Answer:

Step-by-step explanation:
We are given the trigonometric equation of:

Let u = 4x then:

Find a measurement that makes sin(u) = √3/2 true within [0, π) which are u = 60° (π/3) and u = 120° (2π/3).

Convert u-term back to 4x:

Divide both sides by 4:

The interval is given to be 0 ≤ 4x < π therefore the new interval is 0 ≤ x < π/4 and these solutions are valid since they are still in the interval.
Therefore:


is conservative if we can find a scalar function
such that
. This would require



Integrating both sides of the first equation wrt
gives

Differentiating both sides of this wrt
gives

but we assumed
was a function of
and
, independent of
. So there is no such
and
is not conservative.
To find the work, first parameterize the path (call it
) by

for
. Then

and the work is given by the line integral,


Answer:
46784
Step-by-step explanation:
You would multiply all.the numbers and then your w
answer will be 46784
Answer:
First choice A : 2x +3y is <em>less than or equal </em><em>to</em><em> </em>18. x <em>greater</em><em> </em><em>than</em><em> </em><em>or</em><em> </em><em>equal</em><em> </em><em>to</em><em> </em>0. y <em>greater</em><em> </em><em>than</em><em> </em><em>or</em><em> </em><em>equal</em><em> </em><em>to</em><em> </em>0.