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Bad White [126]
2 years ago
9

Prove or disprove that the point (√5, 12) is on the circle centered at the origin and containing the point (-13, 0). Show your w

ork.
Mathematics
1 answer:
pav-90 [236]2 years ago
6 0

Using the equation of the circle, it is found that since it reaches an identity, the point (√5, 12) is on the circle.

<h3>What is the equation of a circle?</h3>

The equation of a circle of center (x_0, y_0) and radius r is given by:

(x - x_0)^2 + (y - y_0)^2 = r^2

In this problem, the circle is centered at the origin, hence (x_0, y_0) = (0,0).

The circle contains the point (-13,0), hence the radius is found as follows:

x^2 + y^2 = r^2

(-13)^2 + 0^2 = t^2

r^2 = 169

Hence the equation is:

x^2 + y^2 = 169

Then, we test if point (√5, 12) is on the circle:

x^2 + y^2 = 169

(\sqrt{5})^2 + 12^2 = 169

25 + 144 = 169

Which is an identity, hence point (√5, 12) is on the circle.

More can be learned about the equation of a circle at brainly.com/question/24307696

#SPJ1

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The equation of the perpendicular line is y + 7 = -1/7(x - 3)

<h3>How to determine the line equation?</h3>

The equation is given as

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The equation of a line can be represented as

y = mx + c

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By comparing the equations, we have the following

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This means that the slope of y = 7x + 14  is 7

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The equation of the perpendicular lines is then calculated as

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Read more about linear equations at

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8 0
11 months ago
Help xd <br> -(x+2) =<br> 3(7x - 8)=<br> 7(x+3)-5x<br><br> Thank you!
HACTEHA [7]

Answer:

x=−5

x= 31/21 =1 10/21= 1.476190476

2x+21

hope this helps

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