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>.
For f(t) = 3 sin(t), the minimum is -3, the maximum is 3.
For additional information:
The period T of the function y=3sin(4t) is;
The period of y = 3 sin (4t) is (2pi) / 4 = pi / 2
The sine function with amplitude A = 0.75 and period T = 10, is
y = 0.75 sin( (2pi / 10) x )
= 0.75 sin( (pi/5) t)
y(4) = 0.75 sin( (pi/5) (4) )
= 0.75 sin ( (4/5) pi ) = .4408
Drawing the sine and cosine function on the same plot shows
that they are identical except for a horizontal shift.
The cosine
function leads the sine function by a shift of (2pi/4) = pi/2.
For the last part, y(4) = 0.75 cos( (2pi/10) (4) ) = - 0.6067
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Answer:
20
Step-by-step explanation:
2x - 5 = x + 15
2x - x = 5 + 15
x = 20
