we know that
The volume of a cylinder is given by the formula
where
r is the radius of the cylinder
h is the height of the cylinder
in this problem
Substitute the values in the formula above
therefore
<u>the answer is</u>
The volume of the can is equal to
The answer is C.379---------
The rule is x+2, y-3 so you apply it to each of these coordinates. Right 2, down 3.
A(-2,2) -> A'(0,-1)
B(-2,4) -> B'(0, 1)
D(2,2) -> D'(4,-1)
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 
.
Abcissa is x, and ordinate is y.
In this case the x is positive, and the y is negative.
Therefore, the answer is "positive, negative"
