Question
If 0.125x3 +0.216y3 = 210 and
0.25x2 + 0.36y2 = 10.30 then find the value of 0.5x + 0.6y.Solution
Given: 0.125x3 + 0.216y3 = 210 (0.5x)3 + (0.6y)3 = 210 ---(1) 0.25x2 + 0.36y2 = 10.30 0.25x2 + 0.36y2 = 10 + 0.30 (0.5x)2 + (0.6y)2 – 0.30 = 10 ---(2) As, a3 + b3 = (a + b)(a2 – ab + b2) ---(3) By analysing the (1), (2) and (3) a = 0.5x and b = 0.6y (3) => (0.5x)3 +(0.6y)3 = (0.5x + 0.6y)[(0.5x)2 + (0.6y)2 – 0.30] Put the respective values in the above expression 210 = (0.5x + 0.6y)(10) (0.5x + 0.6y) = 21
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I. Q > Z
II. B ˃ J
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II). Â W > R
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I. O > K
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I. J ≤ P
II. S > T
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Conclusions:
I)Â O > Q
II) L < G
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