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1 year ago

Which of the following best describes a hotspot as

Computers and Technology

1 answer:

Ierofanga [76]1 year ago

5 0

A statement which best describes a hotspot as used in the Action feature of PowerPoint is: B. an invisible hyperlink embedded in a slide.

<h3>What is a hotspot?</h3>

A hotspot can be defined as a visual effect that can be applied in Microsoft PowerPoint to content on a slide, so as to make elements (objects) interactive with the end users.

This ultimately implies that, an invisible hyperlink embedded in a slide is a statement which best describes a hotspot as used in the Action feature of PowerPoint.

Read more on PowerPoint here: brainly.com/question/26404012

#SPJ1

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In your own words, describe the structure and function of both the stack and queue data structure and discuss how they are diffe

Kobotan [32]

Answer:

Following are the answer to this question:

Explanation:

Stack:

The stack data structure is used to create method, parsing, analysis of expression, and tracking. This data type uses the LIFO system, which stands from the Last In First Out. This data structure operates on the top side. In this data structure, it performs two methods, that are "push() and pop()" method and the time complexity of the stack data structure O(1).

Example of Stack:

import java.util.*;//import package

public class Main //defining class Main

{

public static void main (String[] ag)//defining main method

{

ArrayDeque<Character> stack = new ArrayDeque<Character>();//creating Character array ArrayDeque

stack.push('A'); //add value in stack

stack.push('B');//add value in stack

stack.push('B'); //add value in stack

System.out.println("First Insert:"+stack); //print value stack.pop();//remove value from stack

System.out.println("After removing element: "+stack); //print value stack.push('B'); //add value in stack

stack.pop();//remove value from stack

System.out.println("final value: "+stack); //print value

}

Output:

First Insert:[B, B, A]

After removing element: [B, A]

final value: [B, A]

Queue:

The queue data structure is an abstract type, it is similar to the stacks data structure. It is open at both ends when opposed to lines. It follows the FIFO method, which stands for the First-In-First-Out method, At one end data is inserted and at the other end, it deletes the data and the time complexity of the queue data structure O(1).

Example of Queue:

import java.util.*;//import package

public class Main//defining a class queue

{

public static void main(String[] ars)//defining main method

{

LinkedList<Integer> que= new LinkedList<Integer>();//defining integer array LinkedList

for (int i = 0; i < 5; i++)//defining fo loop to add value in queue

{

que.add(i); //use add method to insert value

}

System.out.println("Queue value: "+ que); //print queue value

int remove= que.remove(); //remove value from the queue System.out.println("after removing value from queue: "+ remove);//removing element from queue

System.out.println(que); //after removing the element print value

}

Output:

Queue value: [0, 1, 2, 3, 4]

after removing value from queue: 0

[1, 2, 3, 4]

6 0

3 years ago

Write a C program that reads two hexadecimal values from the keyboard and then stores the two values into two variables of type

sattari [20]

Solution :

#include $$$$

//Converts $\text{hex string}$ to binary string.

$\text{char}$ * hexadecimal $\text{To}$ Binary(char* hexdec)

{

long $\text{int i}$ = 0;

char *string = $(\text{char}^ *) \ \text{malloc}$ (sizeof(char) * 9);

while (hexdec[i]) {

//Simply assign binary string for each hex char.

switch (hexdec[i]) {

$\text{case '0'}:$

strcat(string, "0000");

break;

$\text{case '1'}:$

strcat(string, "0001");

break;

$\text{case '2'}:$

strcat(string, "0010");

break;

$\text{case '3'}:$

strcat(string, "0011");

break;

$\text{case '4'}:$

strcat(string, "0100");

break;

$\text{case '5'}:$

strcat(string, "0101");

break;

$\text{case '6'}:$

strcat(string, "0110");

break;

$\text{case '7'}:$

strcat(string, "0111");

break;

$\text{case '8'}:$

strcat(string, "1000");

break;

$\text{case '9'}:$

strcat(string, "1001");

break;

case 'A':

case 'a':

strcat(string, "1010");

break;

case 'B':

case 'b':

strcat(string, "1011");

break;

case 'C':

case 'c':

strcat(string, "1100");

break;

case 'D':

case 'd':

strcat(string, "1101");

break;

case 'E':

case 'e':

strcat(string, "1110");

break;

case 'F':

case 'f':

strcat(string, "1111");

break;

default:

printf("\nInvalid hexadecimal digit %c",

hexdec[i]);

string="-1" ;

}

i++;

}

return string;

}

int main()

{ //Take 2 strings

char *str1 =hexadecimalToBinary("FA") ;

char *str2 =hexadecimalToBinary("12") ;

//Input 2 numbers p and n.

int p,n;

scanf("%d",&p);

scanf("%d",&n);

//keep j as length of str2

int j=strlen(str2),i;

//Now replace n digits after p of str1

for(i=0;i<n;i++){

str1[p+i]=str2[j-1-i];

}

//Now, i have used c library strtol

long ans = strtol(str1, NULL, 2);

//print result.

printf("%lx",ans);

return 0;

}

4 0

2 years ago

True or Flase<br><br> In C++, the body of a for loop may never run even once.

stiks02 [169]

Answer: True

Explanation: For loop is used in the C++ programming is defined as the statement that defines about the flow control .This loop works under some condition that is considered.

For loop is evaluated to execute for one time if the statement condition is true but there are also chances of no execution at all because of the incorrect condition. So, for loop might not run even once in that condition.Thus , the statement given is true.

3 0

3 years ago

D. Convert binary fraction .10112 to Decimal fraction.

Vaselesa [24]

To convert binary fraction to decimal fraction, we first write the given

.1011 (base 2)

The process of conversion is to break down to its decimal constituent,

.1011 (base 2) = (1*2^-1)+(0*2^-2)+(1*2^-3)+(1*2^-4)
.1011 (base 2) = 0.5 + 0 + 0.125 + 0.0625
.1011 (base 2) = 0.6875 (base 10)

<em>ANSWER: 0.6875 (base 10)</em>

6 0

3 years ago

PLS HURRY!!<br> Look at the image below

Crank

The output will be 10.

The while loop runs until numb is equal to or less than 13.

25 - 5 = 20

20 - 5 = 15

15 - 5 = 10, which is less than 13 so the loop stops and 10 is printed to the screen.

4 0

2 years ago