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3 years ago

А B 12 14 8 1 What is: n(An BC). Question Help: Message instructor

Physics

1 answer:

Vesnalui [34]3 years ago

5 0

The set A intersection set B complements $(A \ n\ B^C)$ is {5, 12}.

<h3>What is a Venn diagram?</h3>

Venn diagram is an illustration that uses circles to show the relationships among set of numbers.

<h3>Elements in set A</h3>

A = {3,4,5,12}

<h3>Elements in set B</h3>

B = {3,4,7,14}

<h3>Set B complements</h3>

$B^C$ = {5, 8,12}

Set A intersection set B complements $(A \ n\ B^C)$ is calculated as follows;

$(A \ n\ B^C)$ = {3,4,5,12} ∩ {5, 8,12} = {5,12}

Thus, the set A intersection set B complements $(A \ n\ B^C)$ is {5, 12}.

Learn more about Venn diagram here: brainly.com/question/24713052

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The quartz crystal used in an electric watch vibrates with a frequency of 32,768 Hz. What is the period of the crystal's motion?

Elan Coil [88]

Answer:

Time period, $T=3.05\times 10^{-5}\ s$

Explanation:

Given that,

The quartz crystal used in an electric watch vibrates with a frequency of 32,768 Hz, f = 32768 Hz

We need to find the period of the crystal's motion. The relationship between the frequency and the time period is given by :

$T=\dfrac{1}{f}$

T is the time period of the crystal's motion.

Time period is given by :

$T=\dfrac{1}{32768}$

$T=3.05\times 10^{-5}\ s$

So, the time period of the crystal's motion is $3.05\times 10^{-5}\ s$ . Hence, this is the required solution.

8 0

3 years ago

If javier has a car thats 250,000 kg and a forse of 300 newtons.what would be the accleration.

const2013 [10]

Answer:

The acceleration would be 3.455.

3 0

4 years ago

A 39-foot ladder is leaning against a vertical wall. If the bottom of the ladder is being pulled away from the wall at the rate

Viefleur [7K]

Answer:

The rate of change of the area when the bottom of the ladder (denoted by $b$ ) is at 36 ft. from the wall is the following:

$\frac{dA}{dt}|_{b=36}=-571.2\, ft^2/s$

Explanation:

The Area of the triangle is given by $A=h\times b$ where $h=\sqrt{l^2-b^2}$ (by using the Pythagoras' Theorem) and $b$ is the length of the base of the triangle or the distance between the bottom of the ladder and the wall.

The area is then

$A=\sqrt{l^2-b^2}b$

The rate of change of the area is given by its time derivative

$\frac{dA}{dt}=\frac{d}{dt}\left(\sqrt{l^2-b^2}\cdot b\right)$

$\implies \frac{dA}{dt}=\frac{d}{dt}\left(\sqrt{l^2-b^2}\right)\cdot b+\frac{db}{dt}\cdot\sqrt{l^2-b^2}$

$\implies\frac{dA}{dt}=\frac{1}{2\sqrt{l^2-b^2}}\frac{d}{dt}(l^2-b^2)\cdot b+\sqrt{l^2-b^2}}\cdot \frac{db}{dt}$ Product rule

$\implies\frac{dA}{dt}=-\frac{1}{2\sqrt{l^2-b^2}}\cdot 2\cdot b^2\cdot \frac{db}{dt}+\sqrt{l^2-b^2}}\cdot \frac{db}{dt}$ Chain rule

$\implies\frac{dA}{dt}=-\frac{1}{\sqrt{l^2-b^2}}\cdot b^2\cdot \frac{db}{dt}+\sqrt{l^2-b^2}}\cdot \frac{db}{dt}$

$\implies\frac{dA}{dt}=\frac{db}{dt}\left(-\frac{1}{\sqrt{l^2-b^2}}\cdot b^2+\sqrt{l^2-b^2}}\right)$

In here we can identify $b=36\, ft$ , $l=39$ and $\frac{db}{dt}=8\,ft/s$ .

The result is then

$\frac{dA}{dt}=8\left(-\frac{1}{\sqrt{39^2-36^2}}\cdot 36^2+\sqrt{39^2-36^2}}\right)=-571.2\, ft^2/s$

3 0

3 years ago

A 1.5m wire carries a 7 A current when a potential difference of 87 V is applied. What is the resistance of the wire?

kramer

Answer:

Ohm's law states that I=V/R (Current=volts divided by resistance). Since we're looking for resistance, we'll rewrite it as R=V/I. Then just plug in the numbers; R=84/9, R= 9 1/3 or 28/3. The resistance of the wire is 9.33... or 9 1/3 ohm's, depending on how you wanna write it.

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3 0

3 years ago

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I need help on 54 plz

dalvyx [7]

Answer:

a = 2 m/s²

Explanation:

average acceleration = change of velocity / change of time

a = Δv/Δt = (20 - 10) / 5 = 10/5 = 2

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4 years ago

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