Answer:
larger in a presentation, smaller on a document.
Explanation:
this makes it easier to tell which is which.
Answer:
<h3 />
Explanation:
<h3>Although System analysis offers an extensive range of benefits it might also have some disadvantages. One of the main disadvantages which is mostly overlooked is the risk of too much analysing which may be costly and time consuming. It is therefore part of the analyst's job to find the right balance.</h3>
Answer:
Flowchart of an algorithm (Euclid's algorithm) for calculating the greatest common divisor (g.c.d.) of two numbers a and b in locations named A and B. The algorithm proceeds by successive subtractions in two loops: IF the test B ≥ A yields "yes" or "true" (more accurately, the number b in location B is greater than or equal to the number a in location A) THEN, the algorithm specifies B ← B − A (meaning the number b − a replaces the old b). Similarly, IF A > B, THEN A ← A − B. The process terminates when (the contents of) B is 0, yielding the g.c.d. in A. (Algorithm derived from Scott 2009:13; symbols and drawing style from Tausworthe 1977).
Explanation:
Flowchart of an algorithm (Euclid's algorithm) for calculating the greatest common divisor (g.c.d.) of two numbers a and b in locations named A and B. The algorithm proceeds by successive subtractions in two loops: IF the test B ≥ A yields "yes" or "true" (more accurately, the number b in location B is greater than or equal to the number a in location A) THEN, the algorithm specifies B ← B − A (meaning the number b − a replaces the old b). Similarly, IF A > B, THEN A ← A − B. The process terminates when (the contents of) B is 0, yielding the g.c.d. in A. (Algorithm derived from Scott 2009:13; symbols and drawing style from Tausworthe 1977).
Answer:
Check the explanation
Explanation:
223.1.17/24 indicates that out of 32-bits of IP address 24 bits have been assigned as subnet part and 8 bits for host id.
The binary representation of 223.1.17 is 11011111 00000001 00010001 00000000
Given that, subnet 1 has 63 interfaces. To represent 63 interfaces, we need 6 bits (64 = 26)
So its addresses can be from 223.1.17.0/26 to 223.1.17.62/26
Subnet 2 has 95 interfaces. 95 interfaces can be accommodated using 7 bits up to 127 host addresses can represented using 7 bits (127 = 27)
and hence, the addresses may be from 223.1.17.63/25 to 223.1.17.157/25
Subnet 3 has 16 interfaces. 4 bits are needed for 16 interfaces (16 = 24)
So the network addresses may range from 223.1.17.158/28 to 223.1.17.173/28
