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Vikentia [17]
3 years ago
9

MAHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHH

Mathematics
1 answer:
Natali [406]3 years ago
4 0
I believe it is neither

For one if you plug in either of them the first one is going to be negative 3. But if you use the second one then your answer would be -3+12= -10 but -3+12 is 9 so I would go with neither because I don't see how it's both
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Lady bird [3.3K]
The answer is 28. 7x2=14 and then 14x2=28. Remember PEMDAS
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Help on this math question ‍♀️
noname [10]

Answer:

the 2nd 3rd and 4th answers are the correct ones

Step-by-step explanation:

hope this helps you

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2 years ago
Rashawn read 25 pages of his book each day until he finished the book. His book was 400 pages long.
Anna71 [15]
I think it's the second one.
7 0
3 years ago
Read 2 more answers
I need help asapp!!!!!!!
labwork [276]

Answer:

Step-by-step explanation:

As per midsegment theorem of a trapezoid,

Segment joining the midpoints of the legs of the of the trapezoid is parallel to the bases and measure half of their sum.

Length of midsegment = \frac{1}{2}(b_1+b_2)

3). MN = \frac{1}{2}(18+10)

           = 14

4). MN = \frac{1}{2}(57+76)

           = 66.5

5). MN = \frac{1}{2}(AB+DC)

    7 = \frac{1}{2}(AB+10)

    14 = AB + 10

    AB = 14 - 10

    AB = 4

6). 15 = \frac{1}{2}[(3x+2)+(2x-2)]

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4 0
2 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}

7 0
3 years ago
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