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2 years ago

Air pressure may be represented as a function of height.

SAT

1 answer:

miss Akunina [59]2 years ago

3 0

Answer:

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Explanation:

am i supposed to see something?

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Compare and contrast crewed and uncrewed space missions

dexar [7]

Uncrewed space missions cost less to retrieve the same information from space, it also has little to no risk towards human life.

Crewed space missions have an enormous cost, due to the tons of foolproof equipment to keep the humans alive.

Some scientist complain that there is nothing better in a space mission than human intelligence, as only humans can prepare for the unexpected. However at the cost is not available for all space missions.

Hope this helps, if you need any more information, don’t hesitate to ask

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2 years ago

If the atlas and axis were replaced with typical cervical vertebrae ____ would be affected

cricket20 [7]

if the atlas and axis were replaced with typical cervical vertebrae the ability to pivot the head from side to side would be affected.

<h3>Where is the human atlas?</h3>

The atlas is the superior vertebra that can be seen on the neck of a human being. This is known to provide a support for the entire human head.

If this part of the body is replaces there would be issues with the pivoting of the head from one side to another.

Read more on human atlas here: brainly.com/question/13411436

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2 years ago

Use series to approximate the definite integral i to within the indicated accuracy. i = 1/2 x3 arctan(x) dx 0 (four decimal plac

Valentin [98]

The expression $\int\limits^{1/2}_0 {x^3 \arctan(x)} \, dx$ is an illustration of definite integrals

The approximated value of the definite integral is 0.0059

<h3>How to evaluate the definite integral?</h3>

The definite integral is given as:

$\int\limits^{1/2}_0 {x^3 \arctan(x)} \, dx$

For arctan(x), we have the following series equation:

$\arctan(x) = \sum\limits^{\infty}_{n = 0} {(-1)^n \cdot \frac{x^{2n + 1}}{2n + 1}}$

Multiply both sides of the equation by x^3.

So, we have:

$x^3 * \arctan(x) = \sum\limits^{\infty}_{n = 0} {(-1)^n \cdot \frac{x^{2n + 1}}{2n + 1}} * x^3$

Apply the law of indices

$x^3 * \arctan(x) = \sum\limits^{\infty}_{n = 0} {(-1)^n \cdot \frac{x^{2n + 1 + 3}}{2n + 1}}$

$x^3 * \arctan(x) = \sum\limits^{\infty}_{n = 0} {(-1)^n \cdot \frac{x^{2n + 4}}{2n + 1}}$

Evaluate the product

$x^3 \arctan(x) = \sum\limits^{\infty}_{n = 0} {(-1)^n \cdot \frac{x^{2n + 4}}{2n + 1}}$

Introduce the integral sign to the equation

$\int\limits^{1/2}_{0} x^3 \arctan(x)\ dx =\int\limits^{1/2}_{0} \sum\limits^{\infty}_{n = 0} {(-1)^n \cdot \frac{x^{2n + 4}}{2n + 1}}$

Integrate the right hand side

$\int\limits^{1/2}_{0} x^3 \arctan(x)\ dx =[ \sum\limits^{\infty}_{n = 0} {(-1)^n \cdot \frac{x^{2n + 4}}{2n + 1}} ]\limits^{1/2}_{0}$

Expand the equation by substituting 1/2 and 0 for x

$\int\limits^{1/2}_{0} x^3 \arctan(x)\ dx =[ \sum\limits^{\infty}_{n = 0} {(-1)^n \cdot \frac{(1/2)^{2n + 4}}{2n + 1}} ] - [ \sum\limits^{\infty}_{n = 0} {(-1)^n \cdot \frac{0^{2n + 4}}{2n + 1}} ]$

Evaluate the power

$\int\limits^{1/2}_{0} x^3 \arctan(x)\ dx =[ \sum\limits^{\infty}_{n = 0} {(-1)^n \cdot \frac{(1/2)^{2n + 4}}{2n + 1}} ] - 0$

$\int\limits^{1/2}_{0} x^3 \arctan(x)\ dx = \sum\limits^{\infty}_{n = 0} {(-1)^n \cdot \frac{(1/2)^{2n + 4}}{2n + 1}}$

The nth term of the series is then represented as:

$T_n = \frac{(-1)^n}{2^{2n + 5} * (2n + 4)(2n + 1)}$

Solve the series by setting n = 0, 1, 2, 3 ..........

$T_0 = \frac{(-1)^0}{2^{2(0) + 5} * (2(0) + 4)(2(0) + 1)} = \frac{1}{2^5 * 4 * 1} = 0.00625$

$T_1 = \frac{(-1)^1}{2^{2(1) + 5} * (2(1) + 4)(2(1) + 1)} = \frac{-1}{2^7 * 6 * 3} = -0.0003720238$

$T_2 = \frac{(-1)^2}{2^{2(2) + 5} * (2(2) + 4)(2(2) + 1)} = \frac{1}{2^9 * 8 * 5} = 0.00004340277$

$T_3 = \frac{(-1)^3}{2^{2(3) + 5} * (2(3) + 4)(2(3) + 1)} = \frac{-1}{2^{11} * 10 * 7} = -0.00000634131$

..............

At n = 2, we can see that the value of the series has 4 zeros before the first non-zero digit

This means that we add the terms before n = 2

This means that the value of $\int\limits^{1/2}_0 {x^3 \arctan(x)} \, dx$ to 4 decimal points is

$\int\limits^{1/2}_0 {x^3 \arctan(x)} \, dx = 0.00625 - 0.0003720238$

Evaluate the difference

$\int\limits^{1/2}_0 {x^3 \arctan(x)} \, dx = 0.0058779762$

Approximate to four decimal places

$\int\limits^{1/2}_0 {x^3 \arctan(x)} \, dx = 0.0059$

Hence, the approximated value of the definite integral is 0.0059

Read more about definite integrals at:

brainly.com/question/15127807

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1 year ago

Florence is 45 years old, and she makes $65,000 per year. if florence were to die, how much would the beneficiaries of her life

ki77a [65]

Answer: she would receive 39,000 and that is 60% of her incomeExplanation: Now as you can she is making 65,000 per year. So you multiply 65,000 with 60/100. You make 60% into 60/100 because it's a percent and a decimal. Like I was saying if you multiply 65,000 with 60/100 you will get 39,000 which is 60% of her tax income

if my answer is incorrect or if i made a mistake pls let me know, please and thank you :)

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2 years ago

Hey yall umm if yall remember me from my old umm account oitsjoy umm dey had logged me out so now i got to start all da way ova

Juliette [100K]

Answer:

okay, does have a reason that it logged you off?

Explanation:

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3 years ago

Read 2 more answers