Answer:
ccrccr
Step-by-step explanation:
drcgtgtfttcctt
Answer:
I think the question is wrong check ur question and I will answer
The formula is y=mx+b
To get the slope or m, use this formula
(the second y minus the first y)/(the second x minus the first x)
Now set it up.
(10-7)/(2--1)
3/3=slope is 1.
y=1x+b
Insert one of the points for x and y.
i will do (-1,7)
7=1(-1)+b
7=-1+b
8=b
Insert this into the final equation:
y=1x+8
Try it out. If you're not sure, try both points. If it works, then you set it up correctly.
Neither one of the slopes are going to be minus. It means you are travelling backwards in time, which is wonderful if you are a sci-fi fan, but not so good if you are Sharon. A and D has Sharon going from 70 to 0. That can't be happening so both are wrong.
Now you have to decide between B and C. The intersection point has Sharon going upwards until she is 20. She started out at 70. The graph has John starting at 70. That's not right.
So we've eliminated A,D and now C.
The answer must be B. They meet when Sharon is 90 and John is about 22.5 which is what it should be.
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
.