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White raven [17]
3 years ago
15

Which of these is a simplified form of the equation 8y + 4 = 6 + 2y + 1y?

Mathematics
2 answers:
larisa86 [58]3 years ago
3 0
The first one would be the right one!
  
you just take all the y one side and the numbers on one side 
 so 8y+4= 6+3y

now 8y-3y= 6-4

therefore 5y = 2 
 
!!

Lena [83]3 years ago
3 0
So, with this I'm pretty sure you have to get all variables and all normal numbers on the same sides SO if 8y+4=6+2y+1y, then you must get the 4 over to 6 so then the equation would be, 8y=-4+6+2y+1y and the 2y and 1y (or 3y) must change to the other side so then it would be 8y-3y=-4+6. Then you just simplify to 5y=2. Therefore, your answer is A. 

Steps
8y+4=6+2y+1y
8y=-4+6+2y+1y
8y-3y=-4+6
∴ 5y=2



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2 years ago
Which are correct statements regarding proofs? Select three options.
kompoz [17]

Answer:

In a paragraph proof, statements and their justifications are written in sentences in a logical order.

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Step-by-step explanation:

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  • A paragraph proof is a two-column proof in sentence form.

A paragraph proof is only a two-column proof written in sentences. However, since it is easier to leave steps out when writing a paragraph proof.

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8 0
3 years ago
Select the correct answer.
Sidana [21]

Answer:c

Step-by-step explanation:

when you set (x-6) and (x-1) to zero you get 6 and 1

8 0
3 years ago
How many different ways can you make 82 cents using current u.s. currency
andrew11 [14]
 <span>You can probably just work it out. 

You need non-negative integer solutions to p+5n+10d+25q = 82. 

If p = leftovers, then you simply need 5n + 10d + 25q ≤ 80. 


So this is the same as n + 2d + 5q ≤ 16 

So now you simply have to "crank out" the cases. 

Case q=0 [ n + 2d ≤ 16 ] 

Case (q=0,d=0) → n = 0 through 16 [17 possibilities] 
Case (q=0,d=1) → n = 0 through 14 [15 possibilities] 
... 
Case (q=0,d=7) → n = 0 through 2 [3 possibilities] 
Case (q=0,d=8) → n = 0 [1 possibility] 

Total from q=0 case: 1 + 3 + ... + 15 + 17 = 81 

Case q=1 [ n + 2d ≤ 11 ] 
Case (q=1,d=0) → n = 0 through 11 [12] 
Case (q=1,d=1) → n = 0 through 9 [10] 
... 
Case (q=1,d=5) → n = 0 through 1 [2] 

Total from q=1 case: 2 + 4 + ... + 10 + 12 = 42 

Case q=2 [ n + 2 ≤ 6 ] 
Case (q=2,d=0) → n = 0 through 6 [7] 
Case (q=2,d=1) → n = 0 through 4 [5] 
Case (q=2,d=2) → n = 0 through 2 [3] 
Case (q=2,d=3) → n = 0 [1] 

Total from case q=2: 1 + 3 + 5 + 7 = 16 

Case q=3 [ n + 2d ≤ 1 ] 
Here d must be 0, so there is only the case: 
Case (q=3,d=0) → n = 0 through 1 [2] 

So the case q=3 only has 2. 

Grand total: 2 + 16 + 42 + 81 = 141 </span>
3 0
3 years ago
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