What is the constant of variation in the graph below? 3 12 24 48
@Donblue22
I was never really very good at these :/
k
@SolomonZelman
@ganeshie8 @UnkleRhaukus @freethinker @cram
At first, just by the way the x increases and y decreases you can see that this is an inverse variation. Inverse variation looks like this: \(\huge\color{blue}{ y= \frac{k}{x} }\) where k is just a number (real number) Let's try to solve for ` k ` by plugging in the first coordinate. \(\huge\color{blue}{ y= \frac{k}{x} }\) \(\huge\color{blue}{ 24= \frac{k}{2} }\) \(\huge\color{blue}{ 48= k }\)
48?
At first, just by the way the x increases and y decreases you can see that this is an inverse variation. Inverse variation looks like this: \(\huge\color{blue}{ y= \frac{k}{x} }\) where k is just a number (real number) Let's try to solve for ` k ` by plugging in the first coordinate. \(\huge\color{blue}{ y= \frac{k}{x} }\) \(\huge\color{blue}{ 24= \frac{k}{2} }\) \(\huge\color{blue}{ 48= k }\)
so 48?
I got disconnected from the site... yes !
oh okay! more?
Which ordered pair lies in Quadrant 1? (-7,-1) (-1,0) (3,4) (4,-6)
3,4
the downward slope seem to indicate that there is 9, 12, 14 addition
the one that has a positive x coordinate, and a positive y coordinate,
thanks @ramit.dour
wait there's a graph? laughing out loud
no, it's just a question.
Q1 is always (+,+)
that is positive x value and positive y value
3,4
i wish I could give multiple medals!
the point of intersection of line x=3 and y=4 lies in first quadrant
I don't really like medals
Quadrant 1 \(\normalsize\color{blue}{ (+,+) }\) Quadrant 2 \(\normalsize\color{blue}{ (-,+) }\) Quadrant 3 \(\normalsize\color{blue}{ (-,-) }\) Quadrant 4 \(\normalsize\color{blue}{ (+,-) }\)
so it would be 3,4 right?
you don't even have to graph it it should be an intuition by now
yes, @Monster1234
|dw:1406036159591:dw|
can you tell me the real number range of each quadrants? |dw:1406036188812:dw|
Join our real-time social learning platform and learn together with your friends!