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2 years ago

Im still confused on how to use the distance formula

Mathematics

1 answer:

notsponge [240]2 years ago

5 0

Answer:

The distance formula is a formula that is used to find the distance between two points. These points can be in any dimension.

The formula for it is:

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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

(60 points)

natka813 [3]

Answer:

Step-by-step explanation:

$2.\\(a-b)^2=a^2-2ab+b^2\\\\x^2-16x=-8\\\\x^2-2(x)(8)=-8\\\\\text{We have}\ 2ab=2(x)(8).\ \text{Therefore}\ b=8.\\\\x^2-2(x)(8)=-8\qquad\text{add}\ 8^2=64\ \text{to both sides}\\\\x^2-2(x)(8)+8^2=-8+64\\\\(x-8)^2=56$

$5.\\2x^2+x-1=2\qquad\text{subtract 2 from both sides}\\\\2x^2+x-3=0\\\\2x^2+3x-2x-3=0\\\\x(2x+3)-1(2x+3)=0\\\\(2x+3)(x-1)=0\iff 2x+3=0\ \vee\ x-1=0\\\\2x+3=0\qquad\text{subtract 3 from both sides}\\2x=-3\qquad\text{divide both sides by 2}\\\boxed{x=-\dfrac{3}{2}}\\\\x-1=0\qquad\text{add 1 to both sides}\\\boxed{x=1}$

$6.\\2x^2-4x=0\qquad\text{divide both sides by 2}\\\\x^2-2x=0\\\\x(x-2)=0\iff x=0\ \vee\ x-2=0\\\\\boxed{x=0}\\\\x-2=0\qquad\text{add 2 to both sides}\\\boxed{x=2}$

8 0

3 years ago

What is the slope in the table below

eduard

Answer:

<h2>The table does not show a linear function</h2>

Step-by-step explanation:

$\text{The formula of a slope:}\\\\m=\dfrac{y_2-y_1}{x_2-x_1}\\\\\text{From the table we have:}\\\\(5,\ 24),\ (10,\ 21),\ (15,\ 17),\ (20,\ 13)\\\\\text{Check}\\\\\dfrac{21-24}{10-5}=\dfrac{-3}{5}=-\dfrac{3}{5}\\\\\dfrac{17-21}{15-10}=\dfrac{-4}{5}=-\dfrac{4}{5}\\\\-\dfrac{3}{5}\neq-\dfrac{4}{5}$