Explanation:
A logarithm in one base is a constant multiple of a logarithm in any other base. Any "order of ..." specification does not include the applicable constant multiplier or the smaller order terms that may be required for an exact computation.
The concept of "order of" is similar to the concept of the degree of a polynomial. Knowing the degree of a polynomial tells you something about the "end behavior" as the function argument gets large. The specifics of the scale factor and lower-degree terms become largely irrelevant.
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
.
There are 1130 females. There are 1370 males. What I did was take 2500 and subtracted 240 from it. I got 2260 and then divided that number by 2 and got the number of females. Then I took that number and added 240 to it and got the number of males.
1/3, 1/8, 1/24
I'm pretty sure