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algol13
3 years ago
13

What is the equation of a circle with its center at (0,-2) and passing through the point (3,-5)

Mathematics
1 answer:
Ksivusya [100]3 years ago
7 0
Circle's Standard Form Equation:
r^{2} = (x - xo)^{2} + (y - yo)^{2}

First you need to substitute the C (0,-2) in the equation, where: Xo = 0 and Yo = -2;

r^{2} = (x - 0)^{2} + (y - ( -2))^{2} \\ r^{2} = x^{2} + (y + 2)^{2}

Now you substitute the Point (3,-5) values in the last expression, where: X = 3 and Y = -5;

r^{2} = 3^{2} + (( - 5) + 2)^{2} \\ r ^{2} = 9 + ( - 3)^{2} \\ r^{2} = 9 + 9 \\ {r}^{2} = 18

Now you get the R^2 value and substitute in the equation that you found putting the C values on it:

r^{2} = x^{2} + (y + 2)^{2} \\ x^{2} + (y + 2)^{2} = 18

Here is your standard form equation of the circle.
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Given function, f(x) = 4x-8

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Interchanging the x variable with y,

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Now, replacing y with  f⁻¹(x), we get;

f⁻¹(x) = x/4 + 2

For verifying, you can use

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Therefore, the inverse function of f(x) is x/4 + 2.

To learn more about the inverse functions, visit: brainly.com/question/14965513

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