Answer:
scores.append(6,2)
Explanation:
This is a complicated question because in theory, scores.insert can also add values, but I am sure that the only line of code that would work is scores.append(6,2)
Answer:
q1 chees
q2 web and digital continant
Explanation:
because haaa haaa haa dhum
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.
Using the knowledge in computational language in C++ it is possible to write a code that Find the Average of the sum of prime numbers between 1 to any given number
<h3>Writting in C++ code:</h3>
<em />
<em>#include <iostream></em>
<em>using namespace std;</em>
<em>bool isPrime(int n){</em>
<em> for(int i = 2; i < n/2; i++){</em>
<em> if(n%i == 0){</em>
<em> return false;</em>
<em> }</em>
<em> }</em>
<em> return true;</em>
<em>}</em>
<em>int findPrimeSum(int n){</em>
<em> int sumVal = 0;</em>
<em> for(float i = 2; i <= n; i++){</em>
<em> if(isPrime(i))</em>
<em> sumVal += i;</em>
<em> }</em>
<em> return sumVal;</em>
<em>}</em>
<em>int main(){</em>
<em> int n = 15;</em>
<em> cout<<"The sum of prime number between 1 to "<<n<<" is "<<findPrimeSum(n);</em>
<em> return 0;</em>
<em>}</em>
See more about C++ code at brainly.com/question/19705654
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