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spin [16.1K]
3 years ago
6

Create a list of steps, in order, that will solve the following equation.

Mathematics
2 answers:
labwork [276]3 years ago
8 0

Answer:

Add 1 to both sides

Take the square root of both sides

Subtract 3 from both sides

x = -9, 3

Step-by-step explanation:

(x + 3)² – 1 = 35

Add 1 to both sides

(x + 3)² = 36

Take the square root of both sides

(x + 3) = +/- 6

Subtract 3 from both sides

x = 6 - 3 = 3

x = -6 - 3 = -9

Arturiano [62]3 years ago
3 0

Answer:

- Add 1 to both sides

- Take the square root of both sides

- Subtract 3 from both sides

Step-by-step explanation:

Our equation here is: (x + 3)² - 1 = 35. We need to isolate the variable:

(x + 3)² - 1 = 35

First step is to add 1 to both sides:

(x + 3)² = 35 + 1 = 36

Next, take the square root of both sides:

x + 3 = √36 = 6

Finally, subtract 3 from both sides:

x = 6 - 3 = 3

The order of steps is:

- Add 1 to both sides

- Take the square root of both sides

- Subtract 3 from both sides

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Fantom [35]

Part 1: You can simplify a_n to

\dfrac{3n+1}{3n}-\dfrac{3n}{3n+1} = \dfrac1{3n}+\dfrac1{3n+1}

Presumably, the sequence starts at <em>n</em> = 1. It's easy to see that the sequence is strictly decreasing, since larger values of <em>n</em> make either fraction smaller.

(a) So, the sequence is bounded above by its first value,

|a_n| \le a_1 = \dfrac13+\dfrac14 = \boxed{\dfrac7{12}}

(b) And because both fractions in a_n converge to 0, while remaining positive for any natural number <em>n</em>, the sequence is bounded below by 0,

|a_n| \ge \boxed{0}

(c) Finally, a_n is bounded above and below, so it is a bounded sequence.

Part 2: Yes, a_n is monotonic and strictly decreasing.

Part 3:

(a) I assume the choices are between convergent and divergent. Any monotonic and bounded sequence is convergent.

(b) Since a_n is decreasing and bounded below by 0, its limit as <em>n</em> goes to infinity is 0.

Part 4:

(a) We have

\displaystyle \lim_{n\to\infty} \frac{10n^2+1}{n^2+n} = \lim_{n\to\infty}10+\frac1{n^2}}{1+\frac1n} = 10

and the (-1)ⁿ makes this limit alternate between -10 and 10. So the sequence is bounded but clearly not monotonic, and hence divergent.

(b) Taking the limit gives

\displaystyle\lim_{n\to\infty}\frac{10n^3+1}{n^2+n} = \lim_{n\to\infty}\frac{10+\frac1{n^3}}{\frac1n+\frac1{n^2}} = \infty

so the sequence is unbounded and divergent. It should also be easy to see or establish that the sequence is strictly increasing and thus monotonic.

For the next three, I'm guessing the options here are something to the effect of "does", "may", or "does not".

(c) may : the sequence in (a) demonstrates that a bounded sequence need not converge

(d) does not : a monotonic sequence has to be bounded in order to converge, otherwise it grows to ± infinity.

(e) does : this is true and is known as the monotone convergence theorem.

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3 years ago
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2 years ago
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