if a2 + b2 + c2 = 2(4a -5b -9c)-122 , then a-b-c = ?
a2 + b2 + c2 = 2(4a -5b -9c)-122 So a2 + b2 + c2 -2×4a +2 ×5b + 2×9c +122 = 0 a2 -2×4a +16 + b2 +2 ×5b+25 +c2 +2×9c+81 = 0 So (a-4)2 +(b+5)2+(c+9)2 = 0 As we know that if x2 + y2 +z2 = 0 then x=y=z = 0 So here a-4 = 0 so a =4; b +5 = 0 so b = -5 And c+9 =0 so c = -9 So now a-b-c = 4-(-5)-(-9) = 4+5+9 = 18
5121.3 × 641.8 ÷ 80.5 = 8?
32 of (16/8) of (30/24) of (120/x) = 30
?/4 ÷ 9/? = 15% of 800 + `1(2/3)` × `1(1/5)` × 1/2
134% of 1250 – 46% of 2120 = 4 × ?
18/2 of 3/9 of 2/6 of 69690= ?
?= √(4 × ∛(16 × √(4 × ∛(16 ×…… ∝)) ) )
(2/?) x (3/16) x (2/15) x 60 = 1/3
140% of 9/8 of ? = 108% of 2800
√(82 × 7 × 52 - 175) = ?
166/? = √576 - 3.25
