Computers and Technology, 05.05.2020 05:39 cowboo5000pcl655

What is the amount of work you must input in order to move the stone?

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Computers and Technology, 22.06.2019 00:30, angeline2004

At an open or uncontrolled intersection, yield if a. the cross road has more lanes than yours b. the cross road has fewer land than yours c. you’re on a state highway and the cross road is a secondary road d. do you have three or more passengers in your vehicle

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Computers and Technology, 22.06.2019 05:30, Cassandoo

The total revenues for a company are $150,223 and the total expenses were 125,766. if you are calculating the net income, which of these spreadsheets would you use? insert a spreadsheet with $150,223 in cell b2 and 125, 766 in cell b3. enter a formula =b2-b3. the formula should be showing in the formula bar. insert a spreadsheet with $150,223 in cell b2 and 125, 766 in cell b3. enter a formula =b2+b3. the formula should be showing in the formula bar. insert a spreadsheet with $150,223 in cell b2 and 125, 766 in cell b3. enter a formula =b2/b3. the formula should be showing in the formula bar. insert a spreadsheet with $150,223 in cell b2 and 125, 766 in cell b3. enter a formula =b2*b3. the formula should be showing in the formula bar.

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Computers and Technology, 22.06.2019 06:00, Treezy11

In outlook how can cherie look at the details of an event that appears on the month view of her calendar? check all that apply. by switching to the detail view by switching to the week view by switching to the day view by right-clicking on the event by double-clicking on the event by highlighting the event

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Computers and Technology, 22.06.2019 20:00, Jana1517

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; }