Answer:
Less; $25.90 less than at the beginning of the week
Step-by-step explanation:
162-(63.50+23.25+26.45)=162-113.20=48.80
48.80+87.30=136.10; 162>136.10
162-136.10=Δ$25.90
$25.90 less
Note that (a^b)^c=a^(bc) so in this case we have:
(27^(1/3))^2 the cube root of 27=±3
(±3)^2 and ±3^2=9
So the equivalent expression is 9 or if you need an exponent 9^1 :)
I am assuming that you meant 27^(2/3)
The first derivitive is the slope
a positive 1st deriviive is positive slope or increasing
a negaitve 1st derivitive is negative slope or decreasing
the 2nd derivitive tells about the concavity
a negative second derivitive means it is concave down at that point
a positive 2nd derivitive means it is concave up at that point
so
first one is false
2nd is true
3rd is false because it could possibly be a minimum as well
answer is 2nd one
Slope intercepf is y=mx+b wherem=slope and b= y intercept
slope is found by doing
(y1-y2)/(x1-x2)
points are (6,-1) and (-3,2)
(x,y)
x1=6
y1=-1
x2=-3
y2=2
subsitute
(-1-2)/(6-(-2))=-3/(6+2)=-3/8
slope=-3/8
subsitute
y=-3/8x+b
subsitute and solve for b
(-3,2)
x=-3
y=2
2=-3/8(-3)+b
2=9/8+b
2=16/8
subtract 9/8 from both sides
16/8-9/8=b
7/8=b
y=-3/9x+7/8 is the equation
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 
.
