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...
Dextrorotatory and dextrorotation are terms used in chemistry to describe the direction in which a substance rotates plane-polarized light. When a substance is said to be dextrorotatory, it means that it rotates plane-polarized light to the right or clockwise direction, whereas when it is levorotatory, it rotates plane-polarized light to the left or counterclockwise direction.
The terms "dextrorotatory" and "levorotatory" come from the Latin words "dexter" meaning "right" and "lævus" meaning...
conceptualizing the truth,in hegel's term. A lot of philosophers(like descartes' three kinds of ideas) in his time (and a lot of mindfulness practitioners these days)believed a higher form of truth(or knowledge) exists which cannot be articulated but is to be intuited and felt.Hegel didnt consider this to be scientific.you can never know if your intuition is true or its just you making the stuff up.
Asksionov thought of killing himself because he remembered all those harsh situations that he had faced in his life without committing any crime . Even his wife didn't believe in him . He got frustrated by thinking all these painful situations that he was ready to kill himself.
Program to input any number and display number of odd numbers in it;
#include<stdio.h>
#include<stdlib.h>
#include<conio.h>
#include<string.h>
int main() {
int i, number, num1, num2=0, num3, num4 =0, rem, rem1, rem2, rev = 0;
printf("Enter your number ==> ");
scanf("%d", &number);
num1 = number;
while(num1 != 0) {
rem = num1%10;
rem1 = rem%2;
if (rem1 != 0) {
num2 = num2*10 + rem;
}
num1 /= 10;
}
num3 = num2;
while (num2 != 0) {
...
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