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Crank
3 years ago
13

6x-2y=24 how do u solve for y

Mathematics
2 answers:
Tatiana [17]3 years ago
8 0
It's <span>Diophantine equation. 
First, we need to found gcd(6,(-2)):</span>
6=2*3
(-2)=(-1)*2

So, gcd(6,-2)=2
Now, the question is. Can we dived c=24, by gcd(6,-2) and in the end get integer?
Yes we can.
\frac{24}{2} =12
So, we can solve it.

Now is the formula:
{\displaystyle {\begin{cases}x=x_{0}+n{\frac {b}{\gcd(a,\;b)}}\\y=y_{0}-n{\frac {a}{\gcd(a,\;b)}}\end{cases}}\quad n\in \mathbb {Z} .}

Second, we need the first pair (x0,y0) 
 if
x=0
then
y=(-12)

Third, we gonna use that formula:
{\displaystyle {\begin{cases}x=-n}\\y=-3(4+n)}\end{case}\quad n\in \mathbb {Z} .}
Congratulations! We solve it.


klio [65]3 years ago
5 0
6x - 2y = 24

-2y = 24 - 6x      Multiply both sides by -1

2y = 6x - 24        Divide both sides by 2

y = (6x - 24)/2

y = 6x/2 - 24/2

y = 3x - 12
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3 years ago
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Hey im having trouble understanding 40 divided by 8? could you help me out haha
meriva

Answer:

40/8

Step-by-step explanation:

40 divided by 8 is 5.

The way you can check this is by multiplying 8*5

Basically what number times 8 will equal 40

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2 years ago
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(PLEASE HELP) The table below - does it represent an exponential function?
Aliun [14]
Short answer: I don't know, but that doesn't mean I can't give you something that you can decide for yourself.
y = 4*2^(2n - 2) is the pattern.
Go for broke. Try n = 4. You should get 256. Let's try it.
y = 4 * 2^(2*4 - 2)
y = 4 * 2^(8 - 2)
y = 4 * 2^6
y = 4 * 64
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The other end is just as important. Suppose n = 1
Then y = 4 * 2^(2*1 - 2) = 4 * 2^0 = 4*1 = 4 Both work.

If this formula is correct, we can abbreviate it to make your task easier.
y = 4 * 2^(2n - 2)
y = 2^2 * 2^(2n - 2)
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y = 2^(2n) Now try the two end points again.

n = 4
y = 2^(2*4)
y  =  2^8
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n = 1
y = 2^(2*1)
y = 2^2
y = 4 which again checks.

so y = 2^(2n) I think is an exponential function.
Sorry my explanation is so long.

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Which statements about limiting reactants are correct? more than one answer may be correct?
Lerok [7]
Where is the question?
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