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Vlada [557]
4 years ago
12

If ∑|

x-formula"> | is convergent, is ∑ a_{n} also convergent?
Mathematics
1 answer:
Llana [10]4 years ago
8 0
Not necessarily. Consider the series \displaystyle\sum_{n=1}^\infty\frac{(-1)^n}n and \displaystyle\sum_{n=1}^\infty\frac1n. Here a_n=\dfrac1n.

The first series converges by the alternating series test, which says \sum(-1)^na_n converges if \left|(-1)^na_n\right|=|a_n| is a decreasing sequence and converges to 0. This is the case, as a_n=\dfrac1n\to0 as n\to\infty, and each term is decreasing. (Indeed the series converges to -\ln2.)

On the other hand, the second series is a classic example of a divergent sum.
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Giving Brainliest for right answers :)
VashaNatasha [74]

Answer:

∠ DFE = 15°

Step-by-step explanation:

The inscribed angle DFE is half the measure of its intercepted arc, that is

∠ DFE = \frac{1}{2} DE = \frac{1}{2} × 30° = 15°

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3 years ago
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What is a cubic polynomial function with zeros 3,3 and -3? show your work.​
Brums [2.3K]

Answer:

Cubic polynomial function with zeros 3,3 and -3 is \mathbf{x^3-3x^2-9x+27=0}

Step-by-step explanation:

We need to find a cubic polynomial function with zeros 3,3 and -3.

If zeros of polynomial are: 3,3,and -3

we can write:

x=3, x=3, x=-3

Or

We can write:

x-3=0, x-3=0, x+3=0

Now, we can write them as:

(x-3)(x-3)(x+3)=0

Multiplying the terms, we can find the polynomial:

(x-3)(x-3)(x+3)=0\\(x(x-3)-3(x-3))(x+3)=0\\(x^2-3x-3x+9)(x+3)=0\\(x^2-6x+9)(x+3)=0\\x(x^2-6x+9)+3(x^2-6x+9)=0\\x^3-6x^2+9x+3x^2-18x+27=0\\x^3-6x^2+3x^2+9x-18x+27=0\\x^3-3x^2-9x+27=0

So, cubic polynomial function with zeros 3,3 and -3 is \mathbf{x^3-3x^2-9x+27=0}

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3 years ago
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Answer:

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Elena-2011 [213]

Answer:

Step-by-step explanation:

∠A + ∠B = 90    {∠A & ∠B are complementary}

x + 16 + x - 12 = 90

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        2x = 90 - 4

        2x = 86          {divide both sides by 2

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