Answer:
cos²A - cosA - 6
Step-by-step explanation:
Given
(cosA + 2)(cosA - 3)
Each term in the second factor is multiplied by each term in the first factor, that is
cosA(cosA - 3) + 2(cosA - 3) ← distribute both parenthesis
= cos²A - 3cosA + 2cosA - 6 ← collect like terms
= cos²A - cosA - 6
Answer:
The following are the solution to the given points:
Step-by-step explanation:
for point A:


The set A is not part of the subspace 
for point B:


The set B is part of the subspace
for point C:

In this, the scalar multiplication can't behold

∉ C
this inequality is not hold
The set C is not a part of the subspace
for point D:

The scalar multiplication s is not to hold
∉ D
this is an inequality, which is not hold
The set D is not part of the subspace 
For point E:

The
is the arbitrary, in which
is arbitrary

The set E is the part of the subspace
For point F:

The
arbitrary so, they have
as the arbitrary 
The set F is the subspace of 
Answer:
Whole numbers
Step-by-step explanation:
Its not integers, Irrational numbers, or natural numbers. I can tell you what each one means to prove it.