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2 years ago

Can anyone see now? I really need help

Mathematics

2 answers:

butalik [34]2 years ago

6 0

3. Yes the cost of each item is proportional because in each pack each t-shirt cost $2.33. You can check that by dividing the amount of money each pack cost by the amount of shirts in each pack. For example the pack of 3 is $7, so 7/3 = $2.33. The pack of 5 cost $11.65, so 11.65/5 = $2.33. And the pack of 7 cost $16.31, so 16.31/7 = $2.33

4) Yes, it is proportional because each x is being squared or multiplied by it’s self.

laila [671]2 years ago

5 0

By this comment I mean I am a liar and you don’t need me me a favor to you you don’t want me me you wanna fight sonic or something you don’t know how

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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

3 years ago

If the original quantity is 10 and the new quantity is 6, what is percent increase

Ad libitum [116K]

Answer:

Using the percentage difference formula you can find the percentage decrease.

SO its

100(new-old/new)

So we solve

10-6=4

4/10= 2/5

2/5*100=200/5

40 percent decrease

3 0

2 years ago

Question 5 (1 point) What is the mean of this set of data: 78, 79, 80, 80, 89, 89, 92, 93, 94, 95, 95, 95 Round your answer to t

Len [333]

88.3

add all the numbers, you get 1,059 and divide that by how many numbers there are which is 12 and you get 88.3

7 0

2 years ago

Find the difference of (t-8)-(t-15)

Vsevolod [243]

I think u should try (t-23) because -8 minus - 15 is = to 23 so yeah add the t in front of it

5 0

3 years ago

PLEASE HELP!!!!!!!! ASAP

larisa86 [58]

I think the answer to this question is c

8 0

2 years ago