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2 years ago

Identify whether each value of x is a discontinuity of the function by typing asymptote, hole, or neither. 5x/ x3 + 5x2 + 6x

Mathematics

2 answers:

qwelly [4]2 years ago

6 0

Answer:

Function is discontinuous at

x=0 (Hole)

x=-3 (Vertical asymptote)

x=-2 (Vertical asymptote)

Step-by-step explanation:

Given: $f(x)=\dfrac{5x}{x^3+5x^2+6x}$

We need to identity the discontinuity of the function. As we know function is discontinuous where it is not defined.

So, The function is discontinuous at hole, asymptote and break point.

$f(x)=\dfrac{5x}{x^3+5x^2+6x}$

$f(x)=\dfrac{5x}{x(x^2+5x+6)}$

$f(x)=\dfrac{5x}{x(x+3)(x+2)}$

For hole, we will cancel like factor from numerator and denominator.

At x=0 we get hole.

For vertical asymptote, we set denominator to 0

x+3=0 and x+2=0

Vertical asymptote:

x=-3 and x=-2

Function is discontinuous at

x=0 (Hole)

x=-3 (Vertical asymptote)

x=-2 (Vertical asymptote)

tatiyna2 years ago

3 0

Answer: hole at x = 0, asymptotes at x = -2 and x = -3

Step-by-step explanation:

$\frac{5x}{x^{3} + 5x^{2} + 6x}$

= $\frac{5x}{x(x^{2}+5x + 6)}$

= $\frac{5x}{x(x+2)(x+3)}$

x ≠ 0 ---> the x cancels out so this is a hole

x + 2 ≠ 0 ---> x ≠ -2 ---> asymptote at x = -2

x + 3 ≠ 0 ---> x ≠ -3 ---> asymptote at x = -3

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shutvik [7]

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3 years ago

An apartment complex rents an average of 2.3 new units per week. If the number of apartment rented each week Poisson distributed

masya89 [10]

Answer:

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Step-by-step explanation:

Given

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Average rent in a week = 2.3

Required

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Where y represents λ (average)

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$P(X = 0) = \frac{2.3^0 * e^{-2.3}}{0!}$

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$P(X = 0) = 0.10025884372$

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$P(X = 1) = \frac{2.3^1 * e^{-2.3}}{1!}$

$P(X = 1) = \frac{2.3 * e^{-2.3}}{1}$

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