I hope this helps!! Look at the top and the graph, don't know which answer you prefer :)
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.
Gchdbncxthysshj teethed how
Answer:
The given statement that value 5 is an upper bound for the zeros of the function f(x) = x⁴ + x³ - 11x² - 9x + 18 will be true.
Step-by-step explanation:
Given

We know the rational zeros theorem such as:
if
is a zero of the function
,
then
.
As the
is a polynomial of degree
, hence it can not have more than
real zeros.
Let us put certain values in the function,
,
,
,
,
,
,
,
, 
From the above calculation results, we determined that
zeros as
and
.
Hence, we can check that

Observe that,
,
increases rapidly, so there will be no zeros for
.
Therefore, the given statement that value 5 is an upper bound for the zeros of the function f(x) = x⁴ + x³ - 11x² - 9x + 18 will be true.