Answer:
Both careers speak to an audience; however, people involved in the performing arts speak only to a live audience.
Explanation:
sorry im late but im not sure
Answer:
Table for Area codes are not missing;
See Attachment for area codes and major city I used
This program will be implemented using c++ programming language.
// Comments are used for explanatory purposes
// Program starts here
#include <iostream>
using namespace std;
int main( )
{
// Declare Variable area_code
int area_code;
// Prompt response from user
cout<<Enter your area code: ";
cin<<"area_code;
// Start switch statement
switch (area_code) {
// Major city Albany has 1 area code: 229...
case 229:
cout<<"Albany\n";
break;
// Major city Atlanta has 4 area codes: 404, 470 678 and 770
case 404:
case 470:
case 678:
case 770:
cout<<"Atlanta\n";
break;
//Major city Columbus has 2 area code:706 and 762...
case 706:
case 762:
cout<<"Columbus\n";
break;
//Major city Macon has 1 area code: 478...
case 478:
cout<<"Macon\n";
break;
//Major city Savannah has 1 area code: 912..
case 912:
cout<<"Savannah\n";
break;
default:
cout<<"Area code not recognized\n";
}
return 0;
}
// End of Program
The syntax used for the above program is; om
WHICH QUESTION DO U NEED HELP WITH
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
greatflombles or gr8flombles or go2flombles
