Mathematics: In a murder investigation, the temperature of a corpse was 32.5 degrees c at 1: 30 pm and 30.3 degrees c an hour later. normal body temperature is 37.0 degrees c and the temperature of the surroundings was 20.0 degrees c. when did the murder take place?

In a murder investigation, the temperature of a corpse

Ответ:

arlethgonzales10

14.03.2022

Mathematics

Let

be the temperature of the body

hours after 1:30 PM. Then T(0) = 32.5

and

.

Using Newton's Law of Cooling, $\dfrac{dT}{dt} = k(T - T_s)$ , we have $\dfrac{dT}{dt} = k(T-20)$ . Now let y = T - 20

, so

, so

is a solution to the initial value problem dy/dt = ky

with

.
By separating and integrating, we have $y(t) = y(0)e^{kt} = 12.5e^{kt}$ .

$y(1) = 30.3 - 20\ \Rightarrow\ 12.5e^{k(1)} = e^{k} = \frac{10.3}{12.5}\ \Rightarrow \\
k = \ln \frac{10.3}{12.5}
$

$y(t) = 37 - 20\ \Rightarrow\ 12.5e^{kt} = 17\ \Rightarrow\ e^{kt} = \frac{10.3}{12.5} \ \Rightarrow \\ \\
kt = \ln \frac{10.3}{12.5}\ \Rightarrow\ t = \left( \ln \frac{17}{12.5} \right)/ \frac{10.3}{12.5} \approx -1.588 \mathrm{\ h}$

≈ 95 minutes. Thus the murder took place about 95 minutes before 1:30 PM, or 11:55 AM.

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Ответ:

jodygoodwin40

30.05.2023

Mathematics

An image created by a reflection will always be congruent to its pre-image.

Corresponding angles and segments are always congruent in a reflection of a figure.

An image and its pre-image are always the same distance from the line of reflection.

Step-by-step explanation:

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In a murder investigation, the temperature of a corpse was 32.5 degrees c at 1: 30 pm and 30.3 degrees c an hour later. normal body temperature is 37.0 degrees c and the temperature of the surroundings was 20.0 degrees c. when did the murder take place?

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