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=2πV∗1+(e−J2)==2\pi V*1+\left(\frac{e^{-J}}{2}\right)=

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Composition of linear and quadratic functions: find a valueright chevron icon

Full solution

Q. =2πV∗1+(e−J2)==2\pi V*1+\left(\frac{e^{-J}}{2}\right)=
  1. Plug in and simplify: First, let's plug in the values we know and simplify the expression. 2πV×(1+e−J2)2\pi V \times \left(1 + \frac{e^{-J}}{2}\right)
  2. Distribute 2πV2\pi V: Now, we need to distribute 2πV2\pi V across the terms inside the parentheses.\newline2πV×1+2πV×(e−J2)2\pi V \times 1 + 2\pi V \times \left(\frac{e^{-J}}{2}\right)
  3. Simplify first term: Simplify the first term by multiplying 2πV2\pi V by 11.2πV+2πV×(e−J2)2\pi V + 2\pi V \times \left(\frac{e^{-J}}{2}\right)
  4. Divide by 22: Next, divide 2πV2\pi V by 22 to simplify the second term.\newline2πV+πV⋅e(−J)2\pi V + \pi V \cdot e^{(-J)}
  5. Combine terms: Combine the terms to get the final expression. 2πV+πV⋅e−J2\pi V + \pi V \cdot e^{-J} is the simplified form of the original expression.

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