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:
( 3 x 3 x 3 ) and ( 3 x 3 x 5 ) The GCF of 27 and 45 is 3 x 3 = 9
Step-by-step explanation:
Height:shadow=2.5:3
2.5/3=6.7/x
cross multiply aka times both sides by 3x
2.5x=20.1
divide both sides by 2.5
x=8.04
answer is 8.04ft
First, you have to round the .1, which would be .0. Next, you round 19.0, or just 19, to the nearest 10, which is 20. Therefore, 19.1 rounded to the nearest 10 is 20.
<span>1,234,000 in scientific notation = 1.234 x 10^6
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