Z would be located on 2, since (9+2 =11)
By Stokes' theorem,

where

is the circular boundary of the hemisphere

in the

-

plane. We can parameterize the boundary via the "standard" choice of polar coordinates, setting

where

. Then the line integral is


We can check this result by evaluating the equivalent surface integral. We have

and we can parameterize

by

so that

where

and

. Then,

as expected.
Answer:
30 60 90 triangle
the hypotenuse is twice the length of the shorter leg length.
Answer:
quadrilateral ABCD is not congruent to quadrilateral KLMN. quadrilateral ABCD cannot be mapped onto quadrilateral KLMN through a series of rotations, reflections or translations.
Range R={-3-11,1,17}
domain D={-9,-3,5}