#include<stdio.h>
#include<conio.h>
#include<string.h>
int main() {
int i, nextTerm;
int t1 = 2, t2 = 2;
nextTerm = t1+t2;
printf("%d, %d, ", t1, t2);
for (i = 3; i <= 10; ++i) {
printf("%d, ", nextTerm);
t1 = t2;
t2 = nextTerm;
nextTerm = t1 + t2;
}
return 0;
}
Here, f: A—>B
f(x)= (x-1)/(x+2) ; x -2
A= {-1,0,1,2,3,4}
B= {-2,1,-1/2,0,1/2,1/4,2/5}
Range = {-2,-1/2,0,1/4,1/2,2/5}
As range is not equal to codomain so the given function is not bijective.
We can make it bijective by omitting {1} from set B
4.कमपयुटर हारडवेयर तिन परकारमा विभाजित छ। तिनीहरू हुन :
इनपुट उपकरण ( माउस, किबोरड, सकयानर)
आउटपुट उपकरण ( मनिटर, परिनटर, सपिकर)
परशोधन उपकरण (सि.पी. यु)
सोरेज उपकरण ( सिडि, पेन.डराइभ)
In benzene molecule, carbon and carbon bond can't have polarity because there is not any electronegativity difference and the carbon hydrogen is also not much polar because the electronegativity difference is almost zero and also the small pull due to small difference in Electronegativity is balanced due to even pull from all directions.
In chlorobenzene the electronegativity difference between carbon and chlorine is very high so, it is polar.
(to compare polarity first check electronegativity...
