The gravitational force, f, on a rocket at a distance, from the center of the earth is given by f = k / (r^2), where k = 10 ^3 newton*km^2. when the rocket is 10^4 km from the center of the earth, its moving away at .2 km/sec. how fast is the gravitational force changing at that moment? therefore what is the rate of change of the gravitational force in newton/ second?

2 1/2 feet

step-by-step explanation:

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.