Question
$EN$ of element $A$ is $E_1$ and $IP$ is $E_2$ hence $EA$ will be
✓
Answer
a
$\frac{\mathbb{P}+\mathrm{EA}}{2}=\mathrm{EN} ; \frac{\mathrm{E}_{2}+\mathrm{EA}}{2}=\mathrm{E}_{1}$
$\frac{\mathbb{P}+\mathrm{EA}}{2}=\mathrm{EN} ; \frac{\mathrm{E}_{2}+\mathrm{EA}}{2}=\mathrm{E}_{1}$
hence $\mathrm{EA}=2 \mathrm{E}_{1}-\mathrm{E}_{2}$
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