If A B C are real numbers such that A - 2 2 B - 8 2 C

ANS 16
2(A+B+C-16)=0
A+B+C-16=0
A+B+C=16
10 years agoHelpfull: Yes(5) No(0)
A = 2 makes (A-2) = 0
B = 8 makes (B-8) = 0
C = 6 makes (C-6) = 0

A+B+C = 2 + 8 + 6 = 16 (Answer)
10 years agoHelpfull: Yes(5) No(0)
ans:16
(a-2)+(b-8)+(c-6)=(0/2)
a+b+c-16=0
a+b+c=16
10 years agoHelpfull: Yes(3) No(0)
2A-4+2B-16+2C-12=0
2(A+B+C)=32
so, (A+B+C)=16
10 years agoHelpfull: Yes(2) No(0)
2a+2b+2c=32
a+b+c=16
10 years agoHelpfull: Yes(1) No(0)
so they are separately zero ans is 16
10 years agoHelpfull: Yes(1) No(0)
ans is 16........
10 years agoHelpfull: Yes(1) No(0)
(A - 2)2 + (B - 8)2 + (C - 6)2 = 0
2A-4+2B-16+2C-12=0
2(A+B+C)=32
A+B+C=16........ Choice B
10 years agoHelpfull: Yes(1) No(0)
ans is 16....
10 years agoHelpfull: Yes(1) No(0)
16
dividing the equation by 2 we get,
A-2+B-8+C-6=0
=>A+B+C=16
10 years agoHelpfull: Yes(1) No(0)
2a-4=0; a=4;2b-16=0; b=8; 2c-12=0; c=6; a+b+c=2+8+6 =16
10 years agoHelpfull: Yes(1) No(0)
d) not determinable from the info given
let us consider A,B,C==0;
then A+B+C=0 so option--A) is also possible
if we take A,B,C==2,8,6;
then A+B+c=16;
so we r getting different solution
so the answer is d) not determinable from the info given
10 years agoHelpfull: Yes(0) No(14)
16
2A-4+2B-16+2C-12=0
2(A+B+C)=32
therefore, (A+B+C)=16
10 years agoHelpfull: Yes(0) No(0)

If A, B, C are real numbers such that

(A - 2)2 + (B - 8)2 + (C - 6)2 = 0

then A + B + C is

a. 0

b. 16

c. -16

d. not determinable from the information given

e. 104