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2 years ago

How old are yall?............

Mathematics

2 answers:

Reika [66]2 years ago

7 0

Answer:

Step-by-step explanation:05/08/2010

tekilochka [14]2 years ago

4 0

I am Fourteen. (^^)

Why do you ask?

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Find the radius of convergence, r, of the series. ∞ xn 2n − 1 n = 1 r = 1 find the interval, i, of convergence of the series. (e

Bingel [31]

Assuming the series is

$\displaystyle\sum_{n\ge1}\frac{x^n}{2n-1}$

The series will converge if

$\displaystyle\lim_{n\to\infty}\left|\frac{\frac{x^{n+1}}{2(n+1)-1}}{\frac{x^n}{2n-1}}\right|$

We have

$\displaystyle\lim_{n\to\infty}\left|\frac{\frac{x^{n+1}}{2(n+1)-1}}{\frac{x^n}{2n-1}}\right|=|x|\lim_{n\to\infty}\frac{\frac1{2n+1}}{\frac1{2n-1}}=|x|-\lim_{n\to\infty}\frac{2n-1}{2n+1}=|x|$

So the series will certainly converge if $-1$ , but we also need to check the endpoints of the interval.

If $x=1$ , then the series is a scaled harmonic series, which we know diverges.

On the other hand, if $x=-1$ , by the alternating series test we can show that the series converges, since

$\left|\dfrac{(-1)^n}{2n-1}\right|=\dfrac1{2n-1}\to0$

and is strictly decreasing.

So, the interval of convergence for the series is $-1\le x$ .

6 0

4 years ago

Using the quadratic formula, what is the value of x for x^2-4+5=0

PSYCHO15rus [73]

value of x is: x= i and x= -i

Step-by-step explanation:

We need to find the value of x from $x^2-4+5=0$ using quadratic formula.

The term is: $x^2+1=0$

The quadratic formula is:

$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\\$

Where a = 1, b=0,c=1

Putting values:

$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\\x=\frac{0\pm\sqrt{(0)^2-4(1)(1)}}{2(1)}\\x=\frac{\pm\sqrt{-4}}{2}\\x=\frac{\pm2i}{2}\\x=\pm i$

So, value of x is: x= i and x= -i

Keywords: Quadratic formula

Learn more about quadratic formula at:

#learnwithBrainly

4 0

4 years ago

What is the 2nd equivalent fraction to <br> 1/11

alisha [4.7K]

2/22.
i think
sorry if it’s wrong

6 0

3 years ago

The number 27 is <br> please help!

alina1380 [7]

Answer:

27 is a composite number

5 0

3 years ago

Solve the equation by completing the square.Round to the nearest hundredth if necessary. X^2–x–7=0

lesya [120]

NOTE: $\left(x-\dfrac{1}{2}\right)^2=x^2-x+\dfrac{1}{4}$

$x^2-x-7=0 \\ \\ \underbrace{x^2-x+\frac{1}{4}}_{\left(x-\frac{1}{2}\right)^2}-7\dfrac{1}{4}=0 \\ \\ \\ \left(x-\frac{1}{2}\right)^2=7\frac{1}{4}=\frac{29}{4} \qquad /\sqrt{}\\ \\ \left|x-\frac{1}{2}\right|=\dfrac{\sqrt{29}}{2} \\ \\ x_1=\dfrac{1}{2}-\dfrac{\sqrt{29}}{2} \approx -2.19 \\ \\ x_2=\dfrac{1}{2}+\dfrac{\sqrt{29}}{2} \approx 3.19$

ANSWER: c)

5 0

3 years ago