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3 years ago

So that you get points answer this easy question 2 - 1

Mathematics

2 answers:

Nataly [62]3 years ago

8 0

Answer:

Step-by-step explanation:

krek1111 [17]3 years ago

7 0

Answer:

Step-by-step explanation:

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I need help with this problem!!! It’s confusing me so much and I need it by tomorrow!please tell me the answer and how you got i

ExtremeBDS [4]

I think its 3 - (0,10, 42)

if you put in the values of the given x, then these are the resulting y values.

8 0

3 years ago

What is y-value when x equals -12? y =-200 - 6(x)

Arada [10]

Hi,

Solving:

$x = - 12 \\ y = - 200 - 6( x) \\ y = - 200 - 6( - 12) \\ y = - 200 + 72 = \underline{ \underline{ - 128}}$

Answer: y = -128

Hope this helps.
If you have any questions about the answer, comment below and I'll reply.
r3t40

7 1

4 years ago

Please help me with these.. i am stuck

SCORPION-xisa [38]

Answer:

Both are irrational

12√3 = 20.784

9√6 = 22.045

Hopefully this is what you're looking for :)

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:

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

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

The graph of a system of equations with the same slope and the same y-intercepts will have no solutions.

Rasek [7]

If 2 equations have the same y-intercept, they are overlapping, which means they have infinite solutions. So there is no way that 2 equations with the same y-intercept will have no solution. Thus your answer is: C)Never.

4 0

4 years ago

Read 2 more answers