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3 years ago

Help please !!!! my brain hurts

Mathematics

1 answer:

fiasKO [112]3 years ago

8 0

Answer:

the answer id d.

Step-by-step explanation:

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Write the exponential function.

sertanlavr [38]

Answer:

$f(x) = 10*(b)^x$

Step-by-step explanation:

exponential function : $y=a*(b)^x$

where a = y intercept and b = factor ( what you multiply by )

looking at the table the y intercept is at (0,10) so a = 10

so we now have y = 10 * (b)^x

to find the factor we simply plug in one of the ordered pairs.

note that the ordered pair chosen does not effect the outcome

i chose (2,250)

we have y = 10 * (b)^x

(x,y) = (2,250)

250 = 10 * (b)^2

divide both sides by 10

25 = b^2

take the square root of both sides

5 = b

the final equation would be f(x) = 10 * (b)^x

5 0

2 years ago

Solve for a! 264=π×a×(20-a) Thank you!

jasenka [17]

Answer:

a ≈ 14 or 6

Step-by-step explanation:

264 = π × a × (20 - a)

a(20 - a) = $\frac{264}{Pi}$ ≈ 84 (rounded off to nearest whole number)

Opening the brackets we get;

a² - 20a + 84 = 0

Applying the quadratic formula we get:

a ≈ 14 or 6

4 0

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

Find the mistake in this sentence:

wlad13 [49]

The mistake is “there are”

4 0

2 years ago

Define acute, right angle, obtuse, straight angle, reflex

vova2212 [387]

Answer: here

Step-by-step explanation:

Acute Angle - An angle less than 90 degrees. Right Angle - An angle that is exactly 90 degrees. Obtuse Angle - An angle more than 90 degrees and less than 180 degrees. Straight Angle - An angle that is exactly 180 degrees. Reflex Angle - An angle greater than 180 degrees and less than 360 degrees.

5 0

2 years ago

Read 2 more answers

Help please !!!! my brain hurts ​

Help please !!!! my brain hurts