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Jet001 [13]
2 years ago
11

suppose that $\star(n) = \left\{n - 2, n + 2, 2n, \frac {n}{2}\right\}$. for example, $\star(6) = \{3,4,8,12\}$. for how many di

stinct integers $n$ does $\star(n)$ have exactly three distinct elements?
SAT
1 answer:
kondaur [170]2 years ago
3 0

Answer:

dont know sorry

Explanation:

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mylen [45]

Answer:

abrazar, ik because i know spanish

5 0
3 years ago
Okay 480...... here is the next one
kozerog [31]
B) Texture
hope this helps x
would love brainliest if it’s right and you can give it! :)
8 0
3 years ago
Linear algebra with applications 8th edition solutions
nekit [7.7K]

Answer:

That makes 21

Whats your question?

What do you need help on?

I took algebra a while ago so I may be able to help you.

Explanation:

8 0
2 years ago
How did Hongwu follow the teachings of Confucius? by building chariots of porcelain by encouraging exploration beyond China by m
enot [183]

Answer:

The correct answer is D) by making sure that everyone could earn a living.

Explanation:

Hongwu, was the founder of the Ming Dynasty. He decided to make so reformations in the government of China that centralized much of decisions on him. He believed in the teachings of Confucious and the traditional rituals he taught. He followed the Mandate of Heaven and controlled every aspect of the Empire.

4 0
3 years ago
Read 2 more answers
The functions $f$ and $g$ are defined as follows: \[f(x) = \sqrt{\dfrac{x+1}{x-1}}\quad\text{and}\quad g(x) = \dfrac{\sqrt{x+1}}
AlladinOne [14]

Recall that √x has a domain of x ≥ 0.

So, f(x) is defined as long as

(x + 1)/(x - 1) ≥ 0

• We have equality when x = -1

• Otherwise (x + 1)/(x - 1) is positive if both x + 1 and x - 1 are positive, or both are negative:

\begin{cases}x+1>0 \implies x>-1 \\ x-1>0 \implies x>1\end{cases} \implies x > 1

\begin{cases}x+1

Then the domain of f(x) is

x > 1   or   x ≤ -1

On the other hand, g(x) is defined by two individual square root expressions with respective domains of

• x + 1 ≥ 0   ⇒   x ≥ -1

• x - 1 ≥ 0   ⇒   x ≥ 1

but note that g(1) is undefined, so we omit it from the second domain.

Then g(x) is defined so long as both x ≥ -1 *and* x > 1 are satisfied, which means its domain is

x > 1

f(x) and g(x) have different domains, so they are not the same function.

4 0
2 years ago
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