Holly and her two friends went to the movies and total Holly paid 24$ for 3 tickets. how much did one ticket cost

Assume that AGHSALMN Which of the following congruence statements

gulaghasi [49]

Answer: O A MEZH

OBL31

O CIG=LM

E UNE

FBSMM what is that supposed to mean?

Step-by-step explanation:

5 0

3 years ago

On a number line, what is the distance between -1.75 and 2.5 help meerr

ivanzaharov [21]

Answer:

4.25

Step-by-step explanation:

-1.75+x=2.5

x=4.25

5 0

3 years ago

Pls help this is pretty urgent

laiz [17]

Answer:

(a) See below.

(b) x = 0 or x = 1

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

Step-by-step explanation:

Given rational function:

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

Part (a)

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.

Part (b)

Given interval: [-2, 2]

Logs of negative numbers or zero are undefined. As the numerator is the natural log of an absolute value, 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].

Part (c)

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

A hole 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.