Answer:
iooooooooooooooooooooo
Explanation:
ooooooooooooooo
oooo9ooooo9oooooooooo
ooooooooooooooppppooooooooooo
oooooooooooo
ooooooooooooooooooooo
Answer:
Screenplay
Financing
Theme
cinematography
editing
sound and music
Acting and Genre
Explanation:
There are various essential elements that needs to be considered to become a famous and a good filmmaker. A good filmmaker will take care of small elements and will consider minor details. The audience is very keen when it takes interest in the movie. They pay attention to even minor details in the movie. The filmmaker should take care of all the details and will have to decide on the main things such as screenplay, audio, editing, genre of the movie and similar. These elements are dependent on each other. If the audio of the movie is not appropriate or sound quality is not suitable then the overall impact of the movie will be slowed no matter how strong the content is. There is high dependency of all the elements and every element should be paid attention.
Answer:
q1 chees
q2 web and digital continant
Explanation:
because haaa haaa haa dhum
Please Help! Unit 6: Lesson 1 - Coding Activity 2
Instructions: Hemachandra numbers (more commonly known as Fibonacci numbers) are found by starting with two numbers then finding the next number by adding the previous two numbers together. The most common starting numbers are 0 and 1 giving the numbers 0, 1, 1, 2, 3, 5...
The main method from this class contains code which is intended to fill an array of length 10 with these Hemachandra numbers, then print the value of the number in the array at the index entered by the user. For example if the user inputs 3 then the program should output 2, while if the user inputs 6 then the program should output 8. Debug this code so it works as intended.
The Code Given:
import java.util.Scanner;
public class U6_L1_Activity_Two{
public static void main(String[] args){
int[h] = new int[10];
0 = h[0];
1 = h[1];
h[2] = h[0] + h[1];
h[3] = h[1] + h[2];
h[4] = h[2] + h[3];
h[5] = h[3] + h[4];
h[6] = h[4] + h[5];
h[7] = h[5] + h[6];
h[8] = h[6] + h[7]
h[9] = h[7] + h[8];
h[10] = h[8] + h[9];
Scanner scan = new Scanner(System.in);
int i = scan.nextInt();
if (i >= 0 && i < 10)
System.out.println(h(i));
}
}
Answer:
#include <iostream>
using namespace std;
int * reverse(int a[],int n)//function to reverse the array.
{
int i;
for(i=0;i<n/2;i++)
{
int temp=a[i];
a[i]=a[n-i-1];
a[n-i-1]=temp;
}
return a;//return pointer to the array.
}
int main() {
int array[50],* arr,N;//declaring three variables.
cin>>N;//taking input of size..
if(N>50||N<0)//if size greater than 50 or less than 0 then terminating the program..
return 0;
for(int i=0;i<N;i++)
{
cin>>array[i];//prompting array elements..
}
arr=reverse(array,N);//function call.
for(int i=0;i<N;i++)
cout<<arr[i]<<endl;//printing reversed array..
cout<<endl;
return 0;
}
Output:-
5
4 5 6 7 8
8
7
6
5
4
Explanation:
I have created a function reverse which reverses the array and returns pointer to an array.I have also considered edge cases where the function terminates if the value of the N(size) is greater than 50 or less than 0.
