Circle I was dilated with the origin as the center of dilation to create Circle II.
Write the rule that best represents the dilation that has
been applied to Circle I to create Circle II.

Convert 2000 swiss francs to dutch guilders

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.

How is the equation of this circle written in standard form? x2 + y2 - 6x + 14y = 142 a) (x - 3)2 + (y + 7)2 = 200 b) (x+ 3)2 + (y - 7)2 = 200 (x - 6)2 + (y + 14)2 = 142 d) (x+6)2 + (y- 14)2 = 142