Two submersibles, one red and one blue, started rising toward the surface at the same time. they each rose at a constant speed.
the red submersible started rising from an altitude of 80 meters below the surface. after 30 seconds, it was 60 meters below the surface.
the following equation gives the altitude (in meters relative to the surface) of the blue submersible as a function of time (in seconds).
a=−90+0.6t
which submersible started rising from a higher altitude?
which submersible rose faster?

Solve the system by the elimination method. x + y - 6 = 0 x - y - 8 = 0 when you eliminate y , what is the resulting equation? 2x = -14 2x = 14 -2x = 14

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

Anyone know? will mark brainliest if correct!