We want to find 
such that 
. This means
Integrating both sides of the latter equation with respect to 
tells us
and differentiating with respect to 
gives
Integrating both sides with respect to gives
Then
and differentiating both sides with respect to 
gives
So the scalar potential function is
By the fundamental theorem of calculus, the work done by 
along any path depends only on the endpoints of that path. In particular, the work done over the line segment (call it 
) in part (a) is
and does the same amount of work over both of the other paths.
In part (b), I don't know what is meant by "df/dt for F"...
In part (c), you're asked to find the work over the 2 parts (call them 
and 
) of the given path. Using the fundamental theorem makes this trivial:
Answer:
there are<em><u> 28.3495 grams in one single ounce </u></em>
//hope this helps have a nice day//
Answer:
Divide by 2
q^2+4q=3/2
q^2+4q(4/2)^2=3/2+(4/2)^2
(q+4/2)^2=3/2+16/4
taking the square root of both side
√(q+4/2)^2=√(3/2+16/4)
Note that the square will cancel the square root then you will take LCM on the right hand side
q+4/2=√6+16/4
q+4/2=√22/4
q= -4/2+-√22/4
q=(-4+_√22/4)
Remember that to find the vertical asymptotes of rational functions, we just need to set the denominator equal to zero and solve for the variable, in this case
, so lets do it:
As you can see, our rational functions has
2 vertical asymptotes: 
and
Answer:
Step-by-step explanation:
Point A represents the integer 5.
Point B -8
Point C -15
Point D 18
