Answer:
b) 18
c) 80.78
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