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3 years ago

I would really appreciate the help please

Mathematics

1 answer:

dimulka [17.4K]3 years ago

6 0

Answer:

1 and 2 only

Step-by-step explanation:

mark me as brainliest

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Please help I have searched this question and have not found it. I am bad at math and I'm taking a test and need answers so plea

Vanyuwa [196]

Answer:

C because 18-6=12

Step-by-step explanation:

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3 years ago

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Complete the proof that △TVX~△WUX.

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Answer:

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△TVX ~ △WUX

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2 years ago

Which is bigger:30% or 1/3?

dimaraw [331]

1/3 is 33.3333%, so 1/3 is bigger

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3 years ago

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Use any of the methods to determine whether the series converges or diverges. Give reasons for your answer.

Aleks [24]

Answer:

It means $\sum_{n=1}^\inf} = \frac{7n^2-4n+3}{12+2n^6}$ also converges.

Step-by-step explanation:

The actual Series is::

$\sum_{n=1}^\inf} = \frac{7n^2-4n+3}{12+2n^6}$

The method we are going to use is comparison method:

According to comparison method, we have:

$\sum_{n=1}^{inf}a_n\ \ \ \ \ \ \ \ \sum_{n=1}^{inf}b_n$

If series one converges, the second converges and if second diverges series, one diverges

Now Simplify the given series:

Taking"n^2"common from numerator and "n^6"from denominator.

$=\frac{n^2[7-\frac{4}{n}+\frac{3}{n^2}]}{n^6[\frac{12}{n^6}+2]} \\\\=\frac{[7-\frac{4}{n}+\frac{3}{n^2}]}{n^4[\frac{12}{n^6}+2]}$

$\sum_{n=1}^{inf}a_n=\sum_{n=1}^{inf}\frac{[7-\frac{4}{n}+\frac{3}{n^2}]}{[\frac{12}{n^6}+2]}\ \ \ \ \ \ \ \ \sum_{n=1}^{inf}b_n=\sum_{n=1}^{inf} \frac{1}{n^4}$

Now:

$\sum_{n=1}^{inf}a_n=\sum_{n=1}^{inf}\frac{[7-\frac{4}{n}+\frac{3}{n^2}]}{[\frac{12}{n^6}+2]}\\ \\\lim_{n \to \infty} a_n = \lim_{n \to \infty} \frac{[7-\frac{4}{n}+\frac{3}{n^2}]}{[\frac{12}{n^6}+2]}\\=\frac{7-\frac{4}{inf}+\frac{3}{inf}}{\frac{12}{inf}+2}\\\\=\frac{7}{2}$

So a_n is finite, so it converges.

Similarly b_n converges according to p-test.

P-test:

General form:

$\sum_{n=1}^{inf}\frac{1}{n^p}$

if p>1 then series converges. In oue case we have:

$\sum_{n=1}^{inf}b_n=\frac{1}{n^4}$

p=4 >1, so b_n also converges.

According to comparison test if both series converges, the final series also converges.

It means $\sum_{n=1}^\inf} = \frac{7n^2-4n+3}{12+2n^6}$ also converges.

5 0

4 years ago

HAVE A NICE DAY <br> WILL GIVE BRAINLST :)

Damm [24]

Answer:

6. 40

7. 65

8. 50 and 80

Step-by-step explanation:

180-70-70=40

180-55-60=65

180-50-50=80 so that would have to be 50+50+80= 180

Becuase it all needs to add up to 180. :))

7 0

3 years ago