Drag and drop the git commands from the left into the correct order on the right to create a feature branch from the master and then incorporate that feature branch into the master.
Answer:
Explanation:
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NEW QUESTION: 1
An administrator issues the command: MOVE DRMEDIA *
WHERESTATE=VAULTRETRIEVE. None of the tapes that were in VAULTRETRIEVE
state have been returned to scratch. To which state should the tapes be moved to be recalled?
A. VAULT
B. COURIER
C. NOTMOUNTABLE
D. ONSITERETRIEVE
Answer: C
Explanation:
NOTMOuntable
These volumes are onsite, contain valid data, and are not available for onsite processing.
The values change to COURIER if the TOSTATE parameter is not specified.
Note: VAULTRetrieve
These volumes are at the offsite vault and do not contain valid data.The values change to COURIERRETRIEVE if the TOSTATE parameter is not specified.
NEW QUESTION: 2
Drag and drop the git commands from the left into the correct order on the right to create a feature branch from the master and then incorporate that feature branch into the master.
Answer:
Explanation:
NEW QUESTION: 3
CORRECT TEXT
You have been tasked with configuring multilayer SwitchC, which has a partial configuration and has been attached to RouterC asshown in the topology diagram.
You need to configure SwitchC so that Hosts H1 and H2 can successfully ping the server S1. Also SwitchC needs to be able to ping server S1.
Due to administrative restrictions and requirements you should not add/delete vlansor create trunk links. Company policies forbid the use of static or default routing. All routes must be learned via EIGRP 65010 routing protocol.
You do not have access to RouteC. RouterC is correctly configured. No trunking has been configured on RouterC.
Routed interfaces should use the lowest host on a subnet when possible. The following subnets are available to implement this solution:
- 10.10.10.0/24
- 190.200.250.32/27
- 190.200.250.64/27
Hosts H1 and H2 are configured with the correct IP addressand default gateway.
SwitchC uses Cisco as the enable password.
Routing must only be enabled for the specific subnets shown in the diagram.
Note: Due to administrative restrictions and requirements you should not add or delete
VLANs, changes VLAN port assignments or create trunks. Company policies forbid the use of static or default routing. All routes must be learned via the EIGRP routing protocol.
Answer:
Explanation:
There are two ways to configure interVLAN routing in this case:
+ Use RouterC as a"router on a stick" and SwitchC as a pure Layer2 switch. Trunking must be established between RouterC and SwitchC.
+ Only use SwitchC for interVLAN routing without using RouterC, SwitchC should be configured as a Layer 3 switch (which supports ip routing function as a router). No trunking requires.
The question clearly states "No trunking has been configured on RouterC" so RouterC does not contribute to interVLAN routing of hosts H1 & H2 -> SwitchC must be configured as a Layer 3 switch with SVIs for interVLAN routing.
We should check the default gateways on H1 & H2. Click on H1 and H2 and type the
"ipconfig" command to get their default gateways.
TheAnswer: \>ipconfig
We will get the default gateways as follows:
Host1:
+ Default gateway: 190.200.250.33
Host2:
+ Default gateway: 190.200.250.65
Now we have enough information to configure SwitchC (notice the EIGRP AS in this case is 650)
Note: VLAN2 and VLAN3 were created and gi0/10, gi0/11 interfaces were configured as access ports so we don't need to configure them in this sim.
SwitchC# configure terminal
SwitchC(config)# int gi0/1
SwitchC(config-if)#no switchport -> without using this command, the simulator does not let you assign IP address on Gi0/1 interface.
SwitchC(config-if)# ip address 10.10.10.2 255.255.255.0 ->RouterC has used IP
10.10.10.1 so this is the lowest usable IP address.
SwitchC(config-if)# no shutdown
SwitchC(config-if)# exit
SwitchC(config)# int vlan 2
SwitchC(config-if)# ip address 190.200.250.33 255.255.255.224
SwitchC(config-if)# no shutdown
SwitchC(config-if)# int vlan 3
SwitchC(config-if)# ip address 190.200.250.65 255.255.255.224
SwitchC(config-if)# no shutdown
SwitchC(config-if)#exit
SwitchC(config)# ip routing (Notice: MLS will not work without this command)
SwitchC(config)# router eigrp65010
SwitchC(config-router)# network 10.10.10.0 0.0.0.255
SwitchC(config-router)# network 190.200.250.32 0.0.0.31
SwitchC(config-router)# network 190.200.250.64 0.0.0.31
NOTE: THE ROUTER IS CORRECTLY CONFIGURED, so you will not miss within it in the exam,also don't modify/delete any port just do the above configuration. Also some reports said the "no auto-summary" command can't be used in the simulator, in fact it is not necessary because the network 190.200.0.0/16 is not used anywhere else in this topology.
In order to complete the lab, you should expect the ping to SERVER to succeed from the
MLS, and from the PCs as well.
Also make sure you use the correct EIGRP AS number (in the configuration above it is 650 but it will change when you take the exam) but we are not allowed to access RouterC so the only way to find out the EIGRP AS is to look at the exhibit above. If you use wrong AS number, no neighbor relationship is formed between RouterC and SwitchC.
In fact, we are pretty sure instead of using two commands "network 190.200.250.32
0.0.0.31 and "network 190.200.250.64 0.0.0.31 we can use one simple command "network
190.200.0.0 because it is the nature of distance vector routing protocol like EIGRP: only major networks need to be advertised; even without "no auto-summary" command the network still works correctly. But in the exam the sim is just a flash based simulator so we should use two above commands, just for sure. But after finishing the configuration, we can use "show run" command to verify, only the summarized network 190.200.0.0 is shown.
NEW QUESTION: 4
Which of the following values of x would satisfy the inequality x > 1?
I) x = 1 ½ 23
II) x = 1 -4/3 22
III) x = 1 -1/3 2-2
A. II only
B. I, II, and III
C. I only
D. II and III only
E. I and III only
Answer: D
Explanation:
Explanation/Reference:
Explanation:
Statement I simplifies to 1/8, which is less than 1. Statement II simplifies to 16/9, which is greater than 1. In statement III, you need to take the reciprocal of the fraction inside the parentheses (because the exponent is negative) and then evaluate using an exponent of 2. This results in (-3)2 = 9, which is also greater than
1. Both statements II and III would satisfy the inequality x > 1.