Question

Inverse Variation Written Assignment..In need of help with all of it, if you can please help...don't have much time.

Answer 1

What is the assignment?

Answer 2

btw, inverse variation is just y=k/x

Answer 3

What don't you understand about the graphing? Pick an x value. Calculate the y value. Make a mark at that spot. Repeat.

Answer 4

just plot each point and then draw a curve through all the points that you graphed to best fit it

Answer 5

So would I graph the data from A or B or both? I don't understand...I don't even know what its supposed to look like to be perfectly honest. lol

Answer 6

i would make 2 separate graphs

Answer 7

Two graphs.

Here's a graph of an inverse variation:

Answer 8

when one value gets large, the other gets small, and vice versa.

Answer 9

Okay so do I graph these numbers exactly how they are...like the weight side would be x and the distance would be y or what...?

Answer 10

im sorry I've just never been shown how to do these right..

Answer 11

Yes, graph weight along the x axis and distance along the y axis. Usually, the independent variable goes on the x axis and the dependent variable goes on the y axis.

Answer 12

So I wouldn't need to make these numbers smaller? I'm just trying to picture in my head what this is supposed to look like...cause so far I got nothing. I was looking at this page http://math.tutorvista.com/number-system/inverse-variation.html but all of these numbers are big and I've never graphed from a table..ugh the life of home school...

Answer 13

Look, just make your tick marks be bigger chunks of number. Instead of 1, 2, 3, make them 10,20,30, or 100, 200, 300, or whatever gives you a reasonably sized graph.

Answer 14

Also, there's nothing that says your graph has to go from 0 to 3000 if all of the action in the graph is between 1800 and 2200. Just start the graph at 1700 and go to 2300 in that case.

Answer 15

totally made up numbers, only for illustration of the point, not related to your homework!

Answer 16

The first graph I did has a straight line when I plotted the points, it looks like a direct variation graph...is that okay or did I do something wrong?

Answer 17

It shouldn't be a completely straight line, but it doesn't bend much for those points, either.

Here's what I did. x's mark the data points.

Answer 18

and for the second one:

Answer 19

Ohhh..do the small numbers go on the bottom of the graph?? i wonder if i have my x and y axis confused ...my second one looks like yours though :)

Answer 20

The data in the first graph is showing \[y = \frac{90000}{x}\]I plotted both the data from the table (marked with the x's) and then the function for a much wider range, so you can see the data you have in context:

Answer 21

Oh okay...how'd you get the y to equal 9000?? cause i have 900 and 100 in the table...I'm starting to get something though so thank you for helping :]]

Answer 22

the relationship is \[y = \frac{k}{x}\]If we know that \(y=900\) when \(x = 100\) (or vice versa), then \[900 = \frac{k}{100}\]\[100*900 = 100*\frac{k}{100}\]\[90000 = k\]\[y = \frac{90000}{x}\]

Answer 23

Notice that in that last graph, the y-axis is shown at x = 400, NOT x = 0...

Answer 24

I should have adjusted that for you, here it is in the usual spot:

Answer 25

Oh okay!! thank you...=) alright so is that good for the graphs?? Because then it says Does it look as though there is a relationship between the weight and its distance from the fulcrum and what type of variation is there? I'm curious what is the fulcrum...

Answer 26

Oh well I gotta go, but will definitely be back for more help tomorrow.!

Answer 27

the fulcrum is the pivot on the arm...

Answer 28

the spot where the seesaw is held up.