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steposvetlana [31]
2 years ago
13

12 things that has energy

Mathematics
2 answers:
Lapatulllka [165]2 years ago
8 0

Answer: Kinetic Energy Potential Energy Chemical Energy Elastic Energy Gravitational Energy Magnetic Energy Nuclear Energy  Light Energy Radiant Energy Sound Energy Thermal Energy  Mechanical Energy Electrical Energy

Step-by-step explanation:

kap26 [50]2 years ago
3 0

Answer: what he or she said :)

Step-by-step explanation:

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(a)  See below.

(b)  x = 0 or x = 1

(c)  x = 0 removable, x = 1 non-removable

Step-by-step explanation:

Given rational function:

f(x)=\dfrac{\ln |x-1|}{x}

<u>Part (a)</u>

Substitute x = 2 into the given rational function:

\begin{aligned}\implies f(2) & =\dfrac{\ln |2-1|}{2}\\\\ & =\dfrac{\ln 1}{2}\\\\ & =\dfrac{0}{2}\\\\ & =0\end{aligned}

Therefore, as the function is defined at x = 2, the function is continuous at x = 2.

<u>Part (b)</u>

Given interval:  [-2, 2]

Logs of negative numbers or zero are undefined.  As the numerator is the natural log of an <u>absolute value</u>, the numerator is undefined when:

|x - 1| = 0 ⇒ x = 1.  

A rational function is undefined when the denominator is equal to zero, so the function f(x) is undefined when x = 0.

So the function is discontinuous at x = 0 or x = 1 on the interval [-2, 2].

<u>Part (c)</u>

x = 1 is a <u>vertical asymptote</u>.  As the function exists on both sides of this vertical asymptote, it is an <u>infinite discontinuity</u>.  Since the function doesn't approach a particular finite value, the limit does not exist.  Therefore, x = 1 is a non-removable discontinuity.

A <u>hole</u> exists on the graph of a rational function at any input value that causes both the numerator and denominator of the function to be equal.

\implies f(0)=\dfrac{\ln |0-1|}{0}=\dfrac{\ln 1}{0}=\dfrac{0}{0}

Therefore, there is a hole at x = 0.

The removable discontinuity of a function occurs at a point where the graph of a function has a hole in it.   Therefore, x = 0 is a removable discontinuity.

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