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2 years ago

What is the correct answer please

Mathematics

1 answer:

Verizon [17]2 years ago

7 0

Answer:

Step-by-step explanation:

$\frac{15y-10}{3x}$ = 6 ( multiply both sides by 3x to clear the fraction )

15y - 10 = 18x ( add 10 to both sides )

15y = 18x + 10 ( divide terms by 15 )

y = $\frac{18}{15}$ x + $\frac{10}{15}$ , that is

y = $\frac{6}{5}$ x + $\frac{2}{3}$ → C

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A polynomial asymptote is a function $p(x)$ such that

$\displaystyle\lim_{x\to\pm\infty}(f(x)-p(x))=0$

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Since this equation defines a hyperbola, we expect the asymptotes to be lines of the form $p(x)=ax+b$ .

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$\displaystyle\lim_{x\to\infty}(2x-1+2\sqrt{x^2-x}-4x-b)=0$

$\displaystyle\lim_{x\to\infty}(\sqrt{x^2-x}-x)=\frac{b+1}2$

We write

$(\sqrt{x^2-x}-x)\cdot\dfrac{\sqrt{x^2-x}+x}{\sqrt{x^2-x}+x}=\dfrac{(x^2-x)-x^2}{\sqrt{x^2-x}+x}=-\dfrac x{x\sqrt{1-\frac1x}+x}=-\dfrac1{\sqrt{1-\frac1x}+1}$

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So one asymptote of the hyperbola is the line $y=4x-2$ .

The other asymptote is obtained similarly by examining the limit as $x\to-\infty$ .

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

$\displaystyle\lim_{x\to-\infty}(2x-2x\sqrt{1-\frac1x}-ax-(b+1))=0$

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This time the limit is $\dfrac12$ , so

$\dfrac12=\dfrac{b+1}2\implies b=0$

which means the other asymptote is the line $y=0$ .

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Verizon [17]

Hundredth Place

Your Welcome.

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