<span>Validation of electronic signatures was designed to encourage a paperless society.</span>
Answer:
- <u>1,000W</u> (rounded to one significant figure)
Explanation:
I will answer in English.
The question is:
- <em>What is the power of a filament lamp that connects to the 220 V network, knowing that it has a resistance of 50 ohms?</em>
<em />
<h2>Solution</h2>
<em />
<em>Power</em>, <em>voltage</em>, <em>resistance</em>, and current, are related by either of the following equations:
Where:
- R is resistance in ohms (Ω)
- V is voltage in volts (V), and
- I is current in amperes (A)
Since you know the voltage (<em>220V</em>) and the resistance (<em>50Ω</em>), you can use the last equation:
Since the magnitude 50Ω has one significant figure, your answer should be rounded to one significant figure. That is <u>1,000W.</u>
Answer:
You would put the cursor after the letter v in vermont and press back space and type a capital V to make it Vermont.
Answer:
#include <stdio.h>
int main()
{
int userNum1;
int userNum2;
userNum1 = -1;
userNum2 = 7;
if (userNum1 < 0)
printf("userNum1 is negative. \n");
if(userNum2 > 9)
userNum2 = 0;
else
printf("userNum2 is less than or equal to 9.\n");
return 0;
}
Explanation:
Initialize userNum1 and userNum2.
If userNum1 is less than 0, print 'userNum1 is negative" and end with newline.
if userNum2 is greater than 9, assign 0 to userNum2.
Otherwise, print "userNum2 is less than or equal to 9 and end with newline.
Answer:
#include<bits/stdc++.h>
using namespace std;
int sumOfinteger(int );
// Returns sum of all digits in numbers from 1 to n
int sumOfintegersFrom1ToN(int n) {
int result = 0; // initialize result
// One by one compute sum of digits in every number from
// 1 to n
for (int x = 1; x <= n; x++)
result += sumOfinteger(x);
return result;
}
// A utility function to compute sum of digits in a
// given number x
int sumOfinteger(int x) {
int sum = 0;
while (x != 0) {
sum += x %10;
x = x /10; }
return sum; }
// Driver Program
int main() {
int n ;
cout<<"enter a number between 1 and n : ";
cin>>n;
cout << "Sum of digits in numbers from 1 to " << n << " is " << sumOfDigitsFrom1ToN(n);
return 0; }
