The length of the inclined plane divided by the vertical rise, or you can call it run to rise ratio. The mechanical advantage would increase as the slope of the incline decreases, but problem is that the load will have to go a longer distance. The mechanical advantage would be slope of the incline. I also got confused on a question like this and did some research. Hope this helps!
Answer:
does anybody know this answer?
Explanation:
no nobody does
Answer:
#include <stdio.h>
int fib(int n) {
if (n <= 0) {
return 0;
}
if (n <= 2) {
return 1;
}
return fib(n-1) + fib(n-2);
}
int main(void) {
for(int nr=0; nr<=20; nr++)
printf("Fibonacci %d is %d\n", nr, fib(nr) );
return 0;
}
Explanation:
The code is a literal translation of the definition using a recursive function.
The recursive function is not per se a very efficient one.
Answer:
See explaination for the program code
Explanation:
The code below
Pseudo-code:
//each item ai is used at most once
isSubsetSum(A[],n,t)//takes array of items of size n, and sum t
{
boolean subset[n+1][t+1];//creating a boolean mtraix
for i=1 to n+1
subset[i][1] = true; //initially setting all first column values as true
for i = 2 to t+1
subset[1][i] = false; //initialy setting all first row values as false
for i=2 to n
{
for j=2 to t
{
if(j<A[i-1])
subset[i][j] = subset[i-1][j];
if (j >= A[i-1])
subset[i][j] = subset[i-1][j] ||
subset[i - 1][j-set[i-1]];
}
}
//returns true if there is a subset with given sum t
//other wise returns false
return subset[n][t];
}
Recurrence relation:
T(n) =T(n-1)+ t//here t is runtime of inner loop, and innner loop will run n times
T(1)=1
solving recurrence:
T(n)=T(n-1)+t
T(n)=T(n-2)+t+t
T(n)=T(n-2)+2t
T(n)=T(n-3)+3t
,,
,
T(n)=T(n-n-1)+(n-1)t
T(n)=T(1)+(n-1)t
T(n)=1+(n-1)t = O(nt)
//so complexity is :O(nt)//where n is number of element, t is given sum
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Explanation: please give branliest I only need one more to make ace
