Answer:
Explanation:
1 : Ctrl + X
2 Ctrl + C
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4 Ctrl + V
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End
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18 F12
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23 Ctrl + (right arrow)
24 Ctrl + [
25 Ctrl + Del
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27 F1 pls mark me as brainliest and
check all the short cuts
the answer is electronic screens
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Answer:
The correct answer is 16 bit.
Explanation:
The main role of the transport layer provided the communication channel of the process which are running in a different host. When the information flows there is always traffic that is arisen but the transport layer aimed is that process will go at the correct node and received at the proper destination. The 16-bit number specifies that the process or node is running in the networked traffic.
Answer:
The page field is 8-bit wide, then the page size is 256 bytes.
Using the subdivision above, the first level page table points to 1024 2nd level page tables, each pointing to 256 3rd page tables, each containing 64 pages. The program's address space consists of 1024 pages, thus we need we need 16 third-level page tables. Therefore we need 16 entries in a 2nd level page table, and one entry in the first level page table. Therefore the size is: 1024 entries for the first table, 256 entries for the 2nd level page table, and 16 3rd level page table containing 64 entries each. Assuming 2 bytes per entry, the space required is 1024 * 2 + 256 * 2 (one second-level paget table) + 16 * 64 * 2 (16 third-level page tables) = 4608 bytes.
First, the stack, data and code segments are at addresses that require having 3 page tables entries active in the first level page table. For 64K, you need 256 pages, or 4 third-level page tables. For 600K, you need 2400 pages, or 38 third-level page tables and for 48K you need 192 pages or 3 third-level page tables. Assuming 2 bytes per entry, the space required is 1024 * 2 + 256 * 3 * 2 (3 second-level page tables) + 64 * (38+4+3)* 2 (38 third-level page tables for data segment, 4 for stack and 3 for code segment) = 9344 bytes.
Explanation:
16 E the answer
Answer:
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