√(-(-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: \(\sqrt{-{\left( -1 \right)}^{\left\lfloor \frac{1}{2} + \frac{1}{\pi} \right\rfloor} \cdot \left(-\pi \cdot \left\lfloor \frac{1}{2} + \frac{1}{\pi} \right\rfloor + 1\right)}\)

MathBot Answer:

\[\sqrt{-{\left( -1 \right)}^{\left\lfloor \frac{1}{2} + \frac{1}{\pi} \right\rfloor} \cdot \left(-\pi \cdot \left\lfloor \frac{1}{2} + \frac{1}{\pi} \right\rfloor + 1\right)} = i\]

\[\begin{aligned}\sqrt{-{\left( -1 \right)}^{\left\lfloor \frac{1}{2} + \frac{1}{\pi} \right\rfloor} \cdot \left(-\pi \cdot \left\lfloor \frac{1}{2} + \frac{1}{\pi} \right\rfloor + 1\right)}& = \sqrt{- \cdot \left(-1\right)^{\left\lfloor{\frac{1}{2} + \frac{1}{\pi}}\right\rfloor} \cdot \left(- \cdot \pi \cdot \left\lfloor{\frac{1}{2} + \frac{1}{\pi}}\right\rfloor + 1\right)}\\& = \sqrt{- \cdot \left(-1\right)^{0} \cdot \left(- \cdot \pi \cdot \left\lfloor{\frac{1}{2} + \frac{1}{\pi}}\right\rfloor + 1\right)}\\& = \sqrt{- \cdot 1 \cdot \left(- \cdot \pi \cdot \left\lfloor{\frac{1}{2} + \frac{1}{\pi}}\right\rfloor + 1\right)}\\& = \sqrt{- \cdot 1 \cdot \left(- \cdot \pi \cdot 0 + 1\right)}\\& = \sqrt{- \cdot 1 \cdot \left(- \pi \cdot 0 + 1\right)}\\& = \sqrt{- \cdot 1 \cdot \left(0 + 1\right)}\\& = \sqrt{- \cdot 1 \cdot 1}\\& = \sqrt{- 1}\\& = \sqrt{-1}\\& = i\end{aligned}\]

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