√(-(-1)^⌊1/2+1/π⌋ ×(-π×⌊1/2+1/π⌋+1) ).

asked by guest
on Nov 26, 2024 at 7:46 pm



You asked:

Evaluate the expression: −(−1)⌊12+1π⌋(−π⋅⌊12+1π⌋+1)\sqrt{-{\left( -1 \right)}^{\left\lfloor \frac{1}{2} + \frac{1}{\pi} \right\rfloor} \left(-\pi \cdot \left\lfloor \frac{1}{2} + \frac{1}{\pi} \right\rfloor + 1\right)}

MathBot Answer:

−(−1)⌊12+1π⌋(−π⋅⌊12+1π⌋+1)=i\sqrt{-{\left( -1 \right)}^{\left\lfloor \frac{1}{2} + \frac{1}{\pi} \right\rfloor} \left(-\pi \cdot \left\lfloor \frac{1}{2} + \frac{1}{\pi} \right\rfloor + 1\right)} = i

−(−1)⌊12+1π⌋(−π⋅⌊12+1π⌋+1)=−⋅(−1)⌊12+1π⌋⋅(−⋅π⋅⌊12+1π⌋+1)=−⋅(−1)0⋅(−⋅π⋅⌊12+1π⌋+1)=−⋅1⋅(−⋅π⋅⌊12+1π⌋+1)=−⋅1⋅(−⋅π⋅0+1)=−⋅1⋅(−π⋅0+1)=−⋅1⋅(0+1)=−⋅1⋅1=−1=−1=i\begin{aligned}\sqrt{-{\left( -1 \right)}^{\left\lfloor \frac{1}{2} + \frac{1}{\pi} \right\rfloor} \left(-\pi \cdot \left\lfloor \frac{1}{2} + \frac{1}{\pi} \right\rfloor + 1\right)}&=\sqrt{- \cdot \left(-1\right)^{\left\lfloor{\frac{1}{2} + \frac{1}{\pi}}\right\rfloor} \cdot \left(- \cdot \pi \cdot \left\lfloor{\frac{1}{2} + \frac{1}{\pi}}\right\rfloor + 1\right)}\\&=\sqrt{- \cdot \left(-1\right)^{0} \cdot \left(- \cdot \pi \cdot \left\lfloor{\frac{1}{2} + \frac{1}{\pi}}\right\rfloor + 1\right)}\\&=\sqrt{- \cdot 1 \cdot \left(- \cdot \pi \cdot \left\lfloor{\frac{1}{2} + \frac{1}{\pi}}\right\rfloor + 1\right)}\\&=\sqrt{- \cdot 1 \cdot \left(- \cdot \pi \cdot 0 + 1\right)}\\&=\sqrt{- \cdot 1 \cdot \left(- \pi \cdot 0 + 1\right)}\\&=\sqrt{- \cdot 1 \cdot \left(0 + 1\right)}\\&=\sqrt{- \cdot 1 \cdot 1}\\&=\sqrt{- 1}\\&=\sqrt{-1}\\&=i\end{aligned}

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