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damaskus [11]
3 years ago
6

What is (3-6)/8-2+7/2

Mathematics
1 answer:
Sladkaya [172]3 years ago
5 0
(3-6)/(8-2)+(7/2)
Do what is in the parenthesis first.
-3/6+3.5
-0.5+3.5=3

Therefore, your answer is 3
Hope this Helps!
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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
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Round 4.645 to the<br> nearest whole number.
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5. since .6 is greater than 5, we round up.
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Michael sold his painting for $50. He sold his painting for $10 less than its original price. At what price did Michael buy the
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PLEASE HELPPP MEEEEEEEEEEE I DK THIS-
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Answer:

FOr the first one you can do:

y= -x +4

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Step-by-step explanation:

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8 0
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