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Mathematics

1 answer:

katen-ka-za [31]3 years ago

7 0

Answer:

the answer is a, b, c ( if u can choose more than one)

Step-by-step explanation:

just by looking at the triangle u can tell it has equal sides, so every side is 40 if one is, and by testing 8x8-24 that equald to 40, so they are all correct except for the last one which goes against my everything is equal claim

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Solve this identifying holes, vertical asymptotes, and horizontal asymptotes for

USPshnik [31]

$x^2+7x+12=x^2+4x+3x+12=x(x+4)+3(x+4)=(x+4)(x+3)\\\\-x^2-3x+4=-(x^2+3x-4)=-(x^2+4x-x-4)\\\\=-[x(x+4)-1(x+4)]=-(x+4)(x-1)\\\\f(x)=\dfrac{x^2+7x+12}{-x^2-3x+4}=\dfrac{(x+4)(x+3)}{-(x+4)(x-1)}\\\\\text{Vertical asymptotes:}\\\\(x+4)(x-1)=0\iff x+4=0\ \vee\ x-1=0\\\\\boxed{x=-4\ and\ x=1}\\\\\text{Horizontal asymptotes:}$

$\lim\limits_{x\to\pm\infty}\dfrac{x^2+7x+12}{-x^2-3x+4}=\lim\limits_{x\to\pm\infty}\dfrac{x^2\left(1+\dfrac{7}{x}+\dfrac{12}{x^2}\right)}{x^2\left(-1-\dfrac{3}{x}+\dfrac{4}{x^2}\right)}=\dfrac{1}{-1}=-1\\\\\boxed{y=-1}$