Domain of (x+8)/(x(x+10))
1 answer:
The domain of this, assuming it's a function, is all real numbers such that x does not equal zero or negative ten.
In interval notation, this is (-∞, -10) U (-10, 0) U (0, <span>∞).
In set builder notation, this is {x | x </span>≠ -10, x <span>≠ 0}.
These inputs are excluded from the domain because they would otherwise allow division by zero. Try plugging in one of the restricted inputs. You'll see that it does not work! On a graph of this </span>function, you will see vertical asymptotes.
In interval notation, this is (-∞, -10) U (-10, 0) U (0, <span>∞).
In set builder notation, this is {x | x </span>≠ -10, x <span>≠ 0}.
These inputs are excluded from the domain because they would otherwise allow division by zero. Try plugging in one of the restricted inputs. You'll see that it does not work! On a graph of this </span>function, you will see vertical asymptotes.
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Answer:
Check the explanation
Explanation:
Linear search in JAVA:-
import java.util.Scanner;
class linearsearch
{
public static void main(String args[])
{
int count, number, item, arr[];
Scanner console = new Scanner(System.in);
System.out.println("Enter numbers:");
number = console.nextInt();
arr = new int[number];
System.out.println("Enter " + number + " ");
for (count = 0; count < number; count++)
arr[count] = console.nextInt();
System.out.println("Enter search value:");
item = console.nextInt();
for (count = 0; count < number; count++)
{
if (arr[count] == item)
{
System.out.println(item+" present at "+(count+1));
break;
}
}
if (count == number)
System.out.println(item + " doesn't found in array.");
}
}
Kindly check the first attached image below for the code output.
Selection Sort in JAVA:-
public class selectionsort {
public static void selectionsort(int[] array){
for (int i = 0; i < array.length - 1; i++)
{
int ind = i;
for (int j = i + 1; j < array.length; j++){
if (array[j] < array[ind]){
ind = j;
}
}
int smaller_number = array[ind];
array[ind] = array[i];
array[i] = smaller_number;
}
}
public static void main(String a[]){
int[] arr = {9,94,4,2,43,18,32,12};
System.out.println("Before Selection Sort");
for(int i:arr){
System.out.print(i+" ");
}
System.out.println();
selectionsort(arr);
System.out.println("After Selection Sort");
for(int i:arr){
System.out.print(i+" ");
}
}
}
Kindly check the second attached image below for the code output.
Bubble Sort in JAVA:-
public class bubblesort {
static void bubblesort(int[] array) {
int num = array.length;
int temp = 0;
for(int i=0; i < num; i++){
for(int j=1; j < (num-i); j++){
if(array[j-1] > array[j]){
temp = array[j-1];
array[j-1] = array[j];
array[j] = temp;
}
}
}
}
public static void main(String[] args) {
int arr1[] ={3333,60,25,32,55,620,85};
System.out.println("Before Bubble Sort");
for(int i=0; i < arr1.length; i++){
System.out.print(arr1[i] + " ");
}
System.out.println();
bubblesort(arr1);
System.out.println("After Bubble Sort");
for(int i=0; i < arr1.length; i++){
System.out.print(arr1[i] + " ");
}
}
}
Kindly check the third attached image below for the code output.
Binary search in JAVA:-
public class binarysearch {
public int binarySearch(int[] array, int x) {
return binarySearch(array, x, 0, array.length - 1);
}
private int binarySearch(int[ ] arr, int x,
int lw, int hg) {
if (lw > hg) return -1;
int middle = (lw + hg)/2;
if (arr[middle] == x) return middle;
else if (arr[middle] < x)
return binarySearch(arr, x, middle+1, hg);
else
return binarySearch(arr, x, lw, middle-1);
}
public static void main(String[] args) {
binarysearch obj = new binarysearch();
int[] ar =
{ 22, 18,12,14,36,59,74,98,41,23,
34,50,45,49,31,53,74,56,57,80,
61,68,37,12,58,79,904,56,99};
for (int i = 0; i < ar.length; i++)
System.out.print(obj.binarySearch(ar,
ar[i]) + " ");
System.out.println();
System.out.print(obj.binarySearch(ar,19) +" ");
System.out.print(obj.binarySearch(ar,25)+" ");
System.out.print(obj.binarySearch(ar,82)+" ");
System.out.print(obj.binarySearch(ar,19)+" ");
System.out.println();
}
}
Kindly check the fourth attached image below for the code output
