Answer:
138.23
Step-by-step explanation:
2πrh+2πr^2=2·π·2·9+2·π·2^2
 
        
             
        
        
        
1-1, 1-2, 1-3, 1-4 and u should Change it to middle school
        
                    
             
        
        
        
Answer:
C.
Step-by-step explanation:
WE need to convert the given equations  into the standard form
(x - a)^2 + (y - b)^2 = r^2 where r = the radius.
x^2 + y^2 + 12x - 20y+ 132 = 0
 (x + 6)^2 - 36 + (y - 10)^2 - 100 = -132
(x + 6)^2 + (y - 10)^2 = -132 + 36 + 100 = 4 
giving a radius of 2
 x^2 + y^2 + 12x - 20y + 55 = 0
will give a right hand side of -55 + 36 + 100 = 81  giving radius 9.
x^2 + y^2 - 12x+ 20y + 127 = 0
- gives right hand side of -127 + 36 + 100 = 9 giving a radius of 3.
 
        
                    
             
        
        
        
No, it is impossible. Intuitively, a negative number sits at the left of 0 on the number line, and a positive number sits at the right of 0 on the number line. And a number x is greater than another number y if x sits at the right of y on the number line. So, every positive number is greater than any negative number.
Also, by definition, a positive number is greater than 0, and a negative number is smaller than zero. So, if x is positive and y is negative, you have

and since the relation of order "<" is transitive, this implies 
