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Arturiano [62]
3 years ago
10

Please help me please!!!!!!!!!!​

Mathematics
1 answer:
Elodia [21]3 years ago
8 0

Answer:

shoot, sorry but I don't know

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saul85 [17]

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Step-by-step explanation:

jack is correct

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If lim x-> infinity ((x^2)/(x+1)-ax-b)=0 find the value of a and b
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}

So

\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 <em>x</em> 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 <em>a</em> = 1 and <em>b</em> = -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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Triangle ABC is located at A(1, 1), B(8, 1) and C(1, 5). After being reflected over the y-axis, what are the new coordinates of
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