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3 years ago

To extinguish a fire with a portable extinguisher, a person must:

SAT

2 answers:

kipiarov [429]3 years ago

8 0

Know how to use the extinguisher and have access to it right then and there or have it nearby

nevsk [136]3 years ago

6 0

Answer:

have immediate access to the extinguisher, know how to actuate the unit, and know how to apply the agent effectively. Attempting to extinguish even a small fire carries some risk. Hopefully this is what you were looking for.

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3 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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2 years ago

What's the answer to number 1 on the picture?

Art [367]

Its c because the diversty between all of the species which is the variety im pretty sure

6 0

3 years ago

Read 2 more answers