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=2pi J*(1+(e^(-J))/(3))=

=2πJ⋅(1+e−J3)= =2 \pi J \cdot\left(1+\frac{e^{-J}}{3}\right)=

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

Full solution

Q. =2πJ⋅(1+e−J3)= =2 \pi J \cdot\left(1+\frac{e^{-J}}{3}\right)=
  1. Simplify Expression: First, let's simplify the expression inside the parentheses.\newline1+e(−J)31 + \frac{e^{(-J)}}{3}
  2. Multiply by 2πJ2\pi J: Now, multiply the entire expression by 2πJ2\pi J.2πJ×(1+e−J3)2\pi J \times \left(1 + \frac{e^{-J}}{3}\right)
  3. Distribute 2πJ2\pi J: Distribute 2πJ2\pi J to both terms inside the parentheses.\newline2πJ+2πJ×(e(−J))/32\pi J + 2\pi J \times (e^{(-J)}) / 3
  4. Simplify Second Term: Simplify the second term by multiplying 2πJ2\pi J and e−J3\frac{e^{-J}}{3}.2πJ+2πJ⋅e−J32\pi J + \frac{2\pi J \cdot e^{-J}}{3}
  5. Write Final Expression: Now, we can write the final expression. 2πJ+(23)πJ⋅e−J2\pi J + \left(\frac{2}{3}\right)\pi J \cdot e^{-J}

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