∣π−3∣−∣π−4∣
asked by guest on Nov 24, 2024 at 10:18 pm
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∣π−3∣−∣π−4∣=−7+2π≈−0.71681469282041352307471323344099\left\lvert \pi - 3\right\rvert - \left\lvert \pi - 4\right\rvert = -7 + 2 \pi \approx -0.71681469282041352307471323344099∣π−3∣−∣π−4∣=−7+2π≈−0.71681469282041352307471323344099
∣π−3∣−∣π−4∣=∣π−3∣−∣π−4∣=∣−3+π∣−∣π−4∣=(−3+π)−∣π−4∣=(−3+π)−∣−4+π∣=(−3+π)−(4−π)=−7+2⋅π\begin{aligned}\left\lvert \pi - 3\right\rvert - \left\lvert \pi - 4\right\rvert&=\left|{\pi - 3}\right| - \left|{\pi - 4}\right|\\&=\left|{-3 + \pi}\right| - \left|{\pi - 4}\right|\\&=\left(-3 + \pi\right) - \left|{\pi - 4}\right|\\&=\left(-3 + \pi\right) - \left|{-4 + \pi}\right|\\&=\left(-3 + \pi\right) - \left(4 - \pi\right)\\&=-7 + 2 \cdot \pi\end{aligned}∣π−3∣−∣π−4∣=∣π−3∣−∣π−4∣=∣−3+π∣−∣π−4∣=(−3+π)−∣π−4∣=(−3+π)−∣−4+π∣=(−3+π)−(4−π)=−7+2⋅π