I am better at substitution, but you want to get the equations as similar as possible, but are the opposite, so you can cancel it out.
Answer:
878
Step-by-step explanation:
Answer:
This is because 23 students learn salsa and ballet. The second cell of the first row, Not Salsa and Ballet, will have 15. This is because 15 students do not learn salsa but learn ballet.
Step-by-step explanation:
-4 - 24n
you distribute the 4 to the -1 and -6n
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:

<u>Part (a)</u>
Substitute x = 2 into the given rational function:

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.

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.