Answer:
I honestly don't know if you're being serious or not as this was a popular thing a few years ago, there are a few videos on yt, one particularly from 2008 that answers your question pretty well.
Clipping through a loading area requires at least 400 speed. But if you're just interested in the usual speed, there are multiple glitches to make him go faster than his usual x3.5 increased running speed (which is the speed you get at the LEAST when BLJing) but it seems the average is -200. I've seen people get up to -900 though, so.
Basically, it can vary. In a very specific area he can get max momentum of -9373, which is probably the fastest even though it's against a wall.
Answer:
#include <stdio.h>
void printValues ( unsigned char *ptr, int count) // count is no of cells
{
for(int i=0; i<count; i++) {
printf("%d ", ptr[i]);
}
}
int main ( )
{
unsigned char data[ ] = { 9, 8, 7, 5, 3, 2, 1} ;
printValues( data, sizeof(data)/sizeof(data[0]) );
}
Explanation:
Remember that the sizeof() mechanism fails if a pointer to the data is passed to a function. That's why the count variable is needed in the first place.
<em><u>Answer</u></em>
5 hours
<em><u>Explanation</u></em>
The two working together can finish a job in
Also, working alone, one machine would take one hour longer than the other to complete the same job.
Let the slower machine working alone take x hours. Then the faster machine takes x-1 hours to complete the same task working alone.
Their combined rate in terms of x is
This should be equal to 20/9 hours.
Multiply through by;
Factor to get:
It is not feasible for the slower machine to complete the work alone in 4/9 hours if the two will finish in 20/9 hours.
Therefore the slower finish in 5 hours.
