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pentagon [3]
3 years ago
9

What is the proper mathematical definition of "square root?"

Mathematics
1 answer:
pogonyaev3 years ago
6 0
The square root<span> of a number is a value that, when multiplied by itself, gives the number.</span><span> </span>
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Solve the problem<br> Show your work<br> 16.007<br> - 0.55<br> ----------
Roman55 [17]

Answer:

16.007 - 0.55 is 15.457

5 0
2 years ago
Read 2 more answers
What is an equation of the line that is perpendicular to -x+2y=4 and passes through the point (-2, 1)?
KATRIN_1 [288]
First: Slope of -x+2y=42y=x+4y=x/2+2So, m=1/2.
Second: Slope  of the perpendicular line: mp=-2.
Third: Find the line with slope -2 and passes through the point (−2, 1)y-y1=m(x-x1)y-1=-2(x+2)y=-2x-4+1y=-2x-3 

Read more on Brainly.com - brainly.com/question/7942650#readmore

5 0
2 years ago
To estimate the annual salary for a given hourly pay rate, multiply by 2 and insert “000” at the end.
qaws [65]

Answer:20000

Step-by-step explanation:

7 0
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}

7 0
3 years ago
A die is rolled. Which is the probability of rolling an even number or a number greater than 4?
9966 [12]

Answer:

yy

Step-by-step explanation:iii

6 0
2 years ago
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