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3 years ago

Find the distance between the two points rounding to the nearest tenth (if needed)

Mathematics

1 answer:

Ede4ka [16]3 years ago

3 0

Answer:

6.3 units (nearest tenth)

Step-by-step explanation:

$\boxed{Distance \: between \: 2 \: points = \sqrt{ {(x1 - x2)}^{2} + {(y1 - y2)}^{2} } }$

Using the formula above,

distance between (-8, 6) and (-6, 0)

$= \sqrt{ {[ - 8 - ( - 6)]}^{2} + {(6 - 0)}^{2} }$

$= \sqrt{ {( - 8 + 6)}^{2} + {(6 - 0)}^{2} }$

$= \sqrt{ {( - 2)}^{2} + 6^{2} }$

$= \sqrt{4 + 36}$

$= \sqrt{40}$

= 6.3 units (nearest tenth)

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2 years ago

Tangela bought 12 cans of soda for 5.40 what was the unit rate for each can of soda

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How many 4 digits codes can be formed with odd numbers without repetition? 0 is not allowed.

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Answer:

120

Step-by-step explanation:

The number of permutations of 5 things taken 4 at a time is 120.

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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

sergey [27]

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 algebraic 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 derivative 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 rate of change of the distance between particle P and particle Q ( $\dot r_{Q/P} (t)$ ) is equal to the vectorial difference between respective vectors velocity:

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

Where $\vec v_{P}(t)$ is the vector velocity 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 rate of change 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:

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

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

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

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

To learn more on derivatives, we kindly invite to check this verified question: brainly.com/question/2788760

3 0

2 years ago

Give the equation of the line perpendicular to y +7= -2(x - 6) through the point (6,-3)

lilavasa [31]

Note: this answer assumes the equation of the line can be put in slope-intercept form.

Answer:

$y + 3 = \frac{1}{2}(x-6)$

Step-by-step explanation:

1) First, find the slope of y + 7 = -2 (x - 6). We can see that it's already in point-slope form, or $y-y_1 = m (x-x_1)$ format. Remember that the number in place of the $m$ is the slope. Therefore, -2 is the slope of that equation.

What we need is the slope that is perpendicular to that, though. So, find the opposite reciprocal of -2. To do this, change its sign, convert it into a fraction ( $-\frac{2}{1}$ ), and flip its numerators and denominators. Therefore, the perpendicular slope would be $\frac{1}{2}$ .

2) Now that we have a slope and a point the line passes through, we can write an equation using the point-slope formula, $y-y_1 = m (x-x_1)$ . In order to write an equation, the $x_1$ , $y_1$ , and $m$ have to be substitute for with real values.

The $m$ represents the slope. We already calculated that in the last step, so put $\frac{1}{2}$ in place of the $m$ . The $x_1$ and $y_1$ represent the x and y values of a point the line passes through. We know that the line has to pass through (6, -3), so substitute 6 for $x_1$ and -3 for $y_1$ :

$y - (-3) = \frac{1}{2}(x -(6))\\y + 3 = \frac{1}{2}(x - 6)$

Therefore, (again, assuming that the line can be put in point-slope form) the answer is $y + 3 = \frac{1}{2}(x-6)$ .

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3 years ago