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barxatty [35]
2 years ago
5

Can you make a phone app with angular?.

Computers and Technology
1 answer:
emmainna [20.7K]2 years ago
4 0

Answer:

yes

Explanation:

You might be interested in
The int function can convert floating-point values to integers, and it performs rounding up/down as needed.
omeli [17]

Answer:

Thks is true the int function changes a float value to an integer and will round up or down by default.

5 0
2 years ago
Assume that cell F5 to F10 In a spreadsheet contains numeric value of salary earned by some workers and cell F12 contains the va
snow_tiger [21]

Answer:

=F5*$F$12+F5

Explanation:

If we want to increment the salaries in the cell F5, we must multiply the cell F5 by cell F12, and then we must sum that result.

If we want to drag the formula from the cell F5 to F10, we must use the dollar symbol $ to apply the same percent in our formula.

For example:

F12 = 5% = 0.05

F5 = 10,000

=F5*$F$12+F5

=10,000×0.05+10,000 = 10,500

5 0
3 years ago
Given a floating-point formal with a k-bit exponent and an n-bit (fraction, write formulas for the exponent E, significant M, th
ANEK [815]

Answer:

A) Describe the number 7.0 bit

The exponential value ( E ) = 2

while the significand value ( M ) = 1.112  ≈ 7/4

fractional value ( F )  = 0.112

And, numeric value of the quantity ( V )  = 7

The exponent bits will be represented  as :  100----01.

while The fraction bits will be represented  as : 1100---0.

<u>B) The largest odd integer that can be represented exactly </u>

The integer will have its exponential value ( E ) = n

hence the significand value ( M )

=  1.11------12 = 2 - 2-n

also the fractional value ( F ) =  

0.11------12 = 1 – 2-n

Also, Value, V = 2n+1 – 1

The exponent bits  will be represented  as follows:  n + 2k-1 – 1.

while The bit representation for the fraction will be as follows: 11---11.

<u>C) The reciprocal of the smallest positive normalized value </u>

The numerical value of the equity ( V ) = 22k-1-2

The exponential value ( E )  = 2k-1 – 2

While the significand value ( M )  = 1

also the fractional value ( F ) = 0

Hence The bit representation of the exponent will be represented as : 11---------101.

while The bit representation of the fraction will be represented as : 00-----00.

Explanation:

E = integer value of exponent

M = significand value

F = fractional value

V = numeric value of quantity

A) Describe the number 7.0 bit

The exponential value ( E ) = 2

while the significand value ( M ) = 1.112  ≈ 7/4

fractional value ( F )  = 0.112

And, numeric value of the quantity ( V )  = 7

The exponent bits will be represented  as :  100----01.

while The fraction bits will be represented  as : 1100---0.

<u>B) The largest odd integer that can be represented exactly </u>

The integer will have its exponential value ( E ) = n

hence the significand value ( M )

=  1.11------12 = 2 - 2-n

also the fractional value ( F ) =  

0.11------12 = 1 – 2-n

Also, Value, V = 2n+1 – 1

The exponent bits  will be represented  as follows:  n + 2k-1 – 1.

while The bit representation for the fraction will be as follows: 11---11.

<u>C) The reciprocal of the smallest positive normalized value </u>

The numerical value of the equity ( V ) = 22k-1-2

The exponential value ( E )  = 2k-1 – 2

While the significand value ( M )  = 1

also the fractional value ( F ) = 0

Hence The bit representation of the exponent will be represented as : 11---------101.

while The bit representation of the fraction will be represented as : 00-----00.

8 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
What does foreshortening affect
AlexFokin [52]
Foreshortening is the visual effect or optical illusion that causes an object or distance to appear shorter than it actually is because it is angled toward the viewer. Additionally, an object is often not scaled evenly: a circle often appears as an ellipse and a square can appear as a trapezoid.
3 0
3 years ago
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