Answer:
Following are the answer to this question:
Explanation:
In option 1:
The value of n is= 7, which is (base case)
when n=k for the true condition:
when n=k+1 it tests the value:
since k>6 hence the value is KH>3 hence proved.
In option 2:
when:
for n=1:(base case)
0<=0 \\ condition is true
when the above statement holds value n=1
when n=k
when n=k+1
![[\therefore KH>K \Rightarrow \log(KH>\loK)]](https://tex.z-dn.net/?f=%5B%5Ctherefore%20KH%3EK%20%5CRightarrow%20%20%5Clog%28KH%3E%5CloK%29%5D)
In option 3:
when n=1:
when n=k
![\to (A_1\cap A_2 \cap.....A_k) \cup B\\=(A_1\cup B) \cap(A_2\cup B_2)....(A_k \capB).....(a)\\\to n= k+1\\ \to (A_1\cap A_2 \cap.....A_{kH}) \cup B= (A_1\cup B)\\\\\to [(A_1\cap A_2 \cap.....A_{k}) \cup B]\cap (A_{KH}\cup B)\\\\\to [(A_1\cup B) \cap (A_2 \cup B) \cap (A_3\cup B).....(A_k\cup B)\cap (A_{k+1} \cup B)\\\\ \ \ \ \ \ \ substituting \ equation \ a \\\\](https://tex.z-dn.net/?f=%5Cto%20%28A_1%5Ccap%20A_2%20%5Ccap.....A_k%29%20%5Ccup%20B%5C%5C%3D%28A_1%5Ccup%20B%29%20%5Ccap%28A_2%5Ccup%20B_2%29....%28A_k%20%5CcapB%29.....%28a%29%5C%5C%5Cto%20n%3D%20k%2B1%5C%5C%20%5Cto%20%28A_1%5Ccap%20A_2%20%5Ccap.....A_%7BkH%7D%29%20%5Ccup%20B%3D%20%28A_1%5Ccup%20B%29%5C%5C%5C%5C%5Cto%20%20%5B%28A_1%5Ccap%20A_2%20%5Ccap.....A_%7Bk%7D%29%20%5Ccup%20B%5D%5Ccap%20%28A_%7BKH%7D%5Ccup%20B%29%5C%5C%5C%5C%5Cto%20%20%5B%28A_1%5Ccup%20B%29%20%5Ccap%20%28A_2%20%5Ccup%20B%29%20%5Ccap%20%28A_3%5Ccup%20B%29.....%28A_k%5Ccup%20B%29%5Ccap%20%28A_%7Bk%2B1%7D%20%5Ccup%20B%29%5C%5C%5C%5C%20%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20substituting%20%5C%20equation%20%5C%20a%20%5C%5C%5C%5C)
hence n=k+1 is true.
Answer:
#include <bits/stdc++.h>
using namespace std;
// main function
int main()
{
// variables
double mexico = 114;
double usa = 312;
double mexicoRate = .0101;
double usaRate = .0015;
// calculate population after every year until mexico population exceed the usa populationn
while (usa>mexico)
{
// print the population
cout<<"Mexico's population ::"<<mexico<<" million."<<endl;
cout<<"USA's population ::"<<usa<<" million."<<endl;
// update the population
mexico+=mexico*mexicoRate;
usa-=usa*usaRate;
}
return 0;
}
Explanation:
Declare and initialize mexico and usa with their initial population.Also declare and initial their increase and decrease rate.Find the population of both the country each year until mexico population exceeds the usa population.
Output:
Mexico's population ::114 million.
USA's population ::312 million.
Mexico's population ::115.151 million.
USA's population ::311.532 million.
.
.
.
Mexico's population ::270.546million.
USA's population ::274.213 million.
Mexico's population ::273.278million.
USA's population ::273.802 million.
Answer:
The answer is "Option b".
Explanation:
In the given code, a static method "inCommon" is declared, that accepts two array lists in its parameter, and inside the method two for loop is used, in which a conditional statement used, that checks element of array list a is equal to the element of array list b. If the condition is true it will return the value true, and if the condition is not true, it will return a false value. In this, the second loop is not used because j>0 so will never check the element of the first element.
Answer: I think the answer would be 1/30 or 1/60
Explanation: Anything below 1/90 is considered a soft image so i would go with 1/30 Brainliest ans plz.
