Identify the transformation from ABC to A’B’C’.

Anyone who can help do it now! make sure to explain and draw

AnnZ [28]

Up, right, down, left, up

7 0

3 years ago

Given an arithmetic sequence with a7=19 and a12=54 find a15

Illusion [34]

C 75 that’s what it is on apex

6 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.

3 0

2 years ago

Read 2 more answers

Write the slope-intercept form of the equation of each line

Flura [38]

The answer it’s D :)

7 0

3 years ago

Find an equation of the plane with the given characteristics. The plane passes through the points (4, 3, 1) and (4, 1, -7) and i

Minchanka [31]

Answer:

$3x - 4y + z = 1$

Step-by-step explanation:

Given

$Point\ 1 = (4,3,1)$

$Point\ 2 = (4,1,-7)$

Perpendicular to $8x + 7y + 4z = 18$

Required

Determine the plane equation

The general equation of a plane is:

$a(x-x_1) + b(y - y_1) + c(z-z_1) = 0$

For $n =$

$(x_1,y_1,z_1) = (4,3,1)$

$(x_2,y_2,z_2) = (4,1,-7)$

First, we need to determine parallel vector $V_1$

$V_1 =$

$V_1$ is parallel to the required plane

From the question, the required plane is perpendicular to $8x + 7y + 4z = 18$

Next, we determine vector $V_2$

$V_2 =$

This implies that the required plane is parallel to $V_2$

Hence: $V_1$ and $V_2$ are parallel.

So, we can calculate the cross product $V_1 * V_2$

$V_1 =$

$V_2 =$

$n = V_1 * V_2$

$V_1 * V_2 =\left[\begin{array}{ccc}i&j&k\\0&2&8\\8&7&4\end{array}\right]$

The product is always of the form + - +

So:

$V_1 * V_2 = i\left[\begin{array}{cc}2&8\\7&4\end{array}\right]$ $-j\left[\begin{array}{cc}0&8\\8&4\end{array}\right]$ $+k\left[\begin{array}{cc}0&2\\8&7\end{array}\right]$