The center of a circle whose equation is x^2 +y^2 – 12x – 2y +12 = 0 is (6,1)
<h3>Equation of a circle</h3>
The standard equation of a circle is expressed as:
x^2 + y^2 + 2gx + 2fy + c = 0
where:
(-g, -f) is the centre of the circle
Given the equations
x^2 +y^2 – 12x – 2y +12 = 0
Compare
2gx = -12x
g = -6
Simiarly
-2y = 2fy
f = -1
Centre = (6, 1)
Hence the center of a circle whose equation is x^2 +y^2 – 12x – 2y +12 = 0 is (6,1)
Learn more on equation of a circle here: brainly.com/question/1506955
Answer:-4, -3 1/2, 0, 1/4, 5/4, 7/4, 4
Step-by-step explanation: please give brainliest and thankyou
Hey there! :)
JK ≈ RS
Scale Factor = 9/7
JK = 56
<u>56</u> · <u>9</u> = <u>504</u> = 72
1 7 = 7
Your answer ⇒ B.72
Hope this helps :)
Answer:
x = the wight of the large box
y = the weight of the small box
A delivery of 3 large boxes and 2 small boxes has a total weight of 73 kilograms.
3x + 2y = 73
A delivery of 8 large boxes and 4 small boxes has a total weight of 177 kilograms.
8x + 4y = 177
by solving the system of equations
3x + 2y = 73
8x + 4y = 177
we find
x = 15.5 kg
y = 13.25 kg
the large box weights 15.5 kg.
the small box weights 13.25 kg.
Answer:
it's a
Step-by-step explanation:
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