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Computers and Technology, 22.06.2019 05:20
Write a program called assignment3 (saved in a file assignment3.java) that computes the greatest common divisor of two given integers. one of the oldest numerical algorithms was described by the greek mathematician, euclid, in 300 b.c. it is a simple but very e↵ective algorithm that computes the greatest common divisor of two given integers. for instance, given integers 24 and 18, the greatest common divisor is 6, because 6 is the largest integer that divides evenly into both 24 and 18. we will denote the greatest common divisor of x and y as gcd(x, y). the algorithm is based on the clever idea that the gcd(x, y) = gcd(x ! y, y) if x > = y and gcd(x, y) = gcd(x, y ! x) if x < y. the algorithm consists of a series of steps (loop iterations) where the “larger” integer is replaced by the di↵erence of the larger and smaller integer. this continues until the two values are equal. that is then the gcd.
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Computers and Technology, 22.06.2019 11:10
The total cost of textbooks for the term was collected from 36 students. create a histogram for this data. $140 $160 $160 $165 $180 $220 $235 $240 $250 $260 $280 $285 $285 $285 $290 $300 $300 $305 $310 $310 $315 $315 $320 $320 $330 $340 $345 $350 $355 $360 $360 $380 $395 $420 $460 $460
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Computers and Technology, 22.06.2019 15:50
The file sales data.xlsx contains monthly sales amounts for 40 sales regions. write a sub that uses a for loop to color the interior of every other row (rows 3, 5, etc.) gray. color only the data area, columns a to m. (check the file colors in excel.xlsm to find a nice color of gray.)
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Computers and Technology, 22.06.2019 20:00
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; }
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