The denominator is 14-x.
Since the denominator cannot be equal to zero, you take the denominator, set it equal to zero, and solve.
This ill give you the value or values that must be eliminated (cannot be in the domain).
14 - x = 0
add x to both sides
14 = x
The domain is x ≠ 14
OR
(- ∞, 14) ∪ (14, ∞)
OR
All real numbers except 14
Answer: T⊂U⊂W are subspaces of V
Step-by-step explanation:
Proof: This is the easier direction.
If T⊂U⊂W or W⊂U⊂T then we have U⊂T⊂W = T or T⊂U⊂W = U
orT⊂U⊂W=W respectively.
SoT⊂U⊂W is a subspace as T, U and W are subspaces.
1st case :T⊂U⊂W is true Then the disjunction W⊂U⊂T or U⊂T⊂W is trivially true.
Let x∈W1 and y∈W2−W1.
By the definition of the union, we have x∈W∪T∪C and y∈T⊂U⊂W
As T∪U∪W is a subspace, x+y∈T∪C∪W which, again by the definition of the union, means that x+y∈W∪T∪C
V∈W∪T∪C
As V was arbitrary, as desired.
14.72
is the answer all you had to do was divide
I believe the answer to your question is going to be
D) g(x)=(x+3)^2
Hope I’m correct!
You have to add and divide for the mean(average) so add all of them up and then divide by 3 :)