Here, f(x)=x2 -6 logx-3=0
f(2)=4-6 log2-3=-0.806
f(3)=9-6 log3-3=3.1373
f(2).f(3)=-0.806*3.1373=-2.529422 which is negative.
Hence, the root lies between 2 and 3
c0 =(2+3)/2=2.5
f(2.5)=6.25-6 log 2.5-3=0.8623
Now
| n | a(-ve) | b(+ve) | cn | f(cn) |
| 0 | 2 | 3 | 2.5 | 0.8623 |
| 1 | 2 | 2.5 | 2.25 | -0.050595 |
| 2 | 2.25 | 2.5 | 2.375 | 0.38664 |
| 3 | 2.25 | 2.375 | 2.3125 | 0.1631658 |
| 4 | 2.25 | 2.3125 | 2.28125 | 0.05506 |
| 5 | 2.25 | 2.28125 | 2.265625 | 0.001925 |
From the table,
f(2.265625)=0.001928<10-2
Therefore, the...
Hello Subash!
Here is the solution for the question you are asking for, I solved it in procedural way but if you are among the one who prefer OOP style then you can still ask it for me cause I have solved it from both methods but here I am just going to leave procedural one....
//author:Manish Acharya
import java.util.Scanner;
import java.util.*;
public class idgenerator {
public static void main(String[] args) {
String small_name="", long_name="", new_small_name="", new_long_name="";
char lr='a',...
1. How many beats can be heard by sounding together two tuning forks of frequencies 250Hz and 256Hz?
Number of beats (n) =f2- f1
= 256 - 250
= 6 beats /sec
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Where could be the end? Any thoughts?
The end is when node_modules finishes installing.