Remember that the equation of a circle is:

Where (h, k) is the center and r is the radius.
We need to get the equation into that form, and find k.

Complete the square. We must do this for x² - 6x and y² - 10y separately.
x² - 6x
Divide -6 by 2 to get -3.
Square -3 to get 9. Add 9,
x² - 6x + 9
Because we've added 9 on one side of the equation, we have to remember to do the same on the other side.

Now factor x² - 6x + 9 to get (x - 3)² and do the same thing with y² - 10y.
y² - 10y
Divide -10 by 2 to get -5.
Square -5 to get 25.
Add 25 on both sides.

Factor y² - 10y + 25 to get (y - 5)²

Now our equation is in the correct form. We can easily see that h is 3 and k is 5. (not negative because the original equation has -h and -k so you must multiply -1 to it)
Since (h, k) represents the center, (3, 5) is the center and 5 is the y-coordinate of the center.
Should be 7.5% I have to add more words to this sentence to post it
Answer:
the last one talking about victoria has y dollars
Step-by-step explanation:
x times 12 is equal to y, so the equation is y = 12x
The <em>speed</em> intervals such that the mileage of the vehicle described is 20 miles per gallon or less are: v ∈ [10 mi/h, 20 mi/h] ∪ [50 mi/h, 75 mi/h]
<h3>How to determine the range of speed associate to desired gas mileages</h3>
In this question we have a <em>quadratic</em> function of the <em>gas</em> mileage (g), in miles per gallon, in terms of the <em>vehicle</em> speed (v), in miles per hour. Based on the information given in the statement we must solve for v the following <em>quadratic</em> function:
g = 10 + 0.7 · v - 0.01 · v² (1)
An effective approach consists in using a <em>graphing</em> tool, in which a <em>horizontal</em> line (g = 20) is applied on the <em>maximum desired</em> mileage such that we can determine the <em>speed</em> intervals. The <em>speed</em> intervals such that the mileage of the vehicle is 20 miles per gallon or less are: v ∈ [10 mi/h, 20 mi/h] ∪ [50 mi/h, 75 mi/h].
To learn more on quadratic functions: brainly.com/question/5975436
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