If bob went 14.6 miles away form home. then, bob ran 9.8 miles towards home then 503 mile away from is home. how far is bob away from home

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 are the solutions of the system? solve by graphing. y = -x^2 -6x - 7 y = 2

The architect's side view drawing of a saltbox-style house shows a post that supports the roof ridge. the support post is 8 ft tall. the distance from the front of the house to the support post is less than the distance from the post to the back of the house. how far from the front of the house is the support post positioned?