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ArbitrLikvidat [17]
2 years ago
10

What is the horizontal asymptote of fix)= 242 O y = -2 O y=-1 O y = 0 y = 1

Mathematics
1 answer:
7nadin3 [17]2 years ago
6 0

Answer:

y=-1

perhaps replace x with 1

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If f(3) = 9, what is f(1)<br>​
Vaselesa [24]

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A quadrilateral WXYZ has vertices W(3, −5), X(1, −3), Y(−1, −5), and Z(1,−7). What are the vertices of r(90, O)(WXYZ)?
VashaNatasha [74]

Given:

A quadrilateral WXYZ has vertices W(3, −5), X(1, −3), Y(−1, −5), and Z(1,−7).

Rule of rotation is r_{(90^\circ, O)}(WXYZ).

To find:

The vertices after rotation.

Solution:

We know that, r_{(90^\circ, O)}(WXYZ) means 90 degrees counterclockwise rotation around the origin.

So, the rule of rotation is defined as

(x,y)\to (-y,x)

Using this rule, we get

W(3,-5)\to W'(5,3)

X(1,-3)\to X'(3,1)

Y(-1,-5)\to Y'(5,-1)

Z(1,-7)\to Z'(7,1)

Therefore, the required vertices after rotation are W'(5,3), X'(3,1),Y'(5,-1) and Z'(7,1).

8 0
3 years ago
X² - 6x + y² + 2y = -1<br> find center &amp; Radius
Fofino [41]

Answer:

The answer to your question is Center = (3 , -1), radius = 3

Step-by-step explanation:

Equation

                       x² - 6x + y² + 2y = -1

-Process

1.- Leave a space between the like terms

                     x² - 6x           + y² + 2y          = -1

2.- Divide the middle term of each group by 2 and write the result to the power of 2 in both sides of the equation.

                     x² - 6x + (3)² + y² + 2y + (1)² = -1 + 3² + 1²

3.- Simplification

                     x² - 6x + 9 + y² + 2y + 1 = -1 + 9 + 1

4.- Factor

                     (x - 3)² + (y + 1)² = 9

5.- Find the center and the radius

     Center = (3 , -1)

     Radius = √9 = 3

                   

                     

8 0
3 years ago
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