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3 years ago

8

If 3.485÷5.1 is random long division form so that advisor is rain as a whole number where one the decimal point the place in the

dividend.

1 answer:

saveliy_v [14]3 years ago

6 0

When it comes to division 3.485 would be apart of the Warhol there for it is non targeted

You might be interested in

What is the value of log, 81?<br> N<br> け間<br> w<br> оооо<br> ол

kramer

Answer:

1.908

Step-by-step explanation:

Assuming you are going at a base of 10,

㏒81 = 1.908, just use a calculator

7 0

3 years ago

if a motercycle is moving at a constant speed down the highway of 40 km/hr, how long would it the motorcycle to travel 10 km

ahrayia [7]

Answer:

15 minutes

Step-by-step explanation:

First, the motorcycle goes at a speed of 40 km/hr.

The question asks for how long it would take to travel 10 km.

Well, there are 60 minutes in an hour, since we will be translating into minutes.

Also, 10 km is 1/4 of 40 km, so it would make sense that the time length would be 1/4 of an hour as well.

1/4 of 60 minutes is 15 minutes. So it takes 15 minutes for the motorcycle to travel 10 km.

Now, if all this wordy stuff is too much to comprehend, you can also solve using proportional relationships.

$\frac{40km}{60min}=\frac{10km}{xmin}$

Now cross multiply:

$40km*xmin=10km*60min\\40x=600$

Divide both sides by 40:

$\frac{40x}{40}=\frac{600}{40}\\x=15$

Again, this shows that it wouls take 15 minutes for the motorcycle to travel 10 km.

6 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 <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

Solve for the indicated variable.

Hunter-Best [27]

Answer:

<h3><u>1.</u>)</h3>

3n + 1 = 7n - 5

1 + 5 = 7n - 3n

6 = 4n

3/2 = n

<h3><u>2</u><u>.</u>)</h3>

2[x + 3(x - 1)] = 18

2[x + 3x - 3] = 18

2(4x - 3) = 18

8x - 6 = 18

8x = 12

x = 12/8

x = 3/2

<h3><u>3</u><u>.</u>)</h3>

6(y + 2) - 4 = -10

6y + 12 - 4 = -10

6y = -10 - 8

6y = -18

y = -3

<h3><u>4</u><u>.</u>)</h3>

2x² = 50

x² = 50/2

x = √25

x = 5

<h3><u>5</u><u>.</u>)</h3>

5 + 2(k + 4) = 5(k - 3) + 10

5 + 2k + 8 = 5k - 15 + 10

2k - 5k = -5 - 13

-3k = -18

k = 6

<h3><u>6</u><u>.</u>)</h3>

6 + 2x(x - 3) = 2x²

6 + 2x² - 6x = 2x²

-6x = 2x² - 2x² - 6

x = 1

<h3><u>7</u><u>.</u>)</h3>

2/3x - 18 = x/6

12(2/3x - 18 = x/6)

8x - 216 = 2x

8x - 2x = 216

6x = 216

x = 36

<h3><u>8.</u>)</h3>

x - 2/3 = 2x + 1/4

12(x - 2/3 = 2x + 1/4)

4x - 8 = 6x + 3

4x - 6x = 8 + 3

2x = 11

x = 11/2

<u>-TheUnknownScientist</u><u> 72</u>

3 0

3 years ago

Math<br><br><br> is my answer correct??<br><br><br> pls check!

Luba_88 [7]

Answer:

Domain: (-2 , 3]

Range: [-3, 2)

Step-by-step explanation:

Domain is all the x's on the graph. (If you have an equation, then it's all the numbers that are allowed to be x)

Range is all the y's on the graph (or if you have an equation, all the possible y outcomes)

Open dot endpoint means that point is not included in the set (domain and range are sets of numbers) Closed dot means the point is included.

You didn't get this math question right. But look for all the x's for the domain, and all the y's for the range.

6 0

2 years ago

Read 2 more answers