Question

is the equation a direct linear variation?

Answer 1

y=2x^2

y=x+3

y=3x

Answer 2

if you can rearrange the formula into the form \(y = kx\) where \(k\) is a constant, it is direct linear variation. Otherwise, no.

Answer 3

direct linear variation implies that if you double \(x\), \(y\) will also double, for any value of \(x\).

Answer 4

thank you

Answer 5

which of the 3 equations represent direct linear variation?

Answer 6

i think y=x+3

Answer 7

im bad at this type of math really bad

Answer 8

well, let's check.

say x = 1, then y = x+3 = 1+3 = 4.

now let's double x:
x = 2, y = x+3 = 2+3 = 5.

Did y double when we doubled x?

Or, using my first answer, can we arrange \(y = x+3\) to be in the form \(y = kx\)? I can't see any way to convert the addition of 3 into a multiplication by a constant, can you?

Therefore, \(y = x+3\) is not direct linear variation.

Answer 9

damn

Answer 10

what about A

Answer 11

Okay, well, that doesn't have any addition or subtraction in it, which would be a deal-killer. Let's investigate some more:

\[y = 2x^2\]\[y = kx\]Do those appear to have the same form?

Answer 12

i have no idea

Answer 13

well, y matches with y. 2 is a constant, and matches with \(k\). Does \(x^2\) match \(x\)?

Answer 14

No, it doesn't — \(x^2 = x*x\) so the equation in A is really \[y = 2*x*x\]\[y=k*x\]The two don't match up, there's an extra \(x\) in the first one.

Answer 15

so the answer is C

Answer 16

Also, if we say x=1, then \(y = 2(1)^2 = 2\). Change \(x\) to 2, and what happens to \(y\)? \(y = 2(2)^2 = 8\). That's not doubling, that's quadrupling. Again, not direct linear variation.

Looking at the last choice, \[y=3x\]\[y=kx\]Same form if \(k=3\)

x=1, y = 3x = 3(1) = 3
x=2, y = 3(2) = 6

that was doubling, so it doesn't fail that test either. C is direct linear variation.

Answer 17

Do you want me to ask you a few more, just to make sure you have it down?

Answer 18

no i think i got it but thank you:)