Afenced-in rectangular area has a perimeter of 40 ft. the fence has a post every 5 ft.

how many posts are there?
what is the length and width of the area?
are there any other possible answers to the above question? if so, what are they?

How many different three-digit numbers can be formed using the digits 1 comma 2 comma 9 comma 6 comma 4 comma 3 comma and 8 without repetition? for example, 664 is not allowed.

Cone w has a radius of 8 cm and a height of 5 cm. square pyramid x has the same base area and height as cone w. paul and manuel disagree on how the volumes of cone w and square pyramid x are related. examine their arguments. which statement explains whose argument is correct and why? paul manuel the volume of square pyramid x is equal to the volume of cone w. this can be proven by finding the base area and volume of cone w, along with the volume of square pyramid x. the base area of cone w is π(r2) = π(82) = 200.96 cm2. the volume of cone w is one third(area of base)(h) = one third third(200.96)(5) = 334.93 cm3. the volume of square pyramid x is one third(area of base)(h) = one third(200.96)(5) = 334.93 cm3. the volume of square pyramid x is three times the volume of cone w. this can be proven by finding the base area and volume of cone w, along with the volume of square pyramid x. the base area of cone w is π(r2) = π(82) = 200.96 cm2. the volume of cone w is one third(area of base)(h) = one third(200.96)(5) = 334.93 cm3. the volume of square pyramid x is (area of base)(h) = (200.96)(5) = 1,004.8 cm3. paul's argument is correct; manuel used the incorrect formula to find the volume of square pyramid x. paul's argument is correct; manuel used the incorrect base area to find the volume of square pyramid x. manuel's argument is correct; paul used the incorrect formula to find the volume of square pyramid x. manuel's argument is correct; paul used the incorrect base area to find the volume of square pyramid x.

Which inequalities can be used to find the solution set of the following inequality? check all that apply. |2x – 5| (2x – 5) (2x + 5) –(2x – 5) –(2x + 5)