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jasenka [17]
3 years ago
15

Please help!

Computers and Technology
1 answer:
MaRussiya [10]3 years ago
4 0
Use Easybib.com for your citations. All you have to do is type the name of your sources in to "easybib" and choose if it is a book or a website, then copy it on to a word document.

I hope you can understand this and it helps!!
You might be interested in
What is the next line?
marissa [1.9K]

Explanation:

solução de ácido clorídrico (HCI 6 m e HCI 0,6 m)

7 0
3 years ago
17.8.1: Writing a recursive math function. Write code to complete raise_to_power(). Note: This example is for practicing recursi
Yuliya22 [10]

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))

6 0
3 years ago
Answer quickly!!!
ioda

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

7 0
2 years ago
Find the propagation delay for a signal traversing the in a metropolitan area through 200 km, at the speed of light in cable (2.
Ede4ka [16]

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:

v = \frac{(xf-xo)}{(t-to)}

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 =\frac{xf}{v}  =\frac{2e5 m}{2.3e8 m/s} =8.7e-4 sec.

⇒ t = 8.7*10⁻⁴ sec.

4 0
3 years ago
Write a program to implement problem statement below; provide the menu for input N and number of experiment M to calculate avera
zalisa [80]

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;

}

}

8 0
3 years ago
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