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3 years ago

Converting a fraction to a repeating decimal: Advanced

Mathematics

1 answer:

Nastasia [14]3 years ago

8 0

Answer:

4.63636363636 or 4.63 repeating

Step-by-step explanation:

If you do 51/11 you will get 4.63636363636 which is 4.63 repeating (sry I could not find the bar which indicates that it repeats)

Hope this helps

╰(*°▽°*)╯

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PLEASE HELP ME I WILL GIVE YOU BRAINLIEST!

kicyunya [14]

Answer: D

Step-by-step explanation:

C=πd=π·22≈69.11504m :)

6 0

3 years ago

Read 2 more answers

find two positive real numbers such that they sum to 108 and the product of the first times the square of the second is a maximu

topjm [15]

The two positive numbers are 36 and 72 which gives a sum equal to 108 and the product of 36 and the square of 72 is a maximum.

Any integer greater than zero is considered a positive number. A positive number can either be written as a number or with the "+" symbol in front of it.

Let us consider the two positive real numbers as x and y. Then, their sum is written as,

x+y=108

Then, y=108-x

And the product is written as,

P=xy²

Substitute value of y in the above equation, we get,

$\begin{aligned}P&=x(108-x)^2\\&=x(11664-216x+x^2)\\&=11664x-216x^2+x^3\\&=x^3-216x^2+11664x\end{aligned}$

Now, differentiate P with respect to x, and we get,

$\begin{aligned}\frac{dP}{dx}&=3x^2-432x+11664\\&=x^2-144x+3888\end{aligned}$

Solving the above equation to zero, we get,

$\begin{aligned}x^2-144x+3888&=0\\x^2-36x-108x+3888&=0\\x(x-36)-108(x-36)&=0\\(x-108)(x-36)&=0\\x&=\text{108 or 36}\end{aligned}$

Substitute values of x in y=108-x, to get values of y.

If we substitute x=108, we get the y value as zero which doesn't give the required solution.

But, if we substitute x=36, we get,

$\begin{aligned}y&=108-36\\y&=72\end{aligned}$

Thus, the two positive numbers are 36 and 72.

To know more about positive numbers:

brainly.com/question/1635103

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4 0

1 year ago

Which number line shows all the values of a that make the inequality - 3x +1 < 7 true?

GrogVix [38]

Answer:

Third option

Step-by-step explanation:

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3 years ago

Whole numbers with more digits are ____ than the whole numbers with fewer digits.

Kipish [7]

Whole numbers with more digits are greater than whole numbers with fewer digits.

Unless there is a decimal making the number have more digits then the answer to the blank would be greater than.

3 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

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2 years ago