They are all integers so...
first lets substitute
A.3x=13
X=13/3
b.3x+2=13
3x=11
x=11/3
c.3x+4=13
3x=9
x=3
d.3x+6=13
3x=7
x=7/3
E.3x+8=13
3x=5
x=5/3
therefore the possible is letter c because it must be an integer and the definition of integer states that it is not a decimal nor a fraction
press ask for help i think it my help im also stumped on that
Answer:
A), B) and D) are true
Step-by-step explanation:
A) We can prove it as follows:

B) When you compute the product Ax, the i-th component is the matrix of the i-th column of A with x, denote this by Ai x. Then, we have that
. Now, the colums of A are orthonormal so we have that (Ai x)^2=x_i^2. Then
.
C) Consider
. This set is orthogonal because
, but S is not orthonormal because the norm of (0,2) is 2≠1.
D) Let A be an orthogonal matrix in
. Then the columns of A form an orthonormal set. We have that
. To see this, note than the component
of the product
is the dot product of the i-th row of
and the jth row of
. But the i-th row of
is equal to the i-th column of
. If i≠j, this product is equal to 0 (orthogonality) and if i=j this product is equal to 1 (the columns are unit vectors), then
E) Consider S={e_1,0}. S is orthogonal but is not linearly independent, because 0∈S.
In fact, every orthogonal set in R^n without zero vectors is linearly independent. Take a orthogonal set
and suppose that there are coefficients a_i such that
. For any i, take the dot product with u_i in both sides of the equation. All product are zero except u_i·u_i=||u_i||. Then
then
.
The answer is the third choice they both equal -16/77. Hope this helps:)