In many electronic circuits there is sometimes the need of a lower voltage then the supply voltage; we can achieve this goal simply by using what is usually referred to as "voltage divider".
This solution is valid only if the required current is minimal, as happens for example when you need to supply the correct voltage at the input of an operational amplifier.
The voltage divider, which is usually formed by two resistors, works by distributing the input voltage on each of the two resistors, in proportional parts. Figure 1 shows a classic case: a tension V_IN is applied to the two resistances R1 and R2, and, at the intermediate point, a voltage V_OUT is present, obviously always lower than V_IN.
Calculating the voltages is quite simple; we can first calculate, for example, the current I flowing in the resistors:
I = V_IN / (R1 + R2)
then we can calculate V_OUT as a voltage drop across the resistance R2:
V_OUT = I x R2
In the case shown in figure 1, we have R1 = 2.7 kΩ and R2 = 4.7 kΩ
If we suppose to supply the divider with a voltage V_IN = 5 V, the current will be I = 5 / (R1+R2) = 5 / 7,4 = 0,6756 A
It follows that the output voltage will be V_OUT = 4,7 x 0,6756 = 3,175 V

In practice, we can calculate the V_OUT voltage simply with the formula:
V_OUT = V_IN x R2 / (R1 + R2)

Referring to the resistances of common value (those of the commercial series, 5% tolerance) it can be quite boring to find the right combination of resistances to obtain a determined output voltage; in order to facilitate this choice, tables are shown below for three different input voltage values, and exactly for 5V, 9V and 12V.
You can use the following tables as the "math tables" of good memory.
You have to combine the value of R1, read in the upper horizontal line, with the value of R2, read in the left first column: in corrispondence of the cross point you can read the relative output voltage V_OUT.

Tables for calculating the voltage divider