we can see that the center is (-3, 3) and the radius is 9 units.
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How to find the center and radius of the circle?</h3>
The general circle equation, for a circle with a center (a, b) and radius R is given by:
(x - a)^2 + (y - b)^2 = R^2
Here we have the equation:
x^2 + y^2 + 6x = 6y + 63
Let's complete squares:
x^2 + y^2 + 6x - 6y = 63
(x^2 + 6x) + (y^2 - 6y) = 63
(x^2 + 2*3x) + (y^2 - 2*3y) = 63
Now we can add and subtract 9, (two times) so we get:
(x^2 + 2*3x + 9) - 9 + (x^2 - 2*3x + 9) - 9 = 63
(x + 3)^2 + (y - 3)^2 = 63 + 9 + 9 = 81 = 9^2
(x + 3)^2 + (y - 3)^2 = 9^2
Comparing with the general circle equation, we can see that the center is (-3, 3) and the radius is 9 units.
If you want to learn more about circles:
brainly.com/question/1559324
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Answer:
x=4 and x= - 5
Step-by-step explanation:
In order to solve this we should re-arrange the equation:
This is equal to:
Then we can seperate this into:
So solving for x in both cases we get
One cubic mile represent the volume of a cube with sides length equal 1 mile
hope this helped
Answer:
No. It is a constant function.
Step-by-step explanation:
The function f(x) = e^2 is not an exponential functional. Rather, it is a constant function. The reason for this is that in f(x) = e^2, there is no x involved on the right hand side of the equation. The approximate value of e is 2.718281, and the approximate value of 2.718281^2 is 7.389051. This means that f(x) = e^2 = 7.389051. It is important to note that for any value of x, the value of the function remains fixed. This is because the function does not involve the variable x in it. The graph of the function will be a line parallel to the x-axis, and the y-intercept will be 7.389051. For all the lines parallel to x-axis, the value of the function remains the same irrespective of the value of x. Also, the derivative of the function with respect to x is 0, which means that the value of the function is unaffected by the change in the value of x!!!
I'm not sure what you're asking but i think it can be this
Numbers
Colors
Probability
Hope this helps :)
