6. prove that if 3n^3+13 is odd then n is even for all integers n, using the proof by contradiction.

If 3n^3+13 is odd then n is even for all integers n, using the proof by contradiction.

Step-by-step explanation:

By contradiction method, we need to prove that if is odd then n is even for all integers n.

Proof by contradiction:

Let as assume if is odd then n is odd for all integers n.

Substitute the value of n in the given expression.

Cube of any odd number is an odd number.

Product of two odd numbers is an odd number.

is an odd number and 13 is an odd number. We know that addition of two odd numbers is an even number.

Which is the contradiction of our assumption.

If 3n^3+13 is odd then n is even for all integers n, using the proof by contradiction.

Hence proved.

sinn is the total amount of students now and m were the students before, you subtract them both (n-m) and divide that by m. then, since to get a percentage from a fraction you have to multiply by 100, your answer should look like this: n-m/m*100

step-by-step explanation: