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Raunak Kumar asked a question

BODMAS=Bracket, Order, Division, Multiplication, Addition and Subtraction. In certain regions, PEMDAS (Parentheses, Exponents, Multiplication, Division, Addition and Subtraction) is the synonym of BODMAS. It explains the order of operations to solve an expression.

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Sushil Bhandari asked a question

Recently we're working to degrade accounts with 0 contributions from creator to learner. If you're a learner and very keen to be a creator, you must keep posting interesting questions and contact to admins from the Facebook Group of Mattrab Community. For being an admin, you must be in grade 12, either completed or recently enrolled, your notes, and all your records and contributions will be verified for that

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Ishan asked a question

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.

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Rabin Kalikote asked a question

Well, the propagation of light is fundamentally due to oscillation of electric field and magnetic field perpendicularly, which allows the light waves to propagate in the direction perpendicular to both the existing fields, i.e. light waves propagation solely depends on these fields.

But, sound waves are non other than the transference of energy due to disturbance, as well the propagation of sound wave has already been found to be adiabatic in nature, and through the adiabatic equation,...

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Nikas Ghimire asked a question


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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