Try this solution:
1. if to transform the given equation into the standard form:
x²+y²-4x+6y-3=0;
(x²-4x+4)+(y²+6y+9)-16=0;
(x-2)²+(y+3)²=4²
2. Using the standard form it is possible to find the coordinates of the centre of given circle: x=2; y= -3
answer: b) - (2;-3)
Answer:
30=10×3
Step-by-step explanation:
Answer:
2±1\2x
Step-by-step explanation:
[4±x]1\2
we will start by opening the brackets
[4]1\2=2
[x]1\2=1\2x
2±1\2x
hope am right