The temperature at 6:00 A. M. was -3.8°C. By 3:00 P. M., the temperature had increased 9.6 degrees. What is the temperature now.

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

What is the value of the expression 9+n/3-6 when n = 12? 1. 5. 7. 12.

Which expression is equivalent to (4 +6i)^2? ? -20 + 48i 8 + 12i 16 - 36i 20 + 48i