Mathematics : asked on Kylehc21
 19.11.2020

Are the lines y=0 and x=0 parallel?

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Step-by-step answer

09.07.2023, solved by verified expert
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No, these equations are not parallel.

Step-by-step explanation:

The equation Are the lines y=0 and x=0 parallel?, №18010368, 19.11.2020 10:53 is a vertical line. This equation, in fact, is the y-axis.

The equation Are the lines y=0 and x=0 parallel?, №18010368, 19.11.2020 10:53 is a horizontal line. This equation, in fact, is the x-axis.

These lines are perpendicular, not parallel.

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Faq

Mathematics
Step-by-step answer
P Answered by Specialist

No

Step-by-step explanation:

y=0 looks like |  and x=0 looks like  . This means that they are and not parallel

Mathematics
Step-by-step answer
P Answered by Specialist

No, these equations are not parallel.

Step-by-step explanation:

The equation y=0 is a vertical line. This equation, in fact, is the y-axis.

The equation x=0 is a horizontal line. This equation, in fact, is the x-axis.

These lines are perpendicular, not parallel.

Mathematics
Step-by-step answer
P Answered by Master

No

Step-by-step explanation:

y=0 looks like |  and x=0 looks like  . This means that they are and not parallel

Physics
Step-by-step answer
P Answered by PhD
The force on the tool is entirely in the negative-y direction.
So no work is done during any moves in the x-direction.

The work will be completely defined by

                     (Force) x (distance in the y-direction),

and it won't matter what route the tool follows to get anywhere. 
Only the initial and final y-coordinates matter.

We know that    F = - 2.85 y².  (I have no idea what that ' j ' is doing there.)
Remember that 'F' is pointing down.

From  y=0  to  y=2.40  is a distance of  2.40  upward.

Sadly, since the force is not linear over the distance, I don't think
we can use the usual formula for  Work = (force) x (distance).
I think instead we'll need to integrate the force over the distance,
and I can't wait to see whether I still know how to do that.

        Work  =  integral of (F·dy) evaluated from  0  to  2.40

                  =  integral of (-2.85 y² dy) evaluated from  0  to  2.40

                 =  (-2.85) · integral of  (y² dy)  evaluated from  0  to  2.40 .

Now, integral of (y² dy)  =  1/3  y³ .

Evaluated from  0  to  2.40 , it's  (1/3 · 2.40³) - (1/3 · 0³)

                                            =  1/3 · 13.824  =  4.608 .

And the work  =  (-2.85) · the integral

                     =  (-2.85) · (4.608)

                     =      - 13.133  .

-- There are no units in the question (except for that mysterious ' j ' after the 'F',
which totally doesn't make any sense at all).
If the ' F ' is newtons and the 2.40 is meters, then the  -13.133 is joules.

-- The work done by the force is negative, because the force points
DOWN but we lifted the tool UP to 2.40.  Somebody had to provide
13.133 of positive work to lift the tool up against the force, and the force
itself did 13.133 of negative work to 'allow' the tool to move up. 

-- It doesn't matter whether the tool goes there along the line  x=y , or
by some other route.  WHATEVER the route is, the work done by ' F ' 
is going to total up to be  -13.133 joules at the end of the day.

As I hinted earlier, the last time I actually studied integration was in 1972,
and I haven't really used it too much since then.  But that's my answer
and I'm stickin to it.  If I'm wrong, then I'm wrong, and I hope somebody
will show me where I'm wrong.
Physics
Step-by-step answer
P Answered by PhD
The force on the tool is entirely in the negative-y direction.
So no work is done during any moves in the x-direction.

The work will be completely defined by

                     (Force) x (distance in the y-direction),

and it won't matter what route the tool follows to get anywhere. 
Only the initial and final y-coordinates matter.

We know that    F = - 2.85 y².  (I have no idea what that ' j ' is doing there.)
Remember that 'F' is pointing down.

From  y=0  to  y=2.40  is a distance of  2.40  upward.

Sadly, since the force is not linear over the distance, I don't think
we can use the usual formula for  Work = (force) x (distance).
I think instead we'll need to integrate the force over the distance,
and I can't wait to see whether I still know how to do that.

        Work  =  integral of (F·dy) evaluated from  0  to  2.40

                  =  integral of (-2.85 y² dy) evaluated from  0  to  2.40

                 =  (-2.85) · integral of  (y² dy)  evaluated from  0  to  2.40 .

Now, integral of (y² dy)  =  1/3  y³ .

Evaluated from  0  to  2.40 , it's  (1/3 · 2.40³) - (1/3 · 0³)

                                            =  1/3 · 13.824  =  4.608 .

And the work  =  (-2.85) · the integral

                     =  (-2.85) · (4.608)

                     =      - 13.133  .

-- There are no units in the question (except for that mysterious ' j ' after the 'F',
which totally doesn't make any sense at all).
If the ' F ' is newtons and the 2.40 is meters, then the  -13.133 is joules.

-- The work done by the force is negative, because the force points
DOWN but we lifted the tool UP to 2.40.  Somebody had to provide
13.133 of positive work to lift the tool up against the force, and the force
itself did 13.133 of negative work to 'allow' the tool to move up. 

-- It doesn't matter whether the tool goes there along the line  x=y , or
by some other route.  WHATEVER the route is, the work done by ' F ' 
is going to total up to be  -13.133 joules at the end of the day.

As I hinted earlier, the last time I actually studied integration was in 1972,
and I haven't really used it too much since then.  But that's my answer
and I'm stickin to it.  If I'm wrong, then I'm wrong, and I hope somebody
will show me where I'm wrong.
Mathematics
Step-by-step answer
P Answered by Master
Answer D because it will go straight across and be parallel to the x-axis because it’s a Y number
Mathematics
Step-by-step answer
P Answered by PhD

2x - y = 0

Step-by-step explanation:

The equation of a line in slope- intercept form is

y = mx + c ( m is the slope and c the y- intercept )

y = 2x - 1 ← is in slope- intercept form

with slope m = 2

Parallel lines have equal slopes, thus

y = 2x + c ← is the partial equation

To find c substitute (1, 2) into the partial equation

2 = 2 + c ⇒ c = 2 - 2 = 0

y = 2x ← equation in slope- intercept form

Subtract y from both sides

0 = 2x - y , that is

2x - y = 0 ← equation in standard form

Mathematics
Step-by-step answer
P Answered by Master

D \:  y=6

Step-by-step explanation:

SORRY IF THIS IS WRONG......

Mathematics
Step-by-step answer
P Answered by Specialist
Answer D because it will go straight across and be parallel to the x-axis because it’s a Y number
Mathematics
Step-by-step answer
P Answered by PhD

2x - y = 0

Step-by-step explanation:

The equation of a line in slope- intercept form is

y = mx + c ( m is the slope and c the y- intercept )

y = 2x - 1 ← is in slope- intercept form

with slope m = 2

Parallel lines have equal slopes, thus

y = 2x + c ← is the partial equation

To find c substitute (1, 2) into the partial equation

2 = 2 + c ⇒ c = 2 - 2 = 0

y = 2x ← equation in slope- intercept form

Subtract y from both sides

0 = 2x - y , that is

2x - y = 0 ← equation in standard form

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