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Mathematics

2 answers:

lana [24]3 years ago

6 0

The anwser is d -1, 4

PilotLPTM [1.2K]3 years ago

6 0

Step 1: Find one term from both equations on the left side of the equality sign that, when added together, equal 0 (thus canceling each other out.) The two terms that fit this description is -y and y. Add them together we get 0.

Step 2: Add the remaining like terms from both equations.
3x + x = 4x
-5 + 1 = -4
4x = -4

Step 3: Divide both sides by 4 to isolate the variable x.
4/4x = -4/4
x = -1

Step 4: We have x, now we need to find y. Take an original equation (either one is fine) and substitute -1 for x.
3(-1)⇒ -3 + y = 1

Step 5: Isolate 'y' by doing inverse operations to both sides of the equation.
-3 + 3 + y = 1 + 3 ⇒ y = 4

Answer: (-1,4)

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jarptica [38.1K]

Answer:

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5 0

2 years ago

May someone please help me?

Vesna [10]

Answer:

The 3rd table

Step-by-step explanation:

The x values are dividing by 5 to get the y values. The table is proportional because it's a pattern not just random values.

5 0

2 years ago

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:

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