Answer:
.
Step-by-step explanation:
The given expression is
We need to simplify the expression such that answer should contain only positive exponents with no fractional exponents in the denominator.
Using properties of exponents, we get
We can not simplify further because on further simplification we get negative exponents in numerator or fractional exponents in the denominator.
Therefore, the required expression is .
Answer:
Coefficients are 20, 6, 2 and 7
Your variables are your letters in the equations so x and y are your variables in this case.
Your constants are the same as your coefficients I believe (I will include images so you can be sure)
Your terms are what are in between the signs so 20, 6x, 2y and 7
Here you go henderson :)
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).