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 : The statement is false for any value of because the even root function is always positive or
Algebraic expressions are expressions that use numbers and variables
The value of A is 65/64
<h3>How to determine the value of A</h3>
The expression is given as:

Rewrite the expression as:

Factor out 2^x

Rewrite as product

The expression to compare with is given as: A * 2^x.
So, we have:

Divide both sides by 2^x

Take LCM


Hence, the value of A is 65/64
Read more about expressions at:
brainly.com/question/4344214
Answer:
Yes
Step-by-step explanation:
20:16
Reduce the ratio to lowest terms by dividing both numbers by 4.
20:16 = 5:4
Answer: Yes