<span>An embedded os must be developed specifically for use with embedded systems. true or false? The answer is False.</span>
To determine what would be the most appropriate way to address and greet your teacher during an email, we should eliminate some greetings, such as:
"Yo wassup?", "How u doin?", or any other grammatical and socially inappropriate errors.
Let's look at our first option.
"Hi Joseph, How u doin??". This is incorrect as it is not appropriate to address anyone in such a manner and with grammatical errors.
Let's look at our second option.
"Dear Joseph Herman, how are you doing!!". This was on the right path, but didn't end well. The ending of the message, "how are you doing!!" is incorrect punctuation, and has too much excitement.
How about our third option?
"Dear Mr. Herman, I hope you're doing fine.". This is a great email. It has perfect punctuation, grammar, and is appropriate.
What about our fourth?
"Dear Joseph, i hope you are doing great.". This is a good email, but has incorrect punctuation.
Your answer is C.) Dear Mr. Herman, I hope you're doing fine.
Answer:
<em>I </em><em>think </em><em>the </em><em>answer </em><em>is </em><em>a </em><em>bullet.</em>
<em>Hope </em><em>this </em><em>helps</em>
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;
}
}
Answer:
# Solve the quadratic equation ax**2 + bx + c = 0
# import complex math module
import cmath
a = 1
b = 5
c = 6
# calculate the discriminant
d = (b**2) - (4*a*c)
# find two solutions
sol1 = (-b-cmath.sqrt(d))/(2*a)
sol2 = (-b+cmath.sqrt(d))/(2*a)
print('The solution are {0} and {1}'.format(sol1,sol2))
Hope This Helps!!!
