Ask your own question, for FREE!
Mathematics 23 Online
OpenStudy (anonymous):

Which equation represents a direct linear variation? A. Y=x^2 B. Y=1/x C. y = x + 3 D. Y= 2/5x

OpenStudy (anonymous):

@freckles help me?

OpenStudy (freckles):

linear is when you have something in the form y=mx+b or x=a number or y=a number --- but if have something is directly proportional to something else then that one something=a constant*the other something --- so I guess a direct linear variation would be something of a mixture like y=mx

OpenStudy (anonymous):

So the answer would be B?

OpenStudy (freckles):

well that would not be a linear equation

OpenStudy (freckles):

y=mx <---no x is on bottom

OpenStudy (anonymous):

Hmmm, SO A?

OpenStudy (freckles):

I think you should invest some time into what I said

OpenStudy (freckles):

y=mx+b or x=constant or y=constant are all linear equations

OpenStudy (freckles):

so y=x^2 or y=1/x or y=sqrt(x) or not linear they are not in the form I described

OpenStudy (anonymous):

OHHHHHHHH i get it ,it would be C.

OpenStudy (freckles):

Is D y=2x/5?

OpenStudy (freckles):

\[y=\frac{2}{5}x\]

OpenStudy (anonymous):

Yes, the second one

OpenStudy (freckles):

those are the same equation I just mentioned

OpenStudy (freckles):

anyways d is the exact mixture i was talking about

OpenStudy (freckles):

y=mx is a direct linear variation

OpenStudy (freckles):

y=mx+b is a linear equation

OpenStudy (freckles):

but y=mx has the direct proportion we were looking for

OpenStudy (anonymous):

Ok im going to review this but thank you for helping me understand it a little better.

OpenStudy (freckles):

as well as the linear

OpenStudy (freckles):

so you understand that d is the answer

OpenStudy (anonymous):

Yes

OpenStudy (freckles):

but c I would give partial credit for because it s a linear equation after all

OpenStudy (freckles):

it just didn't have the direct thing going for it

Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!
Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!