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Mathematics

2 answers:

wariber [46]2 years ago

3 0

25/6 as improper
4 1/6 as whole

ANTONII [103]2 years ago

3 0

Answer:

in decimal form it is 25/6 but if you need a fraction form it is 4.1667

Step-by-step explanation:

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$\large\boxed{\dfrac{x+2}{x^2-6x-16}\div\dfrac{1}{9x}=\dfrac{9x}{x-8}}$

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$\dfrac{x+2}{x^2-6x-16}\div\dfrac{1}{9x}=\dfrac{x+2}{x^2+2x-8x-16}\cdot\dfrac{9x}{1}\\\\=\dfrac{(x+2)(9x)}{x(x+2)-8(x+2)}=\dfrac{(x+2)(9x)}{(x+2)(x-8)}\\\\\text{cancel}\ (x+2)\\\\=\dfrac{9x}{x-8}$

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Find the number to which the sequence {(3n+1)/(2n-1)} converges and prove that your answer is correct using the epsilon-N defini

Nat2105 [25]

By inspection, it's clear that the sequence must converge to $\dfrac32$ because

$\dfrac{3n+1}{2n-1}=\dfrac{3+\frac1n}{2-\frac1n}\approx\dfrac32$

when $n$ is arbitrarily large.

Now, for the limit as $n\to\infty$ to be equal to $\dfrac32$ is to say that for any $\varepsilon>0$ , there exists some $N$ such that whenever $n>N$ , it follows that

$\left|\dfrac{3n+1}{2n-1}-\dfrac32\right|$

From this inequality, we get

$\left|\dfrac{3n+1}{2n-1}-\dfrac32\right|=\left|\dfrac{(6n+2)-(6n-3)}{2(2n-1)}\right|=\dfrac52\dfrac1{|2n-1|}$
$\implies|2n-1|>\dfrac5{2\varepsilon}$
$\implies2n-1\dfrac5{2\varepsilon}$
$\implies n\dfrac12+\dfrac5{4\varepsilon}$

As we're considering $n\to\infty$ , we can omit the first inequality.

We can then see that choosing $N=\left\lceil\dfrac12+\dfrac5{4\varepsilon}\right\rceil$ will guarantee the condition for the limit to exist. We take the ceiling (least integer larger than the given bound) just so that $N\in\mathbb N$ .