I need solutions for these problems, I prefer to get the answers for some proble
ID: 3886670 • Letter: I
Question
I need solutions for these problems, I prefer to get the answers for some problems using printf and scanf not cout and cin.
1. Write a program to generate cartesian co-ordinates (z,y), if the polar co-ordinates (r. ) of a point are provided. The transformation holds the following relationship: x = r cos and y-r sin . C's default angles are in radians. The conversion from degrees to radians is given by: degrees radians10 P (x, y) FIG. 1: Question 1. 2. What is the value ofi calculated by each of the following statements? (no program is necessary) (a) int i, j-3, k-6; i-j*2/3+k/4+6-j*j*j/8 (b) int i, j-3; i-j/2*4+3/8+j *j *3%4 3. Evaluate the following expressions if: float a-2.5, b-2.5; (no program is necessary) (a) a+2.5/b+4.5 (b) (a+2.5)/b+4.5 (d) a/2.5/bExplanation / Answer
2) a) int i,j=3,k=6;
i=j*2/3+k/4+6-j*j*j/8;
Answer: Value of i = 6
Explanation:
Program:
#include <stdio.h>
int main(void) {
int i,j=3,k=6;
i=j*2/3+k/4+6-j*j*j/8;
printf("Value of i = %d",i);
return 0;
}
Output:
Value of i = 6
b) int i,j=3;
i=j/2*4+3/8+j*j*j%4;
Answer: Value of i = 7
Explanation:
Program:
#include <stdio.h>
int main(void) {
int i,j=3;
i=j/2*4+3/8+j*j*j%4;
printf("Value of i = %d",i);
return 0;
}
Output:
Value of i = 7
3) a) Given float a=2.5,b=2.5;
a) a+2.5/b+4.5;
Answer: 8.0000
Explanation:
Program:
#include <stdio.h>
int main(void) {
float a=2.5,b=2.5,c;
c=a+2.5/b+4.5;
printf("Value of c = %f",c);
return 0;
}
Output:
Value of c = 8.000000
3) b)(a+2.5)/b+4.5;
Answer: 6.500000
Explanation:
Program:
#include <stdio.h>
int main(void) {
float a=2.5,b=2.5,c;
c=(a+2.5)/b+4.5;
printf("Value of c = %f",c);
return 0;
}
Output:
Value of c = 6.500000
3) c)(a+2.5)/(b+4.5);
Answer: 0.714286
Explanation:
Program:
#include <stdio.h>
int main(void) {
float a=2.5,b=2.5,c;
c=(a+2.5)/(b+4.5);
printf("Value of c = %f",c);
return 0;
}
Output:
Value of c = 0.714286
3) d)a/2.5/b;
Answer: 0.400000
Explanation:
Program:
#include <stdio.h>
int main(void) {
float a=2.5,b=2.5,c;
c=a/2.5/b;
printf("Value of c = %f",c);
return 0;
}
Output:
Value of c = 0.400000
3)
e)b++/a+b--;
Answer: 4.500000
Explanation:
Program:
#include <stdio.h>
int main(void) {
float a=2.5,b=2.5,c;
c=b++/a+b--;
printf("Value of c = %f",c);
return 0;
}
Output:
Value of c = 4.500000
4) float b;
int i;
b=2.54;
b=(b+0.05)*10;
i=b;
b=i;
b=b/10.0;
Answer: Value of b = 2.500000
Explanation:
Program:
#include <stdio.h>
int main(void) {
float b;
int i;
b=2.54; // Here we Replacing value of b 2.56 by 2.54
b=(b+0.05)*10;
i=b;
b=i;
b=b/10.0;
printf("Value of b = %f",b);
return 0;
}
Output:
Value of b = 2.500000