Match the graph with its equation.

In the sport competition, France won more gold medals than Italy, who won more gold medals than Korea. If the total number of

MA_775_DIABLO [31]

Answer:

Korea = 15

Italy = 16

France = 17

Step-by-step explanation:

Let :

Korea = x

Italy = x + 1

France = x + 2

(since the values are consecutive)

x + x + 1 + x + 2 = 48

3x + 3 = 48

3x = 48 - 3

3x = 45

x = 45 / 3

x = 15

Korea = x = 15

Italy = x + 1 = 16

France = x + 2 = 17

4 0

3 years ago

Does anyone know what goes into these blanks

avanturin [10]

yes I know but will not give answer because you have not say that

4 0

3 years ago

You roll a number cube with 6 sides two times. What is the probability of rolling a number greater than 2 and an odd number?

svetoff [14.1K]

Answer:

lol you've asked like 10 probability questions in 20 minutes i wonder what chapter you're studying

Step-by-step explanation:

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:

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