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cupoosta [38]
2 years ago
9

What is the password based off of the clues

Computers and Technology
1 answer:
nevsk [136]2 years ago
3 0

Answer:

i would say Plane, if not then Plane then the person's birthdate.

Explanation:

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Colours can be contrasting if<br>​
abruzzese [7]

.....If the colors are from different segments of the color wheel.

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2 years ago
An access control system that grants users only those rights necessary for them to perform their work is operating on which secu
solmaris [256]

Answer:

B. Least privilege                

Explanation:

  • The principle of least privilege an important principle in computer security.
  • It limits the access rights for users and only grant them with the rights that are sufficient for them to perform their required task.
  • For example a user is granted privilege to execute a file or manipulate data or use only the resources that are required for them to perform a particular task.
  • This principle can be used only to limit and control access rights to system resources and applications.
  • The least privilege is beneficial as it reduces the risk of unauthorized access.
  • For example a user whose task is data entry and has nothing to do with controlling access or granting access rights to users, will only be allowed to enter data to the DB by the principle of least privilege.
7 0
4 years ago
Which term refers to the data stored in computer memory on which the processor applies to program instructions
Fed [463]
The answer is B.) variable.
3 0
3 years ago
How do I persuade my parents to allow me to have social media? I will be 15 in a week or so, so I am old enough. I'm pulling my
Flauer [41]

Answer:

my explanation is above my comment :)

Explanation:

6 0
3 years ago
Translate the following MIPS code to C. Assume that the variables f, g, h, i, and j are assigned to registers $s0, $s1, $s2, $s3
Romashka [77]

Answer:

f = 2 * (&A[0])

See explaination for the details.

Explanation:

The registers $s0, $s1, $s2, $s3, and $s4 have values of the variables f, g, h, i, and j respectively. The register $s6 stores the base address of the array A and the register $s7 stores the base address of the array B. The given MIPS code can be converted into the C code as follows:

The first instruction addi $t0, $s6, 4 adding 4 to the base address of the array A and stores it into the register $t0.

Explanation:

If 4 is added to the base address of the array A, then it becomes the address of the second element of the array A i.e., &A[1] and address of A[1] is stored into the register $t0.

C statement:

$t0 = $s6 + 4

$t0 = &A[1]

The second instruction add $t1, $s6, $0 adding the value of the register $0 i.e., 32 0’s to the base address of the array A and stores the result into the register $t1.

Explanation:

Adding 32 0’s into the base address of the array A does not change the base address. The base address of the array i.e., &A[0] is stored into the register $t1.

C statement:

$t1 = $s6 + $0

$t1 = $s6

$t1 = &A[0]

The third instruction sw $t1, 0($t0) stores the value of the register $t1 into the memory address (0 + $t0).

Explanation:

The register $t0 has the address of the second element of the array A (A[1]) and adding 0 to this address will make it to point to the second element of the array i.e., A[1].

C statement:

($t0 + 0) = A[1]

A[1] = $t1

A[1] = &A[0]

The fourth instruction lw $t0, 0($t0) load the value at the address ($t0 + 0) into the register $t0.

Explanation:

The memory address ($t0 + 0) has the value stored at the address of the second element of the array i.e., A[1] and it is loaded into the register $t0.

C statement:

$t0 = ($t0 + 0)

$t0 = A[1]

$t0 = &A[0]

The fifth instruction add $s0, $t1, $t0 adds the value of the registers $t1 and $t0 and stores the result into the register $s0.

Explanation:

The register $s0 has the value of the variable f. The addition of the values stored in the regsters $t0 and $t1 will be assigned to the variable f.

C statement:

$s0 = $t1 + $t0

$s0 = &A[0] + &A[0]

f = 2 * (&A[0])

The final C code corresponding to the MIPS code will be f = 2 * (&A[0]) or f = 2 * A where A is the base address of the array.

3 0
3 years ago
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