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Zina [86]
2 years ago
12

Resolve 3x-1÷(x+1)^2 into partial fraction​

Mathematics
1 answer:
garri49 [273]2 years ago
4 0

9514 1404 393

Answer:

  3/(x +1) -4/(x +1)^2

Step-by-step explanation:

The partial fraction expansion will be of the form ...

  A/(x+1)^2 +B/(x+1)

We can find the values of A and B by writing the sum of these terms:

  = (A +B(x +1))/(x +1)^2

Then we require ...

  B = 3

  A +B = -1   ⇒   A = -4

So, the desired expansion is ...

  3/(x +1) -4/(x +1)^2

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What is the volume of the right rectangular prism?
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c

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Solve sin theta + 1 = cos 2 theta on the interval 0 less than or equal to theta &lt;2 pi
hjlf

Answer:

Ф = 0 and Ф = π

Step-by-step explanation:

* Lets explain how to solve the problem

∵ sin Ф + 1 = cos²Ф, where 0 ≤ Ф < 2π

- To solve we must to replace cos²Ф by 1 - sin²Ф

∵ sin²Ф + cos²Ф = 1

- By subtracting sin²Ф from both sides

∴ cos²Ф = 1 - sin²Ф

- Lets replace cos²Ф in the equation above

∴ sin Ф + 1 = 1 - sin²Ф

- Subtract 1 from both sides

∴ sin Ф = - sin²Ф

- Add sin²Ф for both sides

∴ sin²Ф + sin Ф = 0

- Take sin Ф as a common factor from both sides

∴ sin Ф(sin Ф + 1) = 0

- Equate each factor by 0

∵ sin Ф = 0

∴ Ф = 0 OR Ф = 2π

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Answer:

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The electric cooperative needs to know the mean household usage of electricity by its non-commercial customers in kWh per day. A
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Answer: (14.4, 17.2)

Step-by-step explanation: We are to construct a 98% confidence interval for mean household usage of electricity.

We have been given that

Sample size (n) = 872

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Population standard deviation (σ) = 1.8

The formulae that defines the 98% confidence interval for mean is given below as

u = x + Zα/2 × σ/√n...... For the upper limit

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Hence the 98% confidence interval for population mean usage of electricity is (14.4kwh, 17.2kwh)

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