Ask question

Ask question

All categories

AleksAgata [21]

3 years ago

15

Please help!

1 answer:

Gemiola [76]3 years ago

4 0

Answer:

test test test test test test test test test test test test test test test test test test test test test test test test test test test test test test

You might be interested in

Find the radius of convergence, then determine the interval of convergence

galben [10]

The radius of convergence R is 1 and the interval of convergence is (-3, -1) for the given power series. This can be obtained by using ratio test.

<h3>Find the radius of convergence R and the interval of convergence:</h3>

Ratio test is the test that is used to find the convergence of the given power series.

First aₙ is noted and then aₙ₊₁ is noted.

For ∑ aₙ, aₙ and aₙ₊₁ is noted.

$\lim_{n \to \infty} |\frac{a_{n+1}}{a_{n} }|$ = β

If β < 1, then the series converges

If β > 1, then the series diverges

If β = 1, then the series inconclusive

Here $a_{k}$ = $\frac{(x+2)^{k}}{\sqrt{k} }$ and $a_{k+1}$ = $\frac{(x+2)^{k+1}}{\sqrt{k+1} }$

Now limit is taken,

$\lim_{n \to \infty} |\frac{a_{n+1}}{a_{n} }|$ = $\lim_{n \to \infty} |\frac{(x+2)^{k+1} }{\sqrt{k+1} }/\frac{(x+2)^{k} }{\sqrt{k} }|$

= $\lim_{n \to \infty} |\frac{(x+2)^{k+1} }{\sqrt{k+1} }\frac{\sqrt{k} }{(x+2)^{k}}|$

= $\lim_{n \to \infty} |{(x+2) } }{\sqrt{\frac{k}{k+1} } }}|$

= $|{x+2 }|\lim_{n \to \infty}}{\sqrt{\frac{k}{k+1} } }}$

= $|{x+2 }|$ < 1

- 1 < ${x+2 }$ < 1

- 1 - 2 < x < 1 - 2

- 3 < x < - 1

We get that,

interval of convergence = (-3, -1)

radius of convergence R = 1

Hence the radius of convergence R is 1 and the interval of convergence is (-3, -1) for the given power series.

Learn more about radius of convergence here:

brainly.com/question/14394994

#SPJ1

5 0

1 year ago

Read 2 more answers

Are the expressions x + x + 1 + x + 2 + x + 1 + x and 5x + 4 equivalent? Why or why not?

egoroff_w [7]

Answer:

yes

they are equivalent because lets say the X = 3 then you would have to answer

3 + 3 + 1 + 3 + 2 + 3 + 1 + 3

and that would = 19

and then on the other equation

5 x 3 + 4 = 19

So the both equal the same.

PLZ MARK ME BRAINLIEST

3 0

3 years ago

If<br> DX/DT= 2cm/s <br> x=-1 <br> y=x^2+1<br> DY/DT=??

kobusy [5.1K]

Answer:

$\frac{dy}{dt}$ = - 4 cm/s

Step-by-step explanation:

Let us revise the chain rule

If $\frac{dy}{dt}$ = a and $\frac{dx}{dt}$ = b, then

$\frac{dy}{dx}$ = $\frac{dy}{dt}$ ÷ $\frac{dx}{dt}$ = $\frac{a}{b}$

∵ y = x² + 1

- Use the differentiation to find $\frac{dy}{dx}$

∴ $\frac{dy}{dx}$ = 2x

∵ x = -1

- Substitute x by -1 in $\frac{dy}{dx}$

∴ $\frac{dy}{dx}$ = 2(-1)

∴ $\frac{dy}{dx}$ = -2

∵ $\frac{dy}{dt}$ = $\frac{dy}{dx}$ × $\frac{dx}{dt}$

∵ $\frac{dx}{dt}$ = 2 cm/s

∴ $\frac{dy}{dt}$ = (-2) × (2)

∴ $\frac{dy}{dt}$ = - 4 cm/s

8 0

3 years ago

F(x)=x+3 ;find f (x+1)

nataly862011 [7]

Answer:

x+4

Step-by-step explanation:

(x+1)+3

8 0

3 years ago

Solve the following system using substitution.

ziro4ka [17]

Answer/Step-by-step explanation:

3. By substitution method, let's substitute $\frac{2}{3}x- 4$ for y in the first equation.

Thus,

$\frac{1}{3}x + 2(\frac{2}{3}x- 4) = 1$

Solve for x

$\frac{x}{3} + \frac{4x}{3} - 4 = 1$

Add 4 to both sides

$\frac{x}{3} + \frac{4x}{3} - 4 + 4 = 1 + 4$

$\frac{x}{3} + \frac{4x}{3} = 5$

$\frac{x + 4x}{3} = 5$

$\frac{5x}{3} = 5$

Multiply both sides by 3

$\frac{5x}{3}*3 = 5*3$

$5x = 15$

Divide both sides by 5

$x = 3$

Now, substitute 3 for x in the equation.

$y = \frac{2}{3}x- 4$

$y = \frac{2}{3}(3) - 4$

$y = 2 - 4$

$y = -2$

The solution of the equation is x = 3, y = -2

4. Solving by elimination, let's try to eliminate the x-variable by adding both equation together.

$3x - 2y = 11$

$-3x - y = 4$

$-3y = 15$ => $(-3x +(-3x) = 0; -2y +(-y) = -3y; 11 + 4 = 15)$

Divide both sides by -3 to solve for y

$\frac{-3y}{-3} = \frac{15}{-3}$

$y = -5$

Substitute -5 for y in the first equation to find x

$3x - 2(-5) = 11$

$3x + 10 = 11$

Subtract 10 from both sides

$3x + 10 - 10 = 11 - 10$

$3x = 1$

Divide both sides by 3

$\frac{3x}{3} = \frac{1}{3}$

$x = \frac{1}{3}$

The solution is $x = \frac{1}{3}, y = -5$

8 0

3 years ago