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3 years ago

Aspiring graphic designers can earn a(n) certification for graphic design software, such as Photoshop and Acrobat.

Computers and Technology

2 answers:

kobusy [5.1K]3 years ago

5 0

Answer is:

Yes, they earn a certification as a graphic designer

Explanation:

And they can do everything that graphic designer do, so that is why they earn a certification of a graphic designer such as Photoshop and adobe illustrator.

I hope you got the answer well. Thanks

HACTEHA [7]3 years ago

5 0

Answer:

Yes, they can earn a certification. For example, I actually got one.

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Explain briefly the use of the computers in the advertising area

olya-2409 [2.1K]

Answer:

Brainly.con

Explanation:

5 0

2 years ago

The Polish mathematician Wacław Sierpiński described the pattern in 1915, but it has appeared in Italian art since the 13th cent

soldi70 [24.7K]

Answer:

/ Sierpinski.java

public class Sierpinski {

// method to find the height of an equilateral triangle with side length =

// length

public static double height(double length) {

// formula= length*sqrt(3)/2

double h = (length * Math.sqrt(3)) / 2.0;

return h;

}

// method to draw a filled triangle (pointed downwards) with bottom vertex

// x,y

public static void filledTriangle(double x, double y, double length) {

// finding height

double h = height(length);

// filling triangle as a polygon

// passing x, x-length/2, x+length/2 as x coordinates

// and y, y+h, y+h as y coordinates

StdDraw.filledPolygon(new double[] { x, x - (length / 2.0),

x + (length / 2.0) }, new double[] { y, y + h, y + h });

}

// method to draw an n level sierpinski triangle

public static void sierpinski(int n, double x, double y, double length) {

// checking if n is above 0 (base condition)

if (n > 0) {

// drawing a filled triangle with x, y as bottom coordinate, length

// as side length

filledTriangle(x, y, length);

// drawing the triangle(s) on the top

sierpinski(n - 1, x, y + height(length), length / 2);

// drawing the triangle(s) on the left side

sierpinski(n - 1, x - (length / 2.0), y, length / 2);

// drawing the triangle(s) on the right side

sierpinski(n - 1, x + (length / 2.0), y, length / 2);

}

public static void main(String[] args) {

// parsing first command line argument as integer, if you dont provide

// the value while running the program, this program will cause

// exception.

int n = Integer.parseInt(args[0]);

// length of the outline triangle

double length = 1;

// finding height

double h = height(length);

// drawing a triangle (pointed upwards) to represent the outline.

StdDraw.polygon(new double[] { 0, length / 2, length }, new double[] {

0, h, 0 });

// drawing n level sierpinski triangle(s) with length / 2, 0 as x,y

// coordinates and length / 2 as initial side length

sierpinski(n, length / 2, 0, length / 2);

}

7 0

2 years ago

Viruses and Malware Complete the case project on page 82 of the textbook by researching two types of viruses, two types of malwa

Naddika [18.5K]

Hi, I provided you a general guide on how to go about your writing your research paper.

Explanation:

Note, the current APA (American Psychological Association) format is the APA 7th Edition. This format details how a researcher should state the findings, sources, conclusion, etc.

Hence, you can begin writing the assigned paper only after gathering the needed data.

6 0

3 years ago

What section in an ethernet frame will you find a Virtual Local Area Network (VLAN) header?

Vadim26 [7]

Answer:

Preamble

Explanation:

In the computer network , the ether-net is the frame of link layer protocol data. This frame is used ether net with physical layer of transport mechanism. The ether net frame is of different type.

The ether-net II
The Novel raw IEEE 802.3
IEEE 802.2 LLC
IEEE 802.2 SNAP

Each of the Ethernet frame started from Ether net header. It contains the source and the destination. The MAC address is called its first two address.

8 0

3 years ago

a cryptarithm is a mathematical puzzle where the goal is to find the correspondence between letters and digits such that the giv

Leokris [45]

Using the knowledge in computational language in C++ it is possible to write a code that cryptarithm is a mathematical puzzle where the goal is to find the correspondence between letters and digits

<h3>Writting the code:</h3>

#include <bits/stdc++.h>

using namespace std;

// chracter to digit mapping, and the inverse

// (if you want better performance: use array instead of unordered_map)

unordered_map<char, int> c2i;

unordered_map<int, char> i2c;

int ans = 0;

// limit: length of result

int limit = 0;

// digit: index of digit in a word, widx: index of a word in word list, sum: summation of all word[digit]

bool helper(vector<string>& words, string& result, int digit, int widx, int sum) {

if (digit == limit) {

ans += (sum == 0);

return sum == 0;

}

// if summation at digit position complete, validate it with result[digit].

if (widx == words.size()) {

if (c2i.count(result[digit]) == 0 && i2c.count(sum%10) == 0) {

if (sum%10 == 0 && digit+1 == limit) // Avoid leading zero in result

return false;

c2i[result[digit]] = sum % 10;

i2c[sum%10] = result[digit];

bool tmp = helper(words, result, digit+1, 0, sum/10);

c2i.erase(result[digit]);

i2c.erase(sum%10);

ans += tmp;

return tmp;

} else if (c2i.count(result[digit]) && c2i[result[digit]] == sum % 10){

if (digit + 1 == limit && 0 == c2i[result[digit]]) {

return false;

}

return helper(words, result, digit+1, 0, sum/10);

} else {

return false;

}

// if word[widx] length less than digit, ignore and go to next word

if (digit >= words[widx].length()) {

return helper(words, result, digit, widx+1, sum);

}

// if word[widx][digit] already mapped to a value

if (c2i.count(words[widx][digit])) {

if (digit+1 == words[widx].length() && words[widx].length() > 1 && c2i[words[widx][digit]] == 0)

return false;

return helper(words, result, digit, widx+1, sum+c2i[words[widx][digit]]);

}

// if word[widx][digit] not mapped to a value yet

for (int i = 0; i < 10; i++) {

if (digit+1 == words[widx].length() && i == 0 && words[widx].length() > 1) continue;

if (i2c.count(i)) continue;

c2i[words[widx][digit]] = i;

i2c[i] = words[widx][digit];

bool tmp = helper(words, result, digit, widx+1, sum+i);

c2i.erase(words[widx][digit]);

i2c.erase(i);

}

return false;

}

void isSolvable(vector<string>& words, string result) {

limit = result.length();

for (auto &w: words)

if (w.length() > limit)

return;

for (auto&w:words)

reverse(w.begin(), w.end());

reverse(result.begin(), result.end());

int aa = helper(words, result, 0, 0, 0);

}

int main()

{

ans = 0;

vector<string> words={"GREEN" , "BLUE"} ;

string result = "BLACK";

isSolvable(words, result);

cout << ans << "\n";

return 0;

}

See more about C++ code at brainly.com/question/19705654

#SPJ1

3 0

1 year ago