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GarryVolchara [31]
2 years ago
14

Can someone please help I’m kinda confused

Mathematics
1 answer:
Sophie [7]2 years ago
4 0

it can't have a nice day

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How do you find the median of the set of numbers that are not listed in order, like 1, 2, 4, 3, 5, 3, 6, 7, 8?
nydimaria [60]
First you would put them in order, then cross off one from one side, then another number from the other side, and repeat that until you get to the middle, and thats your answer. If there is 2 numbers, add them and divide by 2
5 0
3 years ago
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Divide 24x2y + 8xy2 + 8xy by -4xy i need this fast a f bois
sweet-ann [11.9K]

Dividing term by term  its:-

-6x - 2y - 2

4 0
3 years ago
Point P is located at (−2, 7), and point R is located at (1, 0). Find the y value for the point Q that is located two thirds the
BARSIC [14]

Answer:

5.1

Step-by-step explanation:

Before we calculate  the y value for the point Q that is located two thirds the distance from point P to point R, we need to get the distance of point p from point R using the formula for calculatingf the distance between two points

D = √(x2-x1)²+(y2-y1)²

Given P(−2, 7), and R(1, 0)

RP = √(1-(-2))²+(0-7)²

RP = √3²+(-7)²

RP = √9+49

RP =√58

To get the y value for point Q that is located two thirds the distance from point P to point R, this will give

PQ = y = 2/3 of √58

= 5.1

6 0
3 years ago
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Find the perimeter of a circle with radius 35cm <br>Take(Π=22/7)​
olasank [31]
Perimeter = 2*pi*r
= 2*22/7*35
=220cm

I HOPE IT HELPED
7 0
3 years ago
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A group issimpleif it has no nontrivial proper normal subgroups. LetGbe a simple group of order168. How many elements of order 7
stepladder [879]

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:

168 = 8*3*7= 2^3 *3*7

And for this case we can use the Sylow theorems, given by:

Let G a group of order p^{\alpha} m  where p is a prime number, with m\leq 1 and p not divide m then:

1) Syl (G) \neq \emptyset

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

3 0
3 years ago
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