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lora16 [44]
3 years ago
10

If you add 12 marshmallows to each of 5 bags of marshmallows, and each bag started with the same number of marshmallows, the tot

al number of marshmallows is given by the expression ­­­5(x + 12). Identify an equivalent expression.
A.) x + 12
B.) 2x − 60
C.) 5x + 60
D.)2x − 60
Mathematics
1 answer:
ss7ja [257]3 years ago
5 0
The answer is a 
because you would add 12 to x x being the bag
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aalyn [17]

Answer:

x+y=1+3

Step-by-step explanation:

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3 years ago
. A segment has a midpoint at (2, -7) and an endpoint at (8, -5). What are the coordinates of the other endpoint?
Sliva [168]
Ok, you can refer to the midpoint formula to find the endpoint.  Here goes...


MP=(2,-7) and EP=(8,-5)
Let x represent the missing endpoint.
(8+x)/2=2      NOTE: =2 represents first number of MP and the representation of number 8 is self explanatory.  You have two endpoints but need to identify the other endpoint so you divide by 2.  Then, multiply by two on both sides.
2(8+x)/2 = 2*2
16+x/2=4 do the next step (simplify) on the left side of equation 16x/2=8
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4 0
3 years ago
Read 2 more answers
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Andrej [43]

Answer:

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b)surface area

c)volume

Step-by-step explanation:

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3 years ago
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Step-by-step explanation:

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3 years ago
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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