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3 years ago

This question is worth 100 points please answer now. Algebra 2. I will mark you brainliest.

Mathematics

1 answer:

anastassius [24]3 years ago

8 0

Answer:

The vertical asymptotes occur at x = 4 and x = -4

The zero of the function is (0, 0)

Step-by-step explanation:

The given function is $f(x) = \dfrac{4\cdot x}{x^2 - 16}$

Therefore, we have;

$f(x) = \dfrac{4\cdot x}{x^2 - 16} = \dfrac{4\cdot x}{(x - 4) \cdot (x + 4)}$

The asymptote is given at the value of x for which the denominator = 0, therefore, the asymptote are the values of x that make the denominator of the equation equal to zero, which are given as follows

For the asymptote, the denominator = x² - 16 = (x - 4)·(x + 4) = 0

Therefore, the vertical asymptotes are the lines x = 4 and x = -4.

The zero of the function is given as follows;

$f(x) = \dfrac{4\cdot x}{x^2 - 16} = 0$

Therefore, 4·x = (x - 4)·(x + 4) × 0 = 0

x = 0/4 = 0

Therefore, when f(x) = 0, x = 0 and the zero of the function = (0, 0).

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$\bf \begin{array}{cccccclllll}
\textit{something}&&\textit{varies directly to}&&\textit{something else}\\ \quad \\
\textit{something}&=&{{ \textit{some value}}}&\cdot &\textit{something else}\\ \quad \\
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and also

$\bf \begin{array}{llllll}
\textit{something}&&\textit{varies inversely to}&\textit{something else}\\ \quad \\
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\end{array}
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now, we know that V varies directly to T and inversely to P simultaneously
thus $\bf V=T\cdot \cfrac{k}{P}$

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\begin{cases}
V=42\\
T=84\\
P=8
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\\\\\\
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3 years ago

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Problems of normal distributions can be solved using the z-score formula.

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The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.

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