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2 years ago

13

Find the value of x ?

1 answer:

Digiron [165]2 years ago

3 0

Answer: 60 please give brainliest

Step-by-step explanation:

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3 years ago

Which function does not have a vertical asymptote? A) y=(x) /(1-x²) . B) y=(5x) /(1-2x²) . C) y=(5x-1) /(3+x^2) . D) (5x) /(x+x²

uysha [10]

A function has a vertical asymptote $x=a$ at point a, where the denominator becomes equal to 0.

A. The denominator of the function $f(x)=\dfrac{x}{1-x^2}$ turns into 0 at $x=1$ or $x=-1.$ Then $x=1$ and $x=-1$ are two vertical asymptotes of this function.

B. The denominator of the function $f(x)=\dfrac{5x}{1-2x^2}$ turns into 0 at $x=\sqrt{\frac{1}{2}}$ or $x=-\sqrt{\frac{1}{2}}$ Then $x=\sqrt{\frac{1}{2}}$ and $x=-\sqrt{\frac{1}{2}}$ are two vertical asymptotes of this function.

C. The denominator of the function $f(x)=\dfrac{5x-1}{3+x^2}$ never turns into 0, then this function hasn't any asymptotes.

D. The denominator of the function $f(x)=\dfrac{x}{x+x^2}$ turns into 0 at $x=0.$ Then $x=0$ is vertical asymptote of this function.

Answer: correct choice is C.

5 0

3 years ago