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Ashok Shah • almost 2 years ago

Hey everyone, Today, I want to take a moment to revisit something special that happened recently. I was invited to my college where I completed my 11th and 12th to share my experiences from my +2...

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Puspak Sah • asked a question

Explain Dalton's atomic theory with its postulates. Write down limitation of theory and merits of atomic theory.


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Rabin Kalikote • over 2 years ago

Since, I don't have any past experience with theatre, I want to recall one of my favorite pastime. I will take you to my childhood when I was in my village in Nepal.

In our corn field, my grandfather used to plough the land with oxes. I used to be fascinated by watching him and had great desire to plough the land. However, the oxes were too big for me to control. I wouldn't be able to even raise and move the plough that my granny...

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Nikas Ghimire • over 3 years ago

NASA has recently released the sharpest and deepest image of the universe. Making us realize again, How tiny we are.

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gaurab giri

routine of nikas banda

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Ashok Shah • over 2 years ago

When Years of Academic Success Became a Tale of Failure!!

Aahan, a student, has always been deeply ingrained withacademic successin his life. From a young age, he was taught that hard work and dedication were the keys to achieving the goals. As a result, he dedicated countless hours to studying, completing assignments, and striving for top grades.The thrill of academic achievement was intoxicating. Each high grade was a validation of...

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Rabin Kalikote • 11 months ago

If we follow each other at Ask Mattrab, you will read this at some point. How are you?

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Surakshya Pandey

very fine!

Reply 7mo
Rabin Kalikote

Glad to hear, miss!

7mo
Arya Jha • asked a question

Using bisection method, find the root of the equation f(x)=x2 -6logx-3=0(2<=x<=3)'correct to 4 decimal places with an accuracy of 10 -2 .

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...

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