1. To make storing and finding data fast, a disk drive deals with a … which is a group of sectors. a) track b) bin c) cluster d) junction
2. You had better … cryptic abbreviations in files and folders.
a) give b) avoid c) access d) accept
3. You can avoid growing your list of files to an unmanageable size by … unnecessary files and folders.
a) saving b) deleting c) renaming d) moving
4. File … facilitates the process of organizing and manipulating your computer-based files.
a) location b) extension c) performance d) management
5. The operating system revises the index file to … the new file name and its location.
a) erase b) include c) number d) exclude
6. … a file is stored on a disk, CD, etc., the storage medium should be formatted.
a) before b) after c) while d) since
7. The pattern of … tracks and sectors is dependent on the storage medium and your PC’s operating system.
a) dividing b) selecting c) numbering d) arranging
Задача 1
var
x1, y1, x2, y2, l: real;
begin
Writeln('Введите значения (x1,y1,x2,y2)');
Read(x1, y1, x2, y2);
l := Sqrt(Sqr(x1 - x2) + Sqr(y1 - y2));
Writeln('Расстояние: ', l:3:2);
end.
Тестовое решение:
Введите значения (x1,y1,x2,y2)
2 3 4 5
Расстояние: 2.83
Задача 2
var
a, b, c, p, s: real;
begin
Writeln('Введите ABC');
Read(a, b, c);
P := (a + b + c)/2;
S := Sqrt(p * (p - a) * (p - b) * (p - c));
Writeln('Площадь по Герону: ', S:3:2);
end.
Тестовое решение:
Введите ABC
2 3 4
Площадь по Герону: 2.90
Заменяем каждый разряд на код из таблицы.
Двоичная СС Восьмеричная СС
000 0
001 1
010 2
011 3
100 4
101 5
110 6
111 7
Получаем числа:
1024757520 (7) = 001000010100111101111101010000 (2)
1232400565 (7) = 001010011010100000000101110101 (2)
Складываем (0 в начале числа можно убрать):
1000010100111101111101010000 + 1010011010100000000101110101 = 10010101111011110000011000101
Меняем 1 на 0, 0 на 1:
10010101111011110000011000101 = 01101010000100001111100111010
Двоичная СС Шестнадцатеричная СС
0000 0
0001 1
0010 2
0011 3
0100 4
0101 5
0110 6
0111 7
1000 8
1001 9
1010 A
1011 B
1100 C
1101 D
1110 E
1111 F
Получаем число:
1101 0100 0010 0001 1111 0011 1010 (2) = D421F3A (16)