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2 years ago

10

Frankie wants to build a path from one corner of his yard to the opposite corner. his yard measures 20 ft. x 32 ft. what will be

the length of his path to the nearest tenth of a foot?
21.5 ft
25.0 ft
32.6 ft
37.7 ftfrankie wants to build a path from one corner of his yard to the opposite corner. his yard measures 20 ft. x 32 ft. what will be the length of his path to the nearest tenth of a foot?
21.5 ft
25.0 ft
32.6 ft
37.7 ft.

1 answer:

ira [324]2 years ago

5 0

The length of his path to the nearest tenth of a foot is 37.7 ft option fourth is correct.

<h3>What is the Pythagoras theorem?</h3>

The square of the hypotenuse in a right-angled triangle is equal to the sum of the squares of the other two sides.

We have:

Frankie wants to build a path from one corner of his yard to the opposite corner. his yard measures 20 ft. x 32 ft.

From the Pythagoras theorem:

Perpendicular(P) = 32 ft

Base(B) = 20 ft

H = √(32²+20²)

H = √1424

H = 37.7 ft

Thus, the length of his path to the nearest tenth of a foot is 37.7 ft option fourth is correct.

Learn more about Pythagoras' theorem here:

brainly.com/question/21511305

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Answer:

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Step-by-step explanation:

7(x+3)=-2(x-5)

Distribute

7x+21 = -2x+10

Add 2x to each side

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Subtract 21 from each side

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3 years ago

Particle P moves along the y-axis so that its position at time t is given by y(t)=4t−23 for all times t. A second particle, part

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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:

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

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

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

$x(k) = \lim_{k \to 0} \frac{\pi\cdot \sin (2\pi-k)}{k}$

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

$x(k) = -\pi$

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.

If we know that $\vec v_{P}(t) = (0, 4)$ , $\vec v_{Q}(t) = \left(\frac{2\pi\cdot \cos \pi t - \pi\cdot t \cdot \cos \pi t + \sin \pi t}{(2-t)^{2}}, 0 \right)$ and $t = \frac{1}{2}$ , then the vector rate of change of the distance between the two particles:

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

$\dot r_{Q/P}\left(\frac{1}{2} \right) = \left(\frac{2\pi\cdot \cos \frac{\pi}{2}-\frac{\pi}{2}\cdot \cos \frac{\pi}{2} +\sin \frac{\pi}{2}}{\frac{3}{2} ^{2}}, -4 \right)$

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

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:

<em>Particle </em> $P$ <em> moves along the y-axis so that its position at time </em> $t$ <em> is given by </em> $y(t) = 4\cdot t - 23$ <em> for all times </em> $t$ <em>. A second particle, </em> $Q$ <em>, moves along the x-axis so that its position at time </em> $t$ <em> is given by </em> $x(t) = \frac{\sin \pi t}{2-t}$ <em> for all times </em> $t \ne 2$ <em>. </em>

<em />

<em>a)</em><em> As times approaches 2, what is the limit of the position of particle </em> $Q?$ <em> Show the work that leads to your answer. </em>

<em />

<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>

<em />

<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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