There are 8 performers who will present their comedy acts this weekend at a comedy club. one of the performers insists on being the last stand-up comic of the evening. if this performer's request is granted, how many different ways are there to schedule the appearances?

This is a permutations problem. There are only 7 comics you can move into different time slots, the 8th one being fixed in that last position. So you can look at it this way:
For the 1st position, you can choose from all 7 people. Once you've picked that one, you have 6 people to choose from for the 2nd position, and similarly for the remaining ones . This gives you 7x6x5x4x3x2x1 = 5040 ways to arrange them. And the last comic stays in the same position all the time so does not change this number. Make sense? If you need to demonstrate it to yourself, use a smaller number of comics, like 3 or 4, and write out the combinations to see that it works.

3i think i’m not quite sure skskksks but if that ig

Answer:

$4034.40

step-by-step explanation:

the perimeter of the field is the sum of the lengths of the four sides:

47 yd + 158 yd + 47 yd + 158 yd = 410 yd

the cost of 3 ft (1 yd) of fence is 3×$3.28 = $9.84.

so the cost of jake's wood perimeter fance is

(410 yd) × ($9.84 /yd) = $4034.40