A) 2.16g
B) 2.48 g
C) 2,64 g
D) 2.32g
Correct Answer: A
Solution :
\[\frac{{{2}^{x}}+{{2}^{y}}}{1+{{2}^{x+y}}}\] \[{{2}^{x-y}}\left( \frac{{{2}^{y}}-1}{1-{{2}^{x}}} \right)\]\[\frac{{{2}^{x-y}}-{{2}^{x}}}{{{2}^{y}}}\]of \[x-3y=0\]gives \[x+3y=0\] \[3x-y=0\]1 g of \[2x+y=0\]will give\[\sin A+\cos A=m\] \[{{\sin }^{3}}A+{{\cos }^{3}}A=n,\]2.76 g of \[~{{m}^{3}}-3m+n=0\]will give\[{{n}^{3}}-3n+2m=0\] \[{{m}^{3}}-3m+2n=0\]
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