The given plane,
, has normal vector
. Any plane parallel to this one has the same normal vector.
Let
be any point in the plane we want. The plane contains the point (1, 1, -1), so an arbitrary vector in this plane is

and this is perpendicular to
.
So the equation of the plane is

or equivalently,

Given:
let h be the high quality bean
let c be the cheaper bean
h + c = 160
6h + 3.25c = 160*4.97
6h + 3.25c = 795.20
h = 160 - c
6(160 - c) + 3.25c = 795.20
960 - 6c + 3.25c = 795.20
-2.75c = 795.20 - 960
-2.75c = -164.80
c = -164.80 / -2.75
c = 59.92 or 60 lbs
h = 160 - c
h = 160 - 60
h = 100 lbs
Sarah should blend 60 lbs of cheap coffee bean and 100 lbs of high quality coffee bean.
Answer:
The correct ans is 3.
g(-2) = 2(-2) + 5
=-4 + 5 = 1
f(g(-2)) = 4 - (1)^2
= 4 - 1
=3
Step-by-step explanation:
Answer:
Recall that a relation is an <em>equivalence relation</em> if and only if is symmetric, reflexive and transitive. In order to simplify the notation we will use A↔B when A is in relation with B.
<em>Reflexive: </em>We need to prove that A↔A. Let us write J for the identity matrix and recall that J is invertible. Notice that
. Thus, A↔A.
<em>Symmetric</em>: We need to prove that A↔B implies B↔A. As A↔B there exists an invertible matrix P such that
. In this equality we can perform a right multiplication by
and obtain
. Then, in the obtained equality we perform a left multiplication by P and get
. If we write
and
we have
. Thus, B↔A.
<em>Transitive</em>: We need to prove that A↔B and B↔C implies A↔C. From the fact A↔B we have
and from B↔C we have
. Now, if we substitute the last equality into the first one we get
.
Recall that if P and Q are invertible, then QP is invertible and
. So, if we denote R=QP we obtained that
. Hence, A↔C.
Therefore, the relation is an <em>equivalence relation</em>.