What is the output of the following code? syms y; dsolve('y*diff(y)+5=x*exp(3)','y')
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- To correct the following code syms y(t); dsolve(diff(y,t)=2*t); * there is no error in the code = should be replaced by == None of the choices Replacing diff(y,t) by diff(y) will solve the issueFind the error in the following codeQ. modify the Code Using set and test or swap? #include <stdio.h>#include <stdlib.h>#include <pthread.h> long long sum=0,num=2000000; pthread_mutex_t mutex = PTHREAD_MUTEX_INITIALIZER; void *mySumFun(void *vargp){int step=*(int *) vargp; for(int i=0;i<num;i++){//pthread_mutex_lock(&mutex);sum+=step;// pthread_mutex_unlock(&mutex);}return NULL;}int main(){pthread_t tid1,tid2;int step1=1, step2=-1;pthread_create(&tid1, NULL, mySumFun, &step1); pthread_create(&tid2, NULL, mySumFun, &step2);pthread_join(tid2, NULL);pthread_join(tid1, NULL);printf("The sum is = %lld \n", sum);exit(0);}
- Q2} Write a user-define function that add or subtracts two polynomials of any order. Use the function to add and subtract the following f₁(x)=x5-7x4+11x³-4x²-5x-2 and f₂(x)=9x²-10x+6 polynomials:Q. use set and test or swap #include <stdio.h>#include <stdlib.h>#include <pthread.h> long long sum=0,num=2000000; pthread_mutex_t mutex = PTHREAD_MUTEX_INITIALIZER; void *mySumFun(void *vargp){int step=*(int *) vargp; for(int i=0;i<num;i++){ //pthread_mutex_lock(&mutex);sum+=step;// pthread_mutex_unlock(&mutex);}return NULL;}int main(){pthread_t tid1,tid2;int step1=1, step2=-1;pthread_create(&tid1, NULL, mySumFun, &step1); pthread_create(&tid2, NULL, mySumFun, &step2);pthread_join(tid2, NULL);pthread_join(tid1, NULL);printf("The sum is = %lld \n", sum);exit(0);}6.use c code to Develop a code that gets two integers from the user and calculates and prints their least common multiplier or LCM. LCM is the smallest number that divides both given integers. Example. lcm (10, 15) = 30, lcm (5,7) = 35 and lcm (12, 24) = 24Hint: we know that lcm(x,y) = xy/gcd(x,y) where gcd is the greatest common divisor (discussed in class). Develop a gcd function in your code, get the two numbers from the user, call your function to calculate their gcd and then use the formula above to calculate and display their lcm GOOD LUCK! Reference This is a list of function prototypes of the C library functions presented in class. You may use any of these functions in your solutions (unless the requirements explicitly indicate otherwise). As you have been provided the function prototypes, you are expected to use the functions correctly in your solutions. double atof(char *string); int atoi(char *string); long atol(char *string); int fclose(FILE *filePointer); char *fgets(char…
- #5.Incorrect gurantee downvote. Euler's totient function, also known as phi-function ϕ(n),counts the number of integers between 1 and n inclusive,which are coprime to n.(Two numbers are coprime if their greatest common divisor (GCD) equals 1)."""def euler_totient(n): """Euler's totient function or Phi function. Time Complexity: O(sqrt(n)).""" result = n for i in range(2, int(n ** 0.5) + 1): if n % i == 0: while n % i == 0: n //= i.25. What is the output of the following code any ([ )? 1 NaN Error// the calling statement is: cout (y)); else return(static_cast(2 * y) - x); } О 40 О 50 30 60
- Examine the following C++ code and determine the output. int s=0, d, x-42; while (x > 0) { d = x 8 10; s = s*10 + d; x = x / 10; } cout « s <« endl; а. 34 b. 4 O c. None of these d. 24(Numerical) Write a program that tests the effectiveness of the rand() library function. Start by initializing 10 counters to 0, and then generate a large number of pseudorandom integers between 0 and 9. Each time a 0 occurs, increment the variable you have designated as the zero counter; when a 1 occurs, increment the counter variable that’s keeping count of the 1s that occur; and so on. Finally, display the number of 0s, 1s, 2s, and so on that occurred and the percentage of the time they occurred.4. Langauge concatentation (a) Let L = {\, a}. Characterize L20. (b) Let L = D* · {aa, bb} · E*. Characterize L. (c) Compute L1 L3 and L2L3 where L1 = {}, L2 = {X}, and L3 = {b}.