You have to get the x to one side so you'll subtract 3x from -6x and 3x from 3x, the 3x will cancel out and -6x - 3x is -9x-7=11, you have to add 7 from both sides the 7+7 cancels out and also add 11+7 which leaves -9x=18, you have to divide -9x/-9 and do 18/-9, which gets you x=-2. Here's the work shown if you didn't what i said...
-6x-7=3x+11
-3x -3x
-9x-7=11
+7 +7
-9x=18
-9 -9
x=-2

6 0

2 years ago

Determine if the following infinite series converges or diverges

Mandarinka [93]

Using limits, it is found that the infinite sequence converges, as the limit does not go to infinity.

<h3>How do we verify if a sequence converges of diverges?</h3>

Suppose an infinity sequence defined by:

$\sum_{k = 0}^{\infty} f(k)$

Then we have to calculate the following limit:

$\lim_{k \rightarrow \infty} f(k)$

If the <u>limit goes to infinity</u>, the sequence diverges, otherwise it converges.

In this problem, the function that defines the sequence is:

$f(k) = \frac{k^3}{k^4 + 10}$

Hence the limit is:

$\lim_{k \rightarrow \infty} f(k) = \lim_{k \rightarrow \infty} \frac{k^3}{k^4 + 10} = \lim_{k \rightarrow \infty} \frac{k^3}{k^4} = \lim_{k \rightarrow \infty} \frac{1}{k} = \frac{1}{\infty} = 0$