Suppose C1 and C2 are physically the same curve, but they are parameterized so that the starting point of C1 is the ending point of C2, and
the ending point of C1 is the starting point of C2.
Express
∫ C1m(x, y)D „ x + n(x, y)D „ y
in terms of
∫ C2m(x, y)D „ x + n(x, y)D „ y.
∫ C1m(x, y)D ‚x + n(x, y)D ‚y = -∫ C2 m(x, y)D ‚x + n(x, y)D ‚y.

step-by-step explanation:

use pothagarean therom

a^2+b^2=c^2 so first ac

acm is a right triangle because c= 90°

and am (the hypotenuse or c) =16+9 =25

but there is not enough info we need another angle or side length

step-by-step explanation:

this can be solved first with pythagorean theorem, which tells us to square both legs of the triangle and sum it. this will be equal to the square of the hypotenuse.

the hypotenuse is the side "opposite" the 90 degree angle, which is x.

so we can write using pythagorean theorem,

simplifying:

to further simplify this radical, we need to use the property  

thus