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sasho [114]
3 years ago
9

It is not possible to use keywords to search for themes. true or false.

Computers and Technology
1 answer:
Drupady [299]3 years ago
3 0
False my dude .........................
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Select the word or phrase from the drop-down menu to complete each sentence. File names consist of a ______ and a file extension
Law Incorporation [45]

Answer:

base file name, file format, bitmap image

Explanation:

8 0
3 years ago
10) ________ objects control the flow of the application. A) Boundary B) Utility C) Control D) All of the above
Cerrena [4.2K]

There are different kinds of applications.  Control objects control the flow of the application.

<h3>What works the flow of control in a program?</h3>

In computer, control flow or flow of control is known to be a type of an order function calls, instructions, and statements.

They are known to be used in the execution or in an evaluation when a specific program is running. Note that a lot of programming languages have the control flow statements, that helps to know the section of code is run in a program at any given time.

Learn more about application from

brainly.com/question/23275071

7 0
2 years ago
Which of the following documents should beprepared before the commencement of a softwareproject?
ludmilkaskok [199]

Answer:

Software requirement specification

Explanation:

requirement specification is the first step before proceeding with any project.

we should validate our output at all phases with software requirement specification

7 0
3 years ago
Develop a plan to design and finally implement a set of functions using C++ that would implement the IEEE standard. Phase 1 will
AnnZ [28]

Answer:

See explaination code

Explanation:

#include<iostream>

#include<math.h>

using namespace std;

typedef union {

float number;

struct

{

// Order is important.

// Here the members of the union data structure

// use the same memory (32 bits).

// The ordering is taken

// from the LSB to the MSB.

unsigned int mantissa : 23;

unsigned int exponent : 8;

unsigned int sign : 1;

} Raw;

} MyFloat;

void printBinary(int n, int i)

{

// Prints the binary representation

// of a number n up to i-bits.

int k;

for (k = i - 1; k >= 0; k--) {

if ((n >> k) & 1)

cout << "1";

else

cout << "0";

}

}

void decToHex(int n){

// char array to store hexadecimal number

char hexaDeciNum[100];

// counter for hexadecimal number array

int i = 0;

while(n!=0)

{

// temporary variable to store remainder

int temp = 0;

// storing remainder in temp variable.

temp = n % 16;

// check if temp < 10

if(temp < 10)

{

hexaDeciNum[i] = temp + 48;

i++;

}

else

{

hexaDeciNum[i] = temp + 55;

i++;

}

n = n/16;

}

// printing hexadecimal number array in reverse order

for(int j=i-1; j>=0; j--)

cout << hexaDeciNum[j];

}

void floatBinary(float f){

long double binaryTotal, binaryFrac = 0.0, frac, fracFractor = 0.1;

long int integer, binaryInt = 0;

long int p = 0, rem, temp;

//separate the integer part from the input floating number

integer = (int)f;

//separate the fractional part from the input floating number

frac = f - integer;

//loop to convert integer part to binary

while (integer != 0) {

rem = integer % 2;

binaryInt = binaryInt + rem *pow(10, p);

integer = integer / 2;

p++;

}

//loop to convert fractional part to binary

while (frac != 0) {

frac = frac * 2;

temp = frac;

binaryFrac = binaryFrac + fracFractor * temp;

if (temp == 1)

frac = frac - temp;

fracFractor = fracFractor / 10;

}

cout << binaryInt + binaryFrac;

}

int findDecimal(float number){

int nfloor = number;

float nfloat = number - nfloor;

int nfloatfloor;

do {

nfloat *= 10;

nfloatfloor = nfloat;

} while (nfloat > nfloatfloor);

return nfloatfloor;

}

void first(float number){

if(number < 0)

cout << "SIGN BIT IS (1) SINCE NUMBER IS NEGATIVE" << endl;

else

cout << "SIGN BIT IS (0) SINCE NUMBER IS POSITIVE" << endl;

}

void second(float number){

cout << "INTEGER PART IN BASE-10:" << int(number) << " AND IN BINARY:";

printBinary(int(number),16);

cout << endl;

}

void third(float number){

cout << "DECIMAL PART IN BASE-10:" << findDecimal(number) << " AND IN BINARY:";

printBinary(findDecimal(number),16);

cout << endl;

}

void fourth(float number){

cout << "ENTERED NUMBER IN BASE-10:" << number << " AND IN BINARY:";

floatBinary(number);

cout << endl;

}

void fifth(MyFloat myfloat){

cout << "MANTISA IN BINARY:";

printBinary(myfloat.Raw.mantissa,32);

}

void sixth(MyFloat myfloat){

cout << "EXPONENT IN BASE-10:" << myfloat.Raw.exponent << " AND IN BINARY:";

printBinary(myfloat.Raw.exponent,8);

cout << endl;

}

void seventh(MyFloat myfloat){

cout << myfloat.Raw.sign << " | ";

printBinary(myfloat.Raw.exponent,8);

cout << " | ";

printBinary(myfloat.Raw.mantissa,32);

cout << endl;

}

void eigth(MyFloat myfloat){

cout << myfloat.Raw.sign << " | ";

decToHex(myfloat.Raw.exponent);

cout << " | ";

decToHex(myfloat.Raw.mantissa);

cout << endl;

}

int main(){

float number;

cout << "PLEASE ENTER A NUMBER TO DISPLAY THE IEEE 754 FLOATING POINT OPTIONS" << endl;

cin >> number;

MyFloat myfloat;

myfloat.number = number;

cout << "PLEASE CHOOSE ONE OF THE FOLLOWING OPERATIONS" << endl;

cout << " 1. DISPLAY THE SIGN BIT VALUE" << endl;

cout << " 2. DISPLAY THE INTEER PART IN BOTH BASE-10 AND CINARY FORMATS" << endl;

cout << " 3. DISPLAY THE DECIMAL PART IN BOTH BASE-10 AND BINARY FORMATS" << endl;

cout << " 4. DISPLAY THE NUMBER ENTERED IN BOTH BASE-10 AND BINARY FORMATS" << endl;

cout << " 5. DISPLAY THE MANTISA IN BINARY FORMATS" << endl;

cout << " 6. DISPLAY THE EXPONENT IN BORH BASEE-10 AND BINARY FORMATS" << endl;

cout << " 7. DISPLAY THE IEEE 754 SINGLE PRECISION BINARY LAYOUT" << endl;

cout << " 8. DISPLAY THE IEEE 754 SINGLE PRECISION BINARY LAYOUT" << endl;

int choice;

cin >> choice;

switch(choice){

case 1:first(number);

break;

case 2:second(number);

break;

case 3:third(number);

break;

case 4:fourth(number);

break;

case 5:fifth(myfloat);

break;

case 6:sixth(myfloat);

break;

case 7:seventh(myfloat);

break;

case 8:eigth(myfloat);

break;

default:cout << "ENTER VALID CHOICE" << endl;

}

}

Refer to attachment please for onscreen look.

6 0
3 years ago
Give an example of a function from N to N that is: Hint: try using absolute value, floor, or ceiling for part (b). (a) one-to-on
Naily [24]

Answer:

Let f be a function  

a) f(n) = n²  

b) f(n) = n/2  

c) f(n) = 0  

Explanation:  

a) f(n) = n²  

This function is one-to-one function because the square of two different or distinct natural numbers cannot be equal.  

Let a and b are two elements both belong to N i.e. a ∈ N and b ∈ N. Then:

                               f(a) = f(b) ⇒ a² = b² ⇒ a = b  

The function f(n)= n² is not an onto function because not every natural number is a square of a natural number. This means that there is no other natural number that can be squared to result in that natural number.  For example 2 is a natural numbers but not a perfect square and also 24 is a natural number but not a perfect square.  

b) f(n) = n/2  

The above function example is an onto function because every natural number, let’s say n is a natural number that belongs to N, is the image of 2n. For example:

                               f(2n) = [2n/2] = n  

The above function is not one-to-one function because there are certain different natural numbers that have the same value or image. For example:  

When the value of n=1, then

                                  n/2 = [1/2] = [0.5] = 1  

When the value of n=2 then  

                                   n/2 = [2/2] = [1] = 1  

c) f(n) = 0

The above function is neither one-to-one nor onto. In order to depict that a function is not one-to-one there should be two elements in N having same image and the above example is not one to one because every integer has the same image.  The above function example is also not an onto function because every positive integer is not an image of any natural number.

5 0
3 years ago
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