Treat

as the boundary of the region

, where

is the part of the surface

bounded by

. We write

with

.
By Stoke's theorem, the line integral is equivalent to the surface integral over

of the curl of

. We have

so the line integral is equivalent to


where

is a vector-valued function that parameterizes

. In this case, we can take

with

and

. Then

and the integral becomes


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For the 1st one we need to solve x. x-y=2 step1: x-y+y=2+y answer: x=y+2
Step-by-step answer:
Please refer to attached image.
1. Quad PQRS is cyclic (all vertices on the same circle), so opposite angles are supplementary, meaning
that
angles QPS and QRS are supplementary =>
QPS+QRS=180 =>
QRS = 180 - 74 = 106
2. Triangle RSQ is isosceles with RS=RQ =>
angles RSQ and RQS are congruent.
3. Angle RSQ = (180 - 106) /2 = 74 / 2 = 37
4. QP is a diameter => angle QSP is a right-angle.
5. From (3) and (4) above,
angle RSP = 37+90 = 127
6. Since PQRS is cyclic, angles RQP and RSP are supplementary =>
RQP+RSP = 180 =>
x + 127 = 180 =>
x = 180 - 127 = 53 degrees.
Thts a nice quote. i dont get wht ur q is
Answer:
5 miles
Step-by-step explanation:
0.5 x 10
= 5