Alan

2 + 41 = 43

11 + 13 + 19 = 43

2 + 11 + 13 + 17 = 43

3 + 5 + 7 + 11 + 17 = 43

"...a. Find the probability that the number on the roll of the die is at most 3.
b. Find the probability that the number on the roll of the die is at least 5."

As follows:

You could also tackle this as follows:

31x + 29y = 1125

30(x+y) + (x-y) = 1125

x+y = 37 + 15/30 - (x-y)/30

let x-y = 15, then

x + y = 37

(Can also find that x = 26, y = 11)

Distance is given by \(d=\sqrt{x^2+y^2+z^2}\)  

Symmetry dictates that x = y = z at the nearest point to the origin, so \(d=\sqrt3\times x\)

Given that \(xyz=1\)  we have \(x^3=1\) or \(x = 1\)  so minimum distance is \(d=\sqrt3\)

EP's answer to the question is correct!  If you think it isn't then you should say why you think that.  If you've just entered the result in some on-line set of questions and the response was that it was incorrect, it is possible that you haven't specified the question correctly here.

You are right!  In fact, if you look at my f(x)+f((x-1)/x) it doesn't equal g(x), there is multiplier of 5 that shouldn't be there!!

Deleted.  See below

(Clearly f(x) = (2 + 3x)/x is incorrect.)

Just set 2x² + 20x = 112

Divide through by 2

x² + 10x = 56

Rearrange

x² + 10x - 56 = 0

This factors as 

(x + 14)(x - 4) = 0

The only positive solution is x = 4 (inches).

If you expect the fourth root of your large number to be an integer you are doomed to disappointment - search "Fermat's last theorem".