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Cengagenowv2 is a comprehensive online learning tool. using cengagenowv2, you may access all of the following except: 2. each time you log in, cengagenowv2 automatically performs a system check and informs you if your computer does not meet the cengagenowv2 system requirements. 3. which tab/page allows you to easily track your assignment scores, number of submissions, time spent, as well as the ability view assign

What is the worst-case complexity of the maxrepeats function? assume that the longest string in the names array is at most 25 characters wide (i.e., string comparison can be treated as o( class namecounter { private: int* counts; int nc; string* names; int nn; public: namecounter (int ncounts, int nnames); int maxrepeats() const; }; int namecounter: : maxrepeats () { int maxcount = 0; for (int i = 0; i < nc; ++i) { int count = 1; for (int j = i+1; j < nc; ++j) { if (names[i] == names[j]) ++count; } maxcount = max(count, maxcount); } return maxcount; }

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