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

=2π⋅1⋅1+e−J2= =2 \pi \cdot 1 \cdot 1+\frac{e^{-J}}{2}=

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

Full solution

Q. =2π⋅1⋅1+e−J2= =2 \pi \cdot 1 \cdot 1+\frac{e^{-J}}{2}=
  1. Calculate Circumference: First, calculate the circumference part 2π⋅1⋅12\pi\cdot1\cdot1. 2π⋅1⋅1=2π2\pi\cdot1\cdot1 = 2\pi
  2. Calculate Exponential: Now, calculate the exponential part (e−J)/(2)(e^{-J})/(2). Since JJ is not a standard mathematical constant and assuming it's meant to be a variable, we can't simplify this further without knowing the value of JJ. So, (e−J)/(2)(e^{-J})/(2) remains as it is.
  3. Add Parts Together: Add the two parts together. 2π+e−J22\pi + \frac{e^{-J}}{2}

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