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ladessa [460]
3 years ago
7

What will be the output after the following code is executed? def pass_it(x, y): z = y**x return(z) num1 = 3 num2 = 4 answer = p

ass_it(num1, num2) print(answer)
Computers and Technology
1 answer:
bearhunter [10]3 years ago
3 0

Answer:

"64" is the output of the above code.

Explanation:

  • The function holds the expression with (**) operator which is used to calculate the power in python language.
  • So if we write 2**4, then this will give 16 as output. It is because when we calculate the 2^{4}, then it will become 16.
  • The above question passes the num1 and num2 value in the function as an argument which is 3 and 4.
  • So 3 will be initialized for x variable and 4 is for y variable.
  • So when we calculate y^{x} then it will become 4^{3} which will become 16.
  • So 64 is the output for the above question.
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Answer:

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Explanation:

8 0
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You should process the tokens by taking the first letter of every fifth word,starting with the first word in the file. Convert t
zhenek [66]

Answer:

See explaination

Explanation:

import java.io.File;

import java.io.IOException;

import java.util.Scanner;

import java.util.StringTokenizer;

public class SecretMessage {

public static void main(String[] args)throws IOException

{

File file = new File("secret.txt");

StringBuilder stringBuilder = new StringBuilder();

String str; char ch; int numberOfTokens = 1; // Changed the count to 1 as we already consider first workd as 1

if(file.exists())

{

Scanner inFile = new Scanner(file);

StringTokenizer line = new StringTokenizer(inFile.nextLine()); // Since the secret.txt file has only one line we dont need to loop through the file

ch = line.nextToken().toUpperCase().charAt(0); // Storing the first character of first word to string builder as mentioned in problem

stringBuilder = stringBuilder.append(ch);

while(line.hasMoreTokens()) { // Looping through each token of line read using Scanner.

str= line.nextToken();

numberOfTokens += 1; // Incrementing the numberOfTokens by one.

if(numberOfTokens == 5) { // Checking if it is the fifth word

ch = str.toUpperCase().charAt(0);

stringBuilder = stringBuilder.append(ch);

numberOfTokens =0;

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System.out.println("----Secret Message----"+ stringBuilder);

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5 0
3 years ago
9. In Electronic Codebook (ECB): a. The message is broken into independent blocks which are encrypted b. The message is broken i
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Answer:

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Explanation:

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saul85 [17]
Capitalized  and properly spelled 

8 0
4 years ago
A DBMS commonly receives data update requests from application programs.A. TrueB. False
Contact [7]

Answer:

A. True

Explanation:

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