Answer:
Marisol must make use of the Shopping campaign for showing the image of her products, as well as advertise her company's online as well as the local inventory of the toys, as well as boost the traffic to their site and the local toy stores. These types of ads are displayed on Google Shopping which comes just after the search results and near the responsive and text ads.
Explanation:
Please check the answer section.
Window is a vary popular operating system because of its runtime and compatible environment .
hope it help
Answer:
The answer to this question is given below in the explanation section.
Explanation:
The question is about writing a C program that prints the initial letter of Name Ferdous.
Therefore, below is the given complete code that prints the first letter of the name Ferdous.
<em>#include <stdio.h> </em><em>/*import strandard input/output library*/</em>
<em>#include<string.h></em><em> /*import string library to handle string type of data*/</em>
<em> </em>
<em>int main(void) </em><em> /*started the program execution- program entry point*/</em>
<em>{ </em>
<em>char *str = "Firdous";</em><em> /*char to pointer str contains string "Firdous"*/</em>
<em>int len = strlen(str); </em><em> /*this line of code is not neccary, but if you print other character for example last character of the name then you can use it*/</em>
<em>printf("First Letter of the name is: %c", str[0]); </em><em> /*print first letter of the name*/</em>
<em>} </em><em> /**program terminated*/</em>
Answer:
- public class Main {
-
- public static void main (String [] args) {
-
- for(int i = 2; i < 10000; i++){
- if(isPrime1(i)){
- System.out.print(i + " ");
- }
- }
-
- System.out.println();
-
- for(int i = 2; i < 10000; i++){
- if(isPrime2(i)){
- System.out.print(i + " ");
- }
- }
- }
-
- public static boolean isPrime1(int n){
-
- for(int i=2; i <= n/2; i++){
- if(n % i == 0){
- return false;
- }
- }
-
- return true;
- }
-
- public static boolean isPrime2(int n){
-
- for(int i=2; i <= Math.sqrt(n); i++){
- if(n % i == 0){
- return false;
- }
- }
-
- return true;
- }
- }
<u></u>
Explanation:
Firstly, create the first version of method to identify a prime number, isPrime1. This version set the limit of the for loop as n/2. The for loop will iterate through the number from 2 till input n / 2 and check if n is divisible by current value of i. If so, return false to show this is not a prime number (Line 22 - 26). Otherwise it return true to indicate this is a prime number.
In the main program, we call the isPrime1 method by passing the i-index value as an argument within a for-loop that will iterate through the number 2 - 10000 (exclusive). If the method return true, print the current i value). (Line 5 - 9)
The most direct way to ensure all the prime numbers below 10000 are found, is to check the prime status from number 2 - 9999 which is amount to 9998 of numbers.
Next we create a second version of method to check prime, isPrime2 (Line 31 - 40). This version differs from the first version by only changing the for loop condition to i <= square root of n (Line 33). In the main program, we create another for loop and repeatedly call the second version of method (Line 13 - 17). We also get the same output as in the previous version.
You can return it for a refund, however with that money you will need to pay the difference, as they are different models and the 64gb will cost more.
