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murzikaleks [220]
2 years ago
7

The diameter of a circle is 5 ft. Find the circumference TO THE NEAREST TENTH

Mathematics
1 answer:
Citrus2011 [14]2 years ago
4 0

Answer:

15.7

Step-by-step explanation:

3.14(5)

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2x+5x=100 x intercept​
Papessa [141]

This is an equation involving x alone, so the most you can do is solve it:

2x+5x=100x \iff 7x=100x \iff 0=93x \iff x=0

3 0
3 years ago
The ratio of students who walk home from school to the students who ride the bus is 2:7
yan [13]
3.5 Ithink that's the answer
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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
2 years ago
5/7 = /14 equivalent fraction answer should be?
Alex777 [14]

Answer:

the answer I got is false

Step-by-step explanation:

5/7=14 convert 14 to fraction 98/7

5/7=98/7 compare 5/7 and 98/7

so the maths is false question

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MatroZZZ [7]

Answer:

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Step-by-step explanation:

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