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3 years ago

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A ball is thrown into the air. The height h, in feet, of the ball after x seconds is given by the function h = −16(x − 2)2 + 72.

What is the equation in standard form and the maximum height of the ball?

1 answer:

Anon25 [30]3 years ago

3 0

Answer:

A. -16x²+64x+8

B. 72 ft.

Step-by-step explanation: A. Expand the perfect square binomial -16(x²-4x+4)+72, dist. -16 (-16x²+64x-64+72), then combine like terms (-16x²+64x+8)

B. Find the abs. max. by looking back at the vertex form. The max is (2,72). Use the y-value for the height.

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Answer:

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2*21/7*21. + 10*7/21*7

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One quarter has a mass of 5.67 grams. Which is the mass of two rolls of quarters each containing 40 quarters? how much is it?

Orlov [11]

Answer:

453.6 grams

Step-by-step explanation:

40 x 5.67 = 226.8

226.8 x 2 = 453.6

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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 <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>

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

3 0

2 years ago

I need help really quick

nasty-shy [4]

Answer:

$162.50

Step-by-step explanation:

I=Prt

I=(5000)(0.065)(0.5)

I=162.50

0.5 = 6 months, which is half of a year

0.065 = the percent (6.5%) in decimal form

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Find the area of the trapozoid.

Bad White [126]

Area =

$\frac{1} {2} \times sum \: of \: parallel \: sides \: \times height$

=》

$\frac{1}{2} \times (6 + 10) \times 5$

=》

$\frac{1}{2} \times 16 \times 5$

=》

$8 \times 5$

=》

$40$

Area = 40 in²

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3 years ago

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