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3 years ago

Circle the expression that give the same product as 6x3/8

Mathematics

1 answer:

Mamont248 [21]3 years ago

5 0

There is not enough information to answer this problem. I wish I could help :(

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Kyra is framing a square painting with side lengths of (x + 6) inches.

mariarad [96]

Answer:

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Step-by-step explanation:

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3 years ago

HURRY!!!! PLS NEED AWMSER IN CLASS I CAN PICK THEM DOWN BELOW

ivann1987 [24]

Answer:

9. 7/2

10. -6

Step-by-step explanation:

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3 years ago

Write a function to represent the set of ordered pairs. {(2, 2), (3,7). (4, 14), (5,23)}

MrMuchimi

Answer:

Step-by-step explanation:

The set of ordered pairs {(1 , 7) , (−1 , 7) , (8 , −6) , (−3 , 2)} is a function

The set of ordered pairs {(4 , 7) , (−1 , 2) , (4 , −6) , (−3 , −2)} is not a

function

The set of ordered pairs {(3 , 7) , (−1 , 2) , (−4 , −2) , (−3 , −2)} is a function

The set of ordered pairs {(−3 , 7) , (1 , 2) , (4 ,−2) , (12,−4)} is a function

An ordered pair represents a relationship between two values

The first value is the input the second value is the output

1. A relation is a set of inputs and outputs

2. A function is a relation with one output for each input

3. All functions are relations

{(1 , 7) , (−1 , 7) , (8 , −6) , (−3 , 2)}

∵ Each x-coordinate in each ordered pair has only one y-coordinate

∴ The relation above is a function

The set of ordered pairs {(1 , 7) , (−1 , 7) , (8 , −6) , (−3 , 2)} is a function

{(4 , 7) , (−1 , 2) , (4 , −6) , (−3 , −2)}

∵ The x-coordinate 4 has two values of y (7 and -6)

∴ The relation above is not a function

The set of ordered pairs {(4 , 7) , (−1 , 2) , (4 , −6) , (−3 , −2)} is not a

function

{(3 , 7) , (−1 , 2) , (−4 , −2) , (−3 , −2)}

∵ Each x-coordinate in each ordered pair has only one y-coordinate

∴ The relation above is a function

The set of ordered pairs {(3 , 7) , (−1 , 2) , (−4 , −2) , (−3 , −2)} is a function

{(−3 , 7) , (1 , 2) , (4 ,−2) , (12,−4)}

∵ Each x-coordinate in each ordered pair has only one y-coordinate

∴ The relation above is a function

The set of ordered pairs {(−3 , 7) , (1 , 2) , (4 ,−2) , (12,−4)} is a function

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 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

I have 3 questions for math

Airida [17]

Answer:

no question content
(x, y) = (1/2, 4)
(x, y) = (2, 10)

Step-by-step explanation:

1. No link, no question

2. Divide the first equation by 2 and substitute for y using the expression in the second equation.

2x +(4x+2) = 5

6x = 3 . . . . . . . . . subtract 2

x = 3/6 = 1/2 . . . . divide by2

y = 4(1/2) +2 = 4 . . . . substitute into the equation for y

The solution is (x, y) = (1/2, 4).

3. The solution using a graphing calculator is (x, y) = (2, 10). (see attached)

7 0

3 years ago