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2 years ago

Consider the sequence: -7, -2, 3, 8…. What is f(7)?

Mathematics

1 answer:

wlad13 [49]2 years ago

5 0

You are adding by 5 in the sequence.
If f(1) is -7
f(2) = -2
f(3)= 3
f(4)=8
f(5)=13
f(6)=18
Therefore, f(7)=23

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Plz help simplify 8/10

Free_Kalibri [48]

$\frac{8}{10}$ → $\frac{4}{5}$

8 and 10 can both be divided by 2. So, 8 ÷ 2 is 4 and 10 ÷ 2 is 5.

Hope this helps :)

3 0

2 years ago

Read 2 more answers

5 less than the quotient of a number y and 2

Marta_Voda [28]

Quotient means division....less then means subtract
(y/2) - 5

5 0

2 years ago

Suppose m = 2 + 6i, and | m + n | = 3√10, where n is a complex number.

Ksju [112]

Answer: a) √50

b) n = 1 + 7i

Step-by-step explanation:

first, the modulus of a complex number z = a + bi is

IzI = √(a^2 + b^2)

The fact that n is complex does not mean that n doesn't has a real part, so we must write our numbers as:

m = 2 + 6i

n = a + bi

Im + nI = 3√10

Im + n I = √(a^2 + b^2 + 2^2 + 6^2)= 3√10

= √(a^2 + b^2 + 40) = 3√10

a^2 + b^2 + 40 = 3^2*10 = 9*10 = 90

a^2 + b^2 = 90 - 40 = 50

√(a^2 + b^2 ) = InI = √50

The modulus of n must be equal to the square root of 50.

now we can find any values a and b such a^2 + b^2 = 50.

for example, a = 1 and b = 7

1^2 + 7^2 = 1 + 49 = 50

Then a possible value for n is:

n = 1 + 7i

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

In a set of data, if there is no clear pattern to the points when graphed on a coordinate plane, we say there is no relative ___

VikaD [51]

Answer: If there is no clear pattern to the points when graphed on a coordinate plane, we say there is no relative correlation.

In a scatter plot, we are looking for a pattern (or correlation) between the points. Sometimes, they increase together. Sometimes, they decrease together.

If we can't see a pattern, we say there is no correlation.

7 0

3 years ago