In a single day more than blank people's lives are affected simply by driving a motor vehicle

Look at the far left lane in the picture. explain what the red car is doing and what it needs to do to travel safely.

1. package newton’s method for approximating square roots (case study 3.6) in a function named newton. this function expects the input number as an argument and returns the estimate of its square root. the script should also include a main function that allows the user to compute square roots of inputs until she presses the enter/return key. 2. convert newton’s method for approximating square roots in project 1 to a recursive function named newton. (hint: the estimate of the square root should be passed as a second argument to the function.) 3. elena complains that the recursive newton function in project 2 includes an extra argument for the estimate. the function’s users should not have to provide this value, which is always the same, when they call this function. modify the definition of the function so that it uses a keyword parameter with the appropriate default value for this argument, and call the function without a second argument to demonstrate that it solves this problem. 4. restructure newton’s method (case study 3.6) by decomposing it into three cooperating functions. the newton function can use either the recursive strategy of project 1 or the iterative strategy of case study 3.6. the task of testing for the limit is assigned to a function named limitreached, whereas the task of computing a new approximation is assigned to a function named improveestimate. each function expects the relevant arguments and returns an appropriate value. 5. a list is sorted in ascending order if it is empty or each item except the last one is less than or equal to its successor. define a predicate issorted that expects a list as an argument and returns true if the list is sorted, or returns false otherwise. (hint: for a list of length 2 or greater, loop through the list and compare pairs of items, from left to right, and return false if the first item in a pair is greater.)

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; }