Help me asap please due in 10 min

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

The weights of players in a basketball league are distributed normally. the distance (d) of each player's weight (w) from the mean of 86 kilograms was computed as follows: d = |w - 86| if d is greater than 7 for approximately 5% of the players, what is the approximate standard deviation of the players' weights? a. 1.75 kilograms b. 3.5 kilograms c. 4.3 kilograms d. 7 kilograms

Δabc is dilated by a scale factor of 3 with the origin as the center of dilation to form δa′b′c′. the slope of is -1.2. the length of is p units, the length of is q units, and the length of is r units. the slope of is . the length of is units.