Look at the quadrilateral shown below: A quadrilateral ABCD is shown with diagonals AC and BD intersecting in point O. Angle AOB is labeled as 1, angle BOC is labeled as 4, angle COD is labeled as 2, and angle AOD is labeled as 3. Clara writes the following proof for the theorem: If the diagonals of a quadrilateral bisect each other, the quadrilateral is a parallelogram: Clara's proof For triangles AOB and COD, angle 1 is equal to angle 2, as they are vertical angles. AO = OC and BO = OD because it is given that diagonals bisect each other. The triangles AOB and COD are congruent by SAS postulate. Similarly, triangles AOD and COB are congruent. By , angle ABD is equal to angle CDB and angle ADB is equal to angle CBD. As the alternate interior angles are congruent, the opposite sides of quadrilateral ABCD are parallel. Therefore, ABCD is a parallelogram.

Samantha swam upstream for some distance in one hour. she then swam downstream the same river for the same distance in only 12 minutes. if the river flows at 4 mph, how fast can samantha swim in still water?

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.

Solve for x. −3/4(x+2)=6 enter your answer in the box