Answer:
Reflected over the x-axis, vertically shrunk by a factor of 1/5, moved right 4 units and up 2 units.
Step-by-step explanation:
Parent Graph: f(x) = a(bx - h)² + k
<em>a </em>is vertical stretch (a > 1) or shrink (a < 1) and reflection over x-axis
<em>b</em> is horizontal stretch or shrink and reflection over y-axis
<em>h</em> is horizontal movement left (if positive) or right (if negative)
<em>k</em> is vertical movement up (if positive) or down (if negative)
Now that we have our rules, we simply apply it to y = -1/5(x - 4)² + 2
<em>a</em> = -1/5, so reflection over x-axis and vertical shrink of 1/5 (1/5 < 1)
Nothing has changed <em>b</em>, so no horizontal stretch/shrink
<em>h</em> = -4, so horizontal movement right 4 units
<em>k</em> = 2, so vertical movement up 2 units
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
.
Answer by BlueSky06
The equation described above can also be written as, y = -x² + 100x + 4000To get the number of notebooks that will give them the maximum profit, we derive the equation and equate to zero. dy/dx = -2x + 100 = 0The value of x from the equation is 50. Then, we substitute 50 to the original equation to get the profit. y = -(50^2) + 100(50) + 4000 = 6500Thus, the maximum profit that the company makes is $6,500/day.
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49: 1 7, 49
84: 1, 2, 3, 4, 6, 7, 12, 14, 21, 28, 42, 84
The GCF is 7
Answer:
Step-by-step explanation:
I must assume that you meant 10(N + 3). Let's not mix { with ).
10(N + 3) is as simple as this expression gets. You could, however, re-write it as 10N + 30.