Do Not Answer If You Are Going To Give Me A Link! DONT WASTE MY POINTS!! Choi is trying out for a reality show on television. The application requires a self-portrait. Choi is setting up a self-portrait but wants to get her whole body. How might Choi BEST take a self-portrait that looks professional enough to submit?

A.
She could stand in front of a mirror holding her phone to take the photo.

B.
She could hold up the camera and take a selfie pointed down toward her body.

C.
She could use a self-timer to allow herself to get into position for the shot.

D.
She could ask her friend to take a photo of her as she stands against a background.

)a grad student comes up with the following algorithm to sort an array a[1..n] that works by first sorting the first 2/3rds of the array, then sorting the last 2/3rds of the (resulting) array, and finally sorting the first 2/3rds of the new array. 1: function g-sort(a, n) . takes as input an array of n numbers, a[1..n] 2: g-sort-recurse(a, 1, n) 3: end function 4: function g-sort-recurse(a, `, u) 5: if u ⒠` ≤ 0 then 6: return . 1 or fewer elements already sorted 7: else if u ⒠` = 1 then . 2 elements 8: if a[u] < a[`] then . swap values 9: temp ↠a[u] 10: a[u] ↠a[`] 11: a[`] ↠temp 12: end if 13: else . 3 or more elements 14: size ↠u ⒠` + 1 15: twothirds ↠d(2 ◠size)/3e 16: g-sort-recurse(a, `, ` + twothirds ⒠1) 17: g-sort-recurse(a, u ⒠twothirds + 1, u) 18: g-sort-recurse(a, `, ` + twothirds ⒠1) 19: end if 20: end function first (5 pts), prove that the algorithm correctly sorts the numbers in the array (in increasing order). after showing that it correctly sorts 1 and 2 element intervals, you may make the (incorrect) assumption that the number of elements being passed to g-sort-recurse is always a multiple of 3 to simplify the notation (and drop the floors/ceilings).

Dave has to create animations for a game. which tool can dave use?

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; }