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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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<h3><em>Hey there today we will solve your problem,</em></h3>

Number 1

_______________

<h3>Definitions</h3>
  • Vertical angles - <em>either of two angles lying on opposite sides of two intersecting lines.</em>

<em />

  • Linear Pair - <em>A linear pair is a pair of adjacent angles formed when two lines intersect.</em>

_______________

Now we can solve Number 1, there is one pair of Vertical angles being ∠5 and ∠3

_______________

Number 2

Since we know the definition of a Linear Pair we can solve this problem also, the only Linear Pair that we can choose out the ones give to us is ∠4 and ∠3, because they are adjacent.

_______________

{\huge{\boxed{\mathfrak{Answer}}}

  1. ∠5 and ∠3

    2. ∠4 and ∠3

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2 years ago
Please help thank you.
MA_775_DIABLO [31]
Question 1:

To start off this question, we can tell that this is a square because it has 4 right angles and 4 congruent sides.

A square has four parallel sides and 4 congruent sides, so a square is a rhombus and parallelogram. 

A square has 4 right angles, so it's also a rectangle.

A square has 4 sides, so it's also a quadrilateral.

The first choice is your answer.

Question 2:

Not all quadrilaterals are rectangles, so A is incorrect.

Not all quadrilaterals are squares, so B is incorrect.

All rectangles are types of quadrilaterals, so C is correct.

Not all quadrilaterals are parallelograms, so D is incorrect.

Thus, C is your answer.

Question 3: 

The first choice will not work because a rhombus will satisfy those conditions, and a rhombus is not always a square.

The second choice will work because only a square will satisfy that condition because only squares have 4 congruent sides along with equal diagonals.

Thus, the second choice is your answer.

Have an awesome day! :)
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