Answer:3
Step-by-step explanation:
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.
Answer:
Complementary angles: the angles whose sum is 90° are called complementary angles.
Supplementary angles: the angles whose sum is 180° are called supplementary angles.
Step-by-step explanation:
hope it might help....
Answer:
Time = 3secs
Max height = 56m
Step-by-step explanation:
Given the height reached by the rocket modeled by the equation;
h(x) = -5(x - 3)² + 56 where;
x is in seconds
The rocket velocity at its maximum height is zero.
Hence dh/dx = 0
dh/dx = 2(-5)(x-3)
dh/dx = -10(x-3)
Since dh/dx = 0
0 = -10(x-3)
0 = -10x + 30
10x = 30
x = 3secs
<em>Hence it takes 3 secs to reach the maximum height.</em>
Get the maximum height reached by the rocket. Substitute x = 3 into the equation given;
Recall that:
h(x) = -5(x - 3)² + 56
If x = 3
h(3) = -5(3 - 3)² + 56
h(3) = 0 + 56
h(3) = 56
<em>Hence the maximum height that the rocket reaches is 56m</em>
