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NEW QUESTION: 1
Refer to the following exhibit.
Assume all links have an equal metric in the IGP. Receiver 2 has joined the multicast group and the source
is sending. After the switchover of Router F, will a branch of the Shared Path Tree still exist between
Router D and Router E?
A. Yes. As it is maintained by the PIM hellos, it will exist until Router D or Router E is PIM disabled
B. Yes, it will exist until Router D prunes itself from the shared path tree
C. No. Only Source Path Trees will remain
D. Yes, it will exist until the source stops sending multicast traffic
Answer: B
NEW QUESTION: 2
What does a network vulnerability assessment intend to identify?
A. Malicious software and spyware
B. Security design flaws
C. 0-day vulnerabilities
D. Misconfiguration and missing updates
Answer: D
Explanation:
Explanation/Reference:
Explanation:
A network vulnerability assessment intends to identify known vulnerabilities based on common misconfigurations and missing updates. 0-day vulnerabilities by definition are not previously known and therefore are undetectable. Malicious software and spyware are normally addressed through antivirus and antispyware policies. Security design flaws require a deeper level of analysis.
NEW QUESTION: 3
In the diagram above, ABCD is a square with an area of 100 cm2 and lines BD and AC are the diagonals
of ABCD. If line EF is parallel to line BC and the length of line CF = 32 cm, which of the following is equal
to the shaded area?
A. 78 cm2
B. 64 cm2
C. 25 cm2
D. 39 cm2
E. 89 cm2
Answer: D
Explanation:
Explanation/Reference:
The area of a square is equal toS2where s is the length of one side of the square. A square with an area
of 100cm2has sides that are each equal to100 = 10 cm. The diagonal of a square is equal to2times the
length of a side of the square. Therefore, the lengths of diagonals AC and BD are 102 cm. Diagonals of a
square bisect each other at right angles, so the lengths of segments OB and OC are each52 cm. Since
lines BC and EF are parallel and lines OC and OB are congruent, lines BE and CF are also congruent.
The length of line OF is equal to the length of line OC plus the length of line CF:52 + 32 = 82 cm. In the
same way, OE =OB + BE =52 + 32 = 82 cm. The area of a triangle is equal to1/2bh, where b is the base of
the triangle and h is the height of the triangle. EOF is a right triangle, and its area is equal to The size of
the
shaded area is equal to the area of EOF minus one-fourth of the area of ABCD:
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