Answer:
Step-by-step explanation:
REcall the following definition of induced operation.
Let * be a binary operation over a set S and H a subset of S. If for every a,b elements in H it happens that a*b is also in H, then the binary operation that is obtained by restricting * to H is called the induced operation.
So, according to this definition, we must show that given two matrices of the specific subset, the product is also in the subset.
For this problem, recall this property of the determinant. Given A,B matrices in Mn(R) then det(AB) = det(A)*det(B).
Case SL2(R):
Let A,B matrices in SL2(R). Then, det(A) and det(B) is different from zero. So
.
So AB is also in SL2(R).
Case GL2(R):
Let A,B matrices in GL2(R). Then, det(A)= det(B)=1 is different from zero. So
.
So AB is also in GL2(R).
With these, we have proved that the matrix multiplication over SL2(R) and GL2(R) is an induced operation from the matrix multiplication over M2(R).
Answer:
Each notebook costs $2
Step-by-step explanation:
We have to find the amount she spent on each notebook.
22-10=12
We know she spent $12 on six notebooks
We need to divide to find the answer
12/6=2
Each notebook costs $12
Part (i)
Plug in theta = pi/3 to get
h(theta) = sin(3theta)
h(pi/3) = sin(3*pi/3)
h(pi/3) = sin(pi)
h(pi/3) = 0 ... use a calculator or the unit circle
<h3>Answer: 0</h3>
=================================================
Part (ii)
Plug in h = pi/2
h(theta) = sin(3theta)
h(pi/2) = sin(3pi/2)
h(pi/2) = -1
<h3>Answer: -1</h3>
Step-by-step explanation:
I assume
SR = 2x + 23
RQ = x + 21
if that is true, then the situation is completely simple :
14 = (2x + 23) + (x + 21) = 3x + 44
3x = -30
x = -10
SR = 2×-10 + 23 = -20 + 23 = 3
RQ = -10 + 21 = 11
Answer:
12⁽¹⁸⁾
Step-by-step explanation:
The expression is given as ;
12³ × 12⁹× 12⁴× 12² -----same base, add the powers according to law of indices
12⁽³⁺⁹⁺⁴⁺²⁾
12⁽¹⁸⁾