Let F be a planar vector field and again consider an annular region A as in the previous problem. Suppose that F has no equilibria and that F points inward along the boundary of the annulus, as before. (a) Prove there is a closed orbit in A. (Notice that the hypothesis is weaker than in the previous problem.)(b) If there are exactly seven closed orbits in A, show that one of them has orbits spiraling toward it from both sides.

step-by-step explanation:

mean or average is 40.6

since no one answered yet i will give it a shot, not completely sure tho.

i think it should be y=cos(x+2pi), (third answer, answer c)

good luck.

parallelogram is the   answer

step-by-step explanation:

consider the parallelogram shown in the attached image.

in the parallelogram hijk, the length of hk is given as,

from the properties of parallelogram we know that the opposite sides of a parallelogram are parallel to each other and are equal in measure.

therefore, the length of the side ij must be equal to the length of the side hk and that is .

hence, the expression for the length of ij confirms that hijk is a parallelogram.