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
Step-by-step explanation:
(Can someone please answer my math question)
Answer:
M = 20
Step-by-step explanation:
Given that M is directly proportional to r³ then the equation relating them is
M = kr³ ← k is the constant of proportion
To find k use the condition when r = 4, M = 160, that is
160 = k × 4³ = 64k (divide both sides by 64 )
2.5 = k
M = 2.5r³ ← equation of proportion
When r = 2, then
M = 2.5 × 2³ = 2.5 × 8 = 20
Answer:
see below
Step-by-step explanation:
side opposite 90 angle = x 
sides opposite 45 angle = x
the multiplication and division by should be self explanatory
1. x = 5, y = 5
2. v = 10, u = 10
3. x = 
, y = 
4. m = 5, n = 
5. a = 
, b =
6. a = , b =
1)5 2)6 3)17 because h(x)+k(x)=3x-1
