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3 years ago

What is the decimal expansion of 7/2

Mathematics

2 answers:

Verdich [7]3 years ago

6 0

Wait sorry, no I answered wrong. It is not 3.5, look at above answer.

valentina_108 [34]3 years ago

5 0

Answer:

0.583

Step-by-step explanation:

you just divide ig

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Is y2=x-1 a function i will give brainliest

adelina 88 [10]

I don’t think it issss

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2 years ago

In the accompanying diagram, PA is tangent to the circle at A and PC is a secant. If

Ganezh [65]

Answer:

c. 12

Step-by-step explanation:

PA² = (BC + PB) × PB => tangent-secant theorem

PA = 8

PB = 4

BC = ?

Substitute

8² = (BC + 4) × 4

64 = 4BC + 16

64 - 16 = 4BC + 16 - 16

48 = 4BC

48/4 = 4BC/4

12 = BC

BC = 12

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2 years ago

How do you explain expanded notation method for 82 times 3

BabaBlast [244]

The answer in expanded form?

4 0

3 years ago

Read 2 more answers

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

3 years ago

2 Which expression is equivalent to 64-*?

Rasek [7]

Answer:

Step-by-step explanation:

(8-x)(8+x)

8×8=64

-x·+x=-x^2

64-x^2

3 0

3 years ago