Ask question

Ask question

All categories

3 years ago

14

A body was found at 6 a.m. in a warehouse where the temperature was 50°F. The medical examiner found the temperature of the body

to be 66°F. What was the approximate time of death? Use Newton's law of cooling, with k = 0.1947.
T(t)=T[A] + (T[O]-T[A])e^-kt

2 answers:

ankoles [38]3 years ago

6 0

<span>It was at Midnight (12 a.m.) </span>

fenix001 [56]3 years ago

4 0

Answer: 12 a.m.

Step-by-step explanation:

You might be interested in

The numerator and denominator of a fraction are in the ratio of 3 to 5. If the numerator and denominator

WARRIOR [948]

Answer: 5n = 3d and 3n + 6 = 2d + 4

Given that the numerator and denominator of a fraction are in the ratio of 3 to 5. When the numerator and denominator are both increased by 2, the fraction is equal to \dfrac{2}{3}.

We are to select the system of equations that could be used to solve the problem.

Since n denotes the numerator and m denotes the denominator of the given fraction, so we have:

n/d = 3/5

5n = 3d

<h3>and</h3>

(n+2)/(d+2) = 2/3

3(n+2) = 2(d+2)

3n + 6 = 2d + 4

Thus, the required system of equations is,

5n = 3d and 3n + 6 = 2d + 4

Brainliest pweaseee if the answer is clear and correct! <3 ~~~

3 0

2 years ago

Find the solution of the initial value problem<br><br> dy/dx=(-2x+y)^2-7 ,y(0)=0

Leokris [45]

Substitute $v(x)=-2x+y(x)$ , so that $\dfrac{\mathrm dv}{\mathrm dx}=-2+\dfrac{\mathrm dy}{\mathrm dx}$ . Then the ODE is equivalent to

$\dfrac{\mathrm dv}{\mathrm dx}+2=v^2-7$

which is separable as

$\dfrac{\mathrm dv}{v^2-9}=\mathrm dx$

Split the left side into partial fractions,

$\dfrac1{v^2-9}=\dfrac16\left(\dfrac1{v-3}-\dfrac1{v+3}\right)$

so that integrating both sides is trivial and we get

$\dfrac{\ln|v-3|-\ln|v+3|}6=x+C$

$\ln\left|\dfrac{v-3}{v+3}\right|=6x+C$

$\dfrac{v-3}{v+3}=Ce^{6x}$

$\dfrac{v+3-6}{v+3}=1-\dfrac6{v+3}=Ce^{6x}$

$\dfrac6{v+3}=1-Ce^{6x}$

$v=\dfrac6{1-Ce^{6x}}-3$

$-2x+y=\dfrac6{1-Ce^{6x}}-3$

$y=2x+\dfrac6{1-Ce^{6x}}-3$

Given the initial condition $y(0)=0$ , we find

$0=\dfrac6{1-C}-3\implies C=-1$

so that the ODE has the particular solution,

$\boxed{y=2x+\dfrac6{1+e^{6x}}-3}$

5 0

3 years ago

Write in expanded form<br>3<br>(-a)

MA_775_DIABLO [31]

Answer:

-3a

Step-by-step explanation:

3(-a)

Expand brackets.

3 × -1a

-3a

3 0

3 years ago

Write the standard form of the line that passes through (-1,-3) and (2,1).

kobusy [5.1K]

Answer:

Step-by-step explanation:

$\mathrm{Line\:passing\:through\:}\left(-1,\:-3\right)\mathrm{,\:}\left(2,\:1\right)\\\\\mathrm{Find\:the\:line\:}\\\mathbf{y=mx+b}\mathrm{\:passing\:through\:}\left(-1,\:-3\right)\mathrm{,\:}\left(2,\:1\right)\\\\\mathrm{Compute\:the\:slope\:}\left(-1,\:-3\right),\:\left(2,\:1\right):\quad \\m=\frac{4}{3}\\\mathrm{Slope}=\frac{y_2-y_1}{x_2-x_1}\\\\\left(x_1,\:y_1\right)=\left(-1,\:-3\right),\:\left(x_2,\:y_2\right)\\\left\\$

$\\m=\frac{1-\left(-3\right)}{2-\left(-1\right)}\\\\Refine\\\\m=\frac{4}{3}\\\\\mathrm{Compute\:the\:}y\mathrm{\:intercept}:\quad b=-\frac{5}{3}\\\\y=\frac{4}{3}x-\frac{5}{3}\\$

Convert equation to standard form

$y - \frac{4}{3} x + \frac{5}{3} = 0\\\\Multiply \:through\: by\: 3;\\\\3\times (y - \frac{4}{3} x + \frac{5}{3} )=0\\\\3y -4x+5 = 0\\Arrange\:n\:form\:of \:;\\Ax +By = C\\\\-4x +3y =-5$

5 0

3 years ago

I need the answer to finding the reciprocal of -72÷(-9)

lbvjy [14]

First lets do the math...
-72 / -9 = 8

the reciprocal of a number is just flipping the number.
example : the reciprocal of 2/3 = 3/2.....the reciprocal of 6/7 = 7/6...and when u have a whole number, put it over 1...

so the reciprocal of 8, or 8/1 is gonna be 1/8 <==

or u could do it this way...
reciprocal of -72 / -9 = -9 / -72, which equals 1/8

5 0

3 years ago