Answer:
BEG
Explanation:
In traditional music theory, pitch classes are typically represented by first seven Latin alphabets (A,B,C,D,E ,F and G) . And in the below music notes attachment we can understand that the answer is option (a) BEG
Answer:
Option A
Explanation:
Because each function has specific amount of space allocated by default on the stack, so it cares about the length of the array being printed by it. By default the space is up to 100 kb but this size can be increased by setting it manually.
Option B is correct answer because scanf takes input from user until first space is hit.
Answer:
Cultural lag
Explanation:
Cultural lag is a situation where by a society experiences technological development at a rate faster than the development that occurs in culture and general way of life.
A very good example is that it takes a longtime before cultural development can match up with technological developments and advancements thus leading to social problems due to lag. The cultural lag theory was formed in 1922 by William F Ogburn a sociologist.
Answer:
#include <stdio.h>
void spaces(int n) {
for(int i=0; i<n; i++) {
putchar(' ');
}
}
void numbersdown(int n) {
for(int i=n; i>1; i--) {
putchar('0'+i);
}
}
void numbersup(int n) {
for(int i=1; i<=n; i++) {
putchar('0'+i);
}
putchar('\n');
}
int main(void) {
int number;
printf("Please enter a digit: ");
scanf("%d", &number);
for(int i=number; i>0; i--)
{
spaces(number-i);
numbersdown(i);
numbersup(i);
}
}
Explanation:
I deliberately separated the solution into different functions for readability. If needed, the code could be made much more compact.
Answer:
Explanation:
The minimum depth occurs for the path that always takes the smaller portion of the
split, i.e., the nodes that takes α proportion of work from the parent node. The first
node in the path(after the root) gets α proportion of the work(the size of data
processed by this node is αn), the second one get (2)
so on. The recursion bottoms
out when the size of data becomes 1. Assume the recursion ends at level h, we have
(ℎ) = 1
h = log 1/ = lg(1/)/ lg = − lg / lg
Maximum depth m is similar with minimum depth
(1 − )() = 1
m = log1− 1/ = lg(1/)/ lg(1 − ) = − lg / lg(1 − )
