Not much can be done without knowing what

is, but at the least we can set up the integral.
First parameterize the pieces of the contour:


where

and

. You have


and so the work is given by the integral


Answer:
b
Step-by-step explanation:
Answer:
(3×20%+2×70%)/3+2=40%
Step-by-step explanation:
Assuming the potions are the same type or do mix then probably the concentration of the potion depends on the type of reaction they have to each other.
Yet we can average the percentage of the active ingredient by the principle mentioned above