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2 years ago

8

Find the dimensions of a deck which will have railings on only three sides. There is 28 m of railing available and the deck must

be as large as possible.

1 answer:

Kobotan [32]2 years ago

8 0

Answer:

2x2x7

Step-by-step explanation:

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(−4)4 is the question. it is in math

Zinaida [17]

-4 * 4 = -16

<h2>This is the answer to math</h2>

4 0

3 years ago

Read 2 more answers

Evaluate using integration by parts

PolarNik [594]

Rather than carrying out IBP several times, let's establish a more general result. Let

$I(n)=\displaystyle\int x^ne^x\,\mathrm dx$

One round of IBP, setting

$u=x^n\implies\mathrm du=nx^{n-1}\,\mathrm dx$

$\mathrm dv=e^x\,\mathrm dx\implies v=e^x$

gives

$\displaystyle I(n)=x^ne^x-n\int x^{n-1}e^x\,\mathrm dx$

$I(n)=x^ne^x-nI(n-1)$

This is called a power-reduction formula. We could try solving for $I(n)$ explicitly, but no need. $n=5$ is small enough to just expand $I(5)$ as much as we need to.

$I(5)=x^5e^x-5I(4)$

$I(5)=x^5e^x-5(x^4e^x-4I(3))=(x-5)x^4e^x+20I(3)$

$I(5)=(x-5)x^4e^x+20(x^3e^x-3I(2))=(x^2-5x+20)x^3e^x-60I(2)$

$I(5)=(x^2-5x+20)x^3e^x-60(x^2e^x-2I(1))=(x^3-5x^2+20x-60)x^2e^x+120I(1)$

$I(5)=(x^3-5x^2+20x-60)x^2e^x+120(xe^x-I(0))$

Finally,

$I(0)=\displaystyle\int e^x\,\mathrm dx=e^x+C$

so we end up with

$I(5)=(x^4-5x^3+20x^2-60x+120)xe^x-120e^x+C$

$I(5)=(x^5-5x^4+20x^3-60x^2+120x-120)e^x+C$

and the antiderivative is

$\displaystyle\int2x^5e^x\,\mathrm dx=(2x^5-10x^4+40x^3-120x^2+240x-240)e^x+C$

8 0

2 years ago

Which expression is equivalent to 5/6 ÷ 4

Ronch [10]

Answer:

5/24 or in decimal form 0.2083

Hope this helped

4 0

3 years ago

The diameter of a circle is 15 in. Which measure below cannot be a chord on the same circle?

diamong [38]

It's c 18, because the diameter is the longest chord so it would be greater than 15.

6 0

3 years ago

Find the value of x in the triangle shown below

Georgia [21]

Answer:

the length of the hypotenuse must be 10.

Step-by-step explanation:

This is a right triangle, so we can apply the Pythagorean Theorem.

6² + 8² = 10² so the length of the hypotenuse must be 10.

6 0

2 years ago