= Write a function ApproxSolver1(a0, a1, b0, b1, m) that takes as input two entries a0, a1 representing the second order dif- ference equation αoxn + a₁xn+1 + Xn+2 0, and two more b0, b1 representing initial conditions (x0,x1) = (bo, b₁), along with a value m, and outputs an integer which is an approximate value of xm by (assuming distinct roots of the auxiliary equation).
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- (Numerical) Heron’s formula for the area, A, of a triangle with sides of length a, b, and c is A=s(sa)(sb)(sc) where s=(a+b+c)2 Write, test, and execute a function that accepts the values of a, b, and c as parameters from a calling function, and then calculates the values of sand[s(sa)(sb)(sc)]. If this quantity is positive, the function calculates A. If the quantity is negative, a, b, and c do not form a triangle, and the function should set A=1. The value of A should be returned by the function.Write a function that takes 4 Inter gets (num1, den1,num2,den2) as parameters which represent the rational numbers num1/den1, num2/den2 and returns the division of these two fractions. A/B ÷ C/D = A/B × D/C = AD/BC • den 1, den2, and num2 can not be zero. • use multiplication_rat function.The least common multiple (lcm) of two positive integers u and v is the smallest positive integer that is evenly divisible by both u and v. Thus, the lcm of 15 and 10, written lcm(15, 10), is 30 because 30 is the smallest integer divisible by both 15 and 10. Write a function lcm() that takes two integer arguments and returns their lcm . The lcm() function should calculate the least common multiple by calling the gcd() function from program 7.6 in accordance with the following identity
- The least common multiple (lcm) of two positive integers u and v is the smallest positive integer that is evenly divisible by both u and v. Thus, the lcm of 15 and 10, written lcm(15, 10), is 30 because 30 is the smallest integer divisible by both 15 and 10. Write a function lcm() that takes two integer arguments and returns their lcm . The lcm() function should calculate the least common multiple by calling the gcd() function from program 7.6 in accordance with the following identity: lcm(u, v) = uv / gcd(u, v) u, v >= 0= Write a function ApproxSolver1(a0, a1, b0, b1, m) that takes as input two entries a0, a1 representing the second order dif- ference equation a0xn + α₁xn+1 + Xn+2 0, and two more b0, b1 representing initial conditions (x0,x1) = (bo, b₁), along with a value m, and outputs an integer which is an approximate value of xm by (assuming distinct roots of the auxiliary equation).1. Fibonacci series can be created starting from 0 to the desired number. A function can be written for this operation and it can take two parameters: start and end. The default fırst parameter 0 is acceptable. 0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 610, 987, 1597, 2584, 4181, 6765..
- Because of the outbreak of the Coronavirus (COVID-19), visitors from infested countries and regions have to be quarantined for 14 days which is 14 X 24 X 60 = 20160 minutes! Write a function Min2DHM(minutes) that takes the number of minutes as a parameter and converts that number of minutes into a specific format below: Day(s): Hour(s): Minutes(s) Note: You can assume the number of minutes is between 0 and 20160 and here is the expected output of your function.Define stubs for the functions get_user_num() and compute_avg(). Each stub should print "FIXME: Finish function_name)" followed by a newline, and should return -1. Each stub must also contain the function's parameters. Sample output with two calls to get_user_num() and one call to compute_avg(): FIXME: Finish get_user_num() | FIXME: Finish get_user_num() FIXME: Finish compute_avg() Avg: -1 354684.2266598.qx3zgy7 2 Your solution goes here ''' 4 user_num1 5 user_num2 6 avg_result get_user_num() get_user_num) compute_avg(user_num1, user_num2) 8 user_num1 9 user_num2 10 avg_result %3D 11 12 print('Avg:', avg_result)Write overloaded function calculate() with following function definitions.void calculate(int x, int y ) – to calculate the sum of all even numbers from xtill y and print the sum .void calculate(char a,char n) – to print all vowels from variable a to n if anyvoid calculate(int x, int y, int z) – to print the middle number from x,y,z ifyou arrange them in ascending order.
- Define stubs for the functions get_user_num() and compute_avg(). Each stub should print "FIXME: Finish function_name()" followed by a newline, and should return -1. Each stub must also contain the function's parameters.Sample output with two calls to get_user_num() and one call to compute_avg(): FIXME: Finish get_user_num() FIXME: Finish get_user_num() FIXME: Finish compute_avg() Avg: -1 ''' Your solution goes here ''' user_num1 = 0user_num2 = 0avg_result = 0 user_num1 = get_user_num()user_num2 = get_user_num()avg_result = compute_avg(user_num1, user_num2) print('Avg:', avg_result)The function doit(a,b,c,d) is a function that multiplies the smallest of its integers arguments by 2. If more than one arguments are equal and are the smallest, they are multiplied by 2. For example given a=5,b=3,c=7,d=9, doit (a,b,c,d) changes b to 6. Write the code of the function doit and test it.Define stubs for the functions get_user_num() and compute_avg(). Each stub should print "FIXME: Finish function_name()" followed by a newline, and should return -1. Each stub must also contain the function's parameters. Sample output with two calls to get_user_num() and one call to compute_avg(): FIXME: Finish get_user_num( ) FIXME: Finish get_user_num() FIXME: Finish compute_avg() Avg: -1 Learn how our autograder works 461710.3116374.qx3zqy7 1234 3 111 Your solution goes here ** 4 user_num1 = 0 5 user_num2 = 0 6 avg_result = 0 7 8 user_num1 = get_user_num() 9 user_num2 = get_user_num() 10 avg_result = compute_avg(user_num1, user_num2) 11 12 print (f'Avg: {avg_result}')