Answer:
The red Mustang should be given the right-of-way at this intersection because:
The red Mustang arrived first before the blue Prius and is closer to the stop sign before the arrival of the blue Prius. This implies that the red Mustang must have waited for its turn, unlike the blue Prius that just arrived at the intersection.
Explanation:
Traffic laws, regulations, and IPDE defensive driving strategy require that drivers always give way to the vehicle that arrived before them at an intersection. Assuming that multiple vehicles reach the intersection simultaneously, then the vehicle must be given the right-of-way. Alternatively, the traffic lights at the intersection should be obeyed and be allowed to regulate the movement of vehicles. IPDE defensive driving strategy requires drivers to be watchful of their driving spaces to determine the appropriate time to move.
Answer:
☆<em>intellectual property</em>☆
<u>intangible </u><u>a</u><u>s</u><u>s</u><u>e</u><u>t</u><u>s</u><u> </u><u> that result from intellectual</u><u> </u><u>p</u><u>r</u><u>o</u><u>p</u><u>e</u><u>r</u><u>t</u><u>y</u><u> </u><u>.</u>
<u>☆</u><u>I</u><u>n</u><u>t</u><u>a</u><u>n</u><u>g</u><u>i</u><u>b</u><u>l</u><u>e</u><u>☆</u><u>;</u>
<u>s</u><u>o</u><u>m</u><u>e</u><u>t</u><u>h</u><u>i</u><u>n</u><u>g</u><u> </u><u>t</u><u>h</u><u>a</u><u>t</u><u> </u><u>c</u><u>a</u><u>n</u><u>n</u><u>o</u><u>t</u><u> </u><u>b</u><u>e</u><u> </u><u>t</u><u>o</u><u>u</u><u>c</u><u>h</u><u>e</u><u>d</u><u>.</u>
<em>P</em><em>l</em><em>e</em><em>a</em><em>s</em><em>e</em><em> </em><em>M</em><em>a</em><em>r</em><em>k</em><em>☆</em>
<em>A</em><em>r</em><em>i</em><em>a</em><em>♡</em>
Answer:
depends on which flash drive
Explanation:
32GB-1000 photos
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