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denpristay [2]
3 years ago
10

I can't find my friend, brazenhmc. Can anyone help me?

Computers and Technology
2 answers:
frosja888 [35]3 years ago
5 0

Answer:

tell your friend to post a question, then look up the question. That way, you can find him.

Explanation:

Hope i helped! :)

aliina [53]3 years ago
4 0
I can help you with anything right here and always like the other guy said
You might be interested in
Write a computer program that determines how many grades are between 0 and 19.
True [87]

Answer:

public class nnn {

   public static void main(String[] args) {

       int [] examScores = {31, 70, 92, 5, 47, 88, 81, 73, 51, 76, 80, 90, 55, 23, 43,98,36,87,22,61, 19,69,26,82,89,99, 71,59,49,64};

       int zeroTo19 = 0;

       int nineteenTo39 = 0;

       int fortyTo59 = 0;

       int sixtyTo79 = 0;

       int eightyTo100 = 0;

       for(int i =0; i<examScores.length; i++){

           if(examScores[i]<=19){

               zeroTo19++;

           }

           else if(examScores[i]>19&&examScores[i]<=39){

               nineteenTo39++;

           }

           else if(examScores[i]>39&&examScores[i]<=59){

               fortyTo59++;

           }

           else if(examScores[i]>59&&examScores[i]<=79){

               sixtyTo79++;

           }

           else {

               eightyTo100++;

           }

       }

       System.out.println("0 - 19 is "+zeroTo19);

       System.out.println("20 - 39 is "+nineteenTo39);

       System.out.println("40 - 59 is "+fortyTo59);

       System.out.println("60 - 79 is "+sixtyTo79);

       System.out.println("80 - 100 is "+eightyTo100);

   }

}

Explanation:

  • This has been solved with Java
  • Create an array of the exam scores
  • Create new variables for each of the score range
  • Use multiple if statements as u loop through the array to determine the range of scores
  • Finally outside the loop print them variables out
7 0
2 years ago
The function below tries to create a list of integers by reading them from a file. The file is expected to have a single integer
Setler [38]

Answer:

Check the explanation

Explanation:

def get_list_of_integers_from_file(filename):

int_list=[]

for line in open(filename).readlines():

try:

int_list.append(int(line))

except:

continue

return int_list

print(get_list_of_integers_from_file('file.txt'))

 

File.txt:

Kindly check the output below.

4 0
3 years ago
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
I just want a refund please. Thank you
Nina [5.8K]
Unfortunately you can’t be refunded points. Sorry :(
7 0
3 years ago
Set the Append Only property to _____ to allow users to add data to a Long Text field but not to change or remove existing data.
diamong [38]

A computer data storage is made of many features. Set the Append Only property to YES to allow users to add data to a Long Text field but not to change or remove existing data.

  • Append-only is known to be a characteristic or component of computer data storage. It is where new data can be appended to the storage, but where existing data is said to be immutable.

A field's data type is very essential property as it shows what kind of data the field can store.

Learn more from

brainly.com/question/24795103

7 0
3 years ago
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