D = 1/2hw
2dhw = 1
h = 1/2dw
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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:
C. (3, 7)
Step-by-step explanation:
plug all the numbers into all of the coordinates and (3, 7) is the lowest number :) pls rate 5 stars and thank me
Answer:
x = 81/4
Step-by-step explanation:
Answer:
Second option
Step-by-step explanation:
It is given that, if the area of the rectangle to be drawn is 10 square units
To find the length of the rectangle
From the figure, we get the length of AB = 2 units
Therefore Area = length * breadth = 10
Therefore length = 10/2 = 5 units
To find the points of C and D
we have B(-1, 3)
Point C is 5 units below the point B
Therefore C(-1, 3-5) = C(-1,-2)
we have A(1, 3)
The point D is 5 units below the point B
Therefore D(1, 3-5) = A(1,-2)
Therefore the correct answer is option 2.