A north-bound ocean liner is approaching Juneau at the rate of 60 km/h and is scheduled to reach the port at 8:21 am. A tugboat leaves Juneau at 7:57 am and heads west at the rate of 75 km/h. What is the smallest distance between the two boats? solve using optimization of distances (calculus)

18.36. give me brainliest!

step-by-step explanation:

you have to multiply 1.53 by 12 so the equation is 12x1.53 and then that is your answer.

intensity of the vector is v= √37 ≈ 6.08 units and make angle ∡α ≈ 9.46° with east direction

step-by-step explanation:

required vector is consists of the two componentsvx= 2+4=6 units   and vy= 1 unit and vx ⊥ vywe will use pythagorian theorem to find intensity of the vector vv∧2 = vx∧2 + vy∧2 => v = √vx∧2 + vy∧2 = √6∧2 + 1∧2 = √36+1 = √37 ≈ 6.08 unitsthe angle ∡α between vector and east direction we wil find with tanαtanα = 1/6 => α = arc tanα = arc 1/6 => α ≈ 9.46°

hope this

(-1,-0.5)

step-by-step explanation:

the answer

b. they are parallel and will never intersect.