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Paladinen [302]
1 year ago
10

P(x) =x and q(x) = x-1Given:minimum x and Maximum x: -9.4 and 9.4minimum y and maximum y: -6.2 and 6.2Using the rational functio

n [y=P(x)/q(x)], draw a graph and answer the following: a) what are the zeroes?b) are there any asymptotes? c) what is the domain and range for this function?d) it it a continuous function?e) are there any values of y= f(x)/g(x) that are undefined? Explain
Mathematics
1 answer:
Arte-miy333 [17]1 year ago
5 0

we have the following function

\frac{p(x)}{g(x)}=\frac{x}{x\text{ -1}}

where x is between -9.4 and 9.4 and y is between -6.2 and 6.2.

We will first draw the function

from the graph, we can see that the zeroes are all values of x for which the graph crosses the x -axis

In this case, we see that that the only zero is at x=0.

Now, we have that the asymptotes are lines to which the graph of the function get really close to. On one side, we can see that as x goes to infinity or minus infinity, the values of the function get really close to 1. So the graph has a horizontal asymptote at y=1. Also, we can see that as x gets really close to 1, the graph gets really close to the vertical line x=1. So the graph has a vertical asymptote at x=1.

Recall that the domain of a function is the set of values of x for which the function is defined. From our graph, we can see that graph is not defined when x=1. So the domain of the function is the set of real numbers except x=1. Now, recall that the range of the function is the set of y values of the graph. From the picture we can see that the graph has a y coordinate for every value of y except for y=1. So, this means that the range of the function is the set of real numbers except y=1.

From the graph, we can see that we cannot draw the graph having a continous drawing. That is, imagine we take a pencil and start on one point on the graph on the left side. We can draw the whole graph on the left side, but we cannot draw the graph on the right side without lifting the pencil up. As we have to "lift the pencil up" this means that the graph is not continous

Finally note that as we have a vertical asymptote at x=1 and horizontal asymptote at y=1 we have that when y is 1 or x is 1, the function y=f(x)/g(x) is undefined

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an ellipse has a center at the origin , a vertex along the major axis at (13,0), and a focus at (12,0). What is the equation of
-BARSIC- [3]

Answer:

The answer to your question is below

Step-by-step explanation:

Data

Center = (0, 0)

Vertex = (13, 0)

Focus = (12, 0)

Process

From the data we know that it is a horizontal ellipse.

1.- Calculate "a", the distance from the center to the vertex.

                  a = 13

2.- Calculate "c", the distance from the center to the focus

                  c = 12

3.- Calculate b

Use the Pythagorean theorem to find it

                  a² = b² + c²

-Solve for b

                  b² = a² - c²

-Substitution

                  b² = 13² - 12²

-Simplification

                  b² = 169 - 144

                  b² = 25

                  b = 5

4.- Find the equation of the ellipse

                       \frac{x^{2} }{13^{2}} + \frac{y^{2}}{5^{2}} = 1    or \frac{x^{2} }{169} + \frac{y^{2}}{25} = 1

7 0
3 years ago
I’ll name brainliest✨
eduard

Answer: -5 1/2 or -5.5

Step-by-step explanation: Negative five and a half. Its technically -5 2/4 but simplify 2/4 its 1/2

8 0
2 years ago
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HELP!!! Andre reads 2 1/2 pages in 1/4 hour. What is the complex fraction and unit rate?
Gre4nikov [31]

Answer:

20

Step-by-step explanation:

2 1/2 = 5/2

5/2 x 4/1 = 20/1 = 20

(We are dividing fraction)

(keep change flip)

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2 years ago
Find the greatest common factor of the terms in the following expression: 32rt – 12t.
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2 years ago
A 2X2 square is centered at the origin. It is dilated by a factor of 3. What are coordinates of the vertices of the square?
Vesna [10]

Answer:

The vertices are:

A' = (-3, -3)

B' = (3, -3)

C' = (3, 3)

D' = (-3, 3)

Step-by-step explanation:

Given:

A 2 x 2 square is centered at the origin.

So, the center of the square is (0, 0)

Since it is 2 x 2 square, the side of the square is 2 units.

So, the vertices of the 2 x 2 square are A (-1, -1),  B(1, -1), C(1. 1), D(-1, 1)

The above square is dilated by a factor of 3.

Let's name the dilated square A'B'C'D'

To find the coordinates of the vertices of dilated square, we need to multiply each vertices of ABCD by 3.

A(-1, -1) = 3(-1, -1) = A'(-3, -3)

B(1, -1) = 3(1, -1) = B'(3, -3)

C(1, 1) = 3(1, 1) = C'(3, 3)

D(-1, 1) = 3(-1, 1) = D'(-3, 3)

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