{{NumFormula|:|<math>y=ax+b</math>|Eq. 3}}

${\displaystyle y=ax+b}$

(Eq. 3)

{{NumFormula|:|<math>ax^2+bx+c=0</math>|Eq. 3}}

${\displaystyle ax^{2}+bx+c=0}$

(Eq. 3)

{{NumFormula|:|<math>\Psi(x_1,x_2)=U(x_1)V(x_2)</math>|2}}

${\displaystyle \Psi (x_{1},x_{2})=U(x_{1})V(x_{2})}$

(2)

{{NumFormula||<math>\mathbf{a}(t)=\frac{d}{dt}\mathbf{v}(t)</math>|3.5}}

{{NumFormula|:|<math>\mathbf{a}(t)=\frac{d}{dt}\mathbf{v}(t)</math>|1}}

${\displaystyle \mathbf {a} (t)={\frac {d}{dt}}\mathbf {v} (t)}$

(1)

{{NumFormula|::|<math>\mathbf{a}(t)=\frac{d}{dt}\mathbf{v}(t)</math>|13.7}}

${\displaystyle \mathbf {a} (t)={\frac {d}{dt}}\mathbf {v} (t)}$

(13.7)

{{NumFormula|:::|<math>\mathbf{a}(t)=\frac{d}{dt}\mathbf{v}(t)</math>|1.2}}

${\displaystyle \mathbf {a} (t)={\frac {d}{dt}}\mathbf {v} (t)}$

(1.2)

{{NumFormula|1=:|2=<math>\mathbf{a}(t)=\frac{d}{dt}\mathbf{v}(t)</math>|3=3.5|RawN=.}} [L'aver posto RawN uguale a un valore non nullo fa sì che il numero venga mostrato senza parentesi tonde predefinite ma esattamente come scritto nel parametro "3".]

${\displaystyle \mathbf {a} (t)={\frac {d}{dt}}\mathbf {v} (t)}$

3.5

{{NumFormula|1=:|2=<math>\mathbf{a}(t)=\frac{d}{dt}\mathbf{v}(t)</math>|3=<3.5>|RawN=.}}

${\displaystyle \mathbf {a} (t)={\frac {d}{dt}}\mathbf {v} (t)}$

<3.5>

{{NumFormula|1=:|2=<math>\mathbf{a}(t)=\frac{d}{dt}\mathbf{v}(t)</math>|3=[3.5]|RawN=.}}

${\displaystyle \mathbf {a} (t)={\frac {d}{dt}}\mathbf {v} (t)}$

[3.5]

{{NumFormula|1=:|2=<math>\mathbf{a}(t)=\frac{d}{dt}\mathbf{v}(t)</math>|3='''[3.5]'''|RawN=.}}

${\displaystyle \mathbf {a} (t)={\frac {d}{dt}}\mathbf {v} (t)}$

[3.5]

{{NumFormula|1=:|2=<math>\mathbf{a}(t)=\frac{d}{dt}\mathbf{v}(t)</math>|3=<Big>'''[3.5]'''</Big>}}

${\displaystyle \mathbf {a} (t)={\frac {d}{dt}}\mathbf {v} (t)}$

([3.5])

{{NumFormula|1=:|2=<math>\mathbf{a}(t)=\frac{d}{dt}\mathbf{v}(t)</math>|3=<Big>'''[3.5]'''</Big>|RawN=.}}

${\displaystyle \mathbf {a} (t)={\frac {d}{dt}}\mathbf {v} (t)}$

[3.5]

{{NumFormula|1=:|2=<math>\mathbf{a}(t)=\frac{d}{dt}\mathbf{v}(t)</math>|3=<math>(3.5)</math>|RawN=.}}

${\displaystyle \mathbf {a} (t)={\frac {d}{dt}}\mathbf {v} (t)}$

${\displaystyle (3.5)\,}$

{{NumFormula|1=:|2=<math>\mathbf{a}(t)=\frac{d}{dt}\mathbf{v}(t)</math>|3=<Big>'''(3.5)'''</Big>|RawN=.|stilelinea=1px dashed red}} [I nomi dei parametri "1", "2" e "3" possono essere esplicitati, come in questo caso, o meno, come nei casi precedenti, in cui non venivano indicati.]

${\displaystyle \mathbf {a} (t)={\frac {d}{dt}}\mathbf {v} (t)}$

(3.5)

{{NumFormula|1=:|2=<math>\mathbf{a}(t)=\frac{d}{dt}\mathbf{v}(t)</math>|3=<Big>'''(3.5)'''</Big>|RawN=.|stilelinea=3px dashed #0a7392}}

${\displaystyle \mathbf {a} (t)={\frac {d}{dt}}\mathbf {v} (t)}$

(3.5)

{{NumFormula|1=:|2=<math>\mathbf{a}(t)=\frac{d}{dt}\mathbf{v}(t)</math>|3=<Big>'''[3.5]'''</Big>|RawN=.|stilelinea=3px solid green}}

${\displaystyle \mathbf {a} (t)={\frac {d}{dt}}\mathbf {v} (t)}$

[3.5]

{{NumFormula|1=:|2=<math>\mathbf{a}(t)=\frac{d}{dt}\mathbf{v}(t)</math>|3=<Big>'''[3.5]'''</Big>|RawN=.|stilelinea=5px dotted blue}}

${\displaystyle \mathbf {a} (t)={\frac {d}{dt}}\mathbf {v} (t)}$

[3.5]

{{NumFormula|1=:|2=<math>\mathbf{a}(t)=\frac{d}{dt}\mathbf{v}(t)</math>|3=<Big>'''[3.5]'''</Big>|RawN=.|stilelinea=0px solid green}}

${\displaystyle \mathbf {a} (t)={\frac {d}{dt}}\mathbf {v} (t)}$

[3.5]

{{NumFormula|1=:|2=<math>\mathbf{a}(t)=\frac{d}{dt}\mathbf{v}(t)</math>|3=<Big>'''[3.5]'''</Big>|RawN=.|stilelinea=5px none green}}

${\displaystyle \mathbf {a} (t)={\frac {d}{dt}}\mathbf {v} (t)}$

[3.5]

{{NumFormula|1=:|2=<math>\mathbf{a}(t)=\frac{d}{dt}\mathbf{v}(t)</math>|3=<Big>'''[3.5]'''</Big>|RawN=.|stilelinea=3px double green}}

${\displaystyle \mathbf {a} (t)={\frac {d}{dt}}\mathbf {v} (t)}$

[3.5]

{{NumFormula|:|<math>y=ax+b</math>|Eq. 3|bordo=1}}

${\displaystyle y=ax+b}$

(Eq. 3)