3 years ago

Center (3, -2) Radius = 3

Mathematics

1 answer:

mel-nik [20]3 years ago

3 0

Answer:

x^2-6x+y^2+4y=-4

Step-by-step explanation:

I am assuming you want the equation of the circle. We know that the equation of a circle is (x-h)^2 + (y-k)^2 =r^2, where r is the radius, and h and k are the x and y coordinates of the center respectively. We can solve for this by plugging in the numbers.

(x-3)^2+(y+2)^2=3^2

x^2-6x+9+y^2+4y+4=9

x^2-6x+y^2+4y=-4

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Here you go please show your work

ValentinkaMS [17]

A. F(x)= 3x+6 =12
3x = 12-6
3x = 6
x=2

B. x=10, f(x)= 3*10+6 = 30+6 = 36

C. f(x) = 3x+6 = 6
3x = 6-6
3x=0
x=0

D. x=5, f(x) = 3x+6 = 3*5+6 = 15+6 = 21

5 0

3 years ago

ASAP WILL MARK BRAINLYEST <br><br><br>answer these two please!

Trava [24]

$19.\\y=3x^2\\a=3,\ b=0,\ c=0\\\\\dfrac{-b}{2a}=\dfrac{-0}{2(3)=0\to x=0\\\\3x^2=0\ \ \ \ |:3\\\\x^2=0\to x=0$

$20.\\y=x^2-x-2\\a=1,\ b=-1,\ c=-2\\\\\dfrac{-b}{2a}=\dfrac{-(-1)}{2(1)}=\dfrac{1}{2}\to x=\dfrac{1}{2}\\\\x^2-x-2=0\\\\x^2-2x+x-2=0\\\\x(x-2)+1(x-2)=0\\\\(x-2)(x+1)=0\iff x-2=0\ \vee\ x+1=0\\\\x=2\ \vee\ x=-1$