Explanation:
solução de ácido clorídrico (HCI 6 m e HCI 0,6 m)
Answer:
The recursion function is as follows:
def raise_to_power(num, power):
if power == 0:
return 1
elif power == 1:
return num
else:
return (num*raise_to_power(num, power-1))
Explanation:
This defines the function
def raise_to_power(num, power):
If power is 0, this returns 1
if power == 0:
return 1
If power is 1, this returns num
elif power == 1:
return num
If otherwise, it calculates the power recursively
else:
return (num*raise_to_power(num, power-1))
Answer: Harry should check that font he has used are readable on every page and element of his website
Harry should check that all images have alt texts
Explanation: edmentum
Answer:
t= 8.7*10⁻⁴ sec.
Explanation:
If the signal were able to traverse this distance at an infinite speed, the propagation delay would be zero.
As this is not possible, (the maximum speed of interactions in the universe is equal to the speed of light), there will be a finite propagation delay.
Assuming that the signal propagates at a constant speed, which is equal to 2.3*10⁸ m/s (due to the characteristics of the cable, it is not the same as if it were propagating in vaccum, at 3.0*10⁸ m/s), the time taken to the signal to traverse the 200 km, which is equal to the propagation delay, can be found applying the average velocity definition:
If we choose x₀ = 0 and t₀ =0, and replace v= 2.3*10⁸ m/s, and xf=2*10⁵ m, we can solve for t:
⇒ t = 8.7*10⁻⁴ sec.
Answer:
Explanation:
#include<iostream>
#include<ctime>
#include<bits/stdc++.h>
using namespace std;
double calculate(double arr[], int l)
{
double avg=0.0;
int x;
for(x=0;x<l;x++)
{
avg+=arr[x];
}
avg/=l;
return avg;
}
int biggest(int arr[], int n)
{
int x,idx,big=-1;
for(x=0;x<n;x++)
{
if(arr[x]>big)
{
big=arr[x];
idx=x;
}
}
return idx;
}
int main()
{
vector<pair<int,double> >result;
cout<<"Enter 1 for iteration\nEnter 2 for exit\n";
int choice;
cin>>choice;
while(choice!=2)
{
int n,m;
cout<<"Enter N"<<endl;
cin>>n;
cout<<"Enter M"<<endl;
cin>>m;
int c=m;
double running_time[c];
while(c>0)
{
int arr[n];
int x;
for(x=0;x<n;x++)
{
arr[x] = rand();
}
clock_t start = clock();
int pos = biggest(arr,n);
clock_t t_end = clock();
c--;
running_time[c] = 1000.0*(t_end-start)/CLOCKS_PER_SEC;
}
double avg_running_time = calculate(running_time,m);
result.push_back(make_pair(n,avg_running_time));
cout<<"Enter 1 for iteration\nEnter 2 for exit\n";
cin>>choice;
}
for(int x=0;x<result.size();x++)
{
cout<<result[x].first<<" "<<result[x].second<<endl;
}
}
