In addition to bandages and gauze, it's recommended that a first aid kit also contains

Write a function so that the main0 code below can be replaced by the simpler code that calls function mphandminutes tomiles0. original main0 int main) l double milesperhour-70.0; double minutestraveled = 100.0; double hourstraveled; double milestraveled; hourstraveled = minutestraveled / 60.0; milestraveled = hourstraveled * milesperhour; cout < "miles" 2 using namespace std; 4 /* your solution goes here/ 6 int maino 1 test passed 7 double milesperhour 70.0 all tests passed 8 double minutestraveled 100.0; 10 cout < < "miles: " < < mphandminutestomiles(milesper-hour, minutestraveled) < < endl; 12 return 0; 13

Write a function called checkfactor that takes two arrays of positive numbers, firstnumberrow and secondnumberrow. checkfactor checks if the first entry in firstnumberrow is divisible by the first entry in secondnumberrow, and performs the same operation on the next array elements until all entries have been checked. all the numbers are positive and the number of entries in the arrays are the same. the function should return the identified divisible numbers in two row arrays named firstdivisible and seconddivisible. restrictions: branches or loops should not be used. the code must use the internal mod and logical functions. hint: the mod function should be used to determine if two numbers are divisible. ex: for num1 and num2 if mod(num1,num2) is 0, then the two numbers are divisible. this is matlab

Ac++an of of pi. in , pi is by : pi = 4 – 4/3 + 4/5 – 4/7 + 4/9 - 4/11 + 4/13 - 4/15 + 4/17 . ., to pi (9 ). , if 5 to pi, be as : pi = 4 - 4/3 + 4/5 - 4/7 + 4/9 = 4 – 1. + 0.8 - 0. + 0. = 3.. atoofbe to pi?