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Damm [24]
3 years ago
7

Determine the center and radius of the following circle equation:

Mathematics
1 answer:
Greeley [361]3 years ago
8 0

Answer:

x^2-18x+y^2+18y+137=0

express both x and y term in completed square form

(x-9)^2 -81+(y+9)^2-81+137=0

(x-9)^2+(y+9)^2-25=0

(x-9)^2+(y+9)^2=5^2

here the x coordinate of the centre is +9

y coordinate is -9(the negative value of the value associated with the x and y square brackets)

radius=5

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F(x)= x-2/ x-4
hodyreva [135]

Answer:

The discontinuity is x = 4

There no holes

The equation of the vertical asymptote is x = 4

The x intercept is 2

The equation of the horizontal asymptote is y = 1

Step-by-step explanation:

* Lets explain the problem

∵ f(x)=\frac{x-2}{x-4}

- To find the point of discontinuity put the denominator = 0 and find

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∵ The denominator is x - 4

∵ x - 4 = 0 ⇒ add 4 to both sides

∴ x = 4

* The discontinuity is x = 4

- A hole occurs when a number is both a zero of the numerator

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∵ The numerator is x - 2

∵ x - 2 = 0 ⇒ add 2 to both sides

∴ x = 2

∵ The denominator is x - 4

∵ x - 4 = 0 ⇒ add 4 to both sides

∴ x = 4

∵ There is no common number makes the numerator and denominator

   equal to 0

∴ There no holes

- Vertical asymptotes are vertical lines which correspond to the zeroes  

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∵ The zero of the denominator is x = 4

∴ The equation of the vertical asymptote is x = 4

- x- intercept is the values of x which make f(x) = 0, means the

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∵ f(x) = 0

∴ \frac{x-2}{x-4}=0 ⇒ by using cross multiplication

∴ x - 2 = 0 ⇒ add 2 to both sides

∴ x = 2

* The x intercept is 2

- If the highest power of the numerator = the highest power of the

 denominator, then the equation of the horizontal asymptote is

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∵ The numerator is x - 2

∵ The denominator is x - 4

∵ The leading coefficient of the numerator is 1

∵ The leading coefficient of the denominator is 1

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* The equation of the horizontal asymptote is y = 1

6 0
3 years ago
Thank you to whoever helps!
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Answer:

  • M
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The segment joining an original point with its rotated image forms a chord of the circle of rotation containing those two points. The center of the circle is the center of rotation.

This means you can find the center of rotation by considering the perpendicular bisectors of the segments joining points with their images. Here, the only proposed center that is anywhere near the perpendicular bisector of DE is point M.

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Segment AD is perpendicular to corresponding segment FE, so the angle of rotation is 90°. (We don't know which way (CW or CCW) unless we make an assumption about which is the original figure.)

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