Answer:
Lets a,b be elements of G. since G/K is abelian, then there exists k ∈ K such that ab * k = ba (because the class of ab,
is equal to
, thus ab and ba are equal or you can obtain one from the other by multiplying by an element of K.
Since K is a subgroup of H, then k ∈ H. This means that you can obtain ba from ab by multiplying by an element of H, k. Thus,
. Since a and b were generic elements of H, then H/G is abelian.
Answer:
Darcy and Jason have $124 combined
Step-by-step explanation:
Can’t see the coordinate plan so I can’t answer it
i think the answer is(1,0) im not sure
Hello!
Explanation:
First, you had to do is multiply by the numbers. And don't forget to used variables.




Answer: → 
Hope this helps!
Thanks!
-Charlie