Cooling an Engine. Suppose you're driving your car on a cold winter (20°f outside) and the engine overheats (at about 220°f). When you park, the engine begins to cool down. The temperature T of the engine t minutes after you park satisfies the equation .
• ln (T - 20/ 200) = -0.11t
a.) solve for T
b.) use part (a) to find the temperature of the engine after 20 minutes (t = 20)
With solution please.
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Answer:
a.) To solve for T, we need to isolate T in the equation. We can start by multiplying both sides by 200 to get rid of the denominator:
ln(T - 20/200) = -0.11t
200 ln(T - 20/200) = -22t
ln(T - 20/200) = -0.11t
e^(-0.11t) = T - 20/200
Multiplying both sides by 200, we get:
200e^(-0.11t) = T - 20
Adding 20 to both sides, we get:
T = 200e^(-0.11t) + 20
b.) To find the temperature of the engine after 20 minutes, we can substitute t = 20 into the equation we found in part (a):
T = 200e^(-0.11(20)) + 20
T = 200e^(-2.2) + 20
T ≈ 123.5°F
Therefore, the temperature of the engine after 20 minutes is approximately 123.5°F.
(^^)