A Cube, has all equal sides, namely, the length, width and height
are all equal to each other
so.. notice the picture added here
you have really, 6 squares, stacked up to each other at the edges
so...what is the Area of one of those squares?
well, if the sides are equal, let's say the side is "x" long, then
the Area is
well, you have 6 of those squares, thus
solve for "x", to get one side's length
Answer:
24
Step-by-step explanation:
Answer:
The equation that represents the relation is;
d. y = -16/x
Step-by-step explanation:
The
An inverse variation of an ordered pairs is the set of ordered pairs that is given by the relationship that as one variable in an ordered pair increases, the other decreases and vice versa
Mathematically, an inverse relation can be expressed as follows;
y ∝ 1/x
From which we have;
y = k/x, where, "k", represents the constant of proportionality
Therefore, we have;
k = y × x
The given ordered pair are;
{(-1, 16), (2, -8), (4, -4)}
Therefore, k = y × x = 16 × (-1) = -8 × 2 = -4 × 4 = -16
The given function is y = k/x = -16/x
Therefore, y = -16/x.
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.
