The length of each side of a rectangular picture frame needs to be 15 in. you have only one 48-in. piece of wood to use for this frame. write and solve an inequality that describes the possible widths for this frame

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

Factor the monomial completely 42a3b3

Which inequalities can be used to find the solution set of the following inequality? check all that apply. |2x – 5| (2x – 5) (2x + 5) –(2x – 5) –(2x + 5)