b) 18
c) 80.78
Step-by-step explanation:
While we can solve this equation with separation of variables, it isn't necessary. Β The coffee is originally at 95Β°, and the ambient temperature is 18Β°. Β The top right graph is the only one that shows this trend.
As t approaches infinity, y approaches the ambient temperature of 18Β°.
yβββ = yβ + Ξt F(tβ, yβ)
yβββ = yβ + 2 (-0.02 (yβ β 18))
yβββ = yβ β 0.04 (yβ β 18)
yβββ = 0.96 yβ + 0.72
When n=0:
yβ = 0.96 yβ + 0.72
yβ = 0.96 (95) + 0.72
yβ = 91.92
When n=1:
yβ = 0.96 yβ + 0.72
yβ = 0.96 (91.92) + 0.72
yβ = 88.96
When n=2:
yβ = 0.96 yβ + 0.72
yβ = 0.96 (88.96) + 0.72
yβ = 86.12
When n=3:
yβ = 0.96 yβ + 0.72
yβ = 0.96 (86.12) + 0.72
yβ = 83.40
When n=4:
yβ
= 0.96 yβ + 0.72
yβ
= 0.96 (83.40) + 0.72
yβ
= 80.78