Answer:
#include<iostream>
using namespace std;
int main()
{
int a,b,c;
cout<<"enter the value of a:";
cin>>a;
cout<<"enter the value of b:";
cin>>b;
cout<<"enter the value of c:";
cin>>c;
cout<<"product is:"<<(a*b*c);
return 0;
}
Explanation:
Answer:
Answer explained
Explanation:
From the previous question we know that while searching for n^(1/r) we don't have to look for guesses less than 0 and greater than n. Because for less than 0 it will be an imaginary number and for rth root of a non negative number can never be greater than itself. Hence lowEnough = 0 and tooHigh = n.
we need to find 5th root of 47226. The computation of root is costlier than computing power of a number. Therefore, we will look for a number whose 5th power is 47226. lowEnough = 0 and tooHigh = 47226 + 1. Question that should be asked on each step would be "Is 5th power of number < 47227?" we will stop when we find a number whose 5th power is 47226.
To accomplish this without using a loop,
we can use math on a string.
Example:
print("apple" * 8)
Output:
appleappleappleappleappleappleappleapple
In this example,
the multiplication by 8 actually creates 8 copies of the string.
So that's the type of logic we want to apply to our problem.
<span>def powersOfTwo(number):
if number >= 0:
return print("*" * 2**number)
else:
<span>return
Hmm I can't make indentations in this box,
so it's doesn't format correctly.
Hopefully you get the idea though.
We're taking the string containing an asterisk and copying it 2^(number) times.
Beyond that you will need to call the function below.
Test it with some different values.
powersOfTwo(4) should print 2^4 asterisks: ****************</span></span>
Answer:
False.
Explanation:
If the institute in which you are enrolled have blackboard account for you then you are given a login ID and password which can be used to login onto the website.
But after logging in you are required to enter a separate password given by the instructor to begin the test.So we conclude that the answer to this question is False.
