MATH 100 Exam Solutions Fall 2018: Absolute Value, Quadratic Equation

• School: Simon Fraser University
• Department: Mathematics
• Course: MATH 100
• Professor: All

Document Summary

1: the coordinates of the midpoint of any line segment can be finding the center of the coordinates of the end points. Therefore, in this case the end points are given as: Midpoint of the x coordinates: (-3-1)/2 = -2. Hence midpoint p is (-2,2: the distance between two points ((cid:1876)(cid:2869),(cid:3)(cid:1877)(cid:2869)) and ((cid:1876)(cid:2870),(cid:3)(cid:1877)(cid:2870)) can be calculated using a simple formula, the equation of a circle is given by: Putting the values given in the question we get: (cid:1856)(cid:3404)(cid:3493)(cid:4666)(cid:1876)(cid:2870)(cid:3398)(cid:1876)(cid:2869)(cid:4667)(cid:2870)(cid:3397)(cid:4666)(cid:1877)(cid:2870)(cid:3398)(cid:1877)(cid:2869)(cid:4667)(cid:2870) (cid:1856)(cid:3404)(cid:3493)(cid:4666)(cid:3398)(cid:883)(cid:3397)(cid:885)(cid:4667)(cid:2870)(cid:3397)(cid:4666)(cid:887)(cid:3397)(cid:883)(cid:4667)(cid:2870) Where point ((cid:1860)(cid:481)(cid:1863)) is the center of the circle and (cid:1870) represents the. Therefore, ((cid:1860)(cid:481)(cid:1863)) = (-2,2) (from question 1a) circle ) as calculated in 1b is (cid:958)(cid:886)(cid:882). Putting the values of ((cid:1860)(cid:481)(cid:1863)) and (cid:1870)(cid:3)in equation (1) we get: (cid:4666)(cid:2206)(cid:3397)(cid:2779)(cid:4667)(cid:2779)(cid:3397)(cid:4666)(cid:2207)(cid:3398)(cid:2779)(cid:4667)(cid:2779)(cid:3404)(cid:4678)(cid:958)(cid:2781)(cid:2171)(cid:2779) (cid:4679)(cid:2779) (cid:4666)(cid:2206)(cid:3397)(cid:2779)(cid:4667)(cid:2779)(cid:3397)(cid:4666)(cid:2207)(cid:3398)(cid:2779)(cid:4667)(cid:2779)(cid:3404)(cid:2778)(cid:2777) radius. It is given that the line segment ab is the diameter of the circle which means that the midpoint of ab is also the center of the circle.