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C-S43-2022 pdf
  • Exam Code: C-S43-2022
  • Exam Name: SAP Certified Application Associate - SAP S/4HANA Asset Management
  • Version: V17.95
  • Q & A: 400 Questions and Answers
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  • Exam Code: C-S43-2022
  • Exam Name: SAP Certified Application Associate - SAP S/4HANA Asset Management
  • Version: V17.95
  • Q & A: 400 Questions and Answers
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  • Exam Code: C-S43-2022
  • Exam Name: SAP Certified Application Associate - SAP S/4HANA Asset Management
  • Version: V17.95
  • Q & A: 400 Questions and Answers
  • Software Price: $51.98
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NEW QUESTION: 1
There are times when you want to use Link Selection to manage high-traffic VPN connections.
With Link Selection you can:
A. Assign links to specific VPN communities.
B. Use links based on services.
C. Prohibit Dynamic DNS.
D. Assign links to use Dynamic DNS.
Answer: B

NEW QUESTION: 2
Refer to the exhibit.

A user connects a device to GigabitEthernet 1/0/1 on switch. The device does not have an 802.1X supplicant. One minute passes. How does the switch port handle the client?
A. It blocks the client
B. It assigns the client to VLAN3
C. It assigns the client to VLAN4
D. It assigns the client to VLAN2
Answer: C
Explanation:
Explanation/Reference:
VLAN4 is configured to be a guest vlan.
Reference: ProCurve Network Security solutions, Guest VLANs: the interim solution
http://www.hp.com/rnd/pdf_html/guest_vlan_paper.htm#interim_solution

NEW QUESTION: 3
A network engineer needs to configure load balancing for an FTP site.
Which type of session persistence method can the engineer select for this scenario?
A. Custom Server ID
B. Cookie Insert
C. Rule
D. Source IP
Answer: D

NEW QUESTION: 4
You need to display the first names of all customers from the CUSTOMERS table that contain the character 'e' and have the character 'a' in the second last position.
Which query would give the required output?
A. SELECT cust_first_name FROM customers WHERE INSTR(cust_first_name, 'e')IS NOT NULL AND SUBSTR(cust_first_name, 1,-2)='a';
B. SELECT cust_first_name FROM customers WHERE INSTR(cust_first_name, 'e')<>0 AND SUBSTR(cust_first_name, -2, 1)='a';
C. SELECT cust_first_name FROM customers WHERE INSTR(cust_first_name, 'e')<>0 AND SUBSTR(cust_first_name, LENGTH(cust_first_name),-2)='a';
D. SELECT cust_first_name FROM customers WHERE INSTR(cust_first_name, 'e')<>'' AND SUBSTR(cust_first_name, -2, 1)='a';
Answer: B
Explanation:
The SUBSTR(string, start position, number of characters) function accepts three
parameters and returns a string consisting of the number of characters extracted from the
source string, beginning at the specified start position:
substr('http://www.domain.com',12,6) = domain
The position at which the first character of the returned string begins.
When position is 0 (zero), then it is treated as 1.
When position is positive, then the function counts from the beginning of string to find the
first character.
When position is negative, then the function counts backward from the end of string.
substring_length
The length of the returned string. SUBSTR calculates lengths using characters as defined
by the input character set. SUBSTRB uses bytes instead of characters. SUBSTRC uses
Unicode complete characters.
SUBSTR2 uses UCS2 code points. SUBSTR4 uses UCS4 code points.
When you do not specify a value for this argument, then the function
The INSTR(source string, search item, [start position],[nth occurrence of search item])
function returns a number that represents the position in the source string, beginning from
the given start position, where the nth occurrence of the search item begins:
instr('http://www.domain.com','.',1,2) = 18

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