Revision as of 03:07, 20 January 2024 by Admin (Created page with "'''Answer: C''' <math <math display="block"> \begin{aligned} & V\left[L_{0} \# 1\right]=\left(B_{1}+\frac{P_{1}}{d}\right)^{2}\left({ }^{2} A_{x}-A_{x}^{2}\right)=20.55==>\left(8+\frac{1.25(1.06)}{0.06}\right)^{2}\left({ }^{2} A_{x}-A_{x}^{2}\right)=20.55 \\ & { }^{2} A_{x}-A_{x}^{2}=\frac{20.55}{\left(8+\frac{1.25(1.06)}{0.06}\right)^{2}}=0.0227 \\ & V\left[L_{0} \# 2\right]=\left(12+\frac{1.875(1.06)}{0.06}\right)^{2}\left({ }^{2} A_{x}-A_{x}^{2}\right)=\left(12+\fr...")
Exercise
ABy Admin
Jan 20'24
Answer
Answer: C
[[math]]
\begin{aligned}
& V\left[L_{0} \# 1\right]=\left(B_{1}+\frac{P_{1}}{d}\right)^{2}\left({ }^{2} A_{x}-A_{x}^{2}\right)=20.55==\gt\left(8+\frac{1.25(1.06)}{0.06}\right)^{2}\left({ }^{2} A_{x}-A_{x}^{2}\right)=20.55 \\
& { }^{2} A_{x}-A_{x}^{2}=\frac{20.55}{\left(8+\frac{1.25(1.06)}{0.06}\right)^{2}}=0.0227 \\
& V\left[L_{0} \# 2\right]=\left(12+\frac{1.875(1.06)}{0.06}\right)^{2}\left({ }^{2} A_{x}-A_{x}^{2}\right)=\left(12+\frac{1.875(1.06)}{0.06}\right)^{2}(0.0227)=46.24
\end{aligned}
[[/math]]
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