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2 years ago

What happens as the key length increases in an encryption application?

1 answer:

6 0

The answer is : Increasing the key length of DES would protect it against brute force attacks. Brute force is when the attacker tries every key knowing that one will eventually work. Key length increase proportionally increases the key space, having a keyspace large enough that it takes too much time and money to accomplish a brute force attack.

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Any Body have a zoom class goin on i wanna bomb

Norma-Jean [14]

This site is only for education related questions. You should only answer or ask questions. That's what we are here for. Whoever is reading this answer, please spread it to stop people from zooming, hurting others, or just posting non-related to education questions. Thank you.

6 0

2 years ago

The National Vulnerability Database (NVD) is responsible for actively performing vulnerability testing for every company's softw

V125BC [204]

Answer:

False

Explanation:

The answer to this question is false. This is because the NVD doesn't perform such tests on their own. Instead they they rely on third-party vendors, software researchers, etc to get such reports and do the assignment of CVSS scores for softwares

The National Vulnerability Database (NVD) is the United State governments leading resource for software vulnerability

8 0

2 years ago

Someone pls help me do this and don't ask for brainliest or say random answers not related to this.

Marat540 [252]

I used to use pixlr all the time so i should know.
Ok so

1 is picker
2 is lasso select
3 is crop
4 is Liquefy
5 is clone
6 is dodge/burn
7 is Pen
8 is Eraser
9 is Gradient
10 is Text
11 is Zoom
12 is Switch Colors
13 is Marquee Select
14 is Wand select
15 is Cutout/Mask
16 is Heal
17 is Blur/Sharpen
18 is Sponge Color
19 is draw
20 is Fill
21 is I don’t know I have never seen that before
22 is Picker

So sorry if it took a while i did not know what the star was

3 0

2 years ago

Read 2 more answers

Using Module operator, write a java program to print odd number between 0 and 1000

jolli1 [7]

Answer:

class OddNumber

{

public static void main(String args[])

{

int n = 1000; //Store 1000 in Variable n typed integer

System.out.print("Odd Numbers from 1 to 1000 are:"); // Print headline of output window

for (int i = 1; i <= n; i++) //For loop to go through each number till end

{

if (i % 2 != 0) //check if number is even or odd.Not divisible by 2 without reminder means it is odd number

{

System.out.print(i + " "); //print odd numbers

}

5 0

2 years ago

Illustrate that the system is in a safe state by demonstrating an order in which the threads may complete.If a request from thre

Kamila [148]

Answer:

a. safe sequence is T2 , T3, T0, T1, T4.

b. As request(T4) = Available, so the request can be granted immediately

c. As request(T2) < Available, so the request can be granted immediately

d. As request(T3) < Available, so the request can be granted immediately.

Explanation:

It will require matrix

[i, j] = Max [i, j] – Allocation [i, j]

A B C D

T0 3 3 3 2

T1 2 1 3 0

T2 0 1 2 0

T3 2 2 2 2

T4 3 4 5 4

Available = (2 2 2 4)

1. Need(T2) < Available so, T2 can take all resources

Available = (2 2 2 4) + (2 4 1 3) (Allocation of T2) = (4 6 3 7)

2. Need(T3)<Available so, T3 will go next

Available = (4 6 3 7) + (4 1 1 0) = (8 7 4 7)

Like wise next T0, T1, T4 will get resources.

So safe sequence is T2 , T3, T0, T1, T4.

(Note, there may be more than one safe sequence).

Solution b.

Request from T4 is (2 2 2 4) and Available is (2 2 2 4)

As request(T4) = Available, so the request can be granted immediately.

Solution c.

Request from T2 is (0 1 1 0) and Available is (2 2 2 4)

As request(T2) < Available, so the request can be granted immediately.

Solution d.

Request from T3 is (2 2 1 2) and Available is (2 2 2 4)

As request(T3) < Available, so the request can be granted immediately.

5 0

2 years ago