Simplify. -14x^3/ x^3-5x^4 A) -14/5x-1; where x is not equal to 1/5,0 B) -14x/1-5x; where x is not equal to 1/5 C) 1-5x/-14x; wh

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swat32

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Simplify. -14x^3/ x^3-5x^4 A) -14/5x-1; where x is not equal to 1/5,0 B) -14x/1-5x; where x is not equal to 1/5 C) 1-5x/-14x; wh

ere x is not equal to 0 D) -14/1-5x; where x is not equal to 1/5, 0

Mathematics

1 answer:

Karolina [17] · Answer 1 · 2022-06-28T19:36:47+03:00

It's D. Factor x^3 out of the denominator leaving 1 - 5x, then cancel out the x^3 in the numerator with the one you factored out of the denominator. That leaves you with 1 - 5x cannot equal 0. Solve that for x and you get 1/5. However, since we are talking vertical asymptotes and the x^3 is a removable discontinuity, we have to count it as a "problem". Therefore, the other value of x that is not allowed is x = 0.