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ycow [4]
2 years ago
15

What is the additive inverse of 1/3

Mathematics
1 answer:
VLD [36.1K]2 years ago
6 0

Answer:

The additive inverse of 1/3 is -1/3

Step-by-step explanation:

Since, 1/3+(-1/3) = 0

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Determine+the+percentile+indicating+the+bottom+ranking+20%.+in+other+words,+what+is+the+dollar+number+that+defines+where+the+low
Aloiza [94]

The dollar number that defines where the lowest values falls is

20% percentile value =$20

This is further explained below.

<h3 /><h3>What is the dollar number that defines where the lowest value falls?</h3>

Generally,  Now we need to determine the 20th percentile ($20). It indicates that we need to locate a dollar amount that marks the point at which the 20 % data has a value lower than that number.

i=(p/100)*n

Where i is the position of p^th

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Therefore, 20% percentile value =$20

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Find the maclaurin series for f(x) using the definition of a maclaurin series. [assume that f has a power series expansion. do n
Nady [450]

The equation of f(x) = e^{-6x} by maclaurin series is f(x)=\sum_{i=0}^{\infty} \frac{(-6.x)^{i}  }{i!}.

The maclaurin series for f(x) is defined by the following formula:

f(x) = \sum_{i=0}^{\infty} \frac{f^{(i)} (0)}{i!} .x^{i}--------------(1)

Where f^{i} is the i - th derivative of the function

If f(x) = e^{-6x}, then the formula of the i - th derivative of the function is:

f^{i} =(-6)^{i} .e^{-6x}----------------------(2)

Then,

f^{i}(0) = (-6)^{i}

Lastly, the equation of the trascendental function by Maclaurin series is: f(x)=\sum_{i=0}^{\infty} \frac{(-6)^{i}.x^{i}  }{i!} \\f(x)=\sum_{i=0}^{\infty} \frac{(-6.x)^{i}  }{i!}---------------(3)

Hence,

The equation of f(x) = e^{-6x} by maclaurin series is f(x)=\sum_{i=0}^{\infty} \frac{(-6.x)^{i}  }{i!}.

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1 year ago
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