A) 0.92
B) 0.092
C) 0.0092
D) 0.00092
Correct Answer: B
Solution :
Let 0.051 = x and 0.041 = y \[\therefore \] The given expression \[=\frac{{{x}^{3}}+{{y}^{3}}}{{{x}^{2}}-xy+{{y}^{2}}}\] \[=\frac{(x+y)({{x}^{2}}-xy+{{y}^{2}})}{{{x}^{2}}-xy+{{y}^{2}}}\] \[=x+y=0.051+0.041\] = 0.092You need to login to perform this action.
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