Replace x with π/2 - x to get the equivalent integral
but the integrand is even, so this is really just
Substitute x = 1/2 arccot(u/2), which transforms the integral to
There are lots of ways to compute this. What I did was to consider the complex contour integral
where γ is a semicircle in the complex plane with its diameter joining (-R, 0) and (R, 0) on the real axis. A bound for the integral over the arc of the circle is estimated to be
which vanishes as R goes to ∞. Then by the residue theorem, we have in the limit
and it follows that
(-1-3i)(-6-i)
=6+i+18i+3i^2
=3i^2+19i+6. Hope it help!
Answer:
2nd debate was 3 hours long.
Step-by-step explanation:
We have been given that during the mayoral election,two debates were held between the candidates. The first debate lasted 1 2/3 hours. The second one was 1 4/5 times as long as the first one.
Let us find the estimate of time spent on 2nd debate.
1 2/3 hours would be approximately 2 hours. 1 4/5 times would be equal to 2 times.
Therefore, the estimated time is less 4 hours.
To find the time spent on 2nd debate, we will multiply 1 2/3 by 1 4/5.
First of all, we will convert mixed fractions into improper fractions as:
Now, we will multiply both fractions as:
Therefore, the 2nd debate was 3 hours long.
Answer:
$14.75 per person
Step-by-step explanation:
because there are 4 people in total including Mrs Cats you would add up 33 + 17 + 3 + 6 to get 59 then you would divide that by the amount of people there are in this case 4 to get 14.75.
Step 1) Draw a dashed line through the points (0,6) and (4,7). These two points are on the line y = (1/4)x+6. To find those points, you plug in x = 0 to get y = 6. Similarly, plug in x = 4 to get y = 7. The dashed line indicates that none of the points on this line are part of the solution set.
Step 2) Draw a dashed line through (0,-1) and (1,1). These two points are on the line y = 2x-1. They are found in a similar fashion as done in step 1.
Step 3) Shade the region that is above both dashed lines. We shade above because of the "greater than" sign. This is shown in the attached image I am providing below. The red shaded region represents all of the possible points that are the solution set. Once again, any point on the dashed line is not in the solution set.
