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Nikolay [14]
2 years ago
13

Agape was asked to make a recommendation regarding short-range wireless technologies to be supported in a new conference room th

at was being renovated. Which of the following would she NOT consider due to its slow speed and its low deployment levels today?A. ANTB. BluetoothC. InfraredD. NFC
Computers and Technology
1 answer:
Bess [88]2 years ago
5 0

Answer:

C. Infrared

Explanation:

According to my experience in the field of information technology, I can say that the one technology that she would not consider is infrared. Even though infrared is used in for many applications such as remote controls and sensors it is not that good for wireless communications. This is because compared to the other options available, infrared is only short range, the speeds at which it can transmit are very low, and any physical object can immediately block the signal and cause interference.

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Which type of keyword is "capital"?
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Answer:

a vocabulary term

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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: ???? = (??????
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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
2 years ago
Consider the following scenario. The Current State of One Particular ArrayList The capacity of a particular ArrayList is five (5
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Answer:

The ArrayList initial functionality is 5, there were 4 components.

Then, add 25 more elements by using add()calling 25 times So total number of elements after the completed operation would be 4 + 25= 29.

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So ArrayList's capacity will be 11.

Next 7th to 11th components will also be added automatically once the 12th element is added, it will name ensureCapacity(2* 11+ 1) and make the capacity 23.

While adding 24th dimension, ensureCapacity(2* 23+ 1) will be called again and this will make the capacity 47.

There will therefore be 3 (three) calls to ensure the feature Ability).  

Elements of Rest 25th to 29th will be added seamlessly as ability becomes 47.

3 0
3 years ago
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