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3 years ago

Tery po

Mathematics

1 answer:

prisoha [69]3 years ago

5 0

Answer:

150.72 units³

Step-by-step explanation:

volume = πr²h

= 3.14×4²×3

= 150.72

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A number cube with faces labeled 1 to 6 is rolled once.

Zepler [3.9K]

Answer:

a) A ∪ B = {2, 4, 5, 6}

b) A ∩ B = {6}

c) $\sf B^c$ = {1, 3, 5}

Step-by-step explanation:

Set notation is used in math to list numbers, objects or outcomes.

It uses curly brackets called "braces". Objects placed within the braces are the elements of the set.

The listing method of set notation simply lists the numbers in the set.

The three dots "..." means it is infinite (goes on forever).

Given:

A number cube with faces labeled 1 to 6.

Therefore the set of all possible outcomes is {1, 2, 3, 4, 5, 6}.

Event A

The number rolled is greater than 4.

Therefore, the set of outcomes for event A is {5, 6}.

Event B

The number rolled is even.

Therefore, the set of outcomes for event B is {2, 4, 6}.

Part (a)
Event "A or B" means the outcomes in A or B or both.

⇒ A ∪ B = {2, 4, 5, 6}

Part (b)

Event "A and B" means the outcomes in both A and B.

⇒ A ∩ B = {6}

Part (c)

The complement of event B means everything that is not in B.

$\sf \implies B^c =$ {1, 3, 5}

Learn more about set notation here:

brainly.com/question/27913822

4 0

1 year ago

Convert 3 and 19/20 to a decimal

inysia [295]

3 is converted to 0.03
19/20 is converted to 0.95.

4 0

3 years ago

How do you integrate arctan(x dx? i think that if you simplify the integral you get:?

Temka [501]

Integration by parts will help here. Letting $u=\arctan x$ and $\mathrm dv=\mathrm dx$ , you end up with $\mathrm du=\dfrac{\mathrm dx}{1+x^2}$ and $v=x$ . Now

$\displaystyle\int\arctan x\,\mathrm dx=uv-\int v\,\mathrm du$
$\displaystyle\int\arctan x\,\mathrm dx=x\arctan x-\int\frac x{1+x^2}\,\mathrm dx$

For the remaining integral, setting $y=1+x^2$ gives $\dfrac{\mathrm dy}2=x\,\mathrm dx$ , so

$\displaystyle\int\frac x{1+x^2}\,\mathrm dx=\frac12\int\frac{\mathrm dy}y=\frac12\ln|y|+C=\frac12\ln(1+x^2)+C$

Putting everything together, you end up with

$\displaystyle\int\arctan x\,\mathrm dx=x\arctan x-\frac12\ln(1+x^2)+C$