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DanielleElmas [232]
3 years ago
5

All the following are graphic file formats exceptHTMLBMPpngTiff​

Computers and Technology
1 answer:
kolbaska11 [484]3 years ago
6 0

Answer:

HTML

Explanation:

HTML is not a graphic file, as it doesn't support graphics. It instead supports text-base, and only text.

~

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The conventional wisdom concerning the security frameworks of domains is that it is always preferable for an organization to cre
Valentin [98]

Answer:

False

Explanation:

A Domain Based Security. "Domain Based Security", abbreviated to "DBSy", is a model-based approach that is being used in analyzing information security risks in a business context and offers an apparent, understandable and direct mapping among the risks and the security controls needed to handle them.

While A security domain is the list of items a subject is permitted to access. More largely defined, domains are collections of subjects and objects with related security requirements.

6 0
4 years ago
Write a program that computes the monthly net pay of the employee for a steel factory. The input for this program is the hourly
Mars2501 [29]

Answer:

#include<stdio.h>

int main()

{

float rate_of_pay,regular_hours,overtime_hours,grosspay,netpay,tax;

printf("Enter the Hourly rate of pay : ");

scanf("%f",&rate_of_pay);

printf("Enter the number of Regular hours : ");

scanf("%f",&regular_hours);

printf("Enter the number of Overtime hours : ");

scanf("%f",&overtime_hours);

grosspay=(regular_hours*rate_of_pay)+(1.5*overtime_hours*rate_of_pay);

netpay=grosspay-(grosspay*0.2);

printf("Employee's Gross pay = %f\n",grosspay);

printf("Tax = %f\n",0.2*grosspay);

printf("Employee's Net pay = %f\n",netpay);

return 0;

}

Note: The variables are declared as float, to support partial hours like 0.5,6.5 etc.

Explanation:

6 0
3 years ago
What is the trade-offs in time complexity between an ArrayList and a LinkedList?
AURORKA [14]

Answer:

  In the time complexity, the array-list can easily be accessible any type of element in the the given list in the fixed amount of time.

On the other hand, the linked list basically require that the list must be traversed from one position to another end position.

The Array-List can get to any component of the rundown in a similar measure of time if the file value is know, while the Linked-List requires the rundown to be crossed from one end or the other to arrive at a position.

4 0
3 years ago
You would like the user of a program to enter a customer’s last name. Write a statement thaUse the variables k, d, and s so that
mojhsa [17]

Answer:

1st question:

Use the variables k, d, and s so that they can read three different values from standard input an integer, a float, and a string respectively. On one line, print these variables in reverse order with exactly one space in between each. On a second line, print them in the original order with one space in between them.

Solution:

In Python:

k = input()  #prompts user to input value of k i.e. integer value

d = input()  #prompts user to input value of d i.e. float value

s = input()  #prompts user to input value of s i.e. a string

print (s, d, k)  #displays these variable values in reverse order

print (k, d, s)#displays these variable values in original order

In C++:

#include <iostream>    // to use input output functions

using namespace std;   //to identify objects like cin cout

int main() {    //start of main function

  int k;   //declare int type variable to store an integer value

  float d; //  declare float type variable to store a float value

  string s;   //  declare string type variable to store an integer value

  cin >> k >> d >> s;    //reads the value of k, d and s

  cout << s << " " << d << " " << k << endl;     //displays these variables values in reverse order

  cout << k << " " << d << " " << s << endl;   } // displays these variable values in original order

Explanation:

2nd question:

You would like the user of a program to enter a customer’s last name. Write a statement that asks user "Last Name:" and assigns input to a string variable called last_name.

Solution:

In Python:

last_name = input("Last Name:")

# input function is used to accept input from user and assign the input value to last_name variable

In C++:

string last_name;  //declares a string type variable named last_name

cout<<"Last Name: ";  // prompts user to enter last name by displaying this message Last Name:

cin>>last_name; // reads and assigns the input value to string variable last_name

The programs alongwith their outputs are attached.

6 0
4 years ago
Write a procedure named Str_find that searches for the first matching occurrence of a source string inside a target string and r
kirill115 [55]

Answer: Provided in the explanation section

Explanation:

Str_find PROTO, pTarget:PTR BYTE, pSource:PTR BYTE

.data

target BYTE "01ABAAAAAABABCC45ABC9012",0

source BYTE "AAABA",0

str1 BYTE "Source string found at position ",0

str2 BYTE " in Target string (counting from zero).",0Ah,0Ah,0Dh,0

str3 BYTE "Unable to find Source string in Target string.",0Ah,0Ah,0Dh,0

stop DWORD ?

lenTarget DWORD ?

lenSource DWORD ?

position DWORD ?

.code

main PROC

  INVOKE Str_find,ADDR target, ADDR source

  mov position,eax

  jz wasfound           ; ZF=1 indicates string found

  mov edx,OFFSET str3   ; string not found

  call WriteString

  jmp   quit

wasfound:                   ; display message

  mov edx,OFFSET str1

  call WriteString

  mov eax,position       ; write position value

  call WriteDec

  mov edx,OFFSET str2

  call WriteString

quit:

  exit

main ENDP

;--------------------------------------------------------

Str_find PROC, pTarget:PTR BYTE, ;PTR to Target string

pSource:PTR BYTE ;PTR to Source string

;

; Searches for the first matching occurrence of a source

; string inside a target string.

; Receives: pointer to the source string and a pointer

;    to the target string.

; Returns: If a match is found, ZF=1 and EAX points to

; the offset of the match in the target string.

; IF ZF=0, no match was found.

;--------------------------------------------------------

  INVOKE Str_length,pTarget   ; get length of target

  mov lenTarget,eax

  INVOKE Str_length,pSource   ; get length of source

  mov lenSource,eax

  mov edi,OFFSET target       ; point to target

  mov esi,OFFSET source       ; point to source

; Compute place in target to stop search

  mov eax,edi    ; stop = (offset target)

  add eax,lenTarget    ; + (length of target)

  sub eax,lenSource    ; - (length of source)

  inc eax    ; + 1

  mov stop,eax           ; save the stopping position

; Compare source string to current target

  cld

  mov ecx,lenSource    ; length of source string

L1:

  pushad

  repe cmpsb           ; compare all bytes

  popad

  je found           ; if found, exit now

  inc edi               ; move to next target position

  cmp edi,stop           ; has EDI reached stop position?

  jae notfound           ; yes: exit

  jmp L1               ; not: continue loop

notfound:                   ; string not found

  or eax,1           ; ZF=0 indicates failure

  jmp done

found:                   ; string found

  mov eax,edi           ; compute position in target of find

  sub eax,pTarget

  cmp eax,eax    ; ZF=1 indicates success

done:

  ret

Str_find ENDP

END main

cheers i hoped this helped !!

6 0
3 years ago
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