For this case, what we must do is fill squares in all the expressions until we find the correct result.
We have then:
x2 + y2 − 4x + 12y − 20 = 0 x2 + y2 − 4x + 12y = 20
x2 − 4x + y2 + 12y = 20
x2 − 4x + (12/2)^2 + y2 + 12y + (-4/2)^2 = 20 + (12/2)^2 + (-4/2)^2
x2 − 4x + (6)^2 + y2 + 12y + (-2)^2 = 20 + (6)^2 + (-2)^2
x2 − 4x + 36 + y2 + 12y + 4 = 20 + 36 + 4
(x − 2)2 + (y + 6)2 = 60
3x2 + 3y2 + 12x + 18y − 15 = 0
x2 + y2 + 4x + 6y − 5 = 0
x2 + y2 + 4x + 6y = 5
x2 + 4x + (4/2)^2 + y2 + 6y + (6/2)^2 = 5 + (4/2)^2 + (6/2)^2
x2 + 4x + (2)^2 + y2 + 6y + (3)^2 = 5 + (2)^2 + (3)^2
x2 + 4x + 4 + y2 + 6y + 9 = 5 + 4 + 9
(x + 2)2 + (y + 3)2 = 18
2x2 + 2y2 − 24x − 16y − 8 = 0
x2 + y2 − 12x − 8y − 4 = 0
x2 + y2 − 12x − 8y = 4
x2 − 12x + (-12/2)^2 + y2 − 8y + (-8/2)^2 = 4 + (-12/2)^2 + (-8/2)^2
x2 − 12x + (-6)^2 + y2 − 8y + (-4)^2 = 4 + (-6)^2 + (-4)^2
x2 − 12x + 36 + y2 − 8y + 16 = 4 + 36 + 16
(x − 6)2 + (y − 4)2 = 56
x2 + y2 + 2x − 12y − 9 = 0
x2 + y2 + 2x - 12y = 9
x2 + 2x + y2 - 12y = 9
x2 + 2x + (2/2)^2 + y2 - 12y + (-12/2)^2 = 9 + (2/2)^2 + (-12/2)^2
x2 + 2x + (1)^2 + y2 - 12y + (-6)^2 = 9 + (1)^2 + (-6)^2
x2 + 2x + 1 + y2 - 12y + 36 = 9 + 1 + 36
(x + 1)2 + (y − 6)2 = 46
It would be 224 , the base is 64 and every triangle is 80 but if you follow the rule for area to a triangle you get 40
So 64+40+40+40+40= 224
Answer:
Step-by-step explanation:
now the equation looks like
3-4Cos(x)=y
hmmm i'm wondering if you can figure it out from here??? it's the same as the other two.. plug in that 
/2 .. can you? :)
Answer:
four tens
Step-by-step explanation:
four tens is greater because:
four tens = 40
thirty six and nine thousandths = 36.009
36.009 < 40
Hope it cleared your doubt.
