The gear ratios would look like A:B=1:9 and C:D=1:32 if you are just needing the gear teeth ratio. Not a lot of information to go off of from the question though
<span>The correct answer is Intranet chat rooms</span>
Answer:
Explanation:
Go and edit your profile then go on prefernces you will see that its written delete my account and that's how you delete your account on brainly.com and if you want to delete you account on phone then
Open your phone's Settings app.
Tap Accounts. If you don't see "Accounts," tap Users & accounts.
Tap the account you want to remove Remove account.
If this is the only Google Account on the phone, you'll need to enter your phone's pattern, PIN, or password for security.
Hope this helped you!
Answer:
A - 2
B - 3
C - 5
D - 4
E - 1
Explanation:
A. Virtual private network : 2
This is very useful when you have to work from home or visiting a client and need access to files/applications just like if you were at the office.
B. Email : 3
Yes, this is basically why the Internet was created in the first place.
C. Social networking sites : 5
Social networks are indeed a great marketing tool, allowing to target potential customers very precisely.
D. Search engines : 4
Yes, to find information about new products for the company, how the competitors are doing and so on.
E. Video and web conferencing: 1
Another great tool for employees working from home for example, or for employees located in an office at the other side of the country.
Answer:
// code in C++
#include <bits/stdc++.h>
using namespace std;
// main function
int main()
{
// variables
int sum_even=0,sum_odd=0,eve_count=0,odd_count=0;
int largest=INT_MIN;
int smallest=INT_MAX;
int n;
cout<<"Enter 10 Integers:";
// read 10 Integers
for(int a=0;a<10;a++)
{
cin>>n;
// find largest
if(n>largest)
largest=n;
// find smallest
if(n<smallest)
smallest=n;
// if input is even
if(n%2==0)
{
// sum of even
sum_even+=n;
// even count
eve_count++;
}
else
{
// sum of odd
sum_odd+=n;
// odd count
odd_count++;
}
}
// print sum of even
cout<<"Sum of all even numbers is: "<<sum_even<<endl;
// print sum of odd
cout<<"Sum of all odd numbers is: "<<sum_odd<<endl;
// print largest
cout<<"largest Integer is: "<<largest<<endl;
// print smallest
cout<<"smallest Integer is: "<<smallest<<endl;
// print even count
cout<<"count of even number is: "<<eve_count<<endl;
// print odd cout
cout<<"count of odd number is: "<<odd_count<<endl;
return 0;
}
Explanation:
Read an integer from user.If the input is greater that largest then update the largest.If the input is smaller than smallest then update the smallest.Then check if input is even then add it to sum_even and increment the eve_count.If the input is odd then add it to sum_odd and increment the odd_count.Repeat this for 10 inputs. Then print sum of all even inputs, sum of all odd inputs, largest among all, smallest among all, count of even inputs and count of odd inputs.
Output:
Enter 10 Integers:1 3 4 2 10 11 12 44 5 20
Sum of all even numbers is: 92
Sum of all odd numbers is: 20
largest Integer is: 44
smallest Integer is: 1
count of even number is: 6
count of odd number is: 4
