Answer:
LA15; LA22
LA16; LA31; LA32
LA169; LA126;
LA127; LA141
Explanation:
Given

Required
Course sequence to satisfy the prerequisite
From the course prerequisite, we have:
and 
This means that LA15 and LA22 are the base courses, and they have no prerequisite. So, we have:
![[LA15; LA22]](https://tex.z-dn.net/?f=%5BLA15%3B%20LA22%5D)
LA16 and LA31 have LA15 as their direct course prerequisite. So, the sequence becomes
![[LA15 \to [LA16, LA31]; LA22]](https://tex.z-dn.net/?f=%5BLA15%20%5Cto%20%5BLA16%2C%20LA31%5D%3B%20LA22%5D)
To complete the sequence, we read each course and place them other their prerequisite.
<em>See attachment for complete tree</em>
<em></em>
From the tree, we have the sequence to be:
<em>LA15; LA22</em>
<em>LA16; LA31; LA32</em>
<em>LA169; LA126;</em>
<em>LA127; LA141</em>
<em />
Answer:
Page orientation
Explanation:
Required
What does portrait and landscape represent?
For better understanding, I'll narrow down the question to Microsoft Office Word. The explanation can be applied to other concepts.
Analysing the option one after the other.
Page layout: This is a term that describes the graphical representation of each document page, visually.
Page Size: This is a term that represents the length and width of the page. Common page sizes are LETTER, A4, A3, and so on.
The above (2) do not in any way represent what portrait and landscape represent.
Page orientation: This is a way the document is represented and it's a direct synonym for landscape and portrait.
Since (1) & (2) do not represented the given term, the last option is incorrect.
Hence,
Page orientation correctly answers the question
Answer:
a. Anycast Addresses
b. Broadcast domain
c. Classless Interdomain Routing (CIDR)
d. Extended Unique Identifier (EUI-64) interface ID
e. Intra-Site Automatic Tunnel Addressing Protocol (ISATAP)
f. IP prefix
g. Loopback Address
h. Metric
i. Multicasting
j. Teredo
Explanation:
a. <u>Anycast Addresses</u>: An address type used in IPv6 to allow a one-to-many relationship between source and destination.
b. <u>Broadcast domain</u>: The bounds of a network that defines which devices must receive a packet that's broadcast by any other device.
c. <u>Classless Interdomain Routing (CIDR)</u>: A method of IP addressing in which the network and host IDs are determined by a prefix number that specifies how many bits of the IP address are network bits.
d. <u>Extended Unique Identifier (EUI-64) interface ID</u>: An auto configure IPv6 host address that uses the MAC address of the host plus an additional 16 bits.
e. <u>Intra-Site Automatic Tunnel Addressing Protocol (ISATAP)</u>: An automatic tunneling protocol used to transmit IPv6 packets between dual IP layer hosts across an IPv4 network.
f. <u>IP prefix</u>: A value used to express how many bits of an IP address are network ID bits.
g. <u>Loopback Address</u>: An address that always refers to the local computer. The loop address is primarily 127.0.01.
h. <u>Metric</u>: A value assigned to the gateway based on the speed of the interface used to access the gateway.
i. <u>Multicasting</u>: A network communication in which a packet is addressed so that more than one destination can receive it
j. <u>Teredo</u>: An automatic IPv6-over-IPv4 tunneling protocol that solves the problem of 6to4's requirement of a public IPv4 address and the inability to traverse NAT routers.
If you are writing a client to a problem with a product, you would have a formal tone and use a decorative (yet professional) Word or PowerPoint.
Hope this helps!
Answer:
(a) the number of times the value is performs is up to four cycles. and as such the integer i is executed up to 5 times. (b)The point version of the floating point can have CPE of 3.00, even when the multiplication operation required is either 4 or 5 clock.
Explanation:
Solution
The two floating point versions can have CPEs of 3.00, even though the multiplication operation demands either 4 or 5 clock cycles by the latency suggests the total number of clock cycles needed to work the actual operation, while issues time to specify the minimum number of cycles between operations.
Now,
sum = sum + udata[i] * vdata[i]
in this case, the value of i performs from 0 to 3.
Thus,
The value of sum is denoted as,
sum = ((((sum + udata[0] * vdata[0])+(udata[1] * vdata[1]))+( udata[2] * vdata[2]))+(udata[3] * vdata[3]))
Thus,
(A)The number of times the value is executed is up to 4 cycle. And the integer i performed up to 5 times.
Thus,
(B) The floating point version can have CPE of 3.00, even though the multiplication operation required either 4 or 5 clock.