Answer:
6*8=48 groups with elements of order 7
Step-by-step explanation:
For this case the first step is discompose the number 168 in factors like this:
And for this case we can use the Sylow theorems, given by:
Let G a group of order 
where p is a prime number, with 
and p not divide m then:
1) 
2) All sylow p subgroups are conjugate in G
3) Any p subgroup of G is contained in a Sylow p subgroup
4) n(G) =1 mod p
Using these theorems we can see that 7 = 1 (mod7)
By the theorem we can't have on one Sylow 7 subgroup so then we need to have 8 of them.
Every each 2 subgroups intersect in a subgroup with a order that divides 7. And analyzing the intersection we can see that we can have 6 of these subgroups.
So then based on the information we can have 6*8=48 groups with elements of order 7 in G of size 168
Answer:
Step-by-step explanation:
The square root is possible only if y≤0
if y>0 the expression lose meaning, in fact (-)(+)(+) = (-) and a square root of a negative number doesn’t exist. x^2 is a square so it is ever positive if x is not 0
if y≤0 can be rewritten as:
3(-27x^2y^7)^1/2
or if we want simplify the expression:
^1/2
we have to put the symbol of absolute value between x and y because a x^2 and y^6 are numbers that are ever positive, while x and y^3 can be also negative. Thanks to the simbol of absolute value we can “force” them to be positive
Answer:
d
Step-by-step explanation:
Answer:
b im not sure
Step-by-step explanation:
Answer:
230
Step-by-step explanation:
23% of 45 = 10.35
10.35 / 4.5 = 2.3
2.3 x 100 = 230
