4.2 tons = ? pounds

Which system of linear inequalities is represented by the graph? ax – 3y > 6 and y > 2x + 4 bx + 3y > 6 and y > 2x

igor_vitrenko [27]

The <em><u>correct answer</u></em> is:

bx + 3y > 6 and y > 2x + 4

Explanation:

Looking at the second inequality, the y-intercept is 4 and the slope is 2. This means the graph of the line crosses the y-axis at (0, 4) and the line goes up 2 and over 1. Since it is greater than, this means the graph is shaded above it. Comparing this to the graph, the line for the blue part crosses the y-axis at (0, 4) and goes up 2 and over 1. The graph is also shaded above the line.

For the first inequality, bx+3y > 6, we want to isolate y. To do this, we subtract bx from each side:

bx+3y-bx > 6-bx

3y > 6-bx

Divide both sides by 3:

3y/3 > 6/3 - bx/3

y > 2 - (b/3)x

This means the line for this will have a y-intercept of 2 and decrease 1 while going over 3. The orange section does this. Additionally, since it is greater than, the graph should be shaded above the line. This one is, so this is the correct answer.

3 0

3 years ago

Read 2 more answers

Preston and horation have 49 video games total. Horation has 11 more than preston. How many games does each person have?

Makovka662 [10]

Preston has 19 games and Horation has 30.

4 0

3 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:

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