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vladimir2022 [97]
3 years ago
10

You use an activity book to solve a puzzle. The book directs you to specific pages within it. The puzzle’s start page directs yo

u to page 20. On page 20, you are given one of the alphabets of the answer, and the page number indicating the next page and letter. You continue flipping to each page until you have all the letters that form the answer. Finally, you arrange these letters to form a meaningful word. What kind of challenge has been described here? resource management spatial awareness pattern recognition and matching reaction time
Mathematics
1 answer:
Tomtit [17]3 years ago
5 0

pattern recognition and matching

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100 Points! What are the classifications of each system?
alexdok [17]

Answer:

Part 1) Is a inconsistent system

Part 2) Is a consistent independent system

Part 3) Is a consistent dependent system

see the attached figure

Step-by-step explanation:

we know that

If a system has at least one solution, it is said to be consistent .  

If a consistent system has exactly one solution, it is independent .  

If a consistent system has an infinite number of solutions, it is dependent

If a system not have solution, it is said to be inconsistent

Part 1) we have

x+5y=-2\\x+5y=4

This system of linear equations has two parallel lines (their slopes are equal) with different y-intercept

so

The lines don't intersect

The system has no solution

therefore

Is a inconsistent system

Part 2) we have

y=3x+4\\-2x+y=4

we know that

If two lines have different slopes, then the lines intersect at one point and the system of equations has one solution

In this problem, the lines have different slopes (m=3 and m=2)

so

The lines intersect at one point

The system has one solution

therefore

Is a consistent independent system

Part 3) we have

3x+y=4 ----> equation A

-6x-2y=-8 ----> equation B

Multiply equation A by -2 both sides

-2(3x+y)=-2(4)

-6x-2y=-8 ----> equation C

Compare equation B and equation C

Are identical lines

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3 years ago
Lim (n/3n-1)^(n-1)<br> n<br> →<br> ∞
n200080 [17]

Looks like the given limit is

\displaystyle \lim_{n\to\infty} \left(\frac n{3n-1}\right)^{n-1}

With some simple algebra, we can rewrite

\dfrac n{3n-1} = \dfrac13 \cdot \dfrac n{n-9} = \dfrac13 \cdot \dfrac{(n-9)+9}{n-9} = \dfrac13 \cdot \left(1 + \dfrac9{n-9}\right)

then distribute the limit over the product,

\displaystyle \lim_{n\to\infty} \left(\frac n{3n-1}\right)^{n-1} = \lim_{n\to\infty}\left(\dfrac13\right)^{n-1} \cdot \lim_{n\to\infty}\left(1+\dfrac9{n-9}\right)^{n-1}

The first limit is 0, since 1/3ⁿ is a positive, decreasing sequence. But before claiming the overall limit is also 0, we need to show that the second limit is also finite.

For the second limit, recall the definition of the constant, <em>e</em> :

\displaystyle e = \lim_{n\to\infty} \left(1+\frac1n\right)^n

To make our limit resemble this one more closely, make a substitution; replace 9/(<em>n</em> - 9) with 1/<em>m</em>, so that

\dfrac{9}{n-9} = \dfrac1m \implies 9m = n-9 \implies 9m+8 = n-1

From the relation 9<em>m</em> = <em>n</em> - 9, we see that <em>m</em> also approaches infinity as <em>n</em> approaches infinity. So, the second limit is rewritten as

\displaystyle\lim_{n\to\infty}\left(1+\dfrac9{n-9}\right)^{n-1} = \lim_{m\to\infty}\left(1+\dfrac1m\right)^{9m+8}

Now we apply some more properties of multiplication and limits:

\displaystyle \lim_{m\to\infty}\left(1+\dfrac1m\right)^{9m+8} = \lim_{m\to\infty}\left(1+\dfrac1m\right)^{9m} \cdot \lim_{m\to\infty}\left(1+\dfrac1m\right)^8 \\\\ = \lim_{m\to\infty}\left(\left(1+\dfrac1m\right)^m\right)^9 \cdot \left(\lim_{m\to\infty}\left(1+\dfrac1m\right)\right)^8 \\\\ = \left(\lim_{m\to\infty}\left(1+\dfrac1m\right)^m\right)^9 \cdot \left(\lim_{m\to\infty}\left(1+\dfrac1m\right)\right)^8 \\\\ = e^9 \cdot 1^8 = e^9

So, the overall limit is indeed 0:

\displaystyle \lim_{n\to\infty} \left(\frac n{3n-1}\right)^{n-1} = \underbrace{\lim_{n\to\infty}\left(\dfrac13\right)^{n-1}}_0 \cdot \underbrace{\lim_{n\to\infty}\left(1+\dfrac9{n-9}\right)^{n-1}}_{e^9} = \boxed{0}

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