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 top option is false.
Step-by-step explanation:
Both segments have a <em>rate</em><em> </em><em>of</em><em> </em><em>change</em><em> </em>[<em>slope</em>] of ⅔. It just that their ratios have unique qualities:
Greatest Common Factor: 2
___ ___
<em>BC</em><em> </em>is at a 4⁄6 slope, and <em>AB</em><em> </em>is at a ⅔ slope. Although their quantities are unique, they have the exact same value.
I am joyous to assist you anytime.
Answer:
c. Mutation
Step-by-step explanation:
should be right I hope lol. good luck :)
Answer:
The answer to your question is
a) Poster width = 3 in
lenght = 9 in
b) banner
width = 2 in
lenght = 10 in
Step-by-step explanation:
a) For the poster
width = x
lenght = 3x
Perimeter of a rectangle = 2base + 2 height
= 2(x) + 2(3x) = 24
2x + 6x = 24
8x = 24
x = 24/8
x = 3
width = 3 in
lenght = 3(3) = 9 in
b) For the banner
width = y
lenght = 5y
Perimeter of a banner = 2 base + 2 height
2 (y) + 2(5y) = 24
2y + 10y = 24
12y = 24
y = 24/12
y = 2
width = 2 in
lenght = 5(2) = 10 in
