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lara [203]
3 years ago
14

________________=12÷(2+4)​

Mathematics
2 answers:
spayn [35]3 years ago
7 0

Answer:

2

Step-by-step explanation:

I must assume you want to simplify the given expression.

You must do the work inside the parentheses first:

---------------------- = 12/6, or

----------------------- = 2

vodomira [7]3 years ago
6 0

Answer:

2 = 12/(2 + 4)

Step-by-step explanation:

order of operation.

P E M D A S( parenthesis, exponents, multiplication and division( left to right), and addition and subtraction(left to right) ).

parenthesis:

2 = 12/(2 + 4)

2 = 12/(6)

division(left to right):

2 = 2 √

this is why it equals to 2

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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))

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

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

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and the antiderivative is

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

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Morgarella [4.7K]

Answer:

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Hey there!

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