A) \[3y+1=0\]
B) \[y-3=0\]
C) \[y+3=0\]
D) \[3y-1=0\]
Correct Answer: D
Solution :
We have \[\frac{8y}{3}+{{y}^{2}}=1\Rightarrow (3y-1)(y+3)=0\] \[\therefore y=\frac{1}{3}(\operatorname{Reject}\,\,y=-3)\]You need to login to perform this action.
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