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Mathematics 9 Online
OpenStudy (anonymous):

Need help with this problem!...

OpenStudy (anonymous):

OpenStudy (anonymous):

@Mertsj

OpenStudy (mertsj):

I thought we did this yesterday.

OpenStudy (anonymous):

@Mertsj i don't think we did

OpenStudy (mertsj):

Direct variations are of the form y = k x where k is a constant.

OpenStudy (mertsj):

In the case of a line, we call k the slope and name it m. So we see that a line that is in y = mx+b form and where b is 0 is a direct variation.

OpenStudy (anonymous):

@Mertsj ok

OpenStudy (mertsj):

Is the y intercept of your line 0?

OpenStudy (anonymous):

@Mertsj no?

OpenStudy (mertsj):

Where does your line cross the y axis?

OpenStudy (anonymous):

@Mertsj 0

OpenStudy (mertsj):

The y intercept is the point where the line crosses the y axis. So what is your y intercept?

OpenStudy (anonymous):

@Mertsj it would be 0

OpenStudy (mertsj):

So it is a direct variation since the y intercept is 0 and it is a line.

OpenStudy (mertsj):

What is the slope (which is the constant of variation)?

OpenStudy (anonymous):

@Mertsj 2/3

OpenStudy (mertsj):

The line is falling.

OpenStudy (mertsj):

So that means that the slope is negative

OpenStudy (anonymous):

@Mertsj oh so its-3/2

OpenStudy (mertsj):

the change in y goes on top

OpenStudy (anonymous):

@Mertsj i dont know what you mean

OpenStudy (mertsj):

slope is rise over run

OpenStudy (anonymous):

@Mertsj yes, so then you go down 4 and run 6

OpenStudy (anonymous):

@Mertsj simplified would be -3/2 right?

OpenStudy (mertsj):

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