<img src="https://tex.z-dn.net/?f=5%20%7Bx%7D%5E%7B2%7D%20%20%2B%203%20%7Bx%7D%5E%7B2%7D%20%20%2B%206x%20%3D%203x%20-%206x%20%2B

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vladimir2022 [97]

2 years ago

%209x" id="TexFormula1" title="5 {x}^{2} + 3 {x}^{2} + 6x = 3x - 6x + 9x" alt="5 {x}^{2} + 3 {x}^{2} + 6x = 3x - 6x + 9x" align="absmiddle" class="latex-formula">
find out x please fast please fast

Mathematics

1 answer:

den301095 [7]2 years ago

5 0

Answer:

$5 {x}^{2} + 3 {x}^{2} + 6x = 3x - 6x + 9x \\ 8 {x}^{2} + 6x = - 3x + 9x \\ 8 {x}^{2} + 6x = 6x \\ 8 {x}^{2} = 6x - 6x \\ 8 {x}^{2} = 0 \\ x = 0 \\ thank \: you$

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PLEASE HELP!!

Musya8 [376]

Answer:

x = - 33

Step-by-step explanation:

Given

$\frac{2}{3}$ (x + 6) = - 18

Multiply both sides by 3 to clear the fraction

2(x + 6) = - 54 ( divide both sides by 2 )

x + 6 = - 27 ( subtract 6 from both sides )

x = - 33

6 0

3 years ago

Read 2 more answers

The sum of four consecutive even integers is 256. what are the numbers?

Naily [24]

Answer:

61 , 63 , 65 , 67

Step-by-step explanation:

Let the least even number be denoted by x. The sum of the four consecutive even numbers would be:

(x) + (x + 2) + (x + 4) (x + 6) = 256

First, simplify. Combine all like terms:

x + x + x + x + 2 + 4 + 6 = 256

4x + 12 = 256

Isolate the variable, x. Note the equal sign, what you do to one side, you do to the other. Do the opposite of PEMDAS.

First, subtract 12 from both sides of the equation:

4x + 12 (-12) = 256 (-12)

4x = 256 - 12

4x = 144

Next, divide 4 from both sides of the equation:

(4x)/4 = (144)/4

x = 144/4

x = 61

61 is your first number. Find the next 3 consecutive numbers:

x = 61

x + 2 = 63

x + 4 = 65

x + 6 = 67

Check:

61 + 63 + 65 + 67 = 256

256 = 256

7 0

2 years ago

Round 12502 to nearest thousand

dalvyx [7]

12,500 would be the answer

7 0

3 years ago

Read 2 more answers

Mia hiked 2 1/2 miles farther than Jacob.Which could be the two distances each person hiked?

Kruka [31]

M= J+2 1/2
So J+M would equal there total distance

5 0

2 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