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2 years ago

9

Encrypt the word TO using an alphabetic Caesar shift cipher with shift 10 (mapping A to K). Use only capital letters in your ans

wer.

1 answer:

Lady bird [3.3K]2 years ago

4 0

Answer:

DY

Step-by-step explanation:

TO ENCRYPT TO USING CAESAR ALPHABETIC SHIFT CIPHER OF 10. THIS MEANS WE'RE EXPECTED TO LOOP EVERY ALPHABET BY 10, SKIPPING EVERYONE TILL WE GET TO THE TENTH.

TO ENCRYPT T, WE HAVE

T = U - V - W - X - Y - Z - A - B - C - D

THEREFORE, T WILL BE ENCRYPTED AS D

TO ENCRYPT O, WE HAVE

O = P - Q - R - S - T - U - V - W - X - Y

HENCE, O WILL BE ENCRYPTED AS Y.

THIS MEANS THAT OUR QUESTION, TO, WILL BE ENCRYPTED AS DY

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If a sentence is defined recursively by f(0)=3 and f(n+1) = f(n) +5 for n ? 0, then f(2) is A) 2 B) 3 C) 5 D) 8

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The recursive formula $f(n+1)=f(n)+5$ means that the next term in the sequence is 5 more than the previous one.

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2 years ago

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The given matrix is the augmented matrix for a linear system. Use technology to perform the row operations needed to transform t

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Answer:

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Step-by-step explanation:

As the given Augmented matrix is

$\left[\begin{array}{ccccc}9&-2&0&-4&:8\\0&7&-1&-1&:9\\8&12&-6&5&:-2\end{array}\right]$

Step 1 :

$r_{1}$ ↔ $r_{1} - r_{2}$

$\left[\begin{array}{ccccc}1&-14&6&-9&:10\\0&7&-1&-1&:9\\8&12&-6&5&:-2\end{array}\right]$

Step 2 :

$r_{3}$ ↔ $r_{3} - 8r_{1}$

$\left[\begin{array}{ccccc}1&-14&6&-9&:10\\0&7&-1&-1&:9\\0&124&-54&77&:-82\end{array}\right]$

Step 3 :

$r_{2}$ ↔ $\frac{r_{2}}{7}$

$\left[\begin{array}{ccccc}1&-14&6&-9&:10\\0&1&-\frac{1}{7} &-\frac{1}{7} &:\frac{9}{7} \\0&124&-54&77&:-82\end{array}\right]$

Step 4 :

$r_{1}$ ↔ $r_{1} + 14r_{2}$ , $r_{3}$ ↔ $r_{3} - 124r_{2}$

$\left[\begin{array}{ccccc}1&0&4&-11&:-8\\0&1&-\frac{1}{7} &-\frac{1}{7} &:\frac{9}{7} \\0&0&- \frac{254}{7} &\frac{663}{7} &:-\frac{1690}{7} \end{array}\right]$

Step 5 :

$r_{3}$ ↔ $\frac{r_{3}. 7}{254}$

$\left[\begin{array}{ccccc}1&0&4&-11&:-8\\0&1&-\frac{1}{7} &-\frac{1}{7} &:\frac{9}{7} \\0&0&1&-\frac{663}{254} &:-\frac{1690}{254} \end{array}\right]$

Step 6 :

$r_{1}$ ↔ $r_{1} - 4r_{3}$ , $r_{2}$ ↔ $r_{2} + \frac{1}{7} r_{3}$

$\left[\begin{array}{ccccc}1&0&0&-\frac{71}{127} &:\frac{176}{127} \\0&1&0&-\frac{131}{254} &:\frac{284}{127} \\0&0&1&-\frac{663}{254} &:\frac{845}{127} \end{array}\right]$

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Step-by-step explanation:

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