How do you do b and c?

b) 18

c) 80.78

Step-by-step explanation:

While we can solve this equation with separation of variables, it isn't necessary.  The coffee is originally at 95°, and the ambient temperature is 18°.  The top right graph is the only one that shows this trend.

As t approaches infinity, y approaches the ambient temperature of 18°.

yₙ₊₁ = yₙ + Δt F(tₙ, yₙ)

yₙ₊₁ = yₙ + 2 (-0.02 (yₙ − 18))

yₙ₊₁ = yₙ − 0.04 (yₙ − 18)

yₙ₊₁ = 0.96 yₙ + 0.72

When n=0:

y₁ = 0.96 y₀ + 0.72

y₁ = 0.96 (95) + 0.72

y₁ = 91.92

When n=1:

y₂ = 0.96 y₁ + 0.72

y₂ = 0.96 (91.92) + 0.72

y₂ = 88.96

When n=2:

y₃ = 0.96 y₂ + 0.72

y₃ = 0.96 (88.96) + 0.72

y₃ = 86.12

When n=3:

y₄ = 0.96 y₃ + 0.72

y₄ = 0.96 (86.12) + 0.72

y₄ = 83.40

When n=4:

y₅ = 0.96 y₄ + 0.72

y₅ = 0.96 (83.40) + 0.72

y₅ = 80.78