I WILL fan and give medals! :))) Which equation represents a direct linear variation? A.y = x2 B. y=1/x C.y = x + 2 D.y = ½x
I think its B. but I might be wrong.
@Isaiah.Feynman
check which of the equations is/are not raised to any power? since x is th eindependent variable, I will check this in the RHS!
For example. I can see that C and D only relate to the dependent variable, y with nothing raised to any power!
of as I see... D will be eliminated if x is in d denominator? then only C is linear in x? please check!
@whpalmer4 what do you think it is I think its b. :)
Direct linear variation is like \(y = kx\)
SO then its a.?
linear means that if you double \(x\), you also double \(y\), no matter what value you start with. That would not be true of answer B.
No it cant be since x is not linear in B. Check and convince yourself that a linear relationship is only observed if both x and y vary in a linear fashion! In B, u can see that x is in a reciprocal relationship with y!
A. is \(y = x^2\), right?
yes
Yeah, so if you double \(x\), \(y\) is quadrupled because you are squaring that factor of 2
x = 1, y = 1*1 = 1 double x to 2 x = 2, y = 2*2 = 4 oops, y went from 1 to 4 when x went from 1 to 2. That isn't linear.
ok so its NOT A.
We've already ruled out B. C is intriguing, because it doesn't raise \(x\) to a power: \[y = x+2\] x = 1, y = 1+2 = 3 x = 2, y = 2 + 2 = 4 oops, failed the linear test again — we doubled x and y did not double in response
C IS linear, but it isn't direct linear variation because of that "+2" in there.
So then its D.
We ruled everything else out. :)
D: \(y = \frac{1}{2}x\) x = 1, y = 1/2 x = 2, y = 1 hey, we doubled x and it doubled y! success!
Thanks! Can I tag you in another question please?
now, a word of warning: testing two points like that isn't enough to prove that it is linear, only to prove that it isn't. but in this case, we could use that to rule out all the other answers
ok thanks for the pointer! :)
unfortunately I have to run...
Awww! What will I ever do without you? haha its ok until next time! Bye!
go ahead and post it in a new post, tag me via "@whpalmer4" and I'll answer it next time I'm on, if someone else hasn't beaten me to it
ok thanks for everything!
Join our real-time social learning platform and learn together with your friends!