I was mistaken. He was using a voltage divider.
This is the equivalent piece of the circuit I was concerned about...
simulate this circuit – Schematic created using CircuitLab
Because I didn't understand exactly how current flows, in my mind ONLY THE FIRST RESISTOR had any effect on the voltage of the output wire because logically current had only gotten to that point. The following terms may not be correct, but they suffice to explain the concept.
What I realize now is that the TOTAL RESISTANCE of everything leading back to ground dictates voltage as well. CURRENT RESISTANCE dictates the amount of CURRENT at the point of measurement.
The following is how you would get the output voltage, if it were unknown:
I = Amperage
V = Voltage
R = Resistance
Ohm's Law: \$I = V/R\$
Total Resistance (TR) (The Total Resistance in Your Circuit) = R1 + R2
Current Resistance (CR) (The Resistance in Your Circuit Up To The Point of Measurement) = R1
TR = 4.4k
CR = 2.2k
This will be the current in mA that flows through our circuit:
\$I = 10v/TR\$
\$I ≈ 2.27mA\$
This will be the voltage drop across the resistance up to our measurement point. (I.E. Our Output):
It is completely coincidental that our voltage drop is our desired voltage. At this stage we have out voltage drop not the actual output voltage.
\$V = .00227A * CR\$
\$Voltage Drop (VD) ≈ 4.994v ≈ 5v\$
This will be the voltage on our measurement point. (I.E. Our Output):
\$Output Voltage (OV) = V - VD\$
\$OV = 10v - 5v\$
\$OV = 5v\$