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.

for this case we have by definition, if two lines are perpendicular, then the product of their slopes is -1.

then, given the following line:

so,

we are looking for m_ {2}:

then, the line is given by:

we find "b" replacing the given point:

finally, the equation is:

option c