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, ![[ab]_K](https://tex.z-dn.net/?f=%20%5Bab%5D_K%20)
is equal to ![[ba]_K](https://tex.z-dn.net/?f=%20%5Bba%5D_K%20)
, 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, ![[ab]_H = [ba]_H](https://tex.z-dn.net/?f=%20%5Bab%5D_H%20%3D%20%5Bba%5D_H%20)
. Since a and b were generic elements of H, then H/G is abelian.
Maximum can be determined by taking derivative and set it equal to zero.
Problem
Solution
For this case we know that the vertex is given by (3,6) and the genera equation for a parabola is given by:
y= a(x-h)^2 +k
Where h = 3, k=6 and replacing we have:
y= a(x-3)^2 +6
And we can find the value of a with the point given x= 4, y=4
4= a(4-3)^2 +6
4= a +6
a= 4-6=-2
And the correct equation would be:
d. y= -2(x-3)^2 +6
Answer:
-4
Step-by-step explanation:
plug in 2 for the x, plug in the -5 for the y which will give you the equation 3(2)+2(-5)=
3 x 2 is 6
2 x -5 is -10
then you will have 6 - 10
and your answer is -4
