Answer:
[tex]\bold{h=\dfrac{3Q}{t-r}}[/tex]
Step-by-step explanation:
[tex]\bold{r+\frac{3Q}h=t}\\-r\qquad\ -r\\\bold{\frac{3Q}h=t-r}\\\cdot h\qquad \cdot h\\ \bold{3Q=(t-r)h}\\^{\div(t-r)\quad\div(t-r)}\\\bold{\dfrac{3Q}{t-r}=h}[/tex]
Or, if you mean (r+3Q)/h=t:
[tex]\bold{\frac{r+3Q}h=t}\\{}\ \ \cdot h\quad\ \cdot h\\\bold{r+3Q=ht}\\{}\ \ \div t\qquad \div t\\\bold{\dfrac{r+3Q}{t}=h}\\\\\bold{h=\dfrac{r+3Q}{t}}[/tex]
