming hour's playlist: 2 are rock songs, 2 are reggae songs and 3 are country songs.

SVETLANKA909090 [29]

Answer:

Probability that song selected after first selection is rock = 2/7

Probability that song selected from the remaining after second selection is country rock = 1/2

Step-by-step explanation:

Total number of songs = 7

After first selection, probability that the song selected is rock = 2/7

After second selection, probability that the song selected from the remaining is a country song = 3/6 = 1/2

8 0

3 years ago

7ft = __ inches. Show work

Harrizon [31]

Answer:

Step-by-step explanation:

12 inches is in a foot

10 x 7= 70

2 x 7= 14

70+14=84

6 0

2 years ago

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

2 years ago

A skyscraper has 49 floors there are 26 offices on each of the bottom 31 fours the top 18 floors each have 38 officers about how

Anettt [7]

Answer:

903 offices

Step-by-step explanation:

Total number of officers in the sky scraper

= 26 + 31 + [(49-2)×18]

= 26 + 31 + 846

= 903 offices

3 0

2 years ago

Please help I’m timed

kupik [55]

It’s the second one :)

3 0

2 years ago

Read 2 more answers