9514 1404 393
Answer:
(b) P = 2(S -A) ÷ L
Step-by-step explanation:
Undo what is done to P.
[tex]S=A+\dfrac{1}{2}PL\qquad\text{given}\\\\S-A = \dfrac{1}{2}PL\qquad\text{subtract $A$}\\\\\dfrac{2}{L}(S-A)=P \qquad\text{multiply by $\dfrac{2}{L}$}\\\\\boxed{P=\dfrac{2(S-A)}{L}}[/tex]
