Alan

Then it seems like you should plot the first of the eight values at 15 seconds, the second at 30 seconds, the third at 45 seconds and so on.

 However, you should really have recorded the time at which you took each measurement.  There is likely to be a lot of unnecessary scatter otherwise!

1f:

$$$$\frac{c^2d^4}{c^4d^2}\frac{c^6d}{(cd)^2}=\frac{c^8d^5}{c^6d^4}=c^{8-6}d^{5-4}=c^2d$$$$

6a:

$$$$z=\frac{3-xy}{2}$$
Multiply both sides by 2 $$2z=3-xy$$
Add xy to both sides $$xy+2z=3$$
Subtract 2z from both sides $$xy=3-2z$$
Divide both sides by y$$x=\frac{3-2z}{y}$$

Now let's see your attempt at some of them.

Hmm! Take care!

(2x³y²)³ means everything inside the brackets is cubed. So we have

2³(x³)³(y²)³, which is 8x^3*3y^2*3 which is 8x⁹y⁶

Or, look at it this way:

Your times are the same, but the DO values aren't.  I've just plotted your three values against the first three times.

How did you determine your three values?

With these images if you right-click on them you can choose "open image in new tab".  In the new tab the image isn't obscured.

The first answer is correct (copied directly from Wikipedia by the looks of it).

A truncated cube is one with the corners lopped off.

As long as I'm within 10% or so reinout I'm happy!!

And Rosala is quite right.

This paper essentially plots what I've plotted above (but with more data points).  Where do your DO values come from?  The dissolved oxygen values in the paper are plotted against temperature; the CO₂ values against time. I can't find your DO values in the paper anywhere.

Ah! I think CO₂ is actually C_O2, for concentration of oxygen.  

Since your values of DO are different from those in the paper, does that mean you have measured them yourself?  If so, is your problem that your times are different from those in the paper or what?  I'm still puzzled as to your specific difficulty.  The data in the paper (that in the last table) is easy to plot in the same way that I've done in the graph in my earlier answer.