Question

jump to level 1
compute: $d = (sqrt{b - c})^3$
ex: if the input is 5.00 1.00, then the output is:
8.00
1 import java.util.scanner;
2
3 public class mathmethods {
4 public static void main(string args) {
5 scanner scnr = new scanner(system.in);
6 double b;
7 double c;
8 double d;
9
10 b = scnr.nextdouble();
11 c = scnr.nextdouble();
12
13 / your code goes here /
14
15 system.out.printf(\%.2f\
\, d); // outputs d with 2 decimal places

Explanation:

Step1: Subtract c from b

First, calculate the value inside the square - root, i.e., $x = b - c$.

Step2: Take the square - root

Then, find the square - root of the result from Step 1, $y=\sqrt{x}$.

Step3: Cube the result

Finally, cube the result from Step 2, $d = y^{3}$.

If we assume the input values $b = 5.00$ and $c = 1.00$:

• Step1: $x=b - c=5.00 - 1.00 = 4.00$.
• Step2: $y=\sqrt{4.00}=2.00$.
• Step3: $d = 2.00^{3}=8.00$.

In Java code (filling in the missing part of the given code):

import java.util.Scanner;
public class MathMethods {
    public static void main(String[] args) {
        Scanner scnr = new Scanner(System.in);
        double b;
        double c;
        double d;
        b = scnr.nextDouble();
        c = scnr.nextDouble();
        d = Math.pow(Math.sqrt(b - c), 3);
        System.out.printf("%.2f
", d);
    }
}

Answer:

The Java code to complete the computation is:

import java.util.Scanner;
public class MathMethods {
    public static void main(String[] args) {
        Scanner scnr = new Scanner(System.in);
        double b;
        double c;
        double d;
        b = scnr.nextDouble();
        c = scnr.nextDouble();
        d = Math.pow(Math.sqrt(b - c), 3);
        System.out.printf("%.2f
", d);
    }
}