In jerome's physics class, he had to solve the equation 40 = 12mv2 for m. each step of his work is shown below.

use the drop-down menus to explain each step.

clear check
given equation: 40 = 12mv2

step 1: 40v2 = 12m
on both sides of the equation.

step 2: 80v2 = m
on both sides of the equation.

i need answer asap like fast plaese by the morning or before 9: 00 you very much

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.