(x)^3(-x^3y)^2please help

Mathematics

1 answer:

julsineya [31]3 years ago

4 0

Answer:

$\large\boxed{(x^3)(-x^3y)^2=x^9y^2}$

Step-by-step explanation:

$(x^3)(-x^3y)^2\qquad\text{use}\ (ab)^n=a^nb^n\\\\=(x^3)(-x^3)^2(y^2)\qquad\text{use}\ (a^n)^m=a^{nm}\\\\=(x^3)(x^6)(y^2)\qquad\text{use}\ a^n\cdot a^m=a^{n+m}\\\\=x^{3+6}y^2\\\\=x^9y^2$

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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

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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 $N\in\mathbb N$ .