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DedPeter [7]
3 years ago
6

When troubleshooting firewalls, you should never attempt to repeat the problem because you could do more damage. T/F

Computers and Technology
1 answer:
Zinaida [17]3 years ago
8 0

Answer:

TRUE

Explanation:

the given statement is correct hence TRUE

troubleshooting is problem solving often applied to repair failed product .

Firewall is used for network security for the computer is monitors and controls the incoming and outgoing network traffic.

Basically what Firewall do is it create barrier between trusted networks and not trusted network.

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Explanation:

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You are tasked to calculate a specific algebraic expansion, i.e. compute the value of f and g for the expression: ???? = (??????
melisa1 [442]

Answer:

.data

prompt: .asciiz "Enter 4 integers for A, B, C, D respectively:\n"

newLine: .asciiz "\n"

decimal: .asciiz "f_ten = "

binary: .asciiz "f_two = "

decimal2: .asciiz "g_ten = "

binary2: .asciiz "g_two = "

.text

main:

#display prompt

li $v0, 4

la $a0, prompt

syscall

#Read A input in $v0 and store it in $t0

li $v0, 5

syscall

move $t0, $v0

#Read B input in $v0 and store it in $t1

li $v0, 5

syscall

move $t1, $v0

#Read C input in $v0 and store it in $t2

li $v0, 5

syscall

move $t2, $v0

#Read D input in $v0 and store it in $t3

li $v0, 5

syscall

move $t3, $v0

#Finding A^4

#Loop (AxA)

li $t6, 0

L1:

bge $t6, $t0, quit

add $s1, $s1, $t0 # A=S+A => $s1= A^2

addi $t6, $t6, 1 # i=i+1

j L1

quit:

#Loop (A^2 x A^2)

li $t6, 0

L1A:

bge $t6, $s1, quit1A

add $s5, $s5, $s1

addi $t6,$t6, 1

j L1A

#End of Finding A^4

#Finding 4xA^3

quit1A:

#Loop (4xB)

li $t6, 0

L2:

bge $t6, 4, quit2

add $s2, $s2, $t1

addi $t6, $t6, 1

j L2

quit2:

#Loop (BxB)

li $t6 , 0

L2A:

bge $t6, $t1, quit2A #loop2

add $s6, $s6, $t1 #add

addi $t6, $t6, 1 #add immediate

j L2A #loop2

quit2A: # perform proper program termination using syscall for exit

#Loop (BxB)

li $t6 , 0 #load immediate

L2AA:

bge $t6, $s2, quit2AA #loop2

add $t7, $t7, $s6 #add

addi $t6, $t6, 1 #add immediate

j L2AA #loop2

#End ofFinding 4xA^3

#Finding 3xC^2

quit2AA: # perform proper program termination using syscall for exit

#3 Loop (3 x (C x C)) FOR S3

li $t6 , 0 #load immediate

L3:

bge $t6, $t2, quit3 #loop3

add $s3, $s3, $t2 #add

addi $t6,$t6, 1 #add immediate

j L3 #loop3

quit3: # perform proper program termination using syscall for exit

#3 Loop (3 x (C x C)) FOR S3

li $t6 , 0 #load immediate

L3A:

bge $t6, 3, quit3A #loop3

add $s0, $s0, $s3 #add

addi $t6,$t6, 1 #add immediate

j L3A #loop3

#End of Finding 3xC^2

#Finding 2xD

quit3A: # perform proper program termination using syscall for exit

#4 Loop (2 x D) FOR S4

li $t6 , 0

L4:

bge $t6, 2, quit4 #loop4

add $s4, $s4, $t3 #add

addi $t6, $t6, 1 #add immediate

j L4 #Loop4

#End of Finding 2xD

#Finding AxB^2

quit4:

li $t6, 0

li $s1, 0

L5:

bge $t6, $t1, quit5

add $s1, $s1, $t1

addi $t6, $t6, 1

j L5

quit5:

li $t6, 0

li $s2, 0

L6:

bge $t6, $t0, quit6

add $s2, $s2, $s1

addi $t6, $t6, 1

j L6

#End of Finding AxB^2

#Finding C^2XD^3

quit6: #finds C^2

li $t6, 0

li $s1, 0

L7:

bge $t6, $t2, quit7

add $s1, $s1, $t2

addi $t6, $t6, 1

j L7

quit7: #finds D^2

li $t6, 0

li $s6, 0

L8:

bge $t6, $t3, quit8

add $s6, $s6, $t3

addi $t6, $t6, 1

j L8

quit8: #finds D^3

li $t6, 0

li $s7, 0

L9:

bge $t6, $t3, quit9

add $s7, $s7, $s6

addi $t6, $t6, 1

j L9

quit9: #finds C^2XD^3

li $t6, 0

li $s3, 0

L10:

bge $t6, $s1, end

add $s3, $s3, $s7

addi $t6, $t6, 1

j L10

#End of Finding C^2XD^3

end: # perform proper program termination using syscall for exit

#f is $t8

li $t8 , 0

sub $t8, $s5, $t7 # addition

add $t8, $t8, $s0 # subract

sub $t8,$t8, $s4 # subract

#g is $t9

li $t9 , 0

add $t9, $s2, $s3 # addition

#Display

#1st equation

li $v0,4 # display the answer string with syscall having $v0=4

la $a0, decimal # Gives answer in decimal value

syscall # value entered is returned in register $v0

li $v0, 1 # display the answer string with syscall having $v0=1

move $a0, $t8 # moves the value from $a0 into $t8

syscall # value entered is returned in register $v0

li $v0,4 # display the answer string with syscall having $v0=4

la $a0, newLine # puts newLine in between answers

syscall # value entered is returned in register $v0

li $v0,4 # display the answer string with syscall having $v0=4

la $a0, binary # Gives answer in binary

syscall # value entered is returned in register $v0

li $v0, 35

move $a0, $t8 # moves the value from into $a0 from $t8

syscall # value entered is returned in register $v0

li $v0,4 # display the answer string with syscall having $v0=4

la $a0, newLine # puts newLine in between answers

syscall # value entered is returned in register $v0

#2nd equation

li $v0,4 # display the answer string with syscall having $v0=4

la $a0, decimal2 # Gives answer in decimal value

syscall # value entered is returned in register $v0

li $v0, 1 # display the answer string with syscall having $v0=1

move $a0, $t9 # moves the value from $a0 into $t8

syscall # value entered is returned in reg $v0

li $v0,4 # display the answer string with syscall having $v0=4

la $a0, newLine # puts newLine in between answers

syscall # value entered is returned in register $v0

li $v0,4 # display the answer string with syscall having $v0=4

la $a0, binary2 # Gives answer in binary

syscall # value entered is returned in register $v0

li $v0, 35

move $a0, $t9 # moves the value from into $a0 from $t8

syscall # value entered is returned in register $v0

li $v0,4 # display the answer string with syscall having $v0=4

la $a0, newLine # puts newLine in between answers

syscall # value entered is returned in register $v0

#end the program

li $v0, 10

syscall

8 0
3 years ago
___ is the amount of data that a storage device can move from the storage medium to the computer per second.
Sindrei [870]

Answer:

The answer is "data transfer rate".

Explanation:

The rate of information transfer is the sum of digital information, that travels from one destination to another at a particular time. This can also be viewed as transportation frequency from one place to another one of a given quantity of information. It is the processing power of the larger and specified map, which significantly lowers the data transfer rate.

6 0
4 years ago
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