Isolate "y" by using the reverse of PEMDAS
Answer: [tex]\bold{y=\dfrac{m}{b}}[/tex]
Step-by-step explanation:
[tex]b{y} = m\\\\\dfrac{b\boxed{y}}{b}=\dfrac{m}{b}\\\\y=\dfrac{m}{b}[/tex]
Answer: [tex]\bold{y=\dfrac{c-b}{a}}[/tex]
[tex]ay+b=c\\\\\boxed{ay}+b=c\\.\underline{\ \quad-b}\quad \underline{-b}\\ay\qquad=c-b\\\\\dfrac{a\boxed{y}}{a}=\dfrac{c-b}{a}\\\\y=\dfrac{c-b}{a}[/tex]
Answer: [tex]\bold{\pm \sqrt{\dfrac{v}{\pi h}}=y}[/tex]
[tex]v=\pi y^2h\\\\\dfrac{v}{\pi h}=\dfrac{\pi \boxed{y^2}h}{\pi h}\\\\\\\dfrac{v}{\pi h}=y^2\\\\\\\sqrt{\dfrac{v}{\pi h}}=\sqrt{y^2}\\\\\\\pm \sqrt{\dfrac{v}{\pi h}}=y[/tex]
Answer: [tex]\bold{\dfrac{a}{pm}=y}[/tex]
[tex]a=pmy\\\\\dfrac{a}{pm}=\dfrac{pm\boxed{y}}{pm}\\\\\\\dfrac{a}{pm}=y[/tex]
