A ship X sailing with velocity (21 km/h, 052°) observes a light from the lighthouse due North. The bearing of the lighthouse from the ship, 20 minutes later, is found to be 312°, calculate correct to three significant figures: (i) the original distance, in kilometres, when the lighthouse is due west of the ship from the time when it is due North of the ship.
(ii) the time, in minutes, when the lighthouse is due West of the ship from the time when it is due North of the ship.
(iii) the distance, in km, of the ship from the lighthouse when the lighthouse is due West of the ship.

6/20

step-by-step explanation:

option (b) is correct

step-by-step explanation:

to find the solution of this system, we need to convert it into reduced row-echelon form. we have,

after applying row operations r2 -> r2/2 and r3 -> r3/3, we get,

step-by-step explanation:

we have

using a graphing tool

see the attached figure

this is a quadratic equation (vertical parabola) open up

the vertex is a > vertex

the domain of the function is all real > interval (-∞,∞)

the range of the function is the > [-6.125,∞)

the function has two x->   and

the function is increasing over the > (0.25,∞)

the function is decreasing over the > (-∞,0.25)

x< 2

step-by-step explanation: