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).
Let her monthly salary be x
1/8x = $65
x = $65 × 8 = $520
Her weekly pay is $520 ÷ 4 = $130
Answer:
Step-by-step explanation:
The total number of marbles is 20
Question A
P(Red) = 4/20 = 1/5
P(yellow) = 9/20
P(Red then Yellow) = 1/5 * 9/20 = 9 / 100
Question B
P(B) = 6/20
P(B again) = 5/19 because you have 1 less total plus 1 less blue.
P(B and B) = 6/20 * 5/19
P(B and B) = 30/ 380 = 3/38
Answer:
A. (9,-5)
Step-by-step explanation:
Translation down is -2 to the y coordinate, which is (5,-5). Then you translate it right 4 units, which is +4 to the x coordinate, so the final coordinate is (9,-5).