Rom

$x+1 = 4 \Rightarrow x=3\\ \text{however 3 is not in the domain of either side of the equation (no division by 0)}$

$\text{again the point is no dividing by 0}\\ x^2-6x+8 = (x-4)(x-2) = 0 \text{ when }x=4, \text{ or }x=2\\ \text{Thus neither 4 nor 2 are included in the domain}$

That would be the object's potential energy

capital E usually means the number following it should be treated as an exponent.

You'll see something like 2.5E6 which means $2.5 \times 10^6$

$\text{There are }\dbinom{4}{1}19^3 \text{ different words containing a single instance of the letter A} \\ \dbinom{4}{1} \text{ spots for the A and then a choice of 19 letters, B-T, for each of the 3 remaining spots}\\ \text{Similarly there are }\dbinom{4}{k}19^{4-k} \text{ words containing }k \text{ instances of the letter A}\\ \text{Thus we have}\\ N=\sum \limits_{k=1}^4 ~\dbinom{4}{k}19^{4-k} = 29679$

It strikes me that perhaps a simpler way of doing this is to subtract all the words with no instances of the letter A

from the total number of words possible.

$\text{There are }19^4 \text{ words that have no A in them}\\ \text{There are }20^4 \text{ total possible words}\\ N=20^4-19^4 = 29679$

I did two of these for you... from what class are you getting these questions?

I was able to see the trick in the first two but this one seems truly complicated.

Is your class teaching any strategies for dealing with these?

u wot mate?

disadvantages

1) they make me pay attention in class/to my online reading

2) they make me read my textbook

advantages

1) they make me pay attention in class/to my online reading

2) they make me read my textbook

If you are strictly enforcing 5 digits, as in no number can start with any number of zeros.

Then we have 

$n = 9 \cdot 10^4 = 90000$

I'm guessing the items in red are what they want you to solve for.

There appear to be 10

$\{1,64,729,4096,15625,46656,117649,262144,531441,1000000\}$