A cyclist from city A was greeted at 5:36, and a cyclist from city B was greeted at 7:15 came out. They met at 12 o'clock, and the speed of cyclist II was 0.75 km faster than that of cyclist I.

Find the speed. Please help me. and no link​

sine

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.

y∈r, 8> y≥0

hope i !

p.s the first bit only matters if you know what that means.

p.p.s brainliest would really