hw3_solution

# hw3_solution - EME5 Homework 3 Solutions 3 3.1 Chapter 4...

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EME5 Homework 3 Solutions October 15, 2010 3 Chapter 4, Problem 1 3.1 Part a False. Some operators are operated from left to right and others are evaluated from right to left depending on their associativity. 4 Chapter 4, Problem 2 4.1 Part e a = (b >> (2 + 4)); 4.2 Part g (a == b) || (x != y) 1

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5 Chapter 4, Problem 4 5.1 Solution Code #include <stdio.h> int main() { float f1 = 0.2, f2 = 0.5; printf("(f1 * f1 == 0.1) = %d\n", f1*f1 == 0.1); printf("(f1 + f1 == 0.07) = %d\n", f1+f1 == 0.07); return 0; } 5.2 Solution Output (f1*f1 == 0.1) = 0 (f1+f1 == 0.07) = 0 6 Chapter 4, Problem 11 6.1 Part a 4 3 πr 3 (1) 6.2 Part b 4 3 πr 3 (2) 6.3 Part c 9 5 c + 32 (3) 2
6.4 Part d 9 5 c + 32 (4) 7 Chapter 4, Problem 12 7.1 Part c (4/3.0)*3.14*r*r*r 7.2 Part f (x >= y) && (x >= 0) 7.3 Part i ((1 < x) && (x <= 20)) || ((1 <= y) && (y <= 3)) 8 Chapter 4, Problem 16 8.1 Part b 8.1.1 Solution Code #include <stdio.h> #include <math.h> int main() { /* declare and initialize variables */ double x1 = 4, y1 = 3, x2 = 6, y2 = 5; double sidex, sidey, distance; /* calculate the distances of two sides in X and Y direction */ sidex = x2-x1; sidey = y2-y1; /* calculate the straight-line distance */ distance = sqrt(sidex*sidex + sidey*sidey);

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