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SIZIF [17.4K]
3 years ago
7

The graph of an even function contains the point (3,-5). Which point must also be on the graph of the function?

Mathematics
1 answer:
Paul [167]3 years ago
3 0

An even function is symmetrical about the y-axis. The mirror image of the point across the y-axis will also be on the graph. That point is found by changing the sign of the x-coordinate: (-3, -5).

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The equation of the circle is (x+3)^2+(y-5)^2 = 17

Step-by-step explanation:

The complete question is

If the coordinates of the endpoints of a diameter of the circle are​ known, the equation of a circle can be found.​ First, find the midpoint of the​ diameter, which is the center of the circle. Then find the​ radius, which is the distance from the center to either endpoint of the diameter. Finally use the​ center-radius form to find the equation.

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Consider that, if both points are the endpoints of a diameter, the center of the circle is the point that is exactly in the middle of the two points (that is, the point whose distance to each point is equal). Given points (a,b) and (c,d), by using the distance formula, you can check that the middle point is the average of the coordinates. Hence, the center of the circle is given by

(\frac{1-7}{2}, \frac{4+6}{2}) = (-3,5).

We will find the radius. Recall that the radius of the circle is the distance from one point of the circle to the center. Recall that the distance between points (a,b) and (c,d) is given by \sqrt[]{(a-c)^2+(b-d)^2}. So, let us use (1,4) to calculate the radius.

r = \sqrt[]{(1-(-3))^2+(4-5)^2} = \sqrt[]{17}.

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

In our case, (x_0,y_0)=(-3,5) and r=\sqrt[]{17}. Then, the equation is

(x-(-3))^2+(y-5)^2 = (x+3)^2+(y-5)^2 = 17

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