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3 years ago

10

Mr. Vestas sells electricity from a wind turbine to his local utility. He receives 3¢ per kilowatt hour for the first 500 kilowa

tt hours, 2¢ per kilowatt hour for the next 1000 kilowatt hours, and 1¢ per kilowatt hour for each kilowatt hour thereafter. Mr. Vestas sold 1800 kilowatt hours last month. What was the average price per kilowatt hour the utility paid to Mr. Vestas last month, rounded to the nearest hundredth of a cent?

1 answer:

Grace [21]3 years ago

5 0

Answer:

<h2> 2.11¢</h2>

Step-by-step explanation:

We are expected to solve the weighted average price per kilowatt hour.

Step one:

The formula is

W =∑wX/∑w

W=weighted average

w=weights applied to x values

X=data values to be averaged

Step two:

given the total killowatt hour is 1800

the 1st = 500 at 3¢

2nd= 1000 at 2¢

3rd = (1800-1000+500) at 1¢ =(1800-1500) at 1¢ = (300) at 1¢

Step two:

Substituting into the expression we have

W =(500*3)+(1000*2)+(300*1)/500+1000+300

W=1500+2000+300/1800

W=3800/1800

W=2.11¢

The average price per kilowatt hour the utility paid to Mr. Vestas last month is 2.11¢

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a) The limit of the position of particle $Q$ when time approaches 2 is $-\pi$ .

b) The velocity of particle $Q$ is $v_{Q}(t) = \frac{2\pi\cdot \cos \pi t-\pi\cdot t \cdot \cos \pi t -\sin \pi t}{(2-t)^{2}}$ for all $t \ne 2$ .

c) The rate of change of the distance between particle $P$ and particle $Q$ at time $t = \frac{1}{2}$ is $\frac{4\sqrt{82}}{9}$ .

<h3>How to apply limits and derivatives to the study of particle motion</h3>

a) To determine the limit for $t = 2$ , we need to apply the following two <em>algebraic</em> substitutions:

$u = \pi t$ (1)

$k = 2\pi - u$ (2)

Then, the limit is written as follows:

$x(t) = \lim_{t \to 2} \frac{\sin \pi t}{2-t}$

$x(t) = \lim_{t \to 2} \frac{\pi\cdot \sin \pi t}{2\pi - \pi t}$

$x(u) = \lim_{u \to 2\pi} \frac{\pi\cdot \sin u}{2\pi - u}$

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$x(k) = -\pi\cdot \lim_{k \to 0} \frac{\sin k}{k}$

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The limit of the position of particle $Q$ when time approaches 2 is $-\pi$ . $\blacksquare$

b) The function velocity of particle $Q$ is determined by the <em>derivative</em> formula for the division between two functions, that is:

$v_{Q}(t) = \frac{f'(t)\cdot g(t)-f(t)\cdot g'(t)}{g(t)^{2}}$ (3)

Where:

$f(t)$ - Function numerator.
$g(t)$ - Function denominator.
$f'(t)$ - First derivative of the function numerator.
$g'(x)$ - First derivative of the function denominator.

If we know that $f(t) = \sin \pi t$ , $g(t) = 2 - t$ , $f'(t) = \pi \cdot \cos \pi t$ and $g'(x) = -1$ , then the function velocity of the particle is:

$v_{Q}(t) = \frac{\pi \cdot \cos \pi t \cdot (2-t)-\sin \pi t}{(2-t)^{2}}$

$v_{Q}(t) = \frac{2\pi\cdot \cos \pi t-\pi\cdot t \cdot \cos \pi t -\sin \pi t}{(2-t)^{2}}$

The velocity of particle $Q$ is $v_{Q}(t) = \frac{2\pi\cdot \cos \pi t-\pi\cdot t \cdot \cos \pi t -\sin \pi t}{(2-t)^{2}}$ for all $t \ne 2$ . $\blacksquare$

c) The vector <em>rate of change</em> of the distance between particle P and particle Q ( $\dot r_{Q/P} (t)$ ) is equal to the <em>vectorial</em> difference between respective vectors <em>velocity</em>:

$\dot r_{Q/P}(t) = \vec v_{Q}(t) - \vec v_{P}(t)$ (4)

Where $\vec v_{P}(t)$ is the vector <em>velocity</em> of particle P.

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$\dot r_{P/Q}(t) = \left(\frac{2\pi \cdot \cos \pi t - \pi\cdot t \cdot \cos \pi t + \sin \pi t}{(2-t)^{2}}, -4 \right)$

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$\dot r_{Q/P} \left(\frac{1}{2} \right) = \left(\frac{4}{9}, -4 \right)$

The magnitude of the vector <em>rate of change</em> is determined by Pythagorean theorem:

$|\dot r_{Q/P}| = \sqrt{\left(\frac{4}{9} \right)^{2}+(-4)^{2}}$

$|\dot r_{Q/P}| = \frac{4\sqrt{82}}{9}$

The rate of change of the distance between particle $P$ and particle $Q$ at time $t = \frac{1}{2}$ is $\frac{4\sqrt{82}}{9}$ . $\blacksquare$

<h3>Remark</h3>

The statement is incomplete and poorly formatted. Correct form is shown below:

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<em>b) </em><em>Show that the velocity of particle </em> $Q$ <em> is given by </em> $v_{Q}(t) = \frac{2\pi\cdot \cos \pi t-\pi\cdot t \cdot \cos \pi t +\sin \pi t}{(2-t)^{2}}$ <em>.</em>

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<em>c)</em><em> Find the rate of change of the distance between particle </em> $P$ <em> and particle </em> $Q$ <em> at time </em> $t = \frac{1}{2}$ <em>. Show the work that leads to your answer.</em>

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