Given that :- a2 + b2 + c2 = 2 (a ? b ? c) ? 3
? a2 + b2 + c2 - 2 (a ? b ? c) + 3 = 0
? a2 + b2 + c2 - 2a + 2b + 2c + 3 = 0
? ( a2 + 1 - 2a ) + ( b2 + 1 + 2b ) + ( c2 + 1 + 2c ) = 0
? ( a - 1 )2 + ( b + 1 )2 + ( c + 1 )2 = 0
This is possible when ( a - 1 )2 = 0, ( b + 1 )2 = 0, and ( c + 1 )2 = 0
? a = 1, b = ?1, c = ?1
Thus, 2a ? 3b + 4c = 2 (1) ? 3 (?1) + 4 (?1)
2a ? 3b + 4c = 2 + 3 ? 4 = 1.
47/10000 = .0047
.02 =(2/100) x 100% =2%
Let N / 11 = 233
Then, N = 233 x 11 = 2563
? Missing digit is 5.
121012 = 12 x 10084 + 4
? remainder = 4
We have the important relation, More work, More time (days)
? A piece of work can be done in 6 days.
? Three times of work of same type can be done in 6 x 3
= 18 days
? = 750.0003 ÷ 19.999
? ? ? 750 ÷ 20
? ? ? 375 ? 38
Subtract 20, 25, 30, 35, 40, 45 from successive numbers. So 0 is wrong.
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