Distance Between Two Ships Two ships leave the same port at noon. Ship A sails north at 18 mph, and ship B sails east at 17 mph. How fast is the distance between them changing at 1 p. m.? (Round your answer to one decimal place.)

24.8 mph

Step-by-step explanation:

as we know if one sailed north and another east, the angle with respect to the port will be 90 degrees

This means that the distance of each ship to the pier can be the legs of a rectangular triangle

The distance of a ship from the other would be represented by the hypotenuse

h: Hipotenuse

c1: leg 1

c2: leg 2

By Pythagoras we know that the hypotenuse is equal to

h = √(c1^2 + c^2)

if we replace with the values we have left

h = √(18^2 + 17^2)

h = √(324 + 289)

h = √613

h = 24.758

we round a decimal place

h = 24.8

answer:   the paragraph proof is mentioned below.

step-by-step explanation:

given: in a δxwy where

such that, ∠wzx ≅ ∠yzx and zw≅zy

we have to prove that: segment zx is a perpendicular bisector of segment wy.

prove: since, zw≅zy ⇒ ∠wzy is a supplementary angle.

then, ∠wzy= 180°

⇒∠wzx + ∠yzx = 180° ( because xz divides ∠wzy into two parts)

but, ∠wzx ≅ ∠yzx ( given)

⇒∠wzy+∠wzy=180° ( by substitution)

⇒2∠wzy=180°

⇒∠wzy=90°

⇒∠wzy=∠yzx =90°

therefore, xz⊥wy where wz=zy

⇒zx is a perpendicular bisector of wy.