a(rl - 1) SUm in C++ code is sum= (a * (pow(r,n)-1))/(r-1); O True O False
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- function iterFunc(x) {return x^3 + x - 1 }oldEst = 0 tol = 0.00005 error = 1 while (error > tol) newEst = iterFunc(oldEst) print(newEst)error = abs(newEst - oldEst) oldEst = newEst } Edit this psuedocode to make it work in python, the output should be 0.6823 as this is a method of root finding!The following code segment is taken (modified) from the open source Barcode system. int Barcode_128b_verify(unsigned char *text) { // if character string length = 0, error if (text[0] == ‘\0’) return -1; // check the last element in the character string is \0 while (*text && *text>=32 && !(*text&0x80)) { text++; } if (*text != 0) { return_code = -1; } else { return_code = 0; } return return_code; } i) Draw a flowgraph for the code segment. ii) Calculate the Cyclomatic complexity metric for this code segment, showing how you calculated the value.FIX ME: In C language Update this function to display the string of 16 bits using the following pattern: 11111 111111 11111 (5) (6) (5) */ void printbits(unsigned short wrd) { int testbit(unsigned short wrd, int bit_to_test); int i; for (i = 15; i >= 0; i--) { printf("%1d", testbit(wrd, i)); if (!(i % 4)) printf(" "); } printf("\n"); }
- Computer Science Implement the Histogram Equalization according to the explanation in Wikipedia in C# Starting code: public static bool ConvertToGray(Bitmap b) { return true; }Write a O(n) time complexity C-function void maxmin(int data[], int n, int *min, int *max); finds and returns max/largest and min/smallest element.IN THE C PROGRAMMING LANGUAGE j = 15; k = 9 // what are the values for n,m,and p after these operations: --j ; k++ ; n = j - ++k; m = j-- + k--; p = k + j;
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- The factorial of a nonnegative integer n is written n! (pronounced as n factorial) and is defined as follows: n! = n * (n-1) * (n-2)….*1 (for values of n greater than or equal to 1) and n! = 1 (for n=0). Write a C++ program that reads nonnegative integer and compute and prints its factorial.Using C Language, trace the following: (c)#include<stdio.h>#include<conio.h>void kar1(char *c, char b, char *a) {*a = 'c'; b = 'a'; *c = 'b';printf("%c %c %c\n", *a, b, *c);}void kar2(char *b, char *a, char *c) {*a = 'b'; *b='c'; *c ='a';printf("%c %c %c\n", *a, *b, *c);}main() {char a = 'a', b = 'b', c = 'c';clrscr();printf("%c %c %c\n", a, b, c);kar1(&a,b,&c);printf("%c %c %c\n", a, b, c);kar2(&a,&b,&c);printf("%c %c %c\n", a, b, c);kar1(&c,b,&a);printf("%c %c %c\n", a, b, c);78kar2(&c,&a,&b);printf("%c %c %c\n", a, b, c);getch();return 0;}Using C language, trace this: void trace(int x, int *y, int z) {x = 1; *y=2;z=4;printf("%2d %2d %2d\n", x, *y, z);}main() {int x=1, y=3,z=4;clrscr();printf("%2d %2d %2d\n",x,y,z);trace(y,&x,z);printf("%2d %2d %2d\n",x,y,z);trace(x,&z,y);printf("%2d %2d %2d\n",x,y,z);trace(z,&y,x);printf("%2d %2d %2d\n",x,y,z);getch();return 0; }(c)#include<s