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3 years ago

What method of the Math object can be used to return the largest value from the values that are passed to it?

Computers and Technology

1 answer:

miss Akunina [59]3 years ago

7 0

Answer:

the Math.max() method

Explanation:

The java.lang.* package contains all the essential components needed for a JAVA program which involves numbers and all the operations done on the numericals.

The java.lang.math class contains all the methods needed to perform mathematical operations on numerical values.

All the methods in the question are discussed below with examples.

Math.max() method

java.lang.Math.max(double/ int/ float a, double/ int/ float b)

This method returns the highest value from among the parameters passed to it.

The syntax for the above is

Numbers can be integers, float and double or long. The return type is the same as the parameters.

Two values to be compared are passed as arguments and the higher value of them is returned.

This function also accepts negative integers as parameters.

Math.max(-9, -2);

Gives -2 as the output.

Math.max(0.34, 0.45)

Gives 0.45 as the output.

Math.ceil() method

java.lang.Math.ceil(double/ int/ float a)

This method returns the nearest greater whole number for the given number.

Integer parameter returns an integer number.

Math.ceil(7);

Gives 7 as the output.

Decimal number will return the nearest higher integer.

Math.ceil(8.7);

Gives 9.0 as the output.

Math.ceil(-0.34)

Gives -0 as the output.

Math.pow() method

java.lang.Math.pow(double/ float/ int a, double/ float/ int b)

The above method returns the result of a raise to the power of b.

The first parameter is the base while the second parameter forms the exponent.

Math.pow( 2.3, 4.5 );

This calculates (2.3)^4.5.

Gives 42.44‬ as the output.

Math.pow( 2, 3 );

This calculates the value of 2^3.

Gives 8 as the output.

Math.abs() method

java.lang.Math.abs(double a)

This method returns non-negative value of the numerical parameter passed to it. Absolute of a number does not changes its value but changes the sign of a negative number. It returns the positive or negation of the negative input number.

Math.abs(-0.1234)

Gives 0.1234 as the output.

Math.abs(12.34)

Gives 12.34 as the output.

These methods require java.lang.Math to be imported in the program.

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Answer:

Here is the JAVA program:

import java.util.Scanner; //to accept input from user

public class Main { // class name

public static void main(String [] args) { // start of main method

Scanner input = new Scanner(System.in); // creates Scanner class object to take input from user

System.out.println("Please Enter the Cost of the Item: "); // prompts user to enter the cost of item

double itemCost = input.nextDouble(); //scans and reads the input value of item cost

System.out.println("Please Enter the Amount Paid: "); // prompts user to enter the amount paid

double amount = input.nextDouble(); //scans and reads the value of amount from user

int change = (int)(amount * 100 - itemCost * 100); //compute the remaining amount i.e. change

System.out.println("Change Owed: " + change / 100.0); // displays the owed change

int quarters = change / 25; // computes the value for quarters

change = change % 25; // compute the quarters remaining

int dimes = change / 10; // computes dimes

change = change % 10; // computes dimes remaining

int nickels = change / 5; // computes nickels

change = change % 5; // compute nickels remaining

int pennies = change; // computes pennies

System.out.println("Quarters: " + quarters); // displays computed value of quarters

System.out.println("Dimes: " + dimes); // displays value of dimes

System.out.println("Nickels: " + nickels); // displays value of nickels

System.out.println("Pennies: " + pennies); }} //displays value of pennies

Explanation:

I will explain the program with an examples.

Suppose the user enters 4.57 as cost of the item and 5.00 as amount paid. Then the program works as follows:

Change is computed as

change = (int)(amount * 100 - itemCost * 100);

This becomes;

change = (int)(5.00 * 100 - 4.57 * 100)

= 500 - 457

change = 43

Now the change owed is computed as:

change / 100.0 = 43/100.0 = 0.43

Hence change owed = 0.43

Next quarters are computed as:

quarters = change / 25;

This becomes:

quarters = 43/25

quarters = 1

Now the remaining is computed as:

change = change % 25;

This becomes:

change = 43 % 25;

change = 18

Next the dimes are computed from remaining value of change as:

dimes = change / 10;

dimes = 18 / 10

dimes = 1

Now the remaining is computed as:

change = change % 10;

This becomes:

change = 18 % 10

change = 8

Next the nickels are computed from remaining value of change as:

nickels = change / 5;

nickels = 8 / 5

nickels = 1

Now the remaining is computed as:

change = change % 5;

This becomes:

change = 8 % 5

change = 3

At last the pennies are computed as:

pennies = change;

pennies = 3

So the output of the entire program is:

Change Owed: 0.43 Quarters: 1 Dimes: 1 Nickels: 1 Pennies: 3

The screenshot of the output is attached.

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(Display four patterns using loops) Ask the user to enter an integer to

fomenos

Answer:

Hi There was small mistake. It is working fine for me. When you run from command line - use LoopPattern, not looppatern

import java.util.Scanner;

public class Looppattern {

public static void main(String[] args) {

Scanner sc = new Scanner(System.in);

System.out.println("Enter how man levels you need: ");

int levels = sc.nextInt();

System.out.println("\n---------------Pattern A-----------------\n");

for (int p = 1; p <= levels; p++) {

for (int k = 1; k <= p; k++) { // increasing each level printing

System.out.print(k);

}

System.out.println();

}

System.out.println("\n---------------Pattern B-----------------\n");

int r = levels;

for (int p = 1; p <= levels; p++) {

for (int k = 1; k <= r; k++) {

System.out.print(k);

}

r--; // decreasing levels

System.out.println();

}

System.out.println("\n---------------Pattern C-----------------\n");

for (int p = 1; p <= levels; p++) { // here incresing

for (int k = p; k > 0; k--) { // and here decreasing pattern to

// achieve our required pattern

System.out.print(k);

}

System.out.println();

}

System.out.println("\n---------------Pattern D-----------------\n");

r = levels;

for (int p = 1; p <= levels; p++) {

for (int k = 1; k <= r; k++) {

System.out.print(k);

}

r--; // decreasing levels

System.out.println();

}

Explanation:

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