Ohms question
 

1 - 8 ohm cab run with 1 - 4 ohm cab + a 6 ohm load. Correct or not?

Not.

2.67

That's right 2.67 ohms. The lower the ohm number the less resistance. Less resistance draws more power. That why amps rated at 4 ohms start to over heat and/or burn up when paired with a 2 ohm speaker load.

yep, 2.67 ohms (or 2 2/3 ohms if you prefer)

1 - 8 = -7
1 - 4 = -3

- 10 + 6 = -4

Correct or not?

In terms of ohms, that^^^^^^is totally wrong.

Appro 3 Ohm

Unlikely the "average" amount runs below 3 Ohm

2.67 Ohms. Reciprocal formula. More here.

http://www.tpub.com/neets/book1/chapter3/1-26.htm

No Parallel: RT = (R1R2)/(R1 + R2)
4X8 / 4+8= 2.6666.......... Rounded to 2.67

As already mentioned 2.67 ohms.

but

SQRT(-1) = j [ohm]

If you would like to rewrite the books be my guest, won't make it correct. Formula and links have been offered.

I'm not at school but thinking about the real physics of speakers.

You can't predict the impedance unless you measure the impedance with a proper frequency band measuring voltage and current.

Most of the time the impedance calculates a bit above the nominal.
Depending on EQ settings the impedance varies upwards to higher values (most of the time).

But, ..so what.

That's right, but understanding Formulas is a very different animal

edit:
Sorry I was wrong, I tried to tell:

Adaptation of Formulas

The OP stated this:

1 - 8 ohm cab run with 1 - 4 ohm cab + a 6 ohm load. Correct or not?

But I think they meant this:

1 - 8 ohm cab run with 1 - 4 ohm cab = a 6 ohm load. Correct or not?

2.67 is correct.

That's not wrong (theoretical) but does not reflect the behavior of a real terms 8 Ohm and 4 Ohm speaker.

No.
one 8ohm cab with one 4 ohm cab = 2.67 ohms, as has been said many times in this thread already.

Speaker are rated for nominal impedance, the point in the pass-band they present the maximum load on the "amp". This is the figure of concern and a simple resistance formula can apply. In reality speakers have inductive and capacitive reactance (to a much smaller degree) as well.

IF you limit the "speakers" to a frequency much higher than lowest part of the impedance curve then you will need to calculate that new impedance. From there you can use the same resistance formula to obtain the new load presented.

That was meant to be a joke

You forgot the :)