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2 years ago

Help please.......Write a system of equations to describe the situation below, solve using any method, and fill in the blanks.

Mathematics

1 answer:

maxonik [38]2 years ago

4 0

Answer:

The cost of 1 dozen doughnut is $6 and the cost of a dozen croissant is $10.

Step-by-step explanation:

We know that:

D = Cost of a dozen doughnut
C = Cost of a dozen croissant
10D + 7C = 130
6D + 4C = 76

Work:

10D + 7C = 130

6D + 4C = 76

4(10D + 7C = 130)

7(6D + 4C = 76)

40D + 28C = 520

42D + 28C = 532

40D - 42D = 520 - 532
=> -2D = -12
=> 2D = 12
=> D = 6

Now, let's substitute the value of D into the first equation.

10D + 7C = 130
=> 10(6) + 7C = 130
=> 60 + 7C = 130
=> 7C = 70
=> C = 10

Hence, the cost of 1 dozen doughnut is $6 and the cost of a dozen croissant is $10.

Hoped this helped.

$BrainiacUser1357$

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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.

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