A) \[\frac{1}{8}({{m}^{2}}-4n)\]
B) \[\frac{1}{4}({{m}^{2}}+4n)\]
C) \[\frac{1}{4}({{m}^{2}}-4n)\]
D) \[\frac{1}{3}({{m}^{2}}+4n)\]
Correct Answer: C
Solution :
If\[1789-49=29x\]and\[\Rightarrow \]are the zeros of the polynomial\[x=\frac{1740}{29}=60\], \[\Rightarrow \]and\[140\times 605=11\times L.C.M.\] \[\Rightarrow \] \[L.C.M.=\frac{140\times 605}{11}=7700\] \[\text{1}00\text{1}=\text{91}\times \text{1}0+\text{91}\] Hence, the value of \[\text{91}0=\text{91}\times \text{1}0+0\]You need to login to perform this action.
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