Answer:
So these are the RTL representation:
MAR<----X
MBR<-----AC
M[MAR]<------MBR
Control signal sequence are:
P3T0:MAR<----X
P2P3 P4T1:MBR<-----AC
P0P1 P3T2:M[MAR]<------MBR
Explanation:
STORE X instruction is used for storing the value of AC to the memory address pointed by X. This operation can be done by using the Register Transfers at System Level and this can be represented by using a notation called Register Transfer Language RTL. Let us see what are the register transfer operations happening at the system level.
1. First of all the address X has to be tranfered on to the Memory Address Register MAR.
MAR<----X
2. Next we have to tranfer the contents of AC into the Memory Buffer Register MBR
MBR<-----AC
3. Store the MBR into memory where MAR points to.
M[MAR]<------MBR
So these are the RTL representation:
MAR<----X
MBR<-----AC
M[MAR]<------MBR
Control signal sequence are:
P3T0:MAR<----X
P2P3 P4T1:MBR<-----AC
P0P1 P3T2:M[MAR]<------MBR
Answer:
import java.util.Scanner;
public class InputExample {
public static void main(String[] args) {
Scanner scnr = new Scanner(System.in);
int birthMonth;
int birthYear;
birthMonth = scnr.nextInt();
birthYear = scnr.nextInt();
System.out.println(birthMonth+"/"+birthYear);
}
}
import java.util.Scanner;
public class InputExample {
public static void main(String[] args) {
Scanner scnr = new Scanner(System.in);
int birthMonth;
int birthYear;
// Get input values for birthMonth and birthYear from the user
birthMonth = scnr.nextInt();
birthYear = scnr.nextInt();
// Output the month, a slash, and the year
System.out.println(birthMonth + "/" + birthYear);
}
}
When the program is tested with inputs 1 2000, the output will be:
1/2000
And when tested with inputs 5 1950, the output will be:
5/1950
Know more about java program:
#SPJ3
A. The heat transfer rate from natural gas is 2105.26 MW
B. The heat transfer rate to river is 1305.26 MW
Efficiency = (power output / power input) × 100
Power input = Power input / efficiency
Power input = 800 / 38%
Power input = 800 / 0.38
Power input = 2105.26 MW
Thus, the heat transfer from natural gas is 2105.26 MW
Heat to the river = 2105.26 – 800
Heat to the river = 1305.26 MW
Learn more about efficiency:
Answer:
heat transfer from natural gas is 2105.26 MW
heat transfer to river is 1305.26 MW
Explanation:
given data
power output Wn = 800 MW
efficiency = 38%
solution
we know that efficiency is express as
......................1
put here value we get
38% =
Qin = 2105.26 MW
so heat supply is 2105.26
so we can say
Wn = Qin - Qout
800 = 2105.26 - Qout
Qout = 2105.26 - 800
Qout = 1305.26 MW
so heat transfer from natural gas is 2105.26 MW
and heat transfer to river is 1305.26 MW
b.600x10-6V
c.800x10-6V
d.1000 x 10-6 V
Answer:
a.400 x 10-6 V
Explanation:
Kindly check the attached image below to see the step by step explanation to the question above.
Web application 3’s “Where’s the Beef” exploit can be seen in the screenshot provided. In order to mitigate this vulnerability, developers should take measures to ensure that input validation is performed on all data.
Additionally, developers should enforce strict rules on which characters are allowed in user inputs and reject all requests that don’t follow these rules. Finally, developers should implement best practices such as password hashing to ensure that user data is secure.
Developers should use secure coding techniques, such as sanitizing user input and properly escaping HTML output, to protect against injection-based attacks. Also, developers should implement authentication and authorization techniques to ensure that only authorized users have access to sensitive data. Furthermore, developers should use an up-to-date web application firewall to protect against known exploits, and use secure protocols such as HTTPS to protect data in transit. Finally, developers should ensure that software is kept up-to-date and patched to prevent exploitation of known vulnerabilities.
Learn more about HTML
#SPJ11
Answer:
Explanation:
The annual worth of land application = $ 110,000 ( A/P , 10% , 5 years ) + $ 95,000 - $ 15,000 (A/F , 10% , 5 years )
The annual worth of land application = $ 110,000 X 0.263797 + $ 95,000 - $ 15,000 X 0.163797
The annual worth of land application = $ 29,017.54 + $ 95,000 - $ 2,456.95
The annual worth of land application = $ 121,560.59
The annual worth of land Incineration = $ 800,000 ( A/P , 10% , 6 years ) + $ 60,000 - $ 250,000 X (A/F , 10% , 6 years )
The annual worth of land Incineration = $ 800,000 X 0.229607 + $ 60,000 - $ 250,000 X 0.129607
The annual worth of land Incineration = $ 211,283.85
The annual worth of contract = $ 190,000
The annual worth of contract = $ 190,000
The land application has the least cost , hence it is preferred .
Answer:
A primary function of NCEES is to prepare standardized, confidential examinations that are used by the state and territorial boards to help determine the competency of individuals seeking to become licensed to practice as professional engineers and land surveyors.
Explanation:
i looked it up
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