2 years ago

Adam is subtracting 708 – 10 mentally. He thinks

1 answer:

8 0

Adam is correct because when the hundreds and tens digit changes in 708, that means that the 7 and 0 will change into 6 and 9

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Montano1993 [528]

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UkoKoshka [18]

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Debora [2.8K]

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MAXImum [283]

We have

$\dfrac{x^2}{x+1}=\dfrac{(x+1)^2-2(x+1)+1}{x+1}=(x+1)-2+\dfrac1{x+1}=x-1+\dfrac1{x+1}$

$\displaystyle\lim_{x\to\infty}\left(\frac{x^2}{x+1}-ax-b\right)=\lim_{x\to\infty}\left(x-1+\frac1{x+1}-ax-b\right)=0$

The rational term vanishes as x gets arbitrarily large, so we can ignore that term, leaving us with

$\displaystyle\lim_{x\to\infty}\left((1-a)x-(1+b)\right)=0$

and this happens if a = 1 and b = -1.

To confirm, we have

$\displaystyle\lim_{x\to\infty}\left(\frac{x^2}{x+1}-x+1\right)=\lim_{x\to\infty}\frac{x^2-(x-1)(x+1)}{x+1}=\lim_{x\to\infty}\frac1{x+1}=0$

as required.

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frutty [35]

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5 bud i think it this but i might be wrong

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