In planning for a trip, Joan estimated both the distance of the trip, in miles, and her average speed, in miles per hour. She accurately divided her estimated distance by her estimated average speed to obtain an estimate for the time, in hours, that the trip would take. Was her estimate within 0.5 hour of the actual time that the trip took? (1) Joan’s estimate for the distance was within 5 miles of the actual distance. (2) Joan’s estimate for her average speed was within 10 miles per hour of her actual average speed.

I'm so close to finishing all these.

Refer to the table below if needed. second quadrant third quadrant fourth quadrant sin(1800- - cos(180° -) tan(180°-e) =- tane cot(1800-0) 10 it to solo 888 sin(180° +c) = - sine cos(180° +) =- cose tan(180° +c) = tane cot(180° +o) = cote sec(180° + c) = - seco csc(180° +2) = - csce sin(360° -) =- sine cos(360° -) = cose tan(360° - e) =- tane cot(360° -) = -cote sec(360° -) = seco csc(360° -) = csco sec(180° -) = csc(180° -) = csca 1991 given that sine = 3/5 and lies in quadrant ii, find the following value. tane

Since opening night, attendance at play a has increased steadily, while attendance at play b first rose and then fell. equations modeling the daily attendance y at each play are shown below, where x is the number of days since opening night. on what day(s) was the attendance the same at both plays? what was the attendance? play a: y = 8x + 191 play b: y = -x^2 + 26x + 126 a. the attendance was never the same at both plays. b. the attendance was the same on day 5. the attendance was 231 at both plays on that day. c. the attendance was the same on day 13. the attendance was 295 at both plays on that day. d. the attendance was the same on days 5 and 13. the attendance at both plays on those days was 231 and 295 respectively.