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3 years ago

Please help to fill in the fractions 2/8=.../4

2 answers:

6 0

1/4 you just divide the denominators and then divide the numerators by the number you divided the denominator by. They are equivalent fractions

mote1985 [20]3 years ago

4 0

1 because if you divide 2/8 by 2, you will get 1/4.

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Katarina [22]

Step-by-step explanation:

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2 years ago

What is 78.867 ÷0.9 ANSWER FAST!!

xeze [42]

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4 0

3 years ago

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Find the radius of sphere with a volume of 500π3 cubic feet.

gulaghasi [49]

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8 0

3 years ago

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

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5 0

1 year ago

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irga5000 [103]

I think it's b but i'm not sure!

5 0

3 years ago