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3 years ago

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A thermometer is removed from a room where the temperature is 70° F and is taken outside, where the air temperature is 40° F. Af

ter one-half minute the thermometer reads 60° F. What is the reading of the thermometer at t = 1 min? (Round your answer to two decimal places.)

1 answer:

fgiga [73]3 years ago

5 0

Answer:

53.3324

Step-by-step explanation:

given that a thermometer is removed from a room where the temperature is 70° F and is taken outside, where the air temperature is 40° F.

By Newton law of cooling we have

T(t) = $T+(T_0-T)e^{-kt}$

where T (t) is temperature at time t,T =surrounding temperature = 40, T0 =70 = initial temperature

After half minute thermometer reads 60° F. Using this we can find k

$T(0,5) = 40+(70-40)e^{-k/2} = 60\\e^{-k/2} =2/3\\-k/2 = -0.4055\\k = 0.8110$

So equation is

$T(t) = 40+(30)e^{-0.8110t}\\$

When t=1,

we get

$T(1) = 40+(30)e^{-0.8110}\\\\=53.3324$

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3 years ago

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Segment EF: y = -x + 8

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Step-by-step explanation:

Given the two similar right triangles, ΔABC and ΔDEF, for which we must determine the slope-intercept form of the side of ΔDEF that is parallel to segment BC.

Upon observing the given diagram, we can infer the following corresponding sides:

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We must determine the slope of segment BC from ΔABC, which corresponds to segment EF from ΔDEF.

<h2>Slope of Segment BC:</h2>

In order to solve for the slope of segment BC, we can use the following slope formula:

$\displaystyle\mathsf{Slope\:(m)\:=\:\frac{y_2 \:-\:y_1}{x_2 \:-\:x_1}} }$

Use the following coordinates from the given diagram:

Point B: (x₁, y₁) = (-2, 4)

Point C: (x₂, y₂) = ( 1, 1 )

Substitute these values into the slope formula:

$\displaystyle\mathsf{Slope\:(m)\:=\:\frac{y_2 \:-\:y_1}{x_2 \:-\:x_1}}\:=\:\frac{1\:-\:4}{1\:-\:(-2)}\:=\:\frac{-3}{1\:+\:2}\:=\:\frac{-3}{3}\:=\:-1}$

<h2>Slope of Segment EF:</h2>

Similar to how we determined the slope of segment BC, we will use the coordinates of points E and F from ΔDEF to find its slope:

Point E: (x₁, y₁) = (4, 4)

Point F: (x₂, y₂) = (6, 2)

Substitute these values into the slope formula:

$\displaystyle\mathsf{Slope\:(m)\:=\:\frac{y_2 \:-\:y_1}{x_2 \:-\:x_1}}\:=\:\frac{2\:-\:4}{6\:-\:4}\:=\:\frac{-2}{2}\:=\:-1}$

Our calculations show that segment BC and EF have the same slope of -1. In geometry, we know that two nonvertical lines are <u>parallel</u> if and only if they have the same slope.

Since segments BC and EF have the same slope, then it means that $\displaystyle\mathsf{\overline{BC}\:\: | |\:\:\overline{EF}}$ .

<h2>Slope-intercept form:</h2><h3><u>Segment BC:</u></h3>

The <u>y-intercept</u> is the point on the graph where it crosses the y-axis. Thus, it is the value of "y" when x = 0.

Using the slope of segment BC, m = -1, and the coordinates of point C, (1, 1), substitute these values into the <u>slope-intercept form</u> (y = mx + b) to solve for the y-intercept, <em>b. </em>

y = mx + b

1 = -1( 1 ) + b

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Add 1 to both sides to isolate b:

1 + 1 = -1 + 1 + b

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Therefore, the linear equation in <u>slope-intercept form of segment BC</u> is:

⇒ y = -x + 2.

<h3><u /></h3><h3><u>Segment EF:</u></h3>

Using the slope of segment EF, <em>m</em> = -1, and the coordinates of point E, (4, 4), substitute these values into the <u>slope-intercept form</u> to solve for the y-intercept, <em>b. </em>

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4 + 4 = -4 + 4 + b

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Therefore, the linear equation in <u>slope-intercept form of segment EF</u> is:

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3 years ago

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olga55 [171]

We know that:

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When we substitute x and y (from the pairs we have), we'll get a system of equations:

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and all we have to do is solve it for a, b and r.

There will be:

$\begin{cases}(-1-a)^2+(2-b)^2=r^2\\(0-a)^2+(1-b)^2=r^2\\(-2-a)^2+(-1-b)^2=r^2\end{cases}\\\\\\
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\begin{cases}a^2+b^2+2a-4b+5=r^2\\a^2+b^2-2b+1=r^2\\a^2+b^2+4a+2b+5=r^2\end{cases}\\\\\\
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From equations (II) and (III) we have:

$\begin{cases}a^2+b^2-2b+1=r^2\\a^2+b^2+4a+2b+5=r^2\end{cases}\\--------------(-)\\\\a^2+b^2-2b+1-a^2-b^2-4a-2b-5=r^2-r^2\\\\-4a-4b-4=0\qquad|:(-4)\\\\\boxed{-a-b-1=0}$

and from (I) and (II):

$\begin{cases}a^2+b^2+2a-4b+5=r^2\\a^2+b^2-2b+1=r^2\end{cases}\\--------------(-)\\\\a^2+b^2+2a-4b+5-a^2-b^2+2b-1=r^2-r^2\\\\2a-2b+4=0\qquad|:2\\\\\boxed{a-b+2=0}$

Now we can easly calculate a and b:

$\begin{cases}-a-b-1=0\\a-b+2=0\end{cases}\\--------(+)\\\\-a-b-1+a-b+2=0+0\\\\-2b+1=0\\\\-2b=-1\qquad|:(-2)\\\\\boxed{b=\frac{1}{2}}\\\\\\\\a-b+2=0\\\\\\a-\dfrac{1}{2}+2=0\\\\\\a+\dfrac{3}{2}=0\\\\\\\boxed{a=-\frac{3}{2}}$

Finally we calculate $r^2$ :

$a^2+b^2-2b+1=r^2\\\\\\\left(-\dfrac{3}{2}\right)^2+\left(\dfrac{1}{2}\right)^2-2\cdot\dfrac{1}{2}+1=r^2\\\\\\\dfrac{9}{4}+\dfrac{1}{4}-1+1=r^2\\\\\\\dfrac{10}{4}=r^2\\\\\\\boxed{r^2=\frac{5}{2}}$

And the equation of the circle is:

$(x-a)^2+(y-b)^2=r^2\\\\\\\left(x-\left(-\dfrac{3}{2}\right)\right)^2+\left(y-\dfrac{1}{2}\right)^2=\dfrac{5}{2}\\\\\\\boxed{\left(x+\dfrac{3}{2}\right)^2+\left(y-\dfrac{1}{2}\right)^2=\dfrac{5}{2}}$

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3 years ago