A) \[1.4\]
B) \[0.054\]
C) \[0.8\]
D) \[1.0\]
Correct Answer: A
Solution :
Let \[0.9\] be a, \[0.2\]be b & \[0.3\] be c Then \[\frac{{{a}^{3}}+{{b}^{3}}+{{c}^{2}}+3abc}{{{a}^{2}}+{{b}^{2}}+{{c}^{2}}-ab-ac-bc}\] \[\frac{(a+b+c)({{a}^{2}}+{{b}^{2}}+{{c}^{2}}-ab-bc-ac)}{({{a}^{2}}+{{b}^{2}}+{{c}^{2}}-ab-bc-ac)}\] \[a+b+c\] \[\therefore \]\[a+b+c=0.9+0.2+0.3=1.4\]You need to login to perform this action.
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