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4 years ago

when a certain positive number is multiplied by 2 the result is the square of the original number. what is the original number

Mathematics

2 answers:

galben [10]4 years ago

6 0

The original number is 2.

(2)*2=4

Please vote my answer branliest! Thanks.

Arada [10]4 years ago

5 0

Wouldn't it be 2? 2x2=4 which is the square of 2:the original number

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PLEASE HELP IT’S DUE TMR!!

zepelin [54]

Answer:

(a) 1.0792 x 10³

(b) 3.769 x 10³

This should be the answer to the problems, the rounding part got me confused for a second.

Significant digits are numbers after the decimal that aren't 0. So the 1.0792, the 0 isn't a significant digit.

6 0

3 years ago

(a) Suppose anxn has finite radius of convergence R and an ≥ 0 for all n. Show that if the series converges at R, then it also c

valina [46]

Answer:

a) See the proof below.

b) $\sum \frac{(-x)^n}{n}$

Step-by-step explanation:

Part a

For this case we assume that we have the following series $\sum a)n x^n$ and this series has a finite radius of convergence $R$ and we assume that $a_n \geq 0$ for all n, this information is given by the problem.

We assume that the series converges at the point $x= R$ since w eknwo that converges, and since converges we can conclude that:

$\sum a)n R^n < \infty$

For this case we need to show that converges also for $x=-R$

So we need to proof that $\sum a_n (-R)^n < \infty$

We can do some algebra and we can rewrite the following expression like this:

$\sum a_n (-R)^n = \sum (-1)^n a)n R^n$ and we see that the last series is alternating.

Since we know that $\sum a_n x^n$ converges then the sequence { $a_n R^n$ } must be positive and we need to have $lim_{n\to \infty} a^n R^n = 0$

And then by the alternating series test we can conclude that $\sum a_n (-R)^n$ also converges. And then we conclude that the power series $a_n x^n$ converges for $x=-R$ ,and that complete the proof.

Part b

For this case we need to provide a series whose interval of convergence is exactly (-1,1]

And the best function for this $\frac{(-x)^n}{n}$

Because the series $\sum \frac{(-x)^n}{n}$ converges to $-ln(1+x)$ when $|x|$ using the root test.

But by the properties of the natural log the series diverges at $x=-1$ because $\sum \frac{1}{n} =\infty$ and for $x=1$ we know that converges since $\sum \frac{-1}{n}$ is an alternating series that converges because the expression tends to 0.