Y 3x 4

Y 3x 4

Parallel: y = 3x 4 (2, 7)?

y = mx + b

Where M is, in this case your number is 3, so for your lines to be parallel, your new equation must also have 3.

Now enter: 7 = 3 (2) + b = 6 + b

So B is equal to one.

So your answer is y = 3x + 1

Hey, if the lines are parallel, the error is like the lines.

That is, y = 3x 4 here m = 3 [think of the common form y = mx + c]

So meter = 3

And the points are (2, 7)

So the equation is in terms of a point

(y7) = 3 (x2)

y = 3x + 1

For the linear equation, the intersection type (to the right of the moment) is y = mx + b. Dance Le Cass O س Vos Obtniz + 1 e Tout en Sutilisent pig zone du sign equivalent, Alvars IC les information vra es are: M, C East l'Mumble è M de la Ligni, ST Lee Confidence of XB East Lanterction Y, that's it . The y-axis situation where the lines intersect with the time of your query, both equations are already abnormal intersection type y = (4/3) x + (7/3) for m = 4/3 Are Multiply whatever you want and return with four / 3 so that 4/3 and = (4/3) x9 / 4 have one m = 4/3 which is 4/3 for both hits. Because your positions are the same, the tracks are parallel.

Y 3x 4

Y 3x 4

The parallel causes (3x in this case) are the same. Draw your equation and see where the intersection point should be and it connects the point (2,7).

y = 3x + c, the same for parallel lines

Roots (2.7),

7 = 3 * 2 + c => c = 1

Answer: y = 3x + 1

Points formula:

aa1 = m (xx1)

Parallel lines have the same thing, m = 3

y1 = 7

x1 = 2

y7 = 3 (x2)

Divide to get: y7 = 3x6

Add 7 on both sides and your answer will be y = 3x + 1

Help for example!

Y 3x 4

Y 3x 4

Parallel: y = 3x 4 (2, 7)?

Should I write an equation that is parallel to y = 3x 4 and contains (2, 7)? I mean, our ally gurus are not asking us to solve this ... it would be good if the necessary steps were taken.

y = mx + b

Where M is, in this case your number is 3, so for your lines to be parallel, your new equation must also have 3. B is the y-intercept (where the line touches the y-axis).

Now enter: 7 = 3 (2) + b = 6 + b

So B is equal to one.

So your answer is y = 3x + 1

Hey if the lines are parallel then the error is the same as the line

That is, y = 3x 4 here m = 3 [think of the common form y = mx + c]

So meter = 3

And the points are (2, 7)

So the equation is in terms of a point

(y7) = 3 (x2)

y = 3x + 1

For the linear equation, the intersection type (to the right of the moment) is y = mx + b. Deniz Le Cass O س Vos Obtniz + 1 e tout en Sutilisent pig zone du sign equivalent, Alvars IC les information vra es are: M, C East l'Mumble; that's it. A situation on the y axis where the lines meet at the time of your question, both equations already exist, type y = (4/3) x + (7/3) has an unusual intersection m = 4 / 3. Multiply if you want and invert 4/3 y = (4/3) x9 / 4 to get 4/3 is one m = 4/3 which is 4/3 for both hits. Because they are like you, the tracks are parallel.

y = 3x + c, the same is true for parallel lines

Roots (2.7),

7 = 3 * 2 + c => c = 1

Answer: y = 3x + 1

Y 3x 4

Y 3x 4

Parallel: y = 3x 4 (2, 7)? ۔

Should I write an equation that is parallel to y = 3x 4 and contains (2, 7)? Look, our seafaring instructor didn't tell us to fix it ... it would have been nice if steps had been taken.

y = mx + b

Where m is, then tu is 3. In order for the lines to be parallel, your new equation must also have 3..

Now enter: 7 = 3 (2) + b = 6 + b.

So b is equal to one.

So your answer is y = 3x + 1.

Hey, if the lines are parallel, the error is like the lines.

I mean y = 3x 4 here m = 3 [think of the common form y = mx + c]

So m = 3

And there are points (2, 7)

So the equation is by the point

(y7) = 3 (x2)

y = 3x + 1.

The type of intersection for the linear equation (to the right of this movement line) is y = mx + b. In this case you will find + 1y tout in the utilizer on a zone of equal sign, alors ici les information are vraies: m, c'est l'emblème de la ligne, est le coefficient of xb est l'intersection y, It means. The y-axis situation where the straight lines in your question meet, the two equations are already within the unusual intersection of type y = (4/3) x + (7/3) m = 4/3 ۔ Multiply what you want and invert 4/3 to get 4/3 and = (4/3) x9 / 4 has m = 4/3 which is 4/3 for both moves. Since you are the same, the tracks are parallel.

Parallel sources (3x in this case) are the same. Draw your equation and see where the intersection point should be and it intersects the point (2,7).

y = 3x + c, the same goes for parallel lines.

Roots (2.7),

7 = 3 * 2 + c => c = 1.

Answer: y = 3x + 1.

Formula of points:

aa1 = m (xx1)

Parallel lines have the same thing, m = 3.

y1 = 7.

x1 = 2.

y7 = 3 (x2)

Divide to get: y7 = 3x6.

Add 7 on both sides and your answer will be y = 3x + 1.

Help for example!

Y 3x 4

Y 3x 4

Y 3x 4

Parallel to: y = 3x 4 (2, 7)? 3

Should I write an equation parallel to y = 3x 4 and containing (2, 7)? The problem is that our guru doesn't ask us to solve it ... it would be nice if the steps were yes.

y = mx + b

Where m is, then tu is 3. In order for the lines to be parallel, your new equation must also have 3. B is y-intercept (where the line touches the y-axis).

Now enter: 7 = 3 (2) + b = 6 + b

So b is equal to one.

So your answer is y = 3x + 1

If the lines are parallel then the fault lies with the lines.

That is, y = 3x 4 here m = 3 [think of the common form y = mx + c]

So m = 3

And there are points (2, 7)

So the equation is per point.

(y7) = 3 (x2)

y = 3x + 1

Y 3x 4

Y 3x 4

The type of intersection for the linear equation (to the right of this momentum line) is y = mx + b. In this case you will find + 1y tout in the utilizer on a zone du signe equivalent, alors ici les information are vraies: m, c'est l'embleme de la ligne, est le coefficient of xb est l'intersection y, It means. The situation of the y axis where the lines meet at the time of your question, the two equations already exist, in an unusual intersection of type y = (4/3) x + (7/3) m = 4/3 Is. Multiply if you want and invert 4/3 to get 4/3 y = (4/3) x9 / 4 m is 4/3 which is 4/3 for both hits. Since your s are identical, the tracks are parallel.

The parallel meanings (3x in this case) are the same. Draw your equation and see where the intersection point should be and this stops point (2,7).

y = 3x + c, the same for parallel lines

Roots (2.7),

7 = 3 * 2 + c => c = 1

Answer: y = 3x + 1

Y 3x 4