Answer
we can decrease risk for an entrepreneur by creating a team of trusted advisors to rely on.
there are some points that we can take in mind know how to position our company as a safe position.
revenue streams and additional technology, paving the way for future growth.
Entrepreneurs can use all these unique strategies to decrease those risk
Answer:
#include <stdio.h>
#include <string.h>
int main(){
char number[100];
printf("Number: ");
scanf("%s", number);
int sum = 0;
for(int i =0;i<strlen(number);i++){
sum+= number[i] - '0';
}
printf("Sum: %d",sum);
return 0;
}
Explanation:
This declares a c string of 100 characters
char number[100];
This prompts user for input
printf("Number: ");
This gets user input
scanf("%s", number);
This initializes sum to 0
int sum = 0;
This iterates through the input string
for(int i =0;i<strlen(number);i++){
This adds individual digits
sum+= number[i] - '0';
}
This prints the calculated sum
printf("Sum: %d",sum);
return 0;
The answers are as follows:
a) F(A, B, C) = A'B'C' + A'B'C + A'BC' + A'BC + AB'C' + AB'C + ABC' + ABC
= A'(B'C' + B'C + BC' + BC) + A((B'C' + B'C + BC' + BC)
= (A' + A)(B'C' + B'C + BC' + BC) = B'C' + B'C + BC' + BC
= B'(C' + C) + B(C' + C) = B' + B = 1
b) F(x1, x2, x3, ..., xn) = ∑mi has 2n/2 minterms with x1 and 2n/2 minterms
with x'1, which can be factored and removed as in (a). The remaining 2n1
product terms will have 2n-1/2 minterms with x2 and 2n-1/2 minterms
with x'2, which and be factored to remove x2 and x'2, continue this
process until the last term is left and xn + x'n = 1
Sometimes I just meet some Unintelligen ppl
