John got 20% discount at a sale .He got a new shirt for $12 .What is the original price of the shirt ?

Mathematics

1 answer:

Arte-miy333 [17]3 years ago

3 0

The original price of the shirt must be $15 if a 20% discount resulted in the final price becoming $12

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Inessa [10]

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Determine the horizontal vertical and slant asymptote y=x^2+2x-3/x-7

lilavasa [31]

Answer:

<h2>A.Vertical:x=7</h2><h2>Slant:y=x+9</h2>

Step-by-step explanation:

$f(x)=\dfrac{x^2+2x-3}{x-7}\\\\vertical\ asymptote:\\\\x-7=0\qquad\text{add 7 to both sides}\\\\\boxed{x=7}\\\\horizontal\ asymptote:\\\\\lim\limits_{x\to\pm\infty}\dfrac{x^2+2x-3}{x-7}=\lim\limits_{x\to\pm\infty}\dfrac{x^2\left(1+\frac{2}{x}-\frac{3}{x^2}\right)}{x\left(1-\frac{7}{x}\right)}=\lim\limits_{x\to\pm\infty}\dfrac{x\left(1+\frac{2}{x}-\frac{3}{x^2}\right)}{1-\frac{7}{x}}=\pm\infty\\\\\boxed{not\ exist}$

$slant\ asymptote:\\\\y=ax+b\\\\a=\lim\limits_{x\to\pm\infty}\dfrac{f(x)}{x}\\\\b=\lim\limits_{x\to\pm\infty}(f(x)-ax)\\\\a=\lim\limits_{x\to\pm\infty}\dfrac{\frac{x^2+2x-3}{x-7}}{x}=\lim\limits_{x\to\pm\infty}\dfrac{x^2+2x-3}{x(x-7)}=\lim\limits_{x\to\pm\infty}\dfrac{x^2+2x-3}{x^2-7x}\\\\=\lim\limits_{x\to\pm\infty}\dfrac{x^2\left(1+\frac{2}{x}-\frac{3}{x^2}\right)}{x^2\left(1-\frac{7}{x}\right)}=\lim\limits_{x\to\pm\infty}\dfrac{1+\frac{2}{x}-\frac{3}{x^2}}{1-\frac{7}{x}}=\dfrac{1}{1}=1$

$b=\lim\limits_{x\to\pm\infty}\left(\dfrac{x^2+2x-3}{x-7}-1x\right)=\lim\limits_{x\to\pm\infty}\left(\dfrac{x^2+2x-3}{x-7}-\dfrac{x(x-7)}{x-7}\right)\\\\=\lim\limits_{x\to\pm\infty}\left(\dfrac{x^2+2x-3}{x-7}-\dfrac{x^2-7x}{x-7}\right)=\lim\limits_{x\to\pm\infty}\dfrac{x^2+2x-3-(x^2-7x)}{x-7}\\\\=\lim\limits_{x\to\pm\infty}\dfrac{x^2+2x-3-x^2+7x}{x-7}=\lim\limits_{x\to\pm\infty}\dfrac{9x-3}{x-7}=\lim\limits_{x\to\pm\infty}\dfrac{x\left(9-\frac{3}{x}\right)}{x\left(1-\frac{7}{x}\right)}$

$=\lim\limits_{x\to\pm\infty}\dfrac{9-\frac{3}{x}}{1-\frac{7}{x}}=\dfrac{9}{1}=9\\\\\boxed{y=1x+9}$

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