ccna 4 final exam v4 0 answers 2014
CCNA 2 Final Exam New docx
Ngày tải lên :
22/06/2014, 16:20
... 0. 0 .0. 0 0. 0 .0. 0 Fa0 /0
ip route 0. 0 .0. 0 0. 0 .0. 0 Fa0/1
ip route 0. 0 .0. 0 0. 0 .0. 0 10. 1.1.1
ip route 0. 0 .0. 0 0. 0 .0. 0 10. 1.1.2
ip route 209 .165. 202 .0 255.255.255 .0 10. 1.1.1
ip route 209 .165. 202 .0 ... 172.16.1 .0/ 24 [ 90/ 219 545 6] via 192.168. 200 .1, 00 :00 :09 , Serial0 /0/ 0
O 172.16.1 .0/ 24 [1 10/ 101 2] via 192.168. 200 .1, 00 :00 :22, Serial0 /0/ 0
R 172.16.1 .0/ 24 [1 20/ 1] via 192.168. 200 .1, 00 :00 :17, Serial0 /0/ 0
I ... 192.168. 200 .1, 00 :00 :09 , Ethernet0
ã O 192.168.168 .0/ 24 [1 10/ 101 2] via 192.168. 200 .1, 00 :00 :22, Ethernet0
ã R 192.168.168 .0/ 24 [1 20/ 1] via 192.168. 200 .1, 00 :00 :17, Ethernet0
ã S 192.168.168 .0/ 24 [1 /0] ...
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Tài liệu CCNA Semester 4 - Skills Based Final Exam - Student Training Instructor Guidelines - Exam Overview and Administration doc
Ngày tải lên :
11/12/2013, 14:15
...
Ip route 172. 50. 2 .0 255.255.255 .0 172. 50. 1.2
Ip route 172.16 .0. 0 255.255 .0. 0 202 .168. 10. 2
ip route 0. 0 .0. 0 0. 0 .0. 0 202 .168. 10. 1
ip access-list 101 deny tcp any 172. 50. 1 .0 0 .0. 0.255 eq telnet ... Semester 4 Skills Based Final – Part 1 Router Configuration
Student Training – Frame Relay Scenario – ( 60 pts)
HeadQuarters
MainOffice
Serial
202 .168. 10. 0/ 24
172. 50. 1 .0/ 24
HQ Network
S0/.1
(DCE)
S0/.2
E0
S1
Frame ... Network
S0/.1
(DCE)
S0/.2
E0
S1
Frame Relay
Network
RemoteOffice
172.16.1.99/ 24
DLCI 201
172.16.1.88/ 24
DLCI 101
Configuring the
Headquarters Router
SalesOffice
E0 E1
172. 50. 2 .0/ 24
Sales Network
SwitchSwitch
E1
Switch
172.16 .0. 0/16
MainOffice...
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Tài liệu Take Assessment - ENetwork Final Exam - Network Fundamentals (Version 4.0) pdf
Ngày tải lên :
13/12/2013, 19:16
... allows 5 10 hosts given the IP address 172. 30. 0 .0?
255.255 .0. 0
255.255. 248 .0
255.255.252 .0
255.255.2 54. 0
255.255.255 .0
255.255.255.128
30
file:///C:/Users/Martinos/Desktop/cisco/HTML /exam. htm
14 van ... 255.255.255. 248 ; Default Gateway: 192.168. 100 .18
IP address: 192.168. 100 .22; Subnet Mask: 255.255.255. 2 40 ; Default Gateway: 10. 1.1.5
IP address: 192.168. 100 . 30; Subnet Mask: 255.255.255. 2 40 ; Default ... 192.168.1.1
IP address: 192.168. 100 .31; Subnet Mask: 255.255.255. 2 40 ; Default Gateway: 192.168. 100 .18
9
file:///C:/Users/Martinos/Desktop/cisco/HTML /exam. htm
11 van 15 10/ 01/ 200 8 20: 03
Refer to the exhibit....
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CCNA Semester 3 - Skills Based Final Exam - Student Training Instructor Guidelines - Exam Overview and Administration
Ngày tải lên :
19/10/2013, 02:15
...
1 171.17 .0. 65 - 171.17 .0. 126 16 171.17 .4. 1 - 171.17 .4. 62
2 171.17 .0. 129 – 171.17 .0. 1 90 17 171.17 .4. 65 - 171.17 .4. 126
3 171.17 .0. 193 – 171.17 .0. 2 54 18 171.17 .4. 129 - 171.17 .4. 1 90
4 171.17.1.1 ...
Interface S0: 171.17.255.66 –
171.17.255.126
Interface S0: 171.17.2 54. 1 94 –
171.17.2 54. 2 54
Interface S1: 171.17.255.2 –
171.17.255.62
Interface S1: 171.17.2 54. 1 30 –
171.17.2 54. 1 90
4. LAN ... Interface Ethernet 0 IP addr: 171.17.255.1 30
CCNA Semester 3 - Skills Based Final Exam - Student Training
Instructor Guidelines - Exam Overview and Administration – Cont.
Exam Part 2 - Troubleshooting...
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CCNA Lab Guide v4 0 (Cisco Certified Network Associate)
Ngày tải lên :
19/10/2013, 11:25
... chỉ IP
Ví dụ 1: 172.16 .0. 0 0. 0.255.255
172.16 .0. 0 = 101 01 100 .00 0 100 00. 000 000 00. 000 000 00
0. 0.255.255 = 00 000 000 .00 000 000 .11111111.11111111
result = 101 01 100 .00 0 100 00. xxxxxxxx.xxxxxxxx
172.16.x.x ... IP nằm trong khoảng từ 172.16 .0. 0 và
172.16.255.255 sẽ tương ứng với ví dụ này)
Ví dụ 2: 172.16.8 .0 0 .0. 7.255
172.168.8 .0 = 101 01 100 .00 0 100 00. 000 0 100 0 .00 000 000
Created by: Dương Văn Toán ... cost 15 64 Thay đổi giá trị cost là 15 64
* Chú ý: Cost của một kết nối được tính
bằng công thức: 100 000 000 / BW (bps)
Bandwidth của một interface nằm trong
khoảng từ 1 đến 100 000 00. Đơn vị...
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Tài liệu CCNA Semester 1 - Skills Based Final Exam - Student Training Guidelines - Exam Overview and Administration docx
Ngày tải lên :
24/01/2014, 19:20
... routing protocol service
network 152. 20. 0 .0 Enables RIP advertisements to be broadcast and received
for network 152. 20. 0 .0 (on interface E1)
network 193.5.5 .0 Enables RIP advertisements to be ... 193.5.5 .0 (on interface E0)
int e0 Selects interface E0 for configuration
ip address 193.5.5.1 255.255.255 .0 Sets IP address and subnet mask for E0
no shutdown Brings up interfaces E0
int ... Scored: ____________
CCNA Semester 1 - Skills Based Final Exam - Student Training
Guidelines - Exam Overview and Administration
This exam is for use with regular CCNA classes taught by...
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Tài liệu CCNA Semester 2 - Skills Based Final Exam - Student Training Instructor Guidelines - Exam Overview and Administration doc
Ngày tải lên :
24/01/2014, 19:20
... ____________
CCNA Semester 2 - Skills Based Final Exam - Student Training
Instructor Guidelines - Exam Overview and Administration
This exam is for use with regular CCNA classes taught ... CCNA Semester 2 - Skills Based Final Exam - Student Training
Instructor Guidelines - Exam Overview and Administration – Cont.
Exam Part 2 – Troubleshooting ... working on Part 3, the Network
Concepts portion of the exam.
Exam Part 1 – Router Configuration – (Practical hands-on, Max. time 40 min., 60 Points)
Objective: Configure the standard 5-router...
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CCNA 2 FINAL 2012 EXAM – V1 potx
Ngày tải lên :
16/03/2014, 07:20
... 172.16.1 .0/ 24 [ 90/ 219 545 6] via 192.168. 200 .1, 00 :00 :09 , Serial0 /0/ 0
O 172.16.1 .0/ 24 [1 10/ 101 2] via 192.168. 200 .1, 00 :00 :22, Serial0 /0/ 0
R 172.16.1 .0/ 24 [1 20/ 1] via 192.168. 200 .1, 00 :00 :17, Serial0 /0/ 0
I ... 172.16.1 .0/ 24 [1 10/ 101 2] via 192.168. 200 .1, 00 :00 :22, Serial0 /0/ 0
R 172.16.1 .0/ 24 [1 20/ 1] via 192.168. 200 .1, 00 :00 :17, Serial0 /0/ 0
I 172.16.1 .0/ 24 [ 100 /1192] via 192.168. 200 .1, 00 :00 :09 , Serial0 /0/ 0
17. ... 172.16.2.1, 00 :00 : 24, Serial0 /0/ 1
R 192.168.9 .0/ 24 [1 20/ 1] via 172.16.1.2, 00 :00 : 24, Serial0 /0/ 0
R 192.168. 100 .0/ 24 [1 20/ 1] via 172.16.1.1, 00 :00 : 24, Serial0 /0/ 0
R 192.168.2 .0/ 24 [1 20/ 1] via 172.16.1.2,...
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REVISION for The final exam
Ngày tải lên :
01/09/2013, 15:10
... woken D. knew/ woke
34. In July, most students in Vietnam take the National Entrance Examinations to enter
higher education institutions.
A. Vocational schools B. high schools
4
Teacher : Quỳnh ... took d. drove
4. a. vehicles b. cars c. motorbikes d. traffic.
5.a. eventually b. lastly c. final d. right.
6.a. up b. dressed c. changed d. away.
7.a. luckily b. fortunately c. finally d. unfortunately
8.a. ... asked him whether he could explain another way to it
d. she asked him could he explain it again.
40 . I remember telling you about the deadline for the assignment.
a. I remember to tell you when...
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final exam first term grade 10
Ngày tải lên :
07/11/2013, 17:11
... each sentence = 0, 25 mark
01 . C; 02 . A; 03 . D; 04 . A;
II. Grammar and vocabulary (5 marks) each sentence = 0, 5 mark
05 . C; 06 . C; 07 . B; 08 . D; 09 . B; 10. A; 11. B; 12. C; 13. B; 14. D;
III. ... each sentence = 0, 25 mark
01 . B; 02 . B; 03 . A; 04 . A;
II. Grammar and vocabulary (5 marks) each sentence = 0, 5 mark
05 . A; 06 . C; 07 . D; 08 . B; 09 . A; 10. B; 11. D; 12. C; 13. D; 14. C;
III. ... each sentence = 0, 25 mark
01 . C; 02 . D; 03 . D; 04 . D;
II. Grammar and vocabulary (5 marks) each sentence = 0, 5 mark
05 . D; 06 . B; 07 . B; 08 . B; 09 . D; 10. D; 11. B; 12. B; 13. D; 14. B;
III....
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De thi thu trieu son 4 tuyen tap de thi toan 2014 moi nhat
Ngày tải lên :
03/01/2014, 15:38
...
∆
:
'
0
0 0
0
1
( )( )
2( 1)
x
y f x x x
x
−
= − +
+
( )
0
0
2
0
0
1
1
( )
2( 1)
1
x
y x x
x
x
−
⇒ = − +
+
+
Gọi A = ∆ ∩ox ⇒A(
2
0 0
2 1
2
x x− −
− ;0)
B = ∆ ∩oy ⇒ B (0;
2
0 0
2
0
2 1
2( ... 2
0 0 0 0
2
0
2 1 2 1
;
6 6( 1)
x x x x
x
− − − −
−
+
.
Do G
∈
đường thẳng:4x + y = 0
⇒
2 2
0 0 0 0
2
0
2 1 2 1
4. 0
6 6( 1)
x x x x
x
− − − −
− + =
+
⇔
( )
2
0
1
4
1x
=
+
...
( )
)02 1(112 301
123
1
2
≠−⇒≤=−+−⇔=
−
−+−
−⇔
xxxx
xx
x
1
=⇔
x .Suy ra nghi
ệ
m c
ủ
a h
ệ
là (x; y) =(1; 0)
0, 5
Gi
ả
i b
ấ
t ph
ươ
ng trình
ĐK:
2
0
0
0
10
2 0
2 10 0
x
x
x
x
x...
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Đề dự bị 4 - Môn Toán cho Sỹ tử 2015
Ngày tải lên :
22/03/2014, 17:13
... 24( 1+2+3 +4) 100 00= 2 40 00 00
Vậy tổng 96 số n là 1 80+ 1 800 +1 800 0+1 800 00+ 2 40 00 00= 25999 80
Cách 2
: Có 24 soá vôùi soá k ( k = 1, 2, 3, 4 ) đứng ở vị trí a
4
.
Có 18 soá vôùi soá k ( k = 1, 2, 3, 4 ) ...
xy−− =48 0
()
,
xy
B
xy
++=
⎧
⇒−
⎨
−−=
⎩
30
41
48 0
2/ Gọi
n aaaaa a a a a a==++++
43 21
43 2 10 4 3 2 1 0
10 10 10 10 10
0
là số cần lập
Ta có 4 cách chọn a
4
4 cách chọn a
3
3 cách chọn a
2
... 1 800 0;tổng các chữ số hàng ngàn là: 1 800 00.
Coù 24 soá
n aaaa=
32 10
1 ; Coù 24 soá naaaa=
32 10
2 ;Coù 24 soá
naaaa=
32 10
3 ; Coù 24 soá naaaa=
32 10
4
Tổng các chữ số hàng chục ngàn 24( 1+2+3 +4) 100 00= 2 40 00 00...
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![ccna security official exam certification guide [exam 640-553]](https://media.store123doc.com/images/document/14/rc/fp/medium_fpp1395720519.jpg)
ccna security official exam certification guide [exam 640-553]
Ngày tải lên :
25/03/2014, 11:08
... Ciphers 44 4
Block Ciphers 44 4
Stream Ciphers 44 5
Exploring Symmetric Encryption 44 5
Functionality of Symmetric Encryption Algorithms 44 6
Key Lengths 44 6
Features and Functions of DES 44 7
Working ... DES Key 44 7
Modes of Operation for DES 44 7
Working with DES Stream Cipher Modes 44 9
Usage Guidelines for Working with DES 44 9
Understanding How 3DES Works 45 0
Encrypting with 3DES 45 0
AES 45 1
The ... Encryption Algorithms 44 0
Symmetric and Asymmetric Encryption Algorithms 44 1
Encryption Algorithms and Keys 44 1
Symmetric Encryption Algorithms 44 1
Asymmetric Encryption Algorithms 44 3
The Difference...
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Đề thi và đáp án Olympic 30-4 môn toán khối 11 năm 2014
Ngày tải lên :
01/06/2014, 10:36
... giả thiết, ta có :
mn
20 14 20 14 100 −
( )
n mn 2 2
20 14 20 14 1 2 .5
−
⇔−
0. 5 đ
Do đó :
n2
20 14 2 n 2⇒≥
0. 5 đ
www.VNMATH.com
mn 2 mn 2 p
20 14 1 5 ( 200 0 14) 1 5 14 1 25
−−
− ⇔ + − ⇒−
... Giả sử (1) thỏa.
(1) ⇒ P
2
(0) = P (0) ⇒ P (0) = 0 hoặc P (0) = 1.
Nếu P (0) = 0 thì P(x) = x
m
Q(x) với m nguyên dương và
Q(x) là đa thức hệ số thực thỏa Q (0) ≠ 0 ………………………………………………… (a)
Khi ... b 3 4a b≥ −+ +
⇔
( )
2
22 22 22
ab 6(ab)4ab 80 + − + +≤
⇒
( )
2
2 2 22 2 2 2 22
6b 6a 4a b 8 a b 6a 4a b 8 8≥+ ++− ≥+ +≥
⇒ b
2
≥
4
3
⇒
2
22
16a
4a b
3
≥
⇒ 6b
2
≥ 6a
2
+8 + 4a
2
b
2
...
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